diff --git "a/stack-exchange/math_overflow/shard_2.txt" "b/stack-exchange/math_overflow/shard_2.txt" deleted file mode 100644--- "a/stack-exchange/math_overflow/shard_2.txt" +++ /dev/null @@ -1,42765 +0,0 @@ -TITLE: Characteristic Complexes in Iwasawa theory -QUESTION [6 upvotes]: For all that follows, $p$ is a fixed odd prime. In the formulation of the Noncommutative Main Conjecture of Iwasawa theory one uses étale cohomology to define an algebraic object analogous to Iwasawas 'charakteristic ideal' in $\Lambda(G)$ for $G=Gal(k_\infty/k)$, with $k^{cyc}\subset k_\infty$ and $\mu=0$: -Let $M_\Sigma$ be the maximal abelian, pro-$p$, outside of $\Sigma$ unramified extension of $k_\infty$ for a finite set of primes of $k$, $\Sigma$. Then $X:=Gal(M_\Sigma/k_\infty)$ is a $\Lambda(G)$ module. If $G$ has elements of order $p$ we are prevented from seeing this $X$ a relative $K_0$ group, associated to a denominator set, $S$, of $\Lambda(G)$. -Now étale cohomology enters the picture: We use it to define a complex $C$ of $\Lambda(G)$-modules which is $S$-acyclic and quasi-isomorphic to a bounded complex of finitely generated $\Lambda(G)$-modules. Although it looks quite technical, I will give here for the sake of quick reference the definition of -$C=RHom(R\Gamma_{ét}(Spec(\mathcal{O_{k_\infty}}[\frac{1}{\Sigma}]),\mathbb{Q}_p/\mathbb{Z}_p),\mathbb{Q}_p/\mathbb{Z}_p)$. -This $C$ is strongly correlated to $X$, namely $H^0(C)=\mathbb{Z}_p$ and $H^{-1}(C)=X$. -Now my question: An expert in the field told me that this $\mathbb{Z}_p$ is 'moraly' related to the pole of a zetafunktion. How is this? -Is this even related to the Main Conjecture, where evalutations at representations of $G$ and $p$-adic interpolation play the lead role? As far as I understand it, the trivial representation, leading to the zeta function, is left undealt with. -I apologize for my ignorance on this basic question of the field. - -REPLY [5 votes]: The reason why one cannot take the class of $X$ in the relative $K_0$ when $G$ has $p$-torsion is because $X$ may not have finite resolution by finitely generated projective $\Lambda(G)$-modules. This is necessary even if $G$ is abelian. Hence we take the complex $C$ above. $\mathbb{Z}_p$ appearing there is indeed interpreted as a pole. One way thinking about this in the commutative Iwasawa theory is as follows- if $G$ is of the form $H \times \mathbb{Z}_p$, with $H$ a finite abelian group, then for each one dimensional character $\chi$ of $H$ there is a $p$-adic $L$-function, say $L_p(\chi)$, constructed by Deligne-Ribet, Cassou-Nogues, Barsky. $L_p(\chi)$ is expected to have a simple pole only when $\chi$ is the trivial character. On the algebraic side one has to make sense of what a characteristic ideal for $X$ means when order of $H$ is divisible by $p$. One can either use the $K$-theory formulation as suggested by Fukaya-Kato or one can do something more directly but only get weaker information. Look at $V:=X \otimes \overline{\mathbb{Q}}_p$. This is a finite dimensional $\overline{\mathbb{Q}}_p$ vector space. For each $\chi$ we can consider $V^{(\chi)}$, the $\chi$ isotypic part. Let $f(\chi)$ be the characteristic polynomial of $1-\gamma$ acting on $V$ (Here $\gamma$ is a fixed topological generator of $\mathbb{Z}_p$). Then for every non-trivial $\chi$ the polynomial $f(\chi)$ generates the same ideal as $L_p(\chi)$ in the ring $\mathbb{Z}_p[\chi][[\mathbb{Z}_p]] = \mathbb{Z}_p[\chi][[T]]$. However, when $chi$ is the trivial character, we have that $f(\chi)$ and $TL_p(\chi)$ generate the same ideal. Or $f(\chi)/T$ is essentially same as the $p$-adic $L$-function. So apart from the information about the characteristic element of $X$, the $p$-adic $L$-function has this additional pole (the $T$ in the denominator) corresponding to the module $\mathbb{Z}_p$ with the trivial action. This is exactly what happens in the noncommutative situation as well. It is not a new phenomenon of the noncommutative theory- it was always there. -The answer to your next question is yes, it is related to the main conjecture. We conjecture that there is an element $\zeta \in K_1(\Lambda(G)_S)$ which maps to the class of $C$ in $K_0(\Lambda(G), \Lambda(G)_S)$ and which is related to values of $L$-functions at Artin representations of $G$ at odd negative integers i.e. evaluations at even positive Tate twists of Artin representations of $G$. We now have this $\zeta$ and the main conjecture in the noncommutative situation and again one expects $\zeta$ a simple pole when evaluated at the trivial representation of $G$.<|endoftext|> -TITLE: Root systems and sums of squares -QUESTION [9 upvotes]: It is easy to see that the quadratic form for the root system $A_n$ is a sum of $n+1$ squares of integral linear forms: -$$q_{A_n} = 2 \sum_{i=1}^n x_i^2 - 2 \sum_{i=1}^{n-1} x_i x_{i+1} = -x_1^2 + (x_1-x_2)^2 + \dots (x_{n-1}-x_n)^2 + x_n^2$$ -It is equally easy to see that $q_{D_n}$ is a sum of $n$ squares. It is a little harder to see but I think is true that $q_{E_n}$ ($n=6,7,8$) is not a sum of $\ge n$ squares of integral forms. -Question: is this a standard fact, well-known to experts? Is there a standard reference? (I hate to reinvent a bycicle.) And has this fact been used for something interesting? (I have an interesting application in mind, so I am looking for connections...) - -REPLY [5 votes]: Just saw this thanks to a "Related" link from -Question 154928. -Yes, it is known that the $E_6$ form cannot be written as a sum of -integral squares, and thus (by specialization) that the same is true of -$E_7$ and $E_8$; moreover the representations of $D_n$ ($n>2$) and $A_n$ -as the sum of $n$ and $n+1$ squares respectively are the only ways -(up to isomorphism) to write these forms as the sum of any number of -nonzero squares, with the exception of $D_3 \cong A_3$ which has -both a three-square and a four-square representation. -Or at least it is known once one makes Will Jagy's -key observation that writing a form as a sum of $m$ integral squares is -tantamount to embedding the corresponding lattice into ${\bf Z}^m$. -(As it happens I was just asked a few days ago whether any integral -positive-definite lattice can be embedded in some ${\bf Z}^m$, -so this question was very familiar.) -I don't know a reference, but the result is not hard starting from the -Coxeter diagrams -and the fact that the vectors of norm $2$ in ${\bf Z}^m$ are exactly -the vectors $e+e'$ for some $\pm$ unit vectors $e,e'$ with $e' \neq \pm e$. -For $A_n$ we need a sequence of $n$ such vectors any two of which -are orthogonal except that consecutive vectors have inner product $-1$. -The first two must be $e'-e, \, e''-e'$ with $e,e',e''$ orthogonal -unit vectors. The third could be either $e'''-e''$ or $e+e'$. -The latter choice does not extend to $A_4$, -and the former extends uniquely to $e''''-e'''$, and then by induction to -$\{ e^{(i)} - e^{(i-1)} \}_{i=1}^n$ with all $n+1$ unit vectors orthogonal. -For $D_n$ we need an $A_{n-1}$ configuration together with a norm-$2$ vector -orthogonal to all but the second vector, with which it has inner product $-1$. -For $n=3$ we've done this already because the $D_3$ and $A_3$ diagrams -are isomorphic. For $n=4$ either of our two $n=3$ solutions extends uniquely -and both give $e'-e, e''-e', e'''-e'', e+e'$. For all $n > 4$ the unique -$A_{n-1}$ diagram extends uniquely, again with extra vector $e+e'$. -We can now obtain the impossibilty of the $E_6$ configuration -by trying to extend either $D_5$ at a short end or $A_5$ at the middle vertex -(or by trying to overlap $D_5$ with $A_5$ or with another $D_5$). -Since $E_7$ and $E_8$ contain $E_6$, they are impossible too. -Hence none of the $E_n$ lattices are contained in any ${\bf Z}^m$, -whence the corresponding quadratic forms are not sums of integral squares, -QED.<|endoftext|> -TITLE: Groups of Hodge type, hodge structure on Lie algebra -QUESTION [5 upvotes]: Hi, -Let $W$ be a real algebraic group, and $G$ the associated complex group. Then $W$ is of Hodge type if there is a $\mathbb{C}^*$ action on $G$ such that $U(1)$ preserves $W$ and the action of $-1$ on $G$ is a Cartan involution. -I have trouble understanding the definition, I guess because I don't understand very well the definition of Cartan involution : it should be something like the complex conjugation relative to a compact real form of $G$. -Examples: $SU(p,q)$, $SO(2n)$ ($n\geq 3$), $Sp(n)$, $Sp(p,q)$, $SO(p,2q)$ ($q\geq 2$) and some other classical Lie groups are apparently of Hodge type. But I don't see the action of $U(1)$, the compact real form etc. -Also, I was told that a group is of Hodge type if the Lie algebra has a Hodge structure. Is it easy to see the Hodge structure in the examples given by Simpson ? -Thanks for your help. - -REPLY [2 votes]: That $-1$ is a Cartan involution is the same as saying that the subgroup $K$ of points in G fixed under the action of the adjoint of this copy of $-1$ is a maximal compact subgroup of $G$. -For example, in $W=U(p,q)$ the subgroup $K=U(p)\times U(q)$ is the centraliser of the diagonal matrix $k$ whose first $p$ entries are $-1$ and whose last $q$ entries are $+1$. This element $k$ may be viewed as an element of the diagonal ${\mathbb C}^*$ embedded in $GL_{p+q}({\mathbb C})$ (the latter is the complexification of $W$), where $z\in {\mathbb C}^*$ is embedded $G$ as the diagonal matrix whose first $p$ entries are $z$ and the last $q$ entries are $1$. It is clear that $K$ is a maximal compact of $W$. -Similar constructions exist for $Sp(2n)$ with $K=U(n)$. In general, you may like to replace your semi-simple group by a reuctive group, to see the ${\mathbb C}^*$ action more clearly; for example, $U(p,q)$ was more convenient that $SU(p,q)$ in the above example. -You can look up Milne's articles on "Shimura varieties" for detailed descriptions of these Hodge groups.<|endoftext|> -TITLE: When can one extend a flat family from a subscheme to the whole scheme? -QUESTION [8 upvotes]: Is there a nice condition on a closed subscheme $Y$ of $X$ such that for every flat family $Z\to Y$, there is a flat family $W\to X$ whose restriction to $Y$ is $Z$? In particular, I'm interested in the case when the closed subscheme is two lines in $\mathbb{P}^2$, or three planes in $\mathbb{P}^3$, or generally $n$ hyperplanes in $\mathbb{P}^n$. - -REPLY [5 votes]: I'm pretty sure the answer is not in general. -Take a Hilbert scheme which is reducible, for example, that of the Hilbert polynomial $3t + 1$ in $\mathbb{P}^3$. This Hilbert scheme has two irreducible components - the one corresponding to twisted cubics and the one corresponding to degree 3 plane curves union a point. The intersection of the these two components corresponds to degree 3 nodal plane curves with an embedded point "pointing out of the plane". It turns out that both components are rational (actually this is pretty easy to see) and even that the intersection is rational. -Anyhow, take a line in the twisted cubic component that meets the other component (see III.9.8.4 in Hartshorne's Algebraic Geometry) at a point $[C]$ and a line through $[C]$ contained in the other component. This gives you a flat family over the union of two intersecting lines in $\mathbb{P}^2$, but you will not be able to "fill it in" because the Hilbert scheme in question is not irreducible. -EDIT: I realized that, in this example, the family extended to $\mathbb{P}^2$ doesn't have to be as a family of closed subschemes in $\mathbb{P}^3$. Still, it cannot extend to a flat family in this example - just for a different reason. The general fiber of the extension would be a rational curve, but over a line in $\mathbb{P}^2$ you would have a disconnected family. This is impossible. - -REPLY [3 votes]: This is not going to help too much in your case, but we do have the following I think. -Suppose that we have a map $f : X \to Y$ which has a section (whose image we call $Y$ as well, as above), in other words $Y \subseteq X \to Y$ is an isomorphism. -Given a flat family $Z \to Y$, we can form $Z \times_Y X \to X$. I think this should do what you want. Of course, it won't help with $Y = $ two lines in $\mathbb{P}^2$.<|endoftext|> -TITLE: What is the outer automorphism group of SU(n)? -QUESTION [18 upvotes]: All the automorphisms of $SU(2)$ seem to be inner, which would mean that $\mathrm{Out}$ $SU(2)$ is trivial. Is that correct? Is this true in general $SU(n)$? I can't quite see -- any thoughts would be helpful. - -REPLY [3 votes]: I found the previous answers somewhat unsatisfying. If one takes a complex semisimple Lie group (e.g. $G=GL(n,\mathbb{C})$), then $Out(G)$ is huge, since it is an algebraic group over $\mathbb{Q}$, and therefore acted on by $Aut(\mathbb{C}/\mathbb{Q})$, which is huge. In fact, I don't know what the full automorphism group is. The same applies to $\mathbb{H}^{\ast}$, the non-zero quaternions, when viewed as the Cayley-Dickson construction applied to $\mathbb{C}$. However, taking the unit quaternions $\mathbb{H}^1\cong SU(2)$ requires complex conjugation, which must preserve $\mathbb{R}$, and therefore gets rid of all of $Aut(\mathbb{C}/\mathbb{Q})$ except for complex conjugation. However, this argument doesn't exclude other outer automorphisms. So I think one can reduce your question to (modulo the previous answers): when does the real form of a Lie group have only continuous (and therefore analytic) automorphisms?<|endoftext|> -TITLE: Is there a percolation threshold in the hard discs model? -QUESTION [11 upvotes]: Take a random configuration of $n$ non-overlapping discs of radius $r$ in the unit square $[0,1]^2$. (You could think of this as taking $n$ points uniform randomly in $[r,1-r]^2$ and then restricting attention to the case that every pair of points $\{ x,y \}$ satisfies $d(x,y) \ge 2r$.) -From the point of view of statistical mechanics, the interesting case is when $r^2 n \to C$ for some constant $C > 0 $ as $n \to \infty$, and various kinds of phase transitions have been studied experimentally, but little seems known mathematically. -What I am wondering about is whether anyone has studied the following kind of percolation. Set $\lambda >1$ to be a fixed parameter to measure proximity. Define a graph by considering the centers of the $n$ discs to be the vertices, and connecting a pair $\{ x,y \}$ by an edge whenever $d(x,y) < 2 \lambda$. - -My question is: given some choice of - $\lambda$ does there exist a critical - threshold $C_t$ (depending on $\lambda$) such that whenever $C -> < C_t$ all the connected components of this - graph are likely to be small, of order - $O(\log n)$ or even $o(n)$, - and whenever $C > C_t$ there is a giant - component, of order - $\Omega(n)$? - -What I know about is that for geometric random graphs on i.i.d. random points, percolation is known to occur for fairly general distibutions, and that this is closely related to bond percolation on a lattice. But in the hard spheres distibution points are far from being independent. -I would also be interested to hear about percolation on other kinds of repulsive point processes -- Matern, Strauss, etc. - -REPLY [4 votes]: Perhaps you are already aware of asymptotics for component sizes in the continuum percolation model. It is there in the book of M. Penrose. -Since you asked about percolation on repulsive point processes, here is a compressed version of our results of non-trivial phase transition on sub-Poisson point processes , point processes that are less clustered than a Poisson in a certain sense. Our main example is a perturbed lattice. We also show that stationary determinantal point processes have a non-zero critical intensity as regards percolation. I think these point processes can be considered as repulsive point processes.<|endoftext|> -TITLE: Variants of Martin's axiom -QUESTION [8 upvotes]: Let P be the statement: Every ccc partial order has $\omega_1$-precaliber; i.e., every uncountable subset $X$ of a ccc partial order $P$ has an uncountable subset $Y$ such that for every finite subset $F$ of $Y$, there is a member in $P$ below every member of $F$. -Let Q be the statement: Product of ccc partial orders is ccc. -It is known that $MA(\omega_1)$ (Martin's axiom at $\omega_1$) implies $P$ and that $P$ implies $Q$. Do these implications reverse? - -REPLY [7 votes]: Todorcevic and Velickovic (Martin's axiom and partitions) proved $P$ implies $\text{MA}_{\aleph_1}$.<|endoftext|> -TITLE: Which covers of Lie groups will I get -QUESTION [5 upvotes]: Here is a question I get from sitting in my Lie algebra class: -Fix a Lie algebra $\mathfrak{h}$, we know there is a unique simply connected Lie group $H$ which serves as the universal cover of other connected Lie groups with the same Lie algebra. Now assuming $H$ is a $n_1$ sheeted cover of $H_1$, and a $n_2$ sheeted cover of $H_2$, (we are not restricting the Deck groups yet). we know $\phi: \mathfrak{h_1}\cong \mathfrak{h_2} (\cong \mathfrak{h})$ but we only cares about the isomorphism between the first two factors. Now as we travel along all $a\in \mathfrak{h_1}$, $(a, \phi (a))$ will form a Lie subalgebra in $\mathfrak{h_1}\bigoplus\mathfrak{h_2}$, denoted by $\mathfrak{h'}$. By the existence theorem of Lie subgroups we have a uniqueLie subgroup $G\subset H_1\times H_2$ corresponding to $\mathfrak{h'}$. Note $\mathfrak{h_1},\mathfrak{h_2}, \mathfrak{h'}$ are isomorphic Lie subalgebras in $\mathfrak{h_1}\bigoplus\mathfrak{h_2}$, however since they sort of "sit in different directions" in the ambient Lie algebra when we form the exponential map we are producing non isomorphic Lie subgroups. Anyhow, Project down from $G$ to $H_1$ and $H_2$ induce an iso on the Lie algebra therefore we know $G$ actually covers $H_i$. That's the set up of the question. -Now if I know $n_1$ and $n_2$ are coprime to each other, then automatically $G$ has to be the universal cover $H$. However, if say the maximal common divisor of $n_1$ and $n_2$ is 2, and that there are two non isomorphic Lie groups that could cover both $H_1$ and $H_2$, namely H and the one doubly covered by H donoted $\tilde{H}$, (here I need some compatible condition on the Deck group, but let assume $\tilde{H}$ covers $H_i$). So my question is which one is isomorphic to $G$ when I draw the graph? I am worried a little bit that the there may not a canonical way of choosing this isomorphism $\phi$ to make this question make sense, (or is there)? My guess is that if there is a "canonical choice", then the $G$ we get from the graph should be the smaller $\tilde{H}$, while my colleage thinks that by choosing different $\phi$, you can get both. When $H_1$ and $H_2$ have a lot of non isomorphic common covers, then by choosing different $\phi$ you can produce all of them as subgroups of $H_1\times H_2$? (fixing $H_1$ and $H_2$.) I really hope someone ccan shed lights on this, thanks very much! - -REPLY [2 votes]: If we fix universal covering maps $H \to H_1$ and $H \to H_2$, then $G$ is uniquely defined as the image of the diagonally embedded $H \subset H \times H$ under the covering map to $H_1 \times H_2$. If you transform one of the covering maps from $H$ using the deck group, you will get a group isomorphic to $G$.<|endoftext|> -TITLE: An elementary lemma of commutative algebra -QUESTION [12 upvotes]: Let $R$ be a commutative ring, $M$, $N$ $R$-modules, and $f: M\rightarrow N$ a homomorphism. It is known that $f$ is injective (surjective) if and only if $f_m$ is injective (surjective) for all maximal ideal $m$. But I don't know whether $f$ is split if and only if $f_m$ is split? Maybe it is true for finitely generated modules over noetherian rings, I suspect. - -REPLY [8 votes]: If you want to avoid the use of Ext-groups, you could prove it like this (which is basically the same proof): -Let $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ be a short exact sequence of $R-$modules with $C$ finitely presented and assume it splits after localisation at every maximal ideal. -Use the natural isomorphism $\operatorname{Hom}_{R_m}(C_m,A_m)=R_m\otimes\operatorname{Hom}_R(C,A)$ (which uses the flatness of localisation and the fact that $C$ is finitely presented) and the assumption to see that the map $\operatorname{Hom}_R(C,B)\rightarrow\operatorname{Hom}_R(C,C)$ is surjective since all of its localisations are. -Since the giving a splitting is equivalent to this map being surjective we are done. -P.S.: Of course one could prove "if and only if" in the statement like this. -Counterexample in the general case: Take $R=\prod_{\mathbb N} \mathbb F_2$, $I=\sum_{\mathbb N} \mathbb F_2$ and $C$ the cokernel of the inclusion. -Now I claim three things: -1.) $R$ has dimension zero -Proof: Every element of $R$ is an idempotent and every prime ideal in $R$ has to contain exactly one of $e$ or $1-e$ for every idempotent $e$ in R. If we had a chain of prime ideals, the larger one would necessarily have to contain $e$ and $1-e$ for one such idempotent and so couldn't exist. -2.) The localisation of $R$ at every maximal ideal is a field. -Proof: It is a zero-dimensional local ring by 1.) and reduced since $R$ doesn't contain nilpotent elements. Thus this localisation is a field. -3.) $I$ is not a direct summand of $R$. -Proof: Direct summands correspond to idempotents in $R$. Since every element in $R$ is idempotent we need to analyse all principal ideals. If an element has only finitely many non-zero entries the ideal created by it has only finitely many elements, thus can't be $I$. -If on the other hand it contains infinitely many non-zero entries, the ideal created by it has uncountably many elements. -Since $I$ contains countably many elements, it can't be a direct summand of $R$. -This establishes the counterexample, since the short exact sequence $0\rightarrow I\rightarrow R\rightarrow C\rightarrow 0$ splits in every localisation at a maximal ideal.<|endoftext|> -TITLE: Reference request: 2-Grothendieck Construction -QUESTION [7 upvotes]: Hi Folks, -i'm looking for a reference on the 2-grothendieck construction for a functor $F:\mathcal{I}\to \mathcal{B}\mathrm{icat}$ from a bicategory $\mathcal{I}$ to the tricategory of bicategories. Actually for my purposes it would be sufficient to consider functors going only to $\mathcal{C}\mathrm{at}$. - -REPLY [9 votes]: I. Bakovic, Grothendieck construction for bicategories.<|endoftext|> -TITLE: Base locus of divisors on blowings up of the projective space -QUESTION [8 upvotes]: Let $X$ be the blowing-up of $\mathbb{P}^n$ in $r$ points in general position. -Let $\{H,E_1,...,E_r\}$ be the standard basis of $Pic(X)$. Further let $D=dH-\sum m_j E_j$, be an effective divisor, with $m_j \in \mathbb{Z}$. -Is it true that if $m_1\geq 0$ then $E_1 \not\subseteq Bs(|D|)$? -If we suppose all the $m_i$'s are non negative, this corresponds to say that -$dim |dH-m_1E_i...-m_rE_r|>dim |dH-(m_1+1)E_1-...-m_rE_r|$. -In other words this means that if we consider all the hypersurfaces of $\mathbb{P}^n$ of degree $d$ passing through $r$ general points $\{p_1,...,p_r\}$ with multiplicities $m_1,...,m_r$, then they do not all pass through $p_1$ with multiplicity $m_1+1$. -My intution brings me to say this is obvious but I cannot prove it. -In fact everything is trivial for nonspecial linear systems, but, a priori, special linear systems may be a problem. -Note that the generality of the points is necessary: for example you can simply take $\mathbb{P}^2$ blown up in 3 collinear points and note that $E_1 \subseteq Bs(|H-E_2-E_3|)$. - -REPLY [2 votes]: It seems to me that my question is easier than SHGH conjecture. -I think that prop. 2.3 in "Weakly defective varieties" by Chiantini-Ciliberto is a positive answer to my question.<|endoftext|> -TITLE: Algebraic integers on the unit circle -QUESTION [5 upvotes]: Consider a set of algebraic integers which lie on the unit circle, they will generate a multiplicative subgroup of $\mathbb S^1$. Do these objects have a name? -I would guess they contain useful arithmetic/number theoretic information, for example if the generating set is the set of roots of an irreducible polynomial, what kind of information would they contain? -Has the group structure of the elements of a number field which lie on the unit circle $\mathbb S^1$ and the subgroup of it which is made up of algebraic integers been studied. -Would greatly appreciate if you could suggest a reference. -Regards -Vagabond -PS -It would be real nice if you could answer keeping in mind that I do not know much Algebra/ commutative algebra/ algebraic geometry. But I hope that does not stop you from answering. - -REPLY [3 votes]: The question is a bit ambiguous because "lie on the unit circle" is ambiguous. A closely worded question is: - -Let $S$ be a set of archimedean places of a number field $K$. What is the subgroup $K_S \subset K^{\times}$ of nonzero numbers with $|x|_v = 1$ for all $v \in S$ ? - - -If we call the complement of $S$ (in the set of places) $\bar{S}$, then the above group is called $\bar{S}$-units. The notion is usually applied to finite $\bar{S}$ containing all archimedean places, so $S$ contains none. But we don't have to be so restrictive. So to answer the first question: the name should probably be "$\bar{S}$-units$\cap O_K^\times$" -Assume $S$ is non empty. For a non-cyclotomic quadratic field the group seems to be empty. For the two exceptional cyclotomic fields the group is of course the roots of unity. I would guess that for most fields the group is finite, and in fact consists of the roots of unity in the field (for most notions of "most fields" that most of us have). Of course this doesn't have to be the case - for example if the field has a Salem number.<|endoftext|> -TITLE: Does a referee have to check carefully the proof ? -QUESTION [64 upvotes]: I have always checked very carefully the papers I was refereeing when I wanted to suggest "accept". Actually I spend almost as much time checking the maths of a paper I referee than checking the maths of a paper of mine (and this is very long !). But I have some doubts. Is it really my job as a referee ? -This question is related to Refereeing a Paper but only a few comments were made on that point in the above cited thread (mainly Refereeing a Paper). - -REPLY [11 votes]: I believe that the referee should check the proof is correct, or at least that the logical deduction of its claimed results from the assumed results it quotes is correct. It probably is unreasonable to expect referees to check the correctness of cited results in a paper- this could easily become a never-ending task. While there is probably a general consensus that responsibility for correctness rests primarily with the author(s), the community needs safeguards to ensure that genuine mistakes by authors are caught before they become assimilated into the body of accepted mathematical knowledge. -However, I also believe that authors who quote previous results in their own proof have a responsibility to ensure as far as possible that they fully understand the results they are quoting, and their correctness. If a quoted result turns out to be wrong, the person who quoted it can't evade their own responsibility by blaming the referee of the paper quoted. -Of course there is a conflict between idealised behaviour, and what is attainable in the "real" world, but in the long run, Mathematics, perhaps above all other human endeavours, strives for an idealized perfection, and abandoning that striving would be fatal for the subject.<|endoftext|> -TITLE: Presentation of the cohomology of generalized flag varieties as graded ranks of rings of symmetric polynomials -QUESTION [8 upvotes]: Hello! -Let $n,m\geq 0$ be integers. If I understand it correctly, there is the following description of the cohomology of the complex Grassmannian $\text{Gr}(m+n;m)$: denote by $\text{Sym}(n,m)$ the subring of the polynomial ring in $n+m$ variables consisting of polynomials which are symmetric in the first $n$ and the last $m$ variables, and by $\text{Sym}(n+m)$ the ring of symmetric polynomials in $n+m$ variables. Then $\text{Sym}(n,m)$ is a free $\text{Sym}(n+m)$ module, and its graded rank (written as a $q$-polynomial) equals the Poincare polynomial of the Grassmannian $\text{Gr}(m+n;m)$. -I'd like to know if this is true for more general flag varieties: Is it true that the graded rank of $\text{Sym}(m_1,...,m_k)$ over $\text{Sym}(m_1+...+m_k)$ is the Poincare polynomial of the variety of flags $(\{0\}=U_0,U_1,...,U_k=V)$ in $V := {\mathbb C}^{m_1+...+m_k}$ such that $\text{dim}(U_{i}) - \text{dim}(U_{i-1}) = m_i$ for $i=1,...,k$? If yes: how to prove it? :-) -On the other hand, there is a presentation of the cohomology of the full flag variety of ${\mathbb C}^N$ as the quotient of ${\mathbb C}[X_1,...,X_N]$ by the ideal generated by the symmetric polynomials of positive degree. How do these two descriptions of the cohomology of flag varieties relate? -Thank you! -Hanno - -REPLY [4 votes]: Yes, this is all true. The cohomology of the partial flag variety $F(m_1,\dots,m_k)$ is the quotient of the polynomials that are symmetric under $S_{m_1}\times \cdots \times S_{m_k}$ by the positive degree ones which are fully symmetric (a special case of this is the presentation you mention for the full flag variety). Since partially symmetric polynomials are a free module over full symmetric ones, the q-rank of the free module is the same as the q-dimension of the quotient by all positive degree fully symmetric polynomials. -The "reason" this presentation works is that $F(m_1,\dots,m_k)$ carries tautological vector bundles $U_i/U_{i-1}$ (these are subquotients of the trivial bundles of rank $m_1+\cdots m_k$). The Chern classes of these bundles are the elementary symmetric polynomials in the variables corresponding to $S_{m_i}$; thus, the fully symmetric polynomials, written in terms of these are the Chern classes of the the trivial bundle of rank $m_1+\cdots m_k$ by the Whitney sum formula (which says that if I think of $c_i(V)$ as elementary symmetric functions in one set of variables, and $c_j(W)$ as elementary symmetric functions in another, the Chern classes $c_k(V\oplus W)$ are the elementary symmetric functions in the union of the variables) and the splitting principle (which says that when I have a filtration $ V_1\subset V_2\subset \cdots\subset V_n=V$, then $c_i(V)=c_i(V/V_{n-1}\oplus \cdots \oplus V_2/V_1\oplus V_1)$) are thus 0. Admittedly, you then have to do some actual geometry to see that these Chern classes generate and that these are all the relations, but to me, this is the heart of the argument. - -REPLY [2 votes]: One way is to use the Eilenberg-Moore spectral sequence applied to the fibration -$$ -\mathrm{Fl}_{m_1,\ldots,m_k} \rightarrow BU_{m_1}\times\cdots\times -BU_{m_k}\rightarrow BU_{m_1+\cdots+m_k} -$$ -which is a spectral sequence -$$ -\mathrm{Tor}_*^{\mathrm{Sym}(m_1+\ldots+m_k)}(\mathrm{Sym}(m_1,\ldots,m_k),\mathbb - Z) \implies H^*(\mathrm{Fl}_{m_1,\ldots,m_k}). -$$ -However, as $\mathrm{Sym}(m_1,\ldots,m_k)$ is projective as module over -$\mathrm{Sym}(m_1+\ldots+m_k)$ (by for instance the fact that is a finite -Cohen-Macaulay module over a regular ring of the same dimension) the sequence -collapses and gives -$$ -H^*(\mathrm{Fl}_{m_1,\ldots,m_k}) = -\mathrm{Sym}(m_1,\ldots,m_k)\bigotimes{}_{\mathrm{Sym}(m_1+\ldots+m_k)}\mathbb Z -$$ -which gives what you want.<|endoftext|> -TITLE: Is every connected scheme path connected? -QUESTION [40 upvotes]: Every (?) algebraic geometer knows that concepts like homotopy groups or singular homology groups are irrelevant for schemes in their Zariski topology. Yet, I am curious about the following. -Let's start small. Consider a local ring $A$ with maximal ideal $M$; is the affine scheme $X=Spec(A)$ connected? Sure, because every open subset of $X$ containing $M$ is equal to $X$ itself. Or because the only idempotents of $A$ are $0$ and $1$. But is it path connected? Yes, because if you take any point $P$ in $X$ the following path $\gamma$ joins it to $M$ (reminds you of the hare and the tortoise...): -$ \gamma(t)=P \quad for \quad 0\leq t < 1\quad , \quad \gamma (1)=M $. -The same trick shows that the spectrum of an integral domain is path connected: join the generic point to any prime by a path like above. More generally, in the spectrum of an arbitrary ring $R$ you can join a prime $P$ to any bigger prime $Q$ $(P \subset Q)$ by adapting the formula above: -$ \gamma(t)=P \quad for \quad 0\leq t < 1\quad , \quad \gamma (1)=Q $. -[Continuity at $t=1$ follows from the fact that every neighbourhood of $Q$ contains $P$ and so its inverse image under $\gamma$ is all of $[0,1]$ ] -The question in the title just asks more generally: -Is a connected scheme path connected ? -Edit (after reading the comments) If an arbitrary topological space is connected and if every point has at least one path connected open neighbourhood, then the space is path connected. But I don't see why the local condition holds in a scheme, affine or not, even after taking into account what I proved about local rings. - -REPLY [25 votes]: There exist connected affine schemes which are not path connected. Let E be a compact connected metric space* which is not path connected (e.g., the closed topologist's sine curve) and consider the following. - -$X={\rm Spec}(A)$ where $A$ is the ring of continuous functions $f\colon E\to\mathbb{R}$. - -Then X is connected, since any idempotent f satisfies $f(x)\in\{0,1\}$ and, by connectedness of E, $f=0$ or $f=1$. The maximal ideals of A are -$$ -\mathcal{m}_x=\left\{f\in A\colon f(x)=0\right\} -$$ -for $x\in E$. There will also non-maximal primes (see this question for example) but, every prime ideal will be contained in one and only one of the maximal ideals**. So, we can define $\pi\colon X\to E$ by $\pi(\mathcal{p})=x$ for prime ideals $\mathcal{p}\subseteq\mathcal{m}_x$. -In fact, $\pi$ is continuous, using the following argument. For any open ball $B_r(x)$ in E, choose $f\in A$ to be positive on $B_r(x)$ and zero elsewhere. Then $D_f=\left\{\mathcal{p}\in X\colon f\not\in \mathcal{p}\right\}$ is open and $\pi^{-1}(B_r(x))\subseteq D_f\subseteq \pi^{-1}(\bar B_r(x))$. -Writing $B_r(x)=\cup_{s < r}B_s(x)=\cup_{s < r}\bar B_s(x)$, this shows that there are open sets $U_s$ lying between $\pi^{-1}(B_s(x))$ and $\pi^{-1}(\bar B_s(x))$. So, $\pi^{-1}(B_r(x))=\bigcup_{s < r} U_s$ is open, and $\pi$ is continuous. -So, $\pi\colon X\to E$ is continuous and onto. If X was path connected then E would be too. -It may be worth noting that ${\rm Specm}(A)$ is also connected but not path connected, being homeomorphic to E. - -(*) I assume that E is a metric space in this argument so that the open balls give a basis for the topology, and there are continuous $f\colon E\to\mathbb{R}$ which are nonzero precisely on any given open ball. Actually, it is enough for the topology to be generated by the continuous real-valued functions. So the argument generalizes to any compact Hausdorff space (+ connected and not path connected, of course). -(**) Maybe I should give a proof of the fact that every prime $\mathcal{p}$ is contained in precisely one of the maximal ideals $\mathcal{m}_x$. Let $V(f)=\{x\in E\colon f(x)=0\}$ be the zero set of f. Then, $V(\mathcal{p})\equiv\bigcap\{V(f)\colon f\in\mathcal{p}\}$ will be non-empty. Otherwise, by compactness, there will be $f_1,f_2,\ldots,f_n\in\mathcal{p}$ with $V(f_1)\cap V(f_2)\cap\cdots\cap V(f_n)=\emptyset$. Then, $f=f_1^2+f_2^2+\cdots+f_n^2\in\mathcal{p}$ would be nonzero everywhere, so a unit, contradicting the condition that $\mathcal{p}$ is a proper ideal. Choosing $x\in V(\mathcal{p})$ gives $\mathcal{p}\subseteq\mathcal{m}_x$. -On the other hand, we cannot have $\mathcal{p}\subseteq\mathcal{m}_x\cap\mathcal{m}_y$ for $x\not=y$. Letting $f,g\in X$ have disjoint supports with $f(x)\not=0, g(y)\not=0$ gives $fg=0\in\mathcal{p}$ and, as $\mathcal{p}$ is prime, $f\in\mathcal{p}\setminus\mathcal{m}_x$ or $g\in\mathcal{p}\setminus\mathcal{m}_y$.<|endoftext|> -TITLE: how to recognize subgroups through Dynkin diagram? -QUESTION [5 upvotes]: Fix $\mathbb{C}$ as the base field, and reductive groups are assumed to be connected. -Consider the example $SO_N\subset SL_N$. $SO_N$ is its own normalizer in $SL_N$, and the rank is much smaller than the rank of $SL_N$. How can this be reflected in the Dynkin diagram of $SL_N$? -In Reductive subgroup corresponding to a subdiagram of the Dynkin diagram of a simple group one sees that Levi subgroups $L$ of a given semi-siimple group $G$ can be recovered from Dynkin subdiagrams of the diagram of $G$, and subgroups of maximal rank can be recovered similarly from extended Dynkin diagrams. As is commented below, having maximal rank means being of the same rank as $G$ is, namely containing a maximal torus of $G$. -Contrary to the discussions in loc.cit, one considers a reductive subgroup $H\subset G$, such that the normalizer $N(H,G)$ of $H$ in $G$ is of lower rank than $G$. Its connected component, denoted as $N^\circ$, is not covered in loc.cit. Say $SO_{2n}\subset SL_{2n}$, the extended Dynkin diagram of $SL_{2n}$ looks like a loop with dots, while the one for $SO_{2n}$ contains branching vertices. It is not clear that the latter is produced from the former by removing vertices. -And furthermore when one passes to a general base field, say perfect or of characteristic zero for simplicity, with separable closure $\bar{k}$, it becomes more complicated if one restricts to the notion of $k$-rank. Consider a reductive $k$-subgroup $H\subset G$ that is self-normalizing, in the sense that it equals the neutral connected component of $N(H,G)$. By comparing the $k$-ranks and $\bar{k}$-ranks of $H$ and $G$, one is led to several different cases. Does one still have arguments similar to the operations on Dynkin diagrams as is in Reductive subgroup corresponding to a subdiagram of the Dynkin diagram of a simple group ? -Sorry for any misunderstanding about the cited mathoverflow discussions above, and references on extended Dynkin diagrams are also welcome. - -REPLY [4 votes]: Not easy!. If you want to know classification of $A_1$-subalgebras then it is the same as classification of nilpotent elements and it can be reduced to classifying all distinguished parabolics in all your Levi subalgebras. -For bigger subalgebras, you should read original Dynkin's paper that contains more or less complete answer if I understand things correctly...<|endoftext|> -TITLE: "Averaging" in von Neumann algebras -QUESTION [8 upvotes]: I've been reading recently about various "averaging" tricks in von Neumann algebras. For example, Christensen and Sinclair, in "On von Neumann algebras which are complemented subspaces of $B(H)$." J. Funct. Anal. 122 (1994), no. 1, 91--102. Here we have a von Neumann algebra $M\subseteq B(H)$ and a completely bounded projection $P:B(H)\rightarrow M$. Consider a family $(u_i)$ in $M$ with $\sum_i u_i^* u_i=1$ strongly. Then it's shown that there is a cb map $Q:B(H)\rightarrow M$ which is a right $M$-module map, such that $Q(x)$ is in the w*-closed convex hull of $\sum_i P(xu_i^*)u_i$, for each $x\in B(H)$. Then doing this on the right yields an $M$-bimodule projection, from which it follows that $M$ is injective (and so this solved the problem of whether having a cb projection-- which isn't contractive-- onto $M$ implies that $M$ is injective). -A vaguely similar notion occurs in Dixmier's approximation theorem: For each $a\in M$, the norm closed convex hull of $\{ u^*au: u\in U(M)\}$ meets the centre of $M$ (here $U(M)$ is the unitary group of $M$). -We might wonder if for each $x\in B(H)$, the norm (or even ultraweak) closed convex hull of $\{u^*xu:u\in U(M)\}$ intersects $M'$. But I think, it you consider $M$ acting on $H\otimes K$ for a sufficiently large $K$, this would imply a projection from $B(H)$ onto $M'$, which is proving too much. -So my, slightly vague, question is: - -What are some other, similar, examples of "averaging" in this fashion? - -I'm not interested in the case when M is assumed injective: that's too much. But I would be interested in special cases, like, if $M$ is a $II_1$-factor. I'd be quite happy just to have references, not detailed answers. - -REPLY [5 votes]: As far as the Dixmier property, there is a relative version that Popa has shown holds for for certain inclusions. (i don't have the paper in front of me and i don't quite remember exactly what he shoes). He also has one for $C^*-algebras.$ -The papers are -S. Popa, The relative Dixmier property for inclusions of von Neumann algebras of ¯nite index, Ann. -Sci. Ec. Norm. Sup., 32 (1999), 743-767. -and -S. Popa, On the relative Dixmier property for inclusions of C¤-algebras, Journal of Funct. Analysis, 171 -(2000), 139-154. -Secondly, (and i would say more interestingly for me) is the averaging involved in Popa's intertwining by bi-modules technique. Without going into too much detail, if two subalgebras are close in norm on their unit balls, then by averaging, one can get a partial isometry that intertwines a corner of one into a corner of the other. This is an absolutely crucial element of all the progress in the past 8 years or so in the classification of von Neumann algebras coming from groups, or ergodic group actions. -For references prob the best place to start would be Popa's ICM talk -http://www.math.ucla.edu/~popa/ICMpopafinal.pdf -or appendix C of Stefaan Vaes's Bourbaki seminar. -http://arxiv.org/PS_cache/math/pdf/0603/0603434v2.pdf<|endoftext|> -TITLE: Langlands in dimension 2: the Yoshida conjecture -QUESTION [44 upvotes]: Background: -One prominent part of the Langlands program is the conjecture that - all motives are automorphic. - It is of interest to consider special cases that are more precise, if less - sweeping. The idea is to formulate generalized modularity conjectures - that are as concrete as the Shimura-Taniyama-Weil conjecture - (now the elliptic modularity theorem). The latter gained much in - precision, for example, by Weil's experimental observation of the link between - the conductor of the elliptic curve and the level of the weight two - modular form. As a next step up the dimension ladder it is natural to consider - abelian surfaces over ${\mathbb Q}$, in which case one encounters -Yoshida's conjecture: - Any irreducible abelian surface $A$ defined over ${\mathbb Q}$ and with End$(A)={\mathbb Z}$ is modular in the sense that associated to each is a holomorphic Siegel modular - cusp eigenform $F$ of genus 2, weight 2, and some level $N$, such - that its spinor L-function $L_{\rm spin}(F,s)$ agrees with that of the - abelian surface - $$ - L(H^1(A),s) ~=~ L_{\rm spin}(F,s). - $$ -Questions: - -Has Yoshida's conjecture been proven for some classes of - abelian surfaces over ${\mathbb Q}$? -Are there lists of abelian surfaces and associated Siegel modular forms that extend the very useful lists constructed by Cremona and Stein for elliptic curves and their associated modular forms? - -REPLY [5 votes]: And so it turns out that I was in the audience of a seminar talk just today on this very subject. The opinion I expressed in comments is apparently not too far from the truth: V.Pilloni and B.Stroh now have some variants of Coleman's criterion for classicality for $\textrm{GSp}_{4}$, and so full modularity for some abelian surfaces is now accessible. So the answer to your question 1 seems to be yes, but you need to rely on cutting edge results to really get these kind of results. -Check out the joint works of V.Pilloni and B.Stroh.<|endoftext|> -TITLE: What is the (Koszul? derived?) interpretation of a pair of Lie algebras with the same cohomology? -QUESTION [7 upvotes]: There are many words and sentences in mathematics that I basically completely don't understand, including the words "Koszul" and "derived". But rather than ask for a complete description of such words, I will ask about a particular example, and hope that an MOer can spell out concretely what's going on. (Of course, links to good write-ups of Koszul, etc., are always welcome.) -There is a relatively easy theorem, probably due to Chevalley and Eilenberg, with many generalizations due to Koszul: - -Let $L$ be a (finite-dimensional) vector space (in characteristic $0$); let $[1] L^*$ denote the dual vector space, "shifted" to a graded vector space supported in degree $1$; and let $([1]L^*)^{\vee \bullet}$ denote the free graded-commutative algebra generated by $[1]L^*$ (I impose the Koszul rule for signs, so that as an algebra $([1]L^*)^{\vee \bullet} = (L^*)^{\wedge\bullet}$ is classically an alternating algebra). Then to give a Lie algebra structure to $L$ is the same as giving to $([1]L^*)^{\vee \bullet}$ a square-zero degree-$1$ derivation. - -The construction is (contravariantly) functorial and full and faithful, and so embeds the category of Lie algebras fully-faithfully into the (opposite to the) category of commutative dgas. -Here are two examples. I will call my characteristic-$0$ field $\mathbb R$. The one-dimensional Lie algebra corresponds to the dga $\mathbb R \overset 0 \to \mathbb R \xi$, where $\xi$ is the coordinate function on the one-dimensional vector space, and the algebra is graded-commutative, so $\xi^2 = 0$. The two-dimensional non-commutative Lie algebra corresponds to the dga $\mathbb R \overset 0 \to (\mathbb R \xi \oplus \mathbb R \upsilon) \overset{\xi \mapsto 0, \, \upsilon \mapsto \xi\upsilon}\longrightarrow \mathbb R\xi\upsilon$. -Now, once we're in the land of dgas, it makes sense to talk about their (co?)homology. In the above examples, the cohomology of the one- and two-dimensional Lie algebras are isomorphic as algebras; the isomorphism on cohomology is given in one direction by the "abelianization" map from the two-dimensional Lie algebra to the one-dimensional Lie algebra. - -Question: How should I interpret "geometrically" the fact that the cohomologies agree? - -An idea that I've heard is that somehow dgas-up-to-? should correspond to some notion of "stack". Now, it is certainly not the case that "point mod one-dimensional" and "point mod two-dimensional" present the same stack. So that's not quite what's going on. Perhaps the problem is that these algebras are not the same as ${\rm A}_\infty$ algebras; we should probably add the word "${\rm A}_\infty$" to the list of words I don't really know. If this is the answer, I hope that some MOer will spell it out. -Another idea that I've heard is that the passage "Lie algebra to dga to cohomology" remembers exactly the "derived category of representations of the Lie algebra". Again, I don't really know what that is, but that's OK. Somehow, a representation of a Lie algebra should be a "quasicoherent sheaf on point mod the Lie algebra", and I know that people like "derived categories of quasicoherent sheaves". So should I understand Koszul duality as remembering not all the data of some "stack-like object" but just some "derived data"? If so, again I hope that some MOer will spell it out pedantically in this example. - -REPLY [6 votes]: You (and "everyone") may already know this, but there is a really simple answer to a variant of your question, that is when instead of starting with a Lie algebra you start with a connected differential graded Lie algebra over a field of characteristic zero. In that case, Quillen's work on rational homotopy theory says that data determines a rational homotopy type, and Lie algebra cohomology is the cohomology of that homotopy type. So that answer to your question is that two (connected differential graded) Lie algebras with the same cohomology have the same kind of relationship to one another as two spaces which have isomorphic cohomology rings... which is not necessarily much unless you know something more about them. To some small extent Lie algebras when viewed as differential graded Lie algebras in degree zero (with trivial differential) are analogous to K(G,1)'s, but general theory says little there, so I don't think this addresses your main question. -Those interested in an elementary, expository topologist's view of Koszul duality might want to look at this.<|endoftext|> -TITLE: What's known about the relationship about EQP and BQP? -QUESTION [12 upvotes]: EQP is the class of problems solvable deterministically using a quantum computer in polynomial time - that seems to me to be a good analogue to P, whereas BQP is the quantum analogue of BPP. -It doesn't seem like much is known about EQP! Just like BPP is not known to be contained in NP, is it known whether EQP \subseteq BQP \subseteq QMA? Is there a corresponding "Derandomization" of BQP into EQP? -What about the relationship of EQP and P? - -REPLY [4 votes]: Scott gives the good answer that at the technical level, EQP is not uniquely defined. However, there are deeper issues of interpreting quantum complexity classes, and deciding which ones are contrived and which ones are natural. -To a computer scientist who first learns finite quantum mechanics, it may look like BQP is to BPP as EQP is to P. That was indeed the motivation for defining EQP. However, the analogy treats quantumness and randomness and separate generalizations of deterministic computation, and this isn't really true. A better viewpoint is that quantumness is a strengthened form of randomness. So the question "BQP is to BPP as (blank) is to P" is like the question "motorcycling is to bicycling as (blank) is to walking". It's a strange question. -The question "BQP is to EQP as BPP is to (blank)" is actually less strange, and my answer would not be P. Rather, you could define "EPP" to be the class of problems where, if the true answer is yes, the algorithm accepts with probability exactly 2/3; if the true answer is no, the algorithm accepts with probability exactly 1/3. This might not be a perfect analogy either, but it does show you how EQP is a contrived combination of probability and exactitude. If you use binary gates, then this definition of EPP is empty! The precise behavior of EPP is depends on algebraic coincidences. In order to have a viable definition (never mind natural), you should either change the probabilities to 1/4 and 3/4 (say), or you should use ternary circuits. -However, it dawned on me once that non-uniform EQP, or EQPn.u.,, is a well-defined complexity class. Because, in the non-uniform setting, you can allow all gates on two qubits. EQPn.u., is not necessarily a natural complexity class, but it isn't particularly less natural than non-uniformity itself. Similarly to a non-uniform classical circuit, you can show that a non-uniform quantum circuit of exponential size can express any unitary operator. A more robust interpretation of Mosca's and Zalka's theorem is that factoring is in EQPn.u.. It is also an interesting question to compare EQPn.u. to BQP/qpoly, and I don't know the answer. -Per Ricky Demer's remark, ZQP as defined in the Complexity Zoo is another EQP-like class that has the same shortcomings. However, I penciled into the Zoo a more interesting operational equivalent of ZPP that I called ZBQP. It's not a great name, but ZQP is already taken. My definition of ZBQP is that a quantum computer produces a certificate for either yes or no that can be checked by a deterministic verifier in polynomial time. And, the quantum computer succeeds in its task in polynomial time with high probability. If you use a classical randomness-enhanced computer instead of a quantum computer, you get the class ZPP. For example, any factoring-related decision question, such as whether a number is square-free, is in ZBQP, and this fact neither implies nor is implied by Mosca-Zalka.<|endoftext|> -TITLE: history of the topological dissection problem -QUESTION [5 upvotes]: I am trying to locate a modern account of the problem of determining the number of pieces into which a certain geometric set is divided by given subsets. An example of such a problem could be to count the chambers of a real hyperplane arrangements which was solved by Zaslavsky. -I am trying to find an article which has some history of related problems. What I have right now are accounts by Grunbaum written 70's. - -REPLY [3 votes]: Maybe look over the Table of Contents of the 1992 book Arrangements of Hyperplanes by Peter Orlik, Hiroaki Terao to see if what you seek is there. -The paper "On the number of arrangements of pseudolines" by Stefan Felsner uses some very nice analysis to improve the known upper bound at the time (1996), and cites earlier relevant work. -This may be more algorithmic than you care about, but an authoritative -survey as of 2000 was "Arrangements and their applications" by Pankaj Agarwal and Micha Sharir in the -Handbook of Computational Geometry. If you are more interested in the combinatorics of, -say, pseudosphere arrangements, then -perhaps you should look at oriented matroids, e.g., Jurgen Richter-Gebert, and Günter Ziegler, "Oriented Matroids" in the Handbook of Discrete and Computational Geometry, 2004. -Perhaps none of this helps, but it may lead to a sharpening of your question.<|endoftext|> -TITLE: Finite covers of complex varieties (all but two questions answered!) -QUESTION [14 upvotes]: EDIT: Thanks to several people I almost have a complete answer. BCnrd pointed out that the generalized Riemann Existence Theorem shows that $\tilde{Y}$ can be uniquely given the structure of a complex variety $\tilde{Y}'$. Also, Georges pointed out that the Kodaira embedding theorem implies that $\tilde{Y}'$ is projective if $Y'$ is projective. -This leaves two parts open to my original question. If $Y'$ is quasiprojective, then must $\tilde{Y}'$ be quasiprojective? Also, if $Y'$ is affine, then must $\tilde{Y}'$ be affine? -The impression I get from reading the comments is that the answers are "yes" and that everyone but me is able to easily prove them given what has already been said. Can anyone give me a hint or a reference as to how to proceed? -Thanks to everyone for all your help so far. - -ORIGINAL QUESTION: -Let $Y$ be a complex manifold that can be given the structure of a complex variety $Y'$. Let $\pi:\tilde{Y} \rightarrow Y$ be a finite, unramified cover of $Y$. Can $\tilde{Y}$ be given the structure of a complex variety $\tilde{Y}'$ such that there is a finite map $\pi' : \tilde{Y}' \rightarrow Y'$ making the obvious diagram commute? If the answer is yes, then can we take $\tilde{Y}'$ to be projective/quasiprojective/affine if $Y'$ is projective/quasiprojective/affine? -This kind of thing is true for Riemann surfaces, but even there I don't know how to prove it except by going through the whole machinery showing that all compact Riemann surfaces are projective varieties. Since such things are not available in higher dimensions, I'm stuck. -I should maybe remark that I don't even know how to do the above for affine varieties. -Thanks! - -REPLY [2 votes]: A finite unramified map mapping to an algebraic variety is finite in the algebraic sense and hence affine, so the preimage of an affine subset is affine. (I suppose this assumes that by affine you mean affine algebraic. I am not entirely sure what happens if the target is an affine analytic subset of $\mathbb C^n$. I assume there may be an analytic equivalent of a finite map being affine, but I don't know). -A finite map is also projective, hence if the target is quasi-projective, then the source is projective over something quasi-projective so itself is quasi-projective.<|endoftext|> -TITLE: When is the time evolution of a Hamiltonian system described by the geodesic flow on a Riemannian manifold? -QUESTION [21 upvotes]: Here is my precise question. Let $M, \omega$ be a symplectic manifold and let $H: M \to \mathbb{R}$ be any smooth function. The symplectic form gives rise to an isomorphism between the tangent bundle and cotangent bundle of $M$, and in this way we can associate to the 1-form $dH$ a vector field $X_H$ which is characterized by the property that $\omega(X_H, Y) = Y(H)$ for any vector field $Y$. The one parameter group of diffeomorphisms associated to $X_H$ is the "Hamiltonian flow" associated to $H$. -An interesting special case of this construction is furnished by Riemannian geometry. For any manifold $M$, there is a canonical symplectic structure on $T^*M$ (regarded as a manifold in its own right) defined as follows. Given a tangent vector $X \in T(T^*M)$ sitting over a covector $p \in T^*M$, define $\eta_p(X) = p(d\pi_p(X))$ where $\pi: T^*M \to M$ is the natural bundle projection. Then $\eta$ is a 1-form on $T^*M$, and one checks that $d\eta$ is a symplectic form. If $M$ is equipped with a Riemannian metric $g$ then the metric yields an isomorphism between $TM$ and $T^*M$, and the construction of the previous paragraph produces a Hamiltonian flow associated to any smooth function on $TM$. If we consider the smooth function $H: TM \to \mathbb{R}$ given by $H(V) = g(V, V)$, then it is a fact that the resulting Hamiltonial flow $F_t$ is precisely the geodesic flow for $M$. In other words, given a tangent vector $W \in T_p M$, $F_t(W)$ is the velocity vector at time $t$ of the unique geodesic $\gamma$ with $\gamma(0) = p$, $\gamma'(0) = W$. -So I am wondering if there are interesting invariants - dynamical, geometric, topological, or otherwise - which help to determine whether or not a given Hamiltonian system is secretly the geodesic flow on some Riemannian manifold. This is kind of a screwy question from a geometric point of view, because it essentially asks if given a smooth function $H: M \to \mathbb{R}$ on a symplectic manifold there is a submanifold $N$ of $M$ such that there is a diffeomorphism $M \to TN$ which carries $H$ to a positive definite quadratic form on each fiber. But dynamically it boils down to a fairly natural question: how can one characterize geodesic flows among all Hamiltonian dynamical systems? -If this question has any sort of reasonable answer, I can think of half a dozen follow-up questions. Is there a natural notion of equivalence up to which $N$ is unique? To what extent does $H$ constrain the geometry and topology of $N$? If a Lie group acts on the pair $M, H$ can we choose $N$ which is invariant under the group action? For example, one idea along these lines that comes to mind immediately is the assertion that if the Hamiltonian flow for $H$ is not ergodic relative to a prescribed smooth invariant measure then $N$ cannot have nonpositive curvature. If you have an answer to this question and you can elaborate on the relationship between $H$ and the geometry of $N$, please do so. - -REPLY [2 votes]: There is the paper ARE HAMILTONIAN FLOWS GEODESIC FLOWS? -by CHRISTOPHER MCCORD, KENNETH R. MEYER, AND DANIEL OFFIN which treats this problem and gives examples of Hamiltonian flows (most cases of the planar $N$-body problem, for instance) that cannot be seen as geodesic flows.<|endoftext|> -TITLE: How does one compute invariants of certain Grassmannians inside the regular representation? -QUESTION [25 upvotes]: Barry Mazur and I have come across the question below, motivated by (but independent -of) issues regarding the Leopoldt conjecture. -Suppose that $\mathbf{C}$ is the complex numbers. -Let $H$ be a finite set, and -let $S$ be a subset of $H$. The vector space $X = \mathbf{C}^{H}$ -has a canonical basis consisting of the generators $[h]$ for $h \in H$. Let -$X_S$ denote the subspace generated by $[h]$ for $h \in S$. -Suppose we consider a subspace $U$ of $X$, and ask for the -dimension of $U \cap X_S$. The answer will depend on $U$, but - we expect the codimension of the intersection will generically -be the sum of the codimensions of each space. In other words, let -$G(X,U)$ denote the set of vector spaces $V$ -inside $X$ which -are abstractly isomorphic to $U$ - that is, $\dim(V) = \dim(U)$. -Then -$$\min \{ \dim(V \cap X_S), \ | \ V \in G(X,U)\} = \min\{0,\dim(U) + |S| - |H|\}\}.$$ -Indeed, $G(X,U)$ is a Grassmannian, and equality holds on an open set in $G(X,U)$. -Suppose we now assume that $H$ is a group. We continue to assume that $S$ is -a subset of $H$ (not necessarily a subgroup), -and that $X_S$ is the subspace generated by $[h]$ for $h \in S$. -The vector space $X$ now has extra structure --- it has a left and right action of $H$, -indeed, $X \simeq \mathbf{C}[H]$ is just the regular representation of $H$. -Suppose that $U$ is a subspace of $X$, and let us additionally assume that -$H.U = U$, that is, $U$ is a representation of $H$, and the inclusion $U -\subseteq X$ is an inclusion of left $X = \mathbf{C}[H]$-modules. We now suppose -that $G_H(X,U)$ is the set of left $X$-modules $V$ inside $X$ which are abstractly -isomorphic to $U$ as representations of $H$ --- this is contained in but generally -much smaller than the space of vector subspaces isomorphic to $U$. The question is: can one compute -$$\delta(H,S,U) := \min \{ \dim(V \cap X_S), \ | \ V \in G_H(X,U)\}.$$ -That is: what is the expected dimension of this intersection given the structure -of $U$ as an $H$-module? -$G_H(X,U)$ can be identified with a product of Grassmannians, namely -$G(v_i,u_i)$ where $V_i$ are the irreducible representations of $H$, -$v_i = \dim(V_i)$, and $u_i = \dim \mathrm{Hom}_H(V_i,U)$. A lower bound for -$\delta(H,S,U)$ is given by $\min\{0,\dim(U) + |S| - |H|\}$, but this is -not optimal in general. -Remark: If $u_i \in \{0,v_i\}$ for all $i$, then $G_H(X,U)$ consists of a single -point. -Example: Let $H = \langle \sigma \rangle$ be cyclic of order four. Let $U$ -consist of the subspace generated by $[1] + [\sigma^2]$ and $[\sigma] + [\sigma^3]$. -Then $G_H(X,U) = \{U\}$ is a point. If $S = \{[1],[\sigma^2]\}$ then -$\dim X_S \cap U = 1$, even though $\dim(U) + |S| -|H| = 0$. -The most general question is: Can one compute $\delta(H,S,U)$ in a nice way? -A more specific question is: Can one compute $\delta(H,S,U)$ when $H$ has an -element $c$ of order two and $U = X^{c = 1}$, that is, the elements of $X$ -which $c$ acts by $1$ on the right. $u_i = \dim \mathrm{Hom}_H(U,V_i)$ is equal to -$\dim(V_i,c = 1)$ in this case. -An even more specific question is: Can one compute $\delta(H,S,U)$ when -$H = S_4$ (with representations of dimension $1$, $1$, $2$, $3$ and $3$, and -$U = X^{(12) = 1}$ has corresponding multiplicities $1$, $0$, $1$, $1$, and $2$? -What about $H = D_{8}$ or $D_{10}$, and $c$ is a reflection? -Example: Suppose that $U = X^{c = 1}$, and suppose that $c$ is central in $H$. -This is exactly the condition on $c$ to ensure that $G_H(X,U)$ is a point. Then -$$\delta(H,S,U) = \frac{1}{2} | \{ s \in S \ | \ cs \in S\}|.$$ -This generalizes the previous Example, where $c = \sigma^2$. - -Let me make the following remark. As I mentioned in the question, One can obtain the complete answer when $c$ is central. In this case, one obtains generic intersections that are "larger" than what one expects from the linear algebra, at least when $|G| > 2$. It follows that a similar thing happens whenever $G$ admits a quotient $G/H$ of order $> 2$ where $c$ is central. This explains why one would expect degenerate answers in dihedral groups $D_{2n}$ of order divisible by $4$. Having done $D_3 = S_3$ by hand, the next "interesting" case is $D_5$. If I understand Greg's answer correctly, the generic intersection is always as small as possible for $D_5$ and $D_7$ and any $S$. Thus the most optimistic conjecture is that this is always the case providing that $|G^{ab}| \le 2$, for example, if $G = S_4$. - -REPLY [7 votes]: This is not a complete solution, but it is a way to compute things in sort-of a nice way. The module $U$ has a complementary module $U^\perp$, and the kernel of a generic module map $F:X \to U^\perp$ is isomorphic to $U$. Then you can look at the restriction $F:X_S \to U^\perp$ and try to compute its generic rank. This puts things as close as possible to linear algebra, because $F$ comes from a linear family of linear maps. The first question is whether there is any anomalous intersection, which amounts to asking whether $F$ has maximal rank, which amounts to computing the full minors of $F$. Each maximal minor simply corresponds to restricting to $X_T$ for a subset $T \subseteq S$ such that $|T| = \dim U^\perp$. This is simply the statement that -$$\dim (X_S \cap V) - \dim (X_T \cap V) \le \dim X_S - \dim X_T,$$ -which is true solely because $X_T$ is a subspace of $X_S$. -You ask about certain special cases in which $U$ comes from a permutation representation $H/c$, and $U^\perp$ is thus induced from the sign representation $c = -1$. In such a case, the matrix of $F$ has a simple form in which each entry is a single variable, and the variables are negated or repeated across entries. For example, if $H = S_3$ and $c$ is a transposition, then I get -$$F = \begin{pmatrix} -x & -y & -x & z & y & -z \\ -y & -x & -z & x & z & -y \\ -z & -z & -y & y & x & -x -\end{pmatrix}$$ -for all of $X$. We are interested in determinants and ranks of minors of this matrix, and evidently we can remove all of the minus signs without changing any answers. This is explained by the fact that $c$ is complemented by the alternating group and we can tensor $X$ by the sign representation to switch $U$ with $U^\perp$. For the same reason, we can switch $U$ and $U^\perp$ for the dihedral cases as well, because $c$ is complemented by a subgroup of index 2. Then $F$ becomes a generic module map between two permutation representations of $H$; each matrix entry of such an $F$ is a single variable. -It is easy to check that for the example of $S_3$ and $c$ a transposition, every full minor has the property that one of three variables ($x$, $y$, or $z$) is a transversal. This is a sufficient condition for that minor to be generically non-zero. So, there are no anomalous generic intersections for this $H$ and this $U$. -If $c$ is central, then the matrix of $F$ has repeated columns. Obviously, if a minor of $F$ has repeated columns, then it vanishes. But this is not the only way to get an anomalous generic intersection. For instance, if $H = D_4$ and $c$ is a reflection, then (if you switch from $U = X^{c=1}$ to $U = X^{c=-1}$ as I argued you can), you get the matrix -$$F = \begin{pmatrix} x & y & z & w & x & y & z & w \\ -w & x & y & z & y & z & w & x \\ -z & w & x & y & z & w & x & y \\ -y & z & w & x & w & x & y & z \end{pmatrix}$$ -The odd-numbered columns are a vanishing minor. The reason is that the vector -$$(1, 0, 1, 0,-1, 0, -1, 0)$$ -is always in $V$. -So, although I certainly don't have a complete solution, I can find examples of $S$ where there isn't an anomalous intersection (because one of the variables is a transversal), and examples where there is an anomalous intersection (because every admissible $V$ contains the same vector in $X_S$). And, this formulation of the problem makes it easier to check a given $S$ by computer. - -In other words, the question is essentially a matroid problem. The set $F(H)$ is a spanning set in the module $U^\perp$, and you are interested in the matroid structure of $F(H)$ for a generic $H$, because a circuit or linearly dependent subset $F(S)$ gives you an $S$ such that $V \cap X_S$ is larger than expected. I wrote a short program in Sage to compute this matroid for $D_n$, where the element $c$ is a reflection, using a convenient matroid package written by David Joyner. -n = 6 # For the dihedral group D_n -gen = primes(100,200) # Generic parameters - -# http://boxen.math.washington.edu/home/wdj/sagefiles/matroid_class.sage -load 'matroid_class.sage' - -R. = PolynomialRing(ZZ,2) -names = [x^k for k in range(n)] + [x^k*y for k in xrange(n)] -a = list(gen)[:n] -F = matrix(QQ,[[a[(i-j)%n] for i in range(n)] + [a[(i+j)%n] for i in range(n)] - for j in range(n)]) -for S in vector_matroid(F).circuits(): - if len(S) <= n: print [names[k] for k in S] - -For $D_4$ and $D_6$, the program found various matroid circuits of size 4 and 6, respectively. For $D_5$ and $D_7$, it did not find any non-trivial circuits. I do not understand their full structure, but there they are. For $D_8$, it eventually reporteed that there are 362 non-trivial circuits of size 8 and none that are smaller. The first one listed is $S = \{1,x,x^2,x^3,y,xy,x^2y,x^3y\}$, if the group is -$$D_8 = \langle x,y | x^8 = y^2 = xyxy = 1 \rangle$$ and $y=c$. -Although this package is convenient, it is not efficient. With a more efficient code, it should be possible to find all of the non-trivial circuits in $D_{10}$ and maybe $S_4$.<|endoftext|> -TITLE: Representation of vectors in $\mathbb{R}^2$ via differences of small vectors. -QUESTION [15 upvotes]: Is the following fact true? - -Let $v_1,\ldots, v_k \in \mathbb{R}^2$, $\|v_i\|\leq 1$, be vectors that add up to zero. Does there exist a permutation $\sigma\in S_k$ and vectors $w_1,\ldots, w_k \in \mathbb{R}^2$, $\|w_i\|\leq 1$, such that $v_{\sigma(i)}=w_i-w_{i-1}$? (here, I assume $w_0=w_k$) - -Edit. Related (known) facts: -1. The same fact in $\mathbb{R}^1$ is true. (can be easily proven by choosing $w_0=0$ and $\sigma(i)$ such that $\|w_i\|\leq 1$ for $w_i:=w_{i-1}+v_{\sigma(i)}$) -2. For each $\epsilon>0$ there exists a family of vectors $v_i$, such that $\max_i\|w_i\|>1-\epsilon$. See my comment below for the proof. - -REPLY [8 votes]: I have a counterexample in $\mathbb R^2$. -Here's how it goes. -Pick two numbers $n$ and $N$, with $N>>n>>1$. -The collection {$v_i$} consists of: - -$N(n-1)$ times the vector $(\frac{n-2}{n},\frac{1}{N(2n-3)})$ -$N(n-2)$ times the vector $(-\frac{n-1}{n},\frac{1}{N(2n-3)})$ -The vector $(0,-1)$ once. - -The smallest ball into which those vectors can be fit back-to-back has diameter $\sqrt{5}-\varepsilon$.<|endoftext|> -TITLE: Computation of a Drazin inverse -QUESTION [6 upvotes]: I need to compute the Drazin inverse $A^D$ of a singular M-matrix $A$, i.e., a matrix in the form $A=\lambda I -P$, where $P$ has nonnegative entries and $\lambda$ is the spectral radius (Perron value) of $P$. I already know the right and left eigenvectors $v$ and $u^T$ of $P$ with eigenvalue $\lambda$ (that is, the vectors in the left and right kernel of $A$). -1) Is there a Matlab subroutine around for computing Drazin inverses? I can't seem to find any, so I had to create my own (which is probably very inefficient) -2) Is there a way to exploit the knowledge of the two nullspaces (and the fact that they are 1-dimensional) to speed up this computation? - -REPLY [2 votes]: If you have access to Matlab, and the matrix is not gigantic -you can use the following to calculate the Drazin inverse: - -Determine the index of $A$, i.e, does $\mathrm{rank}(A^k) = \mathrm{rank}(A^{k+1})$ -Solve the linear system $A^{k+1} X = A^k$ -Use the fact that $A^D$, the Drazin inverse, is given by - -$$A^D = A^k X^{k+1}$$ -Jim Shoaf, N. C. Central University<|endoftext|> -TITLE: What is the name for the composition of a functor with a natural transformation? -QUESTION [12 upvotes]: In the second half of the section "Operations with Natural Transformations" of the wikipedia article on natural transformations, they define the operation taking a natural tranformation $\eta:F\to G$ in the functor category $Cat(C,D)$ and functor $H:D\to E$ which produces a new natural transformation $(H\circ F)\to (H\circ G)$ in $Cat(C,E)$ by applying the functor to every coordinate of the original natural transformation. This operation frequently comes up in the definition of an adjunction. -Does this operation have a name? It's not the "Godement multiplication", but might be related to it. -Is there an agreed-upon symbol for this operation, other than juxtaposition? -(Minor rant: I think that using juxtaposition for this operation is a horribly cruel thing to do to people learning category theory. Juxtaposition is already used for composition of morphisms -- and therefore for composition of functors-with-functors and naturaltransformations-with-naturaltransformations, since they too are the morphisms in $Cat$ and functor categories, respectively. Recycling juxtaposition yet again for this (non-associative!) operation on functors and natural transformations is just asking for trouble.) - -REPLY [22 votes]: It's called "whiskering" -- the 1-cells/functors composed on either side of the 2-cell/transformation look like "whiskers". See for example page 24 of this paper. This terminology is pretty widespread in the categorical community.<|endoftext|> -TITLE: fibonacci series mod a number -QUESTION [10 upvotes]: I'm trying to write a program with an input of numbers $n$ and $k$ (where $n<10^{1000}$ and $k<10^9$), where I compute fib[n] % k. -What is a good FAST way of computing this? -I realize that the resulting series is periodic, just not sure how to find it efficiently. - -REPLY [3 votes]: Perhaps Elsenhans, Jahnel, "The Fibonacci sequence modulo $p^2$ – -An investigation by computer for $p < 10^{14}$" http://www.uni-math.gwdg.de/tschinkel/gauss/Fibon.pdf will be interesting for you. There are sections about the algorithm.<|endoftext|> -TITLE: The difference of two sums of unit fractions -QUESTION [6 upvotes]: I had this question bothering me for a while, but I can't come up with a meaningful answer. -The problem is the following: -Let integers $a_i,b_j\in${$1,\ldots,n$} and $K_1,K_2\in$ {$1,\ldots,K$}, then how small (as a function of $K$ and $n$), but strictly positive, can the following absolute difference be. -$\biggl|(\sum_{i=1}^{K_1} \frac{1}{a_i})-(\sum_{j=1}^{K_2} \frac{1}{b_j})\biggr|$ -As an example for $K_1=1,$$K_2=1$ choosing $a_1=n,$$b_1=n-1$gives the smallest positive absolute difference, that is $\frac{1}{n(n-1)}$. What could the general case be? - -REPLY [3 votes]: We can prove Christian Elsholtz's conjecture using only linear functions. -Let $m_1, ..., m_{2K}$ and $\alpha_1, ..., \alpha_{2K}$ be integers satisfying $\sum_i \frac{\alpha_i^j}{m_i} = 0$ for $j = 0, ..., 2K-2$. Then for large $x$ we have -$\sum_i \frac{1}{m_ix - m_i\alpha_i} = \frac{1}{x}\sum_i\frac{1}{m_i}\sum_j\frac{\alpha_i^j}{x^j} = \sum_j \frac{1}{x^{j+1}} \sum_i \frac{\alpha_i^j}{m_i} \approx \frac{C}{x^{2K}}$, -where $C = \sum_i \frac{a_i^{2K-1}}{m_i}$. -For instance, we can pick $\alpha_i = i-1, m_i = \frac{(-1)^iD}{\binom{2K-1}{i}}$, where $D$ is a multiple of $\binom{2K-1}{i}$ for each $i$. Then we will have the same number of positive and negative $m_i$s, so for large $x$ half of the fractions will be positive and half will be negative. -Examples: -$\frac{1}{3x} + \frac{1}{x-2} - \frac{1}{x-1} - \frac{1}{3x-9} = \frac{-2}{x(x-1)(x-2)(x-3)}$, and -$\frac{1}{10x} + \frac{1}{x-2} + \frac{1}{2x-8} - \frac{1}{2x-2} - \frac{1}{x-3} - \frac{1}{10x-50} = \frac{-12}{x(x-1)(x-2)(x-3)(x-4)(x-5)}$.<|endoftext|> -TITLE: Existence of an $\omega$-nonstandard model of ZFC from compactness -QUESTION [5 upvotes]: I have read several times that assuming Con(ZFC), and using compactness it can be proved the existence of a model of ZFC with an ill-founded $\omega$. How is that? Any reference will be welcome. - -REPLY [6 votes]: I'm not sure, but maybe the question is why a non-ω-model of ZFC must have infinite decreasing $\in$ chains. This follows from two facts: - -Viewed externally, any nonstandard model of Peano arithmetic is non-well-founded. Given a nonstandard number $n$, consider the sequence $n-1$, $n-2$, ... -The standard construction of a model of Peano arithmetic within ZFC uses $\in$ for the order relation on the natural numbers. So if the natural numbers within a model of set theory are not well founded, then the model's set inclusion relation is also not well founded. These are, formally, different meanings of "well founded". - -It's interesting to ask how the model can think that its natural numbers are well founded, given that the descending sequence above was completely concrete. The answer is that if that sequence is defined within the model, as $a_k = n-k$, then $k$ can be any number in the model, even a nonstandard one, and the model verifies that this sequence reaches $0$ after $n$ steps. Only from an external viewpoint can we limit $k$ to an external set of "standard" natural numbers.<|endoftext|> -TITLE: Is there a group-scheme equivalent of the theorem that any Lie group is diff. to a compact one cross R^n? -QUESTION [5 upvotes]: I'm rather ignorant in both fields, but I would still like to endeavor asking this question. I've just learned that any Lie group is diffeomorphic to a compact Lie group cross $\mathbb{R}^n$. While there is no (to my knowledge) equivalent to diffeomorphic in the group-schemes language, this does have obvious implications about the cohomology of Lie groups (which has an analog in the group-scheme language.) -So: Is there an analog of this theorem for group-schemes? What is it? What can we say? - -REPLY [3 votes]: For algebraic groups over a perfect field $k$, one has Chevalley's Theorem. It says that every algebraic group $G$ over $k$ contains a unique closed normal linear subgroup $H$ such that $G/H$ is an abelian variety. The abelian variety is the analogue of the compact Lie group, and the linear group $H$ is the analogue of affine space.<|endoftext|> -TITLE: $\kappa$-scales and the continuum -QUESTION [7 upvotes]: In Jech's Set Theory he defines a $\kappa$-scale as a family of functions $\langle f_\alpha\colon\omega\to\omega | \alpha < \kappa \rangle$ for which: - -$f_\alpha < f_\beta$ except maybe for a finite set -For any $g\colon\omega\to\omega$ there is some $f_\alpha>g$ (again for all but perhaps a finite set) - -(the definition can be found in chapter 10 which deals with measurable cardinals) -Clearly if a $\kappa$-scale does exist then $\kappa \le 2^{\aleph_0}$ (as there are only that many functions to begin with). -If we look at the quasi-order on $\omega^\omega$ defined by as above, that is $f < g$ if $f(n) < g(n)$ for all but a finite number of $n\in\omega$, then we can immediately say that: - -There are no maximal elements (for any $f$ we can define $g(n) = f(n) +1$ and clearly $f< g$) -For any given $f,g$ there exists an upper-bound for both (namely $\max (f(n), g(n))$) -If there exists a $\kappa$-scale then there is a cofinal subset of the quasi-order of cardinality $\le \kappa$ (cofinal in the sense that for every $f$ you can find some $g$ in the cofinal subset for which $f< g$) - -My question is, if so, assuming $\kappa$ is the least cardinal number for which a $\kappa$-scale exists, and $A\subseteq \omega^\omega$ some $ < $-cofinal subset. Does it follow that $|A|\ge\kappa$? Does the assumption that $2^{\aleph_0} = \lambda$ holds some property (e.g. regularity) affects the existence of such $\kappa$ or $A$? - -REPLY [5 votes]: This is a question regarding two famous cardinal characteristics: - -$\mathfrak{b}$ the minimal cardinality of an unbounded family in $(\omega^\omega,{<^*})$. -$\mathfrak{d}$ the minimal cardinality of a cofinal family in $(\omega^\omega,{<^*})$. - -It is not difficult to show that a scale exists if and only if $\mathfrak{b} = \mathfrak{d}$, and then the minimal length of a scale is the common value of these two cardinal characteristics (which is also the cofinality of any scale). However, this equality need not hold. In fact, the only inequalities that must hold are -$$\aleph_1 \leq \mathfrak{b} = cf(\mathfrak{b}) \leq cf(\mathfrak{d}) \leq \mathfrak{d} \leq \mathfrak{c}.$$ -Using Hechler's Theorem (see this answer of mine) any pattern of cardinals consistent with the above is possible to attain by a forcing. See Andreas Blass's handbook of set theory chapter (available here) for more on this topic.<|endoftext|> -TITLE: Is the mapping cylinder of a Serre fibration also a Serre fibration? -QUESTION [8 upvotes]: If we have a Serre fibration $p: E \rightarrow B$ with fiber of homotopy type $S^{k-1}$, then we can create a fibration with contractible fiber by first taking the mapping cylinder $M_p$ of $p$ to get a map $M_p \rightarrow B$ (not necessarily a fibration) with contractible "fiber" and then applying the "space of paths" construction to get a fibration $M_p \times_B B^I \rightarrow B$. My question is, is this last part of the construction necessary, or is the mapping cylinder $M_p$ already a Serre fibration? -I tried lifting a homotopy $f_t: X \times I \rightarrow B$ with starting point $\tilde f_0: X \rightarrow M_p$ by cutting $X$ into the closed preimage $C$ of $B \subset M_p$ and the open preimage $U$ of $E \times [0,1) \subset M_p$. On $C \times I$ we set $\tilde f_t(x) = f_t(x) \in B \subset M_p$. On $U \times I$ we lift $f_t|U: U \times I \rightarrow B$ to $g_t: U \times I \rightarrow E$ and then set $\tilde f_t(x) = (g_t(x),(1-t)(t$ coordinate of $\tilde f_0(x)) + t(1))$. This defines a continuous lift on $C$ and on $U$ separately. If the continuous lift on $U$ extends to the closure of $U$ then we're done. The map $U \rightarrow E$ could be nasty though near the boundary of $U$. Perhaps a better approach is to first construct a map from $X \times I$ that is only "close to" a lift, then use obstruction theory (I'm not an expert on this) to show that it is homotopic to some lift. - -REPLY [9 votes]: Waldhausen, Jahren and myself proved a -fiber gluing lemma for Serre fibrations, -in the context of simplicial sets, that -may be useful. In Propositions 2.7.10 -and 2.7.12 of -"Spaces of PL manifolds and categories -of simple maps" -http://folk.uio.no/rognes/papers/plmf.pdf -we prove that given: - -a diagram of -simplicial sets $Z_1 \leftarrowtail -Z_0 \to Z_2$, where one map is a -cofibration, -a sufficiently -nice base simplicial set $B$ (a -simplicial complex will do), and -compatible maps $Z_i \to B$ that become -Serre fibrations upon geometric -realization, - -then the pushout map -$Z_1 \cup_{Z_0} Z_2 \to B$ becomes -a Serre fibration upon geometric -realization. -Mapping cylinders are a special -case of pushouts. If $p \colon E \to B$ -becomes a Serre fibration upon -realization, then so do -the obvious map $E \times \Delta^1 -\to B$ and the identity map -$B \to B$. The pushout map is -your map $M_p \to B$, and our -conclusion is that its realization -is a Serre fibration. -Our proof depends on working with -simplicial sets. The technical -condition on $B$ is that each -nondegenerate simplex is embedded. - -John<|endoftext|> -TITLE: Which groups can be recovered from their unitary dual? -QUESTION [9 upvotes]: Note: in this post, every topological group under consideration is assumed to be Hausdorff. -Given a locally compact abelian group, one can construct its dual group, i.e. its group of (unitary) characters. This turns out to be a locally compact abelian group with respect to the compact-open topology. Pontryagin duality then tells us that there is a natural isomorphism of topological groups between the original group and its double dual. Thus, if we know only the unitary characters of the group (along with their algebraic and topological structure), we can recover the original group (up to isomorphism) by taking its dual. -Similarly, given the irreducible unitary representations of a compact group, one can construct a compact group out of them which turns out to be isomorphic to the original group (the Tannaka duality theorem). Thus the irreducible unitary representations of a compact group contain enough information to recover the original group. -How much of this remains true for general locally compact groups? By the Gelfand-Raikov theorem, such groups have many irreducible unitary representations (enough to separate points). The question is: can one associate a canonical group structure and/or a topological structure to the irreducible representations of the group (or their equivalence classes) so that one recovers the original group (along with its topology) up to isomorphism? -If the answer is "no", can this be remedied by considering more general representations of the group, e.g. representations which aren't necessarily unitary? -I'm not sure if these questions have an easy answer, so I should mention that I'm mostly interested in the cases where the group is a connected nilpotent Lie group or a connected semisimple Lie group. - -REPLY [8 votes]: The nicest way of phrasing it is the following. Let $\mathcal H$ be the category of Hilbert spaces with unitary maps between them. For each locally compact group $G$, one can define a functor -$$Rep_G : {\mathcal H} \to Top$$ -with $Rep_G(H) = \hom(G,U(H))$, where the space of homomorphisms is endowed with the compact-open topology (with respect to the strong operator topology on $U(H)$). Obviously, the functor $Rep_G$ is compatible with sums and tensor products of Hilbert spaces. Note that -$Rep_{\mathbb Z}(H)=U(H).$ -Consider now any such functor $F: {\mathcal H} \to Top$ -and set -$$D(F) = Nat_{\otimes,\oplus}(F,Rep_{\mathbb Z}),$$ -i.e. all natural transformations of functors which are compatible with the tensor-product and the sum. $D(F)$ is a group since $Rep_{\mathbb Z}(H)=U(H)$ is a group for each Hilbert space $H$. It is also a topological group in a natural way. -Now, there is a natural map $\iota_G : G \to D(Rep_G)$ which is given by -$\iota_G(g)(\pi) = \pi(g)$, where $\pi \in hom(G,U(H))$. So just as in the case of Pontrjagin duality, there is a natural bi-dual. A bit of work (relying on results of Takesaki and Gel'fand-Raikov (which you have mentioned)) shows that $\iota_G$ is a topological isomorphism for all locally compact topological groups. -I studied the analogous question which arises if one restricts everything to the category of finite-dimensional Hilbert spaces (see here). This sometimes goes under the name Chu duality, but is not so extensively studied. Everything works for locally compact abelian groups and compact groups by Pontrjagins result and the Tannaka-Krein theorem. However, for finitely generated discrete groups, interesting things happen. First of all, it is trivial to observe that the analogous map -$$\iota_G : G \to D_{fin}(Rep_G)$$ -is injective if and only $G$ is maximally almost periodic (by a result of Mal'cev iff $G$ is residually finite). Moreover, and this is more difficult, $\iota_G$ is an isomorphism if and only if $G$ is virtually abelian. In particular, for $G={\mathbb F_2}$, the map -$\iota_{\mathbb F_2}$ from $\mathbb F_2$ to $D_{fin}(Rep_{\mathbb F_2})$ -is not surjective. This is a bit surprising as there are no natural candidates of elements in -$D_{fin}(Rep_{\mathbb F_2})$, which do not lie in the image.<|endoftext|> -TITLE: Arrow's theorem and the postseason -QUESTION [18 upvotes]: There are a number of instances of sports teams intentionally losing matches in order to secure a more favorable situation in a playoff round. While this doesn't happen terribly often, when it does it's usually pretty disappointing even for fans of teams who both win and benefit from the win -- and certainly for fans of the sport generally. -So it would be useful if there was a system for seeding the playoff round which was not susceptible to "tactical losing." Unfortunately I can't think of any such rule which seems fair, as long as there is more than one team in the playoffs. -So the question, albeit ill-defined, is this: Is there an analogue of Arrow's theorem for sports tournaments/leagues? (Or perhaps more appropriately an analogue of the related Gibbard-Satterthwaite theorem.) -I can make this a little bit more precise, but probably not totally (and there may be other ways of making the question more rigorous). We'll model the results of the regular season as a directed multigraph on the set of teams $S$, with an edge from $u$ to $v$ for each time team $u$ defeated team $v$. Is there a function from such multigraphs to ordered lists of size $n$ (N.B. that the order isn't meant to represent the relative strength of the teams, but is just a proxy for the extra structure of the playoffs) which satisfies the following (roughly defined) conditions: - -Path-independence: If there is a directed edge from $u$ to $v$ and a directed edge from $v$ to $u$, then the function is invariant under swapping the directions of these two edges. -Universality. At weakest, this condition ought to state that for each underlying multigraph $G$ and each team, there's some orientation $G'$ of the multigraph such that that team makes the playoffs. -Weak independence of irrelevant alternatives. Suppose $G, G'$ differ only in the orientations of edges between $u$ and $v$. Then, if any team $w \in S$ is in exactly one of $f(G), f(G')$, one of $u$ or $v$ must be in exactly one of $f(G), f(G')$. (Intuitively, this says that the only way that changing the result of an individual game changes who's in the playoffs is if it causes one of the teams playing the game to drop out of or enter the playoffs.) -No tactical losing. This is hardest to define, and the big reason why this is a soft question. Is there a reasonable way to make this condition rigorous that leads to an Arrow-type theoreM? - -REPLY [5 votes]: I’m writing my doctoral thesis exactly on seeding in playoffs. -First of all, strictly speaking, a system completely eliminating "tactical losing" can exist. I exclude the example given - a case when only one team advances to the playoffs. The system eliminating “tactical losing” is the one where all the teams participating in the regular season also participate in the postseason and the rule in the postseason is a “random seeding”, eg. all pairs can be formed with the same probability independent of the performance in the regular season. Of course, it is unacceptable. It is unfair and all the matches in the regular season are of a kind of friendly matches with no stake. -The important insight from the above example is such that if we allow only some top teams to advance to the playoffs (as is always a case), even a random seeding cannot eliminate “tactical losing”. There’s always a chance that one team will try to knock the other team out of the postseason. I think the match San Francisco 49ers vs Los Angeles Rams from 1988 is a good example here – see http://en.wikipedia.org/wiki/Match_fixing#Better_playoff_chances. -In my thesis I try to minimize the risk of temptations to “tactical losing”. I propose a measure of this risk (it takes into account the number of such temptations as well as their strength). I suggest a new method of seeding in the playoffs. The comparison of my method to different proposals is performed with Monte Carlo simulations.<|endoftext|> -TITLE: Polygons arising from knot diagrams -QUESTION [9 upvotes]: This question - so far as I know - has no broader mathematical significance, but it occurred to me a while ago and I haven't been able to make any headway. -Any knot diagram $D$ splits the plane into a finite number of pieces. For example, in a standard diagram for a trefoil (e.g., http://en.wikipedia.org/wiki/File:Trefoil_knot_left.svg), the plane is split into five pieces: one lying "outside the diagram," three "lobes" of the knot, and a central region around which the three "lobes" are arrayed. The outer region has three crossings on its boundary, as does the inner region; each of the lobes only have two crossings each on their boundaries. -A diagram for a more complicated knot may split the plane into many more regions, and one of these regions may have many crossings on its boundary. My question is: Does there exist some $n$ such that for any knot $K$, there is a diagram $D$ of $K$ such that any region created by $D$ has at most $n$ crossings on its boundary? -If the answer is no, a (very soft) question that then arises is: given a knot $K$, how hard is it to find the least $n$ ($=n_K$) such that for some diagram $D$ of $K$, every region created by $D$ has at most $n$ crossings on its boundary? For example, for any knot $K$, we can easily see that $n_K>2$. - -REPLY [10 votes]: Colin Adams, Reiko Shinjo and Kokoro Tanaka have a paper (http://arxiv.org/abs/0812.2558) that shows that for any knot you can find a diagram which has only regions with 2, 4 and 5 sides.<|endoftext|> -TITLE: Explicit map for Scholz reflection principle -QUESTION [7 upvotes]: The question is about the specific case of reflection theorems (copied straight from Franz Lemmermeyer's "Class Groups of Dihedral Extensions"): - -Let $k^+ = \mathbb{Q}(\sqrt{m})$ with $m\in \mathbb{N}$, and put $k^- = \mathbb{Q}(\sqrt{-3m})$; then the 3-ranks $r_3^+$ and $r_3^-$ of $Cl(k^+)$ and $Cl(k^-)$ satisfy the inequalities $r_3^+ \le r_3^- \le r_3^+ + 1$. - -The proofs I have seen either use p-adic arguments or galois actions. -Question -Is there an explicit surjective map from $Cl(k^-)[3]$ to $Cl(k^+)[3]$ that might, as the theorem suggests, have kernel of size 3? -At the least, an algorithm for such a map? - -REPLY [2 votes]: Leopoldt's reflection theorem, of which Scholz's result (discovered independently by Reichardt) is a special case, bounds the sizes of certain eigenspaces of the class group -of abelian extensions. I see the main reason for their existence in the fact that abelian -extensions over fields containing the appropriate roots of unity are Kummer extensions. -Anyway what you are looking for is not so much a map between class groups of different fields as a map between eigenspaces of the class group of one single field, which can be pulled down to subfields in some cases. -If you're interested in an explicit map in Scholz's case, you should have a look at Bhargava's -excellent Higher composition laws. I: A new view on Gauss composition, and quadratic generalizations. There he mentions a map defined in terms of binary quadratic forms studied already by Eisenstein, which I think might have something to do with the question -you're asking. I always wanted to study this part in detail, but haven't yet found time to do so. If you come to understand Eisenstein's result before I do let me know -)<|endoftext|> -TITLE: What if Current Foundations of Mathematics are Inconsistent? -QUESTION [97 upvotes]: The title of the question is also the title of a talk by Vladimir Voevodsky, available here. -Had this kind of opinion been expressed before? -EDIT. Thanks to all answerers, commentators, voters, and viewers! --- Here are three more links: -Question arising from Voevodsky's talk on inconsistency by John Stillwell, -Nelson's program to show inconsistency of ZF, by Andreas Thom, -Pierre Colmez, La logique c’est pas logique ! -EDIT. Here the link to the FOM list discussing these themes. - -REPLY [8 votes]: Mathematics is too big to fail.<|endoftext|> -TITLE: A faithful representation "generates" all the other representations -QUESTION [9 upvotes]: Consider the following two similar situations. - -Let $G$ be an algebraic group (namely, of finite type) over a field $k,$ and let $\rho$ be a faithful representation of $G$ over $k.$ Then $\rho$ generates $Rep_k(G)$ as a Tannakian category. -Let $G$ be a real compact Lie group, and let $\rho$ be a faithful complex representation of $G,$ with entries $g\mapsto a_{ij}(g).$ Then the $a_{ij}$'s and $\bar{a}_{ij}$'s generate the set of representative functions on $G$ (cf. Bröcker and tom Dieck, Representations of Compact Lie groups, GTM 98) as a $\mathbb C$-algebra. - -Is there any reason for this similarity? I guess the name of Tannaka was initially involved in the second situation (reconstructing $G$ from its representative functions), and the fact that compact Lie groups (resp. representations of compact Lie groups) complexify to complex algebraic groups (resp. algebraic representations of complex algebraic groups) gives the correspondence between 1 (when the base field is the complex numbers) and 2. I would like to hear more comments from you. - -REPLY [9 votes]: These statements are equivalent; you've made things more mysterious than they need to be by stating 1 in terms of the algebraic group instead of the compact group, where it's also true (and presumably easier to prove). -The equivalence follows immediately from the easy fact that: - -The functions generated by matrix coefficients of V under multiplication are exactly the span of the matrix coefficients of all the tensor powers of V. - -Thus, if your tensor powers contain all irreps, you get all representative functions. Like wise, if you get all representative functions, then you must have any irrep somewhere in a tensor power.<|endoftext|> -TITLE: reference help needed on a fact about poles of L-functions -QUESTION [11 upvotes]: Suppose $\pi$ and $\rho$ are cuspidal automorphic representations on $GL(n)$ and $GL(m)$ respectively. Then the L-function $L(s,\pi \times \rho)$ has a pole iff and $m=n$ and $\pi$ is isomorphic to the contragradient of $\rho$ by some twist. Does anyone know some reference containing the proof of this fact? -I checked Rankin-Selberg convolution paper by Jacquet-P.S-Shalika. It mentioned this result and said the proof would appear somewhere. -Many thanks. - -REPLY [3 votes]: See also -C. Mœglin and J.-L. Waldspurger, Poles des fonctions L de paires pour GL(N), Appendice, Ann. Sci. École Norm. Sup. (4) 22 (1989), 667–674.<|endoftext|> -TITLE: On Noetherian and Japanese rings -QUESTION [10 upvotes]: Let $R$ be a Noetherian ring all of whose local rings are Japanese. Is $R$ necessary Japanese? - -REPLY [9 votes]: No, $R$ is not necessarily Japanese. What follows is a explicit example from a note that Rankeya Datta and I wrote, now available on his website. -Example. We use Hochster's example [Hochster, Ex. 1]; see one of my other answers for the relevant results therein. Let $I$ be the set of positive integers, and set -$$R_i := k[x_i^2,x_i^3] \qquad\text{and}\qquad P_i := (x_i^2,x_i^3) \subseteq R_i$$ -for every $i$, where $k$ is a fixed algebraically closed field. Then, setting $R' := \bigotimes_{i \in I} R_i$, the ring -$$R := \Bigl(R' \smallsetminus \bigcup_{i \in I} P_iR'\Bigr)^{-1}R'$$ -is a domain such that all of its local rings are excellent (hence Japanese), and such that the regular locus in $\operatorname{Spec}(R)$ is not open [Hochster, Prop. 1]. -We claim that $R$ is not Japanese. By Hochster's construction (see [Datta–M, Rem. 4]), the ring $R$ is one-dimensional. But the regular locus is open for any one-dimensional Japanese domain [Datta–M, Lem. 5], and hence $R$ cannot be Japanese.<|endoftext|> -TITLE: Is the model category of Complete Segal Spaces right proper? -QUESTION [13 upvotes]: Well, the title is self-explaining, I guess - I am referring to the complete Segal space model structure of Theorem 7.2 in Rezk's article "A model for the homotopy theory of homotopy theories". -Has anyone tried to find out at all, whether this model category is right proper? And, if yes, then produced a reference? -Thanks! -P.S.: I know that I can get a Quillen equivalent right proper model catgory by the techniques of Thomas Nikolaus, but in that setting it is hard to tell what a general cofibration looks like, which lets me shy away from that step... - -REPLY [19 votes]: The model structure for complete Segal spaces is not right proper. To see this, one can first prove that the model structure for quasi-categories is not right proper: for instance, the map $\delta^1_2:\Delta_1\to\Delta_2$ is a fibration between fibrant objects in the Joyal model category (because it is the nerve of a fibration of the canonical model structure on the category of small categories), but its pullback along the inner horn $\Lambda^1_2\to\Delta_2$ is the boundary $\partial\Delta_1$. Now, given a quasi-category $X$, there is a canonical complete Segal space $N(X)$ associated to it (denoted like this because this is an homotopic version of the classical nerve): the space of $n$-simplices of $N(X)$ is the Kan complex $Map(\Delta_n,X)$, where $Map$ means the mapping space for the Joyal model category structure. A nice explicit model for $Map(\Delta_n,X)$ is just $k(\underline{Hom}(\Delta_n,X))$, where $\underline{Hom}$ is the internal Hom in simplicial sets, and where $k(A)$ denotes the maximal Kan complex contained in the quasi-category $A$. For this explicit model of $N(X)$, if $X=\Delta_m$, then we get that $N(X)$ is just the classical nerve of the poset corresponding to $\Delta_m$ (because there are no other isomorphisms than the identity in $\Delta_m$). In other words, $\Delta_m$ is a complete Segal space already. Therefore, the counter-example for right properness given above for quasi-categories gives a counter-example for complete Segal spaces. For the same reason, the model structure for Segal categories is not right proper. -Edit: Just to avoid any further hope: I considered implicitely that we worked with the model structures for which the cofibrations are the monomorphisms. However, the notion of (right) properness only depends on the class of weak equivalences, so that there just no hope to get right properness for these homotopy theories of $(\infty,1)$-categories, even if we change the notion of fibration.<|endoftext|> -TITLE: Is this finite surjective flat morphism of 2 dimensional schemes a local complete intersection -QUESTION [8 upvotes]: Let $X$ be a regular integral noetherian scheme of dimension 2 and let $D$ be a simple normal crossings divisor in $X$. -EDIT: Let $U = X-D$. -Consider a finite etale morphism $V\longrightarrow U$ with $V$ connected. Let $\pi:Y\longrightarrow X$ be the normalization of $X$ in the function field of $V$. So $Y$ is a $2$-dimensional normal noetherian scheme and $\pi$ is finite. -One can show that $\pi$ is surjective, flat and that $Y$ is CM. -Q: Is $\pi$ a local complete intersection? - -REPLY [7 votes]: Let $k$ be a field of characteristic prime to $n$ and $\zeta\in k$ a primitive $n$-th root of unity. Let $a\in \mathbb{Z}$ and let $Y$ be the quotient of $\mathop{\rm{Spec}}k[u,v]$ by the automorphism $(u,v)\mapsto (\zeta u,\zeta^a v)$. -Then $Y$ is a tame cyclic cover of $X=\mathop{\rm{Spec}}k[u^n,v^n]$ etale away from $D=\{uv=0\}$. Typically $Y$ is not an lci. For example, if $(n,a)=(3,1)$ then -$$Y=\mathop{\rm{Spec}} k[u^3,v^3,uv^2,u^2v]=\mathop{\rm{Spec}}k[x,y,z,t]/(xy-zt,xz-t^2,yt-z^2).$$ -If you allow wild ramification I suspect you can produce an example in which $D$ is even a smooth divisor.<|endoftext|> -TITLE: Ranks of iterated extensions of a group by free groups. -QUESTION [5 upvotes]: Let $G_0$ be a finitely generated group, and suppose there are groups $G_i$ and $K_i$ as in the following short exact sequences -$1\to K_i\to G_{i+1}\to G_i\to 1$ -with $K_i$ free and nonabelian (you may assume finitely generated), and $G_i$ commutative transitive. (If $a$ is nontrivial and $b$ and $c$ both commute with $a$, then $b$ and $c$ commute.) Does it follow that $\mathrm{rank}(G_i)\to\infty$ as $i\to\infty$? Are there examples of extensions of this sort where the rank doesn't increase? - -REPLY [11 votes]: I assume that you consider the infinite cyclic group to be free. Then take the free nilpotent group $G_n$ of class $c\gg 1$ with 2 generators. It has an infinite cyclic central subgroup $K_n$, the factor-group $G_{n-1}=G_n/K_n$ is nilpotent and torsion-free, it has an infinite cyclic subgroup $K_{n-1}$, and so on. Every group $G_i$ is 2-generated, the chain can be arbitrary long ($n$ depends on $c$). -Edit 1: Since you do not want to consider $\mathbb Z$ free enough, here is another example. Take the Baumslag-Solitar group $BS(2,3)=\langle a,t | ta^2t^{-1}=a^3\rangle$. It is non-Hopfian, and has a free normal subgroup $K$ such that $BS(2,3)/K$ is isomorphic to $BS(2,3)$. So all $G_i$ are isomorphic to $BS(2,3)$, all $K_i$ are infinitely generated free groups. Is that what you want? -Edit 2: Since you have another condition now, "commutative transitive", then here is another example. Take a non-elementary torsion-free hyperbolic group $G$. It has a free normal subgroup $N$ such that $G/N$ is still non-elementary torsion-free and hyperbolic (that is proved by Olshanskii, and also by several others, including Delzant). You can continue as long as you wish. All $G_i$ will be hyperbolic and torsion-free (hence commutative-transitive), all $K_i$ will be infinitely generated free groups. -Just to anticipate a future change in the formulation of the question: if you really insist that $K_i$ are finitely generated, the question becomes harder and I am not sure the answer is still the same. -Edit 3: Since you want to have an infinite sequence, here is what to do. For every hyperbolic group $G$ with 2 generators, there exists another 2-generated hyperbolic group $G'$ and a free normal subgroup $N \le G'$ such that $G'/N=G$. This can be done using Rips' construction. In the original Rips' construction (and in all modifications) $N$ was not free, because he wanted $N$ to be finitely generated. But if you do not want $N$ to be finitely generated, it is easy to modify Rips' construction to make $N$ free. Using this you can construct your sequence $G_0=G, G_1=G', G_2=G_1', \ldots$. -Edit 4:In fact Rips' construction does not quite work because the number of generators increases. Certainly if $G$ is free of rank 2, $G'$ cannot be of rank 2. But here is another construction. Take a (torsion-free) lacunary hyperbolic group given by an infinite presentation satisfying a small cancellation condition (see https://arxiv.org/abs/math/0701365). Let $r_1,r_2,...$ be the presentation of $G$. Then $G$ is commutative transitive (it is easily deduced from the fact that $G$ is an inductive limit of hyperbolic groups and surjective homomorphisms). Now the group $G'$ given by the same presentation but without $r_1$ is again lacunary hyperbolic, $G$ is a factor-group of $G'$ over the normal subgroup $N$ generated by $r_1$. It is possible to prove that $N$ is free. Indeed, if some product of conjugates of $r_1$ is equal to 1 in $G$, consider the corresponding van Kampen diagram. The boundary of that diagram has parts labeled by $r_1$ and parts labeled by the conjugators. By Greendlinger lemma, if the diagram has cells, it must have a cell with more than, say, $90\%$ of its boundary common with the boundary of the diagram (take the small cancelation condition $C'(1/300)$). Then more than a half of that part of the boundary must be inside a conjugator, the conjugator can be shortened, and a shorter product of conjugates of $r_1$ is equal to 1 in $G'$. Since $G'$ is lacunary hyperbolic again and satisfies the same small cancellation condition as $G$, we can repeat the construction. Since the presentation is infinite, the process will continue indefinitely. -Edit 5: A more clean way to prove that $N$ is free in Edit 4 is the following. suppose that some product of conjugates of $r_1$ is equal to 1 in $G'$. Consider the corresponding van Kampen diagram $\Delta$. Its boundary label is equal to 1 in the group given by 1 relator $r_1$. Consider the diagram $\Psi$ corresponding to that equality. Now identify the boundaries of $\Psi$ and $\Delta$. We get a diagram over the presentation of $G$ on a sphere: $\Delta$ occupies the northern hemisphere, $\Psi$ occupies the southern hemisphere, and the product of conjugates of $r_1$ labels the equator. Reduce that diagram. Since we can assume that $\Psi$ is reduced, and the $r_1$-cells are only in the south, the $r_1$-cells won't cancel. Hence we shall have a reduced non-empty diagram over the presentation of $G$ on a sphere which is impossible because of the Greendlinger lemma (the boundary of the spherical diagram is empty).<|endoftext|> -TITLE: Compelling evidence that two basepoints are better than one -QUESTION [76 upvotes]: This question is inspired by an answer of Tim Porter. -Ronnie Brown pioneered a framework for homotopy theory in which one may consider multiple basepoints. These ideas are accessibly presented in his book Topology and Groupoids. The idea of the fundamental groupoid, put forward as a multi-basepoint alternative to the fundamental group, is the highlight of the theory. The headline result seems to be that the van-Kampen Theorem looks more natural in the groupoid context. -I don't know whether I find this headline result compelling- the extra baggage of groupoids and pushouts makes me question whether the payoff is worth the effort, all the more so because I am a geometric topology person, rather than a homotopy theorist. - -Do you have examples in geometric topology (3-manifolds, 4-manifolds, tangles, braids, knots and links...) where the concept of the fundamental groupoid has been useful, in the sense that it has led to new theorems or to substantially simplified treatment of known topics? - -One place that I can imagine (but, for lack of evidence, only imagine) that fundamental groupoids might be useful (at least to simplify exposition) is in knot theory, where we're constantly switching between (at least) three different "natural" choices of basepoint- on the knot itself, on the boundary of a tubular neighbourhood, and in the knot complement. This change-of-basepoint adds a nasty bit of technical complexity which I have struggled with when writing papers. A recent proof (Proposition 8 of my paper with Kricker) which would have been a few lines if we hadn't had to worry about basepoints, became 3 pages. In another direction, what about fundamental groupoids of braids? -Have the ideas of fundamental groupoids been explored in geometric topological contexts? Conversely, if not, then why not? - -REPLY [4 votes]: The most convincing example I have found of "two basepoints being better than one" is the incorrect statement of the main result of the following paper: - -Garoufalidis, Stavros, and Andrew Kricker. "A surgery view of boundary links." Mathematische Annalen 327.1 (2003): 103-115. arXiv:math/0205328 - -and its corrected version here: - -Habiro, Kazuo, and Tamara Widmer. "On Kirby calculus for null-homotopic framed links in 3–manifolds." Algebraic & Geometric Topology 14.1 (2014): 115-134. arXiv:1302.0612 - -The result implies a Kirby theorem for framed links in certain classes of link complements. The condition for the statement to be true is that a certain commutative diagram commutes. The diagram for fundamental groups fails to commute in general, but it does commute for fundamental groupoids, and this implies the desired Kirby Theorem. -Further details are provided below. -Quantum topology of knots, links, and 3-manifolds is the study of diagrammatically-constructed topological invariants. The hard kernel of any such construction is a theorem that translates from topology to the combinatorics of a class of diagrams. In 3-manifold topology this is the Kirby Theorem. The Kirby Theorem states that two 3-manifolds are homeomorphic if and only if they are obtained from Dehn surgery on links $L$ and $L^\prime$ correspondingly such that $L^\prime$ can be obtained from $L$ by a sequence of so-called Kirby moves: Stabilization and handle-slide. -To obtain a quantum-topological $3$--manifold invariant, the recipe is to define a quantum topological invariant for a framed link diagram, and to mod out by the relations induced by stabilization and and handle-slide. This turns out to be achievable and this procedure has given rise to interesting invariants, such as the LMO invariant. -For links in general $3$--manifolds, Fenn and Rourke proved that the analogous result holds if we allow a third move: circumcision. From the quantum-topological perspective, this doesn't help us because circumcision is too violent a move- if we mod out by it, the resulting invariants are usually killed. Fenn and Rourke showed that we could do without circumcision when a certain diagram of fundamental groups of 4-manifolds (cobordisms defined by the respective surgeries) commutes. The Fenn--Rourke results was generalized to $3$--manifolds with boundary by Roberts. -Kricker and Garoufalidis considered a certain restricted class of $3$--manifolds with boundary---- complements of so-called boundary links. They argue that the Fenn-Rourke diagram commutes- but it doesn't unless the boundary link happens to be a knot. As shown by Habiro and Widmer, it commutes only when we place a basepoint on each component of the boundary.<|endoftext|> -TITLE: Is the normalization map bijective? -QUESTION [8 upvotes]: Let $X$ be an integral scheme of finite type over a field. Then there is a surjective finite map $\tilde{X} \to X$ from the normalization $\tilde{X}$ of $X$. -Is this going to be bijective? -In the simplest non-normal case, namely the spectrum of $k[x^2, x^3] = k[t,u]/(t^3 -u^2)$, the map is bijective, because the curve is geometrically just a cubic curve (the set of all $(v^2, v^3)$ in affine 2-space, geometrically). I can't find it asserted anywhere that the map is necessarily bijective, though. - -REPLY [11 votes]: If $X$ is reduced and finite type over an algebraically closed field of characteristic zero, then there is a largest finite map $Y \to X$ which is bijective (where $Y$ is also reduced). This map is the seminormalization, $Y = X^{SN}$. In fact, it always factors the normalization map $X^N \to X^{SN} \to X$ where $X^N$ is the normalization.<|endoftext|> -TITLE: How many products specify a sum? -QUESTION [6 upvotes]: Suppose that I have $n$ unknown variables $x_1,\ldots,x_n$. I wish to compute their sum: -$$Sum(x) = \sum_{i=1}^nx_i$$ -However, the only access to these variables is through products: that is, for any subset $S \subset [n]$ I may compute: -$$P(S) = \prod_{i \in S}x_i$$ -That is, I wish to find some number of subsets $S_1,\ldots,S_k$, compute $P(S_1),\ldots,P(S_k)$, and then apply some postprocessing $f$ to find the sum of the variables: -$$f(P(S_1),\ldots,P(S_k)) = Sum(x)$$ -My question is: How large must $k$ be? Clearly, $k = n$ suffices, since with $k$ subsets I may uniquely identify each $x_i$ and then sum the values myself. Is it possible to do with $k < n$? With $k = O(1)$? - -REPLY [30 votes]: Here is an extreme case: I will tell you that either every variable is zero or possibly a single one of them is 1. So your task is to decide if the sum is 0 or it is 1. Any product of more than one term gives no information at all. To rule out all zero you need to check each variable.<|endoftext|> -TITLE: Fast computation of multiplicative inverse modulo q -QUESTION [15 upvotes]: Given a large number $q$ (say, a prime) and a number $a$ between 2 and $q-1$ what is the fastest algorithm known for computing the inverse of $a$ in the group of residue classes modulo $q$? - -REPLY [6 votes]: Another good reference is -http://www.loria.fr/~zimmerma/mca/pub226.html -Other than the Euclidean algorithm as described in other answers, you can get time $O(M(n))$ if the modulus is of the form $p^k$, by working successively mod $p$, $p^2$, etc. See the link for details, section 2.5. -If you need to invert several numbers, say $x$ and $y$, it is often faster to invert $xy$ and then calculate $x^{-1} = (xy)^{-1}y$, $y^{-1} = (xy)^{-1}x$.<|endoftext|> -TITLE: Is there a bound on the length of the longest Morse trajectory? -QUESTION [7 upvotes]: Given a compact Riemannian manifold (with a fixed metric) and a Morse function on it (also fixed). Is there a bound (depending on the metric and the Morse function) on the length of the Morse trajectories? (You can assume the Morse-Smale condition if helpful.) -EDIT (In response to Dick's answer): The Morse function and the metric are fixed. I am just looking for something like $\int_{-\infty}^{\infty}\| \nabla f(\phi_t(p))\|dt\leq C$ and $C=C(f,g)$ where $f$ is the Morse function in question, $g$ stands for the metric and $\phi$ denotes the flow of $-\nabla f$. Note here that the constant is independent of the starting point $p$, as it is easy to see that such a constant additionally depending on $p$ exists (you use the hyperbolicity of $\nabla f$ to deduce exponential convergence towards a critical point). Furthermore it is also easy to see that the above integral is bounded if you include a $2$ in the exponent of the norm (a.k.a. $L^2$), as $\| \nabla f(\phi_t(p))\|^2=-\frac{d}{dt}f(\phi_t(p)).$ -EDIT2 (In response to Bill's answer): I changed the statement to make it abundantly clear that the bound may depend on the metric and the Morse function. Maybe I was somewhat unclear in my formulation - sorry for that. - -REPLY [2 votes]: If everything in sight is real-analytic you can use Łojasiewicz's results to bound the Morse trajectories. See e.g. Andrzej Lenarcik, On the Lojasiewicz exponent of the gradient of a polynomial function.<|endoftext|> -TITLE: What is the indefinite sum of tan(x)? -QUESTION [20 upvotes]: What is the indefinite sum of the tangent function, that is, the function $T$ for which -$\Delta_x T = T(x + 1) - T(x) = \tan(x)$ -Of course, there are infinitely many answers, who all differ by a function of period 1. Ideally, I would like the solution to be of the form -$T(x) = $ nice_function$(x)$ + possibly_ugly_periodic_function$(x)$, -where nice is at least piece-wise continuous. -If any of the following sums can be found, the sum of tan can also be found: - -$\sum \sec x$ -$\sum \csc x$ -$\sum \cot x$ -$\sum \frac{1}{e^{ix} + 1}$ - -I have tried several methods without success, including using a newton series (which does not converge for non-integer $x$), and trying to guess possible functions. -I would also appreciate lines of attack if a solution is not known. - -REPLY [33 votes]: I add more details for the solution in the distinguished answer due to Anixx. First, we need the digamma function -http://en.wikipedia.org/wiki/Digamma_function -which we will call $\Psi(x)$. Important properties (from that web page) are: $\Psi(x)$ is analytic in the complex plane except at the nonpositive integers where it has simple poles. $\Psi(x+1)-\Psi(x) = 1/x$. $\Psi(x) > 0$ for $x>2$. Asymptotics: -$$ - \Psi(x) = \log x - \frac{1}{2x} - \frac{1}{12x^2} + \frac{1}{120x^4} + O(x^{-6}) -\qquad\text{as } x \to \infty . -$$ -So, define $T(z) ={}$ -$$ - -\sum_{k = 1}^{\infty} \Biggl[\Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) - z + 1\Biggr) + \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) + z\Biggr) - \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) + 1\Biggr) - -\Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr)\Biggr)\Biggr] -$$ -For any fixed $z$, only finitely many preliminary terms involve $\Psi$ evaluated at a nonpositive argument, and the asymptotics of the remaining terms are computed (from the asymptotics given above) as -$$ -\Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) - z + 1\Biggr) + \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) + z\Biggr) - \Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr) + 1\Biggr) - -\Psi \Biggl(\pi \biggl(k - \frac{1}{2}\biggr)\Biggr) -$$ -$=z(1-z)/(k^2\pi^2) + o(k^{-2})$ as $k \to \infty$. So the series converges absolutely except when we are at a pole of one of the preliminary terms. Now, because of absolute convergence, we may subtract term-by-term and simplify to get -$$ -T(z+1)-T(z) = \sum_{k=1}^\infty\Biggl[\frac{8z}{(-\pi+2\pi k-2z)(-\pi+2\pi k+2z)}\Biggr] = \tan z . -$$<|endoftext|> -TITLE: Is there an amenabilization of groups ? -QUESTION [7 upvotes]: Given any group $G$, is there an amenable group $A(G)$ together with a morphism $G\rightarrow A(G)$, such that every other morphism $G\rightarrow A'$ to another amenable group $G'$ uniquely factorizes through $A'$? -That is the question. My approach would be to consider the set of normal subgroups with amenable quotient $S:=\{N\unlhd G|G/N $ is amenable $\}$. Then $\bigcap S$ is a normal subgroup of $G$. But I don't know, whether $G/\bigcap S$ is amenable. It embeds into the group $\prod_{H\in S}G/H$. It is not clear, that a infinite product of amenable groups is amenable again. But maybe one can embed $G/\bigcap S$ in a smaller group. - -REPLY [12 votes]: Infinite cartesian product of amenable groups is not necessarily amenable. For example, the free group is a subgroup of the infinite (cartesian) product of finite groups. In general, there are finitely generated residually finite just infinite groups (for example, lattices in semi-simple Lie groups of higher ranks, say, $SL_3({\mathbb Z})$, by the Margulis normal subgroup theorem). These groups do not have amenabelization in your sense. -Update: As Andreas pointed out in a comment below, any residually finite non-amenable group does not have an amenabelization. Indeed, if $G$ is this group. $A$ is the amenabelization, then every hom. $G\to S$ from $G$ to a finite group should factor through $A$. Since the homomorphisms of $G$ onto finite groups separate all elements, the natural homomorphism $G\to A$ is an embedding, but $A$ is amenable while $G$ is not, a contradiction.<|endoftext|> -TITLE: Can dividing out a group action can increase the Lebesgue dimension ? -QUESTION [11 upvotes]: Given any space $X$ of Lebesgue dimension at most $n$. Suppose a group $G$ acts on $X$ continuously. Can the dimension of the quotient $G\backslash X$ exceed the dimension of $X$? -I know examples, where quotient maps increase the dimension. But I don't know an example, where this quotient is given by dividing out a group action. - -REPLY [9 votes]: There are examples of dimension-raising orbit maps arising from actions of p-adic groups. Quoting from the MathSciNet review of the following paper: Raymond, Frank; Williams, R. F. Examples of $p$-adic transformation groups. Ann. of Math. (2) 78 1963 92--106, review by P. Conner: "The authors of this paper show that if $A_p$, $p$ prime, is the compact 0-dimensional $p$-adic group, then for any integer $n\geq 2$ there is a compact $n$-dimensional metric space $X$ together with an action of $A_p$ upon $X$ as a group of transformations so that the dimension of the quotient space, $X/A_p$, is $n+2$." -It is unknown whether a $p$-adic group can act effectively on an $n$-manifold. It is known that if such an action exists, then the orbit space necessarily has higher dimension. The conjecture that no such action exists is known as the Hilbert-Smith Conjecture.<|endoftext|> -TITLE: Formula for n-th iteration of dx/dt=B(x) -QUESTION [5 upvotes]: Let $B(x)$ be infinitely differentiable with respect to $x$. -Drop the use of parentheses on $B$ to delimit the argument $x$ -and use them instead to hold the order of the derivative with respect to $x$. -i.e. $B(0) = B$, -$B(1) = dB/dx$, etc. -Let parentheses on $x$ hold the order of the derivative of $x$ with respect to $t$. -So -\begin{align*} -x_1 & {}= B(x) = B \\ -x_2 & {}= BB_1 \\ -x_3 & {}= BB_1^2 + B^2 B_2. -\end{align*} -Is there a "nice" formula for the integer coefficient of an arbitrary monomial -$B(0)^u(0) \cdot B(1)^u(1) \cdot\dotsb\cdot B(n-1)^u(n-1)$ in $x(n)$? -The first few terms are: -\begin{align*} -x(1) & {}= B, \\ -x(2) & {}= B\cdot B(1), \\ -x(3) & {}= B\cdot B(1)^2 + B^2\cdot B(2), \\ -x(4) & {}= B\cdot B(1)^3 + 4\cdot B^2\cdot B(1)\cdot B(2) + B^3\cdot B(3). -\end{align*} -One of my (many) approaches involved defining $A = 1/B$ so that -$A(x)dx/dt = 1$. Then integrating and solving with the Lagrange Inversion Formula yields -$x(n)$ = sum over all sequences $S=(s(2),s(3),\dotsc)$ of nonnegative integers such that -$\sum (i-1)\cdot s(i)$ equals $n-1$ of -$$(-1)^{T(S)} \cdot (T(S)+n-1)!A^{(-n-T(S))}\cdot \prod_{i=2}^n \frac1{s(i)!}\left(\frac1{i!}\cdot\frac{d^{(i-1)}A}{dx^{(i-1)}}\right)^{s(i)} $$ -where $T(S) = \sum_{i=2}^n i\cdot s(i)$. -I know I can simply make the upper limits in these products be infinity, because all but finitely many of the terms in the products are 1, because all but finitely many of the -$s(i)$ in any sequence $S$ are zero. -But, for future computational purposes, I want to drag around the finite limits to remind myself when it comes time to implement on a computer. -So then I tried substituting the Faa da Bruno formula for -$$ -A(n) = d^A/dx^n = \text{sum of $(-1)^k\cdot k!\cdot B^{-1-k}\cdot\sum_{V=(v(1),v(2),\dotsc)}\prod \left(\frac{B(i)}{i!}\right)^{v(i)}\frac1{v(i)!}$} -$$ -into the equation above and expanding and collecting all similar monomials in the $B$s. -But, I cannot visualize a simple formula for the way all the terms combine. - -So now I tried computing the terms of this sequence directly. -Homogeneity immediately tells you that any monomial product $B(i)^u(i)$ from $i = 0$ to $n-1$ -appearing with nonzero coefficient in $x(n)$ satisfies $u(0) = 1 + \sum_{i = 2}^{n - 1} (i-1)\cdot u(i)$ and $u(1) = n-1 - \sum_{i = 2}^n i\cdot u(i)$. -I solved for $u(0)$ and $u(1)$ in terms of the "slack" (?) variables $u(2), \dotsc, u(n-1)$ -because the terms whose coefficients I CAN compute are most easily expressed in this form. -So, all I've got so far is -\begin{align*} -x(n) ={} & B\cdot B(1)^{n-1} + (2^{n-1} - n)\cdot B^2\cdot (B(1))^{n-3}\cdot B(2) +{} \\ -& {}(1/4)(3^n - 3 - 2^{n+1}\cdot(n-1) + 2(n-1)^2)\cdot B^3\cdot(B(1))^{n-5}\cdot(B(2))^2 \\ -& {}+ (1/4)\cdot(3^{n-1} - 2^{n+1} + 2\cdot n+1)\cdot B^3\cdot (B(1))^{n-4}\cdot B(3) \\ -& {}+ \cdots? +{} \\ -& {}+ \left(\frac{n^2-3\cdot n+4}2\right)\cdot B^{n-2}\cdot B(1)\cdot B(n-2) + B^{n-1}\cdot B(n-1). -\end{align*} -I could keep going, deriving longer and longer formulae for more of the terms, -and then HOPE that I can guess the general pattern for all of them. -$B(i)^{u(i)}$ means $B(i)$ raised to the $u(i)$-th power. - -REPLY [5 votes]: Caveat emptor (July 5, 2021, expanded on the 12th), the devil is in the details: -As I point out in the addendum below, the OP's polynomials provide solutions to a (the) classic flow equation, and, although this ODE has Lie written all over it and a history extending beyond him, I have not found these polynomials in any publication prior to 2008 (Edit, Feb 12, 2022: found, see this MO-Q). The OPs polynomials, which I call the refined Eulerian partition polynomials (REPs), presented for the first time as far as I know in my 2008 OEIS entry A145271, are introduced in the addendum below. The REPs are the expansions of $(g(x)D)^n g(x)$ in terms of the $D^k g(x) = \frac{d^k g(x)}{dx^k}$. If you should find them in any publication other than those I've already compiled there, please point it out to me and I will include it in the entry (or you can--it's up to you). (I have included refs in the entry to publications prior to 2008 that present perspectives from those on flow equations, renormalization, and functional iteration dating from Charles Graves in 1850-53, Abel, and Schroeder, but they do not contain the REPs.) -I have found in the literature predating 2008 two other closely related manifestations of iterated derivatives--the Butcher elementary differentials (B-diffs), enumerated by the Connes-Moscovici weights (CMW) of A139002, and the Comtet diff ops, characterized by the A-polynomials of A139605, for the normal-ordering expansion of $(g(x)D)^n$, i.e., in terms of summands with all $D^k$ to the right. 'Naturally grown' forests of nonplanar rooted trees (and other types of trees) can be associated with these three sets of iterated/nested derivatives. 'Naturally grown" is explained in the addendum below. Pages 8 and 10 of "Lie–Butcher series and geometric numerical integration on manifolds" by Lundervold give examples of the B-diffs and associated trees for a vector function. -Hairer, Lubich, and Wanner, in "Geometric Numerical Integration - Structure-Preserving Algorithms for Ordinary Differential Equations" (2nd Ed., Springer, 2006) on p. 51, consider the flow equation -$$\dot{y} = f(y)$$ -with initial condition $y(t_0)=y_0$. -The derivatives of $y$ in terms of the derivatives of $f$ are presented in Equation 1.4 on page 52 (decidedly NOT the Bell partition polynomials). Table 1.1 on p. 53 lists the first few corresponding forests of trees; $\alpha(\tau)$, enumerating the non-planar rooted tree types $\tau$; and the corresponding elementary differentials $F(\tau)$, in bijection with the trees. The $n$'th derivative of the solution is given as -$$y^{(n)}(t_0) = \sum_{|\tau|=n} \; \alpha(\tau) \; F(\tau)(y_0),$$ -where the cardinality $|\tau|$ is the number of vertices/nodes/knots of the nonplanar rooted tree $\tau$. This is a key identity of the theory of Butcher series. -The tree types $\tau$ are enumerated by the CMW of A139002, i.e., $\alpha(\tau)$, and the trees are in bijection with the B-diffs as denoted in Hairer et al.; however, this bijection dissolves when the tree types are related to the ops $(g(x)D)^n$ of the Comtet iterated Lie ops, with many weights merging for the larger forests. That is, more than one tree type can be associated to the components of the Comtet ops. -This contraction continues when the Comtet ops act on $g(x) = f(x)$, giving the polynomials of A145271, the set of polynomials the OP is trying to characterize. So, there is a refinement proceeding from the array of coefficients of the REPS to those of the Comtet ops to the array of CMW. -For example, in the notation of Hairer et al., with the CMW $(1,\;3,\;1,\;1)$, -$$y^{(4)} = f'''(f,f,f) + 3f''(f'f,f) + f'f''(f,f) + f'f'f'f$$ -$$ = f^3f''' + 3 f^2f'f'' + f^2f'f''+ (f')^3 f $$ -$$ = (fD)^3 \; f = [\;(f(f')^2 + f^2 f'')\; D + 3f^2f' \; D^2 + f^3 \; D^3\;] \;f$$ -$$ = f^3f''' + 4 f^2f'f'' + (f')^3 f.$$ -The last partition polynomial is a (not so) refined Euler partition polynomial (A145271), the diff op in the line just before it is a Comtet op (A139605), and the first line contains the B-diffs with the Connes-Moscovici weights (A139002), all corresponding to the naturally grown forest of six trees with four vertices. -Page 165 of the 3rd edition of "Numerical Methods for Ordinary Differential Equations" by Butcher contains a table similar to the one in Hairer et al. Theorem 311(B) on p. 167 is equivalent to the equation for $y^{(n)}$ above. -Original posting (2011): -You might also try following the references and links in A145271 ["Coefficients for expansion of $[g(x)d/dx]^n g(x)$; refined Eulerian numbers for calculating compositional inverse of $h(x)= (d/dx)^{-1} 1/g(x)$"] and A139605 ["Weights for expansion of $(f(x)D_x)^n$: coefficients of A-polynomials of Comtet"] of the Online Encyclopedia of Integer Sequences for some ideas. Note that the coefficients of A145271 (your expansion coefficients) are embedded in A139605. -An explicit formula for the coefficients are given in the Comtet reference in A139605 on page 166 as egn. (8) with l=1. -Addendum (July 4, 2021): -The OP is essentially asking for the solution to the classic Lie autonomous ODE, the flow equation discussed in A145271, -$$y' = f'(x) = g(f(x)) = g(y),$$ -in terms of derivatives of $g$. The solution is given by the exponential of the iterated vector, or Lie infinitesimal generator (infinigen) $g(x) D_x$ as -$$y = f(x) = e^{(x \; g(w) \; D_w)} \; w \; |_{w=0} = f[ \; x + f^{(-1)}(w)\;] = |_{w=0} $$ -with -$$g(w) = \frac{1}{D_w \; f^{(-1)}(w)},$$ -where $f^{(-1)}(w)$ is the compositional inverse of $f(z)$ about the point $(z,w)$ and $f(0) =f^{(-1)}(0) =0$. -This is easily derived noting that, with $(z,w) = (f^{(-1)}(w),f(z)),$ -$$ g(w) \; D_w = \frac{d}{(f^{(-1)})'(w)dw}= \frac{d}{df^{(-1)}(w)} = \frac{d}{dz} = D_z ,$$ -so, acting on a function $H$, -$$e^{(x \; g(w) \; D_w)} \; H(w) = e^{(x \; D_z)} \; H(f(z)) = H[f(x+z)] = H[\;f(x + f^{(-1)}(w))\;].$$ -This is a chimera of the inverse function and the Taylor series expansion theorems. (One of the Graves brothers, collaborators of Cayley, published this result in the 1850s.). -Specifically, the OP is seeking methods of constructing the polynomials in $D_x^k g(x)$ of $(g(x)D_x)^{n-1} \; g(x)$. Examples of these refined Eulerian polynomials are given in A145271 along with a matrix formula for computing them and references (see also refs 1 and 2 below). One reference is to the 1857 paper of Cayley on operandators and their depiction as trees via which these polynomials may be generated or represented (see the 'normally grown' trees depicted in the "Mathemagical Forests" link in the OEIS entry, in my blog post "Lagrange a la Lah", and in 4 below). Cayley considers multivariable functions and operators. -The relation to rooted trees follows from the product formula -$$D \; f(x)h(x) = f'(x)h(x) + f(x)h'(x)$$ -and pre-Lie action (PLA) of the infinigen -$$g(x) D \; f(x) \; D \to g(x)f'(x) \; D.$$ -(This is called a connection or covariant derivation of vector -fields in "Butcher series: A story of rooted trees and numerical methods for evolution equations" by McLachlan, Modin, and Verdier.) -This PLA of the infinigen allows the iterative action $(g(x)D)^{n-1} \;g(x)$ to be illustrated diagrammatically as PLA action down the branches of forests of rooted trees with initially a copy of the infinigen attached to each node of each tree except for the root, which has only $g(x)$ attached to it. Each iteration corresponds to growing a grove of $n$ trees with $n+1$ nodes, or vertices, from each tree $T_n$ with $n$ nodes by attaching a branch to each node of $T_n$ with each attachment resulting in a new tree. The resultant total action follows from the flow of PLA down along the branches from the leaves to the root for all the trees of this new 'naturally grown' forest. -For example, the 'seed' or root, $T_1$, with only one node, represents the function $g(x)$. Grow the tree $T_2$ from $T_1$ by attaching a branch to its single node, giving a tree with two nodes-- a tree with a single trunk--with $g(x)$ at the root and the infinigen $g(x)D$ at the top node, or leaf. $T_2$ then represents $g(x)D\;g(x) = g(x)\;g'(x)$. Grow a forest from this single tree by attaching a branch to each node of $T_2$, one at a time, giving two rooted trees with three nodes each, i.e., two $T_3$ trees. One is a tree with two trunks with $g(x)$ attached to the root and an infinigen to each of the other two nodes (leaves). Then the simultaneous independent PLA directed down the trunks gives $g^2(x) \; g''(x)$. The second tree with three nodes is a taller one with a single trunk and a single leaf with again $g(x)$ at the root and an infinigen at the other two nodes. This tree represents the action $g(x)\;(g'(x))^2 $--the PLA of the top infinigen on the next lower one gives $g(x) \;g'(x)D$ and then its action on the root $g(x)$ gives $g(x)\;(g'(x))^2$. Then the forest of two $T_3$ trees represents -$(g(x)D)^2 \; g(x) = g(x)D\; g(x) \; g'(x) = (g(x))^2\; g''(x) + g(x)\; (g'(x))^2$. -The next iteration of this 'natural growth' generates a new forest of six trees each with four nodes, and, in general, the number of trees in a forest of type $T_n$ trees is $(n-1)!$. -To grok more complicated trees, consider the example of the simultaneous PLA of three nodes acting via three branches on the node directly below with $F_1(x) \; D$ attached to one of the upper nodes; $F_2(x) \;D$, to another; and $F_3(x)\; D$, to the last. The three functions $F_n$ are generated by PLA directed along branch paths to these nodes from the leaves. With only these three nodes attached directly to the lower node, the PLA action on that node gives -$F_1(x)\;F_2(x)\;F_3(x)\; g'''(x)\; D$. -Continue growing more forests ad infinitum with an infinigen at each node other than the root and direct PLA down along the branches in the manner described above to generate $(g(x)D)^n g(x)$ for each $n$. -Note that the 'raising/creation' operation generating the next younger larger forest from a older smaller forest is the 'natural growth' operation. The 'lowering/annihilation/destruction' operation generating the the older smaller forest from younger larger forest is to burn down all the trees with more than one trunk, i.e., more than one branch attached to the root, and then remove the root from the remaining planted trees, i.e., trees with only a single trunk. -For additional history and refs, see the relevant contribution to the MO-Q "In 'splendid isolation'". -Related MO-Qs: - -"Expansions of iterated, or nested, derivatives, or vectors--conjectured matrix computation" - -"Closed formula for $(g \partial)^n$" - -"differential operator power coefficients" - -"Who invented diagrammatic algebra?" (Abdelmalek Abdesselam's answer and my comments to it)<|endoftext|> -TITLE: unipotent groups, their forms and representations -QUESTION [6 upvotes]: For simplicity fix a base field $k$ of characteristic zero, and consider smooth affine algebraic $k$-groups. (It is understood that unipotent groups in positive characteristic are more complicated, as one might have interesting non-smooth ones.) -Question 1: forms of unipotent groups -If $k$ is algebraically closed, then it is clear that every connected unipotent $k$-group is a successive extension of $\mathbb{G}_\mathrm{a}$'s. Then what about the case $k$ not algebraically closed? Is there non-trivial $K$-forms of, say, the upper trangular unipotent $k$-group of $GL_n$, and of the unipotent radicals of Levi $k$-subgroups of $GL_n$, etc? -If $K$ is a finite Galois extension of $k$ of Galois group $\Gamma$, then a $K$-form of the split $k$-torus is the same as a $\Gamma$-module structure on the group of characters $\mathbb{Z}^d$. Is there analogous results for $K$-forms of unipotent $k$-groups? -For example, with $K$ a finite extension field of $k$. the scalar restriction $$Res_{K/k}\mathbb{G}_{\mathrm{m}}$$ -is not split as $k$-torus, but -$$Res_{K/k}\mathbb{G}_\mathrm{a}$$ -splits into a direct sum of $\mathbb{G}_\mathrm{a}$, becasue $K$ is a finite vector space over $k$ viewed additively. It is from this example that I want to know if there are interesting examples of forms of unipotent groups. -Question 2: representations of unipotent groups -If one has the one dimensional unipotent group $U=\mathbb{G}_\mathrm{a}$, then an algebraic representation of $U$ on $V$ a finite dimensional $k$-vector space is the same as a unipotent operator on $V$. One can then extend this description naturally to obtain the Tannakian category of finite dimensional algebraic representations of $U$. -And what about general unipotent $k$-group $U$? By the theorem of Lie-Engel, we know that such a representation of $U$ on $V$ is upper-triangular: it stabilizes a full flag of $V$, and acts trivially on the successive quotients (because of unipotence). Is there more precise information one can find about these representations so as the determine the Tannakian category of representations of $U$? -Again, let $K$ be a finite Galois extension of $k$ with Galois group $\Gamma$, and $U$, $W$ two connected unipotent $k$-groups that are isomorphic over $K$. Then how can one distinghuish the representations of the two groups by some "action" of $\Gamma$ one the representations spaces, in the spirit one finds in representations of the $k$-torus $$Res_{K/k}\mathbb{G}_\mathrm{m}$$ -thanks! - -REPLY [3 votes]: As Torsten points out, unipotent groups correspond naturally to nilpotent Lie algebras in characteristic 0. This is dealt with nicely on the scheme level, for example, in IV.2.4 of Demazure-Gabriel Groupes algebriques. They also treat in Chapter IV some questions about prime characteristic, which get quite tricky outside the commutative case. Both of your questions are more conveniently studied in the Lie algebra framework, I think, where standard Lie algebra methods for discussing forms in are available and where there is quite a bit of literature on structure, representations, and (in small dimensions) classification in characteristic 0. See for example Jacobson's 1962 book Lie Algebras. -Representation theory is potentially very complicated for nilpotent Lie algebras (say over the complex or real field), even in the finite dimensional situation: unlike the semisimple case, there is no nice general structure based on highest weights, etc. Dixmier and others have studied infinite dimensional representations extensively in connection with Lie groups. Classification of nilpotent Lie algebras is just about impossible in general, but up to dimension 7 or so there are lists. Anyway, there is a lot of literature out there. (Tori are on the other hand also studied a lot over fields of interest in number theory. They have at least the advantage of being commmutative.) -[ADDED] An older seminar write-up might be worth consulting, especially in prime characteristic, along with the relatively sparse literature published since then and best searched through MathSciNet: Unipotent Algebraic Groups by T. Kambayashi, M. Miyanishi, M. Takeuchi, Springer Lecture Notes in Math. 414 (1974). But as their treatment suggests, the main research challenges have occurred in prime characteristic. Over finite fields, there has been quite a bit of recent activity in studying the characters of finite unipotent groups related to the unipotent radical of a Borel subgroup.<|endoftext|> -TITLE: What is a special parahoric subgroup? -QUESTION [23 upvotes]: Let me take this question again from the top. -I would like to know what a special parahoric subgroup is. -I think this is a "real" question, though not an especially good one -- it indicates my complete lack of expertise in this area, but it is certainly a question of interest to research mathematicians (who else?). -The reason that I want to know -- and that I would prefer a short, relatively simple answer rather than an invitation to read the original paper of Bruhat and Tits -- is that I have been asked to write a MathReview for a paper using this concept. Now, perhaps I shouldn't have agreed to review this paper, and I didn't, exactly, but it was sent to me about three months ago so it seems a little late to object. Anyway, the introduction is clear, so I think my final product will be a reasonable specimen of the form. It's just that, for my own benefit, when I write sentences like -"The stabilizer of a self-dual periodic lattice chain is a parahoric subgroup (or, in some cases, contains a parahoric subgroup of index 2, which one recovers by intersecting with the kernel of the Kottwitz homomorphism). The author restricts to subsets $I$ for which the parahoric so obtained is special in the sense of Bruhat-Tits theory." -it would be nice if I understood a little better what that meant. -I am looking ideally for a quick answer that increases my knowledge at least a little bit together with a place to read up on it when I get the time / inclination to deepen my knowledge. - -REPLY [18 votes]: I'd recommend first that you and your friend spend more time with Tits :), "Reductive groups over local fields", from the Corvallis volume (free online, last time I checked). Undoubtably there are other references, like papers of Prasad-Raghunathan mentioned by Greg Kuperberg, and any paper that treats Bruhat-Tits theory for unitary groups. I'll try to provide a background/basic treatment here. -You're certainly used to Bruhat-Tits theory for $SL_2$, or at least for $PGL_2$, over $Q_p$, where one encounters the $(p+1)$-regular tree. As you know, $PGL_2(Q_p)$ acts on this tree, and the stabilizer of a point is a maximal compact subgroup that is conjugate to $PGL_2(Z_p)$, and the stabilizer of an edge is an Iwahori subgroup. This I assume is a familiar picture. -To understand the "special point" subtleties, you should first think about a group like $G=SU_3$ -- the $Q_p$-points of a quasisplit form of $SL_3$, associated to an unramified quadratic field extension $K/Q_p$ with $p$ odd. Let $S$ be a maximal $Q_p$-split torus in $G$. Then $S$ has rank one, though the group $G$ has absolute rank two. It follows that the Bruhat-Tits building for $SU_3$ is again a tree, though not as simple as the $SL_2$ case. In fact, in this unramified situation, the building can be seen as the fixed points of the building of $G$ over $K$ (which is the building of $SL_3$), under the Galois involution. -Now, one must think about the relative roots of $G$ with respect to $S$ -- i.e., decompose the Lie algebra of $G$ with respect to the adjoint action of $S$. There are four eigenspaces with nontrivial eigenvalue -- these are the root spaces for the relative roots which I'll call $\pm \alpha$ and $\pm 2 \alpha$. The root spaces ${\mathfrak g}_{\pm \alpha}$ are two-dimensional. -Now, let $A$ be the apartment of the building associated to $S$ -- $A$ is a principal homogeneous space for the one-dimensional real vector space $X_\bullet(S) \otimes_Z R$. After choosing a good base point (a hyperspecial base point, using the fact that $K/Q_p$ is unramified), $A$ may be identified with $X_\bullet(S) \otimes_Z R$ and the affine roots are functions of the form $\pm \alpha + k$ and $\pm 2 \alpha + k$, where $k$ can be any integer. -Let $h$ be a generator of the rank 1 $Z$-module $X_\bullet(S)$, so that $\alpha^\vee = 2 h$, and $A = R \cdot h$. The affine roots are given by: -$$[\pm \alpha + k](r h) = r + k, [\pm 2 \alpha + k](r h) = 2r + k.$$ -The vanishing hyperplanes of these affine roots are the points: -$$r h : r \in \frac{1}{2} Z.$$ -These are the vertices of the building, contained in the apartment $A$. -Now consider a vertex $nh$, where $n$ is an integer. The affine roots $\pm (\alpha - n)$ and $\pm (2 \alpha - 2n)$ vanish at the vertex $n$. The gradients of these affine roots are the roots $\pm \alpha$ and $\pm 2 \alpha$. These are all of the roots in the original (relative) root system. That's why these vertices are hyperspecial vertices. -On the other hand, consider a vertex $(n + \frac{1}{2}) h$, where $n$ is an integer. The affine roots vanishing at this vertex are $\pm (2 \alpha - 2n - 1)$. The gradients of these affine roots are the roots $\pm 2 \alpha$. These are not all of the original roots, but all original roots are proportional to these roots. You can see how this phenomenon requires the setting of a non-reduced root system to happen. These "half-integral" vertices are special points, since the original root system does not occur in the system of gradients, but it does up to proportionality. At these special (but not hyperspecial) points, the Bruhat-Tits group scheme underlying the parahoric has special fibre with reductive quotient isomorphic to $PGL_2$ (I think... or is it $SL_2$) over the residue field. At the hyperspecial points, the group would be a quasisplit $SU_3$ over the residue field. -If it's not clear from above, a special point in the building occurs where the set of gradients of affine roots vanishing at that point is equal, modulo proportionality, to the set of relative roots. That's the general definition. -Hope this helps - see Tits for more.<|endoftext|> -TITLE: Categories First Or Categories Last In Basic Algebra? -QUESTION [35 upvotes]: Recently, I was reminded in Melvyn Nathason's first year graduate algebra course of a debate I've been having both within myself and externally for some time. For better or worse, the course most students first use and learn extensive category theory and arrow chasing is in an advanced algebra course, either an honors undergraduate abstract algebra course or a first-year graduate algebra course. -(Ok, that's not entirely true, you can first learn about it also in topology. But it's really in algebra where it has the biggest impact. Topology can be done entirely without it wherareas algebra without it beyond the basics becomes rather cumbersome. Also, homological methods become pretty much impossible.) -I've never really been comfortable with category theory. It's always seemed to me that giving up elements and dealing with objects that are knowable only up to isomorphism was a huge leap of faith that modern mathematics should be beyond. But I've tried to be a good mathematican and learn it for my own good. The fact I'm deeply interested in algebra makes this more of a priority. -My question is whether or not category theory really should be introduced from jump in a serious algebra course. Professor Nathanson remarked in lecture that he recently saw his old friend Hyman Bass, and they discussed the teaching of algebra with and without category theory. Both had learned algebra in thier student days from van der Waerden (which incidently, is the main reference for the course and still his favorite algebra book despite being hopelessly outdated). Melvyn gave a categorical construction of the Fundamental Isomorphism Theorum of Abelian Groups after Bass gave a classical statement of the result. Bass said, "It's the same result expressed in 2 different languages. It really doesn't matter if we use the high-tech approach or not." Would algebracists of later generations agree with Professor Bass? -A number of my fellow graduate students think set theory should be abandoned altogether and thrown in the same bin with Newtonian infinitesimals (nonstandard constructions not withstanding) and think all students should learn category theory before learning anything else. Personally, I think category theory would be utterly mysterious to students without a considerable stock of examples to draw from. Categories and universal properties are vast generalizations of huge numbers of not only concrete examples,but certain theorums as well. As such, I believe it's much better learned after gaining a considerable fascility with mathematics-after at the very least, undergraduate courses in topology and algebra. -Paolo Aluffi's wonderful book Algebra:Chapter 0, is usually used by the opposition as a counterexample, as it uses category theory heavily from the beginning. However, I point out that Aluffi himself clearly states this is intended as a course for advanced students and he strongly advises some background in algebra first. I like the book immensely, but I agree. -What does the board think of this question? Categories early or categories late in student training? - -REPLY [5 votes]: A little preliminary: I'm an undergraduate student and I started to study category theory as self-taught at the beginning of second year of university, mostly because of my interest in logic and foundations. Since then I've enjoyed of this fact because knowing some category theory helped me to understand lots of concepts that I've learned more quickly then what I would have done without it, also category theory move me to study some branch of maths like algebraic topology and algebraic geometry. -Now I would distinguish between "category theory" and "the language and instrument of category theory": while the first is an abstract and too specific branch of math, so not adequate to be considered in a undergraduate courses, the second is the very useful conceptual tool that should be taught also to undergraduate students. -What I mean here is that (the language of) category theory shouldn't be teached in a specific course but it should be taught during the regular courses. -I believe that some basic concepts like the ones of category and functor could be taught since first courses of algebra, that's because these concepts are not more abstract than those of groups-group homomorphism,ring-ring homomorphism, vector space-linear map which are taught in the first year's courses. Categories and functors can be easily shown to a young public respectively as graphs with structure (i.e. operations) and as graph morphisms preserving the structure. Many example can be given to those concepts which can be understood by undergraduates: the categories of graphs' points and graphs' paths, the category of sets and functions, the category of groups and group homomorphisms, vectorial spaces and linear maps, but also monoids, groups and poset as categories. -In particular its very useful made these last example in first courses because they help in familiarizing with abstraction before mind is corrupted by concrete (I remember that after having done some basic algebra I found a lot of difficulties to understand why monoids should be categories with one object). -Another good set of examples of category the are quite easy to understand and (in my personal opinion cool) are those of objects (which can be molecules, automaton's states, dynamic system states,...) and processes transforming one object into another. These examples are pretty cool because they open the way to application of category theory also to other science, besides giving really concrete examples of categories. -Obviously categorical concepts should be introduced in a very gradual way, for instance its useless teaching natural transformation before having seen homotopies and groups' representation (or equivalently groups' actions), same apply for other more complex concepts: every thing need to be introduced at right time. -Many would object that probably concepts should be presented every time when they are needed. To those people I would say that probably they right, anyway no-one have ever introduced to me abstract concepts like the ones of groups and rings with some motivation, same apply to topological spaces, the motivations for introducing these objects came late, when where introduced some results which gives us a more abstract framework in which some kind of problems tend to simplify and generalize. -Last motivations of teaching category theory early is that many times seeing thing from an abstract point of view helps when we want to switch constructions from categories, where these constructions are build naturally, to other categories (it comes to my mind the example of homotopies of complexes in homological algebra) and also shows deep unity of lots of mathematical objects that maybe at first seem unrelated. -Before to end I would also like to add some motivation to why not waiting to teach category theory in advanced courses: if you do so usually happens that these categorical concepts are presented in very fast way that make difficult to take familiarity with said concepts and that doesn't allow to deeply understand the meaning and usefulness of categorical results. -One last comment: I don't know why but every time I think to those people which consider category theory too abstract and useless they remind me of what Kronecker said about Cantor, and this make me smile.<|endoftext|> -TITLE: Are packing-homogeneous spaces homogeneous? -QUESTION [10 upvotes]: Given a metric space (M,d) define the packing function P(x,R,r) to be the maximum number of non-intersecting balls of radius r with centers in the ball B(x,R). Let’s call M packing-homogeneous if the packing function is independent of the base point x. -Conjecture A complete connected packing-homogeneous space M is homogeneous. That is, the group of isometries acts transitively on M. -Completeness is necessary as Ricky Demer pointed out. -Connectedness is necessary. It is not difficult to construct finite graphs that are packing-homogeneous, but are not homogeneous (vertex-transitive). -Remark 1 It follows from the work of Gleason, Montgomery, Zippin and others on Hilbert’s 5th problem that a complete connected finite-dimensional homogeneous space is a manifold. Thus, a complete connected finite-dimensional packing-homogeneous space that is not a manifold (for example, has fractional Hausdorff dimension) would provide a counter-example to the conjecture. -Remark 2 I hoped to settle this question for length spaces (or inner metric spaces), i.e. when the distance between two points is equal to the length of a shortest path. Every length space is a Gromov-Hausdorff limit of grpahs. I was hoping to establish a connection between some "almost packing-homogeneous" and "almost homogeneous" properties of graphs that in the limit would give me the desired result (in the spirit of what Gromov does in his paper on groups of polynomial growth). Or, to prove the opposite, constructing a convergent sequence of graphs that is "almost packing-homogeneous" but "increasingly inhomogeneous". After some attempts I thought that the problem for graphs may be as difficult as the original conjecture. -Remark 3 As far as I know, the conjecture is open even for Riemannian manifolds of dimension greater than 2. -I was not able to find any papers that would mention this question or try to attack it. -How plausible is it that the conjecture is true? -Where one would look for counterexamples? -I would also be grateful for any comments that would put this problem in a broader context, or any guesses as to how difficult this problem might be. - -REPLY [2 votes]: Let $d_e$ be the usual metric on $\mathbb{R}^2$ and $S = \{(\frac1n,0) : n \operatorname{is} \operatorname{a} \operatorname{positive} \operatorname{integer}\}$. Let $(X,d)$ be $\mathbb{R}^2-S$ with the metric inherited from $(\mathbb{R}^2,d_e)$. -Clearly, $P_{(X,d)}(x,R,r)\leq P_{(\mathbb{R}^2,d_e)}(x,R,r)$ and $(\mathbb{R}^2,d_e)$ is packing-homogeneous. Consider a ball $B$ in $(\mathbb{R}^2,d_e)$ with center $C$, and disjoint balls of radius $r$ with centers $\{c_i : i\in I\}$ such that $\{c_i : i\in I\} \subseteq B_0$. Bounded subsets of $(\mathbb{R}^2,d_e)$ are totally bounded, so $|\{c_i : i\in I\}| < \infty$. Define $R(\theta)$ as the rotation of $(\mathbb{R},d_e)$ by $\theta$ radians about $C$, clearly these are all isometries which fixe $C$ and $B$ is invariant under. Since the points in $S$ are colinear, $|\{s\in S : d_e(C,c_i) = d_e(C,s)\}| \leq 2$, so if $C$ is not a member of $S$, then $|\{\theta\in (-\pi,\pi] : (r(\theta))(c_i)\in S\}| \leq 2$. This gives us $|\{\theta\in (-\pi,\pi] : (\exists i\in I)((r(\theta))(c_i)\in S)\}| = $ -$|\displaystyle\bigcup_{i\in I} \{\theta\in (-\pi,\pi] : (r(\theta))(c_i)\in S\}| \leq |I|\cdot 2 < \infty\cdot 2 = \infty \leq |(-\pi,\pi]|$, so there exists $\theta\in (-\pi,\pi]$ such that $\neg (\exists i\in I)((r(\theta))(c_i)\in S)$. Let $\phi$ be such a $\theta$. Then $(\forall i\in I)((r(\phi))(c_i)\not\in S)$. This shows that if $C$ is not a member of $S$, then there is a packing in $B$ which does not use any points of $S$, so $P_{(\mathbb{R}^2,d_e)}\leq P_{(X,d)}(x,R,r)$. Let $x,y$ be members of $X$, then $P_{(X,d)}(x,R,r)\leq P_{(\mathbb{R}^2,d_e)}(x,R,r) = P_{(\mathbb{R}^2,d_e)}(y,R,r) \leq P_{(X,d)}(y,R,r)$ -$\leq P_{(\mathbb{R}^2,d_e)}(y,R,r) = P_{(\mathbb{R}^2,d_e)}(x,R,r)\leq P_{(X,d)}(x,R,r)$, which shows that the packing function on $(X,d)$ is independent of basepoint. Therefore $(X,d)$ is packing-homogeneous. -$(X,d)$ is not locally compact near $(0,0)$, but it is locally compact near $(0,1)$, so $(X,d)$ is not even topologically homogeneous. To get to $(0,0)$ from any point in $(X,d)$, if you start below the x axis travel down to $y=-1$, otherwise travel up to $y=1$, then travel horizontally to $x=0$, then travel vertically to $y=0$. This shows that every point has a path to $(0,0)$, so $(X,d)$ is connected. Therefore $(X,d)$ is a connected packing-homogeneous space which is not homogeneous. QED - -Demanding completeness might be enough. It would certainly stop my idea.<|endoftext|> -TITLE: Gosper's Mathematics -QUESTION [18 upvotes]: Sometimes I bump into more of the astonishing results of Gosper (some examples follow) and I gather that a lot of them come from hypergeometrics and special functions. - -Have there been any attempts to try and collect together these? -What papers/books and such are there which study collections of results like this? - - -$$\prod_{n=1}^{\infty} \left(\begin{matrix} -\frac{n}{2(2n+1)} & \frac{1}{2n(2n+1)} & \frac{1}{n^4} \\\\ 0 & -\frac{n}{2(2n+1)} & \frac{5}{4n^2} \\\\ 0 & 0 & 1 \end{matrix}\right) = \left(\begin{matrix} 0&0&\zeta(5)\\\\0&0&\zeta(3)\\\\0&0&1 \end{matrix}\right)$$ -from A third-order Apery-like recursion for $\zeta(5)$ -$$W(x)=a+\sum_{n=0}^\infty \left\{{\sum_{k=0}^n {S_1(n,k)\over \left[{\ln\left({x\over a}\right)-a}\right]^{k-1}(n-k+1)!}}\right\} \left[{1-{\ln\left({x\over a}\right)\over a}}\right]^n$$ -from a page about Lambert's W-Function -$$\prod_{n=1}^{\infty} \frac{1}{e}\left(\frac{1}{3n}+1\right)^{3n+1/2}= -\sqrt{\frac{\Gamma(\frac{1}{3})}{2 \pi}} \frac{3^{13/24}\exp\left(1+\frac{2\pi^2-3\psi_1\left(\frac{1}{3}\right)}{12 \pi \sqrt{3}}\right)}{A^4}$$ -from Mathworld -$$\sum_{n=1}^{\infty} \frac{(-1)^2}{n^2}\cos(\sqrt{n^2 \pi^2 - 9}) = - \frac{\pi^2}{21 e^3}$$ -from On some strange summation formulas - -REPLY [7 votes]: The obvious answer to your question is to search for papers authored by Gosper, or papers with Gosper's name in the title, such as Pages from the computer files of R. William Gosper. But surely this answer is so obvious that you already know it. -A good starting point for understanding modern methods for handling hypergeometric and other special function identities (as opposed to just a list of spectacular identities) is the famous book A=B. -Many such identities begin life by being discovered experimentally and then proved afterwards. A good entry point into the experimental mathematics literature is Bailey and Borwein's website. - -REPLY [7 votes]: Some of Gosper's results are studied, and proved, in the article "Pages from the Computer Files of R. William Gosper," by Mourad E. H. Ismail, Yu Takeuchi and Ruiming Zhang (Proceedings of the American Mathematical Society, Volume 119, Number 3, November 1993). Gosper's work is also discussed in Wolfram Koepf's book Hypergeometric Summation: An Algorithmic Approach to Summation and Special Function Identities. See also Wikipedia's list of hypergeometric identities.<|endoftext|> -TITLE: Continuity of the mutual information -QUESTION [10 upvotes]: The mutual information $I(\mathfrak A_1;\mathfrak A_2)$ of two complete $\sigma$-algebras $\mathfrak A_1$ and $\mathfrak A_2$ in a Lebesgue probability space $(X,m)$ is the integral of the logarithm of the Radon-Nikodym derivative $dP/d(P_1\otimes P_2)$, where $P_i$ are the quotient measures on the factor-spaces $X_i$ determined by the $\sigma$-algebras $\mathfrak A_i$, respectively, and $P$ is the joint distribution on $X_1\times X_2$. If $\mathfrak A_i$ correspond to countable measurable partitions $\alpha_i$ of the space $X$, then $I(\mathfrak A_1;\mathfrak A_2)$ are expressed in terms of the associated entropies as $H(\alpha_1)+H(\alpha_2)-H(\alpha_1\vee\alpha_2)$. I need an explicit reference for the following fact (I know how to prove it): if $\mathfrak B_n$ is a decreasing sequence of $\sigma$-algebras converging to a limit $\mathfrak B$, and $\mathfrak A$ is another $\sigma$-algebra such that $I(\mathfrak A;\mathfrak B_1)<\infty$, then $I(\mathfrak A;\mathfrak B_n)$ converges to $I(\mathfrak A;\mathfrak B)$. - -REPLY [5 votes]: The last proof of this continuity property (and of its analogue for increasing sequences) is given in this paper by Harremöes and Holst. It also contains a pretty comprehensive list of references to earlier work. Apparently, first this property was established by Pinsker in his 1960 book "Information and Information Stability of Random Variables and Processes".<|endoftext|> -TITLE: Congruences mod primes in Galois extensions -QUESTION [9 upvotes]: I have the following situation: let $m,n$ be integers such that $m|n$ and let $\zeta_m$, $\zeta_n$ denote primitive $m$ and $n$th roots of unity. Then we have the inclusion of fields -$$\mathbb{Q}\subset \mathbb{Q}(\zeta_m) \subset \mathbb{Q}(\zeta_n)$$ -Now suppose we also have primes (where $(p,n)=1$) -$$(p)\subset \mathbb{Z}$$ - and then -$$\mathfrak{p}\subset \mathbb{Q}(\zeta_m)$$ - lying over $(p)$ and -$$\mathfrak{P}\subset \mathbb{Q}(\zeta_n)$$ - lying over $\mathfrak{p}$. -I have a congruence in $\mathbb{Q}(\zeta_n)$ of the form $a\equiv b \pmod{\mathfrak{P}}$, where $a,b$ are actually elements of $\mathbb{Q}(\zeta_m)$. -What can I say about the congruence properties of $a,b$ in $\mathbb{Q}(\zeta_m)$? More importantly, if I take the trace or the norm down to $\mathbb{Q}$, can I say anything about their congruence properties there? Ideally I'd like a congruence of something in the integers. -Thanks! -Edit: Are there any assumptions that you can make that might give congruences mod a prime power? - -REPLY [5 votes]: Sure. $a\equiv b\pmod{\mathfrak{P}}$ just means $a-b\in\mathfrak{P}$. Taking norms to any subfield $K$ of $\mathbb{Q}(\zeta_n)$ (e.g., $\mathbb{Q}$ or $\mathbb{Q}(\zeta_m)$) gives you $N_{\mathbb{Q}(\zeta_n)/K}(a-b)\in N_{\mathbb{Q(\zeta_n)}/K}\mathfrak{P}.$ -For $K=\mathbb{Q}$, the latter norm is just $p^f$ where $f$ is the order of $p\pmod{n}$. -For $K=\mathbb{Q}(\zeta_m)$, the former norm is $(a-b)^{\phi(n)/\phi(m)}$ and the latter is $\mathfrak{p}^{f'}$, where $f'$ is the easily-calculated relative residue degree. -This doesn't give you an explicit congruence between $a$ and $b$, but given Gerry's answer, that might have been too much to ask for anyway. On the other hand, if $\phi(n)/\phi(m)$ is small or (as in Alex's answer) if $p$ has few factors in $\mathbb{Q}(\zeta_m)$, you get something at least slightly non-stupid out.<|endoftext|> -TITLE: How can I generate the simulated time series -QUESTION [6 upvotes]: I am curious how one can generate simulated time series data. I found a list of simulated series here and a similar tool for stock market. What is the best way to generate domain specific time series data with some desired patterns? How should one approach this problem? I know this question is not very complete yet so please feel free to suggest modifications. - -REPLY [4 votes]: An example: Simulate an AR1 process in MATLAB by -alpha = 0.8; % smaller than 1 for stationarity` -sigma = 1.3; - -M = 1e3; X = zeros(M, 1); X(1) = randn; % Initialize - -for k = 2:M - X(k) = alpha*X(k-1) + randn*sigma; -end - -This generates M points in time of the model: -$$X_k = \alpha X_{k-1} + \sigma u_k,$$ -where $u_k$ is a standard normal variable. A simple model and a good starting point in simulating time series.<|endoftext|> -TITLE: Axiom of Replacement in Category Theory -QUESTION [12 upvotes]: Does the usual development of category theory (within Goedel-Bernays set theory, for example) require the axiom of replacement? I would have asserted that this was obviously true, but it seems to be common wisdom that the axiom of replacement is an exotic axiom not used outside of axiomatic set theory. Also, sadly the overlap between things that I have thought were obviously true and were in fact false is larger than I would like to admit... -The axiom of replacement basically says that if a class is the same size as a set, then it is a set. This allows us to identify classes that are sets as being small and those that are proper classes as large. Without replacement, you could have countable classes that not sets. -I don't see how to construct most limits and colimits in familiar categories such as Set or Top without the use of replacement. Already, for an infinite set $X$, I don't see how you construct -$$ -\coprod_{i=0}^\infty P^i X, -$$ -where $P^i$ is the $i$-th power set of $X$ (i.e. $P^0 X = X$ and $P^i X = P (P^{i-1}X$)). -Using replacement, it's easy to construct. You form the set -$$ -\bigcup_{i=0}^\infty P^i X, -$$ -and then the coproduct is isomorphic to the set of pairs $(i, x)$, where $x \in P^i X$. Am I completely misunderstanding the issue here, and this coproduct can be proven to exist without replacement? -I suppose you could cook up a definition of diagram so that in the absence of replacement, you cannot even form the diagram for this particular coproduct, but that would be sort-of unsatisfying. - -REPLY [2 votes]: The other answers and comments so far indicate that the axiom of replacement is needed to show the existence of coproducts in Set. This is not true if we adjust the definitions properly ;). -The existence of coproducts in a category $C$ may be formulated that if $F$ is a set of objects in $C$, then there is a object $\coprod F$ with the usual universal property. Note that the usual formulation uses a function from a set to the class of objects of $C$ (a "family"), which almost always requires an application of the axiom of replacement, which is somehow artificial. -Now in the category of sets, let $F$ be a set of sets. Then $x \mapsto x \times \{x \}$ is a function $F \to F \times P(F)$. By comprehension, the image exists and this will be $\coprod F$. -I believe that most of (all?) category theory can be developed with definitions as above and then we do not have to use replacement to show the standard category theoretic properties of the categories in practice. -There is a nice exercise in Kunen's Set theory (IV, Ex. 9), in which the reader has to develope 99 % of modern mathematicss in ZC, which is ZFC without replacement ;-).<|endoftext|> -TITLE: What is the best known estimate for the place of the prime gap with length 1.609*10^18? -QUESTION [7 upvotes]: Suppose $p_n$ is $n$-th prime, $g_n=p_{n+1}-p_n$ is the corresponding prime gap. What is the highest number $C$, such that $p_N>C$ can be proven for $N=\min\{n\mid g_n\geq 1.609\cdot 10^{18}\}$. -Motivation: I've read about Goldbach's weak conjecture. The number $C$ above is obvious lower bound for the first odd number, which does not admit a representation as a sum of three primes, which follows from check of Goldbach's conjecture up to $1.609\cdot 10^{18}$, which is done already by computers. I just want to know, how big is it. - -REPLY [8 votes]: Well, Wikipedia's page on Bertrand's postulate which Scott referred to in his answer does not cite the strongest estimates on this problem. The paper of Olivier Ramaré and Yannick Saouter MR 2004a:11095 "Short effective intervals containing primes. -J. Number Theory 98 (2003), no. 1, 10–33 yield stronger result. They prove that -the interval $[x(1-1/\Delta), x]$ contains a prime for $x \geq 10 726 905 041$ and $\Delta=28 313 999$. In fact they have a table that gives even stronger result in the relevant range for this problem. If we look at Table 1 from their paper we get that for $x \geq e^{60}$ we can choose $\Delta=209 267 308$. We thus need to determine $x$ so that -$$ \frac x {209 267 308} <1.609 \cdot 10^{18}$$ -Computation shows that $x<3.367\cdot 10^{26}$. Since $\log(3.367\cdot 10^{26})=61.1>60$ this is permissable. This gives a better estimate than Charles and Scott's answers above. -It should be remarked that already Ramaré-Saouter used these estimates for the Ternary Goldbach problem. However due to a shorter range where the Goldbach problem had been checked at that time they did get the shorter range $1.13256\cdot 10^{22}$ (Corrollary 1) than using the recent computer checked bounds for the Goldbach problem.<|endoftext|> -TITLE: Local homology of degenerate critical points -QUESTION [5 upvotes]: Given a smooth function $f:M\rightarrow \mathbb R$ on a manifold, its local homology at a critical point $x$ is the group -$$ C_\star(x) := H_\star ( M_{ < c} \cup \{ x \} , M_{ < c} ) ,$$ -where $H_\star$ denotes singular homology (with any coefficient group), $c=f(x)$, and $M_{ < c}$ is the space of those points $x\in M$ such that $f(x) < c$. -If $x$ is a non-degenerate critical point, then $C_\star(x)$ is completely determined by the Morse index of $f$ at $x$: the group $C_j(x)$ is equal to the coefficient group of the homology for $j=\mathrm{ind} (x)$, and is trivial for other values of $j$. -If $x$ is degenerate, the knowledge of $\mathrm{ind}(x)$ and $\mathrm{nul}(x)$ (this latter being the nullity of $f$ at $x$) is not enough to determine $C_\star (x)$. It is easy to build examples of functions on $\mathbb R^2$ having a critical point $x$ with local homology $C_1(x)= G\oplus ...\oplus G$ ($k$ times, where $k>1$ and $G$ is the coefficient group) and $C_j(x)=0$ for $j\neq 1$. For instance, consider the function $f:\mathbb R^2\rightarrow\mathbb R$ given by -$$ f(x,y)=(y-2x^2)(y-x^2)(y+x^2)(y+2x^2). $$ -Here, the origin is a critical point whose local homology (say, with $\mathbb Z_2$ coefficients) should be $C_1(0)=\mathbb Z_2\oplus\mathbb Z_2\oplus \mathbb Z_2$ and $C_j(0)=0$ for $j\neq 1$. -Does anybody know examples of functions having critical points whose local homology is nonzero in more then one degree? -If the answer to the previous question is yes (as I would expect), is it true that given $(n_1,d_1), ..., (n_r,d_r)$ there exists a function $f:M\rightarrow\mathbb R$ with a critical point $x$ whose local homology is given by $C_{d_j}(x)=G^{ \oplus n_j }$ and $C_d(x)=0$ for $d\neq d_1,...,d_r$? - -REPLY [5 votes]: $(x^2+y^2)z^2-c(x^2+y^2+z^2)^2$ for small positive $c$. -More generally $f-cr^{2d}$ where $f\ge 0$ is a homogeneous polynomial function of degree $2d$ in $n$ variables. The local homology at the origin should be essentially the homology of the set of points in $S^{n-1}$ where $f=0$.<|endoftext|> -TITLE: A Boolean function that is not constant on affine subspaces of large enough dimension -QUESTION [5 upvotes]: I'm interested in an explicit Boolean function $f \colon \{0,1\}^n \rightarrow \{0,1\}$ with the following property: if $f$ is constant on some affine subspace of $\{0,1\}^n$, then the dimension of this subspace is $o(n)$. -It is not difficult to show that a symmetric function does not satisfy this property -by considering a subspace $A=\{x \in \{0,1\}^n \mid x_1 \oplus x_2=1, x_3 \oplus x_4=1, \dots, x_{n-1} \oplus x_n=1\}$. Any $x \in A$ has exactly $n/2$ $1$'s and hence $f$ is constant on the subspace $A$ of dimension $n/2$. -Cross-post: https://cstheory.stackexchange.com/questions/1948/a-boolean-function-that-is-not-constant-on-affine-subspaces-of-large-enough-dimen - -REPLY [3 votes]: Unless I am misreading it, the paper Affine dispersers from subspace polynomials by Ben-Sasson and Kopparty gives an explicit construction which is nonconstant on any affine subspace of dimension less than $6 n^{4/5}$.<|endoftext|> -TITLE: Geometric interpretation of Universal enveloping algebras -QUESTION [13 upvotes]: Given a complex Lie algebra $\mathfrak g$, we can form its universal enveloping algebra and interpret it as a noncommutative space. -Is this perspective useful? What does this space "look like"? -How is it related to the space $\mathbb C[G]$ where $G$ is an algebraic group with Lie algebra $\mathfrak g$? -How is it related to the flag variety (in the case of a semisimple Lie algebra)? -This question has also an incarnation which asks about -group algebras. - -REPLY [7 votes]: This remark is just a caution for when one tries to think about what the Spec of $U\mathfrak g$ might mean. -Suppose that $\mathfrak g$ is semisimple, and let $I$ be the kernel of the natural augmentation $U\mathfrak g \to \mathbb C$. (Alternatively, $I$ is the kernel of the action of $U\mathfrak g$ on the trivial representation of $\mathfrak g$.) From the semisimplicity one finds that $I^2 = I$, and hence that $I^n = I$ for every $n \geq 1$. -One also checks (e.g. by a grading argument) that $U\mathfrak g$ is a non-commutative domain, in the sense that it has no zero divisors, and that $U\mathfrak g$ is Noetherian. -Now one easily checks that if $A$ is a commutative Noetherian domain, and if $I$ is -a non-unit ideal such that $I^2 = I$, then necessarily $I = 0$. The point is that if -$I = I^2$, the zero-locus of $I$ in Spec $A$ is "formally" isolated from its complement, -and so morally provides a disconnection of Spec $A$ (and this moral argument is made rigorous via an application of the Krull intersection theorem). If $A$ is a domain, on the other hand, then Spec $A$ is irreducible, and so in particular connected, and so the only possibility -is that $I = 0$. -So $U\mathfrak g$ has various properties that would be mutually incompatible in a commutative ring, and hence one has to be careful in importing geometric intuition naively -in thinking about some kind of non-commutative Spec $U\mathfrak g$. -(Of course, the other answers here give a very non-naive discussion of various geometric points of view on $U\mathfrak g$, but I hope that the above remark will be useful to some people. Let me note that it is also somewhat related to this answer.) - -REPLY [4 votes]: Orbit method -As Mariano and Ben have already mentioned, $U(\mathfrak{g})$ quantizes coadjoint orbits of $\mathfrak{g}.$ -In general, for a real algebraic group $G$ with Lie algebra $\mathfrak{g},$ the following three spaces are closely related: - -The space of primitive ideals of $U(\mathfrak{g}).$ -The set of isomorphism classes of irreducible unitary representations of $G.$ -The set of coadjoint orbits of $G.$ - -The original result is due to Kirillov and states that if $G$ is nilpotent and simply-connected then spaces 2 and 3 are in a natural bijection. Later this was extended in various directions by Auslander and Kostant, Gabriel, Borho and Rentschler, Duflo, and many others. For the case $\mathfrak{g}=\mathfrak{gl}_n,$ the Jacobson topology on the space of primitive ideals was determined by Oshima in the Advances Math paper. It is related to the topology on the space consisting of Zariski closures of the conjugacy classes of $n\times n$ matrices, but there are some subtle differences. - -Relation to flag variety In the case when $G$ is a compact semisimple Lie group, the coadjoint orbit $\mathcal{O}_\lambda=G\cdot\lambda$ is equivariantly isomorphic to generalized flag variety $G_{\mathbb{C}}/P_\lambda;$ moreover, this isomorphism is compatible with symplectic structures and line bundles. Here $\lambda$ is an integral weight (after natural identification between $\mathfrak{g}$ and its dual) and when $\lambda$ is a regular integral dominant weight, $P_\lambda=B,$ so that the coadjoint orbit $\mathcal{O}_\lambda$ may be identified with the full flag variety $G_{\mathbb{C}}/B$ polarized by the line bundle $\mathcal{L}_\lambda.$<|endoftext|> -TITLE: Are flat morphisms of analytic spaces open? -QUESTION [19 upvotes]: Let $f:X\to Y$ be a morphism of complex analytic spaces. Assume $f$ is flat (or, more generally, that there is a coherent sheaf on $X$ with support $X$ which is $f$-flat). Is $f$ an open map? -The rigid-analytic analogue is true (via Raynaud's formal models): see Corollary 7.2 in S. Bosch, Pure Appl. Math. Q., 5(4) :1435–1467, 2009. I don't know about the Berkovich side. -In the algebraic case it's also true (that's what led me to the question, see http://arxiv.org/abs/1010.0341). Specifically, if $K$ is an algebraically closed field with an absolute value, and $f:X\to Y$ is a universally open morphism of $K$-schemes of finite type, the the induced map on $K$-points is open (for the strong topology). -Note that in the complex analytic case, I don't know any reasonable substitute for "universally open". If I believe in the analogy, the result ($f$ is open) should be true assuming for instance that $Y$ is locally irreducible and $f$ is "equidimensional" in some sense (e.g. surjective, $X$ irreducible and the fiber dimension is constant). In this setting the case of fiber dimension 0 is known. - -REPLY [19 votes]: The answer is yes. -In fact, there is the following result, see -Banica-Stanasila, Algebraic methods in the global theory of Complex Spaces, Theorem 2.12 p. 180. - -Theorem. - Let $f \colon X \to Y$ be a morphism of complex spaces and let $\mathscr{F}$ be a coherent analytic sheaf on $X$, which is flat with respect to $f$. Then the restriction of $f$ to supp($\mathscr{F}$) is an open map. - -In particular, every flat morphism is open.<|endoftext|> -TITLE: Classifying stacks and homotopy type of a point -QUESTION [10 upvotes]: Suppose we are working in a category of schemes over a scheme $S$. The scheme $S$ itself is geometrically a ``point''. Let $G$ be a group scheme that acts on a scheme $X$. The quotient stack $[X/G]$ looks in a way as a factor of $X$ by a free action of $G$, i.e. if the action is indeed free and there is a scheme $Y$ such that $X$ is a principal bundle over $Y$, isomorphic as a $G$-scheme to $X$, then $Y$ is isomorphic to $[X/G]$. -In particular, if we take $X=S$, we obtain a classifying stack. There is a surjective map $S \to [S/G]$ and every principal $G$-bundle can be obtained as a pullback of it. This strongly resembles the construction of the classifying space in topology, where we have a space $EG$ with a homotopy type of a point and a cover $EG \to BG$ obtained as a factor by a free action of $G$ on $EG$. -Here is my question (which might sound a bit naive): how this similarity to algebraic topology can be explained? I am mostly curious about how it happens that the construction of the classifying stack is not related to the notion of the homotopy type, whereas the construction of the classifying space relies on the fact that EG has the homotopy type of a point. - -REPLY [16 votes]: This is all best explaied by working with groupoids in schemes. By this, I mean, groupoid objects in the category of schemes over $S$. A groupoid object $\mathcal G$ consists of two schemes $\mathcal G_0$ (the objects) and $\mathcal G_1$ (the morphisms) and a whole bunch of maps between them: -• a map $s:\mathcal G_1\to \mathcal G_0$ that sends an arrow to its source. -• a map $t:\mathcal G_1\to \mathcal G_0$ that sends an arrow to its target. -• a map $e:\mathcal G_0\to \mathcal G_1$ that sends an object to its identity arrow. -• a map $i:\mathcal G_1\to \mathcal G_1$ that sends an arrow to its inverse. -• a map $m:\mathcal G_1\times_{\mathcal G_0}\mathcal G_1\to \mathcal G_1$ that composes arrows. -subject to even more axioms. -From now on, I will simply say "groupoid" instead of groupoid object in schemes. - -Let $G$ be a group over $S$. -Then, there is a groupoid called $EG$, whose objects are $G$, and whose arrows are $G\times_S G$. The groupoid $EG$ is equivalent to $S$, viewed as a groupoid with only identity morphisms. The group $G$ acts freely on $EG$ (this is all happening in the category of schemes over $S$), and the quotient $EG/G$ is $BG$. Here, $BG$ is the groupoid with objects $S$ and morphisms $G$. -If $X$ is a scheme over $S$, then $[X/G]$ is the groupoid whose objects are $X$, and whose morphisms are $X\times _S G$. -This groupoid can also be described as the quotient of the (free) diagonal action of $G$ on the groupoid $X\times EG$. - -You're asking: -"why is the construction of the classifying stack not related to the notion of the homotopy type, whereas the construction of the classifying space relies on the fact that $EG$ has the homotopy type of a point?". -There are two ways of answering your question: -- One is to make the algebraic-geometric story look a little bit more like what people do in topology. That's what I did above. In particular, the fact that the groupoid $EG$ is equivalent to $S$ is the analog of the fact that $EG$ is contractible in topology. -- The other is to make the topological story a little bit like what people do in algebraic geometry. Namely, instead of defining $BG$ as $EG/G$, define it as the space that represents the functor $X\mapsto$ {iso-classes of $G$-bundles over $X$} where I'm now working in the category of topological spaces and homotopy classes of maps. - - -I should maybe add: -There is a stack (over the category of $S$-schemes) associated to any groupoid object in $S$-schemes. The converse is not true for general stacks. But it is true, essentially by definition, for Artin stacks. If you restrict yourself to Deligne-Mumford stacks, then you can also assume that the groupoid is particularly nice, namely, that the maps $s$, $t$, $e$, $i$ and $m$ are étale.<|endoftext|> -TITLE: What is the precise statement of the correspondence between stable Higgs bundles on a Riemann surface, solutions to Hitchin's self-duality equations on the Riemann surface, and representations of the fundamental group of the Riemann surface? -QUESTION [32 upvotes]: I am trying to find the precise statement of the correspondence between stable Higgs bundles on a Riemann surface $\Sigma$, (irreducible) solutions to Hitchin's self-duality equations on $\Sigma$, and (irreducible) representations of the fundamental group of $\Sigma$. I am finding it a bit difficult to find a reference containing the precise statement. Mainly I'd like to know the statement for the case of stable $GL(n,\mathbb{C})$ Higgs bundles. But if anyone knows the statement for more general Higgs bundles that would be nice too. -Just at the level of say sets and not moduli spaces, I think the statement is that the following 3 things are the same, if I am reading Hitchin's original paper correctly: - -stable $GL(n,\mathbb{C})$ Higgs bundles modulo equivalence, -irreducible $U(n)$ (or is it $SU(n)$?) solutions of the Hitchin equations modulo equivalence, -irreducible $SL(n,\mathbb{C})$ (or is it $GL(n,\mathbb{C})$? $PSL$? $PGL$?) representations of $\pi_1$ modulo equivalence. - -Is this correct? Is there a reference? -Hitchin's original paper (titled "Self duality equations on a Riemann surface") does some confusing maneuvers; for example he considers solutions of the self-duality equations for $SO(3)$ rather than for $U(2)$ or $SU(2)$, which would seem more natural to me. Moreover, for instance, he doesn't look at all stable Higgs bundles, but only a certain subset of them - but I think this is just for the purpose of getting a smooth moduli space. And finally, Hitchin looks at $PSL(2,\mathbb{C})$ representations of $\pi_1$ rather than $SL(2,\mathbb{C})$ representations or $GL(2,\mathbb{C})$ representations, which confuses me as well... -Thanks in advance for any help!! -EDIT: Please note that I am only interested in the case of a Riemann surface. Here it appears that degree zero stable Higgs bundles correspond to $GL(n,\mathbb{C})$ representations. But the question remains: are stable Higgs bundles of arbitrary degree related to representations? If so, which representations, and how are they related? Moreover, I think that general stable Higgs bundles should correspond to solutions of the self-duality equations -- but what's the correct group to take? ("Gauge group"? Is that the correct terminology?) I think it's $U(n)$ but I am not sure. -For example, in Hitchin's paper, he considers the case of rank 2 stable Higgs bundles of odd degree and fixed determinant line bundle, with trace-zero Higgs field (see Theorem 5.7 and Theorem 5.8). As for the self-duality equations, he uses the group $SU(2)/\pm 1$. We get a smooth moduli space. In the discussion following Theorem 9.19, it is shown that this moduli space is a covering of the space of $PSL(2,\mathbb{C})$ representations. It seems that this should generalize... - -REPLY [3 votes]: With regard to $PGL(n,{\mathbb C})$ vs $SL(n,{\mathbb C})$, you're right that the $n=2$ case generalizes. On a Riemann surface, the moduli of rank $n$ degree $d$ semistable Higgs bundles with fixed determinant is a $n^{2g}$ cover of a component of the moduli space of $PGL(n,{\mathbb C})$ representations of the fundamental group. The components are labeled by $d$ mod $n$, $d=0$ corresponding to representations that lift to $SL(n,{\mathbb C})$. Hitchin's original space is a $2^{2g}$ covering of the component of $PGL(2,{\mathbb C})$ representations consisting of representations that do not lift.<|endoftext|> -TITLE: A nonlinear first order ordinary differential equation: $y(t)^n+a(t)\frac{dy(t)}{dt}=ba(t)$ -QUESTION [19 upvotes]: I am stuck on solving an apparently simple ODE. I have checked numerous texts, references, software packages and colleagues before posting this... -$$y(t)^n+a(t)\frac{dy(t)}{dt}=ba(t)$$ -If the RHS had a $y$ term it would simply be Bernoulli's equation. Does the $n$ term prevent a solution? - -REPLY [31 votes]: First, rewrite your equation as -$$ -\frac{dy(t)}{dt}=b+f(t)(y(t))^n, \qquad \qquad \qquad \qquad \qquad (*) -$$ -where $f(t)=-1/a(t)$. -This is a special case of the so-called Chini equation -(Equation 1.55 in the Kamke's book mentioned below) -$$ -\frac{dy(t)}{dt}=f(t)(y(t))^n+g(t)y(t)+h(t) -$$ -which generalizes the Riccati and the Abel equations and is in general not solvable by quadratures -but some of its special cases are, see e.g. the book (in German) -E. KAMKE, Differentialgleichungen: Lösungen und Lösungsmethoden, Band I: Gewöhnliche Differentialgleichungen, Leipzig, 1951, -and this list and references therein. -It is known that if the Chini invariant -$$ -C=f(t)^{-n-1}h(t)^{-2n+1}(f(t) -dh(t)/dt-h(t)df(t)/dt+n f(t)g(t)h(t))^n n^{-n} -$$ -is independent of $t$, there is a straightforward recipe (described in the Kamke's book) -for solving the equation. However, in the case under study (both for general $a(t)$ and for $a(t)$ -linear in $t$ as suggested by the original poster in the comment to this reply) this invariant for (*) does depend on $t$ (unless I messed up the computations :)), so the recipe in question does not apply. -The only case when $C$ is independent of $t$ occurs (again modulo possible errors in computations :)) if $((1/f(t))df(t)/dt)^n=\alpha f(t)$, i.e., when $f=(-\alpha(t+\beta)/n)^n$, where $\alpha$ and $\beta$ are arbitrary constants. -Now let us turn to the particular cases with small values of $n$. -For $n=1$ you have a linear inhomogeneous ODE which is easily solved. -For $n=2$ you get a special case of the so-called general Riccati equation -$$ -\frac{dy(t)}{dt}=b+f(t)(y(t))^2, -$$ -solving which is equivalent to solving a second-order linear ODE. -Indeed, upon introducing a new independent variable $\tau(t)=\int f(t) dt$ -you end up with the "standard" Riccati equation -$$ -\frac{dy(\tau)}{d\tau}=h(\tau)+(y(\tau))^2, -$$ -where the function $h$ is defined so that $h(\tau(t))=b/ f(t)$, -and putting $y(\tau)=-(1/z(\tau))dz(\tau)/d\tau$ yields -$$ -d^2z(\tau)/d\tau^2+h(\tau)z(\tau)=0. -$$ -However, this linear equation in general is not necessarily solvable by quadratures. -For $n=3$ (*) is a special case of the Abel differential equation of the first kind, -see e.g. here and references therein for details.<|endoftext|> -TITLE: index of morse functions and homotopical dimension -QUESTION [11 upvotes]: Is it true that any manifold homotopy equivalent to a k-dimensional CW-complex admits a proper Morse function with critical points all of index <= k? I believe this is not true, so I would like to see a counterexample. - -REPLY [7 votes]: Suppose that $M$ has a proper Morse function $f\ge 0$ with all critical points of index at most $k$. Then homotopically $M$ can be made of low-dimensional cells: it has homotopical dimension at most $k$. But also $M$, relative to its boundary $M^{\ge c}$, can be made by attaching high-dimensional cells. Thus the pair $(M,M^{\ge c})$ is $(dim(M)-k-1)$-connected. So for example if $k\le dim(M)-3$ and $M$ is simply connected then $M$ must also be "simply connected at infinity".<|endoftext|> -TITLE: A coverage question -QUESTION [7 upvotes]: Dirichlet showed that if $q$ is a prime number and $q \equiv 3 \bmod 4$ then the sum of the nonresidues of $q$ minus the sum of the quadratic residues of $q$ in the range $1, 2, …, q − 1$ is an odd multiple of $q$. What coverage of the odd numbers is given by these odd multiples? - -REPLY [10 votes]: It is very likely that every (positive) odd number is covered by a sum of this type. -As Robin Chapman points out, this is equivalent to asking, for a given odd number $h$, whether there exists an odd prime $q\equiv3\pmod{4}$ such that $\mathbb{Q}(\sqrt{-q})$ has class number $h$. Let $N(h)$ be the number of quadratic imaginary number fields with class number $h$ (for odd $h$) -- note that such a field is already necessarily of the form $\mathbb{Q}(\sqrt{-q})$ for $q\equiv3\pmod{4}$ by genus theory. This rephrases your question as "Is $N(h)>0$ for all odd $h$"? -By the calculation of mark Watkins mentioned in Stoppie's answer, it is known that $N(h)>0$ for all $h\leq 100$. More importantly, $N(h)$ gets rather large -- $N(1)=9$ is Heegner-Stark, and this is the smallest value of $N(h)$ in this range. For instance, $N(h)>100$ for all $h>37$, and $N(99)=289$. So we'd like an assurance of some sort of a rough upward trend to guarantee that $N(h)$ does not hit zero at some point down the road. Enter Soundararajan's article "The number of imaginary quadratic number fields with a given class number", which studies asymptotics of $N(h)$. Soundararajan remarks that $N(h)$ should be on the order of $h$, and conjectures more precisely that -$$ -\frac{h}{\log h}\ll N(h)\ll h\log h. -$$ -This is accompanied by a probabilistic argument as to why this is a reasonable conjecture. I have not worked through the analysis, but it seems likely to me that if one's only goal was to prove that $N(h)>0$ for all $h$ (a lower bound was not a goal of the paper), this heuristic argument could be made sufficiently rigorous, especially when combined with the data for $h<100$ above, to rule out pathologies for small inputs.<|endoftext|> -TITLE: Are "almost all" strongly regular graphs rigid? -QUESTION [18 upvotes]: I have heard through the academic rumor mill (my advisor heard from so-and-so about a result they heard from big-name who saw it in some journal, etc.) of the following theorem: -Theorem: Almost all strongly regular graphs have trivial automorphism group. -This contrasts that most known families of strongly regular graphs have high symmetry, due to their constructions using algebraic objects. -Does anyone know the reference for this theorem? Also, what is the measure used to describe "almost all"? - -REPLY [11 votes]: The article Random strongly regular graphs? by Peter Cameron http://www.maths.qmul.ac.uk/~pjc/preprints/randsrg.pdf provides some information about what is known and why someone might make that claim. -First an example: There are 11,084,874,829 strongly regular graphs with parameters SRG(57,24,11,9) which arise from a Steiner triple system with 19 points (and 57 blocks); Of these 11,084,710,071 are rigid. (There might be other SRG(57,24,11,9)) -MR2059752 (2005b:05035) -Kaski, P; Östergård, P -The Steiner triple systems of order 19. -Math. Comp. 73 (2004), no. 248, 2075--2092 -http://www.ams.org/journals/mcom/2004-73-248/S0025-5718-04-01626-6/S0025-5718-04-01626-6.pdf -Cameron explains that the SRG with smallest eigenvalue -m are of 4 types: -1) a complete multipartite graph with km blocks of size m (so $v=km$) -2) Produced from $m-2$ mutually orthogonal $k\times k$ Latin squares (so $v=k^2$, nodes connected if they are in the same row or column, or have the same symbol in one of the squares) -3) The vertices are the blocks of a Steiner system with blocks of size m (so $v={\binom{k}{2}}/{\binom{m}{2}}$. -4) A finite list of exceptions $\mathcal{L}(m)$. -Type 1 has a huge automorphism group, but there are not very many of them. -Type 2: For $m=3$, there are on order of $n^{n^2/6}$ latin squares of order $n$, most with trivial automorphism group. -Type 3: For $m=3$ one has Steiner triple systems as above, there are on order $n^{n^2}$ and most are rigid. -Much less is known about sets of $m$ mutually orthogonal latin squares and about Steiner systems with block size $m$. -There are also graphs whose lower two eigenvalues are irrational conjugates (in some ring). -Any graph with $n$ vertices is an induced subgraph of a SRG with at most $4n^2$ vertices. On the other hand, every finite group is the automorphism group of a SRG (if I recall correctly). So the feeling is that there are lots of SRG with lots of freedom to construct them and most are rigid. -The notion of switching is useful. In a STS a Pasch configuration is a set of 6 points and 4 triples abc ade fbe fcd. This would correspond to a 4-clique in the corresponding graph. Switching these to abe acd fbc fde would still leave a 4 clique in the graph but would shift around the connections to the rest of the graph. There can be more elaborate switches too (I think). With enough room one can probably destroy all automorphisms this way. Of the rigid STS(19) above, 2538 don't have any Pasch configurations but over 1,000,000,000 have 14 (similarly for 15 and 16).<|endoftext|> -TITLE: Are all group monomorphisms regular, constructively? -QUESTION [8 upvotes]: Are all group monomorphisms regular, constructively? - -By "constructive" I mean something that would go through in CZF for example. -[added Oct 6] -A sketch of a standard proof (such as referenced in comment below), which is almost constructive: Let H be a subgroup of G, and consider the (say, left-) cosets of H, plus a distinguished element. There is a homomorphism f from G to the permutation group of these, given by left multiplication (leaving the extra element fixed). Construct an involution t which exchanges the coset of 1 and the distinguished element. Then for all g in G, f(g) = t.f(g).t iff g is in H. -The involution t distinguishes the coset of 1 from the other cosets. In other words t can distinguish elements of G which are in H from those which aren't. So in CZF we have the theorem: every "detachable" subgroup H of G is the equaliser of two group homomorphisms. That is, this seems to require the additional hypothesis that $\forall x \in G (x \in H \vee x \not\in H)$. Of course this is a classical tautology so it disappears when we go to ZF (which is CZF + law of excluded middle). -[end of addition] -In practice, an answer to the following question will probably yield an answer (and if it fails to, it will probably be for an interesting reason) -I call a PER-Group a subset G of natural numbers, and equivalence relation R on G, with multiplication and inverse operations as partial computable functions, totally defined over G, and respecting R. -Note G and R need not even be c.e. [and in fact, if H is a decidable subset of G, then the problem is already solved, as above] For instance, addition over the computable reals is computable, even though the set of codes for computable reals is not c.e., nor is equality of two computable reals. Of course that example is abelian, and the proof that every abelian group monomorphism is normal is constructive. I am interested in the non-abelian case. - -Is every PER-group inclusion the equaliser of two PER-group homomorphisms? - -REPLY [9 votes]: Here is a constructive proof that can be enacted in any topos with a natural numbers object. -Let $i: H \to G$ be monic; let $\pi: G \to G/H$ be the canonical surjective function $g \mapsto g H$. Let $A$ be the free abelian group on $G/H$ with $j: G/H \to A$ the canonical injection, and let $A^G$ denote the set of functions $f: G \to A$, with the pointwise abelian group structure inherited from $A$. This carries a $G$-module structure defined by -$$(g \cdot f)(g') = f(g' g).$$ -For any $f \in A^G$, the function $d_f: G \to A^G$ defined by $d_f(g) = g f - f$ defines a derivation. Passing to the wreath product $A^G \rtimes G$, we have two homomorphisms $\phi, \psi: G \rightrightarrows A^G \rtimes G$ defined by $\phi(g) := (d_{j \pi}(g), g)$ and $\psi(g) := (0, g)$. I claim that $i: H \to G$ is the equalizer of the pair $\phi, \psi$. For, -$$\begin{array}{lcl} -d_{j\pi}(g) = 0 & \text{iff} & (\forall_{g': G})\; g\cdot j\pi(g') = j\pi(g') \\ - & \text{iff} & (\forall_{g': G})\; j\pi(g' g) = j\pi(g') \\ - & \text{iff} & (\forall_{g': G})\; j(g' gH) = j(g' H) \\ - & \text{iff} & (\forall_{g': G})\; g' g H = g' H \\ - & \text{iff} & g H = H \\ - & \text{iff} & g \in H. -\end{array}$$ -(All we needed was some injection $j: G/H \to A$ into an abelian group; I chose the canonical one.) -Edit: Peter LeFanu Lumsdaine asked in a comment whether it is indeed constructively true that one can embed any set (or object in a topos) into an abelian group (object), and then later answered this affirmatively, which supplements this answer. The link to this effort is buried in a comment below; it deserves to be made more visible here: Constructively, is the unit of the “free abelian group” monad on sets injective?<|endoftext|> -TITLE: Does every Frobenius algebra in a monoidal *-category give a Q-system? -QUESTION [13 upvotes]: Suppose that C is a fusion C*-cateogry and that A is an irreducible Frobenius algebra object in C, is there always a Frobenius algebra A' isomorphic to A such that A' is a Longo Q-system (that is the coproduct on A' is an isometry)? In other words, wlog can one assume that the coproduct is the * of the product? -The motivation of this question is to understand whether the theory of irreducible Frobenius algebra objects in monoidal *-categories actually agrees on the nose with the theory of irreducible subfactors or whether there's a small loophole. - -REPLY [2 votes]: This is not a full answer. The answer is yes for weakly group-theoretical fusion categories. The question is equivalent to: let C be a unitary fusion category, does every indecomposable C-module category admit a compatible unitary structure (see GMR, for all definitions). In Theorem 5.20, we prove that a weakly group-theoretical fusion category admits a unique unitary structure and every indecomposable module category also admits a unique compatible unitary structure.<|endoftext|> -TITLE: The Cauchy–Riemann equations and analyticity -QUESTION [14 upvotes]: I would be interested to learn if the following generalization of the classical Looman-Menchoff theorem is true. - -Assume that the function $f=u+iv$, defined on a domain $D\subset\mathbb{C}$, is such that - -$u_x$, $u_y$, $v_x$, $v_y$ exist almost everywhere in $D$. -$u$, $v$ satisfy the Cauchy–Riemann equations almost everywhere in $D$. -$f=f(x,y)$ is separately continuous (in $x$ and $y$) in $D$. -$f$ is locally integrable. - - -Question: Does it follow that $f$ is analytic everywhere in $D$? - - -Remark 1. Condition 3 is essential (take $f=1/z$). -Remark 2. G. Sindalovskiĭ proved analyticity of $f$ under conditions 2-4 when the partial derivatives exist everywhere in $D$, except on a countable union of closed sets of finite linear Hausdorff measure (link). - -REPLY [24 votes]: No. -Let $c$ be the Cantor function on $[0,1]$, so that $c$ is continuous, $c' = 0$ a.e., but $c$ is not constant. Then take $u(x+iy) = v(x+iy)=c(x)c(y)$. We have $u_x=u_y=v_x=v_y=0$ a.e. so the Cauchy–Riemann equations are trivially satisfied, and $f(z)=u(z)+iv(z)$ is bounded and continuous on the unit square, but certainly not analytic. -Almost everywhere differentiability is almost never the right condition for solutions to a PDE. A better condition would be to have $u,v$ in some Sobolev space.<|endoftext|> -TITLE: Has anyone thought about creating a formal proof wiki with verifier? -QUESTION [59 upvotes]: Mathematics has undergone some rather nice developments recently with the adoption of new techologies, things like on-line journals, the arXiv, this website, etc. I imagine there must be many further developments that could be quite useful. -What I'm thinking of is a website where anyone can contribute formal proofs of theorems. In particular there would be many proofs of the same theorem provided the proof is different -- like a constructive proof of Brouwer's fixed point theorem, and non-constructive proof, etc. -The idea would be to build up a large web of formal proofs, one building on another so that one could eventually do searches through this space of formal proofs to find out what the most efficient proofs are, in the sense of how many ASCII characters it would take to write-up the proof using Zermelo-Frankel set theory. One hope would be to have a big, active database of verified formal proofs. Another would be to have a webpage where you could hope to discover whether or not there are simpler proofs of theorems you know, that you may have not been be aware of. -Being a web-page there would be certain useful efficiencies -- the webpage could "compile" your proof and check to see it's valid. Being a wiki would make it relatively easy for people to contribute and build on an existing infrastructure. And you'd be free to use pre-existing proofs (provided they've been verified as valid) in any subsequent proofs. One could readily check what axioms a proof needs -- for example to what extent a proof needs the axiom of choice, and so on. -Is there any efforts towards such a development? Such a tool would hopefully function like the publishing arm of some sort of modern internet-era Bourbaki. - -REPLY [2 votes]: What I'm thinking of is a website where anyone can contribute formal proofs of theorems. - -I'm currently working on openmathematics.org although nothing is live yet. It will allow projects developed with the Occam proof assistant to be packaged and published in a similar vein to http://npmjs.com. The command line tool, similar to the npm tool, is called open and should be at least partially functional later this year. - -In particular there would be many proofs of the same theorem provided the proof is different -- like a constructive proof of Brouwer's fixed point theorem, and non-constructive proof, etc. - -There would be nothing to stop people contributing different proofs of the same theorem and nothing to stop users choosing between them. However, all would be developed with Occam, rather than divers proof assistants. - -The idea would be to build up a large web of formal proofs, one building on another so that one could eventually do searches through this space of formal proofs... - -One thing I've been working on with Occam is to turn all labels and references into hyperlinks. If you click on a reference, you go straight to that theorem or axiom. If you click on a label, you get a list of all the places where it's referenced. This functionality is in place in the Occam proof assistant but could surface on the openmathematics.org site at some point, too. - -Being a web-page there would be certain useful efficiencies -- the webpage could "compile" your proof and check to see it's valid. - -Occam will verify proofs on the fly but there could also be a step in the publishing process that verifies the project with a tool written in a language that can be trusted, rather than JavaScript. This is far in the future, however. - -Being a wiki would make it relatively easy for people to contribute and build on an existing infrastructure. And you'd be free to use pre-existing proofs (provided they've been verified as valid) in any subsequent proofs. One could readily check what axioms a proof needs -- for example to what extent a proof needs the axiom of choice, and so on. - -Actually what I decided to do is allow issues in much the vein as http://github.com. They seem to be a great way to foster communication and collaboration. The openmathematics.org site will leverage the GitHub API to allow users to create issues directly, and wash the text through KaTeX to support mathematical markup. - -Is there any efforts towards such a development? Such a tool would hopefully function like the publishing arm of some sort of modern internet-era Bourbaki. - -Well, yes, myself. At the moment I think I have another two years work ahead of me, it's three years and counting so far. Occam is worth a look now, but isn't verifying anything yet.<|endoftext|> -TITLE: Question arising from Voevodsky's talk on inconsistency -QUESTION [21 upvotes]: This question arises from the talk by Voevodsky mentioned in -this recent MO question. On one of his slides, Voevodsky says that - -a general formula even with one free variable describes a subset of - natural numbers for which one can prove, using an argument similar to the - one which is used in Goedel's proof, that there is not a single number n - which can be shown to belong to this subset or not to belong to it. - -And in his spoken commentary he adds that there is a formula defining - -a subset about which you can prove that it is impossible to say - anything about this subset, whatsoever. - -I interpret this as the claim that there is an arithmetically definable -set $S$ for which there is no theorem of Peano arithmetic of the form -$n\in S$ or $n\not\in S$. Perhaps I am misinterpreting, but can anyone -supply (informally) the definition of such a set? - -REPLY [21 votes]: Let $S$ be a first order definable Martin-Löf random set such as Chaitin's $\Omega$. If Peano Arithmetic, or ZFC, or any other theory with a computable set of axioms, proves infinitely many facts of the form $n\in S$ or $n\not\in S$ then it follows that $S$ is not immune or not co-immune and hence not ML-random after all. -(The set of theorems of our theory of the form $n\in S$ (or $n\not\in S$) is computably enumerable and infinite, hence has an infinite computable subset. Being immune means having no infinite computable subset.) -So only finitely many such facts can be proved. Now using an effective bijection between $\mathbb N$ and $\mathbb N\times \mathbb N$, decompose $S$ into infinitely many "columns", $S=S_0\oplus S_1\oplus\cdots$. Then one of these columns has the required property. -The theory should be strong enough to deal effectively with breaking a definable set up into columns and associating values in a column with values in the original set, but this is certainly doable in PA or ZFC. - -REPLY [12 votes]: (n=n)&(con(ZF))<|endoftext|> -TITLE: number fields with no unramified extensions? -QUESTION [18 upvotes]: Asked by a colleague: Do we believe that there are an infinite number of number fields that have no unramified extensions? The rational field Q is the most salient example of such a field, and a couple of others are known. But what is conjectured, and what is the philosophy here? - -REPLY [15 votes]: The question itself is certainly still open. Mostly as an exercise for myself, I'll coalesce my comments above into an answer, and add in some details about where various pieces of the philosophy come from. -The starting point is the following philosophy: - -The ring of integers in any number field with sufficiently small root discriminant admits no non-trivial unramified extensions. - -This philosophy can occasionally be made precise. For example, Yamamura uses tables of root discriminant bounds from Diaz y Diaz to conclude that for a quadratic imaginary number field $K$ of discriminant $|d|\leq 499$ (or $|d|\leq 2003$ under GRH), the maximal unramified extension of $K$ is a finite extension. This is particularly relevant since each of these maximal unramified extensions clearly has the property that you ask about, that they themselves admit no unramified extensions. -The bad news is that the set of number fields with sufficiently small root discriminant to apply these results (at least, without a tremendous of extra effort analyzing carefully constructed extensions) is finite. In particular, results of Odlyzko imply that that there are only finitely many number fields with root discriminant less than $4\pi e^\gamma\approx 22.3$, where $\gamma$ is the Euler-Mascheroni constant (yeah, that Euler-Mascheroni constant!). Under GRH, this remains true for the larger bound $8\pi e^\gamma$. In fact, as a nice concrete factoid to hold on to, Jones and Roberts have shown that there are exactly 7063 abelian number fields with root discriminant under $8\pi e^\gamma$, and sort these according to their Galois group. -Back to good news: So we now ask ourselves whether or not these numbers $4\pi e^\gamma$ and $8\pi e^\gamma$ can be improved. The answer is a definite yes. If we partition number fields based on their proportion of real and complex embeddings, we can get improvements on fields with increased proportion of complex embeddings, up to an improvement factor of $e$ for totally complex number fields. Further, since Odlyzko's argument stems from work of Stark estimating values of $L$-functions, it seems plausible to believe there are analytic improvements to be made as well. So maybe we can keep pushing these bounds higher and higher, enough so that we find infinitely many number fields with smaller root discriminant. -More bad news: There's an inherent limit to how good we can make these bounds, coming from the study of class field towers. (Okay, so this is actually really good news for those of us who like to study class field towers, but I digress...) Namely, since root discriminants are unchanged when moving up an unramified extension, fields with an infinite class field tower provide a stopping point for any claim of the form "there are finitely many number fields with root discriminant less than such-and-such bound." This is also something that can be partitioned by proportion of real and complex embeddings, and it's been a hot topic recently to see how limited these Odlyzko-type bounds can get. Recently, Hajir and Maire have further refined this line of thought by considering towers of number fields with tame ramification. -So, long story short, from this point of view, the big unknown is whether, once we know optimal bounds on root discriminants, whether or not there will be infinitely many number fields with root discriminants less than that bound. Of course, there's also the possibility that there are other techniques for proving that a number field has no unramified extensions that do not go through root discriminants -- perhaps a form of non-abelian class field theory can come to the rescue, as abelian class field theory can address only the weaker (but still open and fantastically interesting) question of fields with no abelian unramified extensions.<|endoftext|> -TITLE: Torsion in GL_n(Z) -QUESTION [18 upvotes]: Fix some $n \geq 3$. It's hopeless to classify the torsion elements in $\text{GL}_n(\mathbb{Z})$, but I have a couple of less ambitious questions. It's well-known that for any odd prime $p$, the map $\phi_p : \text{GL}_n(\mathbb{Z}) \rightarrow \text{GL}_n(\mathbb{Z}/p)$ is ``injective on the torsion''. In other words, if $F$ is a finite subgroup of $\text{GL}_n(\mathbb{Z})$, then $\phi_p|_F$ is injective. -Fixing an odd prime $p$, I have two questions about the map $\phi_p$. - -Are there any interesting restrictions on elements $y \in \text{GL}_n(\mathbb{Z}/p)$ such that there exists some finite-order $x \in \text{GL}_n(\mathbb{Z})$ with $\phi_p(x) = y$? -Do there exist any non-conjugate finite order elements $x,x' \in \text{GL}_n(\mathbb{Z})$ such that $\phi_p(x) = \phi_p(x')$? The hypothesis that $x$ and $x'$ are non-conjugate is to rule out the silly example where $x' = g x g^{-1}$ with $g \in \text{ker}(\phi_p)$. - -REPLY [2 votes]: Let $m$ be the order of $y$. -If $p-1>n$ then a pro-$p$ Sylow of $\mathrm{SL}_n(\mathbb Z_p)$ is torsion free, see (3.2.7.5) on p.101 of http://www.numdam.org/item?id=ASENS_1954_3_71_2_101_0. Hence, if $p-1>n$ and $p$ -divides $m$, then $y$ cannot be lifted. -If $y$ can be lifted to a torsion element $x$, then the order of $x$ is $m$, by "injection of -torsion", and hence the characteristic polynomial $\chi$ of $y$ can be lifted to a monic polynomial $\tilde{\chi}$ of degree $n$ in $\mathbb Z[X]$, such that all irreducible factors -of $\tilde{\chi}$ divide $X^m-1$. (this is also pointed out by Agol above). -If $p$ does not divide $m$ then the converse is also true. -Proof. Suppose $\tilde{\chi}=f_1^{n_1}\ldots f_k^{n_k}$, with $f_i$ irreducible and distinct. -Let $M$ be a block diagonal matrix, with the companion matrix of $f_i$ coming up $n_i$ times. -The char poly of $M$ is $\tilde{\chi}$, the minimal poly of M is $f_1\ldots f_k$, which divides $X^m-1$. Hence $M^m=1$ and thus $M\in\mathrm{SL}_n(\mathbb Z)$. -The image of $M$ in $\mathrm{SL}_2(\mathbb{F}_p)$ is conjugate to $y$, since -they have the same char poly and are both semisimple, as $p$ does not divide $m$. -Hence one can lift $y$.<|endoftext|> -TITLE: Proof that bases etc. exist in early linear algebra course? -QUESTION [10 upvotes]: I'm currently struggling to teach a 2nd course on linear algebra (in the UK, not at an Oxbridge quality university: the students have done a 1st course which concentrated upon algorithms you can apply to matrices, and calculations involving R^2, R^3 etc.) -The syllabus suggests a very abstract approach, stating and proving the Steinitz exchange lemma, and then working up to showing that every (finite dimensional) vector space has a basis of a fixed size etc. etc. -However, I think I'm killing my students. This is all very abstract, it's going to take me weeks to do, and the end result is: All finite dimensional vector spaces look, well, exactly as you think they do. I'm tempted to skip on to linear maps, matrices etc. which seems more interesting to me (and sort of motivates why we might care about choosing a different basis...) -However, I'm also loathed to just assert these facts without proof: the students saw that before in the previous course, and in a pure maths course, I sort of want to prove things (even if I don't expect the students to understand everything). - -What do people think about doing a linear algebra course in maximal abstraction early on? Is there a good book which takes a very streamlined (if perhaps hard to understand) approach to proving the existence of bases-- at least then I could get it over with quickly without lying to the students. - -I read Pedagogical question about linear algebra which was very useful, but maybe concentrated upon a different problem. - -REPLY [2 votes]: This is really an answer by Yemon Choi (who put this in a comment, which is now buried) but as it's community wiki... -A very cute proof of just the fact that an bases have the same (finite) cardinality is given by Ford in http://www.ams.org/mathscinet-getitem?mr=1328020 -Sadly it doesn't work in characteristic p, which my course is meant to touch upon.<|endoftext|> -TITLE: who fixed the topology on ideles? -QUESTION [61 upvotes]: I am teaching a course leading up to Tate's thesis and I told the students last week, when defining ideles, that the first topology that was put on the ideles was not so good (e.g., it was not Hausdorff; it's basically the profinite topology on the ideles, so archimedean components don't get separated well). You can find this mentioned on the second page of the memorial article Claude Chevalley (1909–1984) by Dieudonné and Tits in Bulletin AMS 17 (1987) (doi:10.1090/S0273-0979-1987-15509-1), where they also say that Chevalley's introduction of the ideles was "a definite improvement on earlier similar ideas of Prüfer and von Neumann, who had only embedded $K$ [the number field] into the product over the finite places" (emphasis theirs). [Edit: Scholl's answer says in a little more detail what Prüfer and von Neumann were doing, with references.] -I have two questions: -1) Can anyone point to a specific article where Prüfer or von Neumann used a product over just the finite places, or at least indicate whether they were able to do anything with it? -2) Who introduced the restricted product topology on the ideles? (In Chevalley's 1940 paper deriving global class field theory using the ideles and not using complex analysis, Chevalley uses a different topology, as I mentioned above.) I would've guessed it was Weil, but BCnrd told me that he heard it was due to von Neumann. Any answer with some kind of evidence for it is appreciated. -Edit: For those wondering why the usual notation for the ideles is $J_K$ and not $I_K$, the use of $J_K$ goes right back to Chevalley's papers introducing ideles. (One may imagine $I_K$ could have been taken already for something related to ideals, but in any event it's worth noting the use of "$J$" wasn't some later development in the subject.) - -REPLY [6 votes]: Perhaps it bears noting that, given a commutative topological ring $R$, the group $R^\times$ of units has natural topology given by the subspace topology under the imbedding $x\to (x,x^{-1})\in R\times R$. In particular, this is the coarsest that makes inversion continuous, etc. -(And from the adeles to the ideles this gives the correct topology, unsurprisingly.) -But, yes, this style of characterization was not the norm in those days.<|endoftext|> -TITLE: maximal ideals of $k[x_1,x_2,...]$ -QUESTION [9 upvotes]: What can be said about the structure of maximal ideals of $R=k[\{x_i\}_{i \in I}]$, or geometric properties of $\text{Spm } k[\{x_i\}_{i \in I}]$? Here $k$ is an arbitrary field and $I$ is an infinite set. Kernels of evaluation homomorphisms yield an injective map -$\overline{k}^I / Aut(\overline{k}/k) \to \text{Spm } k[\{x_i\}_{i \in I}]$. -The image consists of those maximal ideals whose residue field is algebraic over $k$. If $I$ is finite, every residue field is algebraic (Noether Normalization). However, if $I$ is infinite and $|I| \geq |k|$, for example $k(t)$ is a residue field which is not algebraic. What happens if $|k| > |I|$? Is there a description in the general case? - -REPLY [15 votes]: If $|k| > |I|$ then the usual cheap proof of Nullstellensatz still works: let $K$ be a residue field. Then $\dim_k K \le \dim_kR = |I|$, but if $t\in K$ is transcendental over $k$, the elements $1/(t-a)$ for $a\in k$ are $k$-linearly independent. So $K/k$ is algebraic.<|endoftext|> -TITLE: Solving a general two-term combinatorial recurrence relation -QUESTION [9 upvotes]: What is known about explicit (not necessarily closed-form) solutions to the recurrence -$$R^n_k= (\alpha n) R^{n-1}_k + (\alpha' n + \beta' k) R^{n-1} _{k-1},$$ -with initial condition $R_0^0 = 1$ and with $R^n_k = 0$ for $n < 0$ or $k < 0$? Special cases of this are closely related to recurrences satisfied by some interesting combinatorial numbers, such as the binomial coefficients and the Stirling numbers. -The more general recurrence -$$R^n_k= (\alpha n + \beta k + \gamma) R^{n-1}_k + (\alpha' n + \beta' k + \gamma') R^{n-1} _{k-1},$$ -is open Problem 6.94 in Concrete Mathematics (2nd edition, p. 319). -The closest published result I have found thus far is the following formula due to Neuwirth ("Recursively defined combinatorial functions: Extending Galton's board," Discrete Mathematics, 2001) for the case $\alpha' = 0$ of the Concrete Mathematics problem, -$$R^n_k = \prod_{i=1}^k (\beta' i + \gamma') \sum_{i=0}^n \sum_{j=0}^n s^n_i \binom{i}{j} S^j_k \alpha^{n-i} (\gamma - \alpha)^{i-j} \beta^{j-k},$$ -which, of course, gives me an answer to my question when $\alpha'=0$. (Here, $s^n_i$ and $S^j_k$ are unsigned Stirling numbers of the first and second kinds, respectively.) -I have tried generating functions without any success thus far. An answer like Neuwirth's that involves sums and binomial coefficients or Stirling numbers would be fine, as would a partial answer or just another idea to try. - -REPLY [5 votes]: A general solution of the Graham-Knuth-Patashnik 6.94 problem -$$ -R^{n}_k=(\alpha \, n + \beta\, k + \gamma)\, R^{n−1}_k+ (\alpha′\, n+ \beta′\, k +\gamma′)\, R^{n−1}_{k−1} + \delta_{n,0}\delta_{k,0}\,, -$$ -with $R_{n}^k=0$ if $n<0$ or $k<0$, can be found in the paper ``Bivariate generating functions for a class of linear recurrences: General structure'', by J.F. Barbero G., J. Salas, and E.J.S. Villaseñor, published in J. Combin. Theory A 125 (2014) 146-165 (see also -arXiv:1307.2010). -The solution was obtained by using generating functions.<|endoftext|> -TITLE: How does one interpret the naive t-structure on constructible sheaves as a t-structure on D-modules? -QUESTION [10 upvotes]: By the Riemann-Hilbert correspondence, there is an equivalence between -(1) - $\mathcal{D}\operatorname{-mod}(X)$ -, the (derived) category of holonomic D-modules on a complex variety X, and -(2) - $D^b_c(X)$ -, the (derived) category of constructible sheaves on X. -There is a "naive" t-structure we can put on both categories. In - $\mathcal{D}\operatorname{-mod}(X)$ -, we can look at a t-structure whose heart - $\mathcal{D}\operatorname{-mod}^\heartsuit$ - is a complex (of D-modules) concentrated in degree 0. In - $D^b_c(X)$ -, we can look at the naive t-structure whose heart - $D^{b \heartsuit}_c$ - is a complex (of constructible sheaves) concentrated in degree 0. -It's known that if we transfer the naive t-structure on - $\mathcal{D}\operatorname{-mod}(X)$ - to - $D^b_c(X)$ - (using the equivalence above), - $\mathcal{D}\operatorname{-mod}^\heartsuit$ - is identified with "perverse sheaves" on X. -My question is: - -If we map - $D^{b\heartsuit}_c$ - to the category of D-modules using the Riemann-Hilbert correspondence, what subcategory of - $\mathcal{D}\operatorname{-mod}$ - do we get? Does this have a well-known name? -More generally, is there some geometric/nice description of what the naive t-structure on - $D^b_c$ - becomes on - $\mathcal{D}{\operatorname{-mod}}$ - ? - -REPLY [13 votes]: The t-structure is described in -this paper of Kashiwara. - It looks essentially like Donu and t3suji suggest in their comments: defined by conditions that look like middle-perversity support conditions.<|endoftext|> -TITLE: Breaking efficiently a binary sequence into given strings -QUESTION [8 upvotes]: Suppose we are given a finite collection of finite binary strings $\mathcal{S}$, of various lengths. Our task is to express any binary sequence $x\in 2^\mathbb{N}$ as juxtaposition of strings taken from $\mathcal{S}:$ -$$x=\sigma_1\sigma_2\sigma_3\dots$$ -For any such sequence of "bricks", $\sigma\in\mathcal{S}^\mathbb{N},$ we consider -$$L^*(\sigma):=\limsup_{n\to\infty}\frac{1}{n}\sum_{k=1}^n \mathrm{length}(\sigma_k)$$ -as a kind of parameter of quality of the factorization: we appreciate a factorization if the mean length of the composing pieces is frequently high. -Question: - -Given the set $\mathcal{S},$ how to -compute the best $L^*(\sigma)$ which -is always attainable, whatever is -$x\in 2^\mathbb{N}?$ That is, the quantity (depending on $\mathcal{S}$ only) -$$\lambda(\mathcal{S}):=\inf_{x\in2^\mathbb{N}} \max \{L^*(\sigma) : \sigma\in\mathcal{S}^\mathbb{N}, x=\sigma_1 \sigma_2 \sigma_3\dots \}.$$ -Also, are there special assumptions on the collection $\mathcal{S}$ that may simplify the analysis? - -Example. Let $\mathcal{S}:=\{ 0,\ 1,\ 00,\ 01,\ 11 \}.$ Then, any binary sequence $x$ can be broken into a sequence of strings in $\mathcal{S}$, with average length larger than or equal to $3/2$. -Remark. For my purposes, we can assume that the collection $\mathcal{S}$ always enjoys the property of being "stable for extraction of sub-strings", that is, if $\sigma=\epsilon_1 \epsilon_2 \dots \epsilon_n\in\mathcal{S}$, then also $\epsilon_p \epsilon_{p+1} \dots \epsilon_q\in\mathcal{S},$ for any $1\leq p < q\leq n.$ If I am not wrong, this allows quite a simple inductive procedure for a canonical optimal factorization $\sigma$ of a binary sequence $x$: having chosen $\sigma_1,\dots,\sigma_k,$ take $\sigma_{k+1}$ as the longest admissible element of $\mathcal{S}$: by the above property of $\mathcal{S}$ any other factorization $\tau$ of $x$ can be easily compared with $\sigma$, showing $L^ *(\sigma) \geq L^ *(\tau)$. So in this case I'd expect there is some hope of being able of computing the quantity $\lambda(\mathcal{S}).$ - -REPLY [3 votes]: Not an answer, but perhaps of some use: -It would help if S, the set of blocks, were finite. In any case, if S contains all words of length k, you know that for sufficiently long words L will have k as a lower bound. -(Computing k for a given S closed under consecutive substrings is feasible, but may not be -polynomial time in the length of a representation of S.) If S contains almost all words of length k, then the lower bound for L will still be k on those sigma which avoid the missing words, otherwise the lower bound for L will approach -the lower bound for L of those missing words which occur more frequently as you read more of sigma. It may then make sense to compute L for short words not in S, or infinite repetitions of such short words, and use this in aiding the computation of L of sigma. -If S has infinitely many finite blocks (and the alphabet is still finite), then there -is the potential for L to assume values of infinity. If you don't care about that or -if you are focussing on the cases where L will be finite, then you have some mathematics -which may touch upon descriptive set theory, and you may encounter a large cardinal or -other surprising axiom to consider. S being closed under consecutive substrings will -in many cases produce a larger k (k as in the above paragraph), and will have to be -pretty special otherwise to avoid any finite substring. There may be other tools in mathematical logic (e.g. filters on finite subsets of the language) that might tell -you what properties you want S to have. -Another poster referenced Huffman encoding. Outside of symbolic dynamics (and -related fields of automata theory and semigroup theory), that and similar string-processing -algorithms is a first choice of where to look for previous work. If you are considering extra-world (surreal? non-real?) applications using infinite S might you then turn to mathematical logic. -Gerhard "Ask Me About System Design" Paseman, 2010.10.07<|endoftext|> -TITLE: A finite index subgroup of the Mapping Class Group -QUESTION [10 upvotes]: Let $G$ be the mapping class group of a closed surface $S_{g}$. Bestvina-Bromberg-Fujuwara http://front.math.ucdavis.edu/1006.1939 recently constructed a finite index subgroup $B$ of $G$ such that for every essential closed simple curve $\gamma$ on $S_{g}$ and every $h\in B$ either $h\gamma=\gamma$ or $h\gamma$ intersects $\gamma$ (everything is modulo isotopy of $S_{g}$). The construction is not difficult: $B$ is just the subgroup of $G$ fixing $\pi_1(S_{g})/N$ for some characteristic subgroup $N$ of the $\pi_1$ of finite index. The $N$ can be found by intersecting kernels of the first homology mod $6$ with kernels of the first homology mod 2 of all index 2 subgroups of the $\pi_1$. Question: can one find another finite index subgroup $B'$ of $G$ with the same property but with a smaller index. Most probably $B'$ cannot be above Torelli subgroup, but can it be "not far from Torelli". I am interested in $B'$ that does not act non-trivially on a simplicial tree. The motivation is here: http://front.math.ucdavis.edu/1005.5056. - Update: I read somewhere (I do not remember where now) that if $g$ is an element of the Torelli subgroup, $\gamma$ is a simple closed curve, then at least two of the three curves $\gamma, g\gamma, g^2\gamma$ must intersect. From this I deduced that two curves are not enough and that a Bestvina-Bromberg-Fujiwara subgroup cannot contain the Torelli subgroup. Now the question is whether it may contain the Johnson kernel. For this one needs first to answer the following question: if $g$ is the Dehn twist about a separating curve, $\gamma$ is a simple closed curve, is it true that $g\gamma$ intersects $\gamma$ or $g\gamma=\gamma$? - -REPLY [5 votes]: Regarding the new question, you're correct that $f\in \mathcal{I}$ doesn't imply $f(\gamma)$ meets $\gamma$: for example, if $f$ is a product of disjoint bounding pairs and $\gamma$ is a (nonseparating) curve meeting each curve-defining-$f$ in at most 1 point, then $f\in\mathcal{I}$ but $f(\gamma)$ is disjoint from $\gamma$. -For the rest of the question, the following theorems are taken from Farb-Leininger-Margalit, "The lower central series and pseudo-Anosov dilatations", American Jour. Math, 130 (2008), no. 3, 799--827. (PDF) -Lemma 2.2: if $f\in \mathcal{I}$, $\gamma$ is separating, and $f(\gamma)\neq\gamma$, then $i\big(f(\gamma),\gamma\big)\geq 4$. -Lemma 2.3: if $f\in \mathcal{I}$ and $\gamma$ is nonseparating, and $f(\gamma)\neq\gamma$ then $i\big(f(\gamma),\gamma\big)\geq 2$ or $i\big(f^2(\gamma),\gamma\big)\geq 2$. -Proposition 3.2: if $f\in\mathcal{K}$ and $f(\gamma)\neq \gamma$, then $i\big(f(\gamma),\gamma\big)\geq 4$. -In particular 3.2 answers your question about a separating twist. The proof of the proposition is as follows: $\mathcal{K}$ contains no nonseparating bounding pairs (use the Johnson homomorphism). By Lemma 2.2 we can assume $\gamma$ is nonseparating. But if $f(\gamma)$ is disjoint frmo $\gamma$, then $[f,T_{\gamma}]=f T_{\gamma}f^{-1} T_{\gamma}^{-1}=T_{f(\gamma)}T_{\gamma}^{-1}$ is a bounding pair, which lies in $\mathcal{K}$ if $f$ does, yielding a contradiction. - -I do appreciate the nice calming green background, but I'm not sure we're done. If $f\in\mathcal{K}$, and $g\in \mathcal{BBF}$ (the group defined by Bestvina-Bromberg-Fujiwara), how do you know that $fg$ has the desired property? -Here is a sketch of a possible construction of a BBF-subgroup which properly contains the Johnson kernel. Let $G$ be the subgroup generated by the Johnson kernel $\mathcal{K}$ together with ($\mathcal{BBF}\cap\mathcal{I}$). Since $G$ is contained in $\mathcal{I}$, separating curves are taken care of by Lemma 2.2 above. If $\gamma$ is a nonseparating curve such that $f(\gamma)$ is disjoint from $\gamma$, then $[f,T_\gamma]$ is a nonseparating bounding pair; $[f,T_\gamma]$ lies in $G$ if $f$ does, since $G$ is normal. -Now the idea is to show that $G$ contains no nonseparating bounding pairs. Certainly $\mathcal{BBF}$ doesn't, but we need to show that this is still true after throwing in $\mathcal{K}$. Since $G$ contains $\mathcal{K}=\ker \big(\tau\colon\mathcal{I}\to \bigwedge^3 H/H\big)$, this is the same as showing that $\tau(\mathcal{BBF}\cap \mathcal{I})$ does not contain the image under $\tau$ of a genus $k$ bounding pair. We know the image of such a BP, and you should be able to write down a condition on $\tau(f)$ for $f\in\mathcal{BBF}$ coming from the triviality of the action on the homology of the double cover. I have the feeling that this condition will rule out such a bounding pair, which would conclude the proof. I don't know how hard this approach would be if you wanted to actually carry it out, though.<|endoftext|> -TITLE: Rational singularities for fibered surfaces -QUESTION [8 upvotes]: This question consists of two parts. I will try to be as short and clear as possible. -Let $S$ be a Dedekind scheme of characteristic zero. The main examples are $\mathbf{P}^1_k$, with $k$ a field of characteristic 0 and $\textrm{Spec} \ \mathbf{Z}$. -A fibered surface is a projective flat morphism $X\longrightarrow S$ with $X$ an integral scheme of dimension 2. -Suppose $S=\mathbf{P}^1_k$ with $k$ a field of characteristic 0. Then $X$ is an algebraic surface over a field of characteristic zero. The theory of rational singularities for $X$ is then explained in Chapter 5 of Kollar and Mori: Birational geometry of algebraic varieties. -Now, I would like to know if there are analogous results for when $S=\textrm{Spec} \ \mathbf{Z}$ as in loc. cit.. Let me make this more precise. -Assume that $X$ is normal. (This will suffice for my applications.) -Call a resolution of singularities $\rho:Y\longrightarrow X$ rational if it satisfies -$$R^i\rho_\ast \mathcal{O}_Y =0 \ \textrm{for} \ i>0.$$ -We say that $X$ has rational singularities if all resolutions of $X$ are rational. -To have a good theory, we should probably show that $R^i \rho_\ast \omega_{Y/S} = 0$ for $i>0$. -Question 1: Is it true that $R^i \rho_\ast \omega_{Y/S} = 0$ for $i>0$ for any resolution of singularities? -Compare my last question with Theorem 5.10 of Kollar-Mori. -Question 2: Is it true that the following properties -A. $X$ has a rational resolution -B. $X$ has rational singularities -C. For every resolution of singularities $\rho:Y\longrightarrow X$, we have that $\rho_\ast \omega_{Y/S} = \omega_{X/S}$. -are equivalent? -I guess one could try to see if some of the arguments given in Kollar-Mori carry over to this setting. For example, it probably suffices to show that a resolution is rational if and only if it satisfies C. -Here is a possible application: Assume $X$ to be regular from now on. -Let $S$ be $\mathbf{P}^1_{\mathbf{C}}$. Let $\pi:Y\longrightarrow X$ be a finite surjective morphism with branch locus a normal crossings divisor. Then, it is easy to see that the singularities of $Y$ are quotient. (Consider the analytic topology or the etale topology and use Abhyankar's lemma). It is well-known that quotient singularities are rational. -I guess Abhyankar's lemma is still valid and therefore I was hoping to show that $Y$ in case $S= \textrm{Spec} \mathbf{Z}$ still has rational singularities. (The notion of quotient singularities seems only to work over $\mathbf{C}$...) - -REPLY [7 votes]: Ariyan, -EDIT: This contains some substantial edits and added references. -Lipman has defined the following notion (EDIT: twice): -Definition (Lipman ; Section 9 of "Rational singularities with applications to algebraic surfaces and factorization"): If $X$ is 2-dimensional and normal, $X$ has two pseudo-rational singularities if for every proper birational map $\pi : W \to X$ there exists a proper birational normal $Y$ over $W$ where, $R^1 \pi_* \mathcal{O}_Y = 0$ -Definition (Lipman-Teissier ; Section 2 of "Pseudo-rational local rings and a theorem of Briancon-Skoda about integral closures of ideals"): $X$ has pseudo-rational singularities if $X$ is CM (Cohen-Macaulay) and if for every proper birational map $\pi : Y \to X$ with $Y$ normal, $\pi_* \omega_Y = \omega_X$. -If these are the same in dimension 2, this seems pretty close to what you want in dimension 2. -EDIT2: These are the same in dimension 2, I was in Purdue and asked Lipman about question 1, which holds, and certainly implies this. -He also points out that regular schemes are pseudo-rational. In particular, this implies that if $\pi_* \omega_Y = \omega_X$ for one resolution of singularities, it also holds for every resolution of singularities (in fact, for every proper birational map with normal domain). -In dimension 2, he also studies relations between this condition and the local-finiteness of the divisor class group. -On the other hand, I'm pretty sure this is different from the definition of rational singularities you gave above at least in higher dimensions (with the appropriate $R^i$ vanishing instead of just $R^1$). -With regards to your specific questions: -Question #1: That vanishing, called Grauert-Riemenschneider vanishing, is known to fail for $\dim X > 2$ outside of equal characteristic zero. I believe the answer should hold in the two-dimensional case, certainly it should assuming that Lipman's various definitions of pseudo-rational singularities are consistent. -EDIT: This holds in dimension 2, see Theorem 2.4 in Lipman's "Desingularization of two-dimensional schemes". -In any dimension, that vanishing has recently been proven in equal characteristic $p > 0$ over a smooth variety (or a variety with tame quotient singularities), see arXiv:0911.3599. -Question #2: In higher dimensions, I'm pretty confident that the answer is no. In the 2-dimensional case, probably this is done by Lipman? In view of question #1, in order to find such a counter example in higher dimensions, one should look at various cones probably over 3 or 4-dimension schemes with negative Kodaira dimension (probably Fano's) but which fail Kodaira vanishing. -I have some thoughts on some other definitions of rational singularities which might be better in mixed characteristic, but I'm not sure I want to post them on MathOverflow right now. If you email me, I'd be willing discuss it a bit. -Quotient singularities can behave a little different outside of characteristic zero as well (see various papers of Mel Hochster from the 70s for instance). This can also lead one to look at questions like the Direct Summand Conjecture.<|endoftext|> -TITLE: "continuous" and "discontinuous" phase transitions in branching processes. -QUESTION [10 upvotes]: Consider a Galton-Watson branching process, with offspring distribution -$\mathbf{p}=(p_0, p_1, \dots, p_n, \dots)$. -Let $O$ be the root of the branching process. -Write $\eta=P(\text{process survives for ever})$ and $\mathcal{H}=\{\mathbf{p}: \eta>0\}$. -Also write $\beta=P(O \text{ is the root of an infinite binary tree contained in the branching process})$ and $\mathcal{B}=\{\mathbf{p}: \beta>0\}$. -Then it's fairly well known that in some sense the phase transition from $\mathcal H^C$ to $\mathcal H$ is typically "continuous" while the phase transition from $\mathcal B^C$ to $\mathcal B$ is typically "discontinuous". -e.g. suppose the offspring distribution is Poisson with mean $\lambda$ and consider $\eta$ as a function of $\lambda$. Then $\eta(\lambda)=0$ for $\lambda\leq 1$ and $\eta(\lambda)>0$ for $\lambda>1$, and $\eta(.)$ is continuous everywhere, including at 1. -On the other hand, consider $\beta$ as a function of $\lambda$. Then there is a critical point $\lambda_c\approx 3.3509$ with the following property: $\beta(\lambda)=0$ for $\lambda<\lambda_c$, -and $\beta(\lambda)>0$ for $\lambda\geq \lambda_c$. In particular, $\beta(.)$ is discontinuous at $\lambda_c$. (In fact, at $\lambda_c$, $\beta$ jumps from 0 to approximately 0.535). -My question: how has this been written as a general statement? (rather than just for particular parametrised families as above). My guess is that one would want to write it something like the following: -Let $M_0$ be the set of offspring distributions with the topology induced by the metric $d_0(\mathbf{p}, \mathbf{q})=\sum|p_n-q_n|$. -Similarly $M_1$ and $M_2$ induced by $d_1(\mathbf{p}, \mathbf{q})=\sum n|p_n-q_n|$ and -$d_2(\mathbf{p}, \mathbf{q})=\sum n^2|p_n-q_n|$. -Write $\mathbf{p}^*$ for the degenerate distribution with -$p^*_n=0$ for $n=0$ and $n\geq 2$ -and $p^*_1=1$. -Then: -(1) $\mathcal{H}\setminus\{\mathbf{p}^*\}$ is open as a subset of $M_0$. -(2) $\mathcal{B}$ is closed as a subset of $M_2$. (Probably also $M_1$?) -Of course (1) is basically trivial, since $\mathcal{H}$ is just the set of distributions -with mean greater than 1, along with the single point $\mathbf{p}^*$. -(Note that it's NOT the case that $\mathcal{B}$ is closed as a subset of $M_0$. -For example, consider a sequence of distributions indexed by $k$ with $p_0=1-4/k$ -and $p_k=4/k$, all other $p_n=0$. Then -$\beta(k)>0$ for all $k$, but $\beta(k)\to 0$ as $k\to\infty$, and the sequence of distributions converges in $M_0$ to a limit in which -just $p_0=1$, for which of course $\beta=0$.) -I don't think it's hard to write down a proof of (2) above, but I don't want to reinvent the wheel. (I don't need to use the result directly, but I would definitely like to refer to it to illustrate a point). So: anyone know where such things have been nicely developed before? - -REPLY [4 votes]: OK, in my opinion this is a very interesting question and, since it got no answer for three weeks, I feel free to post some comments on it (but not a full answer, I am afraid). -As everybody knows, the property that $\eta > 0$ is equivalent to the fact that the mean of the offspring distribution $p$ is $>1$. This is conveniently expressed if one relies on the generating function $f$ of $p$, defined by $f(s)=\sum p_ns^n$. Then (excluding the case where $p$ is a Dirac mass at $1$), $\eta > 0$ iff $f'(1) > 1$ iff the equation $f(s)=s$ has a solution $s_1 < 1$ and when this is so, $\eta=1-s_1$. -Likewise, the property that $\beta > 0$ is equivalent to the fact that the equation $g(s)=s$ has a solution $s_2 < 1$, where $g(s)=f(s)+(1-s)f'(s)$ (hint: condition on the first generation) and when this is so, $\beta=1-s_2$. -Now, it happens that if $s_1 < 1$ exists, the graph of $f$ on the interval $[0,1]$ (in other words, the curve of equation $t=f(s)$ in the $(t,s)$-square $[0,1]\times[0,1]$) must intersect the line $t=s$ with an order of contact $1$, because $f$ is strictly convex and $f(1)=1$. Hence every small modification of $f$ preserves this intersection and is such that $\eta > 0$ as well. -On the other hand, $g(1)=1$ but $g$ is not in general convex, hence it may well happen that the graph of $g$ intersects the line $t=s$ with an order of contact $2$ (in other words, it is tangent to the line $t=s$ at $s_2$). Then, infinitely small modifications of $f$ may move the graph of $g$ away from the line $t=s$, in which case the only solution of $g(s)=s$ is the obvious one $s=1$, hence $\beta$ jumps to $0$. -In other words, "There exists $s < 1$ such that $f(s)=s$" is an open property (excluding the function $f$ such that $f(s)=s$ for every $s$) but "There exists $s < 1$ such that $f(s)+(1-s)f'(s)=s$" is not. -Finally, note that the geometric distributions provide phase transitions that are located at easily computable points. Namely, assume that $p_n=(1-a)a^n$ for every $n\ge 0$, with $0 < a < 1$. Then, $\eta>0$ iff $a > a_1$ with $a_1=1/2$, in particular $\eta(a_1)=0$, but $\beta > 0$ iff $a\ge a_2$ with $a_2=4/5$ and $\beta(a_2)=1/4$ (I think). -To reiterate: these are just minor remarks on a stimulating question, not an answer to the OP's query.<|endoftext|> -TITLE: L'un des problèmes fondamentaux de la théorie des nombres -QUESTION [17 upvotes]: In his 1951 report Sur la théorie du corps de classes, Weil writes that - -La recherche d'une interprétation de $C_k$ si $k$ est un corps de - nombres, analogue en quelque manière - à l'interprétation par un groupe de - Galois quand $k$ est un corps de - fonctions, me semble constituer l'un - des problèmes fondamentaux de la - théorie des - nombres à l'heure actuelle; il se - peut qu'une telle interprétation - renferme la clef de l'hypothèse de - Riemann …. - -As requested by @PeteL.Clark, a translation (by @TonyScholl): - -The search for an interpretation for $C_k$, where $k$ is a number field—in some way analogous to its interpretation by a Galois group when $k$ is a function field—seems to me to be one of the fundamental problems of number theory today; perhaps such an interpretation contains the key to the Riemann hypothesis …. - -Here, $C_k$ is of course the idèle class group of the number field $k$. -I've heard that some people working in noncommutative geometry have thought about this problem. -Question. What progress has since been made towards such an interpretation? - -REPLY [3 votes]: Connes reformulates Weil's question here as: "Is there a non trivial Brauer theory of central simple algebras over $\mathbb{C}$ ?" and tells his solution of his reformulation in the context of a "cosmic galois group" in renormalization here. -Edit: Morava wrote on Weil group representations coming from algebraic topology. The bibl. list at the end of his article shows Weil's article with the question on $C_k$. It would be great if someone would look at it and tell more about it , please understandable for a non-(algebraic topologist) :-) -Edit: Lieven le Bruyn runs a seminar on "a possible connection between Connes’ noncommutative geometry approach to the Riemann hypothesis and the Langlands program", and will post lecture notes on his blog if enough people are interested.<|endoftext|> -TITLE: The 0th homology of a path-connected space -QUESTION [8 upvotes]: In singular homology one of the first calculations you can make is $H_0(X)=H_0(pt)$ for path-connected $X$. This seems to be a property which does not follow from the axioms for a generalized homology theory. This raises the following question: -Assume $H_* : Top^2 \to Ab^\mathbb{N}$ is a homology theory. Thus we impose homotopy invariance, excision, the long exact sequence, the dimension axiom and if you wish also the disjoint union axiom. Is then $H_0(X)=H_0(pt)$ for every path-connected space $X$? -I believe that there is a counterexample. Of course this can't be homotopy equivalent to a CW-complex. And probably this is the reason why this question is not reasonable at all. It's just my curiosity. - -REPLY [5 votes]: I'm not sure exactly what Cech homology is, but I'll assume that something which may or may not be called Cech homology has the following properties: -It's a generalized homology theory. -It vanishes in negative dimensions. -It satisfies the dimension axiom (so it's a homology theory). -There's a natural map from singular to Cech that is part of a natural triangle, an exact sequence -$\dots \to Cech_{n+1}\to ?_n\to Sing_n\to Cech_n\to \dots$. -The map $Sing_0\to Cech_0$ is always surjective, so that $?_n$ vanishes for $n<0$. -The map $Sing_0(TSC)\to Cech_0(TSC)$ is not injective if TSC is the topologist's sine curve. -(End of list of assumed properties.) -Then $?$ is a generalized homology theory vanishing in negative dimensions and vanishing on a point (therefore on CW complexes) but not vanishing on the path-connected space that you get by attaching a suitable $1$-cell to TSC. -The direct sum of ? and singular will then do the job, because $?_0(TSC\cup cell)=?_0(TSC)\ne 0$<|endoftext|> -TITLE: Classification of Quasi-topoi -QUESTION [14 upvotes]: (Grothendieck) topoi are left-exact reflective subcategories of a category of presheaves. An important class of quasi-topoi (see: http://ncatlab.org/nlab/show/quasitopos) arise as the category of concrete sheaves on a concrete site. Concrete sheaves are those sheaves $X$ such that the induced map $Hom(C,X) \to Hom(\underline{C},\underline{X})$ is injective for all objects $C$, where $\underline{C}$ is the underlying set of $C$ and $\underline{X}$ is the value of $X$ on the terminal object. Concrete sheaves are a reflective subcategory of all sheaves. Concrete sheaves are a particularly nice example of a quasi-topos as the resulting quasi-topos is both complete and cocomplete. My question is, is there a way to represent quasi-topoi (or nice ones) as reflective subcategories of a Grothendieck topos (with some condition on the reflector)? (Of course, for this, you'd need the quasi-topos to be complete, since reflective subcategories of complete categories are again complete). More generally, is there some theorem saying that a category is a (possibly non-complete) quasi-topos if and only if it can be embedded into a topos such that the embedding has such-and-such property? - -REPLY [7 votes]: This is really an answer to the question raised in Mike's reformulation of the question, but is too long for a comment and may be of interest. -Richard Garner and I have considered when a reflective subcategory of a presheaf category has the form considered in condition 2 of his answer. It turns out that preservation of finite products and monomorphisms is not enough: to see this, consider the reflection of directed graphs into preorders, which preserves finite products and monomorphisms but is not of this form. -In fact for a full reflective subcategory E of a presheaf category [C^{op},Set], the following conditions are equivalent: - -there is a topology j and a larger topology k for which E consists of the objects which are sheaves for j and separated for k -the reflection preserves finite products and monomorphisms and is semi-left-exact -the reflection preserves monomorphisms and has stable units - -Here the notions of semi-left-exactness and stable units come from - -Cassidy, Hebert, Kelly, Reflective subcategories, localizations, and factorization systems, J. Austral. Math. Soc. Ser. A, 38:287-329, 1985. - -Let R be the reflection and r the unit of the reflection. Semi-left-exactness says that R preserves each pullback of a component rX:X->RX of the unit along a map A->RX with A in the subcategory. -Stable units says the same thing, but without the requirement that A be in the subcategory. This turns out to be equivalent to R preserving all pullbacks over an object of the subcategory.<|endoftext|> -TITLE: When are the intermediate fields totally ordered? -QUESTION [12 upvotes]: The groups whose subgroups are totally ordered by inclusion are easy to classify; they are subgroups of $\mathbb{Z}/p^{\infty} = \text{colim } \mathbb{Z}/p^k$ for some prime $p$, and thus $\mathbb{Z}/p^{\infty}$ or finite cyclic of prime power order. What about fields? - -Is it possible to classify (or at least, give some properties and examples) the field extensions $E/F$, whose intermediate fields are totally ordered by inclusion? - -If $E/F$ is a Galois extension, then we may rephrase the condition: The closed subgroups of $Gal(E/F)$ are totally ordered. It can be shown that then there is a prime $p$ such that $Gal(E/F)$ is pro-$p$-cyclic and thus isomorphic to $\mathbb{Z}/p^k$ for some $k \geq 0$ or to $\mathbb{Z}_p = \lim_k \mathbb{Z}/p^k$. Thus $E/F$ is built up out of cyclic Galois extensions $F_{i+1} / F_i$ degree $p$ (for example $E = \mathbb{F}_{q^{p^\infty}}, F = \mathbb{F}_q$). In characteristic $p$, cyclic Galois extensions of degree $p$ are characterized by a Theorem of Artin-Schreier. In characteristic $q \neq p$ ($q=0$ allowed), there is a characterization if $F_i$ contains a primitive $p$th root of unity. What can be said if this is not the case? -Now do not assume that $E/F$ is Galois. Here is a simple observation: - -The intermediate fields of $E/F$ are totally ordered iff $E/F$ is algebraic and the finite intermediate fields of $E/F$ are totally ordered. - -Proof: $\Rightarrow:$ If $t$ is a variable, then the intermediate fields of $F(t)/F$ are not totally ordered, consider $F(t^2)$ and $F(t^3)$. $\Leftarrow:$ Let $K,L$ be intermediate fields which are not compatible. Choose $a \in K - L, b \in L - K$. Then $F(a), F(b)$ are finite extensions, which are not compatible, contradiction. -An immediate consequence is, that we may first restrict to the finite case: Namely every $E/F$ as above is a directed union of finite subextensions, whose intermediate fields are totally ordered by inclusion; and vice versa. -Note that we cannot restrict the degree of $E/F$. Namely, $S_{n-1}$ is a maximal subgroup of $S_n$. Since $S_n=Gal(E/F)$ for some Galois extension $E/F$, if $K$ is the fixed field of $S_{n-1}$, the extension $K/F$ has degree $n$ and no nontrivial intermediate fields at all. -What about inseparable extensions? And what happens if we take the normal closure? Perhaps we can reduce everything to the Galois case? - -REPLY [5 votes]: In the separable case, as the question already notes, what you're really asking about is finite extensions and ultimately group theory. A finite, separable field extension is equivalent to transitive permutation group $G$ acting on a finite set $X$. This is in turn equivalent to a pair of groups $H \subseteq G$, with $X = G/H$, up to the equivalence of quotienting by a $G$-normal subgroup that lies in $H$. The invariant partitions of $X$ correspond to intermediate subgroups, which then in the Galois theory setting correspond to intermediate fields. If there are no invariant partitions, then $G$ is called primitive. Also, for every permutation group $G/H$, it is easy to find some field extension modeled by $G/H$, even though it is a famous open problem to always find such an extension of $\mathbb{Q}$. -So the question is when the $G$-invariant partitions of $X = G/H$ are totally ordered. The first remark is that if they are totally ordered, then actually the combinatorics of these partitions is completely determined by their relative indices. In other words, $X$ is divided into $n_1$ parts, then each part is divided into $n_2$ parts, and so on, and there is no structure other than this sequence of integers. In the Galois example where $H$ is trivial and $G = \mathbb{Z}/p^k$, each $n_i = p$. -So, $G$ is some group that acts on this tower of partitions. The second remark is that for a given tower of partitions, there is a maximal choice of $G$, namely all of the automorphisms of the tower of partitions. This is an iterated wreath product of the symmetric groups $S_{n_i}$. This is an example where there are no other invariant partitions, hence they are totally ordered. In fact, any iterated wreath product of primitive groups works just as well. These constructions correspond in Galois theory to starting with a base field $K$ and adjoining one root each of a sequence of primitive polynomials (primitive in the sense that the Galois group acts primitively on the roots), chosen "generically" so that the total Galois group is the wreath product. -If $\mathcal{P}$ is the tower of partitions, then the general example is a group $G \subseteq \text{Aut}(\mathcal{P})$ that happens not to have any other invariant partitions. I think that it's a complicated question to find all examples. From this viewpoint, it seems a bit accidental that -$$\mathbb{Z}/p^k \subseteq S_p \wr S_p \wr \cdots \wr S_p$$ -is still big enough that the invariant partitions are linearly ordered. So I don't think that there is any principle of reducing to the Galois case.<|endoftext|> -TITLE: Any sum of 2 dice with equal probability -QUESTION [9 upvotes]: The question is the following: Can one create two nonidentical loaded 6-sided dice such that when one throws with both dice and sums their values the probability of any sum (from 2 to 12) is the same. I said nonidentical because its easy to verify that with identical loaded dice its not possible. -Formally: Let's say that $q_{i}$ is the probability that we throw $i$ on the first die and $p_{i}$ is the same for the second die. $p_{i},q_{i} \in [0,1]$ for all $i \in 1\ldots 6$. The question is that with these constraints are there $q_{i}$s and $p_{i}$s that satisfy the following equations: -$ q_{1} \cdot p_{1} = \frac{1}{11}$ -$ q_{1} \cdot p_{2} + q_{2} \cdot p_{1} = \frac{1}{11}$ -$ q_{1} \cdot p_{3} + q_{2} \cdot p_{2} + q_{3} \cdot p_{1} = \frac{1}{11}$ -$ q_{1} \cdot p_{4} + q_{2} \cdot p_{3} + q_{3} \cdot p_{2} + q_{4} \cdot p_{1} = \frac{1}{11}$ -$ q_{1} \cdot p_{5} + q_{2} \cdot p_{4} + q_{3} \cdot p_{3} + q_{4} \cdot p_{2} + q_{5} \cdot p_{1} = \frac{1}{11}$ -$ q_{1} \cdot p_{6} + q_{2} \cdot p_{5} + q_{3} \cdot p_{4} + q_{4} \cdot p_{3} + q_{5} \cdot p_{2} + q_{6} \cdot p_{1} = \frac{1}{11}$ -$ q_{2} \cdot p_{6} + q_{3} \cdot p_{5} + q_{4} \cdot p_{4} + q_{5} \cdot p_{3} + q_{6} \cdot p_{2} = \frac{1}{11}$ -$ q_{3} \cdot p_{6} + q_{4} \cdot p_{5} + q_{5} \cdot p_{4} + q_{6} \cdot p_{3} = \frac{1}{11}$ -$ q_{4} \cdot p_{6} + q_{5} \cdot p_{5} + q_{6} \cdot p_{4} = \frac{1}{11}$ -$ q_{5} \cdot p_{6} + q_{6} \cdot p_{5} = \frac{1}{11}$ -$ q_{6} \cdot p_{6} = \frac{1}{11}$ - -I don't really now how to start with this. Any suggestions are welcome. - -REPLY [2 votes]: I've heard this brainteaser before, and usually its phrased that 2-12 must come up equally likely (no comment about other sums). With this formulation (or interpretation) it becomes possible. Namely, {0,0,0,6,6,6} and {1,2,3,4,5,6}. In this case, you can also generate the sum 1, but 2-12 are equally likely (1-12 are equally likely). Without allowing for other sums I do suspect its impossible (and looks like proofs have been given). I arrived at this answer by noting we are asking for equal probability for 11 events that come from 36 (6*6) possible outcomes, which immediately seems unlikely. However equal probability for 12 events from 36 outcomes is far more manageable :) -Phil<|endoftext|> -TITLE: Non-Borel subspace of Banach space -QUESTION [7 upvotes]: Let $X$ be a separable Banach space, $M \subset X$ a linear subspace. Must $M$ be a Borel set in $X$? -I believe the answer is "no," since I have seen authors who are careful to talk about "Borel subspaces". But I have not been able to find a counterexample. -If the answer is indeed "no", does every infinite-dimensional separable Banach space contain a non-Borel dense linear subspace? - -REPLY [3 votes]: Similar question was discussed before and answered (in particular to Bill's question in the comments) -Are proper linear subspaces of Banach spaces always meager?<|endoftext|> -TITLE: Symmetric Proof that Product is Well-Founded -QUESTION [6 upvotes]: This is a fairly minor, technical question, but I'll toss it out in case someone has a good idea on it. -Suppose $(X,<_X)$ and $(Y,<_Y)$ are well-founded orderings (not necessarily linearly ordered, though I don't think it matters). Consider the ordering ${<}$ on $X\times Y$ given by $(x',y') < (x,y)$ if $x'\leq x$ and $y'\leq y$, and either $x' < x$ or $y' < y$. Note that this is not the lexicographic ordering; indeed, it's symmetric. -Obviously $X\times Y$ is well-founded. Suppose I want to prove this carefully (by which I really mean "in the formal theory $ID_1$"); more precisely, let's take $X$ to be a set with two properties: -$$Cl_X:\forall x(\forall x'<_X x. x'\in X)\rightarrow x\in X$$ -and -$$Ind_X: \forall Z[\forall x(\forall x'<_X x. x'\in Z)\rightarrow x\in Z)\rightarrow X\subseteq Z]$$ -and similarly for $Y$. (These just characterize that $X$ is its own well-founded part.) I want to prove that for all $(x,y)\in X\times Y$, $(x,y)$ are in the well-founded part of $X\times Y$ under ${<}$; call the well-founded part of $X\times Y$ $Acc(X\times Y)$. -I know one way to prove this: for each $x\in X$, define $Y_x=\{y\in Y\mid (x,y)\in Acc(X\times Y)\}$. Let $X'$ be the set of $x\in X$ such that $Y\subseteq Y_x$. Then it would be good enough to show that $X'$ satisfies the closure property, so I can apply $Ind_X$. To do this, in turn, I show that, if $Y\subseteq Y_{x'}$ for all $x'<_X x$ then $Y_x$ satisfies the closure property, so I can apply $Ind_Y$. -Of course, that means I know I second way: I could swap $X$ and $Y$ in the above proof. Moreover, when one works through the details, it's clear that I'm really proving that the lexicographic ordering is well-founded, and using the fact that ${<}$ is a subrelation of the lexicographic ordering. -Which brings me to my question: is there a proof that $Acc(X\times Y)=X\times Y$ which is symmetric? - -REPLY [3 votes]: I've found a symmetric proof. -First, every well-founded relation admits an ordinal rank function, an assignment of points to ordinals that respects the relation. For example, in your case $\alpha_x=\sup\{\ \alpha_u+1\mid u\mathrel{\lt_X} x\ \}$ is the canonical rank function for $X$ and $\beta_y=\sup\{\ \beta_w+1\mid w\mathrel{\lt_Y} y\ \}$ is the canonical rank function for $Y$. -Second, the key idea is to use the symmetric version of ordinal addition, called natural sum, an associative and commutative addition operation on ordinals. Specifically, the natural sum $\alpha\mathop{\sharp}\beta$ of two ordinals is the supremum of the order types arising in any linear completion of the disjoint sum partial order $\alpha\sqcup\beta$. Alternatively, if you express $\alpha$ and $\beta$ in Cantor normal form, then $\alpha\mathop{\sharp}\beta$ is the ordinal obtained by mixing the Cantor normal forms together and putting the terms in the correct order. The natural sum is defined in a completely symmetric way, and this is why it is commutative. -In your product order $X\times Y$, let us associate to the point $(x,y)$ the ordinal $\alpha_x\mathop{\sharp}\beta_y$. This ordinal assignment is completely symmetric, since we would assign the same ordinal to $(y,x)$ in $Y\times X$, as the natural sum is commutative. The point now is that this ordinal assignment serves as a rank function in $X\times Y$, since if $(x',y')\lt (x,y)$, then we know that $\alpha_{x'}\leq \alpha_x$ and $\beta_{y'}\leq \beta_y$, and at least one of these is strict. It follows that $\alpha_{x'}\mathop{\sharp}\beta_{y'}\lt \alpha_x\mathop{\sharp}\beta_y$, essentially because $(\alpha+1)\mathop{\sharp}\beta=(\alpha\mathop{\sharp}\beta)+1$, and so this really does rank the product relation, and so it is well-founded.<|endoftext|> -TITLE: A general formula for the number of conjugacy classes of $\mathbb{S}_n \times \mathbb{S}_n$ acted on by $ \mathbb{S}_n$ -QUESTION [14 upvotes]: $\def\S{\mathbb{S}}$ Dear all, -So I have $\S_n$ acting on $\S_n \times \S_n$ via conjugacy. That is: -for $g \in \S_n, (x,y) \in \S_n \times \S_n$: $g(x,y) = (gxg^{-1},gyg^{-1}).$ -Is there a general formula (or a nice combinatorial description) for the number of conjugacy classes produced in this case? -Some thoughts on the problem: for a fixed $n$ one can compute this by hand as follows: -Note that if we consider the action of $\S_n \times \S_n$ on $\S_n \times \S_n$ by conjugation, then we have classes of types $(\lambda, \mu)$ for each configuration $\lambda$ and $\mu$ of the Young diagram. Here we restrict ourselves to the diagonal action, so this would split the $(\lambda,\mu)$ pair further. To compute this splitting for a fixed pair $(x,y)$ where $x$ is of type $\lambda$, y is of type $\mu$, let $C(x)$ denotes the centralizer of $x$. For any $g \in \S_n$, write $g = g_1g_x$ for some $g_x \in C(x)$. Then -$(x,y) \sim (g_1xg_1^{-1}, g_1(g_xyg_x^{-1})g_1^{-1})$ -Then the number of equivalence classes we have for this pair is the number of orbits of the class $\mu$ acted upon by $C(x)$ via conjugation. -But is there a general formula? -Thanks, -Ngoc - -REPLY [3 votes]: I discovered this old question in connection with someone else's similar (and current) question: The Simultaneous Conjugacy Problem in the symmetric group $S_N$ -If anyone still cares, here is a slightly different answer for your question using the character table instead of centralizer sizes. The characters might be more accessible in certain cases (and of interest in their own right). -$$N=\frac{1}{n!} \sum_{\alpha \in G} \left( \sum_{\text{irreps}\ \chi} \chi(\alpha)^2 \right)^2$$ -The inner sum is just the centralizer term in Marty Isaacs answer, but expressed in a different way. It might be of some use in this form. For example, if you want to enrich the group action to allow coordinate swap or inversion, similar class functions can be used. -The OP and others may be interested in the above linked question, which discusses the related decision problem and the generalization to more coordinates.<|endoftext|> -TITLE: Is there a "Hilbert syzygy theorem" for smooth manifolds? Or: does every finitely generated $C^\infty$ module have a finite-length resolution in vector bundles? -QUESTION [8 upvotes]: Let $X$ be a real smooth manifold, and $M$ a locally-finitely-generated sheaf of $\mathcal C^\infty(X)$-modules. (If $X$ is not compact, I will also insist that there be a global bound on the number of generators I might need in different regions; maybe this is part of the usual meaning of the words "locally finitely generated".) -I would like to find finite-dimensional vector bundles $E_1,\dots,E_n$ over $X$ and maps of $\mathcal C^\infty$-modules -$$ 0 \to \Gamma(E_n) \to \dots \to \Gamma(E_1) \to M \to 0$$ -so that the sequence is exact. Can I always do this? And is there an explicit bound on the number of vector bundles needed, e.g. $n = \dim X$ or $n = \dim X+1$? - -REPLY [13 votes]: No. Let $X$ be $\mathbb R$. In the ring $C^{\infty}(X)$ let $I$ be the ideal of all functions vanishing to infinite order at $0$. The module $C^{\infty}(X)/I$ does not have a finite resolution by finitely generated projective modules. -Edit: -Still no if you want the finitely generated module to be contained in a finitely generated projective module. For the same $X$ pick a function $f$ such $f(x)$ vanishes precisely when $x<0$. let $J$ be the ideal generated by $f$. The module $J$ does not have a finite resolution by finitely generated projective modules. -For both of these examples, the method I have in mind is this: If a module $M$ has a finite projective resolution $P_\bullet$ then for every point in $p\in X$ the alternating sum of the $k_p$ vector space dimension of $Tor_n(M,k_p)$ is independent of $p$ because it's the alternating sum of the rank of $P_n$. I believe that in the first example this Euler number comes out to be $1$ if $p$ is the origin and otherwise $0$, and in the second it's $1$ if $p> 0$ and $0$ if $p< 0$.<|endoftext|> -TITLE: non-isomorphic objects with no known nontrivial distinguishing invariants -QUESTION [6 upvotes]: If X and Y are non-isomorphic objects, then "[is / is not] isomorphic to [ X / Y ]" are invariants that distinguish X and Y. You can also do things like take an object Z that is not isomorphic to Y, and then "is isomorphic to X or isomorphic to Z" is another invariant that distinguishes X and Y. Similarly, if W is isomorphic to X, then "is isomorphic to W" works too. - -Are there any objects X and Y known to be non-isomorphic, but all known distinguishing invariants -(a) use the concept or definition of isomorphism - -or - -(b) use the concept or definition of being isomorphic to some object Z, where Z is isomorphic to X or Y - -REPLY [4 votes]: Non-principal ultrafilters on the natural numbers don't have enough invariants to distinguish the non-isomorphic ones. (Two such ultrafilters are called isomorphic if one is sent to the other by a permutation of the underlying set $\mathbb N$.)<|endoftext|> -TITLE: abelian categories vs. additive categories -QUESTION [9 upvotes]: This must be common knowledge. -Where exactly in the development of homological algebra does one need the axiom that makes additivepre-abelian and abelian categories different? (I mean this statement: for every morphism $u: X \to Y$, the canonical morphism $\bar{u}: \mathrm{Coim}\ u \to \mathrm{Im}\ u$ is an isomorphism.) -My gut feeling is that it should be necessary for the Snake lemma, but I couldn't find a step in the proof that would use it. - -REPLY [2 votes]: You might be interested in taking a quick look at the following paper: -A. V. Yakovlev - Homological algebra in pre-Abelian categories (1982) -(see http://www.springerlink.com/content/u2506k2608868457/) -Plenty of examples of pre-abelian categories that are not abelian come from categories of more "analytic" objects (such as the category of topological abelian groups), but it is also common for such analytic examples to fail to have the kernels and cokernels required to be even pre-abelian. Because of this, it may be more common when studying cohomology theories in these analytic cases to develop the relevant homological algebra in the context of (Quillen) exact categories rather than say pre-abelian categories (even though you can perhaps develop homological algebra for pre-abelian categories). But I am no expert.<|endoftext|> -TITLE: Why is this set stationary? -QUESTION [5 upvotes]: Hi -I really need a proof for the following statement by Baumgartner: -There exists a stationary subset of $[\omega_2]^{\omega}$ of size $\aleph_2$. -This is Exercise 38.15. in Jechs Book (2003) and you can find a hint there which goes like this: For each $\alpha < \omega_2$, let $f_{\alpha} : \alpha \to \omega_1$ be one to one. If $\alpha < \omega_2$ and $\xi < \omega_1$ set $X_{\alpha, \xi} =$ { $\beta < \alpha : f_{\alpha} (\beta) < \xi$ }. Then $S:=$ { $X_{\alpha, \xi} : \alpha < \omega_2, \xi < \omega_1$} is our desired stationary subset. -But so far my attempts to proof this didn't work, because the sequence of the $f_{\alpha}$s doesn't have any nice regularity properties. -Thank you. - -REPLY [10 votes]: Given $F:[\omega_2]^{<\omega}\to[\omega_2]^{\aleph_0}$ as above, we first claim the existence of an ordinal $\omega_1\leq\alpha<\omega$ that is closed under $F$, i.e., $s\in [\alpha]^{<\omega}$ implies $F(s)\subseteq\alpha$. For this, let $\alpha$ be the limit of the sequence $\omega_1=\alpha_0<\alpha_1<\cdots$ where $\alpha_{n+1}$ is sufficiently large that $F(s)\subseteq \alpha_{n+1}$ for $s\in [\alpha_{n}]^{<\omega}$. -Given $\alpha$ as above, construct similarly the ordinal $\omega\leq\xi<\omega_1$ so that $f^{-1}_\alpha[\xi]$, that is, $\{\beta<\alpha:f_\alpha(\beta)<\xi\}$, is closed under $F$. This can be done similarly: let $\xi$ be the limit of the sequence $\omega=\xi_0<\xi_1<\cdots$ where $\xi_{n+1}$ is chosen so that if $s$ is a finite subset of $\{\beta<\alpha:f_\alpha(\beta)<\xi_n\}$, then $F(s)$ (which is a subset of $\alpha$) is a subset of $\{\beta<\alpha:f_\alpha(\beta)<\xi_{n+1})\}$. Now $X_{\alpha,\xi}$ is closed under $F$.<|endoftext|> -TITLE: Are the Schur functions the minimal basis of the ring of symmetric functions with the following properties? -QUESTION [5 upvotes]: Let $\Lambda$ denote the ring of symmetric functions in variables $x_1,x_2,\dots$ and with coefficients in $\mathbf{Q}$. Then $\Lambda$ is freely generated as an $\mathbf{Q}$-algebra by $p_1,p_2,\dots$, where $p_n$ denotes the $n$-th power sum function $x_1^n+x_2^n+\cdots$. Let $\Delta^+$ and $\Delta^{\times}$ denote the $\mathbf{Q}$-algebra maps $\Lambda\to\Lambda\otimes_{\mathbf{Q}}\Lambda$ determined by $\Delta^+(p_n)=1\otimes p_n + p_n\otimes 1$ and $\Delta^{\times}(p_n)=p_n\otimes p_n$ for all $n\geq 1$. Let $\mathbf{Q}_+$ denote the sub-semiring $\{a\in\mathbf{Q}| a\geq 0\}$ of $\mathbf{Q}$. -Consider the following properties on subsets $S$ of $\Lambda$: - -$S$ is a $\mathbf{Q}$-linear basis of $\Lambda$. -All finite sums and products of elements of $S$ are contained in the $\mathbf{Q}_+$-linear span of $S$. (That is, the span is a sub-$\mathbf{Q}_+$-algebra.) -The subsets $\Delta^+(S)$ and $\Delta^{\times}(S)$ of $\Lambda\otimes_{\mathbf{Q}}\Lambda$ are contained in the $\mathbf{Q}_+$-linear span of $S\otimes S = \{s\otimes s' | s,s'\in S\}$. -For all $s,s'\in S$, the composition $s\circ s'$ is contained the $\mathbf{Q}_+$-linear span of $S$, where $\circ$ denotes plethysm. - -These properties are satisfied if $S$ is the set of Schur functions or the set of monomial symmetric functions, for example. But the Schur functions give a smaller example in the sense that -they're contained in the $\mathbf{Q}_+$-linear span of the monomial symmetric functions. -I have many imprecise questions about subsets $S$ satisfying these properties, but in the interest of fair play, I'll ask a yes/no one: - -Are the Schur functions the smallest example? That is, if $S$ satisfies the properties above, does its $\mathbf{Q}_+$-linear span contain all the Schur functions? - -(Apologies if this is standard, but I don't know much about Schur functions. I didn't find anything about it in Macdonald's book, and rather than emailing random experts, it's more fun to ask it here.) - -REPLY [4 votes]: There is another simple set of symmetric functions fulfilling your properties 1 to 4: the set of products of power sums, -$$ -p_1, p_2, p_1^2, p_3, p_1 p_2, ... -$$ -The convex cone they generate is not contained in the cone generated by the Schur functions. For instance, $p_2=s_2-s_{1,1}$. -EDIT: and the Schur basis is not contained in the convex cone generated by the power sums: $s_{1,1}=\frac{p_1^2-p_2}{2}$.<|endoftext|> -TITLE: Harmonic analysis on semisimple groups - modern treatment -QUESTION [19 upvotes]: For my finals, I am digging through the book by Varadarajan An introduction to harmonic analysis on semisimple Lie groups. I find it a rather hard read and I feel it's a bit outdated now. Any recommendation of a more modern (and/or) introductory treatment to the topics covered by this book would be greatly appreciated. -I am more interested in the representation theory, although I find the connection to harmonic analysis intriguing. Having learned finite-dimensional representation theory I wanted to move on to the infinite-dimensional one. From harmonic analysis I know only the classical Pontryagin duality, which I thought is enough to get me started, but I've found Varadarajan's approach based on examples difficult to follow. - -REPLY [7 votes]: The first thing one should keep in mind is that harmonic analysis on semisimple Lie groups is very different from the "abstract harmonic analysis" a la Loomis or Hewitt and Ross dealing with locally compact abelian groups. The semisimple case was developed in the large part by Harish-Chandra and his papers (reprinted in his 4 volume Collected papers), while considerably older than Varadarajan's book, are still a good source of results and inspiration for many of us. -For an introduction to the subject, I warmly recommend Howe and Tan's book cited by Jim. It treats from the representation theory point of view the simplest nontrivial case, nonabelian harmonic analysis related to $SL(2,\mathbb{R}).$ The book uses elementary tools, and yet it deals with a wide range of topics. Elements of Harish-Chandra's theory for general reductive Lie groups may be found in Howe's survey article A century of Lie theory. -The theory of special functions and harmonic analysis on classical symmetric spaces and reductive symmetric spaces based on representation theoretic approach with different flavor is exposed in the books by Vilenkin, Helgasson, and Heckman and Schlichtkrull, which provide good complementary accounts of this theory.<|endoftext|> -TITLE: Morphism between projective varieties -QUESTION [8 upvotes]: Let $f:X \rightarrow Y$ be a morphism between two smooth projective varieties $X,Y$ which are defined over an algebraically closed field $k$. I am looking for some criteria which guaranties the projectivity of $f$. -For instance if $f$ is finite it is projective. Here we don't need the projectivity of the varieties $X,Y$. -Is the morphism $f$ projective if -Question1: The fibers of $f$ are finite? -Question 2: $f$ is one-to-one? -Question 3: $f$ is onto? -Does the assumption $k=\mathbb{C}$ make the questions easier? - -REPLY [11 votes]: A morphism $X \to Y$ factorizes as an embedding $X \to X\times Y$ (the graph of $f$) followed by the projection $X\times Y \to Y$. The first is a closed embedding (and hence is projective) if $Y$ is separable and the second is projective if $X$ is projective.<|endoftext|> -TITLE: Statistics for Second order properties of Random graphs -QUESTION [8 upvotes]: Hi! -Let G(N) be the number of graphs with vertices {1, 2, ..., N} and GN(F) be the number of those of them which satisfy graph property F. There is a beautiful result by Glebskii and Fagin that limit of GN(F)/G(N) can be only 0 or 1 for any first order graph property F. The proof essentialy uses compactness and Vaught's test. So, any first order property is either infinitely likely or infinitely unlikely for large graphs. Moreover, there is an algorithm which, given any first order graph property F as input, decides which of this possibilities holds for F (infinitely likely properties can be axiomatized by complete recursive first order theory). -Questions. What about second order properties? -More specificaly, what possibilities can be for probability of second order graph properties? Can we construct for any x from [0, 1] SO property with limit probability exactly x? What is the limit probability for connectedness of graph? -Thanks for any answers and comments. - -REPLY [2 votes]: Lots of work has been done on this topic; you might begin by checking the joint papers of Phokion Kolaitis and Moshe Vardi, for example: -Phokion G. Kolaitis, Moshe Y. Vardi: 0-1 Laws and Decision Problems for Fragments of Second-Order Logic Inf. Comput. 87(1/2): 301-337 (1990) -See Kolaitis's web page or -http://www.informatik.uni-trier.de/~ley/db/indices/a-tree/k/Kolaitis:Phokion_G=.html<|endoftext|> -TITLE: Glueing triangulated categories -QUESTION [13 upvotes]: Hello! -Given a triangulated category, one can look for semiorthogonal decompositions into (simpler?) triangulated subcategories. -I'd like to know if there's a way to attack the opposite problem, i.e. to classify the ways two given triangulated categories can be composed to give a big triangulated category which decomposes semiorthogonally into the given ones. Does anybody know about this? -I hope this question isn't too vague. -Thank you! - -REPLY [3 votes]: There is an article by Kuznetsov and Lunts which lifts the glueing of triangulated categories to differential graded categories: http://arxiv.org/abs/1212.6170.<|endoftext|> -TITLE: Can Deligne-Mumford stacks be characterized by their restriction to a small subcategory? -QUESTION [5 upvotes]: If I have a Deligne-Mumford stack $\Pi : X \to (\mathrm{Sch}/k)$ for some field $k$, can it be reconstructed from $\Pi^{-1}(C) \subset X$ for some "small" subcategory $C \subset (\mathrm{Sch}/k)$? For example, let's say we assume the fppf topology (to be concrete), then does the restriction to the fully faithful subcategory $T$ with objects -$\mathrm{Ob}(T) := $ { $I_n := \mathrm{Spec~} k[\epsilon]/(\epsilon^n) \mid n \in \mathbf{N}$ } -completely determine the stack? I think I've read that the restriction to $(\mathrm{Aff}/k)$ is enough, but since the fibers over the $I_n$ determine the $n^{th}$-order formal neighborhoods ($\mathrm{HOM}(I_n, X) \cong X(I_n)$), I wonder if the restriction to $T$ is enough? - -REPLY [7 votes]: Restricting to Aff is certainly enough, but Aff isn't small (there are e.g., polynomial algebras on arbitrary sets). If your DM stack is finitely presented over $k$ (which is probably good to include in the definition, to avoid these issues), then it is determined by it's restriction to finitely-presented affines (which is essentially small). -Without some finiteness hypothesis, no set of finitely-presented algebras can suffice (even for affine schemes, nevermind DM stacks). (And I suppose no small category of test objects can suffice: Take Spec of a field generated by a set of cardinality larger than that of global sections of any of your test schemes.) -The set you give is insufficient even for smooth varieties over an alg. closed field: you will have a morphism whenever you have an (arbitrary) map on $k$-points.<|endoftext|> -TITLE: An eventually different function adding no Solovay real nor dominating function? -QUESTION [6 upvotes]: Definitions -I believe set theorists have studied all of the following three notions in the context of forcing extensions of a model of ZFC, $M$ (hopefully the terminology is the standard one). - -A function $f:\mathbb N\rightarrow\mathbb N$ is eventually different if for each function $g:\mathbb N\rightarrow\mathbb N$, $g\in M$, the set $\{n: f(n)=g(n)\}$ is finite. -A real $r\in [0,1]$ is a Solovay random real if for each measure-zero subset $S$ of $\mathbb R$ with $S\in M$, we have $r\not\in S$. -A function $f:\mathbb N\rightarrow\mathbb N$ is dominating if for each function $g:\mathbb N\rightarrow\mathbb N$ in the ground model $M$, the set $\{n: f(n)\le g(n)\}$ is finite. - -Motivation - - -An eventually different function that is not too fast-growing is reminiscent of a random real. Can we always use it to construct a random real? The analogous problem in computability theory was quite difficult but has been solved by Kumabe and Lewis (J. LMS, 2009). -Question - -I. Is it possible to add an eventually different function to $M$ while adding neither a Solovay real nor a dominating function? - - -EDIT: Now stating the question in the strongest possible form, which is the one Andrés Caicedo answers below. - -REPLY [6 votes]: Hi Bjørn, and congratulations to you and Bonnie! -The answer to I is yes. In fact, there is a standard way of doing this, with the "eventually different forcing ${\mathbb E}$". This notion does not add random or dominating reals, and adds an eventually different function. -Conditions have the form $(s, A)$ where $s\in\omega^{<\omega}$ and $A\in[\omega^\omega]^{<\omega}$, with $(s, A)\le(s',A')$ iff $s\supseteq s'$, $A\supseteq A'$, and for all $f\in A'$ and $j\in[|s'|,|s|)$, we have $s(j)\ne f(j)$. (For me, $p\le q$ means that $p$ is stronger.) -This is a nice forcing: It is ccc, in fact, $\sigma$-centered, since any two conditions with the same first coordinate are compatible. But no $\sigma$-centered forcing adds random reals. -That ${\mathbb E}$ does not add dominating reals is a tad more work. But you can find a written proof in section 7.4.B of "Set Theory: On the structure of the real line", by Tomek Bartoszy´nski and Haim Judah. Let me know if you do not have access to a copy.<|endoftext|> -TITLE: Nonmetrizable uniformities with metrizable topologies -QUESTION [5 upvotes]: I'm looking for such pathological examples of uniform spaces which are not metrizable, but whose underlying topology is metrizable. Willard in his General Topology text constructs such a uniformity using ordinals. I am asking for examples which do not rely on ordinals. -EDIT: Below is an example by Daniel Tausk using families of pseudometrics. I forgot to mention that, if possible, I would like an example that uses the definition through a diagonal uniformity, not the definition through pseudometrics nor the definition through uniform covers. See Cryptomorphisms for some elaboration on this phenomenon. - -REPLY [5 votes]: The family of all neighbourhoods of the diagonal of $\mathbb{R}$ (with its normal topology) is a uniformity without a countable basis but it generates the normal topology. -The normal topology of $\mathbb{R}$ is the topology generated by the usual distance. -To see that the open sets in the plane that contain the diagonal generate a uniformity let $U$ be such an open set. For each $x$ there is $r_x$ such that $B(x,r_x)^2\subseteq U$. Now let $V$ be the union of the squares $B(x,\frac13r_x)^2$; this is again an open set and it is not hard to show that $V\circ V\subseteq U$. -If $U$ is an open set in the plane that contains the diagonal then the vertical section $U[x]$ is open in $\mathbb{R}$, so this uniformity generates at best a subtopology of the normal topology and hence certainly not the discrete topology. To see that it generates the normal topology consider, given $x$ and $r$, the open set $U_{x,r}=B(x,r)^2\cup(\mathbb{R}\setminus\lbrace x\rbrace)^2$; then $U_{x,r}[x]=B(x,r)$. -The uniformity does not have a countable base: if $\langle U_n:n\in\mathbb{N}\rangle$ is a sequence of neighbourhoods of the diagonal then you can use diagonalisation to produce a neighbourhood such that $U_n\not\subseteq U$ for all $n$: make sure that the square $[n,n+1]^2$ contains a point in $U_n$ but not in $U$.<|endoftext|> -TITLE: Is the spectrum of closed geodesics in a closed hyperbolic 3-manifold asymptotically homogeneously dense? -QUESTION [9 upvotes]: It seems to me that the results in this important paper of Kahn-Markovich imply the following fact. Let $M^3$ be any closed hyperbolic 3-manifold. For every $\epsilon > 0$ there is a natural number $R(\epsilon)$ such that for every $R>R(\epsilon)>0$ there is a closed geodesic whose lenght is between $R-\epsilon$ and $R+\epsilon$. -That is, the more $R$ grows, the more the spectrum of all geodesics having length more than $R$ becomes uniformly crowded, without holes. Am I right that this is a consequence of their results? If so, is there a more direct way to prove this? Does this property generalize to closed hyperbolic manifolds with arbitrary dimension $n$? - -REPLY [14 votes]: I think this should just follow from the exponential mixing of the geodesic flow (due to Pollicott). -Exponential mixing says that there is a constant $q$ such that if you have two smooth functions $f$ and $h$ on the unit tangent bundle, and $g_t$ is the geodesic flow, then there is a constant $C$ depending on $f$ and $g$ (some function of some Sobolev norm) such that -$\Big| \int_{T^1M} (g_t^*f)h - (\int_{T^1M} f)(\int_{T^1M}h ) \Big | \leq C e^{-qt}$ -If you take a vector $v$ and then let $f = h$ be a function with integral $1$ supported on an $\epsilon$-neighborhood of $v$, then you find that there is some constant $T(\epsilon)$ depending only on $M$ and $\epsilon$ such that for any $T \geq T(\epsilon)$, there is a closed path whose length is within $\epsilon$ of $T$ and which is a geodesic except at the basepoint, where it is broken at an angle of $\pi - \epsilon$. If $\epsilon$ is small, this path will be close to a bona fide geodesic, which should give you what you want. -I played fast and loose there with the constants, but that's the idea, I think.<|endoftext|> -TITLE: Explicit isomorphism between distributions and universal enveloping algebra -QUESTION [7 upvotes]: The universal enveloping algebra of a Lie algebra $\mathfrak{g}$ is isomorphic to the algebra of distributions on the Lie group $G$ with support at the identity. The proof I have of this fact uses the universal property of the universal enveloping algebra and hence it is not constructive. I was wondering if there is an explicit map? For example what would happen for $\mathfrak{g} = \mathfrak{sl}(2, \mathbf{C})$ and $G = SL(2, \mathbf{C})$? For an explicit monomial in $\mathscr{U} \ \left ( \mathfrak{sl}(2, \mathbf{C}) \right )$ can we write the corresponding distribution and how it acts on functions on $SL(2, \mathbf{C})$? - -REPLY [3 votes]: It seems to me the question is not fully correct because there is NO canonical identification of U(g) and distribitions on G (sup. at e). -In Victor's answer you can take action LEFT invariant vec. fields and you can take RIGHT - the answer would be different. -It seems if you take "left" than U(g) will act as left invariant differential operator, -if you take "right" it will act as "right" invariant differential operator. -It seems to me that it is general fact which should be checked on R^n - take any differential operator D, consider distribution d= D\delta, then action of this distribution by convolution is exactly the same action of D. -The proof seems obvious - just integration by parts -I have not checked all this, hopefully it is nevertheless true :)<|endoftext|> -TITLE: Shortest morphing between shapes embedded in $\mathbb{R}^3$ -QUESTION [11 upvotes]: I am interested in what in computer graphics is called -morphing between two topologically equivalent shapes $S_0$ -and $S_1$ in 3D. -This is a continuous "path" of shapes $S_t$, each embedded and -all with the same genus, for $t \in [0,1]$. -Let us assume the shapes are closed surfaces; -I am particularly interested in polyhedra, but likely could adapt from -a method for smooth surfaces. -For example, I would like to morph between these -two genus-7 polyhedra: - - -$V{=}30, E{=}84, F{=}42$ - - - -It would make a nice movie to show that the first is truly genus 7. -There is an extensive literature on this topic, as it -is needed in many graphics contexts. -A sample of some work in this area is provided in the references -below. -As far as I know, all have some ad hoc heuristics to obtain "nice" -morphs. -What I am seeking to learn here is whether there might be some -attractive embedding theorems that could lead to a clean, -perhaps more principled morph. -Here is what I have in mind, a simple algorithm for -convex shapes. Embed $S_0$ in the 3-flat of -$\mathbb{R}^4$ with 4-th coordinate $x_4=0$, -and embed $S_1$ in the 3-flat with $x_4=1$. -Let $H$ be the convex hull of $S_0 \cup S_1$ in $\mathbb{R}^4$. -Now intersect $H$ with $x_4 = t$ to obtain $S_t$. -Of course this only works for convex shapes. -Here, finally, is my question: - -Is there some mapping that would send $S_0$ and $S_1$ into - some space, and a definition of a canonical parametrized path between - those shapes as endpoints, so that the intermediate shapes $S_t$ - along the path (a) are all embedded in $\mathbb{R}^3$, - (b) all have the same genus? - -It is likely too much to hope for, but -if there were a mapping/space combination that would map -genus-$g$ shapes naturally to convex genus-0 shapes, -then the above convex algorithm would apply. -Or if there were a construct akin to the convex hull that accommodated -nonconvexity and holes... -I know my question is vague, but I hope its intent is clear. -Any ideas, however speculative or tentative, would be appreciated. -Thanks! -References. -[GSLML98] -Gregory, State, Lin, Manocha, Livingston, -"Feature-based surface decomposition for correspondence and morphing between polyhedra," -1998. -link -[ACL00] -Alexa, Cohen-Or, Levin, -"As-rigid-as-possible shape interpolation," -2000. -link -[KSK00] -Kanai, Suzuki, Kimura, -"Metamorphosis of Arbitrary Triangular Meshes," -2000. -link - -REPLY [2 votes]: Joseph, -I believe that there is not a simple linear smooth mapping. My short explanation for it is that the morph from $S_0$ to $S_1$ does not really consist of a single continuous deformation, but instead of the concatenation of a sequence of multiple piece-wise linear deformations consisting of more than three separate steps: $S_0 \to S_a \to S_b \to S_c... \to S_1$ -Your $S_0$ is the frame of an octahedron. - -I would say that $S_a$ should be the skeleton of your $S_0$, formed by shrinking all of the polygonal faces into a simpler tubular (or line segment) skeleton. -Then, $S_b$ should be the transformation of the octahedral skeleton in $\mathbb{R}^3$ into the equivalent planar graph in $\mathbb{R}^2$. (this $S_a \to S_b$ step can be done as a linear interpolation morph) -then, $S_c$ should the the fattening of the skeleton of $S_{b}$ and its transformation into a disk with the seven holes punched in it corresponding to the seven holes in the planar graph representation $S_b$ of the octahedral skeleton $S_a$. -then, $S_d$ should be the elongation of the disk into an elliptical shape as the seven holes are migrated from their positions in the planar graph $S_b$ into a linear configuration along the long axis of this elliptical elongated disk. -then, $S_e$ can be the transformation of this disc with seven horizontally space holes into a skeleton graph of your $S_1$ -then, $S_f$ can be the tranformation of the $S_e$ skeleton graph into the polyhedral representation of $S_1$ with polygonal faces as you have drawn. - -Note that I have broken up the three steps into a few extra substeps to ease in the understanding of the visualization. Also note that the in toto transformation animation consists of a concatenation of multiple piece-wise linear sub-transformations. -I think that is the best way to visualize or animate it. -Attempting to create a single flowing transformation will falsely blur together the distinction of some of the topological procedures, in my opinion. -A similar problem arises in graphical animation morphing: what is the transform of a square $ABCD$ onto the same square mapping $A \to B, B \to C, C \to D$, and $D \to A$? - -Is it the rotation of the 2-d space around the center of the square? (which preserves the shape and the size of the figure) -Is it the rotation of the 2-d space around the point "A" followed by translation of the resulting rotated square? (which preserves shape and size, but is "clunky" and may be perceived as the transform requiring two distinct steps) or any other combination of rotations and translations -Is it the linear "morph" of the coordinate points A to B, B to C, C to D, and D to A, which would result in a square rotated 45 degrees at $t=0.5$ but consisting of half of the area of the square at $t=0.0$ (ABCD) and the square at $t=1.0$ (the square BCDA) --- preserving the shape of the figure, but changing the size/scale of the square over the time of the morph - -A morph in graphics is not always just the "tweening" (in between interpolation of the coordinates) of corresponding points between the starting shape $S_0$ and the ending shape $S_1$. Sometimes, some tweaking modifications such as rotation, bending, flowing, etc., are required in order to produce a visually satisfying or acceptable and believable transformation. (Of course, the concept of visually satisfying and acceptable are very subjective, contextual, and in the eye of the beholder, as evidenced by the nonobviousness of some of the examples in the question "Proofs without Words"). -In the example you are trying to create, only the $S_a \to S_b$ step can be done as a linear interpolation morph. The rest of the steps require a bit of cleaning up and visual "eye candy" in order to be appreciable and understandable. -I might try playing around with some animation tools to play with this, but I don't think that the transformation of the octahedral skeleton to the overtly obvious genus-7 shape can be done by the type of technique which you're asking for.<|endoftext|> -TITLE: Given a C-star dynamical system and a subgroup of the acting group, is the corresponding map on crossed product algebras necessarily an injection -QUESTION [9 upvotes]: Let $(A,\alpha, G)$ be a $C^*$-dynamical system, where $G$ is a discrete group. Let $\Gamma$ be a subgroup of $G$, then we can form two universal crossed products $A\rtimes_\alpha \Gamma$ and $A\rtimes_\alpha G$. -Question 1: Is the canonical map $A\rtimes_\alpha \Gamma \to A\rtimes_\alpha G$ injective? -Question 2: What about reduced crossed products $A\rtimes_{\alpha,r} \Gamma \to A\rtimes_{\alpha,r} G$? -For the universal case, I guess it's wrong even though I can not find a counterexample. But if we let $\alpha$ be the trivial action of $G$, then we only need to look at $A\otimes_{max} C^* (\Gamma) \subseteq A\otimes_{max} C^* (G) $. -For the reduced case, I guess it's should be the case. Just follows from the facts that the left regular representaion of $G$, restricted to $\Gamma$, is a multiple of the left regular representation of $\Gamma$ and $A\rtimes_{\alpha,r} \Gamma = C^*(\pi(A), 1\otimes \lambda(G))$, where $\pi: A\subseteq B(H) \rightarrow B(H\otimes l^2(G))$ and $\lambda$ is the left regular representaion of $G$. Do I make any mistake? - -REPLY [6 votes]: The universal case is also injective. What one needs to show is that any covariant representation of $(A,\Gamma)$ extends to a covariant representation of $(A,G)$. This can be shown by using induced representation.<|endoftext|> -TITLE: How should I think about delooping? -QUESTION [13 upvotes]: When talking about the Eilenberg-Maclane space $K(G,n)$, we usually restrict our attention to the situation where $G$ is abelian. In that case, we get $\Omega K(G,n)=K(G,n-1)$, so we can call $K(G,n)$ a delooping of $K(G,n-1)$. -Since $\pi_n$ is always abelian for $n>1$, it only makes sense to talk about $K(G,1)=BG$ for $G$ nonabelian anyways. So there definitely shouldn't be delooping of this space, because then it would have $\pi_2=G$, which is impossible. From the previous paragraph, it seems like we should therefore be able to say that the nonabelianity of $G$ (i.e., the nontriviality of the commutator $[G,G]$) is the obstruction to delooping $BG$. But this isn't very satisfying, because I can't quite see what's going on with the actual space. -All of which motivates my (slightly open-ended/up-to-interpretation) question: -How should I think about delooping? Is it nothing more than thing like "for the space $X$ that we care about, it just so happens that we've got $Y$ with $\Omega Y\simeq X$", or is there a definite way to measure obstructions? In the cases where a delooping exists, is there an explicit method for its construction? - -REPLY [15 votes]: I'm not sure whether you'll like this, but my natural response to "how should I think about delooping?" is to invoke (higher) category theory. You may know that a homotopy 1-type, i.e. a space (probably a CW complex) with $\pi_n=0$ for n>1, is uniquely specified up to (weak) homotopy equivalence by its fundamental groupoid. In fact, one has an equivalence of (2- or homotopy-) categories, so we can identify homotopy 1-types with groupoids. Under this identification, the space BG which deloops a discrete group G is identified with the groupoid with one object and G as the automorphism group of that object. So one-step delooping of a discrete group really is just the simple process of considering a group as a one-object groupoid (although in the homotopy theory world it requires a fairly elaborate construction). -At higher levels, the "homotopy hypothesis" in higher category theory (which is a theorem for some definitions of higher category) says that homotopy n-types can be identified with n-groupoids, and arbitrary homotopy types with ∞-groupoids. Moreover, the identification of groups with one-object groupoids is believed to continue to higher categories as well: 2-groups (i.e. groupoids equipped with an extra group structure up to coherent isomorphism) can be identified with one-object 2-groupoids, and similarly an n-group can be identified with a one-object n-groupoid. Thus, deloopability of a space requires that it be equipped with a suitable group structure (up to homotopy, i.e. up to equivalence), and in that case its delooping corresponds on the categorical side to regarding an n-group as a one-object n-groupoid (where possibly n=∞). -Finally, as to the obstruction to delooping BG when G is not abelian, only when G is abelian is BG itself a (2-)group. The reason is the same one that other people have mentioned—Eckmann-Hilton—but I prefer to think about it in these terms.<|endoftext|> -TITLE: How to show a set of polynomials is algebraically independent? -QUESTION [17 upvotes]: Suppose that I have $n$ homogeneous polynomials $f_1, \dots, f_n \in \mathbb{C}[x_1, \dots, x_m]$ and that $n < m$. Is there a well known method or algorithm to determine if these polynomials are algebraically independent? -As far as I know the Jacobian criterion works only for the case where $n=m$. - -REPLY [5 votes]: Hi there. A Groebner basis based algorithm that also produces an annihilating polynomial in case the polynomials are algebraically dependent can be found on the Singular site at -http://www.singular.uni-kl.de/Manual/3-0-2/sing_534.htm .<|endoftext|> -TITLE: How to characterize a Self-avoiding path. -QUESTION [10 upvotes]: I cannot find any answer to that apparently simple problem : -On a square lattice, a path is given by a sequence of relative moves in {"move forward", "turn right" and "turn left"}. -Is there a rule that characterizes if a path is self-avoiding (or not) ? -Edit : Let me precise the kind of rule I am looking for: -Let's imagine a walk described, this time, by absolute moves (N=move North, S=move South, E=move East, W=move West), then the presence of a loop in the sequence is characterized by a subsequence for which nb(N) = nb(S) and nb(E)=nb(W). That's a simple rule. -Is there such a rule in the case of a sequence of relative moves ? Or do we have to translate the sequence to absolute moves ? Thanks. -Example (to make myself clear): -here is a walk (or part of a walk), written in absolute moves {North, East, West, South}: -EENWNNWSSS -=> We immediately know it is a loop, without having to draw anything or keep track of the positions visited, because nb(N)=nb(S) and nb(E)=nb(W). -Now here is the same walk written in relative moves {Forward, Turn right, Turn left}: -FFLFLFRFFLFLFFF -=> Without drawing anything, nor converting to absolute moves. Is there a rule that allows to say it is a loop ? - -REPLY [12 votes]: I'm only now starting to understand the question. Let me formulate it slightly differently, but I think it's easy to translate between this and your formulation. I'll imagine a little machine with three possible operations: take a step (which is always in the direction of an arrow on the machine's head, so would be your "forwards"), turn to the left (which changes the direction of the arrow but doesn't change the square the machine is on), and turn to the right. -If you're prepared to characterize self-avoidance with a statement like "There is no subsequence such that ...," as you seem to be, then it is enough to characterize loops. This is easy to do if we convert into absolute moves, as you say, so the idea is not to do that. It is quite hard to say precisely what it means not to keep track of the direction the machine is pointing in, but here is an attempt. -Now there are various ways that a sequence of moves can be simplified. For instance, RRR=L, LLL=R, and RL=LR="do nothing", so we can always rewrite so that there are at most two turns in a row, when they have to be the same. Also, FRRF can be simplified to RR, so if we ever have RR or LL we can cancel Fs on either side until we have three turns in a row and then cancel those. -After all this cancellation, we may assume that we never have two turns in a row. But that is not all the cancellation that can be done. For instance, $FRF^kRF$ cancels to $RF^kR$. So if we ever have $RF^kR$ we can cancel Fs on both sides until we get two turns in a row and go back to the previous method for getting rid of that. -So now we may assume that the sequence consists of Fs with isolated Rs and Ls that have to alternate. That won't be a loop unless the sequence is null. -I think this can be done without thinking about what the absolute direction is, but one could perhaps argue that the cancellation rules are working mod 4 and keeping track of the direction in an implicit way. But I'm not sure that's right, since we can do the cancellation in the middle of the sequence without caring about the absolute direction at all. So perhaps this counts as a sort of answer to your question.<|endoftext|> -TITLE: Lengths of continued fractions for the numbers with fixed ratio -QUESTION [11 upvotes]: Let $s(x)$ is the length of continued fraction expansion of $x$, and let $l(x)$ be the sum of partial quotients. I can prove that for any rational $\alpha$ ratios $\frac{s(\alpha x)}{s(x)}$ and $\frac{l(\alpha x)}{l(x)}$ (for all rational $x$) are bounded with some constants depending on $\alpha$ only. -Is this result new? - -REPLY [7 votes]: It definitely is not new for the length, and I am nearly sure that is not for height either. -See, for example, -Labhalla, Salah; Lombardi, Henri -Transformation homographique appliqu´ee `a un d´eveloppement en fraction continue fini ou -infini. (French) [Fractional linear transformations applied to finite and infinite continued -fractions] -Acta Arith. 73 (1995), no. 1, 29–41.<|endoftext|> -TITLE: Complex manifolds where bounded holomorphic functions are constant -QUESTION [9 upvotes]: Liouville's theorem states that all bounded holomorphic functions on $\mathbb{C}^n$ are constant. -I'm wondering which connected complex manifolds have this property ? -Connected compact complex manifolds have it since all holomorphic functions there are constant. -There are no simply connected proper open subsets of $\mathbb{C}$ satisfying this because of the Riemann Mapping Theorem. But are there some other open subsets satisfying it ? -In higher dimension there are some, such as the Fatou–Bieberbach domains (open subsets of $\mathbb{C}^n$ biholomorphic to it). -I would be interested in references on this property on complex manifolds. -Incidentally, is it true that any complex manifold where all holomorphic functions are bounded is compact ? - -REPLY [2 votes]: The reference for complex manifolds in dimention 1 (Riemann surfaces) is the -subject known under the title "Classification theory of (open) Riemann surfaces". -It used to be very popular in the 1950-s. Some books are: -M. Tsuji, Potential theory in modern function theory, -L. Ahlfors and Sario, Riemann surfaces. -Look also at the nice paper by P. Doyle, On deciding whether a surface is parabolic or -hyperbolic.<|endoftext|> -TITLE: Teaching proofs in the era of Google -QUESTION [61 upvotes]: Dear members, -Way back in the stone age when I was an undergraduate (the mid 90's), the internet was a germinal thing and that consisted of not much more than e-mail, ftp and the unix "talk" command (as far as I can remember). HTML and web-pages were still germinal. Google wouldn't have had anything to search, had it existed. Nowadays Google is an incredibly convenient way of finding almost anything -- not just solutions to mathematics problems, but even friends you lost track of 20+ years ago. -My question concerns how Google (and to a lesser extent other technological advances) has changed the landscape for you. Specifically, when you're teaching proofs. More details on what I'm getting at: -A "rite of passage" homework problem in the 2nd year multi-variable calc/analysis course at the University Alberta was the Cantor-Schroeder-Bernstein theorem. In the 3rd year there was the Kuratowski closure/14-set theorem. It's not very useful to ask students to prove such theorems on homework assignments nowadays, since the "pull" of Google is too strong. They easily find proofs of these theorems even if they're not deliberately searching for them. The reason I value these "named" traditional problems is primarily that they are fairly significant problems where a student, after they've completed the problem, can look back and know they've proven (on their own) some kind structural theorem - they know they're not just proving meaningless little lemmas, as the theorems have historical significance. As these kinds of accomplishments accumulate, students observe they've learned to some extent how an area develops and what it takes in terms of contributions of new ideas, dogged deduction, and so on. -I'm curious to what extent you've adapted to this new dynamic. I have certainly noticed students being able to look-up not just named theorems but also relatively simple, arbitrary problems. After all, even if you create a problem that you think is novel, it's rather unlikely that this is the case - sometimes students find your problem on a 3-year-old homework assignment on a course webpage half-way around the planet, even if it's new to you. -As Jim Conant mentioned in the comments, this is a relatively new thing. When I was an undergraduate, going to the library meant a 30-minute walk each way, then the decision process of trying to figure out what textbook to look in, frequently a long search that led me to learning something interesting that I hadn't planned on, and frequently not finding what I set out to find. But type in part of your problem into Google and it brings you to the exact line of all the textbooks in which it appears. It brings up all the home-pages where the problem appears and frequently solutions keys, if not Wikipedia pages on the problem -- I've deleted more than one Wikipedia page devoted to solutions to particular homework problems. -Of course there are direct ways to adapt: asking relatively obscure questions. And there's "denying the problem" - the idea that good students won't (deliberately or accidentally) look up solutions. IMO this underestimates how easy it is to find solutions nowadays. And it underestimates how diligent students have to be in order to succeed in mathematics. -Any insights welcome. - -REPLY [18 votes]: In most Italian universities, grades are not based on homework but on a final (written and/or oral) text. There are different levels of what you are allowed to use during the written text (textbooks, notes, a limited amount of notes, or plain nothing), but definitely anything that can connect to Google is not authorized. -This would solve the problem completely, wouldn't it? -In fact, there is no (mandatory) homework at all: students may skip all the lectures, study on their own, and arrive prepared at the final test. -When I first learned how heavily universities abroad rely on homework problems (and how many TA hours are devoted to grade them), I was shocked. To my eyes, this method looked like high school, baby-sitting students through the coursework. -Granted, with the Italian system, some students get lost around the way, like in "drink and play World of Warcraft the whole day and then fail miserably". But those who don't, they learn how to organize themselves and work autonomously towards a goal.<|endoftext|> -TITLE: Recurrence relations whose base case is 'at infinity' -QUESTION [14 upvotes]: I ran across this recurrence relation in -a paper by Medina and Zeilberger [MZ] -(who got it from [CR]): -$$f(h,t) = \max \left( \frac{1}{2} f(h+1,t) + \frac{1}{2} f(h,t+1) ,\frac{h}{h+t} \right) \;.$$ -The "base" condition of the recurrence is that, -for $h+t = \infty$, $f(h,t)=\frac{1}{2}$. -This function $f$ represents the expected gain in a paricular -coin game ($h$ and $t$ are heads and tails), -explained in this MSE posting. -I had not before encountered recurrence relations whose -"initial conditions" are "at infinity," and was surprised to learn -that there is no known explicit solution for $f$. -(However, one can compute particular values -numerically by limiting to $n$ trials and letting $n \rightarrow \infty$. For example, -$f(5,3) =\max ( 0.62361957757, 5/8 )$. -See [W].) -My question is: - -Is there a class of recurrence relations that includes the above - example, and for which some theory has been developed for solving - such equations? - -Thanks for pointers and references! -References -[MZ] -Luis A. Medina, Doron Zeilberger, -"An Experimental Mathematics Perspective on the Old, and still Open, Question of When To Stop?" -arXiv:0907.0032v2 [math.PR] -[CR] Y.S. Chow and Herbert Robbins. On optimal stopping rule for $S_n/n$. -Ill. J. Math., 9:444–454, 1965. -[W] -Julian D.A. Wiseman -web page. - -REPLY [6 votes]: For this specific recurrence, Olle Häggström and I have some results in the paper -arXiv:1201.0626 [math.PR] that we just posted. -The key lemma is based on Olle's trick that I also mentioned in an answer to -this question. -Briefly, the argument is this (see our abstract or the link in the OP for a description of the game): If at some point in the game we condition on the event of ever reaching a proportion of (at least) $p$ of heads, then as long as we haven't, the conditional probability of heads in the next toss is at least $p$. Now, one way of showing that an event is unlikely is to show that strange things happen if we condition on it. If the conditional probability of a long run of heads is very different from the unconditional one, then the probability of ever reaching proportion $p$ of heads must be small. -After pursuing the calculations, our main result (in the notation of the OP) is -$$f(h,t) \leq \max\left(\frac{h}{h+t}, \frac12\right) + \min\left(\frac14\sqrt{\frac{\pi}{h+t}}, \frac1{2\cdot\left|h-t\right|}\right).$$ -This allows us to compute $f(h,t)$ to any desired precision, and to verify rigorously that $f(h,t) = h/(h+t)$ in a number of cases. For instance, $f(5,3) = 5/8$, which means that in the Chow-Robbins coin flipping game, stopping is optimal with 5 heads and 3 tails.<|endoftext|> -TITLE: Transfinite induction, a theorem of Pedersen, and chains of subalgebras of $B(H)$ -QUESTION [11 upvotes]: This post is closely related to this one. (In fact I copied some of its content.) -Let $H$ be an infinite dimensional separable complex Hilbert space. All $C^{\star}$-subalgebras of $B(H)$ are assumed to be non-degenerate. -The spectral projections of a self-adjoint element $T$ of $B(H)$ lie in the weakly closed algebra generated by $T$. In the early 1970s Pedersen proved that if a $C^{\star}$-subalgebra $A$ of $B(H)$ contains all of the spectral projections of each of its self-adjoint elements, then $A$ is weakly closed, i.e. $A=A''$. Now, concerning the consequences of this result, Pedersen says in his book: -"For any $C^\star$-subalgebra $A$ of $B(H)$ define $a(A)$ as the smallest $C^{\star}$-subalgebra of $B(H)$ containing all spectral projections of each self-adjoint element in $A$. [...] Note that [...] a transfinite (but countable) application of the operation $a$ will produce $A''$." - -Question 1: What is the reasoning here? - -I see that $\omega_1$ applications of the operation $a$ produce $A''$ and also that each element of $A''$ appears at the $\alpha$-th application for some $\alpha < \omega_1$ (This is due to the fact that the closure in the norm-topology is a sequential closure and hence, only countably many elements play a role.); but why is $a^{\alpha}(A)=A''$ for some $\alpha < \omega_1$? -Maybe there is something deeper behind: - -Question 2: Is there an $\omega_1$-chain of unital subalgebras of $B(H)$? - -Here $\omega_1$-chain means an $\omega_1$-index family $(A_{\alpha})_{\alpha< \omega_1}$ of subalgebras of $B(H)$ such that for all $\beta < \omega_1$ we have -$$\overline{\cup_{\alpha< \beta}A_{\alpha}}^{\|.\|} \subsetneq A_{\beta}.$$ - -REPLY [7 votes]: The answer to Question 2 is yes; such a chain can be found using only diagonal operators with respect to some fixed basis $(e_n)_{n\in{\mathbb N}}$ for $H$. As a starting point, you can find an $\omega_1$-sequence of subsets $S_\alpha\subset{\mathbb N}$ such that if $\alpha<\beta$, $S_\beta\setminus S_\alpha$ is finite and $S_\alpha\setminus S_\beta$ is infinite (that is, the sequence $(S_\alpha)$ is strictly decreasing "modulo finite sets"). This is easy to do because by diagonalizing, given a countable decreasing sequence of infinite subsets of ${\mathbb N}$, you can always find another infinite set which is strictly contained in all of them modulo finite sets. Now let $A_\alpha$ denote the algebra of diagonal operators such that the eigenvalues of the eigenvectors $e_n$ for $n\in S_\alpha$ form a convergent sequence. These are closed because a uniform limit of convergent sequences is convergent. These are nested because if $\alpha<\beta$, then all but finitely many elements of $S_\beta$ are contained in $S_\alpha$ so if the $S_\alpha$-eigenvalues converge, so do the $S_\beta$-eigenvalues. -Here's a more conceptual explanation of this construction. The algebra of diagonal operators is naturally isomorphic to the algebra $C_b({\mathbb N})$ of bounded continuous functions on $\mathbb N$, which is in turn naturally isomorphic to the algebra $C(\beta{\mathbb N})$ of all continuous functions on the Stone-Cech compactification of $\mathbb N$. Each set $S_\alpha\subset{\mathbb N}$ has a closure $\overline{S_\alpha}\subset\beta{\mathbb N}$, and the statement that the $S_\alpha$ are decreasing modulo finite sets says exactly that the sequence of closed sets $C_\alpha=\overline{S_\alpha}\setminus S_\alpha$ is actually decreasing. The subalgebra $A_\alpha$ is then exactly the set of continuous functions on $\beta{\mathbb N}$ which are constant on the set $C_\alpha$.<|endoftext|> -TITLE: Monotone injection of an ordinal into $[0,1]$ -QUESTION [5 upvotes]: This is related to my recent question and would provide a natural positive answer to Question 2. I am sure this must be known to experts. - -Question: Is there a monotone injection $(\omega_1,<) \to ([0,1],<)$ ? - -REPLY [6 votes]: There is a far reaching generalization of this due to Friedman and Shelah. Suppose $X$ is a Borel set in a Polish space and $<$ is a linear order of $X$ that is a Borel subset of $X\times X$. Then there is no order preserving map from $\omega_1$ into $(X,<)$. -The Friedman-Shelah result follows from a later structure theorem of Harrington and Shelah -who proved that for any such Borel linear order $(X,<)$ there is a Borel measurable order preserving -map into ${\bf R}^\alpha$ for some countable ordinal $\alpha$ where ${\bf R}^\alpha$ is ordered lexicographically. The arguments for $[0,1]$ above can be generalized to show -that ${\bf R}^\alpha$ has no $\omega_1$-chains.<|endoftext|> -TITLE: What is the expected maximum out of a sample (size N) from a geometric distribution? -QUESTION [7 upvotes]: Lets say I have a geometric distribution (of the number X of Bernoulli trials needed to get a success) with parameter p (success probability of a trial). -Assume I randomly sample n elements from this distribution. -My problem is: what is the expected maximum element of such a sample (it should depend on n I guess)? Hopefully it makes sense... -For n=1, e.g. if I only pick a single element from the distribution, the answer would be the mean 1 / p of the distribution. For samples of larger cardinalities? - -REPLY [7 votes]: Bennett Eisenberg's paper "On the expectation of the maximum of IID geometric random variables" (Statistics and Probability Letters 78 (2008) 135-143) gives both my answer already accepted by the OP and mentions the infinite sum you get if you take Michael Lugo's comment on my first answer. However, the author also says, "these expressions are not so useful" and "There is no... simple expression for... the expected value of the maximum of $n$ IID geometric random variables." The point of his paper is "to use simple Fourier analysis to show that $E(M_n^*) - \sum_{k=1}^n \frac{1}{\lambda k}$ is very close to 1/2 not only for moderate values of $\lambda$, but also for relatively small values of $n$ and that this difference is logarithmically summable to 1/2 for all values of $\lambda$." Here, $E(M_n^*)$ is the expectation requested by the OP, and $\lambda$ is defined by $q = 1-p = e^{-\lambda}$. (The $\lambda$ is the parameter for the corresponding exponential distribution.) The practical value of Eisenberg's result, of course, is the fact that $\sum_{k=1}^n \frac{1}{k}$ can be closely approximated by $\log n + \gamma$.<|endoftext|> -TITLE: Have all numbers with "sufficiently many zeros" been proven transcendental? -QUESTION [21 upvotes]: Any number less than 1 can be expressed in base g as $\sum _{k=1}^\infty {\frac {D_k}{g^k}}$, where $D_k$ is the value of the $k^{th}$ digit. If we were interested in only the non-zero digits of this number, we could equivilantly express it as $\sum _{k=1}^\infty {\frac {C_k}{g^{Z(k)}}}$, where $Z(k)$ is the position of the $k^{th}$ non-zero digit base $g$ and $C_k$ is the value of that digit (i.e. $C_k = D_{Z(k)}$). -Now, consider all the numbers of this form $(\sum _{k=1}^\infty {\frac {C_k}{g^{Z(k)}}})$ where the function $Z(k)$ eventually dominates any polynomial. Is there a proof that any number of this form is transcendental? -So far, I have found a paper demostrating this result for the case $g=2$; it can be found here: http://www.escholarship.org/uc/item/44t5s388?display=all - -REPLY [19 votes]: I must apologize and correct what I wrote in an earlier paper: the method does extend to base > 2. Using Liouville's inequality, you lose a bit in the constant c, but the great advantage is that everything can be made fully explicit.<|endoftext|> -TITLE: $\omega$-triviality of knots? -QUESTION [8 upvotes]: From the theory of finite type invariants of knots comes the concept of $n$-triviality. A knot is said to be $n$-trivial if there is some projection of the knot and $n$ pairwise-disjoint sets of crossings $S_1,S_2,\ldots,S_n$, such that changing the crossings in every nontrivial subset of $S=\{S_1,\ldots, S_n\}$ turns the knot into the unknot. -An easy example is the trefoil knot with a three-crossing projection. Let $S_1$ and $S_2$ each contain just one crossing from this projection. Than changing $S_1$, $S_2$ or $S_1\cup S_2$ yields the unknot. -This notion of $n$-triviality is quite interesting because Goussarov first proved that a knot is $n$-trivial if and only if all finite type invariants vanish up to degree $n-1$. My question is about generalizing this notion to $\omega$-triviality. For this purpose, it is probably better to regard the crossing changes in $S_i$ as homotopies supported in neighborhoods of arcs connecting the knot to itself (finger moves.) Then one can define a knot to be $\omega$-trivial if there is a pairwise disjoint collection $\{S_1,S_2,....\}$ where each $S_i$ is a set of finger moves on $K$, such that doing any nonempty subcollection $\{S_i: i\in J\}$ for $\emptyset\neq J\subset \mathbb N$, of the finger moves gives you the unknot. -Question: If a knot is $\omega$-trivial, must it be the unknot? -A knot which is $\omega$-trivial would have vanishing finite type invariants of all degrees, so it shouldn't exist, but this question should be a lot easier than the question of whether finite type invariants detect knottedness. Does anybody have any ideas on this question? - -REPLY [5 votes]: I would guess that you would need to understand the transfinite lower central series of the mapping class group. A knot complement which could not be distinguished from a solid torus by any finite lower central series quotient of the mapping class group Torelli group of a Heegaard surface would be n-trivial for all n, which if I'm not mistaken follows from work of [a subset of] Ted Stanford, Habiro, and Garoufalidis-Goussarov-Polyak. I don't see offhand why such an example could not exist. -Therefore, I think this question is open, and I can't see why it would be easy.<|endoftext|> -TITLE: Exponential sums and differential equations -QUESTION [5 upvotes]: Hi, I have a general question about the relationship between exponential sums and differential equations. In particular, I have been trying to read Katz' work on the subject (his book and his lecture notes) but I am having trouble understanding the big idea and getting confused with all of the algebraic geometry background. Can someone explain the general picture, or direct me towards more elementary works introducing the subject? - -REPLY [2 votes]: In simplistic terms, exponential sums arise in characteristic p geometry because curves have Artin-Schreier coverings. The appearance of (complete) exponential sums in that context is a reflection in cohomology of that kind of Galois covering, and can be understood pretty much in Weil's terms and technology. (Appendix in Basic Number Theory? Unfortunately I have the first edition that doesn't have such conveniences.) And linear differential equations on curves have been understood geometrically since Riemann's time, admittedly in varying languages. The case corresponding to finite coverings (i.e. curve morphisms) is that of finite monodromy. -So far so good? Nick Katz's papers have got easier to read as the years go by (at least 10% per decade I think, as he has moved away from the Grothendieckising style). But are probably still hard going.<|endoftext|> -TITLE: What is π_1(BG) for an arbitrary topological group $G$? -QUESTION [16 upvotes]: The classifying space $BG=|Nerve(G)|$ of an arbitrary topological group $G$ does not necessarily have the homotopy type of a CW-complex but the fundamental group should still be accessible. What is $\pi_{1}(BG)$? A reference on this would be great. My initial guess: $\pi_{1}(BG)$ is the quotient group $\pi_{0}(G)$ for arbitrary $G$ -Motivation: There is a natural way to make $\pi_1$ a functor to topological groups. I am interested in relating the topologies of $G$ and $\pi_{1}(BG)$ but the topology on $\pi_{1}(X)$ is boring (discrete) when $X$ is a CW-complex. - -REPLY [2 votes]: Readers might also like to know a result on the first k-invariant of BG contained in -R. Brown and C.B. Spencer, $\cal G$-groupoids, crossed modules -and the fundamental groupoid of a topological group'', Proc. -Kon. Ned. Akad. v. Wet. 7 (1976) 296-302. -This shows that the associated crossed module to fundamental groupoid of $G$ has $k$-invariant which is exactly the $k$-invariant of $BG$. -Since this query is about non-connected topological groups, another related ressult is on universal covers of non-connected topological groups -R. Brown and O. Mucuk, ``Covering groups of non-connected -topological groups revisited'', Math. Proc. Camb. Phil. -Soc, 115 (1994) 97-110. -which relates the question of the existence of topological group universal covers of $G$ to the theory of ostructions to extensions of abstract groups.<|endoftext|> -TITLE: Intersection of subvarieties versus ranks of Chow groups modulo numerical equivalences -QUESTION [13 upvotes]: A nice property of $\mathbb P^n$ is: - -Property 1: Two subvarieties $U,V$ such that $\operatorname{dim} U +\operatorname{dim} V \geq n$ always intersect. - -(for example, any 2 curves in $\mathbb P^2$ intersect) -There are other smooth varieties $X$ when Properties 1 holds. For example, a sufficient condition is that the ranks of $\text{CH}^i_{num}(X)$ are $1$ for $i\leq n/2$. Here $n = \operatorname{dim} X$ and $\text{CH}^i_{num}(X)$ is the Chow group of codimension $i$ modulo numerical equivalences. -My question is whether some converse is true: - -Question: Let $X$ be a smooth projective variety satisfying Property 1. Does that impose some upper bounds on the ranks of $\text{CH}^i_{num}(X)$ for $i\leq n/2$? - -Let's assume we are over $\mathbb C$, but I am also interested in results over any ground fields. -One can ask the same questions for the ranks of $\text{CH}^i_{hom}(X)$ (I think they are conjectured to be the same). The baby case is $i=1$, where the question asks if Property 1 tells us something about the rank of the Neron-Severi group of $X$. -I am aware that the question is a little vague (upper bound as function of what?), but that was because of my ignorance, so comments to improve the question are welcome. - -REPLY [7 votes]: This is a very interesting question and I guess that a general answer is unknown, already in the case $i=1$. Let me just make the following -Remark. There exists no upper bound on $\textrm{rank } NS(X)$ which is independent on the dimension. -In fact, let us consider a complex Abelian variety $X$ of dimension $g$ such that $End_{\mathbb{Q}}(X)$ is a totally real number field of degree $g$ over $\mathbb{Q}$. These -varieties do exist and the general one is simple, see [Birkenhake - Lange, Chapters 5 and 9]. -Therefore it is known that -$\rho(X) =\textrm{rank } \textrm{NS}(X) = \textrm{rank } End^s_{\mathbb{Q}}(X)=g$, -where $End^s_{\mathbb{Q}}(X)$ denotes the subgroups of elements in $End_{\mathbb{Q}}(X)$ which are symmetric with respect to the Rosati involution. -On the other hand, in a simple Abelian variety any effective divisor is ample, so two effective divisors always intersect and $X$ satisfies Property 1.<|endoftext|> -TITLE: A sheaf-theoretic version of the proj construction? -QUESTION [5 upvotes]: Recall that $\operatorname{Sch}$ can be identified with the subcategory of (Zariski-)locally representable (by an affine) étale sheaves on $\operatorname{CRing^{op}}$. In this case, $\operatorname{Spec}(-):\operatorname{CRing^{op}}\to \operatorname{Sch}$ is simply the co-Yoneda embedding $R\mapsto \operatorname{Hom}(R,-)$ (which makes sense since the étale topology is subcanonical and $\operatorname{Hom}(R,-)$ is obviously locally affine). -Is there a nice "sheaf-theoretic" description of $\operatorname{Proj}:\operatorname{(Gr_{{\mathbf Z}_{\geq 0}}CRing)^{op}}\to \operatorname{Sch}$ (I've only seen $\operatorname{Proj}$ for nonnegative integral grading. If we can use more exotic gradings, I guess I'd be interested in that too). Hoping for it to be as nice as $\operatorname{Spec}$ seems like a bit of a pipe dream, but I'm wondering if there is a nicer way to describe it than Hartshorne's approach, which feels rather arbitrary. -Edit: To clarify, I'm looking for a construction $\operatorname{Proj}:\operatorname{(Gr_{{\mathbf Z}_{\geq 0}}CRing)^{op}}\to \operatorname{Sh}_{\acute{et}}(\operatorname{CRing^{op}})$, which we can then see lands in $\operatorname{Sch}$ by showing that we can cover it with Zariski-open affines. - -REPLY [4 votes]: Try EGA chapter II section 3, especially 3.7 if you want to know the functor of points of proj.<|endoftext|> -TITLE: Regular Morphism From Affine Line -QUESTION [6 upvotes]: Hello, I was looking for an answer to the following question: -Consider an algebraically closed field $K$ and a map $K \rightarrow K^n$ given by $a \mapsto (p_1 (a) , \ldots p_n (a))$ for $p_i \in K[t]$. Is the image of this map an algebraic set? -Certainly this is false for $K^2 \rightarrow K^n$, for example $(x,y) \mapsto (x, xy)$ but I feel like polynomials in $1$ variable aren't complicated enough to give this bad behavior. -Thanks! - -REPLY [5 votes]: Dear Damien, let's show that your morphism $f: \mathbb A^1_K \to \mathbb A^n_K $ is proper, hence closed, hence certainly has closed image. -For that it is enough to prove that each $f_i:\mathbb A^1_K \to \mathbb A^1_K$ is proper. But this follows from the stronger property that $f_i$ is finite or dually that the ring morphism $K[T] \to K[T]: T\to p_i(T)$ is finite. This is elementary: it follows, for example, from the fact that $T$ is (tautologically) integral over $K[p_i (T)]$. -Note that in this proof you needn't assume that the field $K$ is algebraically closed. -Edit: As BCnrd remarks, this proof only works if all polynomials $p_i(T)$ are non-constant. Let me modify the proof to take his judicious comments into account. If all polynomials are constant, your morphism is not proper but its image is clearly closed. If at least one polynomial is non-constant, say the first, then the argument above proves that the corresponding morphism $f_1: \mathbb A^1_K \to \mathbb A^1_K $ -is finite.The obvious closed immersion $j: \mathbb A^1_K \to \mathbb A^n_K $ (last $n-1$ coordinates given by the other polynomials) is finite and the composition, which is your morphism $f=j\circ f_1: \mathbb A^1_K \to \mathbb A^n_K $ , is thus also finite.<|endoftext|> -TITLE: Is the operator norm always attained on a $\{0,1\}$-vector? -QUESTION [5 upvotes]: Given an operator $f\colon R^m\to R^n$, can one always find a non-zero vector -$x\in \{ 0,1 \}^m$ such that $\|f(x)\|/\|x\|\ge0.01\|f\|$? (Here I denote by -$\|\cdot\|$ both the Euclidean norms in $R^m$ and $R^n$ and the induced -operator norm.) The answer may well be negative -- any examples? - -In case the answer to the question above is ``no'' (or unknown), would it -help to assume that the matrix of $f$ with respect to the standard -orthonormal bases of $R^m$ and $R^n$ has all its elements equal to $0$ or -$1$? - -As I see it, this is basically a question in the geometry of numbers, and I -would expect the answer should be known. - -REPLY [9 votes]: The answer is no. First, to understand the question, WLOG $f$ is symmetric and positive definite; a general $f$ has a polar decomposition $f = os$ and the orthogonal factor $o$ has no effect on any of the norms in question. Then, WLOG $f$ is a rank 1 projection. The second and subsequent eigenvalues of $f$ do not increase $||f||$, but they could increase $||f(x)||$ for some specific $x$. So in summary, we can assume that $f = vv^T$ for some vector $v$. The question is whether $v$ must always make a small angle with some binary vector. -Let -$$v = (1,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}},\ldots,\frac{1}{\sqrt{m}}).$$ -If $w$ is a binary vector of weight $k$, then $|v \cdot w|$ is maximized when the non-zero entries of $w$ are at the beginning. However, -$$||w|| = \sqrt{k} \qquad ||v|| = \Theta(\sqrt{\log m}) \qquad |v \cdot w| = O(\sqrt{k}).$$ -This means that the angle between $w$ and $v$ is large, and therefore $||f(w)|| = ||v (v \cdot w)||$ is small compared to $||f||\;||w|| = ||v||^2 ||w||$. -The same proof works if $\{0,1\}$ is replaced by $\{-1,0,1\}$, or indeed by any finite subset of $\mathbb{R}$. On the other hand, there is a variation of the question with a positive answer. -Similar to Pietro Majer's remark, you can interpret the question as a comparison between two norms on $\mathbb{R}^m$. One is the $\ell^2$ norm, and the other is the norm whose unit ball is a polytope whose vertices are at the points in $S = \{0,1\}^m$ and its negative. By the theory of spherical packings on a sphere, for any $c < 1$, there exists a set $S$ of exponential size in $m$ such that the two norms are equal up to a factor of $c$. This is then a positive answer for that sample set of vector, even for constants close to 1. But such a set (coming from the centers of a sphere covering of the sphere) has to be fairly complicated, and I don't know if there are explicit asymptotic examples.<|endoftext|> -TITLE: sum of derivatives in roots of a polynomial of odd degree -QUESTION [15 upvotes]: Given odd positive integer $n$ and a monic polynomial $f(x)=(x-x_1)\dots (x-x_n)$ with $n$ distinct real roots. Is it always true that $\sum f'(x_i) > 0$? I may prove it for $n=3$ and $n=5$ and it looks plausible. - -REPLY [10 votes]: Instead, the sum of reciprocals -$$\sum\frac{1}{f'(x_j)}$$ -vanishes. This is because of the formula -$$\frac{1}{f(x)}=\sum_j\frac{a_j}{x-x_j},\qquad a_j:=\frac{1}{f'(x_j)},$$ -together with the asymptotics as $x\rightarrow\infty$. This is valid for every degree, odd or even. - -When $n=3$, this gives an amazing proof of the property that you quote. Denote $y_j=f'(x_j)$. Then $y_1y_2+y_3y_1+y_2y_3=0$, which means that $y=(y_1,y_2,y_3)$ belongs to a quadric whose intersection with the plane $y_1+y_2+y_3=0$ reduces to $(0,0,0)$, not equal to $y$. By continuity and connexity of the parameter space $x_1< x_2< x_3$, the expression $y_1+y_2+y_3$ must keep a constant sign, which we may calculate with $f(x)=x(x^2-1)$.<|endoftext|> -TITLE: Compactifications of varieties with small complement -QUESTION [9 upvotes]: Let $X$ be a smooth variety over an algebraically closed field $k$. If it makes things easier, $X$ may be assumed to be quasi-projective. By Nagata (or quasi-projectivity) there exists a proper variety $\bar{X}$ which contains $X$ as a dense open subvariety, and by the smoothness of $X$ we may assume $\bar{X}$ to be normal. -Are there any criteria/theorems which give information about the codimensio of the $\overline{X}\setminus X$? -The same question can be asked if we assume resolution of singularities, such that we may assume $\overline{X}$ to be smooth. Under which conditions can a smooth compactification $\overline{X}$ be found such that $X$ has complement of codimension $>1$ in $\overline{X}$? -Finally, to rephrase, how can one detect whether a given smooth variety $X$ arises by removing a codimemsion $>1$ closed subvariety from some proper variety? - -REPLY [6 votes]: I don't think you can hope for any kind of useful conditions, here are just some -comments. -Even if $X$ is smooth, whether or not we demand the compactification -matters. Let for instance $X'$ be a smooth projective surface containing a $(-2)$-curve $C$, -i.e., $C$ is isomorphic to $\mathbb P^1$ with self-intersection $-2$. Put -$X:=X'u C$. Then there is no smooth compactification of $X$ with -complement of codimension $³2$ but there is a normal one. In fact $C$ can be -contracted to a point on a normal surface but that point is singular with a non-trivial -local fundamental group. This fundamental group is an invariant of $X$ at -infinity and would be trivial if we could add a smooth point instead. -I think (haven't checked) that there is also for surfaces a distinction between -proper and projective (non-smooth) compactifications. If one does the usual -thing of blowing up $10$ general points on a smooth cubic, then the complement -of the strict transform of the cubic has a proper compactification by adding -just one point but I don't think it has a projective such compactification. -If $X$ has a projective normal compactification $\overline X$ then there are -some necessary conditions such as the space of global functions being finite -dimensional. In fact for a line bundle on $\overline X$, its space of section -are the same on $X$ as on $\overline X$ and hence using an ample line bundle on -$\overline X$ we see that $\overline X$ is the $\mathrm Proj$ of the algebra of -sections of powers of some line bundle on $X$. If we furthermore assume that -$\overline X$ is smooth, then any line bundle on $X$ extends to $\overline X$ -and will in particular have a finite dimensional space of sections. -Hence a reasonable question is if the condition that all line bundle have finite -dimensional spaces of sections is enough to get a (smooth?) compactification -with complement of codimension $>1$. I think that for surfaces the answer may be -positive but feel dubious about higher dimensions.<|endoftext|> -TITLE: Milnor's isotopy invariant using spectral sequence? -QUESTION [7 upvotes]: I'm reading stalling's article "the augmented ideal in group ring" in Ann. Math. Studies 84(R. H. Fox memorial volume) -In his final remark, he says that Milnor's link invariant could be interpreted by using Spectral sequence.(see Milnor, Isotopy of links, Algebraic geometry and topology, Princeton press) -Are there anybody who knows about further advances in this story?? - -REPLY [4 votes]: Have you read the Wikipedia page on Massey products? -http://en.wikipedia.org/wiki/Massey_product -It mentions Massey products are differentials in the Atiyah-Hirzebruch spectral sequence for a K-theory with local coefficients. The Atiyah-Hirzebruch spectral sequence is to a (co)homology theory what cellular (co)homology is for standard singular (co)homology.<|endoftext|> -TITLE: subtracting greatest possible prime -QUESTION [9 upvotes]: Given an infinite set $A$ of positive integers, $\min A:=a_0$. For $x\geq a_0$ define $f(x)=x-a$, where $a\leq x$, $a\in A$ is greatest possible. Then for positive integer $x$ iterations $x$, $f(x)$, $f(f(x))$, $\dots$ finally come to some element of the set $\{0,1,\dots,a_0-1\}$. Denote this final number $F(x)$. For example, if $A$ is the set of primes, $F(x)$ equals either 0 or 1. Do there always exist frequencies $\lim \frac{|F^{-1}(s)\cap [1,N]|}{N}$ for $s=0,1,\dots,a_0-1$? If not, what is criterion of existing such frequencies? Do they exist, say, for $A$=primes? - -REPLY [2 votes]: If I understand you correctly, this is effectively the question about (greedy) systems of numeration for the natural numbers. These are well understood in the case when $A$ satisfies some recurrence relation -- like the Fibonacci sequence (Zeckendorf) or the denominators $q_n$ of the CF convergents for some irrational $\alpha$ (Ostrowski). -If $A$ grows subexponentially, this is usually not good news for the ``ergodic'' questions like this.<|endoftext|> -TITLE: Combinator logic and unification -QUESTION [6 upvotes]: Summary: if we are trying to use combinator logic to solve first-order logic type problems, is the best method to feed in free variables and use the standard first-order unification algorithm? -In standard first-order logic, consider the problem of deducing that the following is a contradiction -p(x) -~p(5) - -By the rule that we can substitute specific for general, we can derive -p(5) -~p(5) - -and q & ~q => false. -The particular choice of the constant 5 to substitute for x is obtained using the standard unification algorithm, which in the case of first-order logic is computable (and indeed quite efficient). -Translating this into combinator logic gives -p = K true -~(p 5) - -Substituting the value of p gives -~(K true 5) - -Evaluating gives -~true - -Which is false, QED. -But it doesn't look so good when we try a more complex expression -p(f(f(x))) -~p(f(f(5))) - -In standard first-order logic, unification works just as well here as for the simpler case. The translation to CL, however, is -S (K p) (S (K f) f) = K true -~(p (f (f 5))) - -This isn't a problem we can solve with simple substitution. One approach we could try is to feed both sides of an equation the same value; if we take the number 5 as a value, the left-hand side of the equation evaluates as -S (K p) (S (K f) f) 5 -K p 5 (S (K f) f 5) -p (S (K f) f 5) -p (S (K f) f 5) -p (K f 5 (f 5)) -p (f (f 5)) - -So we have -p (f (f 5)) = K true 5 -p (f (f 5)) = true - -Substituting into the second expression from the original problem gives -~true - -So we have solved the problem, but we did it by guessing the correct value to feed both sides of the equation, by inspection. Is there a general method that could be used? -One that suggests itself is to feed in a variable, which would give -p (f (f x)) = true - -From which the standard first-order unification algorithm could be used. -Have we just established that the only way to solve problems in CL is to convert them back into standard logic? Perhaps it's not that bad; at least we have eliminated lambda and the other quantifiers, and their associated bound variables, leaving only free variables. -Still, is there a better method that I'm missing? -(I'm omitting consideration of types here; because combinator logic is Turing complete, presumably type checking will also be needed to avoid the paradoxes of the sort that make simple untyped lambda calculus unsound.) - -REPLY [2 votes]: I think first-order unification cannot replace higher-order unification because: -(1) if a variable occurs at the predicate position, FO unification is not supposed to handle that, though it may be fixed (seems easy). -(2) more importantly, sometimes a variable can be replaced by a combinatory term containing combinators such as S K I, and upon substitution, the new term may be reduced by combinatory rules to something else, and then becomes identical with the other term. By definition that would be a valid unifier. But FO unification is completely powerless to do that. So HO unification seems much more complex than the FO version...<|endoftext|> -TITLE: Why symmetric spaces? -QUESTION [11 upvotes]: In Bar-Natan's "Knots at Lunch" seminar at the University of Toronto, we are currently discussing a talk by Alekseev at Montpellier about Rouvière's expansion of the Duflo isomorphism to the setting of symmetric spaces. -We understand the definition of a symmetric space, and we know that people have written books about them; but we I don't understand in what sense symmetric spaces are useful and interesting mathematical objects. In particular: - -Are there significant (analytic? geometric? algebraic?) techniques which work for symmetric spaces, but not for more general classes of homogenous spaces? - -What I'm trying to understand (at least vaguely) is the role of symmetric spaces in the Kashiwara-Vergne picture, and the conceptual reason one might expect Duflo's isomorphism to generalize to this specific class of mathematical objects. There must be a conceptual explanation why symmetric spaces are the natural class of objects to consider in such contexts. -A closely related question is THIS. - -REPLY [9 votes]: The algebra of invariant differential operators on a symmetric space is commutative, and this is certainly not true for an arbitrary homogeneous space. While it is not true that the commutativity of the algebra $D^{G}$ of $G$ invariant differential operators on a homogeneous space $X = G/H$ implies that $X$ is symmetric, if $G$ is reductive it is true that $D^{G}$ is commutative if and only if $G/H$ is weakly symmetric in the sense defined by Selberg (see E.B. Vinberg's survey in Russ. Math. Surveys 56(1)).<|endoftext|> -TITLE: projective generator in the category of left-exact functors -QUESTION [7 upvotes]: I expect this to be a small understanding problem and not a real interesting question. -In Gabriel's thesis you find a proof of the theorem that every small abelian category $C$ admits a faithful exact functor to the category of abelian groups. The proof goes like this: Consider the abelian category $Sex(C,Ab)$, which is the category of left-exact functors $C \to Ab$. By the way, does anybody know why Gabriel has chosen this notation? Why not "Lex" for left-exact or "Gex" for the french "exacte a gauche"? Anyway, it can be shown that $Sex(C,Ab)$ is a (nowadays called) Grothendieck category and has thus enough injective objects. Besides, every injective object is an exact functor $C \to Ab$. If we embed the generator $U=\sum_{X \in C} Hom(X,-)$ into an injective object, we are done. -Now Gabriels claims that $U$ is actually a projective generator. I doubt that this is used in the proof, nevertheless it is interesting. By Yoneda the functor $F \mapsto Hom(U,F)$ is isomorphic to $F \mapsto \prod_{X \in C} F(X)$, thus we have to prove that every epimorphism in $Sex(C,Ab)$ is actually pointwise an epimorphism of abelian groups. This is not clear to me since a cokernel in $Sex(C,Ab)$ is defined by the universal left-exact functor associated to the pointwise-defined cokernel (cf. Prop. 5 in II.2). The vanishing does not seem to imply the vanishing of the pointwise-defined cokernel (cf. Lemme 3 b). -Note that Grothendieck has supervised this thesis. Although there are many many typos, I don't think that such a statement will just be wrong. What am I missing? - -REPLY [4 votes]: The object $U$ is a projective generator in the category of additive functors $C\to Ab$. It is also a generator of the category of left exact functors $C\to Ab$. It is not projective in the latter category, though. -Generally, left exact functors are similar to sheaves (on the category opprosite to $C$; epimorphisms in the latter category, i.e., monomorphisms in $C$, play the role of the coverings). There are no projective sheaves, generally speaking. The object $U$ is like the direct sum of the constant (pre)sheaves on all open subsets, extended by zero to the outside. It is a projective presheaf. It is not supposed to be a projective sheaf. -Update: I have been asked to provide a specific counterexample. It suffices to present an example of a morphism of left exact functors whose object-wise cokernel is not left exact. Let us find such a morphism of functors among morphisms of representable functors. So we want a morphism $X\to Y$ in an abelian category such that the object-wise cokernel of the morphism $Hom(Y,{-})\to Hom(X,{-})$ is not left exact. This means that there is a commutative square of morphisms $X\to Y$, $A\to B$, $X\to A$, $Y\to B$ such $A\to B$ is a monomorphism and the morphism $X\to A$ does not factor through the morphism $X\to Y$. Start with any nonsemisimple abelian category and choose a nonsplit monomorphism $X\to Y$. Set $A=X$ and $B=Y$.<|endoftext|> -TITLE: triple with large LCM -QUESTION [24 upvotes]: Does there exist $c>0$ such that among any $n$ positive integers one may find $3$ with least common multiple at least $cn^3$? -UPDATE -Let me post here a proof that we may always find two numbers with lcm at least $cn^2$. Note that if $a < b$, $N$=lcm$(a,b)$, then $N(b-a)$ is divisible by $ab$, hence $N\geq ab/(b-a)$. So, it suffices to find $a$, $b$ such that $ab/(b-a)\geq cn^2$, or $1/a-1/b\leq c^{-1} n^{-2}$. Since at least $n/2$ our numbers are not less then $n/2$, denote them $n/2\leq a_1 < a_2 < \dots < a_k$, $$2/n\geq \sum (1/a_i-1/a_{i+1})\geq k \min (1/a_i-1/a_{i+1}),$$ so -$\min (1/a-1/b)\leq 2/nk\leq 4/n^2$. -For triples we get lower bound about $c(n/\log n)^3$ on this way. Again consider only numbers not less then $n/2$. If all lcm's are less then $c(n/\log n)^3$, then all number itselves are less then some $n^3/2$, so for $2^k < n^3 < 2^{k+1}$ all of them do not exceed $2^k$, hence at least $n/2k$ of them belong to $[2^r,2^{r+1}]$ for the same $r$, then there exist three numbers $a < b < c$ with $c-a\leq 2^r/(n/4k)\leq 4ka/n$. Then -lcm$(a,b,c)\geq abc/((b-a)(c-a)(c-b))\geq (a/(c-a))^3\geq (n/4k)^3$. - -REPLY [11 votes]: See my post on AoPS -Edit: OK, reposting here. -The first step toward the solution, as it often happens, is to generalize the problem. Instead of just one set $A$, we shall consider $3$ sets $A,B,C$ of cardinalities $|A|,|B|,|C|$ and will try to prove that there exist $a\in A,b\in B,c\in C$ such that $[a,b,c]\ge\sigma |A|\cdot|B|\cdot|C|$. -The reason for such generalization is that we are going to employ the usual "minimal counterexample" technique -(a.k.a. "infinite descent", etc.) and we have much more freedom if we are allowed to modify three different sets independently rather than just one of them. -Our first attempt will be to make the reduction modulo $p^k$ where $p$ is a prime and $k\ge 1$ is an integer. Let $A_{p,k}=\{a\in A: v_p(a)=k\}$ where, as usual, $v_p(a)=\max\{v:p^v\mid a\}$. Let us replace $A$ with $A'=\{a'=p^{-k}a: a\in A_{p,k}\}$. For every $b\in B$, define $b'=\frac{b}{p^{\min(k,v_p(b))}}$. The numbers $b'$ form a set $B'$ of cardinality $|B'|\ge\frac{|B|}{(k+1)}$ because each $b'$ can be obtained from at most $k+1$ different $b\in B$. Define $C'$ in a similar way. Note that if $a'\in A', b'\in B', c'\in C'$, and $a,b,c$ are the elements of $A,B,C$ from which $a',b',c'$ were obtained, we have $[a,b,c]=p^k[a',b',c']$. Thus, if we have a minimal counterexample $A,B,C$ to our statement then $A',B',C'$ is not a counterexample, so we can find $a',b',c'$ with $[a',b',c']\ge \sigma |A'|\cdot |B'|\cdot |C'|\ge \sigma (k+1)^{-2}|A_{p,k}|\cdot|B|\cdot|C|$, which will not give us a triple $a,b,c$ with large least common multiple only if $|A_{p,k}|\le (k+1)^2p^{-k}|A|$. Thus, in our minimal counterexample, we must have this inequality for all prime $p$ and all $k\ge 1$. Note that it is trivially true with $k=0$ as well. The same inequality holds for the cardinalities of sets $B_{p,\ell}$ and $C_{p,m}$. -Now we shall try the averaging technique. Since we are dealing with a multiplicative problem, it will be convenient to use geometric means. So, let us consider the identity -$$ -\prod_{a\in A,b\in B,c\in C}\frac{abc}{[a,b,c]}\prod_{a\in A,b\in B,c\in C}[a,b,c]=\prod_{a\in A,b\in B,c\in C}(abc) -$$ -The products have $|A|\cdot|B|\cdot|C|$ factors in them and the product on the right is at least -$$ -(|A|!)^{|B|\cdot|C|}(|B|!)^{|A|\cdot|C|}(|C|!)^{|A|\cdot|B|}\ge e^{-3|A|\cdot|B|\cdot|C|}(|A|\cdot|B|\cdot|C|)^{|A|\cdot|B|\cdot|C|} -$$ -Our main task will be to estimate the first product on the left by $e^{K|A|\cdot|B|\cdot|C|}$ with some absolute $K>0$. If we manage to do that, we will immediately get the desired result "on average" with $\sigma=e^{-K-3}$. In order to do it, we'll estimate the power at which each prime $p$ can appear in this product. So, fix some $p$ and assume that $a\in A_{p,k},b\in B_{p,\ell},c\in C_{p,m}$. Then $p$ appears in the factor $\frac{abc}{[a,b,c]}$ at all only if $k+\ell+m\ge 2$ and its power in this case does not exceed $k+\ell+m+1$ (I know, this is an idiotic bound, but it holds and will allow me to have all factors of the same form). Thus, the total power in which $p$ appears in the first product is at most -$$ -\begin{aligned} -&\sum_{k,\ell,m:k+\ell+m\ge 2}(k+\ell+m+1)|A_{p,k}|\cdot|B_{p,\ell}|\cdot|C_{p,m}| -\cr -&\le -|A|\cdot|B|\cdot|C|\cdot\sum_{k,\ell,m:k+\ell+m\ge 2}(k+\ell+m+1)^7p^{-(k+\ell+m)} -\end{aligned} -$$ -Since there are at most $(M+1)^2$ ways to represent a positive integer $M$ as a sum of three non-negative integers, the last sum is at most $\sum_{M\ge 2}(M+1)^9p^{-M}$. -Now it is time to put all $p$ together. We get $e^{K|A|\cdot|B|\cdot|C|}$ with -$$ -K=\sum_{M\ge 2,p\text{ prime}}(M+1)^9p^{-M}\log p -$$ -and our only task is to show that this double series converges. We can forget that $p$ is prime, just remember that $p\ge 2$. Also for any $\delta>0$, we can estimate $(M+1)^9\le C_\delta p^{\delta M}$, $\log p\le C_\delta p^{\delta}$ with some finite $C_\delta>0$. Thus, our series is dominated by -$$ -\sum_{M,p\ge 2}p^{\delta-(1-\delta)M}=\sum_{p\ge 2} \frac{p^{3\delta-2}}{1-p^{\delta-1}}\ge \frac 1{1-2^{\delta-1}}\sum_{p\ge 2}p^{3\delta-2}<+\infty -$$ -if $\delta<\frac 13$. -This proof can be easily generalized to any number of sets but the constant it gives is rather terrible. It would be nice to get some better bound even for the case of 2 sets. As usual, questions and comments are welcome.<|endoftext|> -TITLE: How does the cokernel of the J-homomorphism count exotic spheres? -QUESTION [10 upvotes]: The wikipedia article on the J-homomorphism says that "the cokernel of the J-homomorphism is of interest for counting exotic spheres". I'd like to think this makes some sort of philosophical sense; as I understand it, the homomorphism comes from a Hopf construction, which isn't really a smooth sort of thing, but on the other hand it still feels like constructions using the (stable) (special) orthogonal group should somehow keep us in the non-exotic world. Is this at all close to the right intuition? - -REPLY [8 votes]: This isn't really different from the other answer, but I love this story so I wanted to try telling it a little slower. (I will also completely elide all technical details.) -Pontryagin-Thom says that any element of any stable homotopy group of spheres is represented by a stably framed manifold. A natural question to ask (and the first question I asked myself when I learned about P-T) is how simple you can make the representing manifold? The simplest possible manifolds are standard spheres with some stable framing. Relative to the standard with the standard stable framing (i.e. coming from the embedding of $S^n$ in $\mathbb{R}^{n+1}$ with the outward going normal), you've chosen some map from the appropriately stabilized trivial bundle to itself, i.e. an element of SO(n+k) for every point on your sphere. That is these are classified by elements of $\pi_n(SO)$. This gives you exactly the J-homomorphism. So the image of J is the part of the homotopy groups of spheres which comes from standard spheres. -Ok, but what about more complicated manifolds? Well in 2-dimensions you already see roughly how to do this: you start with a surface, you try to find a circle whose induced framing is such that you can cut along it and get a smaller genus surface. However, the Arf invariant on homology gives an obstruction to being able to do this, so you end up with either a trivial manifold or the torus with a particular stable framing (the Lie group framing). The nontrivial element of the second stable stem comes from this torus. We want to mimic this in higher dimension, and this is where surgery theory enters. That is we want to do surgery to decrease the complexity of our manifold, but we have to be a bit careful because we're doing stably framed surgery and so we have to make sure we have choose our elements in homology so that we can glue in stably framed discs. The key results from surgery theory are then that you can successively kill off all homology except possibly a single class in middle dimension. This is the Kervaire problem: are there manifolds of a given dimension with no homology except for one class in middle degree which you can't do framed surgery along? (Answer: exactly one such manifold in dimensions 2, 6, 14, 30, 62, and no such manifolds in any other dimension except possibly 126 which is still open. The explicit manifolds are known in dimensions 2 (flat torus), 6, 14, and 30, but not dimension 62.) -Ok, so outside the elements corresponding to the Kervaire manifolds we've done surgery to get the manifold down to a smooth manifold which is topologically a sphere. But it doesn't have to have the standard smooth structure! So such classes correspond to exotic spheres together with a stable framing. Now you need to know how many stable framings you'd expect an exotic sphere to have. First you want to show they all have some stable framing by immersing them into a large Euclidean space, and then second you want to argue that again the changes of framings are just given by composing with elements in the image of J. -All told: up to stably framed cobordism every manifold comes from a standard sphere with an interesting framing connect-sum your favorite framing on some exotic sphere connect-sum one of five or six explicit Kervaire manifolds generalizing the flat torus.<|endoftext|> -TITLE: How random are unit lattices in number fields? -QUESTION [23 upvotes]: I was wondering how random unit lattices in number fields are. To make this more precise: -If $K$ is a number field with embeddings $\sigma_1, \dots, \sigma_n, \overline{\sigma_{r+1}}, \dots, \overline{\sigma_n} \to \mathbb{C}$ (so we have $r$ real embeddings and $2 (n - r)$ complex embeddings), let $\mathcal{O}_K$ be the ring of integers and $\Lambda_K := \{ (\log |\sigma_1(\varepsilon)|^{d_1}, \dots, \log |\sigma_n(\varepsilon)|^{d_n}) \mid \varepsilon \in \mathcal{O}_K^\ast \}$ be the unit lattice, where $d_i = 1$ if $\sigma_i(K) \subseteq \mathbb{R}$ and $d_i = 2$ otherwise. -Then $\Lambda_K$ is always contained in $H := \{ (x_1, \dots, x_n) \in \mathbb{R}^n \mid \sum_{i=1}^n x_i = 0 \}$, and $\det \Lambda_K$ is the regulator $R_K$ of $K$. Let us normalize $\Lambda_K$ by $\hat{\Lambda}_K := \frac{1}{\sqrt[n]{R_K}} \Lambda_K$; then $\det \hat{\Lambda}_K = 1$. -Now my question is: can we say something on how random the lattices $\hat{\Lambda}_K$ are among all lattices in $H$ of determinant 1? (For example, for fixed signature $(r, n-r)$ of $K$.) -Since these lattices are not completely random (they consist of vectors of logarithms of algebraic numbers), it is maybe better to ask something like this: - -Given $\varepsilon > 0$ and a lattice $\Lambda \subseteq H$ with determinant 1, does there exists a number field $K$ of signature $(r, n - r)$ such that there is a basis $(v_i)_i$ of $\hat{\Lambda}_K$ and a basis $(w_i)_i$ of $\Lambda$ such that $\|v_i - w_i\| < \varepsilon$ for all $i$? -And if this exists, can one bound the discriminant of $K$ (or any other invariant of $K$) in terms of $\varepsilon$? - -(Of course, this question is only interesting when $n > 2$.) -I assume that this is a very hard problem, so I'd be happy about any hint on whether something about this is known, whether someone is working on this, how one could proof such things, etc. - -REPLY [16 votes]: This question was certainly discussed over past years, with no proven results though (as far as I am aware). I learned it from M. Gromov about 15 years ago (probably after he discussed it with G. Margulis). Here how I would formulate it: -Let us fix the signature $(r, n-r)$ (for example, $(3,0)$ for totaly real cubic fields -- the simplest non-trivial example which I researched numerically to some extend). Let $F_{(r,n-r)}$ be the set of all such number fields. For a field $K\in F_{(r,n-r)}$, consider the unit lattice $\mathcal{O}_K^*$, and its normalized (by the volume) logarithmic embedding -$\hat{\Lambda}_K\subset \mathbb{R}^n$ as above. -Hence we obtain for each field of fixed signature, a unimodular lattice in $ \mathbb{R}^{n-1}$. -We are interested in the type of such a lattice. It is reasonable to consider lattices up to isometries of the ambient $\mathbb{R}^{n-1}$. -Hence we obtain for each field $K\in F_{(r,n-r)}$, a point $x_K$ in the the moduli space of unimodilar lattices in $ \mathbb{R}^{n-1}$ up to an isometry. I call this the conformal type of the unit lattice of the field. -This moduli space is the familiar space $X_{n-1}=SL(n-1,\mathbb{Z})\setminus SL(n-1,\mathbb{R})/SO(n-1)$. For $n=3$ this is the modular curve. The space $X_{n-1}$ has a distinctive probability measure $\mu$ (coming from the right invariant measure on $SL(n-1,\mathbb{Z})\setminus SL(n-1,\mathbb{R})$; for $n=3$ this is the (volume one) hyperbolic measure on the modular curve). -The natural question would be what is the behavior of the set of points -$x_K\in X_{n-1}$, $K\in F_{(r,n-r)}$ -with respect to the measure $\mu$, or the geometry of $X_{n-1}$ -(recall that the space $X_{n-1}$ is not compact, and there are cusps). -To formulate conjectures/questions, one need to introduce an order on the set of fields of fixed signature. I am aware of $3$ (probably inequivalent) natural orderings: -Arithmetic order: order the the set $F_{(r,n-r)}$ by the discriminant $d_K$ of the field. -Geometric order: order the the set $F_{(r,n-r)}$ by the regulator $R_K$ of the field. -Dynamical order: order the the set $F_{(r,n-r)}$ by the shortest unit $\epsilon_K$ of the field. -Numerical experiments (with tables provided by PARI) suggest the following conjecture: -Conjecture: The set of points $x_K$, $K\in F_{(r,n-r)}$ becomes equidistributed in $X_{n-1}$ with respect to $\mu$ when $F_{(r,n-r)}$ is ordered arithmetically. -As far as I understand, Margulis expects that when ordered dynamically, points in $F_{(r,n-r)}$ escape to infinity (i.e., to the cusp) with the probability $1$ (certainly some points may stay low, e.g., coming from Galois fields). Probably one should expect the same with respect to the geometric ordering (i.e., by the regulator). It is very difficult to have numerical data for these orderings. -The question about density of points $x_K$ in $X_{n-1}$ would follow from equidistribution, but of course is somewhat separate. In particular even if Margulis is right, this would not mean that $x_K$'s are not dense (its quite possible that this is still true). I do not know about (effective) approximation of a lattice by the unit lattice. -I also would like to mention that the additive analog of this question for totally real cubic fields (ordered by the discriminant) was solved by D. Terr (PhD, Berkeley, 1997, unpublished).<|endoftext|> -TITLE: When should a result be made into a paper? -QUESTION [95 upvotes]: I recently posted a short (6 page) note on arXiv, and have more or less decided that I should not submit it to a journal. I could have tacked it onto the end of a previous paper, but I thought it would be somewhat incongruous -- it is an interesting consequence of the key lemma unrelated to the main result. I really liked the concision of the paper and didn't want to spoil it. -This brings to my mind several philosophical and/or ethical questions about the culture of publishing that I find interesting, particularly at this point in my life since I am nearing the point of seeking a permanent position. - -Two things are clear: It is to one's advantage, especially in the early career stage, to have many publications. It is also to one's advantage to have a strong publications. So take a result that is not very difficult to prove, but is interesting mostly because I think it may be useful as a stepping stone to another as-yet-unknown result. Given that I have already put it on arXiv, is "because I think it could be published" a good enough reason to publish it? -One good reason to submit a paper to a journal is to have it in the refereed scientific record. I was tempted to claim the result at the end of the previous paper without proof as a remark, but in the end thought better of making such a claim without giving a proof. How bad is it to make such a claim if the proof (or even the truth) is not obvious, but can be proven as a reasonably straightforward generalization of someone else's proof of a different result? -This is a more pragmatic and frank question. Should young researchers be careful to avoid the impression of splitting their work into MPUs (minimum publishable units)? In this instance my reasons for not including the result in another paper are somewhat complex and non-obvious. -As a counterpoint to 3, how should one balance the purity/linearity of the ideas/results in a paper, versus including as many related results as possible? - -I realize that these are very soft and subjective questions, but I am interested to hear opinions on the matter, even if there is no universally true answer. - -REPLY [8 votes]: I would say that concerning the overwhelmingly large number of mathematical papers that appears, the main question (for the mathematical community, your career is something else) is: will the publishing of your paper add more signal (being MathReviewed, if someone needs the result, she or he is likely to find it) or more noise? For example, concerning a slightly different question (when should one split a paper in two?), I tend to think that if splitting does improves the probability of finding any of the results by a MR search, then splitting is actually good.<|endoftext|> -TITLE: Is every balanced pre-abelian category abelian? -QUESTION [7 upvotes]: Is every balanced pre-abelian category abelian? That is, given an additive category $\mathcal{A}$ in which cokernels and kernels exists, such that every morphism, which is a mono- and an epimorphism, is an isomorphism; does it follow that $\mathcal{A}$ is abelian? Note that it would suffice to prove that the canonical morphism $coim(f) \to im(f)$, where $f$ is an arbitrary morphism, is a mono- and an epimorphism. -Note that the usual examples for non-abelian categories somehow suggest this (filtered modules, topological abelian groups). See also this related question. -After a google search I have found the following theorem (in "Basic homological algebra" by M. Scott Osborne, Cor. 7.18): If $\mathcal{A}$ is a balanced pre-abelian category with a separating class of projectives and a coseparating class of injectives, then $\mathcal{A}$ is abelian. Ok then this does not seem to be true in general. Does anybody know an example? - -REPLY [12 votes]: Let $B$ be the abelian category of 3-term sequences of vector spaces and linear maps $V^{(1)}\to V^{(2)}\to V^{(3)}$ (the composition can be nonzero). There are 6 indecomposable objects in this category; denote them by $E_1$, $E_2$, $E_3$, $E_{12}$, $E_{23}$, and $E_{123}$. Here $dim E^{(i)}_J=1$ for $i\in J$ and $0$ otherwise. Let $A\subset B$ be the full additive subcategory whose objects are the direct sums of all the indecomposables except $E_{12}$ and $f$ be the morphism $E_3\to E_{123}$. -Then one has $Coker_B f=E_{12}$, $Coker_A f=E_1$ and $Im_A f=E_{23}$, while $Coim_A f=E_3$ and therefore $Coker_A(Coim_A f\to Im_A f)=E_2\ne0$. It is easy to see that any morphism in $A$ has a kernel and cokernel. Besides, any morphism in $A$ with zero kernel and cokernel has also zero kernel and cokernel in $B$, hence is an isomorphism. -Further discussion can be found in my paper http://arxiv.org/abs/1006.4343 , Example A.5(7) (pages 59-60 in version 2).<|endoftext|> -TITLE: Integral models of p-adic representations -QUESTION [6 upvotes]: Let $G$ be a compact group and $K$ a finite extension of $Q_{p}$. If $\rho$ is a continuous representation of $G$ on a finite dimensional vector space over $K$, then it is well known that the semi-simplification of the reduction modulo $p$ of an integral model of $\rho$ only depends upon the isomorphism class over $K$ of $\rho$. -Are there results or strategies that, given a $K$-representation $\rho$ as above, allow one to determine properties of all the possible integral models of $\rho$? For example: are there some conditions under which all the integral models of $\rho$ are semi-simple? -Where can I find references? Thanks! - -REPLY [4 votes]: Let be focus on the two-dimensional case first. Finding an integral model for $\rho$ is equivalent to choosing a lattice in $K^2$ that is invariant under $\rho(G)$. So in this case you are asking for invariant points in the tree for $GL_2(K)$ under a compact subgroup -$\rho(G)$ of $GL_2(K)$. There are various possibilities, and there is a nice paper of Bellaiche and Chenevier Sous-groupes de $GL_2$ et arbres discussing the possibilites. -When you say that you would like all integral models to be semi-simple, I'm not sure what you mean; but it say the represenation over $K$ is irreducible, then it will never happen that all the integral models have semi-simple reduction, if they are not themselves irreducible. (This is a proposition of Ribet, and is easily seen in the tree-based picture.) -Just to illustrate how one argues: if two points in the tree are $\rho(G)$-invariant, then so will be all the points on the line segment joining them. If $\rho(G)$ fixes an infinite half-line, this corresponds to an invariant one-dimensional space over $K$, -and so means that $\rho$ is reducible over $K$. So if $\rho(G)$ is irreducible over $K$, -then the fixed set of $\rho(G)$ will be bounded and convex, which already imposes some -restrictions. (One an prove Ribet's result by these sort of considerations: basically, -any extremal point of the fixed set of $\rho(G)$ will have non-simple reduction.) -Once you understand the $GL_2$ case you can imagine how to try and generalize things to -the $GL_n$ case, although the combinatorics becomes more complictated, and (as Bellaiche and Chenevier note) it is harder to say anything nice, because when $n > 2$, $GL_n$ is the full -automorphism group of the building. - -REPLY [2 votes]: For general representations, there are no such strategies and this is a very hard open problem, even for finite groups. Since your usual compact group will have various finite quotients, your problem is at least as hard. This rather older survey by Reiner is a good place to start. Particularly pages 171-173 should be of interest to you. -As for semi-simplicity, you need to be clear about what you mean. In the integral world, the classical notion of semi-simplicity makes little sense, because your representations always have subrepresentations. E.g. if $\Gamma$ is an integral model, then $p\Gamma$ is a subrepresentation. But at least the Krull-Schmidt theorem holds in your setting, even though the indecomposable summands of an integral model might not sit in irreducible $K$-representations. In fact, if the semi-simplification of the reduction of $\rho$ contains two non-isomorphic irreducible components, then I believe that there is almost no hope for all integral models to only have indecomposable summands that sit in irreducible $K$-representations. (Clearly, if the reduction itself is not semi-simple then your integral models will be at least as badly behaved.)<|endoftext|> -TITLE: Maximum distance to nearest-lattice-point on (hyper-)sphere with unit lat-lon lattice. -QUESTION [5 upvotes]: Let $U$ be the set of all non-null $n \times 1$ vectors $\mathbf{\mathrm{u}}$, where $u_i \in \lbrace-1, 0, 1\rbrace$. Let $\mathbf{\mathrm{x}}$ be an $n \times 1$ vector in $\mathbf{R}^n$. Let $\mathbf{\mathrm{u_x}}$ be the element of $U$ that is closest in angle to $\mathbf{\mathrm{x}}$. Then for any $\mathbf{\mathrm{x}}$, the maximum possible angle between $\mathbf{\mathrm{u_x}}$ and $\mathbf{\mathrm{x}}$ is ... -For $n = 2$, I get $\pi / 8$, obviously, but what is the general expression for larger $n$? I tried to think of this as a nearest-lattice-point problem on the (hyper-)sphere with a latitude-longitude lattice, but didn't get very far. -Thanks. - -REPLY [3 votes]: Given your initial answer for $\mathbf R^2,$ to move to $\mathbf R^3$ consider the vectors $(x,y,z)$ such that $x,y,z \geq 0$ that are equiangular between the plane vectors $(1,0,0)$ and $(1,1,0).$ These make up the plane -$$ y = (\sqrt 2 - 1) x $$ with arbitrary $z \geq 0.$ In order to get the same angle with -$(1,0,1)$ we get the ray -$$ (\sqrt 2 - 1) x = y = z. $$ In order to get the same angle with -$(1,1,1)$ we get the ray -$$ y = (\sqrt 2 - 1) x, \; \; z = (\sqrt 3 - \sqrt 2) x. $$ -Note that $ (\sqrt 2 - 1) = 0.4142... $ while $ (\sqrt 3 - \sqrt 2) = 0.317837...$ So in $\mathbf R^3$ the latter comes first while increasing $z,$, and best vector is $(1, \; \;\sqrt 2 - 1, \; \; \sqrt 3 - \sqrt 2) $ where you can work out the angle. -The same process gets you from $\mathbf R^3$ to $\mathbf R^4,$ take this answer for $\mathbf R^3$ and increase the fourth coordinate until you have an equal angle with $(1,1,1,1).$ -And so on. -EDIT: I get it. In $\mathbf R^n$ the optimal vector is -$$ V = (1, \; \;\sqrt 2 - 1, \; \; \sqrt 3 - \sqrt 2, \ldots, \sqrt n - \sqrt {n-1})$$ with equal angles to -$A_1 = (1,0,\ldots,0),$ $A_2 = (1,1,0,\ldots,0),$ $A_3 = (1,1,1,0,\ldots,0), \ldots,$ $A_n = (1,1,1,\ldots,1).$ -EDIT 2: Note that the entries of $V$ are strictly decreasing. As a result, if instead we consider $B_k$ with $k$ entries set to $1$ and the other $n-k$ set to $0,$ then -$$ | A_k | \; = \; | B_k | $$ -but -$$ V \cdot A_k > V \cdot B_k .$$ Therefore the angle between $V$ and $B_k$ is larger than the angle between $V$ and $A_k,$ and the angle we actually constructed is the best we can get away with.<|endoftext|> -TITLE: Hecke-algebras in your field of mathematics -QUESTION [12 upvotes]: (How) do Hecke-algebras arise naturally in your field of mathematics and why are they important? -How would you define them and how do you think about them? -e.g. generators and relations, functions on some space, grothendieck group... - -REPLY [4 votes]: One very interesting application is the proof by Kuhn and Priddy of the Whitehead Conjecture in stable homotopy theory (MR): - -\bib{MR803606}{article}{ - author={Kuhn, Nicholas J.}, - author={Priddy, Stewart B.}, - title={The transfer and Whitehead's conjecture}, - journal={Math. Proc. Cambridge Philos. Soc.}, - volume={98}, - date={1985}, - number={3}, - pages={459--480}, - issn={0305-0041}, - review={\MR{803606 (87g:55030)}}, - doi={10.1017/S0305004100063672}, -} - -The point of contact is that the proof involves classifying spaces of elementary abelian groups $(\mathbb{Z}/p)^r$ and their suspension spectra. The $p$-local group ring $\mathbb{Z}_{(p)}[GL_r(\mathbb{Z}/p)]$ acts on this spectrum, and one can use the Steinberg idempotent to give a splitting. The relationship between the Steinberg idempotent and Hecke operators is pure algebra. -There is also a relationship between Hecke operators and power operations in elliptic cohomology (MR): - -\bib{MR1637129}{article}{ - author={Ando, Matthew}, - title={Power operations in elliptic cohomology and representations of - loop groups}, - journal={Trans. Amer. Math. Soc.}, - volume={352}, - date={2000}, - number={12}, - pages={5619--5666}, - issn={0002-9947}, - review={\MR{1637129 (2001b:55016)}}, - doi={10.1090/S0002-9947-00-02412-0}, -} - -This actually generalises to give a relationship between power operations in Morava $E$-theory (at the prime $p$ and height $n$) and a kind of Hecke algebra for the monoid of $n\times n$ matrices over $\mathbb{Z}_p$ with nonzero determinant.<|endoftext|> -TITLE: Explaining the number field-function field analogy -QUESTION [9 upvotes]: There is a general circle of ideas according to which true statements about number fields should have analogues in function fields. As best I can tell, the fact that this seems to work is pretty mysterious. The only results I know directly relating the two come from logic, such as Ax-Kochen, and these are limited to first-order statements in restricted languages. But the analogy apparently goes well beyond such statements. Are there any theorems/conjectures/observations that would explain why the analogy is a good one? I am looking for statements that allow one to go directly from the function field case to the number field case or vice versa. - -REPLY [8 votes]: I think your statement could usefully be sharpened in a couple of ways. -Firstly, the state-of-the-art is that true statements for function fields are expected to have analogues for number fields. The outstanding example here is naturally the Riemann hypothesis. Where conjectures for number fields are present, the way of working is via the heuristic that the function field analogue may be sought, and then proved. This has been taken a long way for the Langlands philosophy, for example. -The other major point is that historically what came first was analogies between Riemann surface theory and number fields. Hilbert seems to have conjectured the main outlines of class field theory using complex curves and Jacobians as the source of inspiration. Certainly the prior Dedekind-Weber geometric view fed into that. Subsequently we get "global field" as a kind of middle term: curves over finite fields are a little closer to number fields than complex curves. The attitude of Weil's Basic Number Theory is to develop the parallelism to the point of a common vocabulary (which is now widely used). Global fields can be studied successfully using local compactness, would be a succinct summary. Pontryagin duality, for example, can take much of the strain, and this theory is of course of broader application than number theory. -Of course we don't yet know why this works.<|endoftext|> -TITLE: PDEs as a tool in other domains in mathematics -QUESTION [29 upvotes]: According to the large number of paper cited in MathSciNet database, Partial Differential Equations (PDEs) is an important topic of its own. Needless to say, it is an extremely useful tool for natural sciences, such as Physics, Chemistry, Biology, Continuum Mechanics, and so on. -What I am interested in, here, is examples where PDEs were used to establish a result in an other mathematical field. Let me provide a few. - -Topology. The Atiyah-Singer index theorem. -Geometry. Perelman's proof of Poincaré conjecture, following Hamilton's program. -Real algebraic geometry. Lax's proof of Weyl-like inequalities for hyperbolic polynomial. - -Only one example per answer. Please avoid examples in the reverse sense, where an other mathematical field tells something about PDEs (examples: Feynman-Kac formula from probability, multi-solitons from Riemann surfaces). This could be the matter of an other MO question. - -REPLY [4 votes]: Great! Since this has somehow bubbled to the top, I have yet-another occasion for a not-entirely-selfish rant! :) -Although @WillieWong's answer was a few years ago, it does point (if a bit obscurely) in a significant direction, for example. Immediately, one should note that there is no known (to me, and ... I care) proof of RH that immediately/simply uses PDE ideas on modular curves. -But those ideas, from Haas, Hejhal, Faddeev, Colin-de-Verdiere, and Lax-Phillips, do solidly establish a connection of spectral properties of (certain "customized" extensions of) Laplacians on modular curves and related canonical objects. -A more obviously legit example of interaction of number-theoretic "physical objects" and "PDE" was Milnor's example of two tori with the same spectrum for the Laplacian. But, yes, that wasn't really about PDE. -A problem with reporting modern relevance of PDE to number theory is that there are "complicating" additions, ... :) ... often under-the-radar amounting to things stronger than Lindelof Hypothesis, if not actually the Riemann Hypothesis. -Rather than recap things better documented elsewhere (many peoples' arXiv preprints, my own vignettes and talks various places, ...) please forgive my returning to the homily that "PDE" are merely assertions of relations, as Newton intuited for the planets, and others have observed/inferred for many more things. -(Any hysterically provincial remarks about turf, or "specialties-as-ignorance-of-other" are obviously toxic... despite their dangerous prevalence and popularity...) -Thus, srsly, people, "PDE" means "a kind of condition on functions..." ... If people weren't so caught up in ... oop, sorry, the kind of people do get caught up in... :) ... it'd be obvious that "infinitesimal" conditions would be natural... -Thus, an explanatory but not really useful answer is, that we seem to see that This is not related to That because the respective proprietores have no vested interest in letting on that anyone else could ... perform their guild's function. -(Yes, it is informative to review Europe's late-renaissance guild-culture...) -And, as in many rants, I wish to reassure everyone by my disclaimer, "wait, what was the question, ... again...? " :) -(But, yes, this is a serious-and-important issue, in many ways, so, yeah, just some kidding-at-the-end.)<|endoftext|> -TITLE: Roots of permutations -QUESTION [28 upvotes]: Consider the equation $x^2=x_0$ in the symmetric group $S_n$, where $x_0\in S_n$ is fixed. Is it true that for each integer $n\geq 0$, the maximal number of solutions (the number of square roots of $x_0$) is attained when $x_0$ is the identity permutation? How far may it be generalized? - -REPLY [5 votes]: For a general finite group $G$ and an arbitrary element $y \in G,$ there is a general way to use characters to compare the number of solutions of $x^{2} = y$ in $G$ with the number of solutions of $x^{2} = 1$ within $C_{G}(y).$ -For notice that any element $x$ with $x^{2} = y$ already lies in $C_{G}(y).$ -Hence (using the general formula mentioned in Alex's answer, but within the group -$C_{G}(y)$) the number of solutions of $y^{2} = x$ is $\sum_{ \mu \in {\rm Irr}(C_{G}(y)} \nu(\mu) \mu(y),$ where $\nu(\mu)$ is the Frobenius-Schur indicator of $\mu$ ( which Alex denoted by $s(\mu)$). -But now we note (using Schur's Lemma) that unless $y{\rm ker} \mu$ has order $1$ or $2$ in $C_{G}(y)/{\rm ker}(\mu)$ we have $\nu(\mu) = 0$ since $\mu$ is certainly not real-valued. -Hence if $y$ has order $m,$ then we need only consider irreducible characters $\mu$ with $y$ in their kernel when $m$ is odd, and irreducible characters $\mu$ with $y^{2}$ in their kernel when $m$ is even. -In any case, the contribution to the sum from those characters with $y$ in their kernel is the same as the number of square roots of the identity in $C_{G}(y)/\langle y \rangle.$ -I won't give all details, but we may continue to deduce that -the difference between the number of square roots of the identity in $C_{G}(y)$ and the number of square roots of $y$ in $G$ is given by the formula: -$\sum_{\mu \in {\rm Irr}(C_{G}(y)/\langle y^{2} \rangle) \backslash -{\rm Irr}(C_{G}(y)/\langle y \rangle)} 2\nu(\mu)\mu(1).$ -Note that this number is non-negative unless $C_{G}(y)$ has an irreducible character $\mu$ with $\nu(\mu) = -1$ and with $y^{2} \in {\rm ker} \mu$ but $y \not \in {\rm ker} \mu$. -Hence we may conclude in particular that for a general element $y$ of a general finite group $G,$ the number of square roots of $y$ in $G$ is already less that the number of solutions of square roots of the identity in $C_{G}(y)$ unless $C_{G}(y)$ has an irreducible character $\mu$ of Frobenius-Schur indicator $-1$ such that $y{\rm ker} \mu$ has order $2$ in $C_{G}(y)/{\rm ker} \mu.$ -Latter edit: It is perhaps worth remarking that this argument shows that the number of square roots of $y$ in $G$ is the same as the number of square roots of $yM$ in $C_{G}(y)/M$, where $M$ is the normal subgroup $\langle y^{2} \rangle $ of $C_{G}(y).$ This reduces the computation of the number of square roots of $y \in G$ to the case that $y$ has order $1$ or $2$ and that $y \in Z(G).$ -Returning to the (last part of the) original question, it means that each finite group $G$ for which the maximum number of square roots is not attained at the identity has a section (ie a factor group of a subgroup) $H$ such that $Z(H)$ such contains an involution $u$ with the number of elements of order four with square $u$ at least two greater that the number of involutions of $H$ ( here an involtion is considered to be an element of order exactly two). Furthermore, the section $H$ has the form $C_{G}(y)/\langle y^{2} \rangle$ where $y$ is an element of $G$ with the maximum number of square roots. Note also that if the identity element does not have the maximum number of square roots then the element $y$ attaining the maximum number necessarily has even order. -Note that in the essential case that $y \in Z(G)$ is an involution, then $y$ has more square roots in $G$ than the identity does if and only if we have -$\sum_{ \{\chi \in {\rm Irr}(G) \backslash {\rm Irr}(G/\langle y \rangle): \nu(\chi) = -1 \}}\chi(1) > \sum_{ \{\chi \in {\rm Irr}(G) \backslash {\rm Irr}(G/\langle y \rangle): \nu(\chi) = +1 \} } \chi(1).$<|endoftext|> -TITLE: What is the center of Qcoh(X)? -QUESTION [21 upvotes]: The center of a category $C$ is the monoid $Z(C)=\mathrm{End}_{[C,C]}(\mathrm{id}_C)$. Thus it consists of all families of endomorphisms $M \to M$ of objects $M \in C$, such that for every morphism $M \to N$ the resulting diagram commutes. If $C$ is an $Ab$-category, this actually becomes a ring. For example, the center of $\mathrm{Mod}(A)$ is the center of $A$, if $A$ is a (noncommutative) ring. -Now my question is: What is the center of the category of quasi-coherent modules $\mathrm{Qcoh}(X)$ on a scheme $X$? Observe that there is a natural map $\Gamma(X,\mathcal{O}_X) \to Z(\mathrm{Qcoh}(X))$, mapping a global section to the endomorphisms of the quasi-coherent modules which are given by multiplication with this section. Also, there is a natural map $Z(\mathrm{Qcoh}(X)) \to \Gamma(X,\mathcal{O}_X)$, which takes a compatible family of endomorphisms to the image of the global section $1$ in $\mathcal{O}_X$. The composite $\Gamma(X,\mathcal{O}_X) \to Z(\mathrm{Qcoh}(X)) \to \Gamma(X,\mathcal{O}_X)$ is the identity, but what about the other composite? If $X$ is affine, it also turns out to be the identity. -In the end of his thesis about the Reconstruction Theorem, Gabriel proves that $\Gamma(X,\mathcal{O}_X) \to Z(\mathrm{Qcoh}(X))$ is an isomorphism if $X$ is a noetherian scheme (using recollements of localizing subcategories). I'm pretty sure that the proof just uses that $X$ is quasi-compact and quasi-separated. Now what about the general case? -Note that this is about the reconstruction of the structure sheaf $\mathcal{O}_X$ from $X$. Also note that Rosenberg claims in Reconstruction of schemes that this is possible for arbitrary schemes. But if I understand correctly, Rosenberg uses a structure sheaf on the spectrum of an abelian category which avoids the above problems and uses $Z(\mathrm{Mod}(X))=\Gamma(X,\mathcal{O}_X)$, which is certainly true (use extensions by zero), but doesn't yield the result for $\mathrm{Qcoh}(X)$. But I'm not sure because Rosenberg refers to a proof step (a4) which is not there. -Edit: Angelo has proven the result below if $X$ is quasi-separated. Now what happens if $X$ is not quasi-separated? In that case, $\mathrm{Qcoh}(X)$ might have not "enough" objects, which makes the question much harder (and perhaps also less interesting). - -REPLY [2 votes]: I've surely misunderstood your question or made a simple mistake, but it seems to me that if $X$ is allowed to vary then $Z({\rm QCoh}(X))$ is a sheaf in the Zariski topology---call it $\underline{Z}_X$---and there is a morphism of sheaves $\mathcal{O}_X \rightarrow \underline{Z}_X$. It's an isomorphism in general because it's an isomorphism when $X$ is affine. -Edit: Martin's comment explains the simple mistake I made above, but I think it might be possible to salvage the argument. I'm a little nervous about this though... -Let $A_X$ be the smallest additive subcategory of the category of $\mathcal{O}_X$-modules that contains ${\rm QCoh}(X)$ and is closed under arbitrary products and kernels. Any endomorphism of the identity functor of ${\rm QCoh}(X)$ extends uniquely to an endomorphism of the identity functor of $A_X$ (since product and kernel are functors). Therefore $Z(A_X) = Z({\rm QCoh}(X))$. -Since $f^{-1}$ commutes with arbitrary products and kernels, the categories $A_X$ for varying $X$ form a fibered category (actually a stack) over the Zariski site of $X$. -If $f : U \rightarrow X$ and $G \in A_U$ then $f_\ast G \in A_X$ (see the proof that the pushforward of a quasi-coherent sheaf under a quasi-compact, quasi-separated morphism is quasi-coherent in EGA(1971)I.6.7.1, taking into account that $f_\ast$ commutes with arbitrary inverse limits). If $U$ is an open embedding, this implies that $f^\ast : A_X \rightarrow A_U$ is essentially surjective (since $f^\ast f_\ast G = G$). Therefore any endomorphism of the identity functor of $A_X$ can be restricted to an endomorphism of the identity functor of $A_U$. Therefore as $X$ varies, $Z(A_X) = Z({\rm QCoh}(X))$ forms a sheaf in the Zariski topology. -Edit 2: I have to argue that $f_\ast$ carries $A_U$ into $A_X$. I'm mimicking the argument from EGA here, but I can't see Martin's objection. -We can assume that $X$ is affine (since $A$ is a stack). If $f$ is affine, this is because $f_\ast$ commutes with arbitrary inverse limits and $f_\ast$ carries ${\rm QCoh}(U)$ into ${\rm QCoh}(X)$. -Let $U_i$ be a cover of $U$ by open affines, and let $U_{ij}$ be the pairwise intersections. Let $f_i$ and $f_{ij}$ be the restrictions of $f$. Let $G_i = {f_i}_\ast G_i \big|_{U_i}$ and let $G_{ij} = {f_{ij}}_\ast G \big|_{U_{ij}}$. Then -$f_\ast G = \ker( \prod_i G_i \rightarrow \prod_{ij} G_{ij} )$ -and the $G_i$ are in $A_X$ because $U_i \rightarrow X$ are affine. If the $G_{ij}$ are also in $A_X$ then so is $f_\ast G$. This will be the case if $f$ is separated, since then the $U_{ij}$ will be affine. But this implies the general case because the $U_{ij}$ will be quasi-affine, hence separated, over $X$, so $G_{ij} = {f_{ij}}_\ast G \big|_{U_{ij}}$ will be in $A_X$ by the case mentioned above.<|endoftext|> -TITLE: Self homeomorphisms of $S^2\times S^2$ -QUESTION [12 upvotes]: Every matrix $A\in SL_2(\mathbb{Z})$ induces a self homeomorphism of $S^1\times S^1=\mathbb{R}^2/\mathbb{Z}^2$. For different matrices these homeomorphisms are not homotopic, as the induced map on $\pi_1(S_1\times S_1)$ is given by $A$ (w.r.t the induced basis). -So I am wondering, whether a similar construction also works for other spheres than $S^1$. So is there any non-obvious self-homeomorphism of $S^2\times S^2$ (I know i can use degree $-1$ maps in any choice of coordinates / flip the coordinates /compose such maps). - -REPLY [3 votes]: A small comment extending Greg's comments, given an element of $[g] \in \pi_n Diff(S^n)$ you can construct a diffeomorphism -$f : S^n \times S^n \to S^n \times S^n$ by sending the pair $(x,y)$ to $(x,g_x(y))$, so the corresponding matrices would be upper-triangular as in Greg's examples. -$Diff(S^n)$ has the homotopy-type of $O_{n+1} \times Diff(D^n)$. -Greg's examples come from situations where there are non-trivial elements of $\pi_n O_{n+1}$ that you "know" because they're coming from splittings of the fibrations $SO_n \to SO_{n+1} \to S^n$, but you could also get non-trivial elements from finding elements in $\pi_n Diff(D^n)$. My understanding is there's little to nothing known about these homotopy groups. The rational homotopy $\pi_j Diff(D^n)\otimes \mathbb Q$ of $Diff(D^n)$ is only understood in the "Igusa Stable Range" of $j \leq \min\{ \frac{n-4}{3}, \frac{n-7}{2} \}$ so it's not useful for the kind of constructions you'd like.<|endoftext|> -TITLE: Flatly compactifiable morphisms -QUESTION [7 upvotes]: Let $f:U \to S$ be a flat morphism. Let us say that $f$ is flatly compactifiable if there exists a proper morphism $\bar{f}:X \to S$ and a closed subscheme $Z \subset X$ such that -1) $U = X \setminus Z$ and $f = \bar{f}_{|U}$; -2) $\bar{f}$ is proper; -3) BOTH $X$ and $Z$ are flat over $S$. -My question is whether this notion already appeared in the literature and what is the correct name for it? - -REPLY [2 votes]: In general you can not expect that $f$ will be flat on $X-U$, but you have a locally finite stratification of $X$ for which $U$ is the dense stratum, and such that $f$ is flat over each strata. This notion is called "platification" (in French) and it was dealed with great details in the paper of Gruson and Raynaud "Techniques de platification d'un module".<|endoftext|> -TITLE: Vector fields on complete intersections -QUESTION [10 upvotes]: Here is a question the answer to which I've been trying to locate for some time. -Let $X$ be a smooth projective complete intersection over an algebraically closed field $k$; assume that $X$ is not contained in a hyperplane. Is it true that $X$ admits no nonzero tangent vector fields, unless $X$ is a quadratic hypersurface or a genus 1 curve? -Here are some remarks. - -In the case $char(k)=0$ the positive answer to this question is stated as Proposition 2.11 of "Derivations, automorphisms and deformations of quasi-homogeneous singularities" by J. Wahl (Proceedings of symposia in pure mathematics, vol. 40, part 2, 613-625). However, no proof is given there, it is only mentioned that this can be obtained by the same method as theorem 2.8 (whose proof is only briefly sketched). I was wondering if there is a more complete account. -When $X$ is a hypersurface, the answer is positive; two characteristic free proofs of that are given in Katz-Sarnak, Random matrices, Frobenius eigenvalues and monodromy, 11.6 and 11.7. Either proof may in principle generalize to complete intersections, but I don't see a straightforward way to do that. -It is easy to check that when $X$ is a curve, the answer is positive as well. - -REPLY [5 votes]: I haven't checked the indexing completely but I think Thm 1.1 and Prop. 1.3 of SGA 7 II:Exp -II does what you want (with possibly a small number of exceptions if I've got -the indexing a little bit wrong). You have to use that -$T^1_X=\Omega_X^{d-1}\bigotimes \omega_X^{-1}$, $d=\dim X$, and the formula for $\omega_X$ -in terms of the degrees of the defining hyperplanes.<|endoftext|> -TITLE: Metric angles in Riemannian manifolds of low regularity -QUESTION [13 upvotes]: Given three points $a,b,c$ in a (geodesic) metric space $X$, one defines a comparison angle $\angle(a,b,c)$ by the cosine law: -$$ - \angle(a,b,c) = \arccos \frac{|ab|^2 + |ac|^2 - |bc|^2}{2\cdot|ab|\cdot|ac|} -$$ -where $|ab|$, $|ac|$ and $|bc|$ are distances in $X$. In other words, $\angle(a,b,c)$ is the angle of a planar triangle with the same side lengths as the triangle $abc$ in $X$. -Given two paths $\xi,\eta:[0,\varepsilon)\to X$ starting at the same point $p=\xi(0)=\eta(0)$, define a metric angle $\angle(\xi,\eta)$ by -$$ - \angle(\xi,\eta) = \lim_{t,t'\to 0} \angle(p,\xi(t),\eta(t')) -$$ -if the limit exists. -If $X$ is a smooth manifold with a $C^2$ Riemannian metric (which defines a geodesic distance in the usual way) and $\xi,\eta$ are $C^1$ regular curves, then a curvature comparison argument shows that the metric angle is well-defined and equals the Riemannian angle between the velocity vectors $\dot\xi(0)$ and $\dot\eta(0)$. But what if the Riemannian metric is only $C^1$, or even $C^0$? More precisely: - -Let $X$ be a smooth manifold with a $C^1$ Riemannian metric. Does the metric angle exist for every pair of $C^1$ regular curves $\xi,\eta$? -If not, does every $C^1$ regular curve have a well-defined angle with itself? (Of course, this angle is zero if it exists). -If the answer is affirmative, what about $C^0$ Riemannian metrics? - -Remark. -The limit in the definition trivially exists (even in the $C^0$ case) if the ratio $t/t'$ is bounded away from zero and infinity as $t$ and $t'$ go to zero. The trouble is with very "thin" triangles $p,\xi(t),\eta(t')$ where $t\ll t'$ or vise versa. - -REPLY [8 votes]: To Deane Yang, concerning the smoothness of geodesics. These issues have been studied by Hartman -and his coauthors more than 40 years ago. Together with Calabi he claimed to prove -that in a $C^{\alpha}$ continuous metric all geodesics are uniformly $C^{1,\alpha}$. -(I do not have mathscinet access now, but I think the paper was called -"On the smoothness of isometries", or something like that). The paper is not quite correct, -the computations prove that geodesics are only $C^{1,\alpha /2}$ - (this was observed at some point by Reshetnyak). The optimal regularity appears in my paper with Asli Yaman -"On Hoelder continuous Riemannian and Finsler metrics" (I apologize for citing my own work). - The starting point of all these computations is a metric characterization of $C^{1,\alpha}$ -curves in metric terms (in terms of the distance of the midpoint of a pair of points on the curve and the image of the corresponding midpoint of the interval of the definition). It is contained in the work of Hartman, but is probably much older. -To Sergei Ivanov: The questions of extendibility of geodesics have also been studied by Hartman. I again do not remember the correct reference, but can check it on Monday. I think he constructs example of even $C^{1,\alpha}$ Riemannian metrics, for any $\alpha <1$, that do not have extendible geodesics. (For $\alpha =1$, the metric has curvature bounded from both sides). -Now the main point. I am very sorry and have to apologize for my previous short posting. - I slightly misunderstood your question. But I hope that my answer was still -correct. Is the following more detailed explanation correct? - -We assume that $t'>>t$ and want to prove that -$d(\xi (t), \eta (t') )= -d(\eta (t'), p) - cos (\alpha ) t + o(t)$. Here $o(t) /t \to 0$ and $\alpha$ is -the "usual Euclidean " angle between $\xi$ and $\eta$. - -The statement about the uniform "smoothness" of geodesics, show that it suffices to prove the claim in the case, when $\eta$ and $\xi$ are geodesics (since the "usual" angles -between $\eta$ and any geodesic between $\eta (t')$ and $p$ goes to $0$). -Going on the geodesic $\eta$ first to the point $\eta (rt)$, with large, but fixed $r$, and then to $\xi (t)$, one obtains the right upper bound for $d (\eta (t'),p)$. -In a "general" metric space it would be now difficult to obtain the right lower bound -of $d(\eta (t'),p)$. One could do it, if one knew that the geodesic $\eta$ was extendible beyond $p$. Then one could use a traingle inequality and the distance from $\xi (t)$ and $\eta (-rt)$. I hope I understood your question correctly and this was what you had in mind. -Here one can use another trick to obtain a lower bound. Observe namely, that from the -"smoothness" of geodesics the "usual" angle between $\xi$ and any geodesic from -$\eta (t') $ to $\xi (t)$ is very close to $\alpha$ (goes to $\alpha$ , with $t' \to 0$). -Now we just look at the triangle $\xi (t),p, \eta (t')$ from $\xi (t)$ and not from $p$. -The inequality obtained in point 3. above gives us the right upper bound of $d(\eta (t'),p)$. - -Just a side remark. Your question is very closely related to the question about the -correctness of the first formula of variation. I have thought about that question -and proved that the formula holds true in Riemannian manifolds with Hoelder continuous metrics in "Differentiation in metric spaces". I just hope that the above proof is correct. Other wise the corresponding statement in my paper is wrong as well.<|endoftext|> -TITLE: Is every (Artin/DM) algebraic stack fibered in sets an algebraic space? -QUESTION [7 upvotes]: If $X$ is an algebraic stack fibered in sets (and therefore essentially a sheaf), is it an algebraic space? It seems conceivable that at least when $X$ is Deligne-Mumford, it is actually an algebraic space. However, when $X$ is an Artin stack, it is only required to have an atlas of smooth maps of affines, so it seems less likely that this would be the case. - -REPLY [12 votes]: Yes. The criterion for an Artin stack to be Deligne-Mumford is that it should have unramified diagonal (this is somewhere in Laumon Moret Bailly, I don't have it here). If the stack is fibered in sets, the diagonal is a monomorphism, and a monomorphism is certainly unramified.<|endoftext|> -TITLE: Elementary proof of Nakayama's lemma? -QUESTION [27 upvotes]: Nakayama's lemma is as follows: - -Let $A$ be a ring, and $\frak{a}$ an ideal such that $\frak{a}$ is contained in every maximal ideal. Let $M$ be a finitely generated $A$-module. Then if $\frak{a}$$M=M$, we have that $M = 0$. - -Most proofs of this result that I've seen in books use some non-trivial linear algebra results (like Cramer's rule), and I had come to believe that these were certainly necessary. However, in Lang's Algebraic Number Theory book, I came across a quick proof using only the definitions and induction. I felt initially like something must be wrong--I thought perhaps the proof is simpler because Lang is assuming throughout that all rings are integral domains, but he doesn't use this in the proof he gives, as far as I can see. -Here is the proof, verbatim: We do induction on the number of generators of $M$. Say M is generated by $w_1, \cdots, w_m$. There exists an expression $$w_1 = a_1w_1 + \cdots + a_mw_m$$ with $a_i \in \frak{a}$. Hence $$(1-a_1)w_1 = a_2w_2 + \cdots +a_mw_m$$ If $(1-a_1)$ is not a unit in A, then it is contained in a maximal ideal $\frak{p}$. Since $a_1 \in \frak{p}$ by hypothesis, we have a contradiction. Hence $1-a_1$ is a unit, and dividing by it shows that $M$ can be generated by $m-1$ elements, thereby concluding the proof. -Is the fact that $A$ is assumed to be a domain being smuggled in here in some way that I missed? Or is this really an elementary proof of Nakayama's lemma, in full generality? - -REPLY [5 votes]: I think the following proof is valid and avoids both determinants and maximal ideals. The cost is induction over all $A$-modules generated by $n$ elements. -Nakayama. Let $J$ be the Jacobson radical. Let $M$ be a finitely generated $A$-module satisfying $M=JM$. Then $M=0$. -Proof. Induction on size of generating set. If $M$ is generated by zero elements then it is zero. Assume the assertion holds for modules generated by $n-1$ elements and let $M$ be generated by $n$ elements $m_1,\dots ,m_n$. Then $m_n=\sum _{i=1}^n \varepsilon _im_i$ with $\varepsilon _i\in J$ so $(1-\varepsilon _n)m_n=\sum_{i=1}^{n-1}m_i$. Dividing by the unit $1-\varepsilon _n$ we obtain $m_n$ as a linear combination of $m_1,\dots ,m_{n-1}$. Proceeding in this fashion we find $M=0$.<|endoftext|> -TITLE: How many collections of subsets of {1,2,...,n} are closed under the superset operation? -QUESTION [8 upvotes]: Say that I have the set $[n] = \{1,2,...,n\}$ and a collection $\mathcal{C} = \{ S_1, S_2, ..., S_k \}$ of subsets of $[n]$. Say that $\mathcal{C}$ is valid if it is closed under the superset operation; i.e., if $(S \in \mathcal{C} \wedge S \subseteq S' \subseteq [n]) \implies S' \in \mathcal{C}$. How many valid collections $\mathcal{C}$ are there, as a function of $n$? -Without the requirement to be closed under superset, the question is easier. There are $2^n$ subsets of $[n]$, and so there are $2^{2^n}$ ways to choose which of them belong to the collection. But not all collections are valid; for instance, if $n=2$, the valid collections are -$\mathcal{C} = \{ \emptyset, \{ 1 \} , \{ 2 \}, \{ 1,2 \} \}$, -$\mathcal{C} = \{ \{ 1 \} , \{ 2 \}, \{ 1,2 \} \}$, -$\mathcal{C} = \{ \{ 1 \}, \{ 1,2 \} \}$, -$\mathcal{C} = \{ \{ 2 \}, \{ 1,2 \} \}$, -$\mathcal{C} = \{ \{ 1,2 \} \}$, and -$\mathcal{C} = \{ \}$. -So rather than the answer being $2^{2^2} = 16$, there are only 6 valid collections. -Thank you in advance. - -REPLY [15 votes]: This is a famous problem known as the "Dedekind Problem", and it was posed by Dedekind in 1897. The Wikipedea article has some information. -There have been remarkable progress on understanding the asymptotic value of M(n) the number of antichains of sets from {1,2,...,n}. (This paper of Kleitman gives some of the history, and this paper by Kahn gives an updated history.) While $2^{{n}\choose {n/2}}$ is an obvious lower bound there is a beatiful 1966 proof by Hansel for the upperbound $3^{{n}\choose {n/2}}$. -Kleitman & Markowsky (1975). gave the asymptotic behavior of $\log M(n)$. They showed that $log M(n)$ behave asumptotically like ${{n} \choose {n/2}}$. The paper by Kahn that we already mentioned gives a simpler entropy based proof. -Amazingly, the asymptotic behavior of M(n) itself was discovered as well. -in 1981, Korshunov , using an extremely complicated approach, gave -asymptotics for M(n) itself. Simpler, though still difficult, arguments for Korshunov's -and some related results were later given by Sapozhenko. (See eg this book.) -Sapozhenko's method turned out to be very important, e.g., in the result by Galvin and Kahn on the threshold behavior of the d-dimensional hard core model.<|endoftext|> -TITLE: covering a separable metric space by small balls -QUESTION [15 upvotes]: Let $(X,d)$ be a separable metric space. Can $X$ always be covered by a sequence of balls $B(x_i,r_i) (i=1,2,\dots)$ s.t. radii $r_i$ tend to 0? - -REPLY [14 votes]: The answer is no for the Banach space $c_0$. Suppose $B(x_i,r_i)$ is a sequence of balls with $r_i\to 0$ and WLOG $x_i$ is supported in $[1,N_i]$ with -$N_1 -TITLE: Optimal packing of spheres tangent to a central sphere -QUESTION [5 upvotes]: Please consider a central, ordinary 2-sphere $S_1$, of some radius $r_1$, and a second ordinary sphere, $S_2$, of radius $r_2$, where $r_2 \leq r_1$. -My question concerns optimal values for the number of spheres of type $S_2$ that can be packed in three-dimensional space so that they are non-overlapping and tangent to $S_1$. Is there an analytical result for optimal packing a function of the ${r_2 \over r_1}$, or are there subsets of cases that are solved with methods beyond something like simulating annealing? Does it simplify the problem to apply the further restriction that $r_2 << r_1$? -I've been having trouble finding answers with a literature search, particularly for the latter situation where $r_2 << r_1$, and I appreciate everyone's time. - -REPLY [3 votes]: Assuming w.l.o.g. that $r_1=1$, one can guess the asymptotic behaviour of the kissing number as $r_2\to 0$ as a function $f(r_2)$. Consider a radial projection of the smaller balls onto the surface of the large (closed) ball (where the projection is w.r.t. the centre of the large ball). As $r_2\to 0$, this projection will locally approximate a packing of disks in the plane, and the density of the projection of an optimal packing should approximate the density of an optimal packing of unit disks in the plane. This density is well known to be $\frac{\pi}{2\sqrt{3}}$. Since the projection of each small ball covers a surface area of approximately $\pi r_{\kern-1pt2}^2$ of the unit ball, and the unit ball has surface area $4\pi$, we should have for very small $r_2$ -$$ -f(r_2)\pi r_{\kern-1pt2}^2 \approx \frac{\pi}{2\sqrt{3}}4\pi~, -$$ -or -$$ -\lim_{r_2\to 0} ~f(r_2) = \frac{2\pi}{\sqrt{3}\cdot r_{\kern-1pt2}^{2}}~. -$$<|endoftext|> -TITLE: explicit diffeomorphim between open simplex and open ball -QUESTION [22 upvotes]: What's a good reference e.g. textbook for the fact that the open n-ball and the open n-simplex are diffeomorphic? - -REPLY [8 votes]: I'd recommend taking a look at the book by Brocker and Janich, which discusses diffeomorphisms between star-shaped domains by defining a flow along rays from the star point. This might be an exercise in the book, rather than a theorem. (I think I have the book on my desk at school, and I'll try to give a more precise reference tomorrow.)<|endoftext|> -TITLE: Groups with no perfect subgroups -- terminology? -QUESTION [5 upvotes]: Finite groups are solvable if they have no nontrivial perfect subgroup. But I am sure that for infinite groups, the two notions diverge. Is there standard terminology for groups with no perfect subgroups? - -REPLY [11 votes]: In the infinite case, there is a close notion of "locally indicable group", i.e. a group where every finitely generated subgroup maps onto $\mathbb Z$ (see, for example, this paper). Locally indicable groups are left (right) orderable, hence important. Note that in that notion, not all subgroups are considered but only finitely generated, and "non-perfect" is replaced by a stronger property "maps onto $\mathbb Z$". But in the finite case all subgroups are finitely generated, and "maps onto $\mathbb Z$" is an infinite analog of "maps onto a finite cyclic group" (= non-perfect). So "locally indicable" is possibly the infinite analog of the property you consider. -Update: The groups without perfect subgroups are called hypoabelian. See this text.<|endoftext|> -TITLE: Is the distance function from a point to the Mandelbrot set computable? -QUESTION [5 upvotes]: There is at least one result saying that the Mandelbrot set is undecidable, and there might be more, but I think it (or they all) use real computation rather than Turing machines. This makes some sense, as $\mathbb{C}$ is connected, so the only decidable subsets of it are $\{\}$ and $\mathbb{C}$ itself. However, I've been reading about reverse mathematics, and I was wondering if the set is computable in the representations used. The Mandelbrot set is easily seen to be the complement of an effectively open set, and I'd guess asking whether it is separably closed would run into the open problem of are the hyperbolic components dense. This still leaves located closed, so the question follows: -Let $f : \mathbb{Q}[i] \to \mathbb{R} \hspace{.05 in}$ be given by $f(q) := \operatorname{min}(\{d(q,z) : z\in (\operatorname{Mandelbrot set})\})$. Is $f$ computable? - -REPLY [4 votes]: I just want to add one small piece of information to Bjørn's answer. The distance function for the Mandelbrot set $M$ is not known to be computable as a map into the reals, but it is computable as a map into the upper reals. This means that we can calculate upper bounds for the distance: given $z \in \mathbb{C}$, the set of all rational upper bounds for $d(x,M)$ is computably enumerable. In many cases (such as drawing algorithms) this is already quite useful.<|endoftext|> -TITLE: Is there a generalization of Floquet theory to elliptic functions? -QUESTION [5 upvotes]: Hi, -Consider a system of linear differential equations -$$ -{d f \over dz} = A(z) f, -$$ -where $A(z)$ is a matrix-function. If $z \in \mathbb{R}$ and the function is periodic $A(z) = A(z + T)$, Floquet theorem applies. -I am curious to know if there exists a generalization of Floquet theorem to the case, where $z \in \mathbb{C}$ -and $A(z)$ is a doubly-periodic elliptic function of $z$. -Thanks, -Victor - -REPLY [2 votes]: Another, far reaching, aspect of Floquet theory in differential equations -$$a_0(z)f^{(n)}(z)+\cdots+a_nf(z)=0$$ -with holomorphic coefficients is Fuchs theory of monodromy, where the leading coefficient $a_0$ has a zeros at $z_0$. You cannot solve a Cauchy problem at $z_0$, but you can solve it in a pointed disk $D\setminus z_0$. When you follow a circle around $z_0$, the coefficients look periodic.<|endoftext|> -TITLE: Injection of Ext into H^2 -QUESTION [9 upvotes]: Let $G$ be an abelian group, $A$ a trivial $G$-module. We know that $\text{Ext}(G,A)$ classifies abelian extensions of $G$ by $A$, whereas $H^2(G,A)$ classifies central extensions of $G$ by $A$. So we have a canonical inclusion $\text{Ext}(G,A)\hookrightarrow H^2(G,A)$. Is there some naturally arising exact sequence/spectral sequence which realizes this injection? -Usually this kind of thing can be explained by constructing a clever short exact sequence, but here I have no idea how one might compare $R^1\text{Hom}_\mathbb{Z}(G,\underline{\quad})$ with $R^2\text{Hom}_G(\mathbb{Z},\underline{\quad})$. - -REPLY [13 votes]: You get a description from the universal coefficient theorem which gives a (split) exact sequence -$$ -0\to \mathrm{Ext}(H_1(G),A) \to H^2(G,A) \to \mathrm{Hom}(H_2(G),A) \to 0 -$$ -and the fact that $H_1(G)=G$. We have that $H_2(G)=\Lambda^2G$ and the map $H^2(G,A) \to \mathrm{Hom}(H_2(G),A)$ associates to an extension its commutator map.<|endoftext|> -TITLE: Conjugacy classes with elliptic limit points -QUESTION [5 upvotes]: Let $G$ be a reductive algebraic group over $\mathbb R$ and $K$ a maximal compact subgroup. Then we refer to the conjugacy class in $G$ of some $k \in K$ as an elliptic conjugacy class. -Question: Can one characterizes those conjugacy classes in $G$ which contain an elliptic conjugacy class in their closure? -(For $G = GL_n(\mathbb R)$ they are characterized by the fact that all eigenvalues are of modulus one, if I a not mistaken.) - -REPLY [2 votes]: For g in G, write g=gsgu as its Jordan decomposition into semisimple and unipotent parts. I claim that the closure of the conjugacy class of g contains an elliptic element if and only if gs is elliptic. -Let us first suppose that gs is not elliptic. Choose an embedding of G into GLn(ℂ). Then by our assumption, gs has an eigenvalue of norm greater than one, let λ be the absolute value of such an eigenvalue. Suppose for want of contradiction that the conjugacy class of gs contained an elliptic element a in its closure. WLOG a is in the special unitary group SUn. Let h be in the conjugacy class of gs. Then h has an eigenvalue of absolute value λ. Letting v be an eigenvector, we see that |(h-a)v| is at least (λ-1)|v|, so |h-a|≥λ-1, a contradiction. -Now suppose that gs is elliptic. We may replace G by the centraliser of gs is G, which is also reductive. So WLOG, gs is central in G. Now the Zariski closure of the group generated by gu is a one-dimensional unipotent subgroup of G. Let E be a non-zero element in its lie algebra. This is a nilpotent element. Then by the Jacobson-Morozov theorem, we can extend E to a sl2 triple E,F,H in Lie(G). Now consider conjugation by elements of the form exp(tH) with t real. This shows that gs is in the closure of the conjugacy class of g, and we're done.<|endoftext|> -TITLE: geometric Langlands for GL(1) -QUESTION [11 upvotes]: I should tell you from the beggining that I don't know almost anything about what I'm going to ask/write. -Let $X$ be a smooth projective (e.g. elliptic) curve over a finite field $\mathbb{F}_q$. Then (by geometric Langlands for $GL(1)$ I've heard) we have some sort of correspondence between characters of -$\pi_1(X)$ and characters of the group $Pic(X)$. (*) -Here $\pi_1(X)$ is the algebraic fundamental group, i.e. the semidirect product of -$\hat{\mathbb{Z}}$ and $\pi_1^{et}(\overline{X})$. -The question is: how to prove/see (*)? do you know any good reference for it? - -REPLY [2 votes]: You start with a local system $\mathcal L$ on $X$. By taking symmetric powers, you get a local -system $Sym^n \mathcal L$ on $Sym^n X$ for any $n$. Now if you choose $n\geq g,$ -then the natural map $Sym^n X \to Pic^n(X)$ is surjective, and the fibres are all projective spaces, in particular simply connected. Thus $Sym^n \mathcal L$ descends to a local system on $Pic^n(X)$. This is essentially the desired correspondence. [Sorry; after writing this I realized that it is covered by Peter Toth's answer.]<|endoftext|> -TITLE: Computation of inverses modulo p followup -QUESTION [5 upvotes]: In responding to -Fast computation of multiplicative inverse modulo q -I mentioned an algorithm for computing the inverse of $a \mod p$ different from the extended Euclidean algorithm, hoping that someone could tell me how its speed stacks up against other algorithms. Since no one did, I'm asking directly if someone can tell me. -To compute the inverse of a modulo p, you can run the Euclidean algorithm starting with $p^2$ and $ap+1$, comparing the size of each remainder with $p$. The first remainder less than $p$ that appears will be an inverse for $a \mod p$. -This will always take either the same number of steps to reach the inverse as it takes to reach $\gcd(a,p)=1$ using the Euclidean algorithm with $a$ and $p$, or else one additional step, depending on whether the least positive residue of an inverse of $a$ is greater than $p/2$ or less than $p/2$. -Thus, it requires approximately the same number of computations as the first half of the extended Euclidean algorithm (albeit with bigger numbers initially), excepting an extra comparison with $p$ at each step. -Question: How does the speed of this algorithm compare to others? -Aside: Pedagogically, this is nice since the second half of the extended Euclidean algorithm is the one my students tend to mess up. However, assuming our ultimate goal is for students to understand why they're doing what they're doing, perhaps the extended Euclidean algorithm is preferable. - -REPLY [3 votes]: Here is a tidy way to solve $ax+by=gcd(a,b)$. Start with the matrix -$$\left(\begin{array}{ccc} - 1&0&a\\ - 0&1&b\\ - \end{array}\right) -$$ -Suppose $a\ge b$. Then replace row 1 by row 1 minus $t$ times row 2, where $t=\lfloor a/b\rfloor$. Repeat this operation until the last entry in one of the two rows is zero. If the other row is $x,y,d$ then $ax+by=d$ and $d=gcd(a,b)$. This is very simple to program, and avoids the back-substitutions that students find confusing. -Most of the speedups of the Euclidean Algorithm described in Knuth's Art of Computer Programming v2 work here also.<|endoftext|> -TITLE: David Gale's subset take-away game -QUESTION [18 upvotes]: I learned of this problem through Su Gao, who heard of it years ago while a post-doc at Caltech. David Gale introduced this game in the 70s, I believe. I am only aware of two references in print: - -Richard K. Guy. Unsolved problems in combinatorial games. In Games of No Chance, (R. J. Nowakowski ed.) MSRI Publications 29, Cambridge University Press, 1996, pp. 475-491. -J. Daniel Christensen and Mark Tilford. David Gale's Subset Take-Away Game. The American Mathematical Monthly, Vol. 104, No. 8 (Oct., 1997), pp. 762-766; jstor, arxiv. - -The game depends on a finite set $A$, and is played by two players I and II that alternate, with I playing first. Each move is a subset of $A$. We are given a pool of available subsets, and each move should be a set in the pool. At the beginning, the pool consists of all subsets of $A$ except for $A$ and the empty set. Whenever a set is used in a move, it and all its supersets are removed from the pool. The game ends when the pool is empty, and the player that was to move and cannot (because there are no moves left) loses. -For example, if $|A|\le 2$, then II always wins. If $|A|\le 3$, there is an easy winning strategy for II. To see this, list the elements of $A$ as $a,b,c$ in such a way that the first move of I is either $\{a\}$ or $\{b,c\}$, and have II respond the other one. This means that when I comes to move again, only $\{b\}$ and $\{c\}$ remain, and II wins. -One can also check by hand that II has a winning strategy if $|A|\le 5$, and Christensen-Tilford showed the same when $|A|=6$, although they did not include the code used for their computer verification. -The conjecture is that II always has a winning strategy. -My question is whether there have been any further developments towards establishing this, or any additional references. Suggestions or related ideas are also welcome. - -REPLY [7 votes]: The conjecture that this game is always a second player win has recently been disproved by Brouwer and Christensen (already for $n=7$), https://arxiv.org/abs/1702.03018.<|endoftext|> -TITLE: Group which "resembles" the free product of a cyclic group of order two and a cyclic group of order three, but isn't. -QUESTION [18 upvotes]: Can someone give an explicit example of a group with two generators $a$, $b$, such that $a^2 = b^3 = 1$ and $a b$ has infinite order, but which is not isomorphic to the free product of $\mathbb{Z}_2$ and $\mathbb{Z}_3$? - -REPLY [18 votes]: It is straightforward to calculate that the commutator subgroup $G' = D$ of $G = \langle a,b \mid a^2, b^3 \rangle$ is a free group on the generators $x=bab^{-1}a$, $y=b^{-1}aba$, where $|G:D|=6$. -Now $(ab)^6$ is equal to the commutator $x^{-1}yxy^{-1}$, which lies in $D'$ but not in $D''$, so if we add any nontrivial element of $D''$ as an extra relator of $G$, then we will get an example with the required property.<|endoftext|> -TITLE: About isogeny theorem for elliptic curves -QUESTION [18 upvotes]: $K$ a number field, $G_K$ its Galois group, $E_1, E_2$ two elliptic curves defined over $K$. The isogeny theorem says that if for some prime number $\ell$, The Tate modules (tensored with $\mathbb{Q}$) $V_{\ell}(E_1)$ is isomorphic to $V_{\ell}(E_2)$ as Galois modules. Then these two elliptic curves are isogenous. -My question is, when are these two curves isomorphic? Namely, what more invariants are needed to fully characterize the elliptic curve (besides the Tate modules). Thanks! - -REPLY [18 votes]: If all Tate modules (i.e., for all $\ell$) are isomorphic then they differ by -the twist by a locally free rank $1$ module over the endomorphism ring of one of -them. This is true for all abelian varieties but for elliptic curves we only -have two kinds of possibilities for the endomorphism ring; either $\mathbb Z$ or -an order in an imaginary quadratic field. In the first case there is only one -rank $1$ module so the curves are isomorphic. In the case of an order we get -that the numbe of twists is a class number. -Addendum: Concretely, we have that $\mathrm{Hom}(E_1,E_2)$ is a rank $1$ projective module over $\mathrm{End}(E_1,E_1)$ (under the assumption that the Tate modules are isomorphic) and then $E_2$ is isomorphic to $\mathrm{Hom}(E_1,E_2)\bigotimes_{\mathrm{End}(E_1)}E_1$ (the tensor product is defined by presenting $\mathrm{Hom}(E_1,E_2)$ as the kernel of an idempotent $n\times n$-matrix with entries in $\mathrm{End}(E_1)$ and $E_2$ is the kernel of the same matrix acting on $E_1^n$. Hence, given $E_1$ $E_2$ is determined by $\mathrm{Hom}(E_1,E_2)$ and every rank $1$ projective module appears in this way. -Addendum 1: Note that I was talking here about the $\mathbb Z_\ell$ (and not $\mathbb Q_\ell$ Tate modules. You can divide up the classification of elliptic curves in two stages: First you see if the $V_\ell$ are isomorphic (and there it is enough to look at a single $\ell$). If they are, then the curves are isogenous. Then the second step is to look within an isogeny class and try to classify those curves. -The way I am talking about here goes directly to looking at the $T_\ell$ for all $\ell$. If they are non-isomorphic (for even a single $\ell$ then the curves are not isomorphic and if they are isomorphic for all $\ell$ they still may or may not be isomorphic, the difference between them is given by a rank $1$ locally free module over the endomorphism ring. In any case they are certainly isogenous. These can be seen a priori as if all $T_\ell$ are isomorphic so are all the $V_\ell$ but also a posteriori essentially because a rank $1$ locally free module becomes free of rank $1$ when tensored with $\mathbb Q$. -Of course the a posteriori argument is in some sense cheating because the way you show that the curves differ by a twist by a rank $1$ locally free module is to use the precise form of the Tate conjecture: -$$ -\mathrm{Hom}(E_1,E_2)\bigotimes \mathbb Z_\ell = \mathrm{Hom}_{\mathcal G}(T_\ell(E_1),T_\ell(E_2)) -$$ -which for a single $\ell$ gives the isogeny. -Note also that the situation is similar (not by chance) to the case of CM-curves. If we look at CM-elliptic curves with a fixed endomorphism ring, then algebraically they can not be put into bijection with the elements of the class group of the endomorphism ring (though they can analytically), you have to fix one elliptic curve to get a bijection.<|endoftext|> -TITLE: Graphs with the same chromatic symmetric function -QUESTION [10 upvotes]: Does anyone know more examples of two nonisomorphic connected graphs with the same chromatic symmetric function? The only pair I know is the one in Stanley's paper on c.s.f.'s [http://math.mit.edu/~rstan/pubs/pubfiles/100.pdf; p.5 of the PDF file]. I would especially like to have an example in which at least one of the two graphs is triangle-free. - -REPLY [5 votes]: This is quite an old post of mine. Subsequently, Rosa Orellana and Geoffrey Scott gave an infinite family of pairs of unicyclic graphs with the same chromatic symmetric functions [Discrete Math. 320 (2014), 1–14]. That's the best answer to this question that I know of. Stanley's original question for trees is still (frustratingly) open, so far as I know.<|endoftext|> -TITLE: Spin structures on the Grassmannians -QUESTION [5 upvotes]: I am trying to understand spin structures and am looking at the specific case of complex projective space (viewed as the quotient $SU(N)/U(N-1)$) and more generally the Grassmannians (viewed as the quotient $SU(N)/(U(N-k) \times U(k))$. My questions are as follows: -(1) For what values of $N$ does complex projective $N$-space have a spin structure? -(2) For these values, is the canonical spinor bundle equivarient with respect to the $U(N-1)$ action? -(3) If so, what is the associated representation of $U(N-1)$? -(4) All of the above for the Grassmannians? - -REPLY [9 votes]: I think there is a slight mistake in the formulation of the question. $\mathbb{CP}^n$ is the homogeneous space $U(n+1)/(U(n) \times U(1))=SU(n+1)/G$ with $G= SU(n+1) \cap (U(n) \times U(1))$. The right formulation of question (2) is: is the spin structure on $\mathbb{CP}^n$ (for odd $n$, there is unique spin structure on $\mathbb{CP}^n$, see Charles Siegel's answer) $U(n+1)$-equivariant? -The answer is no, for a very elementary reason: if the spin structure were $U(n+1)$-equivariant, then it certainly were $U(n)$-equivariant, -where $U(n)$ embeds into the product in the standard way. But the $U(n)$-action on $\mathbb{CP}^n$ has a fixed point and it is not too hard to see that the $U(n)$-representation on the tangent space to that fixed point is isomorphic to the standard representation of $U(n)$ on $\mathbb{C}^n$. So if the spin structure were equivariant, then the fixed-point representation has to be spin, which is of course wrong. -You can ask the same question for spheres (is the spin structure on $S^n$ $SO(n+1)$-equivariant), and the answer is again no. But the spin structure on $S^n$ is $Spin(n+1)$-equivariant; likewise the spin structure on $\mathbb{CP}^n$ will be equivariant under the double cover of $U(n)$. -What you can guess from these two examples is that the question has something to do with double covers (alias central extensions of your group by $\mathbb{Z}/2$). Here is the precise relation: $M$ a spin manifold, $s$ a spin structure (viewed as a double cover of the frame bundle of $M$), $G$ a topological group acting on M by diffeomorphisms. The spin structure defines a new group $G'$ and a surjective homomorphism $p:G' \to G$ with kernel. $G'$ consists of pairs $(f,t)$, $f \in G$ and $t$ is an isomorphism of spin structures $f^* s \to s$. The spin structure is equivariant under $G'$, and it is $G$-equivariant iff there is $q:G \to G'$, $pq=\operatorname{id}$. If $G$ is a simply-connected topological group, this is always the case, but otherwise not in general. -This discussion implies that the spin structure on $\mathbb{CP}^n$ is indeed $SU(n+1)$-equivariant, if it exists. Grassmannians and other homogeneous spaces can be dealt with in the same way.<|endoftext|> -TITLE: How to solve a generalization of the Coupon Collector's problem -QUESTION [9 upvotes]: The coupon collector's problem is a problem in probability theory that states the following (from wikipedia): - -Suppose that there are $n$ coupons, from which coupons are being collected with replacement. What is the probability that more than $t$ sample trials are needed to collect all $n$ coupons? - -A generalization of this problem was proposed by Newmann & Shepp, by requiring that $k$ samples of each coupon be collected. The answer to this is known. -I, however, need to calculate the answer to an even further generalization, which is: - -How many sample trials are needed to collect at least $k$ coupons of $m$ different types? - -Any help or a point in the right direction would be greatly appreciated. -(Edit#3): removed the misleading example. -(Edit #2): This problem can be stated simply as a balls-and-bins problem. If we have $n$ bins, how many balls need to be thrown so that at least $m$ bins have at least $k$ balls? - -REPLY [3 votes]: This is given in Thm 3.1 of http://algo.inria.fr/flajolet/Publications/FlGaTh92.pdf<|endoftext|> -TITLE: A balls-and-colours problem -QUESTION [21 upvotes]: A box contains n balls coloured 1 to n. Each time you pick two balls from the bin - the first ball and the second ball, both uniformly at random and you paint the second ball with the colour of the first. Then, you put both balls back into the box. What is the expected number of times this needs to be done so that all balls in the box have the same colour? -Answer (Spoiler put through rot13.com): Gur fdhner bs gur dhnagvgl gung vf bar yrff guna a. -Someone asked me this puzzle some four years back. I thought about it on and off but I have not been able to solve it. I was told the answer though and I suspect there may be an elegant solution. -Thanks. - -REPLY [2 votes]: Just wish to add some sense to $f(k)$... -Let: -$X_{i}$ - the number of balls of the color $i$ at time $t=0$. -$A_{i}$ - the event that in the end all the balls are of the color $i$. -Using this notation: -$E(T|X_{1}=\lambda_{1},\ldots ,X_{n}=\lambda_{n}) = -\sum E(1_{A_{i}}T|X_{1}=\lambda_{1},\ldots,X_{n}=\lambda_{n})$. -But $E(1_{A_{i}}T|X_{1}=\lambda_{1},\ldots,X_{n}=\lambda_{n})$ depends only on $\lambda_{i}$ and is denoted by $f(\lambda_{i})$. -Using "first step" analysis we get: -$E(1_{A_{i}}T|X_{i}=k)=dE(1_{A_{i}}(T+1)|X_{i}=k+1)+ -dE(1_{A_{i}}(T+1)|X_{i}=k-1)+(1-2d)E(1_{A_{i}}(T+1)|X_{i}=k)$ -where $d=\frac{k(n-k)}{n(n-1)}$ -One may compute (using Doob's theorem or otherwise) that $E(1_{A_{i}}|X_{i}=k)=\frac{k}{n}$. -And using that we obtain David Speyer's equation (**) for $f(k)$.<|endoftext|> -TITLE: local to global Galois representation -QUESTION [7 upvotes]: Let $\rho_p : \mbox{Gal}(\overline{\mathbb{Q}}_p / {\mathbb{Q}_p}) \to \mbox{GL}_n(\mathbb{Q}_p)$ be a de Rham $p$-adic representation. Can one find a representation $\rho : \mbox{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) \to \mbox{GL}_n(\mathbb{Q}_p)$ such that $\rho$ is geometric (in the sense of Fontaine-Mazur) and such that the restriction of $\rho$ to $\mbox{Gal}(\overline{\mathbb{Q}}_p / {\mathbb{Q}_p})$ is $\rho_p$ ? - -REPLY [7 votes]: No. There are uncountably many unramified representations from the local Galois group to -$\mathbb{Q}^{\times}_p$, since Frobenius can be sent to anything in $\mathbb{Z}^{\times}_p$. -However, there are only countably many global representations of this form, since, by class field theory, they all factor through the Galois group of a (finite) cyclotomic field. -The general picture is pretty much the same --- the local Galois representations form large $p$-adic analytic families, yet, assuming the Fontaine-Mazur conjecture, there are only countably many geometric representations.<|endoftext|> -TITLE: Does any textbook take this approach to the isomorphism theorems? -QUESTION [21 upvotes]: Below, I present an outline of a proof of the first isomorphism theorem for groups. This is how I usually think of the first isomorphism theorem for ______________, but groups will get the points across. My question is whether there is any textbook which takes this basic approach. -Let $\pi: C \to L$ be a surjection. I like to think of the elements $l$ of $L$ as labels for the preimages $\pi^{-1}(l)$. So I call any surjection a "labeling function", and the codomain a "set of labels". -If $f: C \to R$ is function such that $\pi(x) = \pi(y) \implies f(x) = f(y)$ I say that "$f$ respects the labeling of $\pi$". -Theorem 1: $f$ factors uniquely through $\pi$. -For example if $C$ is the set of all candy bars in a store, $L$ is the set of all "types" of candy bars (mars, trix, heath, ...), $R$ is $\mathbb{R}$ and $f$ assigns to each candy bar its price, then it is plain that $f$ respects $\pi$'s labeling, and so I might as well have just assigned a price to each type of candy bar, rather than pricing each individual candy bar. -Theorem 2: If $\pi_1$ and $\pi_2$ respect each other's labeling, then the induced maps between their codomains are inverses. Hence the codomains are isomorphic. -For example if $\pi_1$ gives the English label of a candy bar and $\pi_2$ gives the Spanish label, then there is an obvious bijection induced between the set of English and Spanish labels (In fact, this is the principle which allows us to learn other languages). -Theorem 3 If $C$, $L$, and $R$ are groups, and $\pi$ and $f$ are homomorphisms, then the induced map from $L$ to $R$ is also a homomorphism. -This is just a couple lines of manipulation. -Theorem 4 If two surjective group homomorphisms respect each others labeling, then their codomains are isomorphic. -This is really a corollary of theorem 2 and 3 -Observation: for a group homomorphism $\phi$, $\phi(a) = \phi(b) \Longleftrightarrow \phi(ab^{-1}) = 1$, i.e. $ab^{-1} \in Ker(\phi)$. Thus "$f$ respects $\pi$'s labeling" can be rephrased "$Ker(\pi) \subset Ker(f)$". -So a rephrasing of theorem 4 is -Theorem 4':Any two group surjective group homomoprhisms with the same kernel have isomorphic codomains -or observing that any homomorphism is a surjection onto its image, you can say -Theorem 4'': Any two group homomorphism with the same kernel have isomorphic images. -This is the first isomorphism theorem right? I have not mentioned normal subgroups, cosets, or factor groups. Those only come in as a construction to pick one representative of the collection of all images of homomorphisms having the same kernel, or when you want to start characterizing which subgroups of a group can be the kernel of some homomorphism. -I think that this very basic outline gets lost on students when they see the isomorphism theorem for the first time. Most textbooks defines normal subgroups, cosets, puts a group structure on the cosets, all without any kind of real motivation, and at the end the first isomorphism theorem seems kind of magical, and not at all natural. I know I thought it was magic when I first learned it. But the first isomorphism theorem does not seem surprising at all when you follow my sequence of theorems 1 through 4''. -So I ask: Is this approach taken by any textbook out there? - -REPLY [7 votes]: You might try pages 47-55 of Burris and Sankappanavar (pages 63--71 of the pdf document).<|endoftext|> -TITLE: No simple groups of order 720? -QUESTION [28 upvotes]: I think most students who first learn about (finite) groups, eventually learn about the possibility of classifying certain finite groups, and even showing certain finite groups of a given order can't be simple (I'm pretty sure every beginning algebra text has some exercises like this). Up to order 1000, I think there is one that is considered by far the most difficult: 720. -Does anyone know of a proof that there are no simple groups of order 720, which avoids showing, via contradiction, such a group would be $M_{10}$? [To clarify the avoids part, the proof sketched by Derek Holt here, while very nice, would not qualify.] -I should also disqualify the inevitable reference to Burnside's article on this very topic, which I am fairly confident is flawed (or, at the very least, incomplete). - -REPLY [15 votes]: How about if we follow Derek Holt's beautiful argument until we establish the following two facts? -1) $G$ has 10 Sylow 3-subgroups. -2) Let $P_3$ be a Sylow 3-subgroup of $G$. In the action of $G$ on $Syl_3(G)$, every nonidentity element of $P_3$ has cycle type (3,3,3,1). -Since $G$ is simple, the action on $Syl_3(G)$ embeds $G$ in $A_{10}$. In $A_{10}$, no element of type (3,3,3,1) commutes with an element of order two. Thus if $Q$ is a Sylow 2-subgroup of $N_G(P_3)$, no nonidentity element of $Q$ centralizes any nonidentity element of $P_3$. Since $|Q|=8$, it follows that all eight nonidentity elements of $P_3$ are conjugate in $N_G(P_3)$. Therefore, $G$ contains exactly one conjugacy class of elements of order three. Let $g_3$ be an element of this class. No element of order 5 in $A_{10}$ commutes with $g_3$, and it follows now that $|C_G(g_3)|=9$, so $G$ has 80 elements of order 3. -Since 5 does not divide $|N_G(P_3)|$, we see that every element of order 5 in $G$ has cycle type (5,5) in the action on $Syl_3(G)$. Using Sylow's Theorem (and the fact that $G$ is not $S_6$), we see that G has either 16 or 36 Sylow 5-subgroups. Let $P_5$ be a Sylow 5-subgroup of $G$. By Burnside transfer, we cannot have $|N_G(P_5)|=45$. Therefore, $|N_G(P_5)|=20$. -No element of order two in $A_{10}$ commutes with an element of cycle type (5,5). Therefore, $N_G(P_5)$ must induce all of $Aut(P_5)$, so all nonidentity elements of $P_5$ are conjugate in $N_G(P_5)$ and therefore, $G$ has one conjugacy class of elements of order 5. There are 144 such elements. Let $g_5$ be one such element. -Now let us consider the character table of $G$. By the arguments above, each of $g_3$ and $g_5$ are conjugate to all of their nonidentity powers, and it follows that all irreducible characters of G take integer values on $g_3$ and $g_5$. Moreover, these integers have absolute value at most two, as can be seen using the orthonormality conditions on the character table and the sizes of the given conjugacy classes. -Now $N_G(P_3)$ has one more conjugacy class than $Q$ (as defined above). It follows from this that the irreducible characters of $N_G(P_3)$ are those with kernel containing $P_3$ and one more, call it $\chi$. Using orthonormality conditions on the character table, we get $\chi(1)=8$, $\chi(g_3)=-1$ and $\chi(q)=0$ for all $q$ not of order 1 or 3. Induce $\chi$ up to $G$ to get a character $\psi$ that takes the value 80 on 1, -1 on the class of $g_3$ and 0 elsewhere. For any irreducible character $\eta$, the inner product $\langle\psi,\eta\rangle$ is -$$\frac{80(\eta(1)-\eta(g_3))}{720}$$. -It follows that $\eta(1)-\eta(g_3)$ is divisible by 9. -Now $N_G(P_5)=Z_5.Z_4$ has a character $\alpha$ such that $\alpha(1)=4$, $\alpha(g_5)=-1$ and $\alpha(q)=0$ if $q$ does not have order 1 or 5. Induce $\alpha$ up to $G$ to get $\beta$. Arguing as we did with $\psi$, we see that for every irreducible character $\eta$, $\eta(1)-\eta(g_5)$ is divisible by 5. -Now using basic facts about irreducible characters, we see that for any irreducible character $\eta$ of G, the triple $(\eta(1),\eta(g_3),\eta(g_5))$ is one of -$$(1,1,1),(8,-1,-2),(9,0,-1),(10,1,0),(16,-2,1),(18,0,-2) or (20,2,0)$$. -Any class function $\eta$ satisfying $\eta(1)=18$ and $\eta(g_5)=-2$ has norm larger than 1. If $\eta$ is a class function of norm 1 satisfying $\eta(1)=20$ and $\eta(g_3)=2$ then $\eta(q)=0$ for all q not of order 1 or 3. But then the inner product of $\eta$ and the trivial character is positive. Similarly, if $\eta$ has norm 1 and $(\eta(1),\eta(g_3),\eta(g_5))=(8,-1,-2)$ then $\eta$ is zero on all remaining classes and the inner product of $\eta$ and the trivial character is negative. We see now that all nontrivial irreducible characters of $G$ have degree 9, 10, or 16. We get a contradiction when trying to add up squares of these degrees to get 719. -This proof has a clear disadvantage of greater length when compared with the argument of Holt, as clarified beautifully by Greg Kuperberg. However, it has the advantage of alerting us to the important fact that, given a purported simple group $G$, it can be profitable to consider large subgroups of $G$ whose characters we understand, in particular those that are Frobenius groups.<|endoftext|> -TITLE: How to resolve a wedge product of vector bundles -QUESTION [7 upvotes]: Let $X$ be an algebraic variety. Consider an exact sequence -$$0\to A\to B\to C\to 0$$ -of vector bundles on $X$. I have seen in different papers the following type resolution of wedge product of $C$ (or $A$) -$$0\to S^kA\to S^{k-1}A\otimes B\to S^{k-2}A\otimes \wedge^2B\to \cdots\to \wedge^kB\to \wedge^k C\to 0.$$ -Question: does this resolution come from certain geometric context? Is there a proof which involves certain geometric aspects, for example, using projective bundles associated to the vector bundles? - -REPLY [10 votes]: The resolution in question is the $k$th graded piece of the symmetric algebra of the complex $A \rightarrow B$, where $A$ is considered to lie in even degrees (say degree $2$) and $B$ in odd degrees (say degree $1$). (Also, a kind of Koszul complex.) -The way I think of this: forgetting the differential $A \rightarrow B$ for a minute, we just have a graded vector bundle $A \oplus B$ with $A$ in degree $2$, $B$ in degree $1$. Since the symmetric algebra construction $\mathcal{S}$ takes sums to tensor products, we have $\mathcal{S}(A \oplus B) \simeq \mathcal{S}(A) \otimes \mathcal{S}(B)$. Since $A$ is in degree $2$, $S(A)=S(A)$ (commutative) and since $B$ is in degree $1$, $\mathcal{S}(B) \simeq \bigwedge(A)$ (the Koszul sign rule for graded tensor products makes this super-commutative). -Now you have to bring in the differential. To do this, it is useful to use the bialgera structure on the symmetric algebra construction, where the coproduct comes from applying $\mathcal{S}$ to the diagonal map $X \rightarrow X \oplus X$. To see how to get the differential out of this: given $S^{k-i}A \otimes \bigwedge^{i} B$, comultiply in the first factor to end up in $S^{k-i-1}A \otimes A \otimes \bigwedge^{i}B$, now apply the map $A \rightarrow B$ to the middle factor of $A$ to end up in $S^{k-i-1}A \otimes B \otimes \bigwedge^{i}B$, and finally, multiply the last two factors together to end up in $S^{k-i-1}A \otimes \bigwedge^{i+1}B$ -This is a standard construction and can be found for instance in the paper of Akin, Buchsbaum, Weyman on Schur complexes, or in the book of Weyman on Cohomology of vector bundles and syzygies. If people have other references, I'd be glad to know about them, since the above mentioned ones are from the point of view commutative algebra (which is for me non-optimal). -While the above construction is certainly useful in geometry, I'm afraid the above is not geometric in the way you were looking for.<|endoftext|> -TITLE: Consequence of equidistribution or not? -QUESTION [13 upvotes]: Let $\theta \not\in \mathbb{Q}$. We know that $(n\theta)_{n \geq 1}$ is equidistributed modulo 1. -Let $\epsilon_n = \mathrm{sign}\bigl(\sin(n\pi \theta)\bigr)$ and $S_N= \sum_{n=1}^N \epsilon_n$. -I'm looking for a "good" asymptotic bound for $|S_N|$ (not $|S_N|\leq N$ obviously). -It looks like for any $x>0$, we should have $S_N =o(n^x)$, or even better, that $(S_N)$ is bounded, but is it? - -REPLY [9 votes]: No, you cannot put any better bound than SN = o(N). There is a general technique, using the Baire category theorem of proving the existence of counterexamples to problems like this (which I discovered while trying to find a counterexample to a question by David Speyer, link). I see that Helge's answer is also pointing towards the same result. -First, for θ irrational, -$$ -S_N/N=\frac{1}{N}\sum_{n=1}^N1_{\{0< n\theta/2 <1/2{\rm\ (mod\ 1)}\}}-\frac{1}{N}\sum_{n=1}^N1_{\{1/2< n\theta/2 <1{\rm\ (mod\ 1)}\}} -$$ -By Weyl's equidistribution theorem, both sides on the right hand side tend to 1/2 and SN / N → 0, so SN = o(N). -It is not possible to do better than this. In fact, if f: ℕ → ℝ+ is any function satisfying liminf f(N) / N = 0 then there will be an uncountable dense set of irrational θ for which limsup SN / f(N) = ∞. In particular, using f(n) = nx for x < 1 rules out bounds such as Sn = O(nx). -In fact, we can find a set of such θ as an intersection of countably many open dense subsets of ℝ, so the Baire category theorem shows the existence of uncountably many counterexamples. -Let u(x) = 1{0≤[x/2]<1/2} - 1{1/2≤[x/2]<1} where [x] is the fractional part of x, and SN(θ) = Σn≤N u(nθ). Let UK be the set -$$ -U_K=\left\{\theta\in\mathbb{R}\colon S_n(\theta)>Kf(n){\rm\ for\ some\ }n\ge K\right\}. -$$ -This contains a dense open subset of ℝ. In fact, if θ = 2p/q for q odd then, for 1 ≤ n < q, u((q-n)θ)  = -u(nθ). So, Sq-1(θ) = 0 and Sq(θ) = 1. Then, by periodicity of [nθ/2], Snq (θ) = n and Sn(θ) increases linearly. So, Sn(θ) > Kf(n) for infinitely many n, and θ ∈ UK. By right continuity of u, (θ,θ+ε) ⊆ UK for small enough ε. This shows that (2p/q,2p/q+ε) is contained in the interior of UK and, as such 2p/q are dense, the interior of UK is a dense open subset of ℝ. The Baire category theorem implies that -$$ -U\equiv\bigcap_{K=1}^\infty U_K -$$ -is an uncountable dense subset of ℝ and, by construction, for any θ ∈ U, limsup Sn(θ) / f(n) > K for each K. - -The further question was asked in the comment: are there any irrational θ for which SN = O(Nx) for x < 1. The answer is yes. In fact this holds for almost every θ and every x > 1/2. -The idea is to consider rational approximations to θ, |θ/2 - p/q| ≤ q-2. Then, there will be an integer 1 ≤ a < q such that |1/2 - [ap/q]| ≤ 1/(2q). So, |1/2-[aθ/2]| ≤ 1/q. With u() as above, it follows that u(nθ) + u((n+a)θ) = 0 unless -2/q ≤ nθ ≤ 2/q (mod 1). So, there is a lot of cancellation in SN(θ), -$$ -\begin{array} -\displaystyle -\vert S_N(\theta)\vert &\displaystyle \le a +\sum_{n=1}^N1_{\{-2/q\le n\theta\le 2/q{\rm\ (mod\ 1)}\}}\\\\ -&\displaystyle\le 2q +\sum_{n=0}^{\lfloor N/q\rfloor}\sum_{m=1}^q1_{\{-2/q\le nq\theta+m\theta\le 2/q{\rm\ (mod\ 1}\}}\\\\ -&\displaystyle\le 2q+\sum_{n=0}^{\lfloor N/q\rfloor}\sum_{m=1}^q1_{\{-4/q\le nq\theta+2mp/q\le 4/q{\rm\ (mod\ 1)}\}} -\end{array} -$$ -The points 2mp/q (mod 1) are equally spaced. If q is odd then they have spacing 1/q and no more than 9 of them can lie in an interval of length 8/q. If q is even then the spacing is 2/q and no more than 5 can lie in such an interval. In either case, the final sum over m above is bounded by 10=5*2. -$$ -\vert S_N(\theta)\vert\le 2q+10N/q. -$$ -If θ has irrationality measure less than α then, for large enough N, the rational approximation p/q can be chosen such that N1/2 ≤ q ≤N(α-1)/2, -$$ -\vert S_N(\theta)\vert\le 2N^{(\alpha-1)/2}+10N^{1/2}. -$$ -In particular, if θ has irrationality measure 2 then $S_N=O(N^x)$ for every $x>1/2$. But, almost every real number has irrationality measure 2.<|endoftext|> -TITLE: Intuition behind Alexander duality -QUESTION [31 upvotes]: I was wondering if anyone could offer some intuition for why Alexander duality holds. Of course, the proof is easy enough to check, and it is also easy to work out many examples by hand. However, I don't have any feeling for why it is true. -To give you an example of what I am looking for, when I think of Poincare duality I think of the picture in terms of triangulations and dual triangulations. Is there any picture like that for Alexander duality? Is there at least maybe some kind of obvious bilinear pairing between the two sides of it or something? - -REPLY [8 votes]: Of course Alexander duality is most exciting for arbitrary compact subspaces $X \subset S^n$, Cech cohomology etc. But there is a very direct combinatorial proof for subcomplexes $X \subset \partial \Delta^{n+1}$, due to Blakers and Massey "The homotopy groups of a triad II" (Ann. of Maths. 55, 192-201 (1952)), which harks back to Poincare's proof of his duality using dual cells. For a simplicial complex $K$ let $K'$ denote the barycentric subdivision, and let $\sigma^* \subset K'$ denote the dual cell of a simplex $\sigma \in K$. Define the supplement of the subcomplex $X\subset \partial \Delta^{n+1}$ to be the subcomplex $\overline{X} \subset \Sigma^n=(\partial \Delta^{n+1})'$ consisting of the dual cells $\sigma^* \subset \Sigma^n$ of the simplices $\sigma \in \partial \Delta^{n+1} \backslash X$. The relative cellular chain complex $C^{cell}(\Sigma^n,\bar{X})$ is isomorphic to the $n$-dual ${C^{simp}(X)}^{n-*}$ of the simplicial chain complex $C^{simp}(X)$, inducing the Alexander duality isomorphisms -$H_* (\Sigma^n,\overline{X}) \cong H^{n-*}(X)$ . -I used this combinatorial approach to Alexander duality in my book "Algebraic L-theory and topological manifolds" (CUP (1992)) to construct a combinatorial model for the generalized homology theory with algebraic surgery spectrum coefficients.<|endoftext|> -TITLE: Relationship between Tangent bundle and Tangent sheaf -QUESTION [31 upvotes]: $\newcommand{\Spec}{\mathrm{Spec}\,}$Let $X=\Spec A$ be a variety over $k$, then we have the definition of the tangent bundle $\hom_k(\Spec k[\varepsilon]/(\varepsilon^2),X)$ (note that this has the structure of a variety). On the other hand, we have the definition of a tangent sheaf $\hom_{\mathscr{O}_X}(\Omega_{X/k},\mathscr{O}_X)$. What is the relationship between the two? Also, when $X$ is an arbitrary scheme (not necessarily affine), then does the relationship still hold? - -REPLY [27 votes]: You can always apply the "vector bundle" construction to $\Omega:=\Omega_{X/k}$ (locally free or not). What you get is a scheme $T=\mathrm{Spec\ Sym}(\Omega)\to X$ which deserves to be called "tangent bundle" (albeit not locally trivial); in particular its $k$-points are what you want and, more generally, if, say, $Z=\mathrm{Spec}\ C$ is an affine $k$-scheme, then $T(C)$ is just $\mathrm{Hom}_k (\mathrm{Spec}\ C[\varepsilon]/\varepsilon^2, X)$. -On the other hand consider the ${\cal O}_X$-dual ${\cal T}:={\Omega}^\vee$. For every $X$-scheme $y:Y\to X$ there is a canonical map $\Gamma(Y,y^*\mathcal{T})\to \mathrm{Hom}_X(Y,T)$. If $Y$ is an open subset of $X$ this is bijective. But if $y$ is a point of $X$, then the LHS is $\Omega^\vee \otimes \kappa(y)$ while the RHS is the $\kappa(y)$-dual of $\Omega\otimes \kappa(y)$. Clearly the image consists of those tangent vectors at $y$ which extend to vector fields in a neighbourhood. The computation when $X$ is the union of the two axes in the plane is a good exercise; if $y$ is the origin the above map is zero. -[EDIT] after seeing Unknown's answer (BTW, there are some problems with TeX there). The above argument shows that the "tangent bundle" is always a scheme, if you define it right. Another way of seeing this is that it's just an instance of Weil restriction: if $R$ is a finite-dimensional $k$-algebra you can define the functor $\underline{\mathrm{Hom}}_k (\mathrm{Spec}(R),X)$ in a similar way. This is always an algebraic space, and it is a scheme if $X$ is quasiprojective. But it is also a scheme if $R$ is local, which is the case here with $R=D_1(k)$. The reason is that if you cover $X$ by affines $U_i$, every morphism frome a local scheme to $X$ factors through one of the $U_i$'s, so we can construct the Weil restriction of $X$ by gluing those of the $U_i$'s.<|endoftext|> -TITLE: Smooth bijection between non-diffeomorphic smooth manifolds? -QUESTION [29 upvotes]: The "textbook" example of a smooth bijection between smooth manifolds that is not a diffeomorphism is the map $\mathbb{R} \rightarrow \mathbb{R}$ sending $x \mapsto x^3$. However, in this example, the source and target manifolds are diffeomorphic -- just not by the given map. Is there an example of a smooth bijection $X \rightarrow Y$ of smooth manifolds where $X,Y$ are not diffeomorphic at all? (and if so, what?) -(For instance, is it possible to arrange a smooth bijection from a sphere to an exotic sphere, failing to be a diffeomorphism because of the existence of critical points? or do homeomorphisms between different smooth structures on spheres fail to be everywhere smooth in some catastrophic way?) - -REPLY [22 votes]: Every smooth manifold has a smooth triangulation, which yields a pseudofunctor from the category of smooth manifolds to the category of PL manifolds. (There is no actual functor; that would be crazy.) If two smooth manifolds are PL isomorphic, then the answer is yes. You can start with the PL isomorphism, and then build a homeomorphism that follows it and that has the property that all derivatives vanish in all directions perpendicular to every simplex. You can build the homeomorphism by induction from the $k$-skeleton to the $(k+1)$-skeleton using bump functions. -The PL Poincaré conjecture is true in dimensions other than 4, so all exotic spheres in the same dimension $n \ge 5$ are PL homeomorphic. (High-dimensional examples of exotic spheres start in dimension 7, it was calculated.) In dimension 4, by contrast, every PL manifold has a unique smooth structure, and it is not known whether there are any exotic spheres. -On the other hand, if the manifolds are homeomorphic but not even PL homeomorphic, then I don't know. It is known that every manifold of dimension $n \ge 5$ has a unique Lipschitz structure, but I do not know a Lipschitz version of the above argument. On the positive side, passing from smooth to Lipschitz is an actual functor, so the answer to a modified question, is there a Lipschitz-smooth homeomorphism, is yes, and you can even make it bi-Lipschitz.<|endoftext|> -TITLE: Using mirrors to make a non-convex polygon visible from a fixed interior point -QUESTION [6 upvotes]: Take a point $A$ inside a non-convex polygon $P$. Is it always possible to place a finite set of mirrors given by straight segments (not necessarily along the boundary of $P$, any position inside $P$ is allowed) such that every point of $P$ is visible from $A$? Light-rays propagate along straight lines, are reflected in the usual way on both sides of mirrors and are absorbed by the boundary of $P$. -Is there a position of mirrors which is "universal" in the sense that any pair of points inside $P$ are within view of each other? -These questions have of course obvious higher-dimensional generalizations. - -REPLY [10 votes]: Permit me to repeat this example from a related MO question, based on a paper by -George Tokarsky, "Polygonal Rooms Not Illuminable from Every Point" -[Amer. Math. Monthly, 102:867-879 (1995)]. Here is a figure from the MathWorld article -on the topic: - - - -These are rooms all of whose walls are mirrors but which have a pair of points $x$ and $y$ -such that if a light is placed at $x$, point $y$ is dark. -It is an unresolved conjecture that, at least in rational polygons (angles rational multiples -of $\pi$), the number of dark points for any light position is finite. -If that conjecture were true, then it seems plausible that it would be possible to place additional internal mirrors to -cover those points and still illuminate the remainder of the room. - -Update (13 May 2015). -The "unresolved conjecture" I mention above has been settled. -I found the reference in -Alex Wright's 2015 paper, "From rational billiards to dynamics on moduli spaces." -(arXiv abstract): - -Lelièvre, Monteil and Weiss have shown that if $P$ is a rational polygon, - for every $x$ there are at most finitely many $y$ not illuminated by $x$. - -The reference is: -Samuel Lelievre, Thierry Monteil, Barak Weiss. -"Everything is illuminated." 2014. -(arXiv abstract.)<|endoftext|> -TITLE: exact categories which are not additive -QUESTION [11 upvotes]: There are various notions of exact categories (nlab). In a lecture I've seen the following definition of an exact category, which is basically (exact) = (abelian) - (additive): -A category $C$ is called exact if a) it contains a zero object, b) every morphism has a kernel and a cokernel, c) the canonical morphism $\text{coim}(f) \to \text{im}(f)$ is an isomorphism for every morphism $f$. -So for example, the category of pointed sets is an exact category in this sense. I think also the category of pointed compactly generated hausdorff spaces is an example. -Questions: 1) Which theorems and constructions of homological algebra carry over from abelian categories to exact categories in the above sense? 2) Where can I find literature about these categories? I can only find some about the other definitions. - -REPLY [11 votes]: These categories are called Puppe-exact or p-exact categories. See paragraph 1.1, of Jordan-Hölder, modularity and distributivity in non-commutative algebra, by Francis Borceux and Marco Grandis (JPAA 208 (2007), 665-689 doi:10.1016/j.jpaa.2006.03.004), for non-abelian examples. And see the papers of Marco Grandis (e.g. On the categorical foundations of homological and homotopical algebra (Numdam)) and Mitchell's book “Theory of categories” for homological results in this context (as a general rule, all homological lemmas non involving direct products hold).<|endoftext|> -TITLE: Height of cyclotomic polynomials -QUESTION [13 upvotes]: Recall that the cyclotomic polynomial of order $n$ is -$$\Phi_n(X)=\prod_{gcd(k,n)=1}(X-e^{2ik\pi/n}).$$ -Its degree is $\phi(n)$ (Euler's indicator). Inversion of -$$X^n-1=\prod_{d|n}\Phi_d(X)$$ -by the Moebius formula implies that $\Phi_n\in{\mathbb Z}[X]$. The height of $\Phi_N$ is the maximal modulus of its coefficients. Thus the height of -$$\Phi_p(X)=X^{p-1}+\cdots+X+1$$ -($p$ odd prime) is $1$. A cyclotomic polynomial is flat if its height is $1$. We know that the height of $\Phi_{2n}$ equals that of of $\Phi_n$, and also that if a prime $p$ divides $n$, then the height of $\Phi_{pn}$ equals that of $\Phi_n$. Therefore it is enough to analyse the case where $n=p_1\cdots p_\ell$ is the product of distinct odd primes. When $\ell=2$ it is known that $\Phi_{p_1p_2}$ is flat. It is known that there are infinitely many flat $\Phi_{p_1p_2p_3}$, but $\Phi_{105}$ ($105=3\cdot 5\cdot 7$) is not flat (its height equals $2$). -Question. What is known about the growth of the height of $\Phi_n$ ? Is there a bound of the form $C_\ell$ (thus extending $C_1=C_2=1$) ? - -REPLY [12 votes]: http://oeis.org/classic/A013594 gives the smallest cyclotomic polynomial with a given coefficient, and the paper http://www.ams.org/mathscinet-getitem?mr=0364141 by Vaughan gives lower bounds for the growth of the max coefficient of the form exp(log 2 log n/log log n)<|endoftext|> -TITLE: On polarized (pure) Hodge structures -QUESTION [12 upvotes]: Some simple questions, for which I know no precise reference (and would be deeply grateful for a nice one!): - -Is it true that the category of (pure) polarized Hodge structures is abelian semi-simple, whereas the whole category of pure Hodge structures is not? -Should one only consider those morphisms of polarized Hodge structures that respect polarizations in order to obtain an abelian category? -Is it true that all pure Hodge structures 'that come from geometry' (for example, the graded pieces of the weight filtration of the singular cohomology of varieties and motives) are polarized? - -REPLY [21 votes]: Fortunately, these questions are easy to answer. First of all, it helps to distinguish -between polarizable Hodge structures and polarized structures. For polarizable, we merely -require that a polarization exists, but it is not fixed. Let Hodge structure -mean pure rational Hodge structure below. -Then - -The category of polarizable pure Hodge structures is abelian and semisimple (morphisms are not required to respect polarizations). This is essentially proved in Theorie de Hodge II. -The category of arbitrary Hodge structures is abelian but not semisimple. -To see the nonsemisimplicity, we can use a theorem of Oort-Zarhin [Endmorphism algebras of -complex tori, Math Ann 1995], that -any finite dimensional $\mathbb{Q}$-algebra is the endomorphism algebra of some complex torus, and therefore -of some Hodge structure. -All pure Hodge structures of geometric origin are polarizable. So this is a very reasonable condition to impose. For $Gr_WH^*(X)$, this is explained for example in Beilinson's Notes on absolute Hodge cohomology (although this was already implicit in Deligne's construction).<|endoftext|> -TITLE: Singularities of pairs -QUESTION [12 upvotes]: In the next days I have to give a talk in which I need to explain some of the usual singularities of pairs that one meets when dealing with the minimal model program: KLT, DLT and LC pairs. -In particular I would like to give an intuition (also to non specialists) of the reason why an LC pair is much more difficult to treat than a DLT pair, for example. -In the same context I would also like to give an intuition of what the LC centers of a pair are and why they are "special" subvarieties for a pair. -Note that I'm always considering normal (possibly non-smooth) projective varieties. -Do you have any ideas? - -REPLY [3 votes]: I know very little of this subject, but I can attest to the value of it, since many of us studied the singularities of theta divisors for years without noticing or even conjecturing some basic results that followed from Kolla'r's observation that (A,D) is a log canonical pair if D is a theta divisor on a principally polarized abelian variety A. As Kolla'r put it with extreme simplicity and clarity: "we know a highly singular divisor moves, but a theta divisor doesn't move, so it should not be very singular." He deduced that the points of D of multiplicity ≥ k have codimension ≥ k in A. I particular there are no points on D of multiplicity > dim(A). The generalization of this result by Ein and Lazarsfeld that the extreme case occurs only for a product p.p.a.v., established the normality of an irreducible theta divisor. To me, this is a beautiful example of the concept of l.c. pairs which allowed the characterization of indecomposable p.p.a.v's in terms of the geometry of their theta divisors. This is one of the first cases of a technique that allowed the analysis of theta divisors that could not be understood in terms of curves, e.g. as Jacobians or Pryms.<|endoftext|> -TITLE: Algorithms for finding rational points on an elliptic curve? -QUESTION [21 upvotes]: I am looking for algorithms on how to find rational points on an elliptic curve $$y^2 = x^3 + a x + b$$ where $a$ and $b$ are integers. Any sort of ideas on how to proceed are welcome. For example, how to find solutions in which the numerators and denominators are bounded, or how to find solutions with a randomized algorithm. Anything better than brute force is interesting. -Background: a student worked on the Mordell-Weil theorem and illustrated it on some simple examples of elliptic curves. She looked for rational points by brute force (I really mean brute, by enumerating all possibilities and trying them). As a continuation of the project she is now interested in smarter algorithms for finding rational points. A cursory search on Math Reviews did not find much. - -REPLY [5 votes]: The best practical solution is to have someone else do the work. You can look up the curve in Cremona's tables, if it is not too large. If it is larger than that, you can use mwrank, a free standing C++ program. I believe that SAGE and MAGMA also both have this functionality, although I couldn't find the syntax in a quick search.<|endoftext|> -TITLE: Range of completely positive projection -QUESTION [15 upvotes]: Let $A$ be a C*-algebra. Suppose that $P:A \rightarrow A$ is a contractive completely positive projection. Does the range $P(A)$ is completely order isomorphic to a $C^*$-algebra? -In the case where the answer is false: -Is it true in supposing, in addition, that the range $P(A)$ is an operator system, $A$ unital and $P(1)=1$? - -REPLY [14 votes]: I would add to Taka's answer, since the Choi-Effros paper is not widely available online, that the nontrivial parts of the proof are to show the "C$^*$-identity" $\|x\|^2=\|x^*x\|$, associativity, and to check the completeness of $P(A)$. -Indeed, the algebra $(P(A),+,\circ)$ is a $*$-algebra, normed with the norm inherited from $A$. For the C$^*$-identity, since $P$ is contractive we have -$$ -\|x^*\circ x\|=\|P(x^*x)\|\leq\|x^*x\|=\|x\|^2. -$$ -On the other hand, since $P$ is cp and contractive, it satisfies the Schwarz inequality $P(x)^*P(x)\leq P(x^*x)$, and so, for any $x\in P(A)$, -$$ -\|x^*\circ x\|=\|P(x^*x)\|\geq \|P(x)^*P(x)\|=\|x^*x\|=\|x\|^2. -$$ -Completeness follows from the fact that $P$ is a bounded projection. If $\{x_j\}$ is a Cauchy sequence in $P(A)$, then by the completeness of $A$ the sequence converges to some $x$ in $A$. As $P$ is bounded, $P(x_j)\to P(x)$; but $P(x_j)=x_j$ (since $P$ is a projection and $x_j\in P(A)$) and so $P(x)=x$, that is $x\in P(A)$. So $P(A)$ is closed. -For the associativity of the product, one needs to check that $P(aP(bc))=P(P(ab)c)$ for any $a,b,c\in A$. For this, since $P$ is selfadjoint, it is enough to show that $P(xP(a))=P(xa)$ for any $x\in P(A)$, $a\in A$. The trick that Choi-Effros use (in Paulsen's version here) is to apply the Schwarz inequality to the ccp map $P^{(2)}$ and the operator $y=\begin{bmatrix}x^*&a\\ 0&0\end{bmatrix}$: thus $P^{(2)}(y)^*P^{(2)}(y)\leq P^{(2)}(y^*y)$ reads -$$ -\begin{bmatrix}xx^*&xP(a)\\ P(a^*)x^*&P(a^*)P(a) \end{bmatrix} -\leq -\begin{bmatrix}P(xx^*)&P(xa)\\P(a^*x^*)&P(a^*a)\end{bmatrix} -$$ -If we now apply $P^{(2)}$ to this inequality, since it preserves positivity, we get -$$ -\begin{bmatrix} 0&P(xa)-P(xP(a))\\ P(a^*x^*-P(P(a^*)x^*))& P(a^*a)-P(P(a^*)P(a))\end{bmatrix} -\geq0; -$$ -the 0 in the 1,1 entry forces the off-diagonal entries to be zero, so $P(xa)=P(xP(a))$.<|endoftext|> -TITLE: Finding a point that lies in a majority of polytopes -QUESTION [8 upvotes]: Suppose I have $k$ $n$-dimensional polytopes $P_1,\ldots,P_k$, each explicitly specified as the intersection of a collection of hyperplanes. If there was a point $p \in \mathbb{R}^n$ that lay in the intersection of all of these polytopes ($p \in P_1 \cap \ldots \cap P_k$), I could efficiently find it by solving a linear program. Unfortunately, I have no guarantee that my polytopes have non-empty intersection. However, someone has promised me that there exists a point $p$ that lies in at least $2/3$ of my polytopes: that is, there exist indices $i_1,\ldots,i_{2k/3}$ such that: -$$p \in P_{i_1} \cap \ldots \cap P_{i_{2k/3}}$$ -Does there exist an efficient algorithm (running in time polynomial in k and n) that can find such a point? - -REPLY [4 votes]: This problem is clearly in NP (guess which polytopes) and becomes NP-complete if we replace $2/3$ with $1/2$ and make it a decision problem, dropping the promise that such a $p$ exists. In particular we can reduce integer programming feasibility to this problem. -Say we wish to determine whether $Ax = b$ has a solution $x$ which is a zero-one vector. We can define $k = 2n$ polytopes $P_i^j = \{x \mid x_i = j, Ax = b, 0\leq x\leq 1\}$ for integers $1\leq i\leq n$ and $0\leq j\leq 1$. A zero-one solution of $Ax=b$ is the same as a point which lies in $l = n = k/2$ of these polytopes. -I imagine there is a simple reduction showing that the problem is still hard if we use the fraction $2/3$ (or any other fixed fraction) instead of $1/2$. But I'm guessing that you just gave $2/3$ as an example so I haven't thought about it. Similarly I am guessing that the problem is still hard if you somehow know that such a $p$ exists and merely want to find one, but I haven't thought of how to show this either.<|endoftext|> -TITLE: Why do we want to have orthogonal bases in decompositions? -QUESTION [5 upvotes]: In the decompositions I encountered so far, we all had orthogonal set of bases. For example in Singular Value Decomposition, we had orthogonal singular right and left vectors, in [discrete] cosine transform (or [discrete] fourier transform) we had again orthogonal bases. -To describe any vector $x \in \mathbb{C}^N$, we need to have $N$ independent set of basis vectors but independent doesn't necessarily mean orthogonal. My intentions behind selecting orthogonal vectors are as follows: - -The solution is not unique for $x$ if the basis are not orthogonal. -It is easy to find the solution numerically by projecting $x$ onto each vector and this solution doesn't depend on the order of the bases. Otherwise, it depends on the order. -If we are talking about some set of vectors, they might be correlated in the original space, but uncorrelated in the transformd space which might be important when analyzing the data, in dimensionality reduction or compression. - -I'm trying to understand the big picture. Do you think that I am right with these? Do you have any suggestions, what is the main reason for selecting orthogonal bases? - -REPLY [3 votes]: Maybe some examples of non-orthogonal decompositions are helpful: In the context of wavelet one frequently uses bi-orthonormal systems, i.e. two sets of vectors $(u_i)$ and $(v_i)$ such that $\langle u_i,v_j\rangle = \delta_{i,j}$. In this setting the calculations of coefficients can be done stably and efficiently. In real life, such bi-orthogonal wavelet bases are at the heart of the JPEG 2000 compression standard. -Another important notion in this context is the notion of frames.<|endoftext|> -TITLE: Is there order to the number of groups of different orders? -QUESTION [8 upvotes]: I was always struck by how uncharacteristically erratic the behavior of the following function is: -$f: \mathbb{N} \rightarrow \mathbb{N}$ given by $f(n):=$ number of isomorphism classes of groups of order $n$. -I blame this on the introduction of the unnatural idea of looking at the size of the group. Why would we? The same is true in Algebraic Geometry: curves are nicer than surfaces for an arbitrary reason. Looking at the dimension introduces an unnatural agent in the mix. -My question is this: -What is the most "natural" statement (and less preferably: viewpoint) you've ever seen about this $f$? I bar asymptotic statements as inherently unnatural (unless you can convince me otherwise). To open the floor the widest, I'll put a CW stamp on this question. - -REPLY [2 votes]: The most general statement that I have seen is due to Holder and gives the following result: if $n$ is squarefree, and if $f_p(n)$ is the number of primes $q$ with $q \mid n$ and $p \mid q-1$, then the number of groups of order $n$ is given by $$g(n) = \sum_{d\mid n}\prod_{p\mid d,\;d > 1} \frac{p^{f_p(n/d)}-1}{p-1}.$$ -This formula is nic ein that it is exact; of course, the asymptotics here are very hard to extract, and given that there tend to be more groups of non-squarefree order than of squarefree order it is not terribly useful in asymptotic estimates.<|endoftext|> -TITLE: Negative holomorphic sectional curvature -QUESTION [11 upvotes]: Let X be a complex hermitian manifold with hermitian form $\omega$. How can you prove that if $\omega$ has negative holomorphic sectional curvature, then its scalar curvature is negative, too? - -REPLY [9 votes]: Here is the answer. -Let $(X,\omega)$ be a Kähler $n$-dimensional manifold. Fix a point $x_0\in X$ an choose local holomorphic coordinates $(z_1,\dots,z_n)$ centered at $x_0$ and such that $(\partial/\partial z_1,\dots,\partial/\partial z_n)$ is unitary at $x_0$. Let -$$ -\Theta_{x_0}(T_X,\omega)=\sum_{j,k,l,m=1}^nc_{jklm}\hspace{0.3mm}dz_j\wedge d\bar z_k\otimes\left(\frac\partial{\partial z_l}\right)^*\otimes\frac\partial{\partial z_m} -$$ -be the Chern curvature at the point $x_0$. Consider the induced hermitian form on rank one tensors of $T_X\otimes T_X$ given by -$$ -\theta_{T_{X,x_o}}(v\otimes w)=\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}v_j\bar v_k w_l\bar w_m, -$$ -where -$$ -v,w\in T_{X,x_0},\quad v=\sum v_j\hspace{0.3mm}\frac\partial{\partial z_j},\quad w=\sum w_j\hspace{0.3mm}\frac\partial{\partial z_j}. -$$ -With this notation, the holomorphic sectional curvature in the direction of $v\in T_{X,x_0}\setminus\{0\}$ is given by -$$ -\frac{1}{||v||_\omega^4}\theta_{T_{X,x_o}}(v\otimes v). -$$ -The idea now is to take the average on the $\omega$-unit sphere $S^{2n-1}$ and try to deduce something on the scalar curvature at the point $x_0$ which is given by -$$ -s(x_0)=2\sum_{j,k=1}^nc_{jjkk}. -$$ -So, let's compute the integral -$$ -\int_{S^{2n-1}}\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi), -$$ -where $d\sigma(\xi)$ is the probability Haar measure on $S^{2n-1}$. It is not hard to see that the integral -$$ -\int_{S^{2n-1}}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi) -$$ -vanishes unless $j=k$ and $l=m$ or $j=m$ and $k=l$. Thus, we have to compute -$$ -\int_{S^{2n-1}}|\xi_j|^2|\xi_k|^2\hspace{0.3mm}d\sigma(\xi),\quad j,k=1,\dots,n. -$$ -It is classically known that -$$ -\int_{S^{2n-1}}|\xi_j|^4\hspace{0.3mm}d\sigma(\xi)=\frac 2{n(n+1)},\quad j=1,\dots,n, -$$ -and -$$ -\int_{S^{2n-1}}|\xi_j|^2|\xi_k|^2\hspace{0.3mm}d\sigma(\xi)=\frac 1{n(n+1)},\quad 1\le j\ne k\le n. -$$ -Then, we get -$$ -\begin{aligned} -\int_{S^{2n-1}}\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi) & =\sum_{j,k=1}^nc_{jjkk}\left(\delta_{jk}\frac 2{n(n+1)}+(1-\delta_{jk})\frac 2{n(n+1)}\right) \\ -& = \frac 2{n(n+1)}\sum_{j,k=1}^nc_{jjkk}=\frac 1{n(n+1)}s(x_0), -\end{aligned} -$$ -where we have used the Kähler identity $c_{jklm}=c_{jmlk}$. -Thus, if $\frac{1}{||v||_\omega^4}\theta_{T_{X,x_o}}(v\otimes v)$ is negative, so is its average and we are done.<|endoftext|> -TITLE: Brownian Motion Winding Number -QUESTION [7 upvotes]: Take a simple random walk $\gamma$ in the complex plane conditioned to start at point $a$ and end at point $b$. For this random walk, we can define the winding number $W_\gamma(a,b)$ around $b$ in the usual way for complex curves. -If instead we have a 2D Brownian motion $Z=X+iY$, then this definition becomes more complicated. For example, if we have a Brownian motion starting at the origin, we can talk about the winding number $\theta_t$ at time $t$ around the origin by solving the stochastic differential equation -$d\theta_t=\frac{X_tdY_t-Y_tdX_t}{|Z_t|^2}$ -with initial condition $\theta_0=0$. -The issue is that for Brownian motion, we cannot condition on the path $Z$ to hit a particular point, because this has probability zero. Moreover, by considering annuli around $b$ and the fact that planar Brownian motion moves between concentric annuli with positive probability, it seems to me the situation becomes rather singular. - -Question: Is there a sensible generalization for the winding number of a Brownian motion conditioned to hit a single point? - -For example, can we look at the limit of the winding number around an annulus about point $b$ whose radius shrinks to zero? I would imagine we would require the Brownian motion to be conditioned to hit some region of positive area just outside the shrinking annulus. - -REPLY [2 votes]: I think you need to take the log of the Brownian motion and look at the imaginary coordinate. -$$R + i\theta = \log(X+iY)$$ -Since $z \mapsto \log z$ is conformal, there exists a time change taking one Brownian motion to the other. -Also, notice that $R_t$ is related to the Cauchy distribution.<|endoftext|> -TITLE: A sphere bundle map -QUESTION [8 upvotes]: I think this may all be classical bundle-theory. But I'm trying to read some old papers on classifications of bundles and the following came up as questions I couldn't immediately answer: -Consider the group homomorphism $\theta:\pi_{n-1}(SO(n)) \to H^n(S^n) $ given by sending an element $\alpha$ in $\pi_{n-1}(SO(n))$ to the Euler class of the $n$ - plane bundle classified by $\alpha$. - -Is there a name for this map? -If we let $n=4$, we have $\theta :\pi_{3}(SO(4))=Z^2 \to H^4(S^4)=Z$. What is this map explicitely? - -Now look at the bundle $SO(n) \to SO(n+1) \to S^n$. If we take $id_n \in \pi_{n}S^n$ to be the (class of the) identity map and $\partial(id_n)$ its image in $\pi_{n-1}SO(n)$ (from the homotopy LES of a fibration), then... - -...what is $\theta (\partial(id_n))?$ - -Does it just map to a generator of $H^n(S^n)?$ - -REPLY [8 votes]: The bundle classified by this element of $\pi_{n-1}SO(n)$ is the tangent bundle of $S^n$, so the image under $\theta$ is twice a generator if $n$ is even, zero if $n$ is odd. -To say what $\theta$ is explicitly in the case $n=4$ in terms of the $\mathbb Z$-basis for $\pi_3SO(4)$, we would have to know what basis you have in mind, but for the one I usually think of it takes both basis elements to the same generator. -Edit: Note also that the map $\theta:\pi_{n-1}SO(n)\to H^n(S^n)$ is isomorphic to the map $\theta:\pi_{n-1}SO(n)\to \pi_{n-1}(S^{n-1})$ induced by the bundle projection $SO(n)\to S^{n-1}$.<|endoftext|> -TITLE: Residually finite + torsion free + finite index = finite complex? -QUESTION [7 upvotes]: Suppose $G$ is a residually-finite group and $H < G$ a torsion-free subgroup of finite index. - -What characterizes such $G$ such that $BH$ is homotopic to a finite complex? - -I believe Serre showed that if $G$ is arithmetic, the result holds. But I am sure since then much more must be known. - -REPLY [10 votes]: Here's a result that gives some idea of how hard it is to characterise linear (let alone residually finite) groups of type $F$ (ie with a $K(G,1)$ that's a finite complex). -Theorem: There is a sequence of finite subsets $S_i\subseteq GL_{n_i}(\mathbb{Z})$ with the property that: - -for every $i$, either $G_i=\langle S_i\rangle$ is of type $F$ or $G_i$ is not finitely presentable (in particular not of type $F$); -the set of $i$ such that $G_i$ is of type $F$ is recursively enumerable but not recursive. - -So there is no algorithm to determine whether or not $G_i$ is of type $F$. -I can give details of the proof if anyone's interested. Basically, it's an easy application of the Haglund--Wise version of the Rips Construction. - -Details -The first ingredient is a sequence of finite presentations for groups $(Q_i)$, with the property that the set $\{i\mid Q_i\cong 1\}$ is recursively enumerable but not recursive. We also want to set things up so the non-trivial $Q_i$ are infinite. Such sequences are quite well known, see for instance Chuck Miller's survey article. -The second ingredient is provided by Haglund and Wise, in one of many variations of a famous construction of Rips. For any finite presentation for a group $Q$, Haglund and Wise construct a short exact sequence -$1\to G\to \Gamma\to Q\to 1$ -with the following properties: - -$\Gamma$ is the fundamental group of a `virtually special', non-positively curved square complex $X$; and -$G$ is finitely generated. - -Non-positive curvature is a local condition on $X$ which ensures that its universal cover is contractible; in particular, $\Gamma$ is of type $F$. Being `special' is a condition on the hyperplanes of $X$. All you need to know is that it ensures that $\Gamma$ is (virtually) a subgroup of a right-angled Coxeter group, from which it follows that $\Gamma$ is a subgroup of $GL_{n}(\mathbb{Z})$ for some $n$. -Everything in this construction is completely explicit. Given a presentation for $Q$, one can write down a presentation for $\Gamma$ and the generators $S$ for $G$. Furthermore, you can also write down an explicit embedding $\Gamma\hookrightarrow GL_n(\mathbb{Z})$. -Finally, we apply this construction to the $Q_i$. If $Q_i\cong 1$ then we have $G_i\cong\Gamma_i$, so in particular it's of type $F$. On the other hand, a result of Bieri ensures that if $Q_i$ is infinite then $G_i$ isn't finitely presentable. (This uses the fact that the $\Gamma_i$ are of cohomological dimension two.)<|endoftext|> -TITLE: Solving the Beltrami Equation for a very simple Beltrami Coefficient -QUESTION [7 upvotes]: Let $\mu$ be a function on the complex plane with the property $\mu(z) = \overline{\mu(\bar{z})}$, such that $\mu(z) = \epsilon e^{-2\pi i \bar{z}}$ on the upper-half plane, where $\epsilon$ is a complex number such that $|\epsilon|<1$. -I'm interested in the solution to the following Beltrami equation with this Beltrami coefficient: -\begin{equation} -\mu \frac{\partial f}{\partial z} = \frac{\partial f}{\partial \bar{z}} -\end{equation} -Now, since $|\mu| = |\epsilon|e^{-2\pi y} < 1$ (since $y >0$), there exists a unique quasiconformal solution $f$, satisfying $f(\bar{z})=\overline{f(z)}$ and fixing the points 0, 1, and infinity. -I'm basically trying to understand how $f_z$ grows. -Now, because $\mu$ is such a simple function, I feel like one should be able to explicitly write a solution in this case. I've tried many different approaches, but I have been very unsuccessful. I tried looking in Ahlfors and a few other references, but I haven't been able to find anything too useful. -Any ideas? - -REPLY [4 votes]: Sorry in advance for responding to an old question, especially since this isn't even a solution. In my defence, this is my first post and I haven't seen anything in the FAQ for etiquette forbidding answering old posts? -I've been thinking about very similar questions lately, and stumbled upon this post during a Google search for ways of solving the Beltrami equation. There is currently a construction (attributed to Gauss) detailed on the "Beltrami equation" Wikipedia entry for locally solving the Beltrami equation at a point (the default is $z=0$) when $\mu$ is (locally) expressible as a power series in $z$ and $\bar{z}$. I've tried the procedure but it hasn't quite produced a correct function, and this has prompted me to try to solve it myself. I believe that I have a working algorithm for finding general solutions to the Beltrami equation for a sufficiently nice $\mu$ - such as the one you've specified. -Caveat: I realize that you're not simply wanting to solve the Beltrami equation, but would like to find the unique normalized solution $f$ such that $f$ is a homeomorphism of the upper half plane, but all that I'm able to furnish is an (explicit) family of solutions which probably contains the desired guy. I have no idea how you might be able to track down the unique normalized quasiconformal solution in this haystack. -A pretty simple non-zero solution to the Beltrami equation for your specified coefficient is: -$$f(z):=z-\frac{\epsilon}{2\pi i}\exp(-2\pi i\bar{z})+\frac{\epsilon}{2\pi i}.$$ -Note that although $f:0\mapsto 0,1\mapsto1,\infty\mapsto\infty$ and is smooth, it doesn't satisfy that $f(\bar{z})=\bar{f(z)}$ and isn't the quasiconformal map we want. -First observe that $f(z)$ satisfies the Beltrami equation over all of $\mathbb{C}$ (not just the upper half plane). -Next observe that post-composing $f$ by a holomorphic function $\varphi(z)$ also results in a solution $\varphi\circ f(z)$ that also satisfies the Beltrami equation. Equivalently, any power series of $f$ is also a solution over its domain of convergence. -Since $f(z)$ contains a $z$ term, it should be reasonably easy to show that any solution to the Beltrami equation defined at $z=0$ that's also expressible as a power series in $z$ and $\bar{z}$ will agree with some power series in $f(z)$ in a neighborhood around $z=0$. -This results in a (countably) infinite dimensional family of solutions to the Beltrami equation, but it's not clear to me how one might be able to extract the quasiconformal solution. In fact, I'm not familiar enough with quasiconformal maps to know if the desired solution can be written as a power series in $z$ and $\bar{z}$ around $z=0$, although this is not, I feel, an unreasonable expectation. -Being overly optimistic, I feel that the $f(z)$ above isn't a terrible approximation for the actual quasiconformal solution, especially if one simply wanted some growth-rate intuition. In particular, the approximation behaves best where the imaginary value of the input is large, which is somewhat unsurprising because the Beltrami coefficient is nearly $0$ when the imaginary part of $z$ is high, so the solution should be roughly the identity map around these parts - the identity function, of course, is the dominant term in $f(z)$ for $\mathrm{Im}(z)\gg0$. -Just in case you want a way of constructing a solution to a given Beltrami equation that is "generative", in the sense that any other solution defined at $z=0$ should be a power series of this one, I believe that the following procedure should work: -Step 1: We adopt the notation that the Beltrami coefficent $\mu(z)$ - which by assumption can be written as a power series in $z$ and $\bar{z}$, is written as $\mu(z,\bar{z})$. Then solve for -$$\frac{\partial F}{\partial z}(z,w)=-\mu(w, F(z,w)),\; F(0,w)=w.$$ -This is purely motivated by the aforementioned Wikipedia entry, although the order of the inputs for $\mu$ do differ from Wikipedia's. -Step 2: Find $G(z,w)$ such that $G(z,F(z,w))=w$. That is, the functions -$$(z,w)\mapsto (z,F(z,w))\text{ and }(z,w)\mapsto (z,G(z,w))$$ -are inverse to each other. -I'd be surprised if $G(z,w)$ always exists over the entire domain of $\mu$, but it should for "nice" examples of $\mu$ (It's existed for all the examples that I've checked by hand, perhaps something along the lines of $\mu$ being defined over a simply connected domain and expressible as a power series in $z$ and $\bar{z}$ might suffice)? -Step 3: A "generative" solution is given by: -$$f(z)=G(\bar{z},z).$$ -Here's hoping that someone'll take this mess and figure out how to build the normalized quasiconformal map. =)<|endoftext|> -TITLE: Best approximation to the Weyl group as a subgroup of a reductive group. -QUESTION [12 upvotes]: Let G be a reductive algebraic group over a field k. Let S be a maximal split torus, Z its centraliser and N its normaliser. The Weyl group W is then defined to be the quotient N(k)/Z(k). Now we cannot hope for W to be realisable as a subgroup of G, but I would like to know how close we can get. -There is a classical result of Tits in the case where G is split which says that for each simple reflection s in the Weyl group, we can find a lift ws in G with the property that these lifts satisfy the braid relations. They do not however square to the identity, instead square to an order 2 element of S, and we get an extension of W by an elementary abelian 2-group embedding in G. -So my question ends up becoming, what generalisation of the above theorem of Tits exists when G is no longer assumed to be split? Ideally I'd get an answer for general reductive G, and if there happens to be a simpler formulation in the quasi-split case, I'd be interested in hearing that too. - -REPLY [8 votes]: Via Theorem 7.2 in Borel-Tits, Groupes reductifs (1965), Tits's lifting result for the Weyl group also applies in the non-split case (for connected groups). -This theorem states that there exists a split subgroup $F$ of $G$ such that $F$ contains the maximal split torus $S$ of $G$ and intersects each relative root group of $G$ non-trivially. In particular, the Weyl groups of $F$ and $G$ are isomorphic.<|endoftext|> -TITLE: Embedding theory for contractible manifolds -QUESTION [6 upvotes]: What is known about spaces of embeddings of contractible manifolds into Euclidean space? I am also curious about the case of small codimension (or even codimension 0). The same question about the configuration spaces in such manifolds. - -REPLY [6 votes]: (This is by far not a complete answer, just an example.) In dimension 4, a paper of Livingstone (build on previous work of Lickorish) constructs some (compact with boundary) contractible 4-manifold which embeds in $\mathbb R^4$ in infinitely many (countable) distinct ways. These are distinguished by the fundamental group of the complement.<|endoftext|> -TITLE: Semicosimplicial totalization -QUESTION [5 upvotes]: Can somebody help me with a reference showing that the homotopy semicosimplicial totalization of a cosimplicial space is homotopy equivalent to its usual homotopy totalization? Is it because the inclusion of the semisimplicial category into the simplicial category is a homotopy left cofinal functor? - -REPLY [2 votes]: I agree this should be standard, but I've only seen the proof in one place. See Lemma 6.5.3.7 of Lurie's Higher Topos Theory.<|endoftext|> -TITLE: Discrete subspaces of Hausdorff spaces -QUESTION [5 upvotes]: does every infinite hausdorff space contains a countable infinite discrete subspace? - -REPLY [15 votes]: In a more general light: -folklore theorem: -Every infinite topological space contains a homeomorphic copy of one (or more) of the following 5 spaces: - -$\mathbf{N}$ in the indiscrete topology (only $\mathbf{N}$ and $\emptyset$ are open). -$\mathbf{N}$ in the co-finite topology (only $\mathbf{N}$ and all finite sets are closed). -$\mathbf{N}$ in the upper topology (the empty set and all sets $U(k) = \{ n \in \mathbf{N} : n \ge k \}$, $k \in \mathbf{N}$, are open). -$\mathbf{N}$ in the lower topology ($\mathbf{N}$, $\emptyset$, and all sets $L(k) = \{ n \in \mathbf{N} : n \le k \}$, $k \in \mathbf{N}$, are open). -$\mathbf{N}$ in the discrete topology (all subsets are open). - -As each of the spaces has the property that every infinite subspace of it is homeomorphic to the whole space, this list is minimal. -And spaces 1-4 are not Hausdorff, which implies what you need, as being Hausdorff is hereditary. -The nicest proof of this I know uses Ramsey's theorem (off hand I do not know a reference, who does?) using a partition of the triples or pairs of X, IIRC. -Reference to the original paper: Minimal Infinite Topological Spaces, John Ginsburg and Bill Sands, -The American Mathematical Monthly -Vol. 86, No. 7 (Aug. - Sep., 1979), pp. 574-576. The Ramsey proof I saw somewhere else, though.<|endoftext|> -TITLE: Sets with equal positive measure in every interval -QUESTION [12 upvotes]: Hi, -I want to write a proof that relies on the fact that: -There are Borel Sets $A$ and $B$ contained in $\mathbb{R}$ such that -$A \cap B = \emptyset$ and $\lambda(A \cap (x,y)) = \lambda(B \cap (x,y)) > 0$. -Note that $x < y \in \mathbb{R}$ are arbitrary. -I'm fairly sure this is true, but am having trouble coming up with a construction of such sets and it's driving me up the wall. Can anyone help? - -REPLY [2 votes]: Let $\mu$ and $\nu$ be two measure on $(E,{\cal E})$. If : - -$\mu(A)=\nu(A)$ for all $A$ in a subset $\pi$ of ${\cal E}$ such that $\pi$ is closed under finite intersection and the $\sigma-$algebra generated by $\pi$ is ${\cal E}$ -there exists an increasing sequence of set $A_n$ in ${\cal E}$ such that $\mu(A_n)=\nu(A_n)<\infty$ and $\cup A_n = E$. - -then $\mu=\nu$. (This is standard generating argument). -If you apply this result with $\mu=1_A \lambda$, $\nu=1_B \lambda$ and ${\cal B}$ the set of open interval, you get that if your condition on $\lambda(A\cap...)$ holds, then $A=B$ almost everywhere. With the other conditions you see that it is not possible.<|endoftext|> -TITLE: Chapters 1--4 of the Artin-Tate notes on Class Field Theory -QUESTION [23 upvotes]: Emil Artin and John Tate held a seminar on class field theory at Princeton University in 1951--1952. Their notes were published in 1967 by Benjamin (New York), but the first four chapters covering (among other things) "the fundamentals of algebraic number theory" and "local class field theory" were omitted from the printed version. -Question. Are the notes of Chapters 1--4 available to you ? -Request. Can you make them electronically accessible to the mathematical community ? -Postscript. Parts of Hasse's Klassenkörperbericht (Bericht über neuere Untersuchungen und Probleme aus der Theorie der algebraischen Zahlkörper) -are available online at the Göttingen library : -Jber. deutsch. Math.-Verein. -Teil I : 35 (1926), 1--55, -Teil Ia : 36 (1927), 233--311. -Ulf Rehmann (Bielefeld) and Keith Dennis (Cornell) have promised to put Teil II (Jber. deutsch. Math.-Verein., Ergänzungsband 6 (1930), 1--204) online soon. -Addendum. In an interview which has appeared recently in the Notices of the AMS, Tate makes some remarks about the genesis of the Artin-Tate notes. - -REPLY [5 votes]: In his preface to "Rapport sur la Cohomologie des Groups", Serge Lang says that those notes "provided missing chapters to the Artin-Tate notes on class field theory". It is available in english translation under the title "Topics in the cohomology of groups". -Edit: So perhaps Lang writing that it providing missing chapters does not mean that it provides ALL missing chapters.<|endoftext|> -TITLE: generalization of Bezout's Theorem? -QUESTION [12 upvotes]: I learned Bezout's Theorem in class, stated for plane curves (if irreducible, sum of intersection multiplicities equals product of degrees). What is the proper general statement, for projective varieties of degree n? -I think it is something like: If finite, the sum of multiplicities equals the product of degrees.. else the (dimension? degree? sums over irreducible components?) of the intersection is less than or equal to the difference in degrees. -Help is appreciated! - -REPLY [10 votes]: In that wonderful book of Fulton, the more general result is at the bottom of page 223, where it says that for r pure dimensional schemes in P^n, whose codimensions add to at most n, the product of their degrees is at least as great as the sum of the degrees of the irreducible components of their intersection. Thus for instance, if three quadric hypersurfaces in P^4 have a common curve of degree 8, their intersection has no further components.<|endoftext|> -TITLE: Self-linkage of the orthogonal group $O_n({\mathbb R})$. -QUESTION [5 upvotes]: In Exercise 153 of my list, it is proved that the connected components $SO_2({\mathbb R})$ and $O_2^-({\mathbb R})$ of the orthogonal group are linked as curves in the three-dimensional sphere defined by $\|M\|=1$ in the operator norm. -Question. Consider the case of $n\times n$ matrices. The real orthogonal group has two connected components $SO_n({\mathbb R})$ and $O_n^-({\mathbb R})$, each one a submanifold of dimension $n(n-1)/2$. Let $s_1(M)\ge\cdots\ge s_n(M)(\ge0)$ be the singular values of a matrix $M$. We have $s_1(M)=\|M\|$. Define the set $V_n$ by the identities $s_1(M)=\cdots=s_{n-1}(M)=1$. Is it a submanifold of dimension $n^2-n+1$ ? In $V_n$, are $SO_n({\mathbb R})$ and $O_n^-({\mathbb R})$ linked ? -Nota. The exercise in $M_2({\mathbb R})$ is included in the forthcoming second edition of my book ``Matrices'' as Exercise 21 in chapter 10. - -REPLY [2 votes]: The dimension of $V_n$ is at most $\binom{n}{2} +n$. For $n>2$, this is less than $2 \binom{n}{2} +1$. -It is convenient to solve a more general problem. Fix $s_1 \geq s_2 \geq \cdots \geq s_n \geq 0$. Let $M(s_1, \ldots, s_n)$ be the space of matrices with singular values $s_i$. Let $d_1$, $d_2$, ..., $d_k$ be the list of multiplicites with which the $s$'s occur, so $\sum d_i = n$. For example, if the $s$'s are $(1,1,1,\ldots, 1, r)$ then $(d_1, d_2) = (n-1, 1)$. -I claim that $\dim M(s) = 2 \binom{n}{2} - \sum \binom{d_i}{2}$. Proof: The group $O(n) \times O(n)$ acts transitively on $M(s)$, so the dimension of $M(s)$ is $\dim ( O(n) \times O(n)) = 2 \binom{n}{2}$ minus the dimensional of the stabilizer of the diagonal matrix $d(s) := \mathrm{diag}(s_1, s_2, \ldots, s_{n-1}, s_n)$. We compute the Lie algebra of the stabilizer. Let $g$ and $h$ be skew-symmetric matrices. Then, to first order, $e^g d(s) e^h = d(s)$ if and only if $g d(s) + d(s) h=0$. -For every $(i,j)$, this gives the equations $g_{ij} s_j + s_i h_{ij}$ and $s_i g_{ij} + s_j h_{ij} =0$. If $s_i \neq \pm s_j$, this forces $g_{ij} = h_{ij}=0$. If $s_i = \pm s_j$ then the sign must be $+$, as the $s_i$ are nonnegative, and we get $g_{ij} = - h_{ij}$. So the Lie algebra of the stabilizer has a basis element for every $(i,j)$ with $s_i = s_j$, and we see that the dimension of the stabilizer is $\sum \binom{d_i}{2}$. -You are interested in $\bigcup_{0 \leq r \leq 1} M(1,1,\ldots,1,r)$. Each space in the union has dimension $2\binom{n}{2} - \binom{n-1}{2} = \binom{n}{2} + n-1$, except for the $r=1$ term which is even smaller. Thus, the union has dimension at most $\binom{n}{2} +n$. I have not been able to work out whether the union is a manifold near the boundary points $r=0$ and $r=1$. -One way to salvage your question would be to look at matrices whose singular values are $(1,1-r/(n-1), 1-2r/(n-1), \ldots, 1-(n-2)r/(n-1), 1-r)$, for $r \in [0,1]$.<|endoftext|> -TITLE: Estimating direction from a distribution on a circle -QUESTION [8 upvotes]: Let there be $n$ points on a unit circle. It is known they come from "normal" distribution around particular unknown direction (i.e. sum of 2 "normal" distributions on circle - one centered at point $p$ and the other at its opposite $-p$). -What is the best way to estimate this direction? By best I mean an algorithm that is a. analytical, b. efficient and c. simple. - -REPLY [7 votes]: Ok, so now I will describe why Niels's estimator works so well. Take a bimodal and symmetric circular density function $f$ with modes $p$ and $-p$ (we will assume that $p$ is positive) such as the one plotted in my previous answer. Let $\Theta_1, \Theta_2, \dots, \Theta_N$ be $N$ observations drawn from $f$. -Niels's estimator first computes the complex numbers $e^{i 2 \Theta_n}$ and takes their average -$$ \bar{C} = \sum_{n=1}^{N} e^{i 2 \Theta_n} .$$ -The estimate, denoted $\hat{p}$, is given by taking the complex argument of $\bar{C}$ and dividing by 2, that is -$$ \hat{p} = \frac{\angle{\bar{C}}}{2}$$ -where $\angle{\bar{C}} \in [0,2\pi)$ denotes the complex argument. The next theorem describes the asymptotic properties of this estimator. I use the notation $\langle x \rangle_{\pi}$ do denote $x$ taken to its representative inside $[-\pi, \pi)$. So, for example, $\langle 2\pi \rangle_{\pi} = 0$ and $\langle \pi + 0.1 \rangle_{\pi} = -\pi + 0.1$. - -Theorem: - Let $\lambda$ denote the difference $\lambda = \tfrac{1}{2}\langle 2\hat{p} - 2p \rangle_{\pi}.$ - Then $\lambda$ converges almost surely to zero as $N \rightarrow \infty$ and the distribution of the normalised - difference $\sqrt{N}\lambda$ - converges to the zero mean normal with - variance - $$ \frac{\sigma_s^2}{c} $$ - where - $$ \sigma_s^2 = \int_{-\pi/2}^{\pi/2}\sin^2(\theta) f(\langle \theta + p \rangle_\pi) d\theta \qquad \text{and} \qquad c = \int_{-\pi/2}^{\pi/2}\cos(\theta) f(\langle \theta + p \rangle_\pi) d\theta. $$ - -The definition of the difference $\lambda$ might seem a little strange at first, but it is actually very natural. To see why note that $p$ and the estimate $\hat{p}$ are both in $[0,\pi)$ but, for example, if $p = 0$ and $\hat{p} = \pi - 0.01$ then the difference between these is not $\pi - 0.01$, because the two modes are actually very close to aligned in this case. The correct difference is $\lambda = \tfrac{1}{2}\langle 2(\pi-0.01) - 2 \times 0 \rangle_{\pi} = 0.01$. -The proof of this theorem follows from a very similar argument to Theorem 6.1 (page 87) from my thesis. The original argument is due to Barry Quinn. Rather than restate the proof I'll just give you some convincing numerical evidence. -I've run some simulations for the case when the noise is a sum of two weighted von Mises circular distributions with concentration parameter $\kappa$. So, when $\kappa$ is large the distribution is concentrated and looks something like the picture on the left below ($\kappa = 20$ in this case) and when $\kappa$ is small the distribution is quite spread out and looks something like the picture on the right below ($\kappa = 0.5$). We obviously expect the estimator to perform better when the distribution is quite concentrated ($\kappa$ is large). -    -(source left) (source right) -Here are the results. The plot below show the simulated variance of $\lambda$ after 5000 trials (the dots) versus the variance predicted in the theorem above for a range of values of $\kappa$ and number of observations $N$. You can see that the theorem does a very good job of accurately predicting the performance if $\kappa$ isn't too small. - (source) -There is still an open question as to whether this is the best estimator (in the sense of maximally reducing the variance of $\lambda$). It would be possible to derive a Cramer-Rao bound for this estimation problem to give an idea of the best possible performance of an unbiased estimator. I suspect that this estimator performs very near the Cramer-Rao bound. So, in that sense it is close to best possible.<|endoftext|> -TITLE: A stochastic process that is 1st and 2nd order (strictly) stationary, but not 3rd order stationary -QUESTION [6 upvotes]: I asked this question on stats.stackexchange.com a little while back but didn't get an answer. It was suggested that I post it here at the time. There appears to be some migratory problems going on over there. Hopefully, this question is seen as appropriate. -Let $X(t)$ be a stochastic process. We say that $X(t)$ is Nth-order stationary if for every set of ''times'' $t_1, t_2, \dots, t_N$ we have that the joint cumulative density functions -$$F_{X(t_1),X(t_2),\dots,X(t_N)} = F_{X(t_1 + \tau),X(t_2 + \tau),\dots,X(t_N + \tau)}$$ -for all ''time shifts'' $\tau$. This is quite a strong condition, it says that the joint statistics don't change at all as time shifts. -For example, a 1st order stationary process is such that $F_{X(t_1)} = F_{X(t_2)}$ for all $t_1$ and $t_2$. That is, the $X(t)$ are all identically distributed. It is quite easy to see that a 1st order stationary process need not be 2nd order stationary. Simply assign a correlation structure to say $X(t)$, $X(t+1)$, $X(t+2)$ that does not correspond to a (symmetric) Toeplitz matrix. That is, in vector form, the covariance matrix of $[ X(t), X(t+1), X(t+3)]$ could be given as -$$\left[\begin{array}{cc} -\sigma^2 & a & b \newline -a & \sigma^2 & c \newline -b & c& \sigma^2 -\end{array}\right]$$ -for $a,b,c$ distinct. This is now not 2nd order stationary because $E[X(t)X(t+1)] = a$ and, time shifting by 1 we have $E[X(t+1)X(t+2)] = c \neq a$. -In a similar way (presumably) a process that is 1st and 2nd order stationary need not be 3rd order stationary and this leads to my question: - -Does somebody have a nice example of a stochastic process that is both 1st and 2nd order stationary, but not 3rd order stationary? - -Motivation: Some of the material I have been expected to teach recently has included stochastic processes and I feel this is a gap in my knowledge. Although no student has actually asked this question yet, I think it is natural enough for it to be asked at some point and I would like to have a neat and simple example. - -REPLY [9 votes]: How about this: -Consider $X_1,X_2$ i.i.d. with $\mathbb{P}(X_i = 1) = \mathbb{P}(X_i = 0) = 0.5$. Now define $X_3 = X_1 + X_2 (\text{mod }2)$. Notice that $X_3$ is independent from $X_1$ and from $X_2$ individually. However the three variables $X_1,X_2,X_3$ are not jointly independent. -Now consider a sequence $X_1,X_2,X_3,\ldots$ such that each triplet $(X_{3n+1},X_{3n+2},X_{3n+3})$ has the same distribution as $(X_1,X_2,X_3)$ and all these triplets are independent. -With this one has that all $X_n$ have the same distribution as $X_1$, so that the process is 1 stationary. Also, all pairs $X_i,X_j$ with $i\neq j$ are independent and hence have the same distribution (this gives 2 stationarity). However the triplet $(X_2,X_3,X_4)$ has a different distribution from $(X_1,X_2,X_3)$ (in fact, $X_2,X_3$ and $X_4$ are jointly independent).<|endoftext|> -TITLE: Kontsevich's conjectures on the Grothendieck-Teichmüller group? -QUESTION [20 upvotes]: Reading Kontsevich's "Operads and Motives in Deformation Quantization", I was wondering about the state of the many conjectures concerning the Grothendieck-Teichmüller group in chapter 4. (Also, where one could read the proof of theorem 6 by M. Nori?) For example I'm interested in the action of GT on deformation quantization mentioned in ch. 4.6., and an update on the issues of chapter 5. Where one could read more about all that? -Edit: The conj.s I'd like to know more about are: That (the algebra generated by) periods of mixed Tate motives come from Drinfeld associators (p.30); That GT comes from the motivic Galois group's action on the Periods (p.30); That GT is $Aut(Chains(C_2))$ (p.31); What is the "universal map $P_{\mathbb{Z},Tate}$$\longrightarrow$$P_{\mathbb{Z},Tate}$" on p. 32?; Where can one read more about the action of GT on deformation quantization (p.32,33)? -Edit: "Theorem 6 (M. Nori)" asked about above is a sheaf version of a theorem by Beilinson. Nori's article about it with the proof (the statement in question is Basic lemma (first form)) and the reference to Beilinson's paper. An article by Morava contains many new ideas, some remarks on a motivic version of the little disk operad, and a possible algebraic topology frame for Kontsevich's ideas on motives and deformation quantization, and how that would fit to Connes' & Kreimer's Galois theory of renormalization. -Edit: Conc. "GT is $Aut(Chains(C_2))$" here and here new articles (communicated by B.V., thanks!). -Edit: Mathilde Marcolli mentions in this article, on her and Connes' work relating renormalization and a "cosmic Galois group", that Kontsevich developed a renormalization theory continuing his article above and relating "in a natural setting" to Connes'/Marcolli's motivic galois action. Her article ends with some remarks on how Beilinson's conjectures may be viewed "extremly suggestive" as something like renormalization, hinting at geometric interpretations of L-values at non-integer points. It would be great if someone knew where to read more about both issues. -Edit: Spencer Bloch's thoughts (video lecture) on motives and renormalization relate to the theme too. - -REPLY [3 votes]: I am late to this discussion, but it seems that briefly after it died out, Vasily Dolgushev established part of what you are looking for (I just heard him advertize this result at GAP XI in Pittsburgh): -that the connected components of the space of (stable) deformation quantizations of a Euclidean space is indeed a torsor over the Grothendieck-Teichmüller group is shown here - -Vasily Dolgushev, Stable Formality Quasi-isomorphisms for Hochschild Cochains I (arXiv:1109.6031) - -with a review in - -Vasily Dolgushev, Exhausting formal quantization procedures (arXiv:1111.2797), - -see theorem 3.1 there. -Aspects of the generalization to spaces other than Euclidean are then discussed in - -Vasily Dolgushev, Christopher Rogers, Thomas Willwacher, Kontsevich’s graph complex, GRT, and the deformation complex of the sheaf of polyvector fields (arXiv:1211.4230) - -I should collect some more pointer on the nLab at deformation quantization -- Motivic Galois group action on the space of quantizations<|endoftext|> -TITLE: Which rights do mathematicians usually have on their published works and how do they use them? -QUESTION [13 upvotes]: Yesterday my first work in mathematics was sent to a publisher, and of course I'm interested in its usefulness. But I know, that sometimes it is hard to get a paper, it is not available for free. I hope my paper will not be of that kind. I hope it will be simple to find it and to read it, I hope a few hundred people will read it. Therefore I have this question. When do mathematicians usually lose their right to share their papers on the internet, why some of them doesn't like to do this and which ways are there to overcome these "sharing difficulties"? - -REPLY [14 votes]: Read any contract you sign carefully, otherwise you may lose rights you wanted to keep. -In 2000, CRC Press sued Eric Weisstein because he posted free updates to the web of a mathematics book he had written and published with them. -Usually non-commercial publishers like the AMS allow you to keep any rights you want, while commercial publishers often try to get the rights for themselves (but will usually allow you to keep them if you make a fuss). - -REPLY [6 votes]: After your paper is accepted, you should politely ask the publisher for a version of the copyright form that either a) places the paper in the public domain or b) allows you to retain copyright. In my experience, a) sometimes works and b) always works.<|endoftext|> -TITLE: Orbits in commutative groups. -QUESTION [6 upvotes]: Let A be finite commutative group say $(Z_m)^h$. I will say that $S \subset A$ is an orbit if exist group $H$ -which acts on A such that $S$ is an orbit of $H$. -Can one give a simple characterization of all orbits of $(Z_m)^h$? -By action on $A$ I mean automorphisms of a group A. - -REPLY [3 votes]: The abelian group in question is the product of its Sylow-$p$ subgroups, which are preserved by automorphisms. Therefore the orbits in it are the products of orbits in the Sylow $p$-subgroups. Therefore, we may consider the case where $m=p^k$ for some prime $p$ and some natural number $k$. -I can answer this question for maximal orbits (orbits under the full automorphism group). I think the more general questions may not have a nice answer. -In $(Z_{p^k})^h$, there are precisely $k+1$ orbits of the full automorphism group, represented by $e, pe, \ldots, p^ke$, where $e=(1,0,\ldots,0)$. The orbit of $p^i e$ consists of those vectors where the gcd of entries divides $p^i$, but not $p^{i+1}$ (except for $i=k$, where the orbit is just the element $0$). -For a general finite abelian group, this problem was solved more than a 100 years ago by Miller, and also discussed by Birkhoff and Baer. For the exact references, as well as a modern treatment see http://arxiv.org/abs/1005.5222.<|endoftext|> -TITLE: Propagation of an error in the LMO invariant? (Revision: I don't think LMO is wrong!) -QUESTION [19 upvotes]: Edit: I think LMO is correct. Massuyeau has a nice explanation here. -Edit: Renaud Gauthier has retracted the claim of an error in the foundations of the LMO construction, and has withdrawn both preprints from arXiv. -Original post follows: - -In two papers posted to the arXiv in the past few days, Renaud Gauthier claims to have discovered an error in the definition of the framed Kontsevich integral used in the construction of the LMO invariant. I have no reason to doubt him. I looked at these papers some years back and recall that something funny was going on with the normalization under handle-slides. I got the wrong multiple of the normalization factor $\nu$, just as Gauthier does. Gauthier fixes the normalization so that it works, but then remarks that subsequent results depending on this construction need to be carefully checked. -My question is whether anyone knows of results that use the fine details of the definition of the framed Kontsevich integral (or LMO invariant or Aarhus integral) which are now thrown into doubt because of this error. -Edit: Here are links to the papers. -On the foundations of the LMO invariant -On the LMO Invariant, the Wheeling Theorem, and the Aarhus Integral -Edit 2: -Moskovich has started a blog post on this. Thanks to Ryan Budney for pointing this out. -A problem with LMO? - -REPLY [17 votes]: Having been a part of the LMO story from its beginning, and having read and checked all relevant papers carefully at the time, and having taken part in many cross-checks that the LMO invariant passed (normalization-compatibility with Reshetikhin-Turaev, various explicit computations), and having consulted on email with my collaborators at the time, and having superficially read through Gauthier, my informed guess is that in this particular case of inconsistency the first place to look for a problem is in Gauthier, not in LMO.<|endoftext|> -TITLE: Degrees of self-maps of aspherical manifolds -QUESTION [19 upvotes]: In "Infinitesimal computations in Topology", Publ IHES, page 318, Dennis Sullivan writes "Recall any self-mapping -of a Riemann surface of genus $g>1$ either has -degree $0$ or degree $\pm 1$." There is a footnote to that sentence, saying this statement is more generally -"[t]rue for a closed $K(\pi,1)$-manifold of non-zero Euler characteristic." Why is that true (even the surface case is a mystery to me)? - -REPLY [9 votes]: Here's an argument that the map on fundamental groups $\phi:M\to M$ is surjective if $deg(\phi)\neq 0$, where $M$ is an $n$-dimensional closed orientable manifold and $\chi(M)\neq 0$. Suppose $deg(\phi)\neq 0$, then there exists a finite-sheeted cover $\tilde{M}\to M$ such that $\phi_{\#}(\pi_1(M))=\pi_1(\tilde{M})$ (as noted by Richard Kent, if $\tilde{M}\to M$ were an infinite cover, then $deg(\phi)=0$). Consider the lift $\tilde{\phi}:M\to\tilde{M}$. Then $deg(\tilde{\phi})\neq 0$ as well. Then $\tilde{\phi}^\ast:H^n(\tilde{M},\mathbb{Q})\to H^n(M,\mathbb{Q})$ is an isomorphism of vector spaces. By Poincare duality, for any $\alpha\in H^k(\tilde{M},\mathbb{Q})$ there exists $\beta\in H^{n-k}(\tilde{M},\mathbb{Q})$ such that $\alpha\cup\beta = [\tilde{M}]$. Then $\tilde{\phi}^\ast(\alpha\cup\beta)=\tilde{\phi}^\ast(\alpha)\cup\tilde{\phi}^{\ast}(\beta) = \tilde{\phi}^\ast[\tilde{M}]\neq 0$, so $\tilde{\phi}^\ast(\alpha)\neq 0$. Thus $\tilde{\phi}^\ast$ is an injection from $H^\ast(\tilde{M},\mathbb{Q})\hookrightarrow H^\ast(M,\mathbb{Q})$. But the covering projection $\tilde{M}\to M$ induces an injection $H^\ast(M,\mathbb{Q})\hookrightarrow H^\ast(\tilde{M},\mathbb{Q})$, so we see that $H^\ast(\tilde{M},\mathbb{Q})\cong H^\ast(M,\mathbb{Q})$ (as graded vector spaces), and therefore $\chi(\tilde{M})=\chi(M)$. Thus the cover $\tilde{M}\to M$ is degree one, and we see that $\phi_{\#}:\pi_1(M)\to \pi_1(M)$ is a surjection. -If $\pi_1(M)$ is Hopfian, then $\phi_{\#}$ is an isomorphism, and we conclude that $\phi$ is a homotopy equivalence when $M$ is a $K(\pi,1)$, and therefore $deg(\phi)=\pm 1$. However, $\pi_1(M)$ might not be Hopfian. Assume $n\geq 4$ (since $n=2$ is Hopfian, and is taken care of in Richard Kent's answer) and $M$ is aspherical. Then $Ker(\phi_{\#})$ is finitely normally generated (since $\pi_1(M)$ is finitely presented). Choose a link $L\subset M$ such that $Ker(\phi_{\#})$ is normally generated by $\pi_1$ of the components of $L$. We surger $M$ by adding 2-handles along the components of $L$ to get $M'$ such that $\pi_1(M')=\pi_1(M)$ ($M'$ might not be aspherical). But we may extend the map $\phi_{|M-\mathcal{N}(L)}:M-\mathcal{N}(L)\to M$ to a map $\phi': M' \to M$ by mapping the attached 2-handles into $M$, which is possible since each component of $L$ maps to a contractible loop in $M$. Since the cores of the 2-handles are codimension $\geq 2$, we see that $deg(\phi')=deg(\phi)$. But since $\phi'_{\#}:\pi_1(M')\to \pi_1(M)$ is an isomorphism, $\phi$ is homotopic to the classifying map, there's a gap here so we conclude that $deg(\phi')=\pm 1$, so $deg(\phi)=\pm 1$.<|endoftext|> -TITLE: Can a module be an extension in two really different ways? -QUESTION [21 upvotes]: (Edit: I've realized that there was an error in my reasoning when I was convincing myself that these two formulations are equivalent. Hailong has given a beautiful affirmative answer to my first question in the case of finite type modules over a noetherian commutative ring. Mariano has given a slick negative answer to the question for non-finite-type modules. Greg has given a beautiful negative answer to my "alternative formulation" even in the finite type case over a noetherian commutative ring. I'm accepting Hailong's answer since that's the one I imagine people will be most immediately interested in if they find this question in the future.) -Suppose we're working the category of modules over some ring $R$. Suppose a module $E$ is an extension of $M$ by $N$ in two different ways. In other words, I have two short exact sequences - -\begin{array}{ccccccccc} - 0&\to &N&\xrightarrow{i_1}&E&\xrightarrow{p_1}&M&\to &0\\ - & & \wr\downarrow ?& & \wr\downarrow ?& & \wr\downarrow ?\\ - 0&\to &N&\xrightarrow{i_2}&E&\xrightarrow{p_2}&M&\to &0 -\end{array} - - -Must there be an isomorphism between these two short exact sequences? - - -Alternative formulation -$Ext^1(M,N)$ parameterizes extensions of $M$ by $N$ modulo isomorphims of extensions. Suppose I'm interested in parameterizing extensions of $M$ by $N$ modulo abstract isomorphisms (which don't have to respect the submodule $N$ or the quotient $M$). One obvious thing to note is that there is a left action of $Aut(M)$ on $Ext^1(M,N)$, and that any two extensions related by this action are abstractly isomorphic. Similarly, there is a right action of $Aut(N)$ so that any two extensions related by the action are abstractly isomorphic. - -Does the quotient set $Aut(M)\backslash Ext^1(M,N)/Aut(N)$ parameterize extensions of $M$ by $N$ modulo abstract isomorphism? - -Note: I'm not asking whether all abstract isomorphisms are generated by $Aut(M)$ and $Aut(N)$. They certainly aren't. I'm asking whether for every pair of abstractly isomorphic extensions there exists some isomorphism between them which is generated by $Aut(M)$ and $Aut(N)$. - -REPLY [20 votes]: It is worth noting some very interesting cases when the answer is yes. An amazing result by Miyata states that if $R$ is Noetherian and commutative, $M,N$ are finitely generated and $E \cong M\oplus N$, any exact sequence -$ 0 \to M \to E \to N \to 0$ must split! -This holds true slightly more generally, when $R$ is (not necessarity commutative) module-finite over a Noetherian commutative ring. Also, the statement holds for finitely generated pro-finite groups, see Goldstein-Guralnick, J. Group Theory 9 (2006), 317–322. -Added: in fact, this paper by Janet Striuli may be useful for you. She addressed the question: if two elements $\alpha, \beta \in \text{Ext}^1(M,N)$ give isomorphic extension modules, how close must $\alpha, \beta$ be? Her Theorem 1.2 extend Miyata's result (let $I=0$).<|endoftext|> -TITLE: In general... (convention in mathematical papers) -QUESTION [6 upvotes]: I often see in papers something like: - 1) This is in general not true -or - 2) This is not true in general -Which I personally would consider to be written formally as something like -1) $\forall x: \neg p(x)$ -2) $\exists x: \neg p(x)$ -But I wonder whether this is generally what is meant and if the mathematical community is careful about how they use the word "general" or if it used in a more colloquial sense. Being somewhat of an outsider I find this hard to judge. Partly as it is often used as an aside and rarely a formalisation of the statement is present to check it against. -It's the sort of thing you just can't look up. - -REPLY [8 votes]: I don't know about what this means in general. I use it as a way of avoiding twisting my prose into horrendously convoluted statements whilst avoiding the possibility that some smart alec is going to pick up on a technicality. -More precisely, I use it when I wish to say something like "Not all snarks are boojums" but the sentence would work much better (either for grammatical reasons or to better convey the intended meaning) if I could just say, "snarks are not boojums". That's false as stated[1], so to avoid either saying anything actually incorrect or that someone's going to say, "But what about ...", I say "in general, snarks are not boojums". -What's important here is that I use it mostly in the prose section of a paper or seminar when I'm trying to focus the reader or listener's attention on the important facets of whatever it is that I'm explaining. So getting in to a long diversion of which snarkss are not boojumss (is it the lesser-spotted or the warbler variety?) would be counterproductive. Saying, "not all snarks are boojums" tends to draw ones attention to that class of snarks which are boojums. Saying "snarks are not boojums" is almost guaranteed to get some smart alec saying, "But what about greater-wrinkled snarks?" (especially in a lecture). So "in general, snarks are not boojums" has the triple benefit of 1) being true, 2) focussing the attention on the key point, and 3) not grammatically convoluted. -[1]: Banker and Carroll, Identifying subspecies of snark (1874)<|endoftext|> -TITLE: Does "compact iff projections are closed" require some form of choice? -QUESTION [38 upvotes]: There are many equivalent ways of defining the notion of compact space, but some require some kind of choice principle to prove their equivalence. For example, a classical result is that for $X$ to be compact, it is necessary and sufficient that every ultrafilter on $X$ converge to a point in $X$. The necessity is easy to prove, but the sufficiency requires a choice principle to the effect that every filter can be extended to an ultrafilter. -Some years ago I heard from a very good categorical topologist that many, perhaps most of the useful properties of compact spaces $X$ readily flow from the fact that for every space $Y$, the projection map $\pi: X \times Y \to Y$ is closed. Of course that is a very classical consequence of compactness which can be left as an exercise to beginners in topology, and I was struck by the topologist's assertion that you could in fact use this as a definition of compactness, and that this is a very good definition for doing categorical topology. (I am still not sure what he really meant by this, but that's not my question.) -My own proof that this condition implies compactness goes as follows. Let $Y$ be the space of ultrafilters on the set $X$ with its usual compact Hausdorff topology, and suppose the projection $\pi: X \times Y \to Y$ is a closed map. Let $R \subseteq X \times Y$ be the set of pairs $(x, U)$ where the ultrafilter $U$ converges to the point $x$. One may show that $R$ is a closed subset, so the image $\pi(R)$ is closed in $Y$. But every principal ultrafilter (one generated by a point) converges to the point that generates it, so every principal ultrafilter belongs to $\pi(R)$. Now principal ultrafilters are dense in the space of all ultrafilters, so $\pi(R)$ is both closed and dense, and therefore is all of $Y$. This is the same as saying that every ultrafilter on $X$ converges to some point of $X$, and therefore $X$ is compact. -I was at first happy with this proof, but later began to wonder if it's overkill. Certainly it uses heavily the choice principle mentioned above, and my question is whether the implication I just proved above really requires some form of choice like that. - -REPLY [7 votes]: This question is related to the notion of proper map, of which there is quite a lot in Bourbaki and also in my book `Topology and groupoids'. Note the elegant Bourbaki definition: a map $f: X \to Y$ is proper if for all spaces $Z$ the map $$f \times 1_Z: X \times Z \to Y \times Z$$ is a closed map. -Once I was teaching a second year course in analysis and realised how nice were the proofs involving sequences. So I started looking at questions like `what is the one point sequential compactification?' (add an extra point to which the non convergent sequences converge!). It all worked out quite well and was published as -[12]. ``Sequentially proper maps and a sequential compactification'', J. London Math Soc. (2) 7 (1973) 515-522.<|endoftext|> -TITLE: Spanning trees in planar graphs -QUESTION [10 upvotes]: Is the 3-connected graph(s) on $n$ vertices with the minimum number of spanning trees always planar? - -REPLY [2 votes]: Edit. As it turned out I was not using the right switch for plantri. -This is therefore not an answer anymore but rather an extended comment for the case $n=11.$ -As it turns out the minimal number of spanning trees of a 3-connected planar graph of order 11 is 3965 and is attained by the graph on the figure bellow. -alt text http://shrani.si/f/3f/IS/jiP8yvQ/pmin.png -As for the non-planar 3-connected graph I am yet to compute the answer. I'll post the result here as soon as it gets computed.<|endoftext|> -TITLE: Preferred embedding of finite metric spaces in riemaniann manifolds of given dimension -QUESTION [8 upvotes]: In search for a Machian formulation of mechanics I find the following problem. In Machian mechanics absolute space does not exists, and the only real entities are the relative distances between the particles. As a consequence, the configuration space of a N-particle system is the set of the distances on a set of N elements. Actually these distances are usually required to be isometrically embeddable in $\mathbb{R}^3$. But if absolute space does not exists, this requirement appears to be not appropriate. The natural generalization it therefore to admit any possible distance as physically acceptable, and to find a preferred way to derive a 3-geometry, possibly non-flat, form a generic distance. -To be more specific, consider the following simple example. Let A be a metric space with 3 elements. There are infinitely many bi-dimensional riemaniann manifolds (surfaces) is which A can be isometrically embedded. There is however a preferred embedding, namely the embedding into a plane. The existence of a preferred embedding defines a preferred value for the angles between the geodetics joining the points, which in this case are simply the angles of the triangle defined by the distance between the points. -Suppose now that A has four point. In general this metric space cannot be isometrically embedded in a 2-plane. The problem therefore is the following: is there a preferred isometric embedding of this metric space in a 2-surface, or equivalently, there is a preferred way for defining the values of the angles between the geodetics? -In more forma way, the problem is the following: is there a preferred isometric embedding of a finite metric space in a riemaniann manifold of given dimension? - -REPLY [5 votes]: Your question is not well stated. -In particular I did not understand why embedding into a plane is "preferred embedding". -Here are some associations... -4-point case. -A generic 4-point metric space can be isometrically embedded into two different model planes (i.e. simply connected surfaces of constant curvature $K$, which is eiter sphere, Euclidean plane or Lobachevsky plane depending on sign of $K$). -Thus you have two values of curvature $K_1\le K_2$ associated to (almost) any 4-point metric space. -In this case the metric space can be isometrically embeded into a model 3-space of curvature $K_1\leq K\leq K_2$. -In fact for any 4-point metric space $M$ there is an subinterval $\mathbb I_M$ of $[-\infty,\infty)$ such that $M$ can be isometrically embedded into a model 3-space of any curvature $K\in \mathbb I_M$. -(We assume that model space of curvature $-\infty$ is an $\mathbb R$-tree.) -Nearly all this was discovered by A. Wald in 1936 or so.<|endoftext|> -TITLE: What are the zero entropy invariant measures for an Anosov geodesic flow? -QUESTION [13 upvotes]: Let $M$ be the double-torus with a hyperbolic Riemannian metric. The geodesic flow on the unit tangent bundle $T^1M$ has many invariant Borel probability measures. In particular there are closed geodesics projecting to each non-trivial homotopy class of $M$, and one can support an invariant probability measure on each of these. Also one can take convex linear combinations of these invariant measures. -My question is the following: Are these the only invariant measures of zero metric (Kolmogorov-Sinai) entropy? -More generally, what are the zero entropy invariant probability measures of an Anosov geodesic flow? -Also I'm interested in the same question for shifts on finitely many symbols (i.e. What are the zero entropy invariant measures?). -Besides references giving an answer, other related references are of course very welcome. - -REPLY [8 votes]: Another type of answer. It can be shown that generically (in the sense of Baire), an invariant probability measure for the geodesic flow is ergodic, of full support, with entropy zero, and not mixing. See the article: - K. Sigmund. On the space of invariant measures for hyperbolic flows, Amer. J.Math. 94 (1972), 31–37 -For shifts, it is due to Oxtoby -J. C. Oxtoby. On two theorems of Parthasarathy and Kakutani concerning the shift transformation, (1963) Ergodic Theory (Proc. Internat. Sympos., Tulane Univ., New Orleans, La., 1961) pp. 203–215 Academic Press, New York. -These results can be generalized to any nonelementary negatively curved manifold, or shifts with infinitely many symbols, see for example " Generic measures for hyperbolic flows on non compact spaces", written with Yves Coudene , Israël Journal of Maths vol 179 (2010). (The proof that in this situation, generic measures have entropy zero is not written in this article, and still unpublished, but works without any difficulty.)<|endoftext|> -TITLE: Does constructing non-measurable sets require the axiom of choice? -QUESTION [21 upvotes]: The classic example of a non-measurable set is described by wikipedia. However, this particular construction is reliant on the axiom of choice; in order to choose representatives of $\mathbb{R} /\mathbb{Q}$. -"Since each element intersects [0,1], we can use the axiom of choice to choose a set containing exactly one representative out of each element of R / Q." - -Is it possible to construct a non-measurable set (in $\mathbb{R}$ for example) without requiring the A.o.C.? - -REPLY [8 votes]: To add to the answer about what is the weakest choice-like principle required: let me take this opportunity to mention Consequences of the Axiom of Choice by Rubin and Howard. This is form 93 in the book and no known exact equivalents are listed for it. An extensive implication table is available. -For instance, as pointed out, the existence of a non-trivial ultrafilter on $\omega$ is sufficient, and BPI (the boolean prime ideal theorem) implies the existence of such an ultrafilter. According to the book, neither of these implications is reversible. -Another intermediate principle the book mentions is the sock selection principle (every family of pairs has a choice function). This is implied by BPI and implies the existence of a non-measurable set, and neither of these is reversible.<|endoftext|> -TITLE: Motivic Cohomology vs. Chow for singular varieties? -QUESTION [21 upvotes]: I'm absolutely new to this stuff I'm asking about, so I hope this is not nonsense. -If X is a smooth scheme over a perfect field k, I can study its motivic cohomology in the sense of Voevodsky and Morel. -More precisely, to $\mathbb{Z}$ there is associated an Eilenberg-MacLane T-Spectrum. I'll write $H^{p,q}(X,\mathbb{Z})$ for the motivic cohomology of $X$ defined by Hom into this Spectrum. -One fact for smooth $X$ now is that -$$ H^{2p,p}(X,\mathbb{Z}) = CH^p(X). $$ -Now I've got two questions: -1) Can I feed something singular into this Voevodsky-Morel machine? -2)If the answer to 1 is yes, are there some maps from $ H^{2p,p}(X,\mathbb{Z})$ to $CH^p(X)$? Maybe defined by some spectral sequence? I'm really interested just in some maps, no isomorphisms. - -REPLY [27 votes]: The answer to question 1) is yes. However, Chow groups do not form what we should call a cohomology theory, but are part of a Borel-Moore homology theory. This ambiguity comes from the fact that, by Poincaré duality, motivic cohomology agrees with motivic Borel-Moore homology for smooth schemes (up to some reindexing), while they are quite different for non-regular schemes. You have the same phenomena in K-theory: K-theory of vector bundles (which defines a kind of cohomology) agrees with K-theory of coherent sheaves (which defines the corresponding Borel-Moore homology) for regular schemes. That said, they are plenty of ways to extend motivic cohomology into a cohomology theory for possibly non-singular schemes; they all agree in char. 0, or if you work with rational coefficients. As you seem to be interested by Chow groups, it seems that what you want is motivic Borel-Moore homology for possibly non-singular schemes (which agrees with Bloch's higher Chow groups, whence gives classical Chow groups for the appropriate degrees). -To define motivic Borel-Moore homology of $X$ (separated of finite type over a field $k$ of char. 0, unless you tensor everything by $\mathbf{Q}$, or admit resolution of singularities in char. $p$), you may proceed as follows: there is a motive with compact support $M_c(X)$ in $DM(k)$, and we define -$H^{BM}_i(X,\mathbf{Z}(j))=Hom_{DM(k)}(\mathbf{Z}(j)[i],M_c(X)).$ -If you prefer the six operations version, for $f:X\to Spec(k)$ a (separated) morphism of finite type, we have -$H^{BM}_i(X,\mathbf{Z}(j))\simeq Hom_{SH(k)}(f_!f^*\Sigma^\infty(Spec(k)_+)(j)[i],H\mathbf{Z})$ -where $H\mathbf{Z}$ denotes the motivic Eilenberg-MacLane spectrum in $SH(k)$. To answer your question 2), the link with Bloch's higher Chow groups is that -$H^{BM}_i(X,\mathbf{Z}(j))\simeq CH^{d-j}(X,i-2j)$ -for $X$ equidimensional of dimension $d$, and that, for any separated $k$-scheme $X$, -$H^{BM}_{2j}(X,\mathbf{Z}(j))\simeq A_j(X)$ -where $A_j(X)$ stands for the group of $j$-dimensional cycles in $X$, modulo rational equivalence.<|endoftext|> -TITLE: What counts as an 'invited' talk? -QUESTION [36 upvotes]: When you write your CV, what do you count as an invited talk? Does it count if you're invited but not provided funding (this has happened to me several times at AMS special sessions)? What if someone runs into you in the hallway and invites you, but never sends you a 'formal' invitation email? What if you're a PhD student and your advisor asks you to give a talk in the department seminar or in a conference they are organizing? -Also, does it even matter? What if I just put a section in my CV called "talks" and not worry about the distinction? -Thanks in advance! - -REPLY [5 votes]: One idea is to include expository talks to undergraduate and graduate students in the "Service" section of the CV.<|endoftext|> -TITLE: Rugged manifold -QUESTION [20 upvotes]: It is well known that any compact smooth $m$-manifold can be obtained from $m$-ball by gluing some points on the boundary. - -Is it still true for topological manifold? - -Comments: - -To proof the smooth case, fix a Riemannian metric and consider exponential map up to cutlocus. -The question was asked by D. Burago. I made a bet that a complete answer will be given in an hour — please help :) - -REPLY [22 votes]: The answer is yes -- Morton Brown's mapping theorem says that for every closed (connected) topological $n$-manifold $M$ there is a continuous map $f$ from the $n$-cube $I^n$ onto $M$ which is injective on the interior of the cube (for manifolds with boundary, see Remark 1 below). This was proved in early sixties and can be found in M.Brown,"A mapping theorem for untriangulated manifolds," Topology of 3-manifolds, pp.92-94, M.K.Fort, Jr.(Editor), Prentice-Hall, Englewood Cliffs, N.J.,1962. MR0158374 -Main idea of the proof is simple: "Use local PL structures to expand a small $n$-cell in $M$ gradually, until it becomes the whole manifold." -This can be realized the "infinite composition" of engulfings of finitely many points at a time. This argument is not too difficult, so I will try to sketch it. -First consider a closed $n$-cell $C$ in $M$ and a finite set $X=\{x_1,\ldots,x_k\}$ of points of $M$ disjoint from $C$ which we want to engulf. Assume that each $x_i$ lies in some open $n$-cell $U_i$ which intersects $C$. For each $i$, fix a PL structure on $U_i$, and join $x_i$ and some point $y_i\in \partial C$ by a PL arc $\alpha_i\subset U_i$ relative to this structure. If we assume $\dim M\geq 3$ (this is not a restriction because the theorem is clear for $\dim M=1$, and easily follows from the surface classification for $\dim M=2$), we may require that $\alpha_i$'s should be disjoint. The regular neighborhood $Q_i$ of $\alpha_i$ within $U_i$ is a closed $n$-cell, and $Q_i$'s can be made disjoint. Let $h$ be a homeomorphism of $M$ pushing $y_i$ towards $x_i$ within $Q_i$ for each $i$ which is identity outside $Q_i$'s. Then $X\subset h(C)$, and we can apply this process again to $h(C)$ and another finite set $X'\subset M\setminus h(C)$. Repeating this process we obtain a sequence of engulfing homeomorphisms $h_1, h_2,\ldots$. -We can arrage that the uniform limit $f\colon M\to M$ of the composition of engulfing homeomorphisms $f_n=h_n\circ\cdots\circ h_1$ exists and $f(C)=M$. If we choose sufficiently small $Q_i$'s in each stage, one can make sure that $f$ is injective on the interior of $C$ (indeed, we can arrange that for each interior point $x$ there is $n$ such that $f_n(x)=f(x)$). -Remark 1: In Brown's paper, one can find a corollary that "If $M$ is a compact -connected manifold with nonempty boundary $B$, then there is a surjection $f\colon B\times [0,1]\to M$ that restricts to the identity on $B\times 0$ and is injective on $B\times[0,1)$". Notice also that the above theorem can be applied to the closed manifold $B$. Then it follows that any compact topological $n$-manifold (possibly with boundary) can be obtained by identifying some points in the boundary of the $n$-ball. -Remark 2: Berlanga's theorem extended the Brown's theorem to noncompact manifolds: -This theorem states that for every (connected) $\sigma$-compact $n$-manifold $M$, there is a nice kind of surjection $I^n\to \overline{M}$ similar to Brown's map, where $\overline{M}$ denotes the end compactification of $M$.<|endoftext|> -TITLE: Errata for Atiyah–Macdonald -QUESTION [98 upvotes]: Is there a good list of errata for Atiyah–Macdonald available? A cursory Google search reveals a laughably short list here, with just a few typos. Is there any source available online which lists inaccuracies and gaps? - -REPLY [2 votes]: Not sure this can be considered an error, but it confused me: -on p. 65 the homomorphisms in "Let $\Sigma$ be the set of all pairs $(A, f)$, where $A$ is a subring of $K$ and $f$ is a homomorphism of $A$ into $\Omega$ " are not supposed to be injective (I'm not sure if this is the universal usage, but I normally understand into = injective).<|endoftext|> -TITLE: Next (Restricted) B-Smooth Number Problem? -QUESTION [9 upvotes]: Given a bound, $B$, and a list of (small) primes $(p_0, p_1, \dots, p_{n-1})$ is there an efficient algorithm to find the next number greater than $B$ that can be expressed as a product of primes from the list? -More formally, given -$$B = 2^\beta$$ -$$(p_0, p_1, \dots, p_{n-1}),\ \ p_i < poly(\beta)$$ -can one find -$$min_{ q - B > 0 }(q-B), \ \ q = \prod_{i=0}^{n-1} p^{\nu_i}_i $$ -in time/space $O( poly(\beta, n) )$? Even pointers to literature would be helpful... -Note: I saw the Polymath paper on deterministic prime finding in an interval ( Deterministic methods to find primes ) and thats what inspired this question. It's superfluous to have the list be primes, but I've kept it in for simplicity. -Also note that this is a slightly more cleaned up version of this post on cstheory.stackexchange.com. - -REPLY [4 votes]: It's not quite the same problem, since it doesn't involve the threshold B, but it is possible to compute the sequence of all smooth numbers (with factors from a given finite set P of primes $p_i$) in sorted order in time $O(|P|)$ per generated value, in the unit cost model of computation in which all arithmetic operations take constant time. Even with a more realistic cost model the time is polynomial in $|P|$ and the number of bits of the outputs. -The technique is easiest to describe as one of functional programming: if S is the generated output sequence, then S can be expressed as the number 1 concatenated in front of the sorted merge of the $|P|$ sequences $p_i S$. The sequences $p_iS$ needed as inputs to this merge process can be generated by feeding them back from the output of the merge-and-concatenate operations. Merging two sorted sequences can be done in constant time per output value (this is the basic principle behind the standard merge sort algorithm) and a binary tree of pairwise merges allows all $|P|$ sequences to be merged. If the sequences being merged were disjoint, the time per generated value would be $O(\log|P|)$ (as my answer originally erroneously stated) but some values may be generated redundantly in all $|P|$ of the merged sequences; the time taken to generate any particular value is proportional to the size of the subtree of the merge tree where that value appears, which can be as large as $O(|P|)$. -The special case where the primes are 2, 3, and 5 is known as "Hamming's problem" and has a large literature in the functional programming community.<|endoftext|> -TITLE: How do Lie groups classify geometry? -QUESTION [8 upvotes]: I have often heard that Lie groups classify geometry. For example that $O(n)$ is about real manifolds, $U$ is about almost complex manifolds, $SO(n)$ about orientable real manifolds and so on. -I have also heard that manifolds can be defined very generally by patching local pieces via a pseduogroup of morphisms. -My questions are -1) Does this pseduogroup relate to the Lie group? -2) How do Lie groups classify geometry? -3) Is there a geometry for every Lie group or only some of them? -and maybe even -4) Does this classification seem work for algebraic (i.e. non-differential) geometry? -I am wondering whether this Erlangen program classification is actually formalized, or only serves as a slogan or principle. - -REPLY [5 votes]: The examples you give of Lie groups associated to geometric objects are the structure groups of the tangent bundles of manifolds. The tangent bundle is locally trivial, so you have a covering of your manifold and isomorphisms over the $U_i$ (i.e. commuting with the projections to the $U_i$) $\varphi_i:TM|_U \rightarrow \mathbb{R}^n \times U_i$. On the intersections this gives you isomorphisms over $U_i \cap U_j$ $\varphi_i \circ \varphi_j^{-1}: \mathbb{R}^n \times U_i \cap U_j \leftarrow TM|_{U_i \cap U_j} \rightarrow\mathbb{R}^n \times U_i \cap U_j$, i.e. isomorphisms $g_{ij}: \mathbb{R}^n \rightarrow \mathbb{R}^n$. These can be seen as gluing data of your bundle. The "structure group" of your bundle tells you what gluing data you allow: You could require the $g_{ij}$ to be in $GL(n), O(n), SO(n), U(n), Sp(n)$ and so on. The smaller this group of gluing data is, the more special is your bundle, e.g. if you can glue your tangent bundle only using maps from $SO(n)$, then your manifold will be orientable. The buzz word to google for is "reduction of structure group". -I am not sure what you have heard about pseudogroups but there is one consisting of the $\varphi_i$ - open patches homeomorphic to $\mathbb{R}^n$ are such that the tangent bundle is trivial over them, so if you know how to glue your manifold from such patches, you get the relation to the Lie group as above. -Of course you can embed every Lie group into some $GL_n$ and require the $g_{ij}$ to lie in that subgroup, but that seems somewhat arbitrary. -Finally some references on the Erlangen program are Sharpe's book and this, where you can look for more ingredients of the big picture. - -REPLY [3 votes]: R.W.Sharpe, Differential Geometry - Cartan's Generalization of Klein's Erlangen Program<|endoftext|> -TITLE: What are the τ-local rings for a subcanonical Grothendieck topology τ on the category of affine schemes of finite type over Spec(Z)? (specifically for τ=fppf) -QUESTION [25 upvotes]: Let $\tau$ be a subcanonical topology on the category of affine schemes of finite type over $Spec(\mathbf{Z})$. Call this site $(S,\tau)$ or just $S$, and call its associated topos $\mathcal{S}$. Recall that given a topos $T$, we have an equivalence of categories $Hom_{Topos}(T,\mathcal{S})\cong Hom_{Sites}(S,T)$, where $T$ is given the canonical topology. It is a theorem of M. Hakim that $Hom_{Sites}(S,T)$ gives the category of commutative ring objects in $T$ when $\tau$ is the chaotic topology, the category of local rings in $T$ when $\tau$ is the Zariski topology, and the category of "strict local rings" in $T$ when $\tau$ is the étale topology. -In particular, when $T$ is the category of sets, it means that the points of $\mathcal{S}$ are precisely the commutative rings, local rings, and Henselian rings with separably closed residue fields (strict Henselian) respectively. It is also well-known that when $\tau$ is the Nisnevich topology, the local rings are precisely the Henselian rings. -There are other subcanonical Grothendieck topologies on the category of affine schemes of finite type. What are the local rings, for example, when we look at the fppf and fpqc topologies? (Just a guess, but fppf-local is going to be complete local rings? (Wrong! See Laurent Moret-Bailly's comment)). -How about for more obscure subcanonical topologies? - -REPLY [6 votes]: It looks like there was a paper published a few years ago by Stefan Schröer that answers this question: - -Points in the fppf topology, Ann. Sc. Norm. Super. Pisa Cl. Sci Vol. XVII, issue 2 (2017) 419–447, doi:10.2422/2036-2145.201408_001, arXiv:1407.5446. - -From the abstract: -"Using methods from commutative algebra and topos-theory, we construct topos-theoretical points for the fppf topology of a scheme. These points are indexed by both a geometric point and a limit ordinal. The resulting stalks of the structure sheaf are what we call fppf-local rings. We show that for such rings all localizations at primes are henselian with algebraically closed residue field, and relate them to AIC and TIC rings. Furthermore, we give an abstract criterion ensuring that two sites have point spaces with identical sobrification. This applies in particular to some standard Grothendieck topologies considered in algebraic geometry: Zariski, etale, syntomic, and fppf."<|endoftext|> -TITLE: The main ideas in choosing the strategy for numbering displayed math -QUESTION [8 upvotes]: There are different strategies for numbering displayed math. The most common are -1. Number only the formulas you reference to. It makes your paper more clean and gives more freedom to the editor (i.e. making this math inline). -2. Number all displayed math. Even if you don't reference your formulas, take care of those who will read your paper and will want to reference them. -3. Number only very important formulas and the formulas you reference to. -I am thinking, which strategy to choose and I want to make an informed choice. - -Which important strategies have I missed? What are pitfalls and benefits of the strategies, described above? How important are these benefits/pitfalls? - -Ideas from the answers below: -strategy 4. Check with the journal you are planning to publish your contribution. -The similar question was asked here. The most popular answer was "Use strategy 2: number all your displayed math to help the reader to reference your formulas". However it is not clear for me if it is ok to have a 40-page paper with 150 numbered formulas. It seems a bit crazy to see "it follows from (146)". Also it was noted, that this strategy violates "Checkov's gun principle", i.e. fills the paper with irrelevant details (numbers). However the question, stated there was a bit different. I think that that question was completely answered, but not this. For me it is still unclear, whether it is ok to have formula (146) (or formula (1353)) or not. The question made community wiki, so you are free to improve it. - -REPLY [3 votes]: Fiktor, many journals explicitly indicate option (1). Some also ask to use single/double numeration, the list of references in alphabetical/appearance order, and many other things. Therefore, - -Strategy (4): check with the journal you are planning to publish your contribution. - -As for (2) and (3), are you happy if somebody else cites your ms in the form [1, Eq. (7.15)]? Wouldn't it be better if there is no number and he needs to reproduce your formula with link to your original source? -Strategy (2) is standard for physical journals, so if you are more on the math physics side, choose (2) or (3) without doubts.<|endoftext|> -TITLE: Choosing the algebraic independent elements in Noether's normalization lemma -QUESTION [14 upvotes]: Given a field $k$ and a finitely generated $k$-algebra $R$ without zero divisors, one knows that there exist $x_1, \ldots, x_n$ algebraically independent such that $R$ is integral over $k[x_1, \ldots, x_n]$. How can one choose actually the $x_i$'s ? -More precisely, if $a\in R$ is transcendent over $k$, can one find $x_2, \ldots x_n$ such that $R$ is integral over $k[a,x_2 \ldots, x_n]$ ? If it is false, can one get the conclusion under stronger assumptions ? In this discussion, I am also interested by geometric explanations. - -REPLY [25 votes]: This is an amplification of Mohan's suggestion to consider the context of projective varieties. I will asume that $k$ is infinite. -Given your $k$-algebra $R$, we get Spec $R$ as a variety $V$ in $\mathbb A^d$ for some $d$. -We may take the projective closure (i.e. Zariski closure) of $V$ in $\mathbb P^d$ to obtain a projective variety $\overline{V}$. -If $V = \mathbb A^d$ there is nothing to say: $R = k[x_1,\ldots,x_d]$ and witnesses its -own Noether normalization. So assume that $V$ is a proper subvariety of $\mathbb A^d$, -so that $\overline{V}$ is a proper subvariety of $\mathbb P^d$. -Because $k$ is infinite, we can (and do) choose a point $P$ (defined over $k$) lying at infinity, and not on $\overline{V}$. We can (and do) also choose a hyperplane $H$ -(defined over $k$) which does not contain $P$. We may then -apply the projection from $P$ to $H$ to obtain a morphism $\pi: \overline{V} -\to H$. Properness of projective morphisms implies that this map has closed image -in $H$, say $\overline{W}$. -Because $P$ was chosen to lie at infinity, this restricts to a morphism -$V \to H \cap \mathbb A^d$, which again has closed image, say $W$. (As the notation -suggests, $\overline{W}$ will then be the projective closure of the affine variety -$W$.) This map is finite, in the sense that $V \to W$ corresponds to a map -$R' \to R$ with $R$ finite over $R'$. -Now if $W = H\cap \mathbb A^d$, we have obtained our Noether normalization of $R$, -since $R'$ is a polynomial ring in this case. If not, we proceed inductively, -replacing $V$ by $W$ and $\mathbb A^d$ by $H\cap \mathbb A^d$ (an affine space of -one dimension less). Eventually we will reach a stage where the projection -to the hyperplane is surjective. (If $V$ has dimension $n$ then we have to -perform $d -n$ projections altogether.) -So you see that you have a lot of flexibility in how to achieve the normalization -(because at each stage there are a lot of choices of $P$ and $H$), but the choices -are not completely arbitrary (because of the condition that $P$ not lie on -$\overline{V}$, and that $H$ not contain $P$). For example, thinking this way, -you will easily see what goes wrong in Mohan's counterexample. (Taking $a = x$ -in his example corresponds to projecting from a point $P$ that does lies on -the projective closure of the hyperbola.)<|endoftext|> -TITLE: Does there exist any massive proper $C^*$-subalgebra? -QUESTION [6 upvotes]: Definition 1: Suppose $B$ is a $C^* $-algebra. $A$ is massive $C^* $-subalgebra of $B$ iff -1. $A$ is a subalgebra of $B$; -2. for each irreducible representation $\pi$ of $B$ representation $\pi|_A$ is irreducible; -3. if representations $\pi$ and $\pi'$ aren't (unitary) equivalent then $\pi|_A$ and $\pi'|_A$ aren't equivalent too. -Definition 2: Suppose $B$ is a $C^* $-algebra. $A$ is massive $C^* $-subalgebra of $B$ iff inclusion $i\colon A\to B$ is epic in the category of $C^* $-algebras, i.e. for any $C^* $-algebra $D$ and any two homomorphisms $g_1,g_2\colon B\to D$ we have $g_1\circ i=g_2\circ i \Rightarrow g_1=g_2$. -It's not hard to prove, that these definitions are equivalent. But I don't know the answer to the following question: - -Does there exist an example of massive proper $C^*$-subalgebra? - -There is a theorem in the book of Dixmier "C*-algebras" (French original was published in 1969), that if $A$ is a massive $C^* $-subalgebra of postliminal $C^* $-algebra $B$ then $A=B$. What's for the general case? Is it an open problem? - -REPLY [7 votes]: It was proved by Karl H. Hofmann and Karl-H.Neeb here (see here for the preprint), that epimorphisms in the category of $C^{\star}$-algebras are surjective.<|endoftext|> -TITLE: Link Repository of International Dissertations -QUESTION [29 upvotes]: This question (cry for help?) grew out of Colin Tan's question: does anyone have a copy of schmid’s effective work on hilbert 17th? which was a request for a copy of an Habilitationsschrift. -We've all been in this situation: somebody works on a dissertation of some kind, obtains interesting results, but either can't be bothered to publish because they're leaving the profession, or do publish the essentials, but have to omit some important details because of space considerations. -Inter-library loan can be an option in these cases, but it's very likely to take a very long time if it works at all. In the US, it appears that most universities require you to sign away your reproduction right to an independent company which will keep your dissertation on microfilm (correct me if I'm wrong, this is CW after all!). So in that case, you can track down a copy of the dissertation in question (if you're willing to pay for it) -- that is, if you can manage to track down the reproduction service in question. -Some countries, on the other hand, feel the need to maintain an online repository of the student work done in their universities. Here again, it involves a non-trivial slog through various library links before finding the right place (and may demand a better command of the language than for math reading). Hence, the question, which is more of a suggestion: - -How about we gather a list of links to various dissertation repositories / reproduction services right here? - -Suggestion: Since no answer is "better" than any other, and to improve readability, it would be best to keep this to a single list answer that can be edited with the various contributions. Thanks in advance! - -REPLY [3 votes]: For Greece, try at the National Archive of PhD Theses.<|endoftext|> -TITLE: When is $L^2(X)$ separable? -QUESTION [44 upvotes]: I have never studied any measure theory, so apologise in advance, if my question is easy: -Let $X$ be a measure space. How can I decide whether $L^2(X)$ is separable? -In reality, I am interested in Borel sets on a locally compact space $X$. I can also assume that the support of the measure is $X$, if it helps... -I cannot even decide at the moment for which locally compact groups $G$ with Haar measure, $L^2(G)$ is separable... - -REPLY [10 votes]: I have just found an old article which contains the following result: for a locally compact group $G$, its topological weight $w(G)$ [the minimal cardinality of a topology base] is equal to the dimension of $L^2(G)$. Whence $L^2(G)$ is separable iff $G$ is second countable. This is Theorem 2 in: de Vries, J. The local weight of an effective locally compact transformation group and the dimension of $L^2(G)$. Colloq. Math. 39 (1978), no.2, 319-323. -Since I spent a considerable time searching for such a reference, I post it here. -Worth noting also that the case of a compact $G$ is contained in Hewitt-Ross, Theorem 28.2.<|endoftext|> -TITLE: Universal definition of Fourier transform -QUESTION [8 upvotes]: Is there a category theoretic definition for the Fourier transform using only its universality properties? I am not looking for the most general definition -- one that works only in some special settings will do. I am looking for a simple definition that will make precise my (possibly incorrect) intuition that Fourier transforms are in some sense extremal among unitary transforms. -Here is another non category-theoretic way to ask this question, which may or may not be equivalent: Give a "natural" optimization problem on the space of unitary transforms whose solution turns out to be the Fourier transform. - -REPLY [9 votes]: For me, the "traditional Fourier Transform" is a change of basis of the algebra of functions from a group to some chosen field: from the canonical basis to something sometimes called the Fourier Basis. Because the Transform is constructed using the representation theory of the group, it has "natural" generalisations to objects with "similar" representation theory, e.g. it is defined for Hopf algebras. -I always think of the FT as this kind of duality. The Fourier Basis have lot's of interesting properties, but I have not seen a definition of it using extremals. It would be very interesting to see that. -A good start would be if someone gives an answer for finite abelian groups and finite non-abelian groups. Though "non-abelian" groups have a notion of FT, it is not uniquely defined and it is hard to work with it. An "extremal" condition would be enlightening. - -Update 14th Dec 2011 - -Sorry that this comes several months later, but I found that there is "a way" to define the quantum Fourier transform for abelian finite groups using an extremal argument. This argument comes from reference [1] where the Fourier transform is studied as a tool to design measurements in Quantum Computation and proven to be optimal to solve the abelian hidden subgroup problem. Unfortunately, this property does not hold for non-abelian quantum Fourier transforms. -More concretely, what it is proven in [1] is the following (all definitions I use are defined in this paper): - - -Consider the hidden subgroup problem defined for an abelian group $G$, where the hidden subgroup $H$ is chosen uniformly at random from all subgroups of $G$. Given $n$ tensored random coset states (cosets of $H$), then the measurement that maximises the probability of correctly identifying the subgroup $H$ is the following: - -Start on a random coset-state $|x+H\rangle$ for unknown $H$ which is just a uniform quantum superposition over the elements of the coset $x+H$. Cf. [1] for details on how to create these quantum states. -Apply the abelian quantum Fourier transform of $G$ on this state. -Perform a projective measurement. - - - -Taking several outcomes of the above procedure one obtains a generating set of the orthogonal group $H^\perp$ from which the original subgroup $H$ can be recovered solving a system of linear modular equations [2]. As far as I know these "orthogonal subgroups" are sometimes called orthogonal complements in Mathematics. -To sum up, they key ingredient of the above quantum algorithm is the abelian Fourier transform which is used to implement a quantum measurement to solve the hidden subgroup problem since it maximises the probability of distinguishing the hidden subgroup. In [1] it is shown that the abelian quantum Fourier transform arises as an optimal POVM which is the solution of a semidefinite program. I guess that maybe you could adopt this kind of extremal property as a definition of the Fourier transform for finite abelian groups. Note: it is not clear to me that the optimal POVM found in [1] is unique (up to permutations).<|endoftext|> -TITLE: What is the Poincare dual of a symplectic form? -QUESTION [12 upvotes]: Every symplectic form on a manifold $M^n$ determines a De Rham cohomology class in $H^2(M)$ (often a nontrivial class), and this in turn determines a class in $H_{n-2}(M)$. What in general can be said about this class? For example, over the rationals this class is represented by a submanifold of $M$; is it possible to explicitly describe such a submanifold in terms of the symplectic structure? -If there is a nice answer to this question, does it also shed light on the Poincare duals of $\omega^2$, $\omega^3$, etc? - -REPLY [3 votes]: Paul Biran has also studied this situation where $(M, \omega, J)$ is Kahler, $[\omega] \in H^2(M, \mathbb{Z})$ and $\Sigma$ is a complex hypersurface Poincare dual to $k[\omega]$. -He has proved that in this setting, $M$ can be decomposed symplectically as the unit normal disk bundle of $\Sigma$ and an isotropic CW-complex. -This decomposition result is then used to investigate various embedding questions into $M$. -Biran: "Lagrangian barriers and symplectic embeddings. Geom. Funct. Anal. 11 (2001), no. 3, 407–464" -Biran and Cieliebak: "Symplectic topology on subcritical manifolds. Comment. Math. Helv. 76 (2001), no. 4, 712–753" -Biran: "Lagrangian non-intersections. Geom. Funct. Anal. 16 (2006), no. 2, 279–326"<|endoftext|> -TITLE: Constants in the Rosenthal inequality -QUESTION [17 upvotes]: Let $X_1,\ldots,X_n$ be independent with $\mathbf{E}[X_i] = 0$ and $\mathbf{E}[|X_i|^t] < \infty$ for some $t \ge 2$. Write $X = \sum_{i=1}^n X_i$. Then we have the family of "Rosenthal-type inequalities": -$$ \mathbf{E}[|X|^t] \le C_1(t)\cdot \left(\sum_{i=1}^n \mathbf{E}[|X_i|^t]\right) + C_2(t)\cdot \left(\sum_{i=1}^n \mathbf{E}[X_i^2]\right)^{t/2} .$$ -Henceforth, $c>0$ is some absolute constant. I have seen in various papers that we can, for example, take $C_1(t) = C_2(t) = (ct/\log(t))^t$. We can also take $C_1(t) = (ct)^t, C_2(t) = (c\sqrt{t})^t$. Another option is $C_1(t) = c^t, C_2(t) = c^t\cdot 2^{t^2/4}$. -Is it known whether there is some non-trivial tradeoff curve for the relationship between $C_1(t)$ and $C_2(t)$? For example, if I'm fine with setting $C_2(t) = (ct^{2/3})^t$, what's the best $C_1(t)$ I can get? - -REPLY [3 votes]: $\newcommand{\de}{\delta} -\newcommand{\De}{\Delta} -\newcommand{\ep}{\varepsilon} -\newcommand{\ga}{\gamma} -\newcommand{\Ga}{\Gamma} -\newcommand{\la}{\lambda} -\newcommand{\Si}{\Sigma} -\newcommand{\thh}{\theta} -\newcommand{\R}{\mathbb{R}} -\newcommand{\X}{\mathcal{X}} -\newcommand{\E}{\operatorname{\mathsf E}} -\newcommand{\PP}{\operatorname{\mathsf P}} -\newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}}$ -According to Theorems 6.1 and 6.2, the best constants $C_1(t)$ and $C_2(t)$ are given by -\begin{equation} - C_1(t)=(c\ga)^t,\quad C_2(t)=(c\sqrt\ga\,e^{t/\ga})^t -\end{equation} -for $\ga\in[1,t]$, up to a universal real constant $c>0$. -E.g., if you want $C_2(t) = (ct^{2/3})^t$, solving the equation $(ct^{2/3})^t=(c\sqrt\ga\,e^{t/\ga})^t$, you get $\ga\sim\dfrac{6t}{\ln t}$ as $t\to\infty$, so that you get $C_1(t)=\Big(\dfrac{ct}{\ln t}\Big)^t$ for some (possibly different) universal real constant $c>0$. -Exact versions of Rosenthal-type bounds for $t\in[2,3]\cup[5,\infty)$ are given in -Theorems 1.3 and 1.5; see also Proposition 1.2.<|endoftext|> -TITLE: Midpoint geodesic polygon / Birkhoff curve shortening -QUESTION [11 upvotes]: I would like to know under what conditions the process -of creating a midpoint piecewise geodesic polygon converges -on a surface $S \subset \mathbb{R}^3$. -$S$ may be assumed smooth, closed, and oriented; -or it may be assumed a Riemannian manifold. -Let $P$ be a closed geodesic polygon -each of whose finite number of edges is a shortest path -connecting its endpoint vertices $v_i$. -Moreover, the edges of $P$ are "short" in a sense specified -below. -$P$ need not be simple, i.e., -it may intersect and cross itself. -Define the midpoint polygon $M(P)$ for $P$ -to have vertices $u_i$, each the midpoint of the geodesic -segment $v_i v_{i+1}$ of $P$, where $u_i$ is connected -to $u_{i+1}$ by a shortest path. -I would like the segment $u_i u_{i+1}$ to be the unique -shortest path between those points. -So assume that $|u_i u_{i+1}|$ is smaller than the injectivity radius -of the manifold. (This might require adding new vertices to ensure -this condition holds for the next iteration.) -In the plane, iterating this -midpoint polygon -construction is well studied. -On a surface (or Riemannian manifold) -it plays a role in curve shortening, -in particular Birkhoff curve shortening, which he -used to prove the existence of a simple closed geodesic [B27]. -Subsequently the technique has many uses. -For example, -Christopher Croke used Birkhoff shortening to find -the length of a shortest closed geodesic on a sphere [C88]. -The uses I have seen of Birkhoff shortening are on the sphere, -not on more general surfaces. -Which brings me to the question: - -Q. Does Birkoff shortening, repeatedly applied to a geodesic polygon -on a closed surface $S$, always converge (perhaps to a point)? -If so, always to a closed geodesic? -If not, under what conditions might it not converge? - - -          - -References. - -[B27] George D. Birkhoff, -Dynamical Systems, AMS, 1927. p.135ff. -[C88] -Christopher B. Croke, -"Area and the length of the shortest closed geodesic." -J. Differential Geom., Volume 27, Number 1 (1988), 1-21. - -REPLY [4 votes]: You may also have a look at the paper -https://arxiv.org/pdf/1502.01741.pdf<|endoftext|> -TITLE: Does regularity of a prime ideal in the fibre imply regularity of the prime? -QUESTION [6 upvotes]: Recall that a prime $\mathfrak{p}$ is called nonsingular (regular) if the localization at that prime is a regular local ring. If all primes of a ring $R$ are nonsingular, $R$ is called regular. Let $S\subseteq R$ be a ring extension, and let $\mathfrak{q}$ be a prime ideal of $S$. Then $R$ is called regular above $\mathfrak{q}$ if any prime contracting to $\mathfrak{q}$ is nonsingular. -Recall that a ring $R$ is said to have regular fibre over $\mathfrak{q}$ if the scheme-theoretic fibre $R\otimes_S \frac{S_\mathfrak{q}}{\mathfrak{q}S_\mathfrak{q}}=R\otimes_S \kappa(\mathfrak{q})$ is regular. -Are these conditions equivalent? There is an obvious implication that regularity above $\mathfrak{q}$ implies regularity of the fibre, but the converse is not obvious to me. If it's not true in general, is it true when $R$ is a subring of a number field and $S=\mathbf{Z}$ (with $\mathfrak{q}$ a nonzero prime)? - -REPLY [6 votes]: How about we try the simple case $R=S$ and the identity map. Let $\mathfrak q$ be a singular prime. The fibre over $\mathfrak q$ is $\kappa (\mathfrak q)$ which is regular (it is a field) but $R_{\mathfrak q}$ is not regular. -Edit: let me add a few more details since my original terse answer hardly deserves the upvotes! Let $P$ be the set of primes in $\text{Spec}(R)$ which contracts to $\mathfrak q $ (the set-theoretic fibre). Your first condition says that: -(1) $R_{\mathfrak p}$ is regular for all $\mathfrak p \in P$. -while the second says: -(2) $R_{\mathfrak p}/\mathfrak qR_{\mathfrak p}$ is regular for all $\mathfrak p \in P$. -So you can see where the problems come from: in general, for a local ring $R$, regularity of $R$ has nothing to do with regularity of $R/I$ for some ideal $I$, unless if $I$ is very special. That is how I think about the counter-examples, hopefully it helps. - -REPLY [6 votes]: First note that it is not true that regularity above $\mathfrak q$ implies regularity of the fibre. -For example, consider the map $\mathbb k[s] \to \mathbb k[t]$ given by $s \mapsto t^2$. -Each prime in $k[t]$ is regular, and so in particular each prime of $k[t]$ above -the prime $(s)$ is reqular. (In fact there is just one of them, namely $(t)$.) On the other hand, the fibre over $(s)$ is the ring $k[t]/(t^2)$, which is a non-regular -local ring. -If you want an arithmetic example instead, consider the inclusion -$\mathbb Z \subset \mathbb Z[i]$, and take the prime $\mathfrak q = (2)$ downstairs, -which has a unique prime $(1+i)$ lying over it. Again, every prime in the PID $\mathbb Z[i]$ -is regular, but the fibre $\mathbb Z[i]/2 = \mathbb Z[i]/(1+i)^2$ is a non-regular local -ring. -(The general phenomenon is that a map $X \to Y$ between smooth spaces can have -non-smooth fibres: these occur at the critical points of the map.) -As for your question, your asking if the fact that a map $X\to Y$ has -a smooth fibres over some point implies that the target is smooth at that point. -This is also false in general. -Consider for example the identity map -$k[t^2,t^3] \to k[t^2,t^3]$. The fibre over $(t^2,t^3)$ is just $k$, which is a regular local ring. But $K[t^2,t^3]$ is not regular at $(t^2,t^3).$ (Of course this example is cheap, but its existence foreshadows the existence of many other counterexamples, for example for any etale map $k[t^2,t^3] \to R$, of which there are many non-trivial examples, -as well as my trivial example.) -On the other hand, when the base is Spec $\mathbb Z$, which is very nicely behaved -(regular, Noetherian, excellent, perfect residue fields, ... ), if the map -$X \to $ Spec $\mathbb Z$ is flat and of finite type (e.g. arising from an inclusion -$\mathbb Z \subset R$ of the form you envisage) then having regular (and hence smooth) fibre -at one point implies being smooth in a neighbourhood of that point, and a smooth map over -a regular base has a regular total space --- thus $X$ will be regular in a neighbourhood of the regular fibre. In particular, the points of the regular fibre will themselves be regular in $X$. -In particular, in your special case $\mathbb Z \subset R$, the answer is "yes". -Edit: I should note that in your situation, where $R$ is a subring of a number field, -this "yes" is easily proved directly: one combines the fact that $R/\mathfrak q$ is -regular with the fact that it is a priori finite (in cardinality) to see that -it is a product of finite field extensions of $\mathbb Z/\mathfrak q$, and hence -that the completion of $R$ at $\mathfrak q$ is a product of DVRs, and hence that -$R$ is a DVR --- and thus regular --- after localization at each prime above $\mathfrak q$. -The point of the more highbrow explanation above is to indicate how one thinks about -such questions geometrically --- which is normally the easiest way to see what should -be true and what should be false for these kinds of questions.<|endoftext|> -TITLE: Poles from the Continued Fraction Expansion of the Tangent Function? -QUESTION [18 upvotes]: Consider the well known continued fraction expansion -$$ z \tan z = \frac{z^2}{1 - \cfrac{z^2}{3- \cfrac{z^2}{5 - \ldots}}} $$ -of the tangent function going back to Euler and Lambert (Lambert used -it for showing that $\tan z$ is irrational for rational nonzero values -of $z$, which implies the irrationality of $\pi$; Legendre later observed -that the same proof gives the irrationality of $\pi^2$). Wall. in his -book on continued fractions, claims that the formula is valid "for all $z$". -Is there a nice way of determining the poles of $\tan z$ from looking at the -right hand side of this expansion? - -REPLY [4 votes]: This verifies Henri Cohen's comment. Since the function is even, it is simpler to look at $\sqrt{x}tan(\sqrt{x})$, the $Q_n(x)$ is given by -$[1,3-x,15-6x,x^2-45x+105,15x^2-420x+945,...]$. Since $deg(Q_n)=\lfloor n/2 \rfloor$, one should look at $p_n(x)=Q_{2n}(x)$, the roots are all real and distinct and interlace so that for each $k>0$ the $k$th smallest root decrease monotononically, the values of the roots -divided by $(\pi/2)^2$ are given by -$$1 [1.215854203708053257326553558]$$ -$$2 [1.000556567684670043780569921, 17.23725648793612881611773345]$$ -$$3 [1.000000264021497302735160628, 9.243071092446397955245651578, 74.86672290309583275487793689]$$ -$$4 [1.000000000033926859936132129, 9.003785153434677150307469375, 27.94131378418484607770210908, 217.3842838410377339506305367]$$ -$$5 [1.000000000000001649799381439, 9.000019418957686887495315571, 25.17287514526503834031255350, 62.70166691403289751607854631, 503.9732693572307379829582146]$$ -$$6 [1.000000000000000000036873888, 9.000000041594544789352719498, 25.00432366956189670320729176, 50.59590302069961836345079636, 121.9182580673728203679597413, 1009.551573112532430359872689]$$ -$$7 [1.000000000000000000000000430, 9.000000000044261664894057822, 25.00004695103961704597815305, 49.10575115766100355065900035, 88.08889135341595365433652728, 215.8656260646040985633268236, 1824.794335221891993855132913]$$ -$$8 [1.000000000000000000000000000, 9.000000000000026169987166719, 25.00000026069484992294309458, 49.00327767012444219447267331, 81.88750001909035955812019217, 142.0164417098733597317941152, 356.4649305639281633307802602, 3056.141713122931766511156386]$$ -which gives very fast convergence to $(2k-1)^2$. -One can also find the zeros by looking at roots of $p_n(x)=P_{2n}(x)$ or $P_{2n+1}(x)$. - In fact for any real $\alpha$, if $P_n/Q_n$ are the convergents to $x tan(x)$, the roots of -$P_{2n}(x)-\alpha xQ_{2n}(x)$ will still be real and distinct and converge monotonically to the root of $tan(x)=\alpha$.<|endoftext|> -TITLE: If the series Σ pᵃ⁽ʷ⁾·xᴵʷᴵ is rational, is Σ a(w)·xᴵʷᴵ also rational (summation over words w in a regular language)? -QUESTION [7 upvotes]: Let $p$ be a prime number and let $a_i$ be a sequence of natural numbers such that the series $\sum_{i=1}^\infty p^{a_i} x^i$ is rational. A warm-up question: - -Question 1. Does it follow that the series $\sum_{i=1}^\infty a_i x^i$ is rational? - -Let $p$ be a prime number and let $L$ be a regular language. Let $a\colon L\to \mathbb N$ be a function such that the series -$$ -\sum_{w\in L} p^{a(w)}x^{|w|} -$$ -is rational. - -Queston 2. Does it follow that the series $\sum_{w\in L} a(w) x^{|w|}$ is rational? - - -Motivation -Let $L$ be a regular language and let $T=T_w$, $w\in L$, be a family of finite dimensional matrices with integral coefficients. Consider the series -$$ -F(T,p)(x) := \sum_{w\in L} \dim_{\mathbb F_p}\ker_{\mathbb F_p} T_w \cdot x^{|w|}, -$$ -and the series -$$ -G(T,p)(x) := \sum_{w\in L} |\ker_{\mathbb F_p} T_w| \cdot x^{|w|}, -$$ -where $|\ker_{\mathbb F_p} T_w|$ is the number of elements in $\ker_{\mathbb F_p} T_w$. The relation between $G$ and $F$ is as in the questions above. In my specific situation the series $G$ is rational, roughly because the elements of $\ker_{\mathbb F_p} T_w$ also form a regular language. I would like to know that the series $F$ is rational because $F(T,p)(\frac{1}{2})$ is a so called $l^2$-Betti number over $\mathbb F_p$ and it would be very nice to know that these are rational in the case at hand. - -Remarks -In the key example I have (section 2-C is the relevant one) the series -$$ -F(T)(x) := \sum_{w\in L} \dim_{\mathbb C}\ker_{\mathbb C} T_w \cdot x^{|w|} -$$ -is transcendental, but all the series $F(T,p)$ are rational, because $a(w)$ are bounded, and in this case the answer to Question 2 is, as explained by Dylan below, positive. Reason for transcendality over complex numbers is that the family of vectors which make it transcendental have rapidly growing coefficients, and so no single prime number can detect this family. Questions above can be informally seen as asking whether "unboundedness of elements in the kernels" can also be a reason for the series not being rational. - -REPLY [2 votes]: There are many papers dealing with the largest prime divisor of elements from a recurrent sequence. See for example the paper "On the greatest prime factor of terms of a linear recurrence sequence" by C.L. Stewart. Since in your case the largest prime factor remains bounded this implies that the sequence is degenerate in some sense (Polya 1921) and I beleieve these cases can be checked to show that the $a_i$'s form a recurrent sequence as well.<|endoftext|> -TITLE: Applications of Stacks -QUESTION [17 upvotes]: I've been aware of stacks since grad school, and I can usually follow in rough lines a discussion about stacks, but I've often wondered what particular (purely!) scheme-theoretic argument or theorem is significantly simplified by the introduction of stacks. I'm sure there are many, but since I don't deal with stacks on a regular basis I don't encounter them as frequently, and I thought maybe some of you can enlighten me. - -REPLY [14 votes]: The classic application is Deligne & Mumford's paper proving the irreducibility of the coarse moduli scheme $\overline{M}_g$ of stable genus g curves over any algebraically closed field. They proved irreducibility first for the moduli stack $\overline{\mathcal{M}}_g$ of curves, and then inferred from this result the irreducibility of the scheme. - -REPLY [10 votes]: The DeRham space is a stack $X_{DR}$ associated to a smooth variety $X$, so that modules on $X_{DR}$ are D-modules on $X$. This is accomplished by declaring the maps from $Y$ into $X_{DR}$ are the same as maps from $Y^{red}$ (the reduced scheme) into $X$. This has the effect of identifying points with their infinitesmal neighborhoods. -The DeRham space is often most useful as a conceptual tool. However, a specific application of it was by Ben-Zvi and Nevins, who used it (and other tools) to show that certain cusped versions $\widetilde{X}$ of $X$ had equivalent categories of D-modules. The idea being, these cusps were identifying some of the infinitesmal neighborhoods of some of the points, and so they should be intermediate between a variety and its DeRham space.<|endoftext|> -TITLE: Rational Isogenies of Prime Degree -QUESTION [10 upvotes]: Dear MO Community, -Let $N$ be a prime, and let $X_0(N)$ be the classical modular curve over $\mathbb{Q}$. We know ([1]) that, if there are noncuspidal points in $X_0(N)(\mathbb{Q})$, then $N \in$ {${ \mbox{primes } \leq 19} $} $ \cup $ {37,43,67,163}. -The basic question of this post is: - -Are there similar lists of primes when $\mathbb{Q}$ is replaced by a number field $K$? That is, if we fix a general number field $K$, can we determine the primes $N$ for which $X_0(N)(K)$ has noncuspidal points? - -Perhaps in this generality the question is hard, so suppose we restrict from now on to imaginary quadratic $K$. Then [1] gives an approach to the question, but with the following snags: - -We need to construct an "optimal" quotient of $J_0(N)$, call it $A$, such that $A(K)$ has Mordell-Weil rank 0; -We must restrict ourselves to primes $N$ which are inert in $K$. - -Actually, I don't think (1) is a big problem; provided $N \> 48h(K)^3 + 1$, we can take $A = \widetilde{J}$, the Eisenstein quotient. [Speculation : if we took the "winding quotient" instead, maybe we can lower that bound...] -When $N$ splits, then we can construct "CM points" on $X_0(N)(\mathbb{C})$, but usually they will not be defined over $K$, and even if they are, there will only be a handful of them. - -Question: For which $N$ that splits in $K$ do we have points on $X_0(N)(K)$ that are neither cuspidal nor CM? Is there a way to systematically find these points? - -By "systematically", I guess I mean something like the 'isogeny character' approach of [1], where the hunt for the $N$s comes down to when certain congruences are satisfied mod $N$. -Many thanks. -[1]: Mazur, B. "Rational isogenies of prime degree", Inventiones Mathematicae, 1978 - -REPLY [9 votes]: You can see also the recent paper of A. David on ArXiv: Critères d'irréductibilité pour une courbe elliptique semi stable sur un corps de nombres. A. David gives a uniform bound for irreducibility of mod p Galois representations of semi stable elliptic curves over number field K provided some condition (namely that there is not any CM elliptic curve with everywhere good reduction on K). -But the general question of isogenies over a number field is hard. The "Mazur-Kamienny" style method to study this should be to look at the morphism from the d-th symmetric power of X_0(p) to an optimal quotient of J_0(p). (See also Kamienny works for torsion points on number fields).<|endoftext|> -TITLE: Has Stirling’s Formula ever been applied, with interesting consequence, to Wilson’s Theorem? -QUESTION [5 upvotes]: Pressing the envelope, presumably the best scenario would be a simple proof of the Prime Number Theorem. After all, Wilson’s Theorem gives a necessary and sufficient condition, in terms of the Gamma Function, for a number to be a prime, and Stirling’s Formula specifies the asymptotic behaviour of the Gamma Function. - -REPLY [36 votes]: Using Robbins' [1] form of Stirling's formula, -$$\sqrt{2\pi}n^{n+1/2}\exp(-n+1/(12n+1))< n!< \sqrt{2\pi}n^{n+1/2}\exp(-n+1/(12n))$$ -we get -$$\left\lceil\sqrt{2\pi}(n-1)^{n-1/2}\exp(-n-1+1/(12n-11))\right\rceil$$ -$$\le (n-1)!\le$$ -$$\left\lfloor\sqrt{2\pi}(n-1)^{n-1/2}\exp(-n-1+1/(12n-12))\right\rfloor$$ -which is accurate enough to distinguish prime from composite for $n\le8$. For larger numbers, the error bound is too large. - -This can be extended further using a modification of Wilson's theorem: for n > 9, -$$\lfloor n/2\rfloor!\equiv0\pmod n$$ -if and only if n is composite. This allows testing 10 through 15, plus (with some cleverness) 17. -With tighter explicit bounds and high-precision evaluation, it might be possible to test as high as 100 with related methods: direct evaluation up to 25 and the 'divide by 4' variant of the above for n > 25. -This is not so much 'using a cannon to swat a fly' (using methods more powerful than needed) as it is 'using the space station to swat a fly': the methods must be extremely powerful and accurate to do very little. - -[1] H. Robbins, "A Remark on Stirling's Formula." The American Mathematical Monthly 62 (1955), pp. 26-29.<|endoftext|> -TITLE: Why certain diophantine equations are interesting (and others are not) ? -QUESTION [59 upvotes]: It is quite clear why certain differential equations, among the jungle of possible diff equations that is possible to conceive, are studied: some come from physical problems, or from "spontaneous" mathematical generalizations thereof, others come from geometry in a variety of ways. -For diophantine equations there seem not to be such a direct link to other areas. -I would like to roughly understand why the attention of number theorists concentrates on some kinds of diophantine equations and not on others. -Why an equation such as -$x^2-ny^2=1$ -or -$x^3+y^3=z^3$ -is (or have been) considered worth studying, and not, say, any other random variant such as (if that specific example is not enough nontrivial for you or if it actually happens to have been studied, feel free to substitute it with your favourite "random" diophantine equation): -$x^3+y^5=z^2$ ? So: - -Are there any reasons why certain diophantine equations are worth attention besides the mere approachability (i.e. being neither trivial nor hopelessly difficult to analyze)? - -REPLY [6 votes]: As has been well told by others, there are many interesting classes of diopantine equations (norm equations, elliptic curves and abelian varieties, curves of genus > 1, S-unit equations, varieties where the rational points are always Zariski dense or are always finite, failure of the Hasse principle (or not), Lang's conjecture, etc.). However, it is a theorem of Wiles that are no more specific diophantine equations of interest. Any particular problem, say, the existence of infinitely many integral solutions to $x^3+y^3+z^3 = 3$, will only be interesting to the extent that the solution sheds light on the general arithmetic properties of surfaces. Fermat was special, for a combination of historic and aesthetic reasons.<|endoftext|> -TITLE: What is the dual concept to "annihilator" called, and do any linear algebra textbooks discuss this concept first? -QUESTION [13 upvotes]: When introducing dual spaces for the first time, most linear algebra textbooks proceed in what seems to me a rather backwards fashion: the annihilator $\{f\in V^*: f(u)=0\quad \forall u\in U\}$ of a subset/space $U$ of a vector space $V$ is introduced before the dual concept of the "joint kernel"(?) $\{v\in V: f(v)=0\quad \forall f\in W\}$ of a subset/space $W$ of the dual space $V^*$. The latter notion, despite corresponding to the intuitive idea of the solution space of a homogeneous linear system of equations, is then introduced indirectly by mapping $V$ into $V^{**}$ and considering the annihilator of the system of equations $W$ of $V^*$, which is $\{\phi\in V^{**}: \phi(f)=0\quad \forall f\in W\}$. This seems a pity as many students find the isomorphism of a finite dimensional vector space with its double dual difficult to grasp! -Worse, as the "(?)" above suggests, there appears to be no common terminology or notation for the concept dual to annihilator. Possibilities include "(joint) kernel, null/zero space, pre-annihilator, solution space". I'd be grateful for any pointers to textbooks which introduce the concept directly, or any suggestions for terminology and notation. -Update: Many thanks for the suggestions so far. I am still rather surprised and disappointed by the lack of references to elementary linear algebra text books which discuss solution spaces (aka null spaces, joint kernels) before or on the same footing as annihilators. I am unconvinced by the arguments that have so far been made for this omission. -Instead it seems to me that the current situation is the result of inertia. The "annihilator" is a term that caught on, and has been carried forward in the absence of a cool name for the dual concept. In this respect I like the invention of the "vanquished". -The question is not moot as I am currently lecturing this material. I am going to stick to "solution/null space" and "joint kernel". I will then discuss duality and row rank = column rank without ever mentioning the double dual (the double dual will come later in the course). -In preparing this, I noticed the fact that solution spaces and annihilators provide a Galois connection between the posets of subsets/subspaces of a vector space and its dual (that is, solution spaces and annihilators are contravariant adjoint functors between these posets). I've seen discussion of this in the context of annihilators in ring theory, but it seems to me to be at the heart of the concept of duality for vector spaces. I hope this rings some bells among MO readers. -Such categorical thinking leads to a fairly clean analysis of solution spaces and annihilators, which has in turn simplified some results subsequently. - -REPLY [7 votes]: In the book "Algebra Lineal y Geometría" by Angel Rafael Larotonda, he introduces the concept right after introducing the annihilator of a set $M\in V$ as $M^o$, he introduces the "left" annihilator of a set $F \in V^*$ as $^oF=${$x \in V: f(x)=0 \; \forall \; f \in F$}, the only problem is that its written in spanish. I have also seen the concept introduced directly as pre-annihilator, but alas, also in spanish books.<|endoftext|> -TITLE: An algorithm to find non-trivial linear dependencies -QUESTION [10 upvotes]: This question is inspired by another MO question about special stratifications of equivariant Grassmannians, that turned out to be a problem of computing non-trivial circuits in a vector matroid. To review, a vector matroid is just a list of vectors or lines in $\mathbb{F}^n$ for a field $\mathbb{F}$, and a "circuit" is a linearly dependent set of vectors. The information in the matroid is exactly the combinatorial information in the set of circuits, or for that matter the set of minimal circuits. (But, matroids are defined by axioms and there are matroids that don't come from a list of vectors in a vector space.) -In response to the question, I computed some vector matroids in Sage. David Joyner's Sage code for this purpose is based on exhaustive searches for the minimal circuits. Of course, every set of $n+1$ vector is a circuit, so you can stop the search at subsets of $n$ vectors. Of course, in searching for minimal circuits, you never need to check a superset of a known circuit. But otherwise I couldn't think of any better algorithm than exhaustive search. Consider especially the difficult case in which there are no circuits of size less than $n$. If there are $v$ vectors total, you would have to search over all $\binom{v}{n}$ sets of $n$ vectors. -I can think of one small acceleration that is still basically an exhaustive search. If you have guessed $k < n$ vectors, you can put the vectors that you have in reduced echelon form, to avoid repeated work when you assume more vectors. You can also use those vectors to reduce the remaining vectors that you haven't yet chosen. This saves a factor of $O(n^2)$ if you had planned to use Gaussian elimination to see if each set of $n$ vectors is linearly dependent. You can even use this method to incrementally compute all $\binom{v}{n}$ determinants. -Is there any better algorithm known? For simplicity suppose that the field is $\mathbb{Q}$, and the vectors are all rescaled to integer vectors. I would guess that it is NP-hard to determine if there are any linear dependencies. If so, then NP-hardness is not the main part of my question. Because, for example, computing the permanent of a matrix is #P-hard, but there is an interesting acceleration: The naive algorithm takes $\tilde{O}(n!)$ time, but there as an important algorithm that works in time $\tilde{O}(2^n)$. - -REPLY [2 votes]: There are results pertaining to the finite field case. According to this paper, for a finite field $F_q$ and a set of $r$ points $x_1,\dotsc,x_r\in F_q^n$ deciding whether there exists a subset of $k$ dependent points among them is NP-complete and requires $poly(n)\min(\binom{r}{k}, q^{\Theta(n+k^2)})$ time (assuming 3-SAT requires exponential time). -The algorithm for $q^{\Theta(n+k^2)}$ part is based on FFT. Moreover, the algorithm not only checks existence, but also counts the solutions (approximates — in the case of minimal dependent $k$-subsets, i.e. circuits). Also, one can find a specific solution by running the algorithm on sequentially increased inputs and paying overhead of $O(rk)$.<|endoftext|> -TITLE: Two questions about finiteness of ideal classes in abstract number rings -QUESTION [21 upvotes]: Let us say that an abstract number ring is an integral domain $R$ which is not a field, and which has the "finite norms" property: for any nonzero ideal $I$ of $R$, the quotient $R/I$ is finite. -(I have taken to calling such rings abstract number rings and have some vague ambitions of extending the usual algebraic number theory to this class of rings. Note that they include the two basic rings $\mathbb{Z}$ and $\mathbb{F}_p[t]$ and are closed under: localization, passage to an overring -- i.e., a ring intermediate between $R$ and its field of ractions -- completion, and taking integral closure in a finite degree extension of the fraction field. In order to answer my questions affirmatively one would have to know something about abstract number rings which are not obtained from the two basic rings via any of the above processes -- if any!) -Note that such a ring is necessarily Noetherian of dimension one, so it is a Dedekind domain iff it is normal, and in any case its integral closure is a Dedekind abstract number ring. - -Question 1: Does there exist an integrally closed abstract number ring with infinite Picard (= ideal class, here) group? - -$\ \$ - -Question 2: Let $R$ be a not-necessarily integrally closed abstract number ring with integral closure $\tilde{R}$. Suppose that the ideal class group of $\tilde{R}$ is finite. Consider the ideal class monoid $\operatorname{ICM}(R)$ of $R$, i.e., the quotient of the monoid of nonzero ideals of $R$ by the submonoid of principal ideals. (Note that the group of units of $\operatorname{ICM}(R)$ is precisely the Picard group, but if $R$ is not integrally closed it will necessarily have non-invertible ideals so that $\operatorname{Pic}(R)$ will not be all of $\operatorname{ICM}(R)$.) Can it be that $\operatorname{ICM}(R)$ is infinite? - -REPLY [14 votes]: To answer Question 1: Yes, there do exist integrally closed abstract number rings with infinite class group. -By factorization of ideals, for $R$ to be an abstract number ring it is enough that it is a Dedekind domain with finite residue field $R/\mathfrak{p}$ at each prime $\mathfrak{p}$. Theorem B of the paper mentioned by Hagen Knaf in his answer actually gives what you ask for (R. C. HEITMANN, PID’S WITH SPECIFIED RESIDUE FIELDS, Duke Math. J. Volume 41, Number 3 (1974), 565-582). - -Theorem B: Let G be a countable abelian torsion group. Then there is a countable Dedekind domain of characteristic 0 whose class group is G, and whose residue fields are those of the integers (i.e. one copy of $\mathbb{Z}/p\mathbb{Z}$ for each prime $p$). - -As such rings have finite residue fields, this gives an integrally closed abstract number ring with class group any countable torsion group you like. -We can do much better than this though. After thinking about your question for a bit, I see how we can construct the following, so that all countable abelian groups occur as the class group of such rings. - -Let G be a countable abelian group. Then, there is a Dedekind domain $R$ with finite residue fields such that $\mathbb{Z}[X]\subseteq R\subseteq\mathbb{Q}(X)$ and ${\rm Cl}(R)\cong G$. - -I see some surprise mentioned in the comments below that it is enough to look at over-rings of $\mathbb{Z}[X]$ to find Dedekind domains with any countable class group. In fact, over-rings of $\mathbb{Z}[X]$ are very general in terms of prime ideal factorization, and can show the following. I'll use ${\rm Id}(R)$ for the group of fractional ideals of $R$ and $R_{\mathfrak{p}}$ for the localization at a prime $\mathfrak{p}$, with $\bar R_{\mathfrak{p}}$ representing its completion (which is a compact discrete valuation ring (DVR) in this case). - -Let $R$ be a characteristic zero Dedekind domain with finite residue fields. Then, there is a Dedekind domain $R^\prime$ with $\mathbb{Z}[X]\subseteq R^\prime\subseteq\mathbb{Q}(X)$ and a bijection $\pi\colon {\rm Id}(R)\to{\rm Id}(R^\prime)$ satisfying - -$\pi(\mathfrak{ab})=\pi(\mathfrak{a})\pi(\mathfrak{b})$. -$\pi(\mathfrak{a})$ is prime if and only if $\mathfrak{a}$ is. -$\pi(\mathfrak{a})$ is principal if and only if $\mathfrak{a}$ is. -If $\mathfrak{p}\subseteq R$ is a nonzero prime then $\bar R_{\mathfrak{p}}\cong\bar R^\prime_{\pi(\mathfrak{p})}$. - -In particular, the class groups are isomorphic, ${\rm Cl}(R)\cong{\rm Cl}(R^\prime)$. - -The idea is that we can construct Dedekind domains in a field $k$ by first choosing a set $\{v_i\colon i\in I\}$ of discrete valuations on $k$ and, letting $k_v=\{x\in k\colon v(x)\ge0\}$ denote the valuation rings, we can take $R=\bigcap_ik_{v_i}$. Under some reasonably mild conditions, this will be a Dedekind domain with the valuations $v_i$ corresponding precisely to the $\mathfrak{p}$-adic valuations, for prime ideals $\mathfrak{p}$ of $R$. In this way, we can be quite flexible about constructing Dedekind domains with specified prime ideals (and, with a bit of work, specified principal ideals and class group). Constructing discrete valuations $v$ on $k=\mathbb{Q}(X)$ is particularly easy. Given a compact DVR $R$ of characteristic 0 and field of fractions $E$, every extension $\theta\colon k\to E$ gives us a valuation $v(f)=u(f(X))$ where $u$ is the valuation in $E$. To construct such an embedding only requires choosing $x\in E$ which is not algebraic over $\mathbb{Q}$ and, if we want the localization $k_v$ to have completion isomorphic to $R$, then we just need $\mathbb{Q}(x)$ to be dense in $E$. There's plenty of freedom to choose $x\in R$ like this. In fact, there's uncountably many $x$, as they form a co-meagre subset of $R$. So, we have many many valuations on $\mathbb{Q}(X)$ corresponding to any given compact DVR. In this way, we have a lot of flexibility in constructing Dedekind domains in over-rings of $\mathbb{Z}[X]$. -I've written out proofs of these statements. As it is much too long to fit here, I'll link to my write-up: Constructing Dedekind domains with prescribed prime factorizations and class groups. Hopefully there's no major errors. I'll also mention that this is an updated and hopefully rather clearer write-up than my initial link (which were very rough notes skipping over many steps). -I think also that my linked proof can be modified to show that you can simultaneously choose any prescribed unit group of the form $\{\pm1\}\times U$ where $U$ is a countable free abelian group.<|endoftext|> -TITLE: How can one express the Dedekind eta function as a sum over the lattice? -QUESTION [7 upvotes]: The Dedekind eta function $\eta(\tau)$ can be regarded as a formula which assigns a number to a lattice $L \subset \mathbb{C}$. The algorithm is: rotate the lattice so that one of its basis vectors lies along the real axis, then pick another basis vector $\tau$ in the upper half plane, then compute the usual formula -$$ - \eta(\tau) = q^\frac{1}{24} \prod_{n=1}^\infty (1-q^n) -$$ -where $q=e^{2 \pi i \tau}$. It would be nice if there were a more "canonical" way to compute it directly from the lattice $L$ (i.e. without this rotate-and-pick-a-basis-vector story which breaks symmetry). I'm looking for a formula similar to that of the Eisenstein series where one sums over all points of the lattice: -$$ - G_n (L) = \sum_{\omega \in L, \omega \neq 0} \frac{1}{\omega^n} -$$ -We have the theorem of Jacobi that the 24th power of $\eta$ computes as the discriminant of the lattice, -$$ -(2\pi)^{12} \eta(\tau)^{24} = 20G_4(\mathbb{Z} + \tau \mathbb{Z})^3 -49 G_6 (\mathbb{Z} + \tau \mathbb{Z})^2, -$$ -which is great, since it shows that the 24th power of $\eta$ can be defined canonically via a sum over the lattice points... but how about $\eta$ itself? - -REPLY [8 votes]: The functions $G_k$ and $\Delta = \eta^{24}$ can be regarded as functions on the set of lattices because they're modular of level 1. The $\eta$ function isn't modular of level 1 (its level is 24) so there's no natural way to regard it as a function on lattices -- it's a function on lattices with additional "level structure". -Similarly, in order to regularise $G_2$ to get something with good convergence, you end up having to make it depend on some level structure as well.<|endoftext|> -TITLE: How to prove that every real number is a zero of some power series with rational coefficients (if true) -QUESTION [17 upvotes]: How would one approach proving that every real number is a zero of some power series with rational coefficients? I suspect that it is true, but there may exist some zero of a non-analytic function that is not a zero of any analytic function. I was thinking about approaching the problem using arguments of cardinality, but I am unsure about how to begin. -Thank you in advance. - -REPLY [4 votes]: From "Note on integral functions" by A.G Walker: - -The following theorem, though perhaps well known, does not appear in any of the elementary text-books of analysis. It may, however, be of interest in connection with the various rational representations of irrational numbers, and an elementary proof is given. -Theorem. Every real number is a zero of a rational integral function, i.e. a function expressible as a power series, with rational coefficients, convergent for all x. - -The proof is basically the same as the one in Gerry's answer.<|endoftext|> -TITLE: Where can I easily look up / calculate (abelian) group cohomology? -QUESTION [11 upvotes]: For an example I'm trying to understand, I need to calculate some cohomology group of some $\mathbb Z$-module with coefficients in some other $\mathbb Z$-module (with no interesting actions). (In particular, letting $\mathbb C^\times$ be the multiplicative group of the complex numbers with the discrete topology, I would like to know ${\rm H}^2(\mathbb C^\times,\mathbb C^\times)$, which is probably trivial but I have no idea how to show it is.) Having never really learned much cohomology theory, I know some basic definitions and I know how to turn an extension of groups into an exact sequence in cohomology. But this doesn't help me do particular computations unless I at least know the necessary cohomologies for pieces of the group. (For example, I may know that $\mathbb C^\times = S^1 \times \mathbb R$ as discrete groups, but I don't know any $\mathbb H^\bullet(S^1,\mathbb C^\times)$ or $\mathbb H^\bullet(\mathbb R,\mathbb C^\times)$.) -What would be best is some reference book or, as I won't be near a good library for at least two days, a good website that lists sufficiently many cohomology groups and pedagogically explains some ways to do explicit computations to extend their list. Something like (but presumably easier than) the Knot Atlas or the OEIS. Does such a thing exist? - -REPLY [16 votes]: This group is best understood in terms of the universal coefficient formula, -i.e., in terms of the homology of the involved group. Hence, if $A$ is any -abelian group we have $H_1(A)=A$ and the addition map $A\times A\rightarrow A$ -induces a Pontryagin product on homology making $H_\ast(A)$ a (graded) -commutative algebra. We also have that the square of an element of $H_1(A)$ is -zero (in $H_2(A)$). This does not directly follow in the presence of $2$-torsion -but it follows by functoriality; given $a\in H_1(A)=A$ we have a group -homomorphism $\mathbb Z\rightarrow A$ taking $1$ to $a$ and we are reduced to -showing that $1\cdot1=0\in H_2(\mathbb Z)$ but $H_2(\mathbb Z)=0$. -Hence we get an algebra map $\Lambda^\ast A\rightarrow H_\ast(A)$. This is an -isomorphism in degrees $1$ and $2$ and an isomorphism in all degrees if $A$ is -torsion free. This is proved easily by noting that both sides commutes with -filtered direct limits so that one is reduced to the case when $A$ is finitely -generated and then using the Künneth formula to reduce to the case when $A$ is -cyclic in which case $H_2(A)=0$ and $H_i(A)=0$ for $i\geq2$ if $A=\mathbb Z$. -Now, using the universal coefficient formula and the fact that the coefficient -group $\mathbb C^\ast$ is injective we get $H^2(\mathbb C^\ast,\mathbb -C^\ast)=\mathrm{Hom}(\Lambda^2\mathbb C^\ast,\mathbb C^\ast)$. Of course as -abelian group $\Lambda^2\mathbb C^\ast$ is humongous as is the coefficient group -$\mathbb C^\ast$ but they appear naturally in some cases. For instance -$H^2(\mathbb C^\ast,\mathbb C^\ast)$ is the group of characteristic classes of -degree $2$ for complex line bundles with integrable connection (in the smooth -case, otherwise think local systems). Also the related group $\mathbb -R\bigotimes S^1$ appears in the solution of Hilbert's third problem. We also -have that $K_2(\mathbb C)$ is a quotient of $\Lambda^2\mathbb C^\ast$ and -similar but worse groups (such as $S^2\Lambda^2\mathbb C^\ast$) appear in the -relation with generalisations of Hilbert's third problem (see for instance -J. Dupont: Scissors congruences, group homology and characteristic classes, Nankai -Tracts in Mathematics). -I may misrepresent the feelings of a lot of people if I say that algebraic -topologists are resigned to the appearance of such large abstract groups -whereas algebraic geometers want to believe that they have more structure (such -as $K_1(-)$ being the multiplicative group scheme) but haven't really been able -to realise this (there are some tantalising results such as Bloch et al's results -on the deformation theory of $K_2(-)$).<|endoftext|> -TITLE: Is a real power series that maps rationals to rationals defined by a rational function? -QUESTION [21 upvotes]: Suppose that the function $p(x)$ is defined on an open subset $U$ of $\mathbb{R}$ by a power series with real coefficients. Suppose, further, that $p$ maps rationals to rationals. Must $p$ be defined on $U$ by a rational function? - -REPLY [38 votes]: No. In fact, $p(x)$ can be a complex analytic function with rational coefficients that takes any algebraic number $\alpha$ in an element of $\mathbb{Q}(\alpha)$. -(And everywhere analytic functions are not rational unless they are polynomials). -The algebraic numbers are countable, so one can find a countable sequence of polynomials $q_1(x), q_2(x), \ldots \in \mathbb{Q}[x]$ -such that every algebraic number is a root of $q_n(x)$ for some $n$. Suppose that -the degree of $q_i(x)$ is $a_i$, and choose integers $b_i$ such that -$$b_{n+1} > b_{n} + a_1 + a_2 + \ldots + a_n.$$ -Then consider the formal power series: -$$p(x) = \sum_{n=0}^{\infty} c_n x^{b_n} \left( \prod_{i=0}^{n} q_i(x) \right),$$ -By the construction of $b_n$, the coefficient of $x^k$ for $k = b_n$ to -$b_{n+1} -1$ in $p(x)$ is the coefficient of $x^k$ in -$c_n x^{b_n} \prod_{i=1}^{n} q_i(x)$. Hence, choosing the $c_n$ to be appropriately -small rational numbers, one can ensure that the coefficients of $p(x)$ decrease sufficiently rapidly and thus guarantee that $p(x)$ is analytic. -On the other hand, clearly $p(\alpha) \in \mathbb{Q}[\alpha]$ for every (algebraic) -$\alpha$, because then the sum above will be a finite sum. -With a slight modification one can even guarantee that the same property holds for all derivatives of $p(x)$. -I learnt this fun argument from the always entertaining Alf van der Poorten (who sadly died recently). - -REPLY [16 votes]: You can map $\mathbb{Q}$ in itself by plenty of entire functions that are not rational functions. Let fix an enumeration of the rationals, $\mathbb{Q}=\{q _ j\ : j=1,2,\dots \} $. Consider a series -$$f(x):=\sum_{n=1}^\infty\ \epsilon _n \ \prod _{j=1}^n (x-q _j)$$ -If $\epsilon _n$ is a sequence of rationals converging to 0 with sufficient velocity, the series converges uniformly on bounded sets to an entire function. -(edit) rmk. of course with some more small care we can even make an entire function $f(x)=\sum_{n=0}^\infty p _n(x)$ invertible over $\mathbb{R}$, and a bijection between two assigned countably infinite dense subsets $A$ and $B$. Start the series with the identity $p_0(x)=x$, then add inductively only odd degree polynomials $p _n$, that vanish on the (finitely many) already settled points, and do not destroy the invertibility on $\mathbb{R}$ (say, keeping all partial sums of the series with derivative greater than $1/2$). The bijectivity between $A$ and $B$ is to be ensured by a standard ping-pong argument.<|endoftext|> -TITLE: How does a tournament's structure affect the likelihood that the best player will win? -QUESTION [8 upvotes]: Background -The origin of this question is a conversation I had with some friends a few years ago. At the time, Roger Federer and Tiger Woods were dominating professional tennis and golf, respectively, and we were comparing and contrasting the two. It occurred to me that there was a mathematical question that was relevant to our discussion; namely, the structure of golf tournaments vs. tennis tournaments. -For example, the Masters is a four-round tournament. After two rounds, roughly the bottom half of the field is sent home. The winner is the person with the lowest total score after four rounds. On the other hand, Wimbledon is a single-elimination tournament; the winner must defeat seven other players in head-to-head competition. From a structural standpoint, if you are the best player in the field, is it harder to win a tournament like the Master's or harder to win a tournament like Wimbledon? - -The mathematics -More generally, how does a tournament's structure affect the likelihood that the best player will win? -This question is a bit fuzzy because there are some modeling issues involved that will affect the answer. Something like the following seems a reasonable place to start. - -Let $F_1, F_2, \ldots F_n$ be independent normal distributions such that a random variable drawn from distribution $F_i$ gives the performance by the $i$th best player in a particular round of competition. So we would have $\mu_{F_i} > \mu_{F_j}$ when $i < j$. Assuming that the $F_i$'s have the same variance seems a reasonable starting point, too, as does the assumption that a given player's performances from round to round are independent. -First question: What is the best way to model the $\mu_{F_i}$'s vs. the $\sigma_{F_i}$'s? The difference $\mu_{F_1} - \mu_{F_2}$ ought to be much larger than $\mu_{F_{99}} - \mu_{F_{100}}$, so maybe something like $\mu_{F_i} = 1/i$ would work, but I'm not sure what makes for a reasonable variance to go with this function. - - -For specificity's sake, let's assume three types of tournament structure: 1) that of the Masters, 2) that of Wimbledon, and 3) that of the World Cup (which has a round-robin stage before moving to a single-elimination stage). - -Second question: Given a satisfactory answer to the first question, what is the probability that Player 1 will win each of these three tournaments? - - -My reasoning so far -It seems to me that the two most important factors involved that would prevent the best player from winning the tournament are - -an unusually poor performance from the best player in a particular round, and -an unusually good performance from someone else in the field in a particular round. - -There's not much that the tournament's structure could do to mitigate factor (1), although a single-elimination tournament would seem to be the most unforgiving. On the other hand, the structure of the tournament probably has a large effect on the impact of factor (2). For instance, an incredible performance from someone in two separate rounds of the Masters raises the bar quite a bit for the best player. On the other hand, in a tournament like Wimbledon two great performances might lead to two upsets of major players but doesn't provide any advantage in later rounds, and, for the best player to be negatively affected, he/she would have to be playing directly against the overperforming player. Also, if there are enough players around (like the early rounds of Wimbledon and all the way through the Masters) there is a high probability that someone in the field will turn in two great performances in two different rounds. -So, if you are the best player in the field it seems to me that contests in which you are essentially playing most of the field simultaneously, like the Masters, would be more difficult to win than single-elimination tournaments like Wimbledon, which in turn would be more difficult to win than those with a round-robin format in the early rounds and single-elimination in the later rounds, like the World Cup. - - -Third question: Are there any known results that address this problem of the effect of tournament structure on the best player's chances of winning? - -I would be happy to see critiques/comments on my modeling and my reasoning as well. - -REPLY [5 votes]: It turns out this problem has been studied extensively in the economics literature. The motivation is to create some sort of competition that will maximize the likelihood of the best candidate for a job or the best application for a grant actually being awarded the job or grant. -For example, "The Predictive Power of Noisy Elimination Tournaments," by Dmitry Ryvkin, examines the effects of seeding under different numbers of players and some different performance probability distributions. -The paper "Three Prominent Tournament Formats: Predictive Power and Costs," (apparently published in Management Science under the title "The Predictive Power of Three Prominent Tournament Formats"), by Ryvkin and Andreas Ortmann, addresses my question exactly, though. They calculate the exact probability (under uniform, normal, and Pareto distributions for player performance) that the best player wins a round robin tournament, a binary elimination tournament, and a contest. (The last involves all players performing simultaneously at once; the winner is the player with the best performance.) By calculating these probabilities for specific values they show numerically that for all but small numbers of players in a tournament, the best player in the tournament has a higher probability of winning a round robin tournament than a binary elimination tournament and a higher probability of winning a binary elimination tournament than a contest. -Given these results, it does appear that (all other things being equal) the structure of golf tournaments makes them more difficult to win than tennis tournaments, and, consequently, that dominating professional golf is even more impressive than dominating professional tennis.<|endoftext|> -TITLE: Derived categories of coherent sheaves: suggested references? -QUESTION [16 upvotes]: I am interested in learning about the derived categories of coherent sheaves, the work of Bondal/Orlov and T. Bridgeland. Can someone suggest a reference for this, very introductory one with least prerequisites. As I was looking through the papers of Bridgeland, I realized that much of the theorems are stated for Projective varieties (not schemes), I've just started learning Scheme theory in my Algebraic Geometry course, my background in schemes is not very good but I am fine with Sheaves. It would be better if you suggest some reference where everything is developed in terms of Projective varieties. - -REPLY [15 votes]: Kapustin-Orlov'a survey of derived categories of coherent sheaves is pretty good, - -A. N. Kapustin, D. O. Orlov, Lectures on mirror symmetry, derived categories, and D-branes, Uspehi Mat. Nauk 59 (2004), no. 5(359), 101--134; translation in Russian Math. Surveys 59 (2004), no. 5, 907--940, math.AG/0308173 - -but more slow/elementary exposition starting with fundamentals of derived categories is in an earlier survey of Orlov - -D. O. Orlov, Derived categories of coherent sheaves and equivalences between them, Uspekhi Mat. Nauk, 2003, Vol. 58, issue 3(351), pp. 89–172, Russian pdf, English transl. in Russian Mathematical Surveys (2003),58(3):511, doi link, pdf at Orlov's webpage (not on arXiv!) - -There are also Orlov's handwritten slides in djvu from a 5-lecture course in Bonn - -djvu, but the link is temporary - -For derived categories per se, apart from Gelfand-Manin methods book and Weibel's homological algebra remember that a really good expositor is Bernhard Keller. E.g. his text - -Bernhard Keller, Introduction to abelian and derived categories, pdf - -...and also his Handbook of Algebra entry on derived categories: -pdf<|endoftext|> -TITLE: Algebraic geometry -QUESTION [6 upvotes]: Possible Duplicate: -A learning roadmap for algebraic geometry - -I am a masters student and I want to study algebraic geometry. -Does there exist good good book for selfstudy of algebraic geometry? - -REPLY [4 votes]: As an introductory book, I will recommend Karen Smith's(plus some other's) book "An invitation to Algebraic geometry". It is very well wrtitten. - -REPLY [4 votes]: In addition to the excellent text by K. Smith et al. recommended above, you could consider -the book -Ideals, Varieties, and Algorithms by David A. Cox, John B. Little and Don O'Shea -which is very accessible, and some older texts which are more elementary and (therefore) quite readable, e.g. --- Rudiments of algebraic geometry by William Elliott Jenner --- Introduction to algebraic curves by Phillip A. Griffiths<|endoftext|> -TITLE: How can one "see" the Hopf fibration in the space of lattices in the plane? -QUESTION [13 upvotes]: This question is inspired from Etienne Ghys's talk on Knots and Dynamics from ICM 2006. -The map $L \mapsto (G_4(L), G_6(L))$ gives a bijection between all lattices $L\subset \mathbb{C}$ (including the degenerate ones) and $\mathbb{C}^2 - \{0\}$. Here $G_4$ and $G_6$ are the Eisenstein series of a lattice, -$$ - G_n (L) = \sum_{\omega \in L, \omega \neq 0} \frac{1}{\omega^n}. -$$ -By scaling the lattice $L \mapsto t L$ we can arrange that these two numbers satisfy $|z_1|^2 + |z_2|^2 = 1$; in other words we have that -$$ -\{ \mbox{lattices in } \mathbb{C} \mbox{ up to rescaling} \} \cong S^3. -$$ -Now $S^3$ carries a nice action of $S^1$ given by sending $(z_1, z_2) \mapsto (e^{i \theta}z_1, e^{i \theta}z_2)$, the quotient being $S^2$; this is the Hopf fibration $S^3 \rightarrow S^2$. -Since the collection of lattices up to rescaling identifies with $S^3$ in such a nice way, it is tempting to try and see this action of $S^1$ at the level of lattices. It would be nice if it were to correspond to rotation of the lattice! But alas, it does not. Firstly, it can't, because the action of $S^1$ on $S^3$ is free, while rotating a lattice might `click' it back into itself before one has rotated a full rotation. In fact we see that if we rotate the lattice via $L\mapsto e^{i\theta} L$, we find that the invariants change as -$$ -(G_4(L), G_6(L)) \mapsto (e^{-4i\theta}G_4(L), e^{-6i\theta}G_6(L)) -$$ -which is not the behaviour we are looking for. -So what does the action of $S^1$ on $S^3$ correspond to in the space of lattices up to rescaling? In other words, what is $L'$ in terms of $L$ if -$$ -(G_4(L'), G_6(L')) = (e^{i \theta} G_4(L), e^{i \theta} G_6(L))? -$$ - -REPLY [3 votes]: The circle action $(z,w) \mapsto (e^{2i\theta}z, e^{3i\theta}w)$ has all orbits, except two, isotopic to the trefoil. -I'm guessing that the exceptional orbits correspond to the square lattice and the hexagonal lattice. Further, I'm thinking that the degenerate lattices land on a single orbit, proving that the "space of lattices" is identical to the trefoil complement. -Ah - looking at Ghys' lecture, I see that he says all of this. Well, it is still pretty cool. -Edit: However, this tells you that there are Hopf orbits that cross the set of degenerate lattices exactly twice. This makes a nice answer to your question unlikely?<|endoftext|> -TITLE: Can the Knaster-Tarski theorem be proved using the Schroeder-Bernstein theorem? -QUESTION [5 upvotes]: The reverse can be done easily and the proof is well known I am wondering if the exact same argument can be used to prove reverse as well. - -REPLY [5 votes]: The -Cantor-Schroeder-Bernstein -theorem admits many proofs of various natures, which have -been extended in diverse mathematical contexts to show that -the phenomenon holds in many other parts of mathematics. So -the question of whether the CSB property holds is an -interesting mathematical question in many mathematical -contexts (and it is particularly interesting in contexts -where it fails). -And the proof of CSB from the Knaster-Tarski -theorem -that you have in mind proceeds as follows: suppose that -$f:A\to B$ and $g:B\to A$ are both injective. Define -$\varphi:P(A)\to P(A)$ on the power set of $A$ by -$\varphi(X)=A-g[B-f[X]]$. It is easy to see by applying the -functions and taking complements (twice) that $X\subset -Y\to \varphi(X)\subseteq \varphi(Y)$, and so $\varphi$ is a -monotone (order-preserving) operation on the power set of -$A$, a complete lattice. Thus, by KT there is a fixed point -$\varphi(X)=X$. From this, it follows that the function -$h=(f|X)\cup(g^{-1}|(A-X))$ is a bijection between $A$ and -$B$, and I leave all the details as a fun exercise. -Meanwhile, the proof of KT itself has a very short direct -proof, which it would seem difficult to improve upon by -using CSB. Namely, if $\varphi:L\to L$ is a -order-preserving function on a complete lattice $L$, then -let $d=\wedge\{e \mathrel{|} \varphi(e)\leq e\}$. Note -that $\varphi(e)\leq e$ implies $d\leq e$ which implies -$\varphi(d)\leq \varphi(e)\leq e$ and so $\varphi(d)\leq d$ -and consequently $\varphi(\varphi(d))\leq \varphi(d)$, and -so $\varphi(d)$ is one of the $e$'s, and so $d\leq -\varphi(d)$ and hence $d=\varphi(d)$, as desired. -Note that the function $\varphi$ arising in the proof of -CSB from KT is not merely monotone, but also continuous, -since if $X=\bigcup_i X_i$, then $f[X]=\bigcup_i f[X_i]$ -and so $B-f[X]=\bigcap_i B-f[X_i]$ and thus -$g[B-f[X]]=\bigcap_i g[B-f[X_i]]$, because $g$ is -injective, and hence $\varphi(X)=A-g[B-f[X]]=A-\bigcap_i -g[B-f[X_i]]=\bigcup_i (A-g[B-f[X_i]])=\bigcup_i \varphi(X_i)$. -In short, $\varphi(\bigcup_i X_i)=\bigcup_i\varphi(X_i)$, -which means that $\varphi$ is continuous. -Thus, the standard argument shows that CSB follows not only from KT, but from the special case of continuous-KT, that is, KT restricted to continuous monotone functions. But the full KT is true for arbitrary monotone -$\varphi$, even when they are not continuous, by the simple argument above. I take this -as suggesting that the ``exact same argument,'' as you -asked in your question, does not establish KT from CSB. Not -every monotone $\varphi$ arises as the $\varphi$ used in -the proof, since not every monotone map is continuous. -Meanwhile, I also observe that the direct proof of KT is so -simple, it would seem difficult to find a substantially -simpler proof of it by using any other principle, including -CSB.<|endoftext|> -TITLE: Motivic characterization of affine spaces -QUESTION [9 upvotes]: Let $k$ be a field and let $X$ be a smooth irreducible variety over $k$. -Suppose that I know that the image of $X$ in the Grothendieck group -of varieties over $k$ is equal to that of -a) ${\mathbb A}^n$ for some $n$ -b) ${\mathbb A}^{n_1}\times {\mathbb G}_m^{n_2}$ for some $n_1, n_2$. -Does it follow that $X$ is isomorphic to the varieties appearing in a) and b)? -If the answer is "yes", then can one drop the smoothness assumption on $X$? - -REPLY [19 votes]: The answer is no: Let $X$ the projective plane $\mathbb P^2$ minus a smooth -quadric $Q$. Then $[X]=[\mathbb P^2]-[Q]=\mathbb L^2+\mathbb L+1-(\mathbb -L+1)=[\mathbb A^2]$ but $X$ is not isomorphic to $\mathbb A^2$ as its Picard group is -$\mathbb Z/2$. For b) just cross $X$ with $\mathbb G_m^n$, the Picard group is still $\mathbb Z/2$. -For the singular case the following is an interesting example: Let the symmetric group -$\Sigma_n$ act on $(\mathbb A^m)^n$ by permuting the factors. Then it is a fact -(proved by Totaro, see Lemma 4.4 of L. Göttsche, On the motive of the Hilbert -scheme of points on a surface, Math. Res. Lett. 8 (2001), no.~5-6, 613--627.) -that the class of $(\mathbb A^m)^n/\Sigma_n$ in the Grothendieck group of -varieties is equal to that of $\mathbb A^{mn}$. However, when $m>1$ then the -quotient is always singular and hence not isomorphic to $\mathbb A^{mn}$. -Addendum: In response to the further question by Alexander, I can't think -of any strengthening that would make it true. For a moment I thought that if one -instead asked that $X$ be smooth and proper and have the same class as $\mathbb -P^n$ would imply that $X$ is isomorphic to $\mathbb P^n$. That however is -counterexampled by a smooth quadric of odd dimension greater than $1$. We can -project from a point to make the blowing up of one point isomorphic to the -blowing up of a smooth quadric of $\mathbb P^n$ which gives that the class of -$X$ is equal to $\mathbb P^n$. On the other hand $X$ is not isomorphic to -$\mathbb P^n$ as its cohomology ring is not generated by its degree $2$ part. -Addendum 1: Some comments related to Alexander's latest question. It is known that the stable birational class of $X$ can be recovered from the class $[X]$. Let us assume that we actually get that it determines the birational class (this may very well always be true and is true in small dimensions). - - -For curves (we shall always assume that $X$ is smooth and irreducible) we get that we can recover the projective model of $X$ but as $[X]$ is equal to the class of the projective model minus a number of points we see that if $[X]$ is not proper we can not recover which points we have removed from the projective model to get $X$. Hence the reconstruction problem is reasonable only when $X$ is also proper which we shall assume. - - Let us consider a ruled surface $X\to C$ with fibres $\mathbb P^1$'s. If we blow up a point on a fibre and then blow down the strict transform of the fibre containing the point, we get a new ruled surface which in general (ever?) is not isomorphic to $X$ but has the same class. - - On the other hand in the case of non-ruled surfaces we have a minimal model such that all other surfaces birational to it is constructed by a succession of blowing ups. Hence we can distinguish the minimal model from all other birational surfaces. However, we can not tell which points we have blown up and by blowing up different points we get different surfaces. - -In higher dimensions things seem worse as there almost never is a unique minimal model. For instance the two small resolutions of an ordinary double point in three dimensions have the same class but are in general non-isomorphic. - -There are a few cases where a minimal model is unique. The one that comes to mind is when $X$ is an abelian variety in which case $X$ can actually be recovered from its class. - -Hence I think that there will only be a very limited number of situations when one can hope to recover $X$ from its class.<|endoftext|> -TITLE: Repairing the Lie operad in characterstic 2? -QUESTION [5 upvotes]: Recall: We present an operad (with $S_n$-action) in $R-Mod$ for any commutative ring $R$ not of characteristic 2 generated by a single element in degree $2$ satisfying the following identities: - -$\theta+\theta\tau=0$ -$\theta(1,\theta)+\theta(1,\theta)\sigma + \theta(1,\theta)\sigma^2$. - -where $\tau$ and $\sigma$ are 2-cycles and 3-cycles respectively. -However, in characteristic $2$, this fails to characterize Lie algebras in the obvious way, since the first equation says that $\theta$ is skew-symmetric (and hence symmetric in characteristic 2). -The proper axiom to include is that $[x,x]=0$, i.e. that $[-,-]$ is alternating rather than skew-symmetric. Can we present this relation operadically? It seems like on the face of it, we can't, but I'd be happy to be surprised. - -REPLY [12 votes]: In fact, several monads can naturally be associated to an operad $P$ and this might be used to answer your question. -In the usual setting, one considers a generalized symmetric algebra $S(P,X) = \bigoplus_n (P(n)\otimes X^{\otimes n})_{\Sigma_n}$ where we form coinvariants under the action of the symmetric groups $\Sigma_n$. But we can also take invariants instead of coinvariants and form another functor $\Gamma(P,X) = \bigoplus_n (P(n)\otimes X^{\otimes n})^{\Sigma_n}$ associated to $P$. The image of the norm map from coinvariants to invariants still gives another functor $\Lambda(P,X)$ associated to $P$. -Under the assumption $P(0) = 0$, we have a monad structure on $\Lambda(P): X\mapsto\Lambda(P,X)$ and $\Gamma(P): X\mapsto\Gamma(P,X)$ inherited from the operadic composition structure of $P$. See (1.2.12-1.2.17) in -http://math.univ-lille1.fr/~fresse/PartitionHomology.pdf -(ref.: http://www.ams.org/mathscinet-getitem?mr=2005g:18015) -For the operad $P = Lie$, the algebra category associated to $\Gamma(Lie)$ can be identified with the category of $p$-restricted Lie algebras (where $p$ is the cateristic of the ground ring), while the algebra category associated to $\Lambda(Lie)$ can be identified with the category of Lie algebras equipped with an alternating Lie bracket.<|endoftext|> -TITLE: How badly can Krull's Hauptidealsatz fail for non-Noetherian rings? -QUESTION [17 upvotes]: Krull's Hauptidealsatz (principal ideal theorem) says that for a Noetherian ring $R$ and any $r\in R$ which is not a unit or zero-divisor, all primes minimal over $(r)$ are of height 1. How badly can this fail if $R$ is a non-Noetherian ring? For example, if $R$ is non-Noetherian, is it possible for there to be a minimal prime over $(r)$ of infinite height? - -REPLY [9 votes]: Valuation rings demonstrate quite clearly the failure of Krull's principal ideal theorem: -take a valuation ring O of finite dimension. The prime ideals then form a chain -$p_0:=0\subset p_1\subset\ldots\subset p_d$ -so that for every $i\in\{1,\ldots ,d\}$ there exists $r_i\in p_i\setminus p_{i-1}$. Obviously $p_i$ is a minimal prime over $r_iO$. -For valuation domains of infinite dimension one has to consider the so-called limit-primes: a prime ideal $p$ of a commutative ring $R$ is called limit-prime if -$p=\bigcup\limits_{q\in\mathrm{Spec} (R): q\subset p}q$. -There exist valuation domains $O$ of infinite Krull dimension such that the maximal ideal $m$ of $O$ is no limit-prime. For example take a valuation ring such that the corresponding value group is -$\mathbb{Z}\times\mathbb{Z}\times\ldots$ (countably many factors ordered lexigraphically). -Then one can find $r\in m$ such that $m$ is minimal over $rO$. -H<|endoftext|> -TITLE: Awfully sophisticated proof for simple facts -QUESTION [247 upvotes]: It is sometimes the case that one can produce proofs of simple facts that are of disproportionate sophistication which, however, do not involve any circularity. For example, (I think) I gave an example in this M.SE answer (the title of this question comes from Pete's comment there) If I recall correctly, another example is proving Wedderburn's theorem on the commutativity of finite division rings by computing the Brauer group of their centers. - -Do you know of other examples of nuking mosquitos like this? - -REPLY [10 votes]: Claim: $\sum\limits_{k=0}^n (-1)^k {n\choose k} = 0$ for all integers $n≥1$ -Proof: Take the $n-1$-dimensional simplex $\Delta_{n-1}$. We can compute it's Euler characteristic by using simplicial homology. -There are exactly $n \choose k+1$ many $k$-sub-simplexes of $\Delta_{n-1}$. Thus we get a simplicial chain complex of the form -$\mathbb{Z}^{n\choose n} \to \mathbb{Z}^{n\choose n-1} \to \cdots \to \mathbb{Z}^{n\choose 2}\to\mathbb{Z}^{n\choose 1}$. -So the Euler characteristic is $\chi(\Delta_{n-1}) = \sum\limits_{k=0}^{n-1} (-1)^k {n\choose k+1}=-\sum\limits_{k=1}^{n} (-1)^k {n\choose k}$ -On the other hand $\Delta_{n-1}$ is contractible, and $\chi$ is homotopy-equivalence-invariant, so $\chi(\Delta_{n-1})=\chi(pt) =1$. -Putting those toghether we obtain: -$0=\chi(\Delta_{n-1})-\chi(\Delta_{n-1})=1+\sum\limits_{k=1}^{n} (-1)^k {n\choose k}=\sum\limits_{k=0}^n (-1)^k {n\choose k}$<|endoftext|> -TITLE: Poincare dual in equivariant (co)homology? -QUESTION [16 upvotes]: Let $G$ be a compact Lie group, $X$ be a (compact, oriented) smooth manifold, with $G$ acts on $X$ smoothly. Then we can talk about the $G$-equivariant homology and cohomology. -My question: In what sense can we have a duality between the equivariant homology and cohomology, in analogue with the Poincare duality between the ordinary homology and cohomology of $X$? In particular, the degree of a nontrivial equivariant (co)homology class could exceed the dimension of the manifold $X$. Then in such case, what does the dual mean, geometrically? - -REPLY [5 votes]: See the paper Equivariant Ordinary Homology and Cohomology by Steve Costenoble and myself for a comprehensive theory of Poincare duality in ordinary equivariant homology and cohomology (as opposed to Borel homology/cohomology) for compact Lie group actions.<|endoftext|> -TITLE: Morse Theory and Exotic Spheres -QUESTION [13 upvotes]: I'm finally at the end of Milnor's "On manifolds homeomorphic to the 7-sphere", and I stumbled upon something I cant figure out... -For those with the reference I'm talking about "lemma 5", it goes something like this, you have two $\mathbb{S}^3$ bundles over $\mathbb{S}^4$, we want to obtain the total space of this bundle, so you glue them via the transition function, one can think of this as having a pair of copies of $(\mathbb{R}^4 \setminus \{0\}) \times \mathbb{S}^3$ and gluing them by identifiying $(u,v) \mapsto (u',v')=(u / \|u\|^2, u^hvu^j/\|u\|)$ where $u$ and $v$ are quaternions, so far so good, now Milnor states that if $h+j =1$ then this manifold is a $7$-sphere, his reason is that the function $f(x) = \mathfrak{R}(v)/(1+\|u\|^2)^{1/2}$ is a morse function, this with the "first" coordinate chart, for the second he defines $u'' = u'(v')^{-1}$ and substitutes $(u',v')$ for $(u'',v')$ stating that the function $f$ is now given by $\mathfrak{R}(u'')/(1+\|u''\|^2)^{1/2}$. -He then says "It is easily verified that f has only two critical points (namely $(u,v) = \pm (0,1)$) and that these are nondegenerate". -That's where I get lost; I don't understand his change of coordinates $(u',v') \mapsto (u'',v')$, nor why he states the function is now the one stated... I tried developing the algebra but I can't get it to work out, I thought maybe he was using the involution $v \mapsto v^{-1}$ somehow but it doesn't add up either... - -REPLY [10 votes]: Milnor didn't explain the formula as much as maybe he should have, but the point is that the real part of a unit-length quaternion is invariant under both conjugation and inversion. Let $$r = ||u|| \qquad \hat{u} = u/r,$$ -so that -$$v' = \hat{u}^h v \hat{u}^j \qquad u' = \hat{u}r \qquad ||u'|| = ||u''|| = 1/r.$$ -Thus -$$v' \hat{u}^{-1} = \hat{u}^h v \hat{u}^{-h}$$ -is conjugate to $v$. Thus -$$\mathfrak{R}(v) = \mathfrak{R}(v'\hat{u}^{-1}) = \mathfrak{R}(\hat{u} (v')^{-1}).$$ -The first equality is conjugation, the second one is inversion. So then you get -$$\frac{\mathfrak{R}(v)}{\sqrt{1 + ||u||^2}} = \frac{\mathfrak{R}(v)}{\sqrt{1+r^2}} = \frac{\mathfrak{R}(\hat{u} (v')^{-1})}{\sqrt{1+r^2}} = \frac{\mathfrak{R}((\hat{u}/r) (v')^{-1})}{\sqrt{1+r^{-2}}} = \frac{\mathfrak{R}(u'')}{\sqrt{1+||u''||^2}}.$$ -Note that, although $(u'',v')$ certainly is a valid parameterization of the second chart, it's enough to think of $u''$ as a convenient function rather than part of a coordinate frame. -The question now in my mind is, how did Milnor think of this algebra? I do not know the answer. Maybe he started with a round 4-sphere with its quaternionic Hopf fibration, and the elementary Morse function that consists of one of the coordinates in $\mathbb{R}^8$. You immediately get that there are two critical points (the north and south pole) and that they lie on the same Hopf fiber, since opposite points on a sphere always do. Apparently this Morse function fits together in a similar way for all of these 3-sphere bundles over the 4-sphere.<|endoftext|> -TITLE: Algorithm for Weierstrass Preparation Theorem for Formal Power Series -QUESTION [5 upvotes]: The Weierstrass preparation theorem for formal power series rings guarantees that if a given formal series $f(z) = \sum a_k z^k \in R[[z]]$ where $R$ is a complete local ring with maximal ideal $M$ has $a_k \in M$ for $k < n$ and $a_n \in R^* = M^c$, then -$$ -f = (z^n + b_{n-1}z^{n-1} + \cdots + b_0)u -$$ -where $b_k \in M$ and $u$ is a unit in $R[[z]]$. -I need an explicit algorithm for calculating this Weierstrass polynomial (or distinguished polynomial) for a given $f$. In my case the coefficient ring is $R = \mathbb Z_3[[x]]$, formal power series over the 3-adics. So any algorithm would have to be robust enough to handle these coefficients. -Does anyone know of such an algorithm for a math software package? - -REPLY [6 votes]: Here's an algorithm that I use. Let's call $S$ the degree-$n$ shift operation, sending $\sum c_kz^k$ to $\sum c_{n+k}z^k$, in other words the quotient when you divide a power series by $z^n$. Step 0: divide $f$ by $Sf$, giving you a power series $f_1$ such that $Sf_1\equiv 1$ modulo $M$. Step $i$, for $i > 0$: repeat. At each stage, you get a power series $f_i$ for which $Sf_i\equiv 1 $ modulo $M^i$. For a quicker variant of Step $i$ (for $i > 0$), instead multiply by $2-Sf_i$. It works because you've constructed a convergent infinite product.<|endoftext|> -TITLE: Examples of ZFC theorems proved via forcing -QUESTION [16 upvotes]: This is an old suggestion of Joel David Hamkins at the end of his answer to this question: Forcing as a tool to prove theorems -I just noticed it while trying to understand his answer. But indeed it would be nice to have a big list of $ZFC$ theorems that were proven first by forcing. -A very well known example is Silver's Theorem about the fact that the $GCH$ can't fail first at a singular cardinal of uncountable cofinality (say for instance $\aleph_{\omega_1}$), I had read somewhere (Jech, maybe) that Silver proved it first using forcing. -Also if anyone knows theorems of pcf theory that were first proven using forcing, please post them. - -REPLY [6 votes]: Another example around a theorem of Gitik and Shelah: If there is a total extension $m:\mathcal{P}([0, 1]) \rightarrow [0, 1]$ of Lebesgue measure then there is an uncountable set of reals all of whose Lebesgue null subsets are countable. -Proof sketch: Force with the measure algebra of $m$. By (a weak form of) Gitik-Shelah theorem (whose original proof uses generic ultrapower arguments too), this forcing adds at least $\omega_1$-many random reals in the generic extension. But this set of randoms is also in the generic ultrapower so by elementarity we have such a set in $V$. -For some open questions along these lines see Fremlin's survey article on real valued measurable cardinals and also his problem list.<|endoftext|> -TITLE: How complicated is infinite-dimensional "undergraduate linear algebra"? -QUESTION [15 upvotes]: The name "undergraduate linear algebra" in the title is a bit of a joke, and so I don't know how widely spread it is. To wit: - -High school linear algebra is the theory of a finite-dimensional vector space — i.e. it consists of the finite-dimensional representation theory of a field $\mathbb F$, or, better, for representation theory over $\mathbb F$ of the (trivial) monoid of size $1$ (the free monoid on zero generators). The complete structure theory is well-known: every finite-dimensional vector space is (noncanonically isomorphic to) a direct sum of copies of the one-dimensional (free) vector space: $V \cong \mathbb F^n$ for some integer $n$. In fact, assuming the axiom of choice, one can extend this result to arbitrary vector spaces: if $\mathbb F$ is a field and $V$ is a $\mathbb F$-module, then $V \cong \mathbb F^\kappa = \coprod^\kappa \mathbb F$ for some cardinal $\kappa$. -Undergraduate linear algebra is the theory of a finite-dimensional vector space with a choice of matrix, which is to say it is the finite-dimensional representation theory of the polynomial ring $\mathbb F[x]$, or equivalently the finite-dimensional representation theory of the free monoid $\mathbb N$ on one generator. When $\mathbb F$ is algebraically closed, the complete structure theory is given by the Jordan Canonical Form theorem. The category of finite-dimensional representations of $\mathbb N$ is not semisimple, but the indecomposables are parameterized by pairs $(\lambda,n)$ where $\lambda \in \mathbb F$ and $n\in \mathbb Z_{>0}$ representing "Jordan blocks", and the irreducibles are precisely the indecomposables with $n = 1$, so that the irreducibles are parameterized by $\lambda \in \mathbb F$. If $\lambda \neq \mu$, then the irreducibles corresponding to $\lambda,\mu$ have no nontrivial extensions; there is a unique nontrivial extension of the irreducible corresponding to $\lambda$ by itself. - -I recently realized, however, that I never learned much of anything about the infinite-dimensional representation theory of the monoid $\mathbb N$, even assuming that my field $\mathbb F$ is algebraically closed (and, if you like, characteristic zero), and even assuming the axiom of choice. There are plenty of vector spaces with endomorphisms that have no eigenvectors — a standard example with $\mathbb F = \mathbb C$ is $V = $ the vector space of smooth $\mathbb C$-valued functions on $\mathbb R$ that vanish outside the interval $[0,1]$, and the endormorphism is differentiation. -With some restrictions (words like "compact operator" and "spectral theorem" come up), there are lots of results that try to reproduce the Jordan theorem. But since I hate all types of analysis, my question is: - -Assuming the axiom of choice, to what extent is it possible to describe the full (infinite-dimensional) $\mathbb C$-representation theory of the monoid $\mathbb N$? (Equivalently, the category of $\mathbb C[x]$-modules.) - -A good answer might consist of a minimal "generating" set, i.e. a set $\mathcal S$ of $\mathbb N$-representations to that every $\mathbb N$-rep is a colimit of a diagram whose objects are all in $\mathcal S$. (E.g. for infinite-dimensional high school algebra, $\mathbb S$ consists of the one-dimensional vector space.) I could also imagine an answer consisting of a "no-go theorem" that the representation theory of $\mathbb N$ is hard; maybe that's why I never learned it. - -REPLY [11 votes]: In the structure theory of an operator acting on an infinite-dimensional vector space over a field $F$, there is some good news and a lot of bad news. The best news (as Matthew Emerton mentions) is the classification of finitely generated modules over a PID. This beautiful theorem immediately generalizes the Jordan canonical form theorem and the classification of finite or finitely generated abelian groups. It says that every f.g. module over a PID $R$ is a direct sum of cyclic modules $R/d$, and the summands are unique in either the primary decomposition (which comes from the proof by isotypic submodules and Jordan blocks) or the invariant factor decomposition (which comes from the Smith normal form proof). You can take $R = \mathbb{Z}$, or $R = F[x]$ for a field $F$, or of course there are other choices. -I suspect that the bad news is likely to be uniformly bad for countably generated modules over any PID. I found a reference for the case of torsion-free abelian groups, so I will concentrate on that case first. Consider the abelian groups $A$ that lie between $\mathbb{Z}^2$ and $\mathbb{Z}[\frac1p]^2$ for a prime $p$. Those choices for $A$ that are free abelian are just lattices in the plane (whose coordinates have denominators that are powers of $p$). There is an important theorem from the theory of affine buildings that you can make an infinite tree of valence $p+1$ out of these lattices. Two lattices are equivalent if they are the same lattice up to homothetic expansion (by a power of $p$). Two lattices are connected by an edge if they are nested and their quotient is $\mathbb{Z}/p$. So, the theorem says that this graph whose vertices are lattices is acyclic and has degree $p+1$ everywhere; the graph is an affine Bruhat-Tits building of type $A_1$. -Given a general $\mathbb{Z}^2 \subseteq A \subseteq \mathbb{Z}[\frac1p]^2$, you can let $A_n = (\frac1{p^n}\mathbb{Z}^2) \cap A$. This is the type of direct limit that Matthew Emerton describes, which looks good, but bad news is coming. For a while, $A_n/A_{n-1}$ can be $(\mathbb{Z}/p)^2$, then it can be $\mathbb{Z}/p$, and then $A_n$ might stop growing. If it stops growing, then $A$ is free with two generators. If it always grows by $(\mathbb{Z}/p)^2$, then $A = \mathbb{Z}[\frac1p]^2$. But in the middle case, if it grows by $\mathbb{Z}/p$, then $A$ is described by a path in the affine building from the origin to infinity. There are uncountably many ($2^{\aleph_0}$) such paths, and they have a $p$-adic topology. -So when are two such modules $\mathbb{Z}^2 \subseteq A \subseteq \mathbb{Z}[\frac1p]^2$ and $\mathbb{Z}^2 \subseteq B \subseteq \mathbb{Z}[\frac1p]^2$ isomorphic? The outer module is obtained by tensoring either $A$ or $B$ with $\mathbb{Z}[\frac1p]$, so the question is when $A$ and $B$ are equivalent under the action of $\text{GL}(2,\mathbb{Z}[\frac1p])$, which is a countable group. After localizing at all of the other primes, this question was studied by Hjorth, Thomas, and other logicians and algebraists. As Brian Conrad suggests, they have several theorems that indicate that the answer is wild, similar in spirit to the classification of infinite graphs up to isomorphism. -Now suppose that $F$ is a field, most conveniently a countable or finite field, and suppose that $V$ is an $F[x]$-module lying between $F[x]^2$ and $F[x,x^{-1}]^2$. Then you get the same geometric picture as before: The $F[x]$-free choices for $V$, up to homothety by powers of $x$, form an affine building which is an infinite tree, such that the neighbors of each vertex are a projective line $FP^1$. The complicated submodules are the ones that correspond to paths in this tree from the origin to infinity. If $F$ is finite or countable, then $\text{GL}(2,F[x,x^{-1}])$ is also countable, but the set of these paths is always uncountable and the group action is always far from transitive. I don't know if Hjorth's and Thomas' results extend to this case (because I am not qualified to read their papers), but my guess is that they would be comparably pessimistic. -There are other cases in which you get good news, basically by making infinite-dimensional operators look like finite-dimensional operators. The module $V$ in the previous paragraph has no $x$-eigenvectors even after passing to the algebraic closure $\bar{F}$. This is the same statement as that $V$ is torsion-free over $F[x]$. I think that a torsion module over a PID $R$ behaves much better, if it is the direct sum of its isotypic summands. For instance, if $x$ is locally nilpotent on $V$, then I think that it is a direct sum of Jordan blocks, possibly including infinite Jordan blocks (like the differentiation operator $x = \partial/\partial t$ on $\mathbb{Q}[t]$).<|endoftext|> -TITLE: homology of a symplectic leaf in GL(n) -QUESTION [13 upvotes]: I would be grateful for any info -concerning the following question. -Given an n-tuple z=(z_1,...,z_n) of -nonzero complex numbers, let X(z) denote the -GL(n,C)-conjugacy class of the -diagonal matrix diag(z_1,...,z_n). -Let Y(z) be an open subset of X(z) formed -by the matrices g in X(z) such that each of the -principal minors d_1(g), d_2(g),...,d_n(g), of g, -is nonzero -(by definition, d_k(g) := determinant -of the upper-left k by k block of the matrix g). -I am interested in the following -Problem. Compute the first Betti number, -i.e. the rank of the first homology group -with rational coefficients, -of the manifold Y(z), for a sufficiently -general n-tuple z=(z_1,...,z_n). - -REPLY [7 votes]: Below now is a proof that for sufficiently generic $z$, $b_1(Y(z))=n-1$. -First we show $b_1(Y(z))\geq n-1$. -Let $d=(d_1,d_2..,d_{n-1}): Y(z)\to({\mathbb C}^*)^{n-1}$. -One can explicitly lift the loops in $H_1(({\mathbb C}^*)^{n-1})$ as follows. Fix $i$ between $1$ and $n-1$. Let $e(t)=exp(2\pi \sqrt{-1} t)$ and $$B(t):=\left(\begin{array}{cc} e(t) & 1\\\ e(t)(z_i+z_{i+1}-e(t))-z_iz_{i+1} & z_i+z_{i+1}-e(t) \end{array}\right).$$ Then the characteristic polynomial of $B(t)$ is $(\lambda-z_i)(\lambda-z_{i+1})$. Form an $n\times n$ matrix $A(t)$ as having $z_1,\dots,z_{i-1}$ -on the main diagonal, then $B(t)$, then $z_{i+2},\dots,z_{n}$ and zero everywhere else. This matrix has the property that $A(t)\in Y(z)$ (as we assumed generic z) and $d_j(A(t))=z_1\cdot\dots\cdot z_j$ unless $j=i$ when $d_i(A(t))=e(t)z_1\cdot\dots\cdot z_{i-1}$. Thus $A(t)$ is a loop in $Y(z)$ covering -the $i$th generator of $H_1(({\mathbb C}^*)^{n-1})$. -Thus we find that $d_*:H_1(Y(z))\to H_1(({\mathbb C}^*)^{n-1})$ is surjective so $b_1(Y(z))\geq n-1$. -To argue that $b_1\leq n-1$ we first show that the divisors $D_i:=d_i^{-1}(0)$ are irreducible on $X(z)$ for sufficiently generic $z$. We note that -$D_i$ is irreducible in $M_{nxn}$ which can be easily seen over a finite field by counting points on $D_i$. This implies that for a sufficiently -generic $z$ the divisor $D_i$ is irreducible on $X(z)$. When $z_i\neq z_j$ for $i\neq j$ we see that $X(z)$ fibers over the full flag variety with -contractible fibers, so it is simply connected. From the homology Gysin sequence we get that removing an irreducible divisor from a smooth variety can increase -$b_1$ by at most one. By induction we can conclude that $b_1(Y(z))\leq n-1$. -As an example take $n=2$. Here we are removing a copy of ${\mathbb C}^*$ from $X(z)$ an affine line bundle over ${\mathbb P}^1$. This will have Betti numbers $b_1=1$ and $b_2=2$ with weights $q,q,q^2$ respectively, to give Serre-polynomial $q^2+1$. We will have the map - $d=d_1:Y(z) \to {\mathbb C}^{*}$. -The fibers of this map will generically be a ${\mathbb C}^{*}$ and will have two singular fibers, which will give the non-trivial fundamental group. The point is that one can lift the loop generating $\pi_1({\mathbb C}^*)$ as generically $d_1$ is a fibration with connected fibers.<|endoftext|> -TITLE: Penrose tilings and noncommutative geometry -QUESTION [19 upvotes]: Are there "elementary" resources on Penrose Tilings in relation to noncommutative geometry? It's all a big blur to me. There are two transformations S and T that can grow the tilings and every tiling corresponds to a sequence of S's and T's. Somehow, there is a C* algebra related to this. -Can the elements of this algebra be interpreted as observables in a quantum mechanical system? What is the "geometry" of this noncommutative space? -(source) - -REPLY [2 votes]: There is a concise description of this in section 1.1 of this nice paper Trees, Ultrametrics, and Noncommutative Geometry by Bruce Hughes: http://arxiv.org/abs/arXiv:math/0605131. -This isn't particularly elementary, I guess, but it is a line of investigation that embarks from Connes's example. -I hope it is helpful!<|endoftext|> -TITLE: Is Fürstenberg's topology useful? -QUESTION [20 upvotes]: It's hard not to be amused and perhaps even amazed when first encountering Fürstenberg's clever "topological" proof that there are infinitely many primes. Closer inspection, however, reveals the disappointing truth that there really isn't anything topological going on there, as pointed out by BCnrd in a comment to this answer. -Nevertheless, the topology on $\mathbb{Z}$ introduced in the proof, where an open set is defined as any union of arithmetic sequences, does seem both natural and interesting. -My question is this: Can anything useful be done with this topology? Useful would include a new theorem, a simplification to a proof of a known result, or even fresh insight into standard material. - -REPLY [6 votes]: A more rigorous version of Scott's answer: If a topology on a group $G$ is translation-invariant, then it also defines a uniformity on $G$, by definition a distinguished set of neighborhoods of the diagonal $G \times G$ that is analogous to a metric. Actually, in the present example with $G = \mathbb{Z}$, the uniformity comes from a metric. Like the metric spaces that they generalize, uniform spaces have completions. The completion of $\mathbb{Z}$ with respect to the uniformity cited by Furstenberg is exactly the adelic profinite completion of $\mathbb{Z}$. Or if $G$ is any group, there is a similar topology generated by finite-index subgroups, and a uniformity, and the completion is the profinite completion.<|endoftext|> -TITLE: Concavity of $\det^{1/n}$ over $HPD_n$. -QUESTION [13 upvotes]: One of my beloved theorems in matrix analysis is the fact that the map $H\mapsto (\det H)^{1/n}$, defined over the convex cone $HPD_n$ of Hermitian positive definite matrices, is concave. This is accurate, if we think that this map is homogeneous of degree one, thus linear over rays. - -it has important applications in many branches of mathematics, -it has many elegant proofs. I know at least three complety different ones. - -I am interested to learn in both aspects. Which is your prefered proof of the concavity ? Is it useful in your own speciality ? In order to avoid influencing the answers, I decide not to give any example. But those who have visited my page may know my taste. - -REPLY [6 votes]: Here is an interesting calculus proof. Let $f:A\mapsto(\det A)^{1/n}$, defined over $SPD_n$. Differentiating twice, we find the Hessian -$${\rm D}^2f_A(X,X)=\frac1{n^2}f(A)\left(({\rm Tr} M)^2-n{\rm Tr}(M^2)\right),$$ -where $M=A^{-1}X$. This matrix, being the product of two symmetric matrices with one of them positive definite, is diagonalisable with real eigenvalues $m_1,\ldots,m_n$. The parenthesis above is now -$$\left(\sum_jm_j\right)^2-n\sum_jm_j^2,$$ -a non-positive quantity, according to Cauchy-Schwarz. We infer that ${\rm D}^2f_A\le0$ and that $f$ is concave.<|endoftext|> -TITLE: discrete subgroups in p-adic Lie groups? -QUESTION [7 upvotes]: It is known, from the works of G.Margulis, etc. that lattices in semi-simple real (algebraic) groups are "often" arithmetic subgroups, as long as the split rank is high enough. Here by a lattice in a Lie group $G$ is understood a discrete subgroup $\Gamma$ in $G$ such that the measure on $\Gamma\backslash G$ induced from the Haar measure on $G$ is of finite volume. -Typical example: for $G$ a simple $\mathbb{Q}$-group, with $\mathbb{Q}$-rank at least 2, then a lattice in $G(\mathbb{R})$ is arithmetic, in the sense that it is commensurable with a congruence subgroup, the latter being the intersection $G(\mathbb{Q})\cap K$ inside $G(\mathbb{A}^f_\mathbb{Q})$ for some compact open subgroup $K$ in the group of adelic points of $G$. -And what should be said about discrete subgroups in $G(\mathbb{Q}_p)$? Write for simplicity $G_p=G(\mathbb{Q}_p)$. At least one knows that for a discrete subgroup $\Gamma$ in $G_p$, the quotient $\Gamma\backslash G_p$ is compact if and only if it is of finite volume with respect to the Haar-induced measure. If $G$ is defined over a global field, say $\mathbb{Q}$, then a conguence subgroup of $G$ "should" be also a lattice in $G_p$. Moreover, for general semi-simple $\mathbb{Q}_p$-group $G$, how to classify the co-compact discrete subgroups in $G_p$? Do they admit explicit constructions like those coming from global fields? -Any comments and references are welcome. -Thanks! - -REPLY [10 votes]: The arithmeticity theorem of Margulis also characterizes lattices in product of simple groups over $R$ as well as $p$-adic fields. All such lattices in the higher rank case are S-arithmetic. When all but one of the factors is compact, the projection onto the non-compact factor (that you can take to be a p-adic field) gives you a lattice there. For instance, take a quadratic form $q$ that is anisotropic over $R$, but isotropic over ${\mathbb Q}_p$ for some $p$. Then $SO({\mathbb Z}[1/p])$ is a lattic in the product $SO(q, {\mathbb R}) \times SO(q, {\mathbb Q}_p)$. The projection onto the first component is dense, and onto the second component is a lattice there. For the precise statement of the (S-)arithmeticity theorem, the best place to look is Margulis's book "Discrete Subgroups of Semisimple Lie Group".<|endoftext|> -TITLE: Is the inclusion of Lebesgue spaces compact? -QUESTION [5 upvotes]: [Disclaimer: this may be a very trivial question; it certainly looks like it ought to have been studied and understood. I started thinking about it this morning when writing some notes for Rellich-Kondrachov, but cannot find a simple counterexample.] -For the time being, let us just work on $\mathbb{R}^d$ with the Lebesgue measure. It is well-known that for an open, bounded domain (hence with finite measure) $\Omega$, the inclusion -$$ L^p(\Omega) \to L^q(\Omega) $$ -is continuous for $\infty \geq p \geq q \geq 1$. -Question: Is the inclusion completely continuous (i.e. compact)? If not, what is a simple counterexample? -I dug around a bit and cannot find any references to a proof (or even the statement). The usual non-compactness mechanisms, of course, do not work. Let $f_i$ be a sequence of functions with $\|f_i\|_{L^p(\Omega)} \leq 1$, so by continuous inclusion it is a bounded sequence in $L^q(\Omega)$. Because $\Omega$ is compact, we cannot have the problem fixed-scale translations: if $f_i(x) = f(x + y_i)$ for a sequence of points $y_i$, since $y_i$ must have a converging subsequence, then so must $f_i$, even in $L^p$. The other usual non-compactness mechanism is dilations. WLOG assume $0\in supp(f) \subset\subset \Omega$ and that the $supp(f)$ is convex. Then the sequence $f_j = 2^{jd/p} f(2^j x)$ is a bounded but non-compact sequence in $L^p(\Omega)$. But in $L^q(\Omega)$ for $q < p$, the sequence converges strongly to 0. - -REPLY [2 votes]: Some further remarks. A nice aspect of this question is that it provides an example of how some general (and non-constructive) theorems of Functional Analysis often show us the way to a constructive answer to concrete problems. Let's focus on the case of the inclusion $L^\infty(\Omega)$ in $L^1(\Omega)$ (any other non-compactness follows from $L^\infty(\Omega)\to L^p (\Omega)\to L^q(\Omega)\to L^1(\Omega)$ by composition). By the Fréchet-Kolmogorov theorem, a bounded set $B\subset L^1(\Omega)$ (with $\Omega$ a bounded subset of $\mathbb{R}^n$) is relatively compact if and only if it is "equicontinuous-$L^1$," meaning that -$$\sup _{f\in B}\\ \omega_f(\delta) \\ =o(1),\qquad \mathrm {as}\\ \delta\to0.$$ -Here $\omega_f$ is the "modulus of continuity-$L^1$" of the function $f$, namely -$$\omega_f(\delta)= \sup _{|h|\leq \delta}\| f-f(\cdot-h) \|_1\\ .$$ -This condition is clearly not satisfied by the unit ball $B$ of $L^\infty,$ because for any $\delta$ there is in $B$ a function whose support is disjoint from a $\delta$ translate of it. In fact $\| f-f(\cdot-h) \|_1=|\Omega|$ holds, for instance, for the characteristic function of a suitable measurable set. So this answers the question, and also suggests an answer independent from the FK thm, by exhibiting directly a non-compact sequence of oscillating functions, like in Denis Serre's answer.<|endoftext|> -TITLE: Functions whose gradient-descent paths are geodesics -QUESTION [33 upvotes]: Let $f(x,y)$ define a surface $S$ -in $\mathbb{R}^3$ with a unique local minimum at $b \in S$. -Suppose gradient descent from any start point $a \in S$ -follows a geodesic on $S$ from $a$ to $b$. -(Q1.) -What is the class of functions/surfaces -whose gradient-descent paths are geodesics? -Certainly if $S$ is a surface of revolution -about a $z$-vertical line through $b$, -its "meridians" -are geodesics, and these would be the paths followed -by gradient descent down to $b$. -So the class of surfaces includes surfaces of -revolution. But surely it is wider than that? -(Q2.) -One could ask the same question about paths followed by -Newton's method, which in general are different from gradient-descent -paths, as indicated in this Wikipedia image: -       - -Gradient descent: green. -Newton's method: red. - -(Q3.) These questions make sense in arbitrary dimensions, -although my primary interest is for surfaces in $\mathbb{R}^3$. -Any ideas on how to formulate my question as constraints on $f(\;)$, -or pointers to relevant literature, -would be appreciated. Thanks! - -REPLY [4 votes]: Here is a function $f(x,y)$ which is 0 inside the square $C=[\pm1,\pm1]$, -and outside that square -has value equal to the Euclidean distance $d( p, C )$ from $p=(x,y)$ to the boundary of $C$. -[I am trying to follow Pietro Majer's suggestion, as far as I understand it.] -It is not a surface of revolution -(but it is centrally symmetric). -Are its gradient descent paths geodesics? -I think so... - - - -Left above: $f(x,y)$. Right above: Level sets of $f$. -Below: $\nabla f$. - - - -And here (below) is a closeup of the function defined using squared distance -$[d( p, C )]^2$, as per Willie Wong's suggestion:<|endoftext|> -TITLE: Are the Stiefel-Whitney classes of a vector bundle the only obstructions to its being invertible? -QUESTION [21 upvotes]: Consider vector bundles on connected paracompact topological spaces. Such a vector bundle $E$ on $X$ is said to be invertible if there exists some other bundle $F$ whose sum with $E$ is trivial: $E\oplus F \simeq \epsilon ^N $. The terminology "invertible" (used by Tammo tom Dieck for example) comes from K-theory and is not so weird as it looks:in $\tilde K(X)$ the class of $F$ is indeed the additive inverse of that of $E$. If all vector bundles on $X$ are invertible, then every class (=virtual bundle) in $\tilde K(X)$ is represented by an actual bundle, which is rather nice. -Now, every vector bundle is invertible if $X$ is compact or is a differentiable manifold or even a topological manifold or even a subspace of some $\mathbb R^n$ or even a space of finite combinatorial Lebesgue dimension or even ... [Please correct me if I'm wrong: this is an interpretation/synthesis of what I read, sometimes between the lines, in several places.] -So one might optimistically hope that every vector bundle on a paracompact connected space is invertible: after all, what could go wrong? Here is what. -Consider $X=\mathbb {RP}^{\infty}$ (infinite dimensional real projective space) and the tautological line bundle $\gamma$ on $X$. Its total Stiefel-Whitney class is -$w(\gamma)=1+x \in H^\ast (\mathbb {RP}^{\infty},\mathbb Z /2)=(\mathbb Z /2)[x]$, where $x$ is the first Stiefel-Whitney class of $\gamma$ . If $\gamma$ had an inverse vector bundle $F$ we would have $w(\gamma) w(F)=1$ and this is impossibl since $w(\gamma)=1+x$ is not invertible in the cohomology ring $H^\ast (\mathbb {RP}^{\infty},\mathbb Z /2)=(\mathbb Z /2)[x]$ ( a polynomial ring in one indeterminate over $\mathbb Z /2)$. -This leads me to ask the question: -If a vector bundle on a connected paracompact space has a total Stiefel-Whitney class invertible in its cohomology ring, does it follow that the bundle itself is invertible? - -REPLY [21 votes]: The answer is no. Let $G$ be a cyclic group of order $n$ not divisible by $2$, let $V$ be an irreducible $2$-dimensional representation of $G$, and consider the associated vector bundle $EG\times_G V\to BG$ (which I'll call $V$ again). Then $V$ has trivial Stiefel-Whitney classes since $H^q(BG;\mathbb{Z}/2)=0$ if $q>0$, but $V$ can have non-vanishing Pontryagin class (we have $H^*(BG;\mathbb{Z})=\mathbb{Z}[[x]]/(nx)$ where $x\in H^2$, and $p(V)=1-a^2x^2$ with $a\in \mathbb{Z}/n$ depending on the original representation.) Since the Pontryagin class satisfies Whitney sum up to $2$-torsion, this gives a counterexample: the virtual bundle $-V$ is does not come from a vector bundle, since $p(-V)=(1-a^2x^2)^{-1}$. -Now you ask, what if we also require that the Pontryagin classes are (finitely) invertible? There's probably a counterexample here too, though I don't have one at hand. -Added later. Here's one. (I hope: I keep needing to fix it.) If $G$ is a finite $p$-group, then the "Borel construction" defines a bijection between of: the set $\mathrm{Rep}_n(G)$ of isomorphism classes of real $n$-dimensional representations of $G$ into the set $\mathrm{Bun}_n(G)$ of isomorphism classes of real $n$-dimensional vector bundles over the classifying space $BG$. In representation theory, there are no additive inverses, so non-trivial bundles over $BG$ which come from representations cannot have inverses. So it's enough to find a non-trivial representation $V$ whose characteristic classes all vanish. -Let $G$ be a cyclic group of order $p^2$, where $p$ is an odd prime, generated by an element $\sigma$. Let $L$ be the $1$-dimensional complex representation given by $\sigma|L=e^{2\pi i/p}$, and write $V$ for the real $2$-dimensional vector bundle underlying the complex line bundle $EG\times_G L\to BG$. It appears that the Pontryagin classes vanish: up to signs, the total Pontryagin class is the total Chern class of $V\otimes \mathbb{C}\approx L\oplus \overline{L}$, and we compute $c(V)=c(L)c(\overline{L}) = (1+px)(1-px) = 1-p^2x^2 = 0$. -(The fact about the bijection $\mathrm{Rep}(G)$ into $\mathrm{Bun}(G)$ is non-trivial; it follows from a theorem of Dwyer and Zabrodsky ("Maps between classifying spaces", LNM 1298). I don't know a more elementary proof, but a condition such as "$G$ is a $p$-group" is probably necessary.) -End addition. -There are obstructions in $K$-theory to "inverting" bundles. There are exterior power operators $\lambda^k:KO(X)\to KO(X)$, such that if $[V]$ is the $K$-class of an actual bundle $V$, we have $\lambda^k([V])=[\Lambda^k V]$, the class of the exterior power bundle. The formal sum $\Lambda_t(x)\in KO(X)[[t]]$ given by -$$ -\Lambda_t(x)= 1+\lambda^1(x)t+\lambda^2(x)t^2+\dots -$$ -has a Whitney formula ($\Lambda_t(x+y)=\Lambda_t(x)\Lambda_t(y)$), coming from the usual decomposition $\Lambda^nV=\sum \Lambda^iV\otimes \Lambda^{n-i}V$. Furthermore, if $x=[V]$ is the class of an honest $n$-dimensional bundle, we must have $\lambda^i([V])=[\Lambda^iV]=0$ for $i>n$, so that $\Lambda_t(x)$ is polynomial. -In this case, trivial bundles don't have trivial $\Lambda$-class; instead, writing "$n$" for the trivial $n$-plane bundle, we have $\Lambda_t(n)=(1+t)^n$. Thus, an isomorphism $V\oplus W= n$ implies an identity $\Lambda_t([V])\Lambda_t([W])=(1+t)^n$. -So an even stronger form of your question is: if $V$ is a vector bundle (over a nice space) such that $\Lambda_t([V])/(1+t)^n \in KO(X)[[t]]$ is a polynomial, must it follow that there exists a bundle $W$ such that $V\oplus W\approx n$? Again, I don't have a counterexample here.<|endoftext|> -TITLE: Slice knots and exotic $\mathbb R^4$ -QUESTION [14 upvotes]: In the http://arxiv.org/abs/math/0606464v1 I read -"If you want to prove existence of exotic smooth structure on $\mathbb R^4$ you can do this if you are in -possession of a knot which is topologically slice but not smoothly slice (slice means zero slice genus). -Freedman has a result stating that a knot with Alexander polynomial 1 is topologically slice. We now -have an obstruction (s being non-zero) to being smoothly slice." (p.28) -What does it mean? Does anyone know the construction of exotic $\mathbb R^4$ using slice knots? Please, give me a references, if there are several constructions. -Added: http://arxiv.org/abs/math/0408379. Does there exist some other construction? - -REPLY [24 votes]: I wrote this up for my own personal edification when I was going through Gompf and Stipsicz. Please, excuse the pedantry. I had to make sure even I understood it. -Definition: (Smoothly Slice Knot) -A knot $K$ in $\partial D^4$, where $D^n$ is the standard closed $n$--dimensional disk, is smoothly slice if there exists a two-disk $D^2$ smoothly embedded in $D^4$ such that the image of $\partial D^2$ is $K$. -Definition: (Flat Topological Embedding) -Let $X$ be a topological manifold of dimension $n$ and $Y$ a topological manifold of dimension $m$ where $n < m$. A topological embedding $\rho : X \rightarrow Y$ is flat if it extends to a topological embedding $\rho : X \times D^{m - n} \rightarrow Y$. -Definition: (Topologically Slice Knot) -A knot $K$ in $\partial D^4$ is topologically slice if there exists a two-disk $D^2$ flatly topologically embedded in $D^4$ such that the image of $\partial D^2$ is $K$. -Lemma 1: -Let $K$ be a knot in $\partial D^4$ and $X_K$ the two-handlebody obtained by attaching a two-handle to $D^4$ along $K$ with framing $0$. $X_K$ has a smooth embedding into $\mathbb{R}^4$ iff $K$ is smoothly slice. -Proof: -First, let us assume there exists a smooth embedding $\rho: X_K \rightarrow \mathbb{R}^4$ and prove $K$ is smoothly slice. -As $X_K$ is compact and $\rho$ a smooth embedding, $\rho(X_K)$ is compact. Thus, the Heine-–Borel Theorem (Theorem 2.41) Rudin implies there exists an open ball $B^4 \subset \mathbb{R}^4$ such that $\rho(X_K) \subset B^4$. Therefore, $\rho$ can be used to create a smooth embedding $\rho: X_K \rightarrow S^4$, mapping $X_K$ to one hemisphere. -The zero-handle of $X_K$ is $D^4$. Thus, $\rho$ smoothly embeds this $D^4$ into $S^4$. The Disk Theorem (Corollary 4.7 Chapter III) Kosinski implies that any two orientation preserving smooth embeddings of $D^4$ into $S^4$ are isotopic. Thus, the closure of the complement of any open ball in $S^4$ is diffeomorphic to $D^4$. Hence, $S^4 - int(\rho(D^4))$ is diffeomorphic to $D^4$, where $int(X)$ denotes the interior of $X$. -By construction, $K$ is smoothly embedded in $\partial(S^4 - int(\rho(D^4)))$ and the $D^2$ core of the $X_K$ two-handle is smoothly embedded in $S^4 - int(\rho(D^4))$ such that the image of $\partial D^2$ is $K$. Thus, we conclude that if there exists a smooth embedding $\rho: X_K \rightarrow \mathbb{R}^4$, then $K$ is smoothly slice. -Now, let us assume $K$ is smoothly slice and prove there exists a smooth embedding $\rho': X_K \rightarrow \mathbb{R}^4$. -We now work backwards. As $K$ is smoothly slice, there exists a smooth embedding of $D^2$ into $D^4$ such that the image of $\partial D^2$ is $K$. As this $D^2$ is smoothly embedded, the Tubular Neighborhood Theorem (Theorem 4.2 Chapter III) Kosinski implies there exists a closed tubular neighborhood $D^2 \times D^2$ of $D^2$ smoothly embedded in $D^4$. (This closed tubular neighborhood will become the two-handle of $X_K$.) -Using the identity map glue a second $D^4$ to the above $D^4$ forming $S^4$ as $D^4 \cup_{\partial D^4} \bar{D}^4$, where $\bar{D}^4$ is $D^4$ with the opposite orientation. This new $D^4$ along with the closed tubular neighborhood above forms $X_K$. This two-handlebody $X_K$ is smoothly embedded into $S^4$. We notate this smooth embedding as $\rho'$. -We can modify the closed tubular neighborhood above, shrinking it if required, such that $S^4 - \rho'(X_K)$ is not empty. Thus, removing a point in $S^4 - \rho'(X_K)$ we obtain the desired smooth embedding $\rho': X_K \rightarrow \mathbb{R}^4$. QED. -Lemma 2: -Let $K$ be a knot in $\partial D^4$ and $X_K$ the two-handlebody obtained by attaching a two-handle to $D^4$ along $K$ with framing $0$. $X_K$ has a topological embedding into $\mathbb{R}^4$ iff $K$ is topologically slice. -Proof: -First, let us assume there exists a topological embedding $\rho: X_K \rightarrow \mathbb{R}^4$ and prove $K$ is topologically slice. -A slight variation of the logic in the previous proof implies $\rho$ can be used to create a topological embedding $\rho: X_K \rightarrow S^4$, mapping $X_K$ to one hemisphere. -Again, a variation of the logic in the previous proof implies that $D^4$ is diffeomorphic to $S^4 - int(\rho(D^4))$, where $\rho$ is acting on $D^4$ the zero-handle of $X_K$. -By construction, $K$ is topologically embedded in $\partial(S^4 - int(\rho(D^4)))$, and $D^2$, the core of the $X_K$ two-handle, is topologically embedded in $S^4 - int(\rho(D^4))$ such that the image of $\partial D^2$ is $K$. -The two-handle of $X_K$, which is homeomorphic to $D^2 \times D^2$, is topologically embedded into $S^4 - int(\rho(D^4))$ such that it extends the topological embedding of $D^2$, the core of the $X_K$ two-handle. Thus, we conclude that if there exists a topological embedding $\rho: X_K \rightarrow \mathbb{R}^4$, then $K$ is topologically slice. -Now, let us assume $K$ is topologically slice and prove there exists a topological embedding $\rho': X_K \rightarrow \mathbb{R}^4$. -Again, we work backwards. As $K$ is topologically slice, there exists a topological embedding of $D^2$ into $D^4$ such that the image of $\partial D^2$ is $K$ and there exists a topological embedding of $D^2 \times D^2$ into $D^4$ that extends the topological embedding of $D^2$. (This topological embedding of $D^2 \times D^2$ will become the two-handle of $X_K$.) -Using the identity map glue a second $D^4$ to the above $D^4$ forming $S^4$ as $D^4 \cup_{\partial D^4} \bar{D}^4$. This new $D^4$ along with the above topological embedding of $D^2 \times D^2$ forms $X_K$. This two-handlebody $X_K$ is topologically embedded into $S^4$. We notate this topological embedding as $\rho'$. -Using the same logic as in the previous proof, we can remove a point from $S^4 - \rho'(X_K)$ to construct the desired topological embedding $\rho': X_K \rightarrow \mathbb{R}^4$. QED. -Definition: (Large Exotic $\mathbb{R}^4$) -A large exotic $\mathbb{R}^4$ is an exotic $\mathbb{R}^4$ containing a four-dimensional compact smooth submanifold that can not be smoothly embedded into $\mathbb{R}^4$. -Thorem: -If there exists a knot $K$ that is topologically slice but not smoothly slice, then there exists a large exotic $\mathbb{R}^4$. -Proof: -As $K$ is topologically slice, Lemma 2 implies there exists $\rho$ a topological embedding of $X_K$ into $\mathbb{R}^4$, where $X_K$ is the two-handlebody obtained by attaching a two-handle to $D^4$ along $K$ with framing $0$. -First let us prove $\mathbb{R}^4 - int(\rho(X_K))$ is a topological manifold. As $\rho$ is a topological embedding, $\rho(X_K)$ is a topological submanifold of $\mathbb{R}^4$. Hence, $int(\rho(X_K))$ is also a topological submanifold of $\mathbb{R}^4$. This implies $\mathbb{R}^4 - int(\rho(X_K))$ is a topological submanifold of $\mathbb{R}^4$. Thus, $\mathbb{R}^4 - int(\rho(X_K))$ is a topological manifold in its own right. -Next let us prove $\mathbb{R}^4 - int(\rho(X_K))$ is not compact. $X_K$, by construction, is compact. Thus, as $\rho$ is a topological embedding, $\rho(X_K)$ is a compact topological submanifold of $\mathbb{R}^4$. As $\mathbb{R}^4$ is not compact, it follows that $\mathbb{R}^4 - \rho(X_K)$ is not compact. Hence, as $\mathbb{R}^4 - \rho(X_K)$ has two ends, this implies $\mathbb{R}^4 - int(\rho(X_K))$ is not compact. -Now we will prove $\mathbb{R}^4 - int(\rho(X_K))$ is connected. In the standard handle presentation, $S^4$ has no one-handles. Thus, as a result of Theorem 3.4 Chapter VII Kosinski, we have $H_1(S^4;\mathbb{Z}) = 0$. As in Lemma 2, we can use $\rho$ to create a topological embedding $\rho: X_K \rightarrow S^4$. Hence, $\rho(X_K)$ is a closed, proper subset of the compact connected orientable manifold $S^4$. Thus, Alexander Duality (Theorem 6.6(a) Chapter XIV) Massey implies that the Cech cohomology group $\check{H}^3(\rho(X_K);\mathbb{Z})$ is isomorphic to reduced homology group $\tilde{H}_0(S^4 - \rho(X_K);\mathbb{Z})$. As $X_K$ is compact, $\rho(X_K)$ is compact. Thus, $\rho(X_K)$ is paracompact. Hence, (Page 372) Massey $\check{H}^3(\rho(X_K);\mathbb{Z})$ is isomorphic to $H^3(\rho(X_K);\mathbb{Z})$. As $\rho$ is a topological embedding, $H^3(\rho(X_K);\mathbb{Z})$ is isomorphic to $H^3(X_K;\mathbb{Z})$. The ``lower'' boundary $\partial_- X_K$ of $X_K$ is empty. Thus, $H^3(X_K;\mathbb{Z})$ is isomorphic to $H^3(X_K,\partial_- X_K;\mathbb{Z})$. Poincare duality (Theorem 5.1 Chapter VII) Kosinski then implies $H^3(X_K,\partial_- X_K;\mathbb{Z})$ is isomorphic to $H_1(X_K,\partial_+ X_K;\mathbb{Z})$. However, as $X_K$ has no three-handles, Theorem 3.4 Chapter VII Kosinski implies $H_1(X_K,\partial_+ X_K;\mathbb{Z}) = 0$. Thus, tracing isomorphisms, we have proven $\tilde{H}_0(S^4 - \rho(X_K);\mathbb{Z}) = 0$; in other words $S^4 - \rho(X_K)$ is connected. Removing a point from a four-manifold will not disconnect it; thus, $\mathbb{R}^4 - \rho(X_K)$ is connected. (Equivalently, one could prove this by applying Alexander Duality again.) Finally, as adding in boundary points also does not disconnect, $\mathbb{R}^4 - int(\rho(X_K))$ is also connected. -So, as $\mathbb{R}^4 - int(\rho(X_K))$ is a non-compact connected topological four--manifold, Theorem 9.4.22 Gompf and Stipsicz implies $\mathbb{R}^4 - int(\rho(X_K))$ admits a smooth structure. -As $\rho$ is a topological embedding, the restriction $\rho: \partial X_K \rightarrow \partial (\mathbb{R}^4 - int(\rho(X_K)))$ is a homeomorphism. Thus, $\partial X_K$ and $\partial (\mathbb{R}^4 - int(\rho(X_K)))$ are homeomorphic. -As $X_K$ is a smooth manifold, it induces a smooth structure on $\partial X_K$. Similarly, as $\mathbb{R}^4 - int(\rho(X_K))$ is a smooth manifold, it induces a smooth structure on $\partial (\mathbb{R}^4 - int(\rho(X_K)))$. A topological three-manifold admits a unique smooth structure Moise. Thus, the smooth structure on $\partial X_K$ is the same as that on $\partial (\mathbb{R}^4 - int(\rho(X_K)))$. Hence, $\partial X_K$ and $\partial (\mathbb{R}^4 - int(\rho(X_K)))$ are diffeomorphic. -The diffeomorphism of $\partial X_K$ and $\partial (\mathbb{R}^4 - int(\rho(X_K)))$ can be used to join $X_K$ and $\mathbb{R}^4 - int(\rho(X_K))$ along their respective boundaries forming a smooth manifold $R$. Obviously, $R$ is homeomorphic to $\mathbb{R}^4$. -We will now prove, by contradiction, that $R$ is exotic. Let us assume $R$ is diffeomorphic to $\mathbb{R}^4$. Thus, there exists a diffeomorphism $\varphi: R \rightarrow \mathbb{R}^4$. The restriction of this diffeomorphism to $X_K$, which is smoothly embedded in $R$, is a smooth embedding of $X_K$ into $\mathbb{R}^4$. However, Lemma 1 implies such a smooth embedding exists if and only if $K$ is smoothly slice. But, by hypothesis, $K$ is not smoothly slice. Thus, our assumption that there exists a diffeomorphism $\varphi: R \rightarrow \mathbb{R}^4$ leads to a contradiction. Therefore, no such diffeomorphism exists. Hence, $R$ is exotic, homeomorphic but not diffeomorphic to $\mathbb{R}^4$. -Finally, we prove $R$ is large. $X_K$, by construction, is compact. Also, by construction, $X_K$ is a smooth submanifold of $R$, an exotic $\mathbb{R}^4$. By hypothesis, $K$ is not smoothly slice. Thus, Lemma 1 implies $X_K$ can not smoothly embed in $\mathbb{R}^4$. Hence, we conclude, $R$ is large. QED.<|endoftext|> -TITLE: Can all n-manifolds be obtained by gluing finitely many blocks? -QUESTION [26 upvotes]: Fix a dimension $n\geqslant 2$. Let $S= \{M_1,\ldots, M_k\}$ be a finite set of smooth compact $n$-manifold with boundary. Let us say that a smooth closed $n$-manifold is generated by $S$ if it may be obtained by gluing some copies of elements in $S$ via some arbitrary diffeomorphisms of their boundaries. -For instance: - -Every closed orientable surface is generated by a set of two objects: a disc and a pair-of-pants $P$, -Waldhausen's graph manifolds are the 3-manifolds generated by $D^2\times S^1$ and $P\times S^1$, -The 3-manifolds having Heegaard genus $g$ are those generated by the handlebody of genus $g$ alone, -The exotic $n$-spheres with $n\geqslant 5$ are the manifolds generated by $D^n$ alone. - -A natural question is the following: - -Fix $n\geqslant 3$. Is there a finite set of compact smooth $n$-manifolds which generate all closed smooth $n$-manifolds? - -I expect the answer to be ''no'', although I don't see an immediate proof. In particular, I expect some negative answers to both of these questions: - -Is there a finite set of compact 3-manifolds which generate all hyperbolic 3-manifolds? - -and - -Is there a finite set of compact 4-manifolds which generate all simply connected 4-manifolds? - -REPLY [12 votes]: I posted a paper on the arXiv, Group Width which answers this question for manifolds of dim $>3$ with sufficiently complicated fundamental group (there will be no finite set of blocks). As Greg Kuperberg said, there are many interesting variations which remain open and are a nice challenge to technique, e.g. the case of simply connected manifolds.<|endoftext|> -TITLE: Exact computation of the null-space basis of an integer matrix -QUESTION [8 upvotes]: Hi all, -Let $\mathbf{A} \in \mathbb{Z}^{M \times N}$. Suppose that $\mathbf{A} \cdot \vec{x} = \vec{0}$, where $\vec{x} \in \mathbb{N}^{N \times 1}$. Does anyone know about a C/C++/Java program that can compute the null-space basis vector(s) exactly? - -REPLY [15 votes]: The question seems to be about programs that can compute the nullspace of an integer matrix exactly. (Though there seems to be the constraint that all the entries in the solution vector are positive, which I'm going to ignore.) The best software you might want to use could depend on the size of your matrices, which is a function of: (1) the parameters $M$, $N$, (2) the number of bits of the entries, and (3) the sparseness of the matrix. - -IML (Integer Matrix Library) -- this is a free C library, but what it does is compute the rational kernel of a system defined over the integers, so it isn't directly useful for your problem. Anyway, in my benchmarks, this is the fastest library available for computing the rational kernel if your matrix is large (both $M$ and $N$), dense, and the entries are also large. When entry sizes get to several hundred bits or larger, IML seemed to me in 2007 to be by far the fastest library in existence (easily beating Mathematica, Magma, NTL, etc.). IML is used part of Sage and used for the kernel function. Getting going with IML standalone can be hard (though I contributed code a few years ago to make it easier), but if you just want a library to link into an existing C program, you'll definitely want to look into IML. -PARI -- this free calculator (which is also usable as a C library) is extremely good for computing kernels of small matrices, e.g., when the entries aren't too big and the number of rows and columns aren't too large either (say, less than 100). It's optimized for this small case, since that's what comes up in class group computations, among other things. The command to use is matkerint, e.g., "matkerint([1,2,3;4,5,6;7,8,9])". -SAGE -- this free software uses some combination of IML, PARI, Linbox, NTL and a bunch of new code to compute kernels of integers matrices. In some cases (e.g., nearly square matrices) it is very good, and it could be good in far more cases but I got lazy and didn't implement fast algorithms for all shapes of matrices. Type "A = matrix(ZZ,3,3,[1..9])" to make an example integer matrix; then "A.kernel()" will find the kernel over ZZ (by default, contrary to what another answer on this page claims). If you type "A._k[press tab key]" you'll see that you can directly call two other routines for computing kernels, one involving PARI, and the other a $p$-adic algorithm. Behind the scenes, really the Hermite normal form of the matrix is being computed using the algorithm described here. The HNF algorithm in Sage makes fundamental use of IML in several ways behind the scenes. -Magma -- non-free and closed source, but has a fast tuned Hermite normal form algorithm presumably similar to the one in Sage, which for certain (not all) ranges of inputs is the best available. -Mathematica -- in my benchmarks (from 2007), as the matrices get very big this gets totally useless compared to Magma or Sage, but it is fine for small matrices. Similar remarks apply about Maple, at least as of 2007. - -Here's a potentially helpful example in Sage, which illustrates some of the computations discussed here and elsewhere, and explains how to work with some matrices and modules in Sage: -sage: A = random_matrix(ZZ,5,2); B = random_matrix(ZZ,5,2) -sage: A.kernel() -Free module of degree 5 and rank 3 over Integer Ring -Echelon basis matrix: -[ 1 60 0 1 30] -[ 0 66 0 1 33] -[ 0 0 1 0 0] -sage: A.kernel().intersection(B.kernel()) -Free module of degree 5 and rank 1 over Integer Ring -Echelon basis matrix: -[ 19 -378 -93 -4 -189] -sage: VA = A.change_ring(QQ).kernel(); VB = B.change_ring(QQ).kernel() -sage: W = VA.intersection(VB); W -Vector space of degree 5 and dimension 1 over Rational Field -Basis matrix: -[ 1 -378/19 -93/19 -4/19 -189/19] -sage: W.intersection(ZZ^5) -Free module of degree 5 and rank 1 over Integer Ring -Echelon basis matrix: -[ 19 -378 -93 -4 -189]<|endoftext|> -TITLE: Square roots of elements in a finite group and representation theory -QUESTION [44 upvotes]: Let $G$ be a finite group. In an an earlier question, Fedor asked whether the square root counting function $r_2:G\rightarrow \mathbb{N}$, which assigns to $g\in G$ the number of elements of $G$ that square to $g$, attains its maximum at the identity element, when $G=S_n$. I gave an affirmative answer using representation theory, which is valid when $S_n$ is replaced by any group that has no symplectic representations. In other words, the argument I gave works whenever any irreducible complex representation of $G$ is either realisable over $\mathbb{R}$ or has non-real valued character. Again in other words (and closest to the actual argument) I need the Frobenius-Schur indicators of all irreducible complex representations to be non-negative. -This is the first of two follow-up questions (the second one being about $n$-th roots), which Pete Clark encouraged me to ask here. I ran a quick computer experiment. There are 1911 groups among the groups of size up to 150 that have a symplectic representation. In 1675 of them, the square root counting function does not attain its maximum at the identity element. Is there a nice (representation theoretic?) criterion that singles out the 300-odd remaining groups? A criterion that includes the previous one as a special case would of course be particularly interesting. in other words, I am asking: - -what does it tell you about the group - (about its representation theory) if - the square root counting function - attains its maximum at the identity? - -Any brain storming ideas or heuristics are welcome. Criteria that catch some of the remaining groups, even if not all, are also of interest. Thank you in advance! - -REPLY [16 votes]: As pointed out in comments and at this question, this answer is not entirely correct because real/quaternionic does not always give a $\mathbb{Z}/2$ grading. Nonetheless in practice this answer seems to be "mostly right." I'm going to leave it up unmodified because changing it would confuse the comments and other answers too much. Sorry for the mistake! - -If $G$ has a quaternionic representation and $G$ has no complex representations (i.e. reps whose character is complex) then for certain nontrivial central elements the square root function is $\sum_\chi \chi(1)$ which is clearly the maximum possible value and clearly larger than the value on the identity. -The key idea in the proof is the fact that $Z(G)^*$ (the group of characters of the center of $G$) is the universal grading group for the category of $G$ representations. -Let me unpack that a little bit. To every irrep of $G$ you can assign it's "central character", that is an element of $Z(G)^*$. This assignment is multiplicative in the sense if $U$ is a subrep of $V \otimes W$ then the central character of $U$ is the product of the central characters of $V$ and $W$ (this is just because central elements act by scalars). In other words, $\operatorname{Rep}(G)$ is graded by $Z(G)^*$. The claim is that an $H$-grading of $\operatorname{Rep}(G)$ is the same thing as a map $Z(G)^* \to H$. -Why is this true? I'm just going to sketch the proof, in particular I'm just going to talk about the case where $Z(G)$ is trivial, but I'll indicate how the general case works. Since the center is trivial you can find a faithful representation $V$. But then let's look at a really high tensor power of $V$ and ask how it breaks up. We can compute that using character theory. Since the rep is faithful and there's no center, the character of a high tensor power of $V$ is dominated by the value on $1$. Hence any high tensor power of $V$ contains all other irreps. In particular, the $n$-th and $(n+1)$-th powers contain the same irreps and this tells you that the grading group is trivial. In general you want to argue that the contribution of the center dominates the values of inner products (but you need to be a bit careful about non-faithful representations). -The "universal grading group" is used by Gelaki and Nikshych to define the upper central series of an arbitrary fusion category and thereby define nilpotent fusion categories. -Ok, how is this relevant to anything? Well, if you have a group with a quaternionic representation but no complex representations, then the Frobenius-Schur indicator gives a $\pm 1$ grading of $Rep(G)$, namely put the real reps in grade $1$, and the quaternionic ones in grade $-1$. Hence the Frobenius-Schur indicator $s(\chi)$ must be given by the central character $\chi(z)/\chi(1)$ for some central element $z$. Clearly these central elements maximize the square root function.<|endoftext|> -TITLE: Rings in which every non-unit is a zero divisor -QUESTION [44 upvotes]: Is there a special name for the class of (commutative) rings in which every non-unit is a zero divisor? The main example is $\mathbf{Z}/(n)$. Are there other natural or interesting examples? - -REPLY [6 votes]: How about calling these things balanced commutative rings? -Recall that a category satisfying "monic + epic --> isomorphism" is said to be balanced. Therefore: - -a monoid is balanced iff every element that is both left-cancellative and right-cancellative is a unit. -a commutative monoid is balanced iff every element that is cancellative is a unit. - -It makes sense to apply this terminology to rings, too: - -a ring is balanced iff every element that is neither a left zero-divisor nor a right-zero divisor is a unit. -a commutative ring is balanced iff every non-zero divisor is a unit.<|endoftext|> -TITLE: Iwasawa Decomposition & Polar Decomposition related how ? -QUESTION [9 upvotes]: In an earlier post (Use Lie Sub-Groups of GL(3, R) for elastic deformation ? here), I mentioned polar decompositions as in F = RU where R in SO(3) & U in symmetric positive-semidefinite matrices. In response, I received the following comment: "The decompositions you mention are well-known in the theory of Lie groups, e.g., F=RU is called Iwasawa decomposition." -I have been trying to understand how a polar decomposition & Iwasawa decomposition are related. I have found the following definitions of both. -Iwasawa: G=KAN for G in GL(n, R) where K = orthogonal matrices (rot'ns ?), A = positive diag. matrices & N = upper triag. matrices with diag. entries = 1. -Polar: G=KAK where G = semi-simple Lie group, K = maximal compact subgroup of G, A = abelian subgroup of G. For GL(n, R), I think K in SO(n) but unsure about A. The only abelian subgroup of GL(n, R) appears to be the set of all nonzero scalar matrices (i.e. scalar multiples of identity matrix). I'm also puzzled by G=KAK having 3 terms but F=RU only 2. -Given all of this, I'm still unsure how these two decompositions are related. - -REPLY [13 votes]: You can obtain the $G=KAK$ decomposition from a decomposition of the type $F=UR$. To avoid unnecessary complications, let's assume that our reductive group $G$ is a selfadjoint subgroup of $\operatorname{GL}(n,\mathbb{R})$. Then the map $g \mapsto g^{-t}$ is an involution of $G$, which is called the Cartan involution and is typically denoted by $\theta$. The first observation to make is that the fixed-point set $K = \{ g \in G \colon \theta(g)=g \}$ of $\theta$ is a maximal compact subgroup of $G$. For example, if $G=\operatorname{GL}(n,\mathbb{R})$, then $K=\operatorname{O}(n)$. -Next we observe that $\theta$ induces an involution (also denoted by $\theta$) at the Lie algebra level: explicitly, this is the map $X \mapsto -X^t$. If $\mathfrak{p}$ denotes the $-1$-eigenspace of this latter involution, then one has the following result. - -The map $K \times \mathfrak{p} \to G$ given by $(k, X) \mapsto k e^X$ is a diffeomorphism. - -In particular, every $g \in G$ can be expressed as $k e^X$ for some $k \in K$ and $X \in \mathfrak{p}$. This decomposition is known as the Cartan decomposition; it is a generalization of the polar decomposition to $G$ (and is, I presume, the $F=UR$ decomposition stated in the OP). Indeed, if $G = \operatorname{GL}(n,\mathbb{R})$, then $\mathfrak{p}$ is just the set of symmetric matrices, and thus the set $\exp \mathfrak{p}$ consists of symmetric, positive semidefinite matrices. -Now let $\mathfrak{a}$ denote a maximal abelian subspace of $\mathfrak{p}$. Then it can be shown that $A = \exp \mathfrak{a}$ is a closed abelian subgroup of $G$ with Lie algebra $\mathfrak{a}$. It can also be shown that $\mathfrak{a}$ is unique up to conjugacy via an element of $K$. That is to say, if $\mathfrak{a}'$ is another maximal abelian subspace of $\mathfrak{p}$, then there is a $k \in K$ such that $\text{Ad}(k) \mathfrak{a} = \mathfrak{a}'$. With this information we can obtain the decomposition $G=KAK$: given $g \in G$, one observes that $p=gg^t \in \exp \mathfrak{p}$, say $p=e^X$. Thus there is a $k \in K$ such that $\text{Ad}(k)X \in \mathfrak{a}$, and then $e^{-\text{Ad}(k)X/2}kg \in K$ (because it is fixed by $\theta$), whence $g \in KAK$. -This hopefully alleviates your 3-terms-vs-2-terms issue. -I'm not aware of any relationship between the Iwasawa decomposition and the $KAK$ (polar) decomposition.<|endoftext|> -TITLE: Number of n-th roots of elements in a finite group and higher Frobenius-Schur indicators -QUESTION [21 upvotes]: This is the second follow-up to this question on square roots of elements in symmetric groups and is concerned with generalisations to $n$-th roots. Let $G$ be a finite group and let $r_n(g)$ be the number of elements $h\in G$ such that $h^n = g$. In other words, -$$r_n(g) = \sum_{h\in G}\delta_{h^n,g},$$ -where $\delta$ is the usual Kronecker delta. In a comment to my answer to the above mentioned question, Richard Stanley notes that if $G=S_m$, then $r_n(g)$ attains its maximum at the identity element of $G$. My question is: how far does this generalise and what exactly does it tell us about $G$? This should be primarily a question about higher Frobenius-Schur indicators. Let me elaborate a bit. -The function $r_n$ is clearly a class function on $G$ and, upon taking its inner product with all irreducible characters of $G$, one finds that -$$r_n(g) = \sum_\chi s_n(\chi)\chi(g),$$ -where the sum runs over all irreducible complex characters of $G$ and $s_n(\chi)$ is the $n$-th Frobenius-Schur indicator of $\chi$, defined as -$$s_n(\chi) = \frac{1}{|G|}\sum_{h\in G}\chi(h^n).$$ -When $n=2$, the Frobenius-Schur indicator is equal to 0,1 or -1 and carries explicit information about the field of definition of the representation associated with $\chi$. - -What do higher Frobenius-Schur - indicators tell us about the - representations and, by extension, - about the group? What do we know about - their values? Have higher Frobenius-Schur indicators been studied in any detail? - -For additional focus: - -Given $n\in \mathbb{N}$, for what groups $G$ do we have $\max_g \; r_n(g) = r_n(1)$? For what groups does this hold for all $n$? - -As noted by Richard Stanley, the latter is true for all symmetric groups. It is also easy to see that the set of groups with this property is closed under direct products, and that all finite abelian groups possess this property. - -REPLY [5 votes]: The concept of higher Frobenius-Schur indicators has been generalized to far broader contexts than just group representations, and these have proven extremely useful. -Linchenko and Montgomery generalized the second indicators to semisimple Hopf algebras, and Kashina, Sommerhauser, and Zhu generalized higher indicators to the same. They were then defined for spherical fusion categories by Ng and Schauenburg, and have been used in problems ranging from classifications of low dimensional (quasi-)Hopf algebras; relating exponents and dimensions; obtaining new gauge invariants; and much more— the intro for this paper by Negron and Ng gives a pretty solid rundown on these generalizations and their uses. Some of the generalizations do not require semisimplicity, which for the most part means the invariants are only of the regular representation (rather than arbitrary fin. dim. reps). Searching for "frobenius-schur indicators" on the math arxiv should pull up quite a number of results. -When applied to the quasi-triangular Hopf algebras D(G), the Drinfeld double of the group G, you get a number of new group invariants which are, loosely speaking, based on how many roots (of a given element) are sent to roots (of the same element) under the action of each element under the regular representation. It then becomes quite interesting to investigate how this depends on the given element. Sometimes it only depends on the cyclic subgroup it generates (such as for all regular p-groups), sometimes it doesn't (such as for some irregular p-groups and the Monster group). Ultimately these can be tied into how the group is pieced together from its one-element centralizers, which if you know much about the representation theory of D(G) is quite sensible for an invariant of Rep(D(G)): representations are given by inducing representations from centralizers to the whole group. -Quite recently, Barter, Jones, and Tucker showed how these indicators effectively govern the structure of certain annular Temperley-Lieb-Jones modules in the fusion categories. Such objects are specified by rotation eigenvalues (which all have a predictable modulus, provided you know the twists), and the indicators govern the coefficients of the polynomials that dictate the multiplicities of these eigenvalues.<|endoftext|> -TITLE: Multiplication by $n$ on commutative algebraic groups -QUESTION [6 upvotes]: Let $K$ be a field of characteristic zero. Let $G/K$ be a group scheme of finite type. -Assume that $G$ is commutative and connected. For a natural number $n$ denote by $n_G: G\to G$ the multiplication by $n$ morphism. Is it true that $n_G$ is surjective with finite kernel? -(I know that the answer is yes provided $G$ is an abelian variety. But the proof of this fact makes use of a very ample invertible sheaf on $G$, so it does not carry over to the general case directly.) - -REPLY [10 votes]: Yes, this is true. The derivative of $n_G$ at the identity element $e$ is multiplication by $n$ on the Lie algebra (tangent space at $e$) which gives that the kernel is finite and the image of $n_G$ has the same dimension as $G$ which implies that it is equal to $G$ as $G$ is connected. -Addendum: Jim may be right: The statements may be checked over an algebraic closure of $K$ so we may assume $K$ algebraically closed. As $dn_G$ is an isomorphism the kernel is finite and the dimension of the image is equal to that of $G$. As $G$ is of finite type the image is constructible by Chevalley's theorem and hence contains an open subset of $G$. As it is also a subgroup it is open and as $G$ is connected (and the cosets are also open) we get that the image equals $G$.<|endoftext|> -TITLE: Alternative for Kadison and Ringrose's book -QUESTION [7 upvotes]: I have read over the book by R. V. Kadison and J. R. Ringrose: Fundamentals of the theory of operator algebras. Vol 1, and have done most of the exercises in it. Now I want to find an alternative book for Vol 2, because I once heard that the content in this book is somewhat out of date and the theories are developed in a rather slow pace. Which book can I choose then? - -REPLY [4 votes]: I'm surprised no one mentioned C${}^*$-algebras and their Automorphism Groups by Pedersen. The later chapters get a little specialized, but no one does basic C*-algebra more elegantly than this book.<|endoftext|> -TITLE: Two reference requests: Pinsker's inequality and Pontryagin duality -QUESTION [12 upvotes]: Sorry for such a newbie post and for asking two unrelated references in one shot. -First, I am interested in any proof of Pinsker's inequality. -Second, I wonder what is the best place to read about Pontryagin duality and harmonic analysis. To clarify, I took only standard functional analysis course. - -REPLY [3 votes]: My favorite proof so far is the one from Theorem 4.5 of Yihong Wu's lecture notes, which uses the data processing inequality to reduce the problem to the binary case: if $S$ is a measurable subset of the domain, then we can consider the two random variable $\mathbf{1}_S$ under $P$ and $Q$, respectively, which has then distribution either $\mathrm{Bern}(P(S))$ or $\mathrm{Bern}(Q(S))$. First -$$ -\operatorname{TV}(\mathrm{Bern}(P(S)), \mathrm{Bern}(Q(S))) = |P(S)-Q(S)| \tag{1} -$$ -so we get the TV between $P$ and $Q$ by taking the supremum over all $S$. By the DPI, however, -$$ -\operatorname{KL}(\mathrm{Bern}(P(S))\ \|\ \mathrm{Bern}(Q(S))) -\leq \operatorname{KL}(P\ \|\ Q) \tag{2} -$$ -This shows that it suffices to prove the binary case of Pinsker's, which is just a matter of proving a simple inequality: -\begin{equation} - 2(p-q)^2 \leq p\log\frac{p}{q} + (1-p)\log\frac{1-p}{1-q} \tag{3} -\end{equation} -The cases where either $p$ or $q$ is in $\{0,1\}$ are easily checked , so we can assume $p,q\in(0,1)$. -To prove (3) in this case, we introduce the function $f\colon(0,1)\to\mathbb{R}$ defined by $f(x) = p\log x + (1-p)\log(1-x)$, and observe that the RHS of (3) is exactly $f(p)-f(q)$. We then can write -$$ - f(p)-f(q) = \int_{q}^p f'(x)dx = \int_{q}^p \frac{p-x}{x(1-x)}dx \geq 4\int_{q}^p (p-x)dx = 4\cdot\frac{1}{2}(p-q)^2 -$$ -which is (3) (in the middle, we used the fact that $x(1-x) \leq 1/4$ for $x\in(0,1)$).<|endoftext|> -TITLE: Tomita-Takesaki theory for a simple class of crossed products -QUESTION [8 upvotes]: This question is inspired by the construction of the time evolution for endomotives as given by Connes and Marcolli in their book http://www.alainconnes.org/docs/bookwebfinal.pdf. -Let $M$ be a monoid (countable and discrete) acting on a locally compact Hausdorff space $X$ and consider the $C^*$-algebra $A$ given by the (semigroup) crossed product $$A = C_0(X) \rtimes M$$ Now let us assume given a state $\mu : A \to \mathbb C$ (for example one could try to use the one induced by the integration map $f \in C_0(X) \mapsto \int_X f $ with respect to the Haar measure on $X$, as in the above reference) such that the associated GNS constrution yields -a faithful representation $\pi : A \to \mathcal B (H)$, further let us assume that the map $\mathbb R \to Aut(\pi(A)'')$ induced by Tomita-Takesaki theory on the von Neumann algebra given by the bicommutant $\pi(A)''$ restricts to a map $$\sigma : \mathbb R \to Aut(A)$$ -My question is now if there are known examples where $\sigma$ is described explicitly in the literature. (Of course my general formulation allows trivial examples which are not asked for...) -In some sense $\sigma$ should depend only on $M$ and its action on $X$ because this is the only source for noncommutativity. Recall that Tomita-Takesaki theory is only visible in the noncommutative case, in the commutative case the $\mathbb R$-action is trivial. -The only cases I know are essentially all given by so called "Bost-Connes type systems", as explained for example in the nice article http://arxiv.org/pdf/0710.3452v2 by Laca, Larsen and Neshveyev. (The best known example is given by the original Bost-Connes system $C(\hat {\mathbb Z}) \rtimes \mathbb N$.) - -REPLY [4 votes]: I'd like to take this opportunity to dispel this wrong beleif: - -"Recall that Tomita-Takesaki theory is only visible in the noncommutative case, in the commutative case the $\mathbb R$-action is trivial" - -As mentioned in this question, -given a C*-algebra and/or von Neumann algebra $A$, the modular flow -is a homomorphism $i\mathbb R\to \text{BIM}^\times(A)$, where the latter refers to the -2-group of invertible $A$-$A$-bimodules. -Now let's take $A$ to be commutative, and let $X$ be the (locally compact/measurable) space on which $A$ is the algebra of functions. -The bimodule correcponding to $it\in i\mathbb R$ is the module (since $A$ is abelian, the left and right actions coincide) of $it$-densities on $X$. For any $\kappa\in \mathbb C$, a $\kappa$-density is a section of a particular line bundle that exists on $X$. -To simplify by exposition, I'll cheat and assume that $X$ is an oriented manifold: -then it's easier to say what a density is: it's a section of $\Lambda^{top}(TX)$. -Let $\mathcal L$ be the total space of the line bundle of densities, and let $\mathcal L_+\subset \mathcal L$ be its positive part. Then $\mathcal L_+$ is a principal $\mathbb (R_+,\cdot)$-bundle -on $X$. -The bundle of $\kappa$-densities is the associated bundle for the representation -$\lambda\mapsto \lambda^\kappa$ of $\mathbb R_+$ on $\mathbb C$, -and a $\kappa$-density is a section of that bundle. - -Now let's go back to the actual question. -If $G$ is a group that acts on $X$ (or monoid, assuming it acts by covering maps), then it induces an action on the bundle of $\kappa$-densities. -The $\big((\text{functions on } X) \rtimes G\big)$-bimodules that describe the modular flow are then given by -$$ -\big((it)\text{-densities on } X\big) \rtimes G. -$$ -[Added later] -Let me add a few words in order to connect the above story to the one that you are familiar with: -First of all, $(\text{densities on } X) \rtimes G$ can be identified with $L^1$ of the algebra, i.e. the predual in the von Neumann algebra setting [I'm ignoring all issues relted to completions]. Indeed, an element -$$\bigoplus_{g\in G}\mu_g\in(\text{densities on } X) \rtimes G$$ -can be identified with the functional that sends an algebra element -$$\bigoplus_{g\in G}f_g\in(\text{functions on } X) \rtimes G$$ -to the number $\sum_{g\in G}\displaystyle\int_X f_g(x) d\mu_{g^{-1}}(x)$. -Given a state $\phi = \bigoplus_{g\in G}\phi_g$, the modular flow is given by -$\sigma_t^\phi(f) = \phi^{it} f \phi^{-it}$ (see p. 1083 of Yamagami's paper Algebraic aspects in modular theory). -For an arbitrary such state $\phi$, the expression $\phi^{it}$ can be quite difficult to compute. -But if we take $\phi$ such that $\phi_g=0$ for all non-trivial elements of $G$, then things become suddenly much simpler: "computing" $\phi^{it}$ is now almost tautological. -We can now go back to the question of computing $\sigma_t^\phi(f)$ and give a complete answer: - -The $g$-component of $\sigma_t^\phi(f)$ is given by $\left(\frac{d(g_\*\phi)}{d\phi}\right)^{it} f_g$, - -where $f_g$ is the $g$-component of $f$, where $g_\*\phi$ is the density obtained by letting $g\in G$ act on $\phi$, and where $\frac{d(g_\*\phi)}{d\phi}$ is the Radon-Nikodym derivative.<|endoftext|> -TITLE: pro-discrete = locally compact and open normal subgroups have trivial intersection? -QUESTION [10 upvotes]: EDIT: After talking to some experts on the subject, I have concluded that a) the answer is not obvious or well-known for locally compact groups in general, b) the answer should be 'no' and I have some idea how to construct examples, but would rather try to write them up properly somewhere. Perhaps this question should be closed? Thanks for the help anyway. -This is a fairly basic question, but I can't seem to find a clear answer. -Let $G$ be a locally compact group. Suppose that the open normal subgroups of $G$ have trivial intersection. Does it follow that every open subgroup of $G$ contains an open normal subgroup of $G$? -If so, can the locally compact condition here be weakened? -Edit: some steps towards an answer: - -The open subgroups of $G$ have trivial intersection, so $G$ is totally disconnected. -Any compact group satisfying the conditions is profinite and in particular pro-discrete. (Profinite = compact totally disconnected.) -A locally compact totally disconnected group has an open compact (indeed profinite) subgroup by van Dantzig's theorem; this compact subgroup is either finite (in which case $G$ is discrete) or uncountable. So any non-discrete example would need to be uncountable. -To show every open subgroup of $G$ contains an open normal subgroup, I think it would suffice to show there is an open compact normal subgroup $K$ say. For then, given $H$ open, then $H$ contains a finite index subgroup of $K$, and so by intersecting $K$ with finitely many suitably chosen open normal subgroups we can obtain an open normal subgroup contained in $H$. - -REPLY [2 votes]: Let $K$ be an infinite profinite group and let $K_n < K$ be a decreasing family of open subgroups with trivial intersection. Thus, $K$ acts continuously on the discrete space $X = \sqcup_n K/K_n$ and this in turn gives rise to a continuous action of $K$ on the free abelian group $\mathbb Z[X]$. Define $G$ to be the semidirect product $G = K \ltimes \mathbb Z[X]$. -$G$ is a locally compact group and if we denote $X_n = \sqcup_{m \geq n} K/K_m \subset X$ then since $K_n$ acts trivially on $X \setminus X_n$ we have that $K_n \ltimes \mathbb Z[X_n]$ is a family of open normal subgroups with trivial intersection in $G$. Also, $K < G$ is an open subgroup but contains no nontrivial normal subgroups since the action of $K$ on $X$ is faithful.<|endoftext|> -TITLE: post correspondence problem variant -QUESTION [7 upvotes]: Is there an algorithm which takes as input two lists of words $v_1,...,v_n$ and $w_1,...,w_n$ over an alphabet $X$ and decides if there is an infinite sequence $(k_i)$ where $1 \leq k_i \leq n$ for all $i$ such that $v_{k_1}v_{k_2}...=w_{k_1}w_{k_2}...$? It seems that undecidability of the original Post Correspondence problem should imply there is no such algorithm. Is there a reference that shows undecidability of this variation of Post? Thanks. - -REPLY [6 votes]: See Halava, Vesa, Harju, Tero, Karhumäki, Juhani -Decidability of the binary infinite Post correspondence problem. If the alphabet consists of $\le 2$ letters, then the problem is decidable, if the number of letters is at least 7, then the problem is undecidable. The latter result is proved in Y. Matiyasevich, G. Sénizergues, Decision problems for semi-Thue systems with a few rules, in: Proceedings, 11th Annual IEEE Symposium on Logic in Computer Science, New Brunswick, NJ, 27–30 July 1996, IEEE Computer Society, Silver Spring, MD, pp. 523–531.<|endoftext|> -TITLE: Calabi - Yau Manifolds -QUESTION [13 upvotes]: I just started reading about Calabi-Yau manifolds and most of the sources I came across defined Calabi-Yau manifold in a different way. I can see that some of them are just same and I can derive one from other. But my question is the following : -"What is the most strict definition of Calabi-Yau Manifolds" -By that I mean the definition from which all the others follow. - -REPLY [6 votes]: For singular varieties , Siu introduced Calabi-Yau varieties by using numerical Kodaira dimension -Let $X_0$ be a projective variety with canonical line bundle $K\to X_0$ of Kodaira dimension $$\kappa(X_0)=\limsup\frac{\log \dim H^0(X_0,K^{\otimes \ell})}{\log\ell}$$ This can be shown to coincide with the maximal complex dimension of the image of $X_0$ under pluri-canonical maps to complex projective space, so that $\kappa(X_0)\in\{-\infty,0,1,...,m\}$. Also since in general we work on Singular K\"ahler variety we need to notion of numerical Kodaira dimension instead of Kodaira dimension. -$$\kappa_{num}(X)=\sup_{k\geq 1}\left[\limsup_{m\to \infty}\frac{\log\dim_{\mathbb C}H^0(X,mK_X+kL)}{\log m}\right]$$ -where $L$ is an ample line bundle on $X$.Note that the definition of $\kappa_{num}(X)$ is independent of the choice of the ample line bundle $L$ on $X$. Siu formulated that the abundance conjecture is equivalent as -$$\kappa_{kod}(X) =\kappa_{num}(X)$$ -The Minimal Model and abundance conjectures would imply that every variety of Kodaira dimension $\kappa (X)=0$ is birational to a Calabi-Yau variety with terminal singularities. -Note that when $K_X$ is psudoeffective vanishing numerical Kodaira dimension is equivalent that $X$ is Calabi-Yau variety<|endoftext|> -TITLE: Question about Hodge number -QUESTION [12 upvotes]: Hi. I am studying Hodge theory on Kahler manifolds. -I have several questions. - -Is Hodge number a topological invariant? (I mean, is it independent of the choice of -Kahler structure?) -If the question 1 is true, then is there any variation formula of Hodge numbers on blowing up(down)? (along Kahler submanifolds) --- please let me know the reference. -I read Huybrechts's book "complex geometry". In there, Hodge-index theorem is explained in -case of Kahler surfaces. Is there Hodge-index theorem in higher dimensional cases? -Is there a symplectic version of Hodge-Riemann bilnear relation? - -I am sorry to ask much questions. -Thank you in advance. - -REPLY [6 votes]: For a Kaehler surface, the Hodge numbers are topological invariants. -By Hodge Index Theorem, the signature of the Poincare pairing is equal -to 2h^{2,0} +2 - h^{1,1}, hence h^{2,0} is a topological invariant, -and for the rest of the numbers it is obvious. -For dimension 3, I think there are counterexamples.<|endoftext|> -TITLE: How do we compare models of ETCS? -QUESTION [6 upvotes]: The elementary theory of the category of sets (nLab) gives axioms on a category such that it is a category of sets. In answering this MO question I realised that we might have trouble comparing different models of ETCS. So here are some questions starting with easy ones (which I should know the answer to!) to the more difficult. - -Are two models of ETCS necessarily equivalent categories? -Are two models of ETCS equivalent as well-pointed topoi with NNO and Choice? -Is there a functor (Edit: a logical functor, as Todd pointed out) between any two models of ETCS? A span? -Do models of ETCS necessarily even belong to the same category? - -I wonder, because the sets that one model of ETCS has as hom-objects may be completely different to the sets that the other model has as hom-objects, and there is a priori no way to map between them. - -REPLY [6 votes]: (Caveat: I come from set theory rather than category theory and know only a little about ETCS.) -The answer to your question is no. The basic reason is that even the models -of set theory themselves can differ vastly. If $M$ is a -model of ZFC, then the category $Set^M$, which is Set as -interpreted in $M$, will be a model of ETCS. But if ZFC is -consistent, then the models of set theory $M$ are diverse. -For example, some have CH and others have $\neg CH$, and -furthermore, by the incompleteness theorem, they can -satisfy different arithmetic statements. Such statements -show up in the category $Set^M$, since every -arithmetic statement (first order statement about natural -numbers) has a translation into the formal language of -ETCS. So in general these categories are not elementary equivalent in the language of ETCS. In particular, the natural numbers objects of such categories will not in general be isomorphic, and so there can be no nice functors between the categories. For example, by the Lowenheim-Skolem theorem, some models of set theory will be countable and others will have an uncountable set of natural numbers, with a different theory, and these aspects will prevent their corresponding Set categories from being equivalent as categories or from having nice functors. In -general, it will not be possible to map the natural number -object from one to the other in any nice way.<|endoftext|> -TITLE: BSD for modular forms -QUESTION [12 upvotes]: Given a modular form, what is the precise formulation of BSD (in particular, the residue formula for the $L$-function at special values)? And what about the special values if the $L$-function is twisted by some character? Does there exist a good reference? - -REPLY [3 votes]: This paper http://wstein.org/papers/motive_visibility/ by Dummigan, me, and Watkins gives a detailed answer to your question. Plus it has some explicit examples of weight 4 modular forms so that the corresponding motive conjecturally has nontrivial (visible) Shaferevich-Tate group.<|endoftext|> -TITLE: Fourier coefficient of a modular form -QUESTION [14 upvotes]: If someone hands you a prime number $p$, and an algebraic number $x$ inside the Hasse-Weil bound, is there a normalized newform (say of weight two) so that $a_p=x$, where $a_p$ is the $p$th Fourier coefficient? - -REPLY [8 votes]: Some Remarks. -I parse the problem in the following way: -Start with a totally real algebraic integer $\alpha$ such that every conjugate of $\alpha$ has absolute value at most -$2 \sqrt{p}$. Then does there exist a normalized cuspidal Hecke eigenform $f$ of weight $2$ with $a_p = \alpha$? -First, here is a heuristic reason why one should expect this to be false. -Suppose we ask a slightly stronger question, namely, that all the coefficients of $f$ are defined over the field $E = \mathbb{Q}(\alpha)$. Then, we are asking for the existence of an abelian variety of $\mathrm{GL}_2$-type with endomorphisms by (some order in) the ring $\mathcal{O}_E$. Such objects (ignoring issues of polarization) correspond to rational points on Shimura curves. But these curves, in general, will have large genus, and so there's no reason to expect that they have any rational points. It will probably be hard to prove anything this way, however. -A second heuristic is to ask the problem in different weights. For example, is there -a weight $12$ normalized cuspidal eigenform $f$ of level co-prime to $p$ -with $a_p = 0$? This sounds tricky. Maybe Serre even conjectured once that this never happened if $p$ was sufficiently large. Let's say he did. Are you going to contradict Serre? -Finally, let me show in a rather cheap way that the answer to the original question is -"not always". Suppose that $\alpha = 2 \sqrt{p}$, which satisfies the Weil bounds. Suppose that $a_p = \alpha$, and let $\epsilon$ denote the nebentypus character of -$f$. Then the characteristic polynomial of Frobenius is -$$x^2 - 2 \sqrt{p} \cdot x + p \cdot \epsilon(p).$$ - I claim that $\epsilon(p)$, which is a root of unity, is actually $1$. (This follows -easily from the fact that the roots of this polynomial are Weil numbers and the triangle inequality.) - In particular, the characteristic polynomial of Frobenius -is actually -$$x^2 - 2 \sqrt{p} \cdot x + p = (x - \sqrt{p})^2.$$ - This doesn't happen! -Losely speaking, one knows that the action of crystalline Frobenius is semi-simple -on Abelian varieties, and yet the Eichler-Shimura relations implies -that $(\mathrm{Frob}_p - \sqrt{p})^2 = 0$, which then implies that -$\mathrm{Frob}_p = \sqrt{p}$ acts as a scalar, which contradicts how -one knows Frobenius to interact with the Hodge Filtration --- all -this is explained in (and is indeed the main point of) a paper -of Coleman and Edixhoven from the groovy 90's.<|endoftext|> -TITLE: Diffeomorphic Kähler manifolds with different Hodge numbers -QUESTION [29 upvotes]: This question made me wonder about the following: -Are there orientedly diffeomorphic Kähler manifolds with different Hodge numbers? -It seems that this would require that those manifolds are not deformation equivalent. -However, there are examples by Catanese and Manetti that that happens already for smooth projective surfaces. - -REPLY [13 votes]: A few years after this post appeared, the question of oriented diffeomorphism invariance of Hodge and Chern numbers of smooth, projective varieties was settled completely by Kotschick and Schreieder in this paper. In fact, they even refer to this post(!).<|endoftext|> -TITLE: Non standard (?) model category structure on (co)chain complexes. -QUESTION [11 upvotes]: Let $\cal{A}$ be an abelian category with enough projectives and $\mathbf{C}_+ (\cal{A})$ the category of bounded below chain complexes. -Since Quillen (Homotopical algebra, 1.2, examples B), there is a well-known "standard" model category structure on $\mathbf{C}_+ (\cal{A})$ taking as weak equivalences the maps inducing isomorphisms on homology, as fibrations the degree-wise epimorphisms in $\cal{A}$ and with cofibrations maps $i$ which are injective and such that $\mathrm{cok}\ i$ is a complex having a projective object of $\cal{A}$ in each degree. -More recently, Hovey (Model categories), proved an analogous result for the category of unbounded chain complexes, but with ${\cal A} = R$-modules, $R$ a ring (but cofibrations are not so easy to characterise). Finally, it's folklore (at least, I don't know if it is published somewhere) that the same holds for $\cal{A}$ an abelian category with a projective generator -the fact that allows the small object argument to work, as Eric Wofsey points out in his answer to this MO question. - -I'm interested in the following variant of this problem: is there a model structure on $\mathbf{C}_+ (R)$ taking as weak equivalences the homotopy equivalences? - -If it's true, I think this should be easy to verify: just taking a look to the classical proof and seing if you can change "homology equivalences" everywhere by "homotopy equivalences". I'm willingly going to do it, but, prior to start, I would like to know if it is already done, much as in the case of topological spaces where, together with the "standard" (Quillen too) model structure with weak homotopy equivalences as weak equivalences, there is the Strom model structure (The homotopy category is a homotopy category), with homotopy equivalences as weak equivalences. - -REPLY [20 votes]: This is well known, but formulated in a slightly different way. -Recall that a Frobenius category is an exact category which has enough injectives as well as enough projectives, and such that an object is projective if and only if it is injective (injectivity (resp. projectivity) is defined with respect to inflations (resp. deflations)). -If you forget about the existence of (finite) limits and colimits, any Frobenius category satisfies all the axioms of a Quillen model category: the cofibrations (resp. fibrations) are the inflations (resp. deflations), while the weak equivalences are the maps which factor through some projective-injective object. Therefore, any Frobenius category which has finite limits as well as finite colimits is a (stable) closed model category in the sense of Quillen. -Now, given an additive category $A$, the category of chain complexes $C^\sharp(A)$ (where $\sharp=\varnothing$ for unbounded chain complexes, $\sharp=b$ for bounded chain complexes, etc) is a Frobenius category: inflations (resp. deflations) are the degreewise split monomorphisms (resp. split epimorphisms), and projective-injective objects are the contractible chain complexes. If $A$ has finite limits as well as finite colimits, this shows that the category $C^\sharp(A)$ is a stable closed model category whose cofibrations (fibrations) are the degreewise split monomorphisms (resp. epimorphisms), and whose weak equivalences are the chain homotopy equivalences.<|endoftext|> -TITLE: Abundance for algebraic surfaces -QUESTION [24 upvotes]: I am currently teaching a course in algebraic geometry where one of the aims is to give an overview of the Enriques-Kodaira classification of surfaces. I am trying to throw in some modern aspects so I formulated the cone theorem and then used it (together with the existence of extremal contractions) to show that one can blow down to either $\mathbb P^2$, a ruled surface or one with $K_X$ nef. That worked quite well (the right amount of detail and non-detail). However, continuing with the nef case one of the major results is abundance (a positive multiple of $K_X$ is basepoint free). The classical Enriques-Kodaira classification does give abundance but only at the end of an almost complete classification of Kodaira dimension $\leq0$ surfaces (with $K_X$ nef). -Hence my question is: Using modern ideas is it possible to give a quicker proof of abundance for surfaces? -(Actually I am not quite sure that abundance for Kodaira dimension $2$ can be considered to be part of the E-K classification but let's ignore that). - -REPLY [5 votes]: I am a novice here, but have you looked at Lazarsfeld's chapter "Lectures on linear series" in the book Complex algebraic geometry, IAS Park City math series vol 3? There he deduces it from Reider's theorem which he deduces from Bogomolov's instability theorem, whose proof he also sketches. This is a rewrite of some parts of his chapter on application of vector bundles techniques in the book Lectures on Riemann surfaces from Trieste, where however he says his argument has an error. I am assuming the result you want is that when K is nef then some positive multiple of K is free, (in fact 4K). -yes you are right he does assume general type, but unfortunately for my understanding he does not state that in the theorem itself but only in a paragraph above the theorem, which is then apparently a blanket assumption not restated later. So since you are presumably dealing with the opposite case this is useless to you. On p. 81 of Kollar and Mori they merely state that for surfaces base point freeness is a "non trivial result". This suggests they did not know an easy proof. -edit: In Miles Reid's chapters on algebraic surfaces, he discusses this point explicitly at the end of his treatment of classification of surfaces with K nef. See E.9.1. "Abundance as a logical bottleneck". He states there that he knows only one proof in the literature not using Enriques' argument at a crucial point, namely that in the book of Barth, Peters, and Van de Ven, where they use Ueno's prof of Iitaka's additivity conjecture C(2,1) via moduli of curves. Does that help?<|endoftext|> -TITLE: minimal diameter of full preimage of torus -QUESTION [8 upvotes]: Given a set $A\subset \mathbb{R}^n$ such that $A\cap (x+\mathbb{Z}^n)\ne \emptyset$ for any $x\in \mathbb{R}^n$ (that is, $p(A)=\mathbb{T}^n$ for the projection $p:\mathbb{R}^n\rightarrow \mathbb{T}^n=\mathbb{R}^n/\mathbb{Z}^n$). Is it true that supremum of Eucledian distances between points of $A$ is not less then $\sqrt{n}$? (equality holds for the unit cube) -Or maybe even two points with $x=(x_1,\dots,x_n)$, $y=(y_1,\dots,y_n)$ with $|x_i-y_i|\geq 1-\epsilon$ for each coordinate? -It is not hard to check both claims for $n=2$, but already for $n=3$ I do not know. - -REPLY [4 votes]: The second claim is false for $n=3$. Choose $\varepsilon$ small and $\delta\ll\varepsilon$. Let $A$ be the set of all points $(x,y,z)\in\mathbb R^3$ satisfying the following inequalities: -$$ -\begin{cases} - -1.5+\varepsilon+\delta &\le x+y+z &\le 1.5+\varepsilon \\ - -1.5+\delta &\le x+y-z &\le 1.5 \\ - -1.5+\delta &\le x-y+z &\le 1.5 \\ - -1.5+\delta &\le -x+y+z &\le 1.5 \\ -\end{cases} -$$ -The integer translates of this set cover the space, but its $\ell_1$-diameter is no greater than $3-\delta$. -Added. The first claim is false too. In the above example, fix $\delta=\varepsilon/10$ and add the inequality -$$\max\{|x|,|y|,|z|\}\le 0.5+10\varepsilon$$ -to the system.<|endoftext|> -TITLE: How indepenedent of a chosen metric is the box-counting dimension? Is there a non-integral dimension which is defined for topological spaces? -QUESTION [5 upvotes]: Question 1. Given a topological space $X$ and two metrics $a$ and $b$ on it, compatible with the topology, what conditions should I impose on them so that box-counting (or other, for example Hausdorff) dimensions of $(X,a)$ and $(X,b)$ are equal? -Question 2. Is there a notion of a dimension for topological (as opposed to metric) spaces which can assume non-integral values? - - -My motivation -Let $G$ be a finitely generated group and let $p$ be a prime number. Consider the compact topological space $X:=\prod_G \mathbb Z/p$, infinite product of copies of the cyclic group of order $p$, indexed by elements of $G$. Let $T$ be an element of the integral group ring of $G$. Note that $T$ gives a map $X\to X$ in a natural way (in this context $T$ is sometimes called cellular automaton). Define $Y$ to be the subset of those points $x$ of $X$ such that $T(x)=0$. -I want to measure "how big" $Y$ is. -If we choose a generating set for $G$ then we get a metric $d$ on $G$, and we also get a metric on $X$: two sequences $x_i$ and $y_i$ of $X$ are $p^{-|B(1,k)|}$ apart if they agree on the ball $B(1,k)$ of radius $k$ around the neutral element $1$ of $G$, but they don't agree on any larger ball. It's straightforward to see that box counting dimension of $X$ is $1$. -Unfortunately the metric we just defined depends on the generators chosen for $G$, so I'd like to know if I have a chance to get any kind of "intrinsic size of $Y$" this way. - -REPLY [5 votes]: Q2. If you want that dimension for topological spaces to agree with Hausdorff dimension (for example) in case of metric space, then NO. For any $0 \le s \le \infty$, there is a metric on the Cantor set so that the Hausdorff dimension is $s$. -Another topological result. Let $X$ be a separable metrizable space. The infimum of the Hausdorff dimensions of all metric spaces homeomorphic to $X$ is the topological dimension of $X$. (Integer valued.) -for the My Motivation comment. See the notion "topological entropy".<|endoftext|> -TITLE: Is the injective model structure on symmetric spectra Bousfield localizable? -QUESTION [9 upvotes]: I am interested in injective model structures on both symmetric spectra as exposed in Hovey/Shipley/Smith and motivic symmetric spectra as in Jardine's article. Both authors take a model structure on the underlying spaces - simplicial sets in Hovey/Shipley/Smith and motivic spaces in Jardine - with monomorphisms as cofibrations. Then both establish two level model structures on symmetric (resp. motivic symmetric) spectra with weak equivalences given levelwise, an injective one with levelwise cofibrations and a projective one with levelwise fibrations. -However, both then proceed using only the projective model structure, and Bousfield localizing it with respect to an appropriate class of maps. Presumably they do so because it is much more straightforward to verify that the projective model structure satisfies the prerequisites for a Bousfield localization, in particular one has easy candidates for generating cofibrations which then turn out to do the job. -My question is: Is the injective level model structure on symmetric spectra (resp. motivic symmetric spectra) known or expected to be cellular or combinatorial? How would you try to prove this? - -REPLY [5 votes]: The answer is yes. As Lennart Meier points out, part of this answer is contained in Schwede´s book project. I´m writing to flesh out those references and discuss the topological injective case, which doesn´t appear in Schwede. Plus, old questions shouldn´t linger around without answers. Let´s start with the case where we build spectra from simplicial sets. Then Theorem 1.9 on page 129 of the Schwede pdf discusses how to do the levelwise injective model structure. Observe that cofibrations are monomorphisms, so all objects are cofibrant, so it´s immediately left proper. Furthermore, because it´s built via sequences of simplicial sets it will remain combinatorial. The same considerations apply for motivic symmetric spectra, i.e. you´ll get something left proper and combinatorial (both cases are actually right proper as well, but this doesn´t matter right now). Thus, Bousfield localization applies and you get the stable injective structures. This is mentioned in Theorem 2.2. -Now let´s think about symmetric spectra built from Top rather than sSet. For motivic symmetric spectra this is the wrong question, since you seem to really need there that everything is built from sSet. For non-motivic one would be tempted to look into the paper Model Categories of Diagram Spectra by Mandell-May-Schwede-Shipley. This paper does a nice job of treating topological symmetric spectra, but doesn´t make mention of the injective structure at all. Similarly Hovey´s Spectra and Symmetric Spectra in General Model Categories also doesn´t mention injective structures. So we turn back to Schwede, but in Theorem 1.13 on page 132, he claims not to know if topological symmetric spectra admit a levelwise injective structure. Of course you can define the classes of maps, and once you have them you can see that if this forms a model structure then it´ll be left proper and cellular (because Top is cellular, see Hovey), so it would admit left Bousfield localization. I see no reason why following Schwede´s proof for the simplicial case would break down here. True, you don´t have combinatoriality, but you should be able to use cellularity to get the necessary smallness to make the small object argument work. This notion doens´t appear in Schwede´s book, which is probably why he didn´t do it. If it´s very important to you, email me and we can try to write down all the details. Otherwise I´ll leave it as an exercise in cellular model categories -Incidentally, if you´re interested in whether or not you can ever get away with fewer hypotheses and still have Bousfield localization, check out this MO question. It´s related to a current project of mine.<|endoftext|> -TITLE: More upper/lower semi-continuous functions in (algebraic) geometry? -QUESTION [12 upvotes]: The notion of upper/lower semi-continuity is sometimes encountered in algebraic geometry. -Here by upper semi-continuity one means a function on a topological space $f:X\rightarrow S$ with value in some ordered topological space (like the field of real numbers), such that $\lim\sup_{x\rightarrow y}f(x)\leq f(y)$. Intuitively, for points $x$ that are close to a given point $y$, may the value $f(x)$ "exceeds" $f(y)$, the difference should be "small" and "vanishes" as $x$ approaches $y$. And lower semi-continuity means the opposite, namely replace "$\leq$" by "$\geq$". -Among the typical examples one thinks of the dimension function: let $k$ be a base field, and $X$ be a locally finite type $k$-scheme. The function $$x\in X \mapsto \dim_xX$$ is upper semi-continuous, where by $\dim_xX$ is understood to be the combinatory dimension of $X$ at $x$, namely the length of chains of inclusions of irreducible closed subschemes $$x\in X_1\subsetneq X_2\subsetneq \cdots\subsetneq X_d=X$$ -There seems to be a lot of examples of such upper/lower semi-continuous functions in geometry counting certain "discrete invariants", especially those related to stratifications of spaces. It'll be great to have the list extended in mathoverflow. -I hope this question is well-posed... - -REPLY [2 votes]: Seshadri constants (see arXiv:0810.0728) are lower semicontinuous. -This fact is of course related to J.C. Ottem's example, but (a priori) it is not "the same". I believe that there is not a single coherent F giving the same stratification that the Seshadri constant gives (you could get it by a sequence of coherent F, or using non-coherent F, trivially.)<|endoftext|> -TITLE: Given a graph embedded on a torus, how many edges are necessary for noncontractible loops to be long? -QUESTION [11 upvotes]: If we are given a graph embedded on a torus, with the following properties, what is the minimum number of edges it can have? - -Any noncontractible loop is comprised of at least n edges. -Any noncontractible dual loop is comprised of at least n edges. -Any noncontractible loop drawn on the torus intersects the graph at least once. - -(The third condition is just to rule out cases where we embed a small planar graph on the torus, and trivially satisfy the first two conditions, there being no noncontractible loops) -We use the following definitions: - -A loop is a series of edges, with each consecutive pair sharing a (different) common vertex, and with the first and last sharing a common vertex. It is noncontractible if the path formed by tracing along these edges is noncontractible on the torus. -A dual loop is a series of edges, with each consecutive pair sharing a (different) common face, and with the first and last sharing a common face. The name is because these edges form a loop on the dual graph. Likewise, it is noncontractible if it is noncontractible on the torus. - -I believe that the answer is $n^2$ for even n, $n^2 + 1$ for odd n. The equality case, I think, is a square lattice on the torus, but rather than identifying horizontal and vertical lines, as is usually done to put a grid on the torus, you identify lines at 45 degrees to the grid. (Or slightly off 45 degrees, if n is odd) -It seems like a simple statement, but I haven't been able to find out whether this is true. -Thanks for any help! -Graham -Edit: Whoops - rather than face-width, the second condition is asking about the edge-width of the dual graph. Apologies for the confusion! - -REPLY [3 votes]: There should be a nice proof, but here is a reference that proves something stronger and weaker. This paper by de Graaf and Schrijver proves that every graph embedded on the torus with face-width at least $n \geq 5$, contains the toroidal $\lfloor 2n/3 \rfloor$-grid as a minor. Note that the toroidal $\lfloor 2n/3 \rfloor$-grid has (almost) $8n^2/9$ edges. So any graph on the torus with face-width at least $n$ has at least (almost) $8n^2/9$ edges, which is pretty close to the conjectured answer of $n^2$.<|endoftext|> -TITLE: A decomposition of the "spin representation" of SL(2) -QUESTION [11 upvotes]: Let us take an N-dimensional (N odd) irreducible representation V of SL(2,R). -It is known that (e.g., Lie groups and Lie algebras III by Vinberg and Onischik, 1994 p. 94) in V there is an invariant symmetric bilinear form $b$ for the action of SL(2,R). Thus, SL(2,R) is embedded into $O(V,b)$ - the orthogonal group of V with respect to the form $b$. -Consider a spin representation of $O(V,b)$, this is a representation in the space of dimension $2^N$. One can restrict this representation to $SL(2,R)\subset O(V,b)$. -The question is: how to decompose this representation of SL(2,R) of dimension $2^N$ into irreducibles? - -REPLY [11 votes]: Yes this question is a bit old, but I can never resist some fun character theory. Maybe you have already figured out a satisfactory answer to your original question, but in case not, here is a purely computational method for finding the decomposition you seek. -By general weight theory, the weights of the representation of $SL_2$ of dimension $2n+1$ are $\langle(2n)\beta, (2n-2)\beta, \ldots,(-2n+2)\beta, (-2n)\beta\rangle$ where $\beta$ is the highest weight of the defining 2-dimensional representation (Although $\omega$ is more standard for weights, I use $\beta$ here to avoid confusion with the $\omega$'s I use for the weights of $SO_{2n+1}$). -The weights of the $2n+1$ dimensional representation of $SO_{2n+1}$ are $\langle\omega_1, -\omega_1+\omega_2, -\omega_2+\omega_3,\ldots, -\omega_{n-2}+\omega_{n-1},-\omega_{n-1}+2\omega_n,0,\ldots,\rangle$ where the weights after the $0$ weight are just the negatives of the weights before the $0$ weight. -The embedding $SL_2\hookrightarrow SO_{2n+1}$ that you are interested in takes weights of $SL_2$ to weights of $SO_{2n+1}$; in both lists above I ordered the weights starting from the highest weight and working down. -Associating corresponding weights on the lists, one gets the following correspondence of the fundamental weights: -$\omega_1 = (2n)\beta$ -$\omega_2 = (4n-2)\beta$ -$\omega_3 = (6n-6)\beta$ -$\ldots$ -$\omega_{n-1} = (n^2+n-2)\beta$ -$\omega_n = (\frac{n^2+n}{2})\beta$ -In general, for $k\neq n$, one gets $\omega_k = (2kn-(k^2-k))\beta$ -So how do we use this to decompose the spinor representation $\Sigma_n$ (with highest weight $\omega_n$) of $SO_{2n+1}$? Let $X_i = exp(\omega_i)$; then the character of the spinor representation of $SO_{2n+1}$ can be written as: -$\chi(\Sigma_n) = (X_1X_2\ldots X_n)^{-1}(1+X_1)(X_1+X_2)\ldots(X_{n-2}+X_{n-1})(X_{n-1}+X_n^2)$. -From the weight correspondence above, letting $Y = exp(\beta)$ (hence $X_1 = Y^{2n}, X_2 = Y^{4n-2}$ and so on) one gets the restriction of this character back to $SL_2$: -$Res^{SO_{2n+1}}_{SL_2}\chi(\Sigma_n) = Y^{-\frac{1}{6}(4n^3+3n^2-n)}(1+Y^{2n})(Y^{2n}+Y^{4n-2})\ldots(Y^{n^2+n-6}+Y^{n^2+n-2})(Y^{n^2+n-2}+Y^{n^2+n})$ -Expanding this product out, one can read off the weights of the restriction as the exponents of $Y$. -For example, when $n=2$ one gets: -$Res^{SO_5}_{SL_2}\chi(\Sigma_2) = Y^{-7}(1+Y^4)(Y^4+Y^6) = Y^3+Y+Y^{-1}+Y^{-3}$ -This has weights $3\beta$, $\beta$, $-\beta$, and $-3\beta$, and hence corresponds to the $4$-dimensional irreducible representation of $SL_2$. -When $n=3$, one gets: -$Res^{SO_7}_{SL_2}\chi(\Sigma_3) = Y^{-22}(1+Y^6)(Y^6+Y^{10})(Y^{10}+Y^{12}) = Y^6+Y^4+Y^2+2+Y^{-2}+Y^{-4}+Y^{-6}$ -This has weights $6\beta$, $4\beta$, $2\beta$, $0$ (multiplicity 2), $-2\beta$, $-4\beta$, and $-6\beta$ and so this is the sum of the $7$-dimensional irreducible representation with the trivial representation. -Continuing this, for small $n$, the restriction of $\Sigma_n$ splits as $11\oplus 5$ (n=4), $16\oplus 10\oplus 6$ (n=5), and $22\oplus 16\oplus 12\oplus 10\oplus 4$ (n=6).<|endoftext|> -TITLE: Quasi-Lie algebras in nature? -QUESTION [8 upvotes]: A Lie algebra over $\mathbb Z$ is defined to be an abelian group with a bracketing operation $[\cdot,\cdot]$, satisfying the Jacobi identity and the relation $[x,x]=0$ for every $x$. On the other hand, a quasi-Lie algebra is defined by replacing the axiom $[x,x]=0$ with the antisymmetry axiom $[x,y]+[y,x]=0$ for all $x$ and $y$. Jerry Levine used the free quasi-Lie algebra to study the group of homology cylinders. See this paper and this paper. My question is whether there are other contexts in which quasi-Lie algebras appear. Lie algebras arise naturally in all sorts of different contexts, -but I don't know of many places in which quasi-Lie algebras appear. For example, if one looks at the associated graded object for the lower central series of a group, one naturally obtains a Lie algebra and not a quasi-Lie algebra. -One possible answer is this MO post, where it is noted that in operad language, the quasi-Lie axioms are necessary, and this is quite germane to why Levine needed to use quasi-Lie algebras, since he was essentially building a group out of the Lie operad. But I'd like to know of any other examples where quasi-Lie algebras occur that people might know. -Added (Moskovich): I wonder also whether Levine was the first person ever to consider quasi-Lie algebras, in 2002. Surely that can't possibly be the case- as a structure, it's surely too natural to be so new! -Added (Conant) This is not just an idle question. My main motivation for asking this question is to find new link concordance invariants. Milnor's mu invariants are associated to the lower central series of a link group, the associated graded object being a Lie algebra. I'm hoping there is some kind of quasi-Lie algebra one can assign to a link complement that generalizes this associated graded Lie algebra. -Edit: Based on the comments and Mariano's answer, I'd like to make some clarifications. First, when working over the integers, maybe it is better to use the term "Lie ring" as opposed to "Lie algebra," but I had just been following Levine's convention. (Unfortunately, Hilton's definition of a quasi-Lie ring, given in the reference in Mariano's answer, does not match Levine's!) Levine's definition means "a graded (or super) Lie algebra where all elements are of even degree." So one could take the even degree part of the Lie algebra formed from homotopy groups under the Whitehead product, as Tom mentions. -Edit: It turns out that Levine's definition of a quasi-Lie algebra is a specific case of Hilton's definition (see Mariano's answer), where the module $M$ is $0$. - -REPLY [2 votes]: Tom Goodwillie: -The homotopy groups of a (say, simply connected) space $X$ form a graded Lie algebra under Whitehead product, in which the even-dimensional part (which is actually the $\pi_n$ for odd $n$ ) can have elements $x$ such that $[x,x]$ has order $2$ -- for example $\pi_n(S^n)$ for most odd values of $n$.<|endoftext|> -TITLE: Deformations of Kähler manifolds where Hodge decomposition fails? -QUESTION [19 upvotes]: This is partly inspired by answers to the question: -Question about Hodge number . -Is there a family of compact complex manifolds, where the general fibres are -Kähler, but for which $E_1$ degeneration of the Hodge to de Rham spectral sequence fails -at the special fibre? Or, even better, such that the special fibre has nonclosed -holomorphic forms? -I feel like I should know the answer, but somehow I don't. All -the examples I know where the spectral sequence doesn't degenerate are nilmanifolds*, -so they aren't even homotopic to Kähler manifolds by standard rational homotopy theoretic obstructions (e.g. they aren't formal). -Also the famous Hironaka example [Ann. Math 1962] won't work either, because -the special fibre is an algebraic variety, so the spectral sequence will degenerate -(by an argument that can found in Deligne [Théorème de Lefschetz...]). -Obviously, I haven't thought about this deeply enough, but perhaps someone else has**. -Footnotes -*I was bit sloppy yesterday, since the examples I have in mind include -solvmanifolds. However, there are still topological obstructions to these being Kähler -due to Nori and myself. -** From the answers, I gather that the work of Popovici suggests that -there may be no counterexample. - -REPLY [4 votes]: Just as a comment. -In general, given a family of sG (i.e., strongly-Gauduchon) manifolds, the limit need not to be still sG: an example was provided by Ceballos, Otal, Ugarte, and Villacampa, http://arxiv.org/abs/1111.5873 (see also http://arxiv.org/abs/1210.0406). -It may be possible that the sG property is preserved to the limit assuming stronger conditions on the fibres: however, the published version of the paper by Popovici does not prove this fact, as noticed also by YangMills. -On the other hand, an example of a family of (non-Kähler) manifolds satisfying the Hodge decomposition (namely, the $\partial\overline{\partial}$-Lemma) and whose limit does not, is studied in http://arxiv.org/abs/1305.6709. -Note that there are several examples of sG manifolds non-satisfying the Hodge decomposition: for example, every nilmanifold admitting a balanced or sG metric.<|endoftext|> -TITLE: How many Hecke operators span the level 1 Hecke algebra? -QUESTION [18 upvotes]: Let $k \ge 4$ be an even integer, and let $d$ be the dimension of the space $M_k(\operatorname{SL}_2(\mathbb{Z}))$ of modular forms of level 1 and weight $k$. Then the space of Hecke operators acting on $M_k$ also has dimension $d$. Is it spanned by $T_1, \dots, T_d$? -Equivalently (more explicitly but also more messily): if $f \in M_k(\operatorname{SL}_2(\mathbb{Z}))$ satisfies $a_i(f) = 0$ for $1 \le f \le d$, where $a_i(f)$ are the $q$-expansion coefficients of $f$, with no assumption on $a_0(f)$, then is it necessarily true that $f = 0$? -(Edit: See also this follow-up question which asks a related question for modular forms of higher level.) - -REPLY [17 votes]: Write $k = 12\ell + k'$, where $k'$ is one of $0, 4, 6, 8, 10, 14$, and let $f_{k, m}$ be the unique weakly holomorphic modular form (poles allowed at cusps) of weight $k$ for $SL_2(\mathbb{Z})$ with Fourier expansion $f_{k, m} = q^{-m} + \sum_{n \geq \ell+1} a_k(m, n) q^n$. The duality of coefficients $a_k(m, n) = -a_{2-k}(n, m)$ between forms of weight $k$ and forms of weight $2-k$ holds (see http://www.math.ucla.edu/~wdduke/preprints/serre.pdf ), so the original question is equivalent to asking whether it is true that the coefficient $a_{k}(0, \ell+1)$ is never zero. By duality, this coefficient is the negative of the constant term in $f_{2-k, \ell+1} = \frac{E_{14-k'}}{\Delta^{\ell+1}} = q^{-\ell-1} + \sum_{n=-\ell}^\infty a_{2-k}(\ell+1, n) q^n$. Siegel's 1969 paper referenced in Robin Chapman's answer proves that this constant term is always nonzero (Theorem 2), so the answer to the original question is yes in all cases. An English translation of Siegel's paper appears as an appendix in his Advanced Analytic Number Theory book, available online at http://www.math.tifr.res.in/~publ/ln/tifr23.pdf .<|endoftext|> -TITLE: Explicit examples of resolution of (projective) 3-folds over k? -QUESTION [6 upvotes]: I'm looking for examples of explicit resolutions of (projective) 3-folds over a field k (char 0), with isolated singularities, or at least with smooth singular locus. I've looked in various books and online, but the examples they present have only been for curves and surfaces, for which resolution of singularities is much less complicated. It would also be nice if the exceptional divisor were sufficiently nice, say, with smooth components that intersect transversally (or if the exceptional divisor were itself smooth). -Are there any well-known/easy examples in dimension 3? Because resolution quickly becomes complicated as dimension increases, I'd imagine that examples become harder to come by, although I'm sure many exist! -In general, I seem to not be very good at finding references/papers relevant to specific things I'm looking for. Search engines typically turn up a lot of irrelevant papers, and thumbing through a bunch of books seems time-inefficient. Search engines have turned up a few papers, but I suspect that I can do better than what I've found thus far. -I realize that adding this second part about looking for references may detract from the specific issue I have right now, but I'd rather get better at finding things than ask here when I can't find something. - -REPLY [3 votes]: Another, curve related, example would be the secant variety to a smooth curve. If the embedding is 'sufficiently ample', which would mean that it separates 5 points (it's early and no coffee yet, so '5' should be double checked with a reference) then the secant variety is smooth away from the curve and singular along the curve, so the singular locus is smooth. There is a well know resolution which is discussed in papers of Bertram. -A curve related example with isolated singularities would be the theta divisor of a general genus 4 curve or non-general curves of lower genus. Geometry of Algebraic Curves is a reference for that. Roy Smith is another excellent reference for information of this sort !<|endoftext|> -TITLE: Stable orthogonalization procedure -QUESTION [6 upvotes]: At a high level, my question is the following: given a set of $k$ vectors in Euclidean space which are pairwise "almost orthogonal", can one find a set of $k$ orthogonal vectors which are pairwise close to the original ones? This could be seen as a stable version of Gram-Schmidt orthogonalization, in which, under the promise that the original set of vectors is not too far from being orthogonal, one has the guarantee that they do not need to be moved too much in order to become orthogonal. -More precisely, assume given vectors $v_i$, $i=1,\ldots,k$ in a real (or complex) $k$-dimensional space, such that $\sum_{i\neq j} \langle v_i,v_j\rangle \leq \epsilon k$ ($\epsilon$ should be understood as an arbitrarily small constant, or it could even be $o(k)$ if needed -- the weaker the assumption the better). Then, does there exist a set of orthogonal vectors $w_i$ such that $\sum_i \|w_i - v_i\|^2 \leq \epsilon' k$? (The norm is the usual Euclidean norm.) The interesting question is whether one can get an $\epsilon'$ which depends on $\epsilon$ only, not on $k$. -A related question (the one I am originally most interested in), is that of orthogonalizing $d$-dimensional projector matrices $P_1,\ldots,P_k$: assume $\frac{1}{d} \sum_{i\neq j} \langle P_i,P_j \rangle \leq \epsilon$ (where now the inner product is the trace inner product), can you find orthogonal projectors $Q_i$ such that $\frac{1}{d}\sum_i \|P_i-Q_i\|_F^2 \leq \epsilon'$? (Here the norm is the Frobenius norm, the sum of squares of the coefficients.) So far using various iterative procedures I can only get a bound where $\epsilon' = poly(\log k) \epsilon$, but I would like to know if the dependence on $k$ can be removed. - -REPLY [4 votes]: Use a Procrustes rotation of the standard basis vectors onto your vectors. This gives the set of orthogonal vectors with the smallest sum of squares of distances to your vectors. -http://en.wikipedia.org/wiki/Orthogonal_Procrustes_problem -"The orthogonal Procrustes problem is a matrix approximation problem in linear algebra. In its classical form, one is given two matrices A and B and asked to find an orthogonal matrix R which most closely maps A to B." -In your case you want to find the orthogonal matrix R which most closely maps the standard basis to your matrix. Something like the columns of R should then be the set of orthogonal vectors which are nearest to your vectors, where 'nearest' is in the sense of sum of squares.<|endoftext|> -TITLE: Good functorial model for BG -QUESTION [23 upvotes]: There are several functorial constructions of the space BG for a topological group (meaning BG plus the universal G-bundle). First, there is the Milnor construction, treated in several textbooks. The Milnor construction is functorial and $EG \to BG$ is locally trivial for all topological groups. The Milnor construction is NOT monoidal in the sense that $B(G \times H) \cong BH \times BG$ ($B1$ is something like an infinite-dimensional simplex and not a point). -On the other hand, there is the nerve-construction $BG:= |N_{\bullet} G|$ (plus a construction of $EG$). This is monoidal, but the map $EG \to BG$ is not always locally trivially (according to Graeme Segal, Classifying spaces and spectral sequences, p. 107). It is locally trivial if G is "locally well-behaved" (Segal gives a precise condition). Segal claims that if G is not locally well-behaved, then local triviality is not an appropriate concept. -I would be happy to exclude groups like the p-adic integers from having a classiying space, but there are other groups which I do not like to throw away, like Homeo (X) for a manifold X (is this locally well-behaved??). Here is my question: -Is there a construction of $BG$, satisfying the following properties: - -functorial, -monoidal, -$EG \to BG$ is locally trivial, -the class of groups to which it applies is "very large", including Homeo of reasonable spaces, -Simple enough to be reasonably presented in a lecture course? - -REPLY [5 votes]: Segal's classifying space $BG$ (the geometric realisation of the nerve of $G$ considered as a one-object topological groupoid) and the associated universal bundle (the geometric realisation of the nerve of the action groupoid of $G$ acting on itself by mulitplication, or equivalently, the codiscrete groupoid with objects $G$) was also studied by May, Milgram and Steenrod. These latter three only require that $G$ be well-pointed: the inclusion of the identity element is a closed cofibration. I note however that this is all done in the category of $k$-spaces. -The nice result of May is that $EG\to BG$ is not just a locally trivial bundle, but a numerable bundle, that is, there is a trivialising cover of $BG$ such that this cover admits a subordinate partition of unity. Thus $EG \to BG$ classifies numerable bundles (over paracompact spaces these are clearly the same as bundles). -Additionally, $EG$ is a topological group. Also since construction this uses ordinary geometric realisation, $E(-) \to B(-)$ preserves products (and indeed pullbacks) - this where the geometric realisation into k-spaces is used.<|endoftext|> -TITLE: Generalization of the positive semidefinite Grothendieck inequality -QUESTION [8 upvotes]: In a recent paper, S. Khot and A. Naor show a natural generalization of the positive semidefinite Grothendieck's inequality. Grothendieck showed that there exists a constant $K > 0$ such that for every $n \times n$, symmetric semidefinite matrix $A=[a_{ij}]$, the following inequality holds: -$$\max_{x_1,\ldots,x_n} \sum_{ij} a_{ij}x_i^Tx_j \le K \max_{\epsilon_1,\ldots,\epsilon_n \in [-1,1]} \sum_{ij}a_{ij} \epsilon_i\epsilon_j,$$ -where each $x_i$ is a unit vector (in Euclidean norm, so that $x_i^Tx_i=1$). -Khot and Naor studied a natural variant of this inequality, where the $n$ numbers $\epsilon_i$ are replaced by $n$ vectors $u_1,\ldots,u_n$ chosen from a set of $k < n$ unit vectors $v_1,\ldots,v_n$. The inequality becomes: -$$\max_{x_1,\ldots,x_n} \sum_{ij} a_{ij}x_i^Tx_j \le C \max_{u_1,\ldots,u_n \in \{v_1,\ldots,v_k\}} \sum_{ij}a_{ij} u_i^Tu_j.$$ -They proved that this inequality actually holds, but the proof looks very complicated; as does the constant $C$. My question is thus in two parts: - -(a) Is there any chance that there is a simpler proof for this "natural" generalization to Grothendieck's inequality? -(b) Does it seem like a feasible project to try and estimate this constant? - -I am not an expert in these inequalities, but this generalization looks so interesting, that I had to seek out expert opinion. - -REPLY [3 votes]: Perhaps it would be beneficial to summarize the proof in words: -We wish to prove a certain Grothendieck-type inequality. We begin with a Lemma. -In Lemma 3.2 of this paper, we find an expression for a certain probability. In some sense, this calculation is really just (Hermite) Fourier analysis. We have a certain function, and we expand it into a (Hermite) Fourier basis. This proves our lemma. -We now use the probabilistic method to calculate the expected value of a certain random inner product. Using positive definiteness of the matrix A, and applying Lemma 3.2, we can throw away all but the "first level" (Hermite) Fourier coefficients. This gives our desired inequality, which ends up being sharp. -By the way of analogy, think about Hurwitz's proof of the Isoperimetric Inequality using Fourier series. We have an arbitrary curve, we expand it into its Fourier components, and we see that the inequality becomes sharp when all but the "first level" Fourier coefficients are set to zero. In a way, the same proof idea is used here. -As for computing the constant, this is indeed a challenge even in the simplest nontrivial cases. I don't think estimating it is very hard (for say, a "rough" estimate). To see what I mean, I would recommend reading the first in the series, "Approximate kernel clustering" (by the same authors).<|endoftext|> -TITLE: Braid groups acting on CAT(0)-complexes -QUESTION [18 upvotes]: Does the braid group $B_n, n\ge 3$, act properly by isometries on a CAT(0) cube complex? - Update 1. During a recent talk of Nigel Higson in Pennstate Dmitri Burago asked whether the braid groups are a-T-menable. I seem to remember that somebody proved that they do act properly by isometries on CAT(0) cube complexes. That would imply a-T-menability by Niblo and Roller or Cherix, Martin and Valette. Hence the question. Note: no co-compactness required. - Update 2. It looks like my question is still an open problem for $n\gt 3$. I think my confusion came from terminology. Dan Farley proved that all braided diagram groups (including the R. Thompson group $V$) act property by isometries on CAT(0)-cube complexes. But braided diagram groups (defined by Victor Guba and myself) are not related to braid groups, at least not explicitly because wires there intersect and do not form braids. One can define the notion of "really braided" diagram groups where wires form braids, but I do not think Farley's method will work. So I got confused by my own terminology. By the way, I do not see an obvious reason that $B_n$ does not embed into $V$. $V$ is a big group with lots of complicated subgroups. - Update 3. As Bruce Hughes pointed out to me, even though the Haagerup property (a-T-menability) is unknown for $B_n, n\ge 4$, all forms of Baum-Connes conjecture have been proved for it by Thomas Schick in Finite group extensions and the Baum-Connes conjecture. - Update 4 Concerning the question from Update 2. Collin Bleak and Olga Salazar-Diaz proved in Free products in R. Thompson's group V that $V$ does not contain subgroups isomorphic to ${\mathbb Z}^2\ast {\mathbb Z}$. Does $B_n$ contain such subgroups? - -REPLY [9 votes]: As Sam Nead says, $B_n$ contains $\mathbb{Z}^2 * \mathbb{Z}$, and you can find an example the way he suggests. -If you'd like something much more explicit, you can simply take the first three standard generators $\sigma_1$, $\sigma_2$ and $\sigma_3$, and then it follows from the solution (by Crisp and Paris) of a conjecture of Tits that the subgroup $\langle \sigma_1^2, \sigma_2^2, \sigma_3^2 \rangle$ is isomorphic to $\mathbb{Z}^2 * \mathbb{Z}$. -See The solution to a conjecture of Tits on the subgroup generated by the squares of the generators of an Artin group by Crisp and Paris. -Also, see Clay, Leininger and Mangahas' The geometry of right angled Artin subgroups of mapping class groups for more info and references about right angle Artin subgroups of mapping class groups.<|endoftext|> -TITLE: A non-standard ergodic limit -QUESTION [7 upvotes]: Suppose $T$ is an ergodic measure-preserving transformation on a measure space $(X,\Sigma,\mu)$, and $f\in L^1(\mu)$. Does the limit -$\lim_{X\to\infty} \pi(X)^{-1}\sum_{p\leq X} f(T^{p}x)$ -exist almost everywhere? Is it constant almost everywhere? Here the sum runs over primes, and $\pi(X)$ is the prime counting function. -When $X=\mathbb{R}/\mathbb{Z}$ with Lebesgue measure, and $T:x\to x+\theta$ is an irrational rotation, the answers to these questions are "yes" and "yes", by Vinogradov. - -REPLY [3 votes]: Patrick LaVictoire shows that the answer is negative if you ask for all of $L^1$. -His paper is Universally L^1-Bad Arithmetic Sequences (to appear in Journal d'Analyse Mathematique). The paper extends results of Buczolich and Mauldin (who showed a negative result if you sum along the squares). The paper can be found online at http://arxiv.org/abs/0905.3865<|endoftext|> -TITLE: Locally a topos -QUESTION [12 upvotes]: What kind of categories $C$ have the property that each slice category $C/c$ is a topos? Obviously, topoi have this property, but, the converse is not true. An example is the category $EtTop$ of topological spaces and only local homeomorphisms. $EtTop/X \cong Sh(X)$, but $EtTop$ is very far from being a topos! It doesn't have an initial or final object, nor pushouts... -Are there other of these "locally a topos" categories which are not topoi? Have they been studied before? - -REPLY [11 votes]: The question of local toposes and similar categories was discussed a couple of years ago on the categories list by Peter Johnstone and others, if I recall correctly. I don't know anywhere that they appear in print, but I don't know the topos theory literature nearly well enough to be an authoratitive source on this. - -On the local toposes themselves: $\newcommand{\Topos}{\mathbf{Topos}_\mathit{slice}}$one other in-some-sense-trivial example, if I'm not mistaken, is the category $\Topos$, with objects all (small, fsvo small) elementary toposes, and with maps just the geometric morphisms that are (up to equivalence) induced by slicing, modulo natural isomorphism. (If we wanted to cover our tracks a little, we could say “the geometric morphisms whose inverse image functors are logical”; the equivalence of this is shown in Mac Lane and Moerdijk in their chapter on logical morphisms, iirc.) -But this is the universal example: any other local topos $\newcommand{\E}{\mathcal{E}}$has a unique-up-to-equivalence “local equivalence” $\E \to \Topos$, sending $A$ to $\E/A$, and any local equivalence into $\Topos$ must come from a local topos.[1] -But local equivalences into a fixed category $\newcommand{\C}{\mathcal{C}} \C$, in turn, correspond (up to equivalence-over-$\C$) to functors $\C^\mathrm{op} \to \mathbf{Sets}$. ($F : \mathcal{D} \to \C$ corresponds to the functor taking $C \in \C$ to the set of isomorphism classes of objects of $\mathcal{D}$ over $C$.) Of course, there's a size consideration: whatever size of $\mathbf{Sets}$ we use constrains the essential size of the fibers. -In terms of local toposes seen as categories in their own right, one point of interest is that you can interpret pretty much all the logic in them that you can in toposes (i.e. higher-order type theory; and geometric logic if you go Grothendieck-y)… except that of course you have to abandon the “empty context”, since you don't have a terminal object. (Everything else in the interpretation of logic is purely local.) In a local topos, there is no “global validity”: all truth is relative :-) -[1] The uniqueness issues here are a little subtle: if I'm not mistaken we need to either assume (large) choice, or use anafunctors instead of functors, or restrict the size of $\E$ enough that each slice will itself literally be an object of $\Topos$.<|endoftext|> -TITLE: An explicit computation in class field theory -QUESTION [15 upvotes]: Let $K$ be the imaginary quadratic field obtained by joining $\sqrt{-1}$ to the field of rational numbers $Q$. I would like to describe the extension $K^{ab}/Q^{ab}$, where for $F$ a number field, $F^{ab}$ denotes its maximal abelian extension (everything is taking place inside a big fixed field...). -More precisely I would like to know the Galois group and the ramification properties of such extension. Is this possible/easy? I suppose one should look at the kernel of the norm map between Idele class groups $N_{K/Q}:I_K\rightarrow I_Q$. But at the moment it is not clear to me how to get the answer. Any hint or comment would be appreciated. Thanks. -EDIT: Probably the idele class group of a number field $F$ should be denoted by $J_F$. Or by anything other than $I_F$... - -REPLY [16 votes]: Given that you want to know the structure of the Galois group and ramification, I think that you are best off working with the kernel of the norm map between connected components of idele class groups, as you yourself suggest. -These groups are very explicit: for $K := \mathbb Q(i)$, one obtains $\hat{\mathcal O}_K^{\times}/\{\pm 1,\pm i\}$, -and for $\mathbb Q$ one obtains $\hat{\mathbb Z}.$ (Here $\hat{}$ denotes the profinite -completion.) Apart from the diagonally embedded $\{\pm 1,\pm i\}$ quotient in the -group for $K$, both groups factor as a product over primes, and the norm map is given -component wise. -So the kernel of the norm map is equal to -$$\bigl(\prod_p (\mathcal O_K\otimes_{\mathbb Z}\mathbb Z_p)^{\times, \text{Norm } = 1}\bigr)/ -\{\pm 1,\pm i\}.$$ -This should be explicit enough to answer any particular question you have.<|endoftext|> -TITLE: Strange question about Hechler -QUESTION [5 upvotes]: Recently during a lecture, my professor mentioned that forcing over any poset which is countable, separative, and atomless, is essentially the same as forcing over the Cohen poset, that is to say results in adding a Cohen real. -My question is: Are there any other similar characterizations of "commonly used" forcing posets? Specifically, is there one for the Hechler poset? -The Hechler Poset/forcing notion $(H,\le)$ is given by setting $H=\omega^{\lt\omega} \times \omega^{\omega} $, and defining the relation $(t,v) \le (s,u)$ iff $ ( t \supset s \wedge (\forall n\in\omega) (u(n) \le v(n)) \wedge (\forall m \in dom(t \backslash s))(t(m) \gt u(m))$. When forcing with this poset, you end up adding an unbounded real to the ground model. -I understand that you cannot produce a model in which $\mathfrak{b}=\omega_2$ using product forcing, and that you need iterated forcing to do so. Moreover, the iterated forcing construction I've seen that produces $\mathfrak{b}=\omega_2$ in the forcing extension, used the finite support iteration of $\omega_2$ many copies of the Hechler poset. Is this evidence for the lack of such a characterization? -(I apologize in advance if this is an ill-stated question, I will change it accordingly if it is.) - -REPLY [8 votes]: Here are a few additional examples of the kind you seek, -and in fact each of them directly generalizes the -characterization you mention of the forcing to add a Cohen -real. However, I know of no such characterization of Hechler forcing. - -The collapse forcing - $\text{Coll}(\omega,\theta)$ is, up to forcing equivalence, the unique forcing - notion of size $|\theta|$ necessarily collapsing - $\theta$ to $\omega$. (Note, this includes the case of - adding a Cohen real, since $\text{Add}(\omega,1)\equiv\text{Coll}(\omega,\omega)$.) -To see this, suppose that $\mathbb{Q}$ is a forcing notion -of size $|\theta|$ that necessarily collapses $\theta$ to -$\omega$. We may assume without loss of generality that -$\mathbb{Q}$ is separative, since the separative quotient -of $\mathbb{Q}$ is forcing equivalent to it and no larger -in size. Observe that below every condition in -$\mathbb{Q}$, there is an antichain of size $\theta$. Since -forcing with $\mathbb{Q}$ adds a function from $\omega$ -onto $\theta$ and $\mathbb{Q}$ has size $\theta$, there is -a name $\dot g$ forced to be a function from $\omega$ onto -the generic filter $\dot G$. We build a refining sequence -of maximal antichains $A_n\subset\mathbb{Q}$ as follows. Begin -with $A_0=\{ 1\}$. If $A_n$ is defined, then let -$A_{n+1}$ be a maximal antichain of conditions such that -every condition in $A_n$ splits into $\theta$ many elements -of $A_{n+1}$, and such that every element of $A_{n+1}$ -decides the value $\dot g(\check n)$. The union -$\mathbb{R}=\bigcup_n A_n$ is clearly isomorphic as a -subposet of $\mathbb{Q}$ to the tree $\theta^{\lt\omega}$, and -so it is forcing equivalent to -$\text{Coll}(\omega,\theta)$. Furthermore, $\mathbb{R}$ is -dense in $\mathbb{Q}$. To see this, fix any condition -$q\in\mathbb{Q}$. Since $q$ forces that $q$ is in $\dot G$, -there is some $p\leq q$ and natural number $n$ such that -$p$ forces via $\mathbb{Q}$ that $\dot g(\check n)=\check -q$. Since $A_{n+1}$ is a maximal antichain, there is some -condition $r\in A_{n+1}$ that is compatible with $p$. Since -$r$ also decides the value of $\dot g(\check n)$ and is -compatible with $p$, it must be that $r$ forces $\dot -g(\check n)=\check q$ also. In particular, $r$ forces -$\check q\in\dot G$, and so by separativity it must be that -$r\leq q$. So $\mathbb{R}$ is dense in $\mathbb{Q}$, as -desired. Thus, $\mathbb{Q}$ is forcing equivalent to -$\mathbb{R}$, which is forcing equivalent to -$\text{Coll}(\omega,\theta)$, as desired. -If $\kappa^{\lt\kappa}=\kappa$, then the forcing $\text{Add}(\kappa,1)$ to add a Cohen subset -to $\kappa$ with conditions of size less than $\kappa$ is, -up to forcing equivalence, the unique ${\lt}\kappa$-closed -necessarily nontrivial forcing notion of size $\kappa$. -For this, one similarly builds up a dense tree of conditions -inside the poset that is isomorphic to $\kappa^{\lt\kappa}$, which -is forcing equivalent to $\text{Add}(\kappa,1)$. -More generally, if $\theta^{\lt\kappa}=\theta$, then the forcing $\text{Coll}(\kappa,\theta)$ -is the unique ${\lt}\kappa$-closed forcing notion -necessarily collapsing $\theta$ to $\kappa$ and having size $\theta$. (The -straightforward generalization uses separativity, but the -separative quotient of a partial order may no longer be $\lt\kappa$-closed, so there is an -issue about it; but my -student Norman Perlmutter found a solution avoiding the issue.)<|endoftext|> -TITLE: Sub-Hopf algebras of group algebras -QUESTION [5 upvotes]: Let $k$ be a field and $G$ a finite group. Is every sub-Hopf algebra over $k$ of the group algebra $k[G]$ of the form $k[U]$ for a subgroup $U$ of $G$ ? - -REPLY [2 votes]: Moreover, finiteness of $G$ is irrelevant as Greg's proof works there as well. BTW, you don't need to go into $H^\ast$ if you are willing to use coalgebras: any subcoalgebra of $k[G]$ is $k[U]$ for some subset of $G$. $U$ must be a subgroup for the subco to be subHopf... -A more interesting question is to ask about forms of $k[G]$, i.e. Hopf algebras over a subfield $m$ such that $k\otimes_m H \cong k[G]$. There are a plenty of those which are not group subalgebras!!<|endoftext|> -TITLE: Categories with products that preserve quotients -QUESTION [5 upvotes]: It is well known that in the category of all topological spaces, quotient maps aren't preserved by products (this follows from the simpler fact that $X\times (-):Top\to Top$ doesn't preserve quotients). The usual solution, if one is needed, is to change to the category of k-spaces and k-continuous maps. There are other categories where products and quotients 'get along' (e.g. $Set$, $Ab$). -Question 1: What is a large class of categories where quotient maps are preserved by products? Topoi? (Semi)abelian categories? Categories of algebras for a monad on a given category with this property? -Now one may be only interested in a certain class of quotient maps (like surjective submersions in $Diff$, the category of finite dimensional smooth manifolds). Say, regular epimorphisms, or maps admitting local sections (assuming we're in a site), or perhaps something like surjective topological submersions, where there are sections through every point in the domain. So in this case it is not a matter of putting restrictions on $X$ such that $X\times(-)$ preserves quotients, or changing the category, but narrowing the scope of the quotient maps one wants to preserve. -Question 2: Is there are large class of quotient maps (in $Top$, or in a general category - with finite products and enough colimits) that are preserved by products? - -REPLY [3 votes]: Note: I have edited this answer further because I was being silly before (unnecessarily restrictive). -I take it you mean categories $E$ for which $- \times -: E \times E \to E$ preserves quotients. The word 'quotient' may be slightly ambiguous because sometimes people use it to mean'coequalizer', and sometimes just 'epi' (as in 'quotient object'), but I take it you mean 'coequalizer'. -A reasonably large class would be regular categories, which includes categories of algebras of monads on $Set$ and semi-abelian categories and toposes. Here quotients = regular epis are stable under pullback and in particular are closed under taking products on either side. Furthermore, in a regular category, every quotient is a reflexive coequalizer, meaning a coequalizer of a pair $f, g: X \to Y$ for which there exists $h: Y \to X$ with $f \circ h = g \circ h = 1_Y$. In particular, the two projections $\pi_1, \pi_2: E \to Y$ of an equivalence relation $E$ on $X$, for example the kernel pair of a quotient, form a reflexive pair by the reflexivity property. So in a regular category, where quotients = coequalizers are necessarily coequalizers of their kernel pairs, quotients are quotients of reflexive pairs. -The reason reflexivity is relevant is a $3 \times 3$ lemma which says that in a (edit: commutative-in-parallel) diagram of $3 \times 3$ objects in which all rows and all columns are coequalizer diagrams of reflexive pairs, the diagonal is a coequalizer diagram. See the first page of Johnstone's Topos Theory. Then apply this lemma to the evident diagram whose rows are of the form -$$X_i \times X_{1}' \stackrel{\to}{\to} X_i \times X_{2}' \to X_i \times X_{3}'$$ -and whose columns are of the form -$$X_1 \times X_{j}' \stackrel{\to}{\to} X_2 \times X_{j}' \to X_3 \times X_{j}'$$ -In the category $Top$, it would therefore be natural to consider quotients by equivalence relations (or even just reflexive relations) which are preserved by taking products on each side. It's that latter condition which needs to be characterized (or at least discussed further), and I may come back to that later after I get the kids off to school. :-) -Edit: For a discussion of topological quotients which are stable under taking a product on either side, see the paper by Day and Kelly, On topological quotient maps preserved by pullback or product, Math. Proc. Cam. Phil. Soc. 67 (1970), 553-558. Or google Day-Kelly maps to find out more.<|endoftext|> -TITLE: The determinant of the sum of normal matrices -QUESTION [16 upvotes]: Given two normal matrices $A,B\in M_n({\mathbb C})$ -whose respective spectra are $(\alpha_{1},\ldots,\alpha_{n})$ and -$(\beta_{1},\ldots,\beta_{n})$, is it true that $\det(A+B)$ belongs to -the convex hull of the set of numbers -$$\prod_{i=1}^n(\alpha_i+\beta_{\sigma(i)}),$$ -as $\sigma$ runs over the set ${\mathfrak S}_n$ of permutations of $\{1,\ldots,n\}$ ? -Nota. It is known (see Exercise 101) that the trace of $AB$ belongs to the convex hull of the -set of numbers -$$\sum_{j=1}^n\alpha_{j}\beta_{\sigma(j)},\qquad\sigma\in {\mathfrak S}_n.$$ - -REPLY [15 votes]: This claim is nothing but the well-known Marcus and de Oliveira conjecture, which has been open since 1973 or earlier. -Reference: Open Problems in Matrix Theory, X. Zhan. -For the simpler case of Hermitian matrices, the claim holds; a slightly more general case seems to be the paper "The validity of the Marcus-de Oliveira conjecture for essentially Hermitian matrices." -PS: you might want to add the "open problem" tag to your question.<|endoftext|> -TITLE: Missing exposes in SGA 5, and the composition of the SGA's -QUESTION [17 upvotes]: Over the past couple of years I had to look in SGA for various results, and I can't but marvel at how poorly constructed it is. In SGA1 expose VII "n'existe pas", SGA 1 references higher SGA's, and so forth. -My first, albeit less urgent, question is: how were these composed, and why the many missing exposes and references to future manuscripts? Were all SGA's written simultaneously? Are the missing exposes a product of type-writing it (so changing the table of contents is too much of a bother?). Were they going to write those exposes later but never did? Why didn't they? -My second question is specifically regarding the missing exposes in SGA5. In the introduction to expose XIII in SGA1 it says that it generalizes the results of SGA5 II -- one of the missing exposes! What happened there? Is there a more appropriate reference for what SGA1 expose XIII generalizes? - -REPLY [7 votes]: According to Grothendieck's "Recoltes et Semailles", SGA 5 was totally butchered by Illusie, in a combined effort with Deligne so that it looked useless in comparison with SGA 4 1/2 (which wasn't a true seminar, and stole some of the missing exposés from SGA 5). This is also the reason why SGA 5 was the last to be published. -Precisely, there were three lost exposés: - -Exposé XI was called "Computation of local terms" or something like that, and was substituted by exposé III-b. -Exposé IX was about Serre-Swan modules and was published elsewhere by Serre (Linear Representations of Finite Groups). -Exposé XIII was deleted. - -Also, important points remarked by Grothendieck in the oral seminar were suppressed in this edition. It's a real pity, and probably the only way to recover the lost ideas would be to find Grothendieck's original notes (probably the archives at Montpellier University contain them or the IHES library). -EDIT: More about what was missing from SGA 5: - -Exposé IV, about "The cohomology class associated with a cycle", was going to be redacted by Deligne, who instead included it in SGA 4 1/2, chapter 4. - -A related theme covered by Grothendieck was "The homology class associated with a cycle". This was discussed throughout many exposés in the seminar, but is absent from the published book. According to ReS, these ideas were published by Verdier in an article with the same name. You can read it here: Verdier, Jean-Louis, Classe d’homologie associee à un cycle, Astérisque 36–37, 101–151 (1976). ZBL0346.14005. - -Exposé II, as is known from the introduction to SGA 5, was reworked by Deligne and included in SGA 4 1/2 as well.<|endoftext|> -TITLE: How does the Constructibility Degree of a real compare with its Turing Degree? -QUESTION [6 upvotes]: Specifically, is it the case that (for $a,b\in\omega^\omega$) $a$ $\leq_T$ $b$ implies $a$ $\leq_c$ $b$? -I suspect it might be trivial, but not knowing much Recursion Theory, it's hard to see how it could. -Thank you in advance. - -REPLY [5 votes]: As Bjorn pointed out, $a\leq_Tb$ implies $a\leq_cb$. But it should also be mentioned that constructible degrees are much coarser then Turing degrees: Suppose $\aleph_1^L$ is the real $\aleph_1$. Then every constructible degree (set of $\leq_c$-equivalent reals) is uncountable (take your real $a\in\omega^\omega$ and consider all coordinate wise sums of $a$ and a constructible real. These sums are all of the same constructible degree as $a$). Every Turing degree is countable since there are only countably many Turing machines. -This observation explains Carl Mummert's statement "in $L$ there are extremely large Turing degrees".<|endoftext|> -TITLE: orders and length functions on finitely generated groups -QUESTION [6 upvotes]: Let $G$ be a finitely generated group with the natural word length function ($|x|$ is the length of the shortest word in generators of $G$ representing $x$). We call a partial left invariant order $\le $ on $G$ word order if whenever $a\le b\le c$ we have $|b|\le C(|a|+|c|)$ for some constant $C$. Say, the standard order on $\mathbb Z$ is a word order. - Question. Is there a group $G$ with a left invariant linear order but without left invariant linear word order? - -REPLY [10 votes]: If $G$ is a finitely generated infinite group and $\leq$ is a linear word order, then for each $a, c \in G$ there are only finitely many elements $b \in G$ such that $a \leq b \leq c$. From this it follows that $(G, \leq)$ is order isomorphic to $\mathbb Z$. If $\leq$ is also left invariant, then this isomorphism must be a group isomorphism as well. - -REPLY [4 votes]: Let $G=\mathbb Z^2$. Every invariant linear order on $\mathbb Z^2$ is either induced from the standard order on $\mathbb R$ (or its inverse) by a linear map of the form -$$ - (x,y)\mapsto x+\alpha y : \mathbb Z^2\to\mathbb R -$$ -where $\alpha\in\mathbb R\setminus\mathbb Q$, or is a composition of the lexicographic order -$$ - (x,y)>(x',y') \quad\iff\quad x>x' \text{ or } (x=x' \text{ and } y>y') -$$ -with a bijective linear transformation $\mathbb Z^2\to\mathbb Z^2$. -Indeed, the set of elements of $\mathbb Z^2$ that are greater or equal to zero is a semi-group $H$ such that $H\cap(-H)=\{0\}$. The closed convex hull of such $H$ must be a half plane. If this half-plane is bounded by an irrational line, we have the first type of a linear order. If this half-plane is bounded by a rational line, then we may assume that it is a coordinate line $\{x=0\}$ (up to a linear change of coordinates in $\mathbb Z^2$), then the order is the lexicographic one (up to a change $y\mapsto -y$ depending on whether $(0,1)$ is "positive" or "negative" in this ordering). -In both cases, there are infinitely many elements between $(0,0)$ and $(1,0)$, hence it is not a word order.<|endoftext|> -TITLE: Suggestions for good notation -QUESTION [182 upvotes]: I occasionally come across a new piece of notation so good that it makes life easier by giving a better way to look at something. Some examples: - -Iverson introduced the notation [X] to mean 1 if X is true and 0 otherwise; so for example Σ1≤n -TITLE: How do I stop worrying about root systems and decomposition theorems (for reductive groups)? -QUESTION [18 upvotes]: I apologize for this being a very very vague question. -Just as personal experience, I never feel that I fully grasped the theory of root systems in Lie algebras and Lie/algebraic groups (I shall call these $\textbf {LAG} $objects for short). My "psychological" response to these objects seem to be--they are really "ugly" and unnatural, where do they come from?? Of course, we can say, OK, let's work out the theory of $sl_2$, then everything is just "natural" generalization of easier cases. But this explanation just doesn't satisfy me. -I've been worrying about such things for really quite a long time. And probably the real question is, do these root systems have some (really nice, really simple) geometrical interpretation? For example, is it possible that they are somehow related to some nice topological space, some sheaf, some cohomology, etc...? And do they have some nice counterparts in other branches of mathematics? (it seems to me they only appear in places like LAG objects), do they show up in some other (probably surprising) places? -Actually, as we know, study of LAG objects themselves are closely related to their representation theories. And their representation theory are also obejects that seem hard to visualize in some "easy" geometrical way. I know we can probably put these things as a category and do some abstract algebraic geomtry with them. But still, we cannot avoid using these ad-hoc techniques with them. (like root systems, Dynkin diagrams, etc, etc) -Perhaps I should stop here. In any case, I hope this question will not be closed as a spam. And hope some one can understand some of my confusion and shed some light (geometry) on it. -Thank you. - -REPLY [3 votes]: I had a problem which may be similar when I encountered root systems for the first time. In retrospect, I think that the problem was that I was reading about specific realizations of root systems (eg, the $A_n$ roots embedded into $\mathbb R^{n+1}$ as $e_i-e_j$ , etc.). Those specific realizations looked very ad hoc to me. The point, I would say, is that, while it's good to spend some time with some of those specific realizations, you should get used to thinking about root systems in a uniform way, at which point their power becomes pretty evident. (But maybe this is really a different problem from yours.)<|endoftext|> -TITLE: Calculating cup products using cellular cohomology -QUESTION [39 upvotes]: Most algebraic topology books (for instance, Hatcher) contain a recipe for computing cup products in singular or simplicial homology. In other words, given two explicit singular or simplicial cocycles, they contain a recipe for computing an explicit cocycle representing the cup product of the cocycles in question. -Is there a similar recipe in cellular cohomology? In other words, if I have a very explicit CW complex and two explicit cellular cocycles, then is there a recipe for computing a cellular cocycle representing their cup product? -Of course, one answer is to subdivide everything up into a simplicial complex, but that is messy (and not always possible). Is there a better way? -I'm especially interested in the special case of 2-dimensional CW complexes, where the only interesting cup products are between elements of $H^1$. - -REPLY [3 votes]: A recipe is given in Whitney's 1937 paper "On Products in a Complex" (Annals of Mathematics, vol. 39, no 2). You can find a copy here. -In section 12 he gives a very explicit construction of the cap product for low-dimensional, regular complexes from which the cup product follows immediately (via Eqn 5.3 in the same paper).<|endoftext|> -TITLE: End point compactification for metric spaces -QUESTION [8 upvotes]: Freundenthal introduced ends of topological spaces and the end point compactification of locally compact topological spaces adding one point for each end of the topological space (see here). -For example the end point compactification of $\mathbb{R}$ is homeomorphic to the unit interval $[0,1]$ because "$\mathbb{R}$ has two ends". -If you take the set $A=\{ (x, 1/x)\ |\ x>0 \}\cup\{(x, 0)\ |\ x\in\mathbb{R}\}$, it is homeomorphic to $\mathbb{R}\sqcup\mathbb{R}$ so its end point compactification will be homeomorphic to $[0,1]\sqcup[0,1]$. -But seeing $A$ as a metric space one wants to say that "the two ends at the right are the same" and to compactify $A$ with only three ends, giving a connected topological space homeomorphic to $[0,1]$. -Is there in the litterature such a notion of "metric end point compactification" which would compactify $A$ with only three ends? -(I’m not asking how to define such a compactification, I already have a definition which seems to work for my purpose, I just want to know if something like that is already known) -Thank you -Edit: -Here is a more precise definition of the compactification I need (I’m not sure this is exactly the correct definition, but it should be something like that): let’s say that a metric space $(E,d)$ is 0-connected if for all $\varepsilon>0$ and $x,y\in E$, there exists a finite sequence $(u_0, \dots, u_n)$ such that $u_0 = x$, $u_n = y$ and $d(u_i,u_{i+1})<\varepsilon$ for all $0\leqslant i\leqslant n-1$ (every connected space is 0-connected, but the space $A$ introduced before is an example of space which is non connected but 0-connected) (by the way, if this notion of 0-connectedness has already a name in the litterature, I will be happy to know it). -Then, take the definiton of the Freudenthal end point compactification (see link above) but replace "$U_i$ is a connected component of …" by "$U_i$ is a 0-connected component of …". -In particular $A$ has three ends, and the complex plane or the hyperbolic space have only one end. - -REPLY [2 votes]: Another possibility is to use proximities - or equivalently (totally bounded) uniformities: in the metric case one defines $A$ and $B$ to be 'close' (usually denoted $A\mathrel\delta B$) if $d(A,B)=0$. The relation $\delta$ satisfies the proximity axioms and it determines a compactification $\tilde X$ of the space $X$ with the property that for subsets $A$ and $B$ of $X$ one has $A\mathrel\delta B$ iff the closures of $A$ and $B$ in $\tilde X$ intersect. -The Freudenthal compactification comes from the proximity defined by declaring $A$ and $B$ to be 'far apart' iff they have disjoint open neighbourhoods $U$ and $V$ respectively such that $X\setminus (U\cup V)$ is compact, that is, they are separated by a compact set.<|endoftext|> -TITLE: Topological spaces whose continuous image is always closed -QUESTION [17 upvotes]: If $X$ a topological space one says that $X$ is universally closed if for every Hausdorff space $Y$ and every (continuous) map $f:X\rightarrow Y$, the image of $X$ is a closed subset of $Y$. -It is clear that every compact space is universally closed, but are there non compact universally closed spaces? - -REPLY [8 votes]: I believe the question whether $X$ can be chosen to be Hausdorff was left open by both existing answers. The solution is provided by the H-closed spaces of Henno Brandsma's comment. I shall answer the question from that comment in the positive. -Proposition. A Hausdorff space $X$ is H-closed iff it is universally closed. -Proof. The "if" direction is trivial. -To prove "only if", we use the well-known fact that a Hausdorff space is H-closed iff every open cover of $X$ contains a finite collection of sets whose closures cover $X$. (See the Wikipedia page cited above, or the section on H-closed spaces in Extensions and Absolutes of Hausdorff Spaces by Jack R. Porter and R. Grant Woods.) -So suppose $f\colon X\to Y$ is continuous, where $Y$ is Hausdorff, and let $y\in Y\setminus f(X)$. Since $Y$ is Hausdorff, for every $x\in X$ there is a closed neighbourhood $U[x]\subset Y$ of $f(x)$ not containing $y$. Since $X$ is H-closed, there is a finite collection $x_1,\dots,x_k$ such that -$$ X = \bigcup_i f^{-1}(U[x_i]). $$ -So -$$ y\in Y\setminus \bigcup_i U[x_i] \subset Y\setminus f(X), $$ -and hence $Y\setminus f(X)$ is a neighbourhood of $y$. So $f(X)$ is closed, as claimed. $\blacksquare$ -Alternatively, you may consider directly the following simple example of a H-closed space from Porter-Woods. This space is given by -$$ X := \{p^-,p^+\} \cup \{(1/n,1/m)\colon n\in\mathbb{N}, m\in\mathbb{Z}\setminus\{0\}\} \cup \{(1/n,0)\colon n\in\mathbb{N}\}.$$ -$Y:=X\setminus\{p^-,p^+\}$ has the usual topology induced from $\mathbb{R}^2$, while a neighbourhood of $p^+$ (resp. $p^-$) should contain all points $(1/n,1/m)$ with sufficiently large $n$ and positive (resp. negative) $m$. -It is easy to verify directly that this space is universally closed (and H-closed). -As noted in Pietro's answer, the map from a universally closed space to its Stone-Cech compactification is surjective. The question remains whether there is a space with the latter property which is not H-closed.<|endoftext|> -TITLE: Relation between tame fundamental group w.r.t. to D, and the fundamental group of the complement of D -QUESTION [6 upvotes]: Motivation -I was re-reading parts of Grothendieck-Murre, and these questions came up naturally. -The situation in chapter 9 is that $S'=Spec(A)$ where $A$ is a complete local noetherian ring of dimension $2$ with an algebraically closed residue field, and $D$ is some divisor in $A$. Then they take a "desingularization" of $S'$, $T'$, such that $T'$ maps isomorphically to $S'$ away from $D$, and the divisor that goes to $D$ is normal crossings. Call the divisor that maps to $D$ also $D$. They first show that $\pi_1^{(p)\,D}(T')$ is isomorphic to $\pi_1^{(p)}$ of the completion of $T'$ with respect to $D$ where this $\pi_1$ is the tame $\pi_1$ with respect to $D$. I assume this goes through even if $D$ is not normal crossings (if I'm wrong about this, I would really like to know). -Then they prove that $\pi_1^{(p)}(S'\setminus D)(=\pi_1^{(p)}(T'\setminus D))$ is isomorphic to $\pi_1^{(p),D}(T')$. -The question that arises is: why did they take this detour through this desingularization? Which step doesn't go through in the case that $D$ is not normal crossings? As I said, I think it's unlikely that it is the step that the tame fundamental groups are the same when completing, but I would like to know if I'm wrong. -Question -Let $S'=Spec(A)$ where $A$ is a complete local noetherian ring of dimension $2$ with an algebraically closed residue field, and $D$ is some divisor in $A$. Could there exist a divisor, $D$ (necessarily not normal crossings), such that $\pi_1^{(p)}(S'\setminus D)$ is not isomorphic to $\pi_1^{(p),D}(S')$? - -REPLY [5 votes]: Your question as stated does not strictly speaking make sense. In Grothendieck-Murre sect.2.4 the tame fundamental group $\pi_1^D(S,\xi)$ is only defined when $S$ is normal and $D$ is a DNC. Although in 2.2.2 they define tamely ramified covers in greater generality, they observe in Remark 2.2.3(4) that this is "certainly not the correct" definition when $D$ is not a DNC. However if you were to use their definition 2.2.2 (which requires in addition that $S'$ be normal) to define the tame $\pi_1$, then the groups in your question would be isomorphic for any $D$. Indeed, you have an equivalence of the relevant categories of coverings, given by restriction to $S'\setminus D$ in one direction and by normalisation in the other. -The aim of Chapter 9 Grothendieck-Murre is to prove that $\pi_1^{(p)}$ of a connected open subscheme of $S'$ (not necessarily normal) is topologically finitely presented. For this you really do need to pass to the desingularisation, so as to bring Kummer theory of tame covers to bear.<|endoftext|> -TITLE: Computing homology of very large posets -QUESTION [12 upvotes]: I'm studying the homology of a couple of very large posets (one has over 4 million vertices, though the dimension is only 3). I want to show the posets are spherical (homology vanishes except in top dimension) or better Cohen-Macaulay. Because of the size, a direct computation of homology seems impossible, even with a clever program like Simplicial Homology for GAP. -Does anyone know of an algorithm for this sort of thing which could make life easier? I have tried to find an explicit recursive atom ordering already, but haven't found one. - -REPLY [3 votes]: Short note for similar problems. -We solved the problem using RedHom software in 2012, 3 days of computations, ~400GB of RAM. -However, now we have better methods and we can check Cohen-Macaulay property of the huge complex in 30 minutes, 340GB of RAM, using 64-cores Intel E7-8837@2.6GHz (assuming hdf5 input file format). It is so fast, because in parallel we can build a data structure for the complex and compute an optimal discrete Morse vector field.<|endoftext|> -TITLE: Varieties as an introduction to algebraic geometry / How do professional algebraic geometers think about varieties -QUESTION [24 upvotes]: This really is two questions, but they are kind of related so I would like to ask them at the same time. -Question 1: -In a question asked by Amitesh Datta, BCnrd commented that it is important to learn about varieties in a classical sense before learning about modern algebraic geometry because it is where much of the intuition in the subject comes from. -I was hoping to get some opinions on how much one should learn about varieties (in the sense of chapter 1 of Mumford's red book) before moving onto more modern formulations of algebraic geometry. -Is one meant to gain a rudimentary understanding of varieties and then start learning about schemes, OR is one meant to have a really good understanding of abstract varieties before learning about schemes. -Question 2: -Do professional algebraic geometers think about varieties from a scheme theoretic perspective or from a classical perspective. -This is a seriously soft question, so I will make it community wiki. I am however half expecting it to be closed. - -REPLY [3 votes]: I don't seem to have the option to edit my answer, but I wanted to correct myself: it is the scheme structure of the limit that determines the possible nearby varieties in the family, not vice versa. The question of whether the nearby varieties determine the limit is that of Hausdorfness of the parameter space. One has to restrict the nature of the possible limits to get uniqueness of the limit. This arises in deciding just how complicated the limit should be when trying to construct a nice compactification of a given family of nice varieties. It is a wonderful fact that for smooth curves one can compactify them without introducing non varieties. All one needs is some simple singular curves. This was first noticed by Alan Mayer and David Mumford in their talks at the Woods Hole conference 1964, whose notes appear on James Milne's webpage at Michigan, (and mine at UGA). I am roy smith, and have been using mathwonk as alias for so many years online I have forgotten it is not my name. Is it sufficient to register and add it to my profile, or do I need to sign posts here as roy smith? That would apparently unlink me from all my mathwonk activity so far.<|endoftext|> -TITLE: counter example for semi direct product of groups -QUESTION [13 upvotes]: Hi, -I have got a very natural question in group theory. -Suppose you have two countable groups $G_1,G_2$, some action of $\mathbb Z$ on them such that -the semi direct products are isomorphic $\phi:G_1\rtimes \mathbb Z\simeq G_2\rtimes \mathbb Z$. -We suppose that $\phi(\mathbb Z)=\mathbb Z$. -Do we have that $G_1\simeq G_2$? -It looks silly but I have not been able to find a counterexample. -Arnaud - -REPLY [8 votes]: Take a compact 3-manifold $M$ with $b_1(M)\geq 2$. Then there are many homomorphisms $\pi_1(M)\to \mathbb{Z}$, since $\mathbb{Z}^{b_1(M)}\leq H_1(M)$. Further, if the manifold fibers over $S^1$ corresponding to a map $\phi:M\to \mathbb{Z}$, then $ker(\phi)$ is finitely generated. If $\phi:M\to \mathbb{Z}$ is not fibered, then a theorem of Stallings implies that the cohomology class is not dual to a fiber. For example, consider the link L4a1: - -The complement is a compact manifold $M$ with $H_1(M)=\mathbb{Z}^2$. Orienting the two circles of the link in two different ways (up to negation) gives two different homomorphisms to $\mathbb{Z}$ (via the linking number). One orientation corresponds to a fibering, while the other does not (there is an annulus running between the two components). Also, the intersection number with the meridian is the same (up to sign) for each choice of orientation, so the cyclic subgroup condition is satisfied. So the kernel of one map is finitely generated (in fact free), while the other is infinitely generated.<|endoftext|> -TITLE: Which monoidal categories are equivalent to their centers? -QUESTION [6 upvotes]: Let $\mathcal C$ be a monoidal category. Recall that the (Drinfel'd) center of $\mathcal C$ is the braided monoidal category $Z(\mathcal C)$ with: - -Objects: pairs $M \in \mathcal C$ and $\mu: M\otimes(-) \overset\sim\to (-)\otimes M$ a natural iso, which is required to satisfy: $\mu: M \otimes \mathbb 1 \to \mathbb 1 \otimes X$ is the canonical isomorphism $X\otimes 1 \to M \to 1\otimes M$, and $\mu_{A\otimes B} = (\operatorname{id}_A \otimes \mu_B)\circ (\mu_A\otimes \operatorname{id}_B)$ as maps $M\otimes A \otimes B \to A\otimes B \otimes M$. (I will drop all associators.) So $\mu$ is tryin to be a "braiding" — the second axiom is some sort of "hexagon" axiom. -Morphisms: a map $(M,\mu) \to (N,\nu)$ is a morphism $f: M \to N$ so that $\nu \circ (f\otimes \operatorname{id}) = (\operatorname{id}\otimes f) \circ \mu$ as natural transformations $M\otimes (-) \to (-) \otimes N$. -(Here $f\otimes \operatorname{id}$ is the natural transformation $M\otimes \to N\otimes$, etc.) -The monoidal structure is $(M,\mu)\otimes (N,\nu) = (M\otimes N, (\mu \otimes \operatorname{id}_N) \circ (\operatorname{id}_M \otimes \nu))$. -The braiding $(M,\mu) \otimes (N,\nu) \to (N,\nu)\otimes (M,\mu)$ is given my $\mu_N : M\otimes N \to N\otimes M$. It is a fun calculation to check that this is a (bi-natural) isomorphism in the category, and satisfies all requirements to be a braiding. - -An important example: when $\mathcal C$ is the representation theory of a finite abelian group $G$, and writing $G^\vee$ for the Pontrjagin dual to $G$ (it is isomorphic for finite abelian groups), then as a monoidal category, but not as a braided monoidal category, $Z(\mathcal C)$ is (equivalent to) the representation theory of $G\times G^\vee$. -When $\mathcal C$ is braided (with braiding $\beta_{M,N} : M\otimes N \to N\otimes M$), there is a full, faithful injection $\mathcal C \hookrightarrow Z(\mathcal C)$, which is usually not an essential surjection, given on objects by $M \mapsto (M,\beta_{M,-})$. (For the finite abelian group case, this map is dual to the projection $G\times G^\vee \to G$.) -In addition to representations of monoids, another important source of categories are as categories of modules of commutative rings. I think I can prove that if $R$ is a commutative ring, then the injection $\operatorname{Mod}(R) \to Z(\operatorname{Mod}(R))$ is an equivalents of categories (the monoidal structure on $\operatorname{Mod}(R)$ is $\otimes_R$). (Although I've read that the center of $\operatorname{Mod}(R)$ is supposed to be more like sheaves on the loop space of $\operatorname{Spec}(R)$, so I might have made an error.) -My question is about the converse. -The weak statement I do not expect to be true: - -Question (weak assumptions): Suppose that $(\mathcal C,\otimes)$ is a monoidal category (abelian, complete, cocomplete, etc. ... maybe also some statement along the "tensor products distribute over filtered limits" variety) and that there exists a monoidal equivalence $\mathcal C \to Z(\mathcal C)$. Must $(\mathcal C,\otimes)$ be equivalent to $(\operatorname{Mod}(R),\otimes_R)$ for some commutative ring $R$? - -A more reasonable question is: - -Question (stronger assumptions): Suppose that $(\mathcal C,\otimes,\beta)$ is a braided monoidal category (with more adjective?) such that the canonical map $\mathcal C \to Z(\mathcal C)$ is an equivalence of braided monoidal categories. Must $(\mathcal C,\otimes,\beta)$ be equivalent as a braided category to $(\operatorname{Mod}(R),\otimes_R,\operatorname{flip})$ for some commutative ring $R$? - -Probably (super)commutative rings in SuperVectorSpaces will provide counterexamples. And so the real question is: - -Real question: What is / is there a nice characterization of those monoidal categories that are equivalent to their centers? - -REPLY [8 votes]: Some of these questions are addressed (in the derived setting) in my paper Integral Transforms and Drinfeld Centers in Derived Algebraic Geometry with John Francis and David Nadler --- for the underived setting you might also want to look at Hinich's Drinfeld double for orbifolds. I think the answer to your main question is no: if you take (quasi)coherent sheaves on any scheme as a symmetric monoidal category, its Drinfeld center is equivalent to itself. A more precise statement is that the derived Drinfeld center of QCoh of a stack is given by sheaves on the derived loop space of the stack. The underlying underived stack is the inertia --- which in the case of a scheme is just the underlying scheme. In this case though you still see something interesting on the derived level -- you get modules over the sheaf of Hochschild chains on your scheme, i.e. in char zero sheaves on the odd version of the tangent bundle [tangent complex if things aren't smooth]. The braided structure on this Drinfeld center is given by convolution over the base -- the free loop space has a homotopical group structure over the space -(with fibers the based loop spaces). For R-R bimodules, aka endofunctors of R-modules [for R a k-algebra], its center is k-mod -- in fact it's 2-Morita equivalent to k-modules. (We can think of it as a matrix algebra with entries given by Spec R, so we expect it to be Morita equivalent to scalars).<|endoftext|> -TITLE: Fundamental groups of topological groups. -QUESTION [20 upvotes]: Let $G$ be a topological group, and $\pi_1(G,e)$ its fundamental group at the identity. If $G$ is the trivial group then $G \cong \pi_1(G,e)$ as abstract groups. My question is: -If $G$ is a non-trivial topological group can $G \cong \pi_1(G,e)$ as abstract groups? -About all I know now is that $G$ would have to be abelian. - -REPLY [2 votes]: [I originally entered the answer below in response to another question. Later other people pointed out that that question was a duplicate of this one, so I joined them in closing it.] -Let $V$ be a vector space over $\mathbb{Q}$ of dimension $2^{\aleph_0}$ (e.g. $V=\mathbb{R}$) and put $G=BV$. Here we use the simplicially defined classifying space functor $B$, so $G\times G=BV\times BV=B(V\times V)$, so we can apply $B$ to the addition map $V\times V\to V$ to get a topological group structure on $BV$. By construction we have $\pi_1(G)\simeq V$, but also $G$ is a vector space over $\mathbb{Q}$ of dimension $2^{\aleph_0}$, so $G\simeq V$ as abstract groups. I don't think that you can do anything much smaller than this.<|endoftext|> -TITLE: Griffiths-positive metric -QUESTION [10 upvotes]: How to find a Griffiths positive metric on an ample vector bundle? - -REPLY [11 votes]: The positivity of holomorphic bisectional curvature is as -same as Griffiths positivity of the holomorphic tangent bundle -Y.T. Siu and Shing Tung Yau gave an affirmative answer(which can be considered as Analytical proof of Hartshorne conjecture due to algebraic proof of Mori) to the Frankel conjecture -saying that every compact K\"ahler manifold with positive holomorphic bisectional curvature is -biholomorphic to the projective space -Campana and Flenner gave a positive answer to the Griffiths conjecture when the base S is a projective curve see -F. Campana and H. Flenner, A characterization of ample vector bundles on -a curve. Math. Ann. 287 (1990), no. 4, 571–575. -Your question need more details, what do you mean of hermitian metric?, since the Griffiths positivity definition for smooth hermitian metric is different with continuous hermitian metric (which may not be smooth in general) or well defined singular hermitian metric -A continuous hermitian metric $h$ on a vector bundle $\pi : E \to X$ is said to be Griffiths positive, if there -exists a smooth positive real (1, 1)-form $ω_X$ on $X$ such that -$$\pi^*\omega_X+\sqrt{-1}\partial\bar\partial \log h\leq 0$$ -in the sense of currents -But a Hermitian vector bundle $E$ with smooth hermitian metric over a complex manifold $M$ to be Griffiths positive if $$\sqrt{-1}\Theta_{u\bar u}(X,\bar X) >0 \Longleftrightarrow R_{X\bar X u \bar u} >0$$ for any nonzero $(1,0)$ tangent vector $X$ of $M$ and nonvanishing section $u$ of the vector bundle $E$. -The good thing for Griffiths positivity of smooth hermitian metric is that $(E, h)$ is semi-positive if and -only if its dual $(E^⋆, h^⋆)$ is semi-negative in the sense of Griffiths(I know it just for smooth hermitian metric and I don't know for continuous hermitian metric on vector bundle )(positivity duality property for other definitions like Nakano positivity is not known ) -Ph. Griffiths in the following paper gave the following important conjecture for smooth hermitian metric -Ph. Griffiths, Periods of integrals on algebraic manifolds, III. Some global differential geometric properties of the period mapping, Inst. Hautes Etudes Sci. Publ. Math. ´ 38 -(1970) 125–180 -Conjecture: Let $E$ be an ample vector bundle, in the sense that $\mathcal O_E (1)$ is -ample on $\mathbb P(E^⋆)$. Then $E$ admits a smooth Hermitian metric $h$ such that $(E, h)$ is positive in the sense of Griffiths -singular Hermitian metric on vector bundle is not well defined in general , but we know some result at least for some cases -A singular Hermitian vector bundle $(E, h)$ which is positively curved in the -sense of Griffiths is weakly positive -We can extend Griffiths conjecture for singular hermitian metric that weak positivity of $E$ implies the Griffiths semi-positivity? -There are some good results for direct image of relative line bundle (note that direct image of relative line bundle is not line bundle in general and is vector bundle if we take its double dual to be reflexive ) -See Paun survey paper -The nice question is that if we choose a continuous (or singular)hermitian metric with Griffiths positivity to run Yau-Donaldson flow -$$\frac{\partial h_t}{\partial t}=-2h_t(\Lambda F_{h_t}-\lambda Id)$$ -to get singular Hermitian-Einstein metric then all the solutions remain positive in the sense of Griffiths? This can help to verify Existence of relative Hermitian Einstein metric on fibrations such that moduli of fibers are stable vector bundles. Since we must choose singular inital Hermitan metric for such flow<|endoftext|> -TITLE: Infinite loop space maps into or out of BAut(F_n) -QUESTION [8 upvotes]: There is an inclusion $S_n \to Aut(F_n)$ from the symmetric group into the automorphism group of a free group. After applying the Quillen +-constriction, both $BS_{\infty}$ and $BF_{\infty}$ become infinite loop spaces and the map above can be promoted to an infinite loop space map $BS_{\infty}^+ \to BF_{\infty}^+$. - -Are there any other infinite loop space maps into or our of $Aut(F_{\infty})?$ - -Some notes: -1) Galatius has shown relatively recently that $BAut(F_{\infty})^+$ is homotopic to the sphere spectrum; the latter has been known for about 4 decades to be homotopic to $BS_{\infty}^+$ (Barratt-Priddy-Segal). So loop maps in or out of the symmetric group can all be defined up to homotopy on $Aut(F_{\infty})$ instead. Also, in all of the above, can "Aut" be replaced by "Out"? -2) Answers involving $\Omega^{\infty}S^{\infty}$ that don't relate to spaces of graphs or free groups don't count - I'm trying to get information on this from a geometric group theory point of view. - -REPLY [9 votes]: Let $Aut(\bigvee^k S^n)$ be the topological monoid of pointed homotopy equivalences from a wedge sum of $k$ copies of $S^n$ to itself. Then $Aut(\bigvee^k S^0) = S_k = \Sigma_k$. The fundamental group functor induces a homotopy equivalence -$Aut(\bigvee^k S^1) \simeq Aut(F_k)$. The $n$-th homology functor induces a homomorphism -$Aut(\bigvee^k S^n) \to Aut(Z^k) = GL_k(Z)$. Suspension induced maps -$$ -B\Sigma_k \to BAut(F_k) \to \dots \to BAut(\bigvee^k S^n) \to \dots \to BGL_k(S) -\to BGL_k(Z) . -$$ -Here $S$ is the sphere spectrum. -Taking the wedge sum with the identity map on $S^n$ allows $k$ to grow. -In the colimit we get -$$ -BS_\infty \to BAut(F_\infty) \to \dots \to BAut(\bigvee^\infty S^n) \to \dots -\to BGL_\infty(S) \to BGL_\infty(Z) . -$$ -Taking the plus-construction and multiplying with $Z$ you get infinite loop maps -$$ -Q(S^0) \to Z \times BAut(F_\infty)^+ \to \dots \to A(\ast) \to K(Z) -$$ -where $A(\ast)$ is Waldhausen's algebraic $K$-theory of a point. The -intermediate infinite loop spaces $Z \times BAut(\bigvee^\infty S^n)^+$ -for $1 < n < \infty$ seem to be poorly understood. -An early reference: - -Friedhelm Waldhausen, Algebraic - $K$-theory of topological spaces. II. - Algebraic topology, Aarhus 1978 (Proc. - Sympos., Univ. Aarhus, Aarhus, 1978), - pp. 356–394, Lecture Notes in Math., - 763, Springer, Berlin, 1979.<|endoftext|> -TITLE: The conditions in the definition of Brownian motion -QUESTION [12 upvotes]: A (standard, real-valued) Brownian motion $W = \{W(t): t \geq 0\}$ is commonly defined by the following properties: 1) $W(0) = 0$ a.s., 2) the process has independent increments, 3) for all $s,t \geq 0$ with $s -TITLE: annihilator/common left multiple of matrix polynomials -QUESTION [5 upvotes]: Let $A_{n,d}$ be the space of polynomials of degree $d$ or less whose coefficients are real $n\times n$ matrices --- or, if you prefer, the space of matrices whose entries are degree-$d$ polynomials. Given $A(z),B(z)\in A_{n,d}$, there are $P(z),Q(z)\in A_{n,d}$ so that $P(z)A(z)=Q(z)B(z)$ (proof: consider the $2dn\times (2d-1)n $ Sylvester resultant matrix of the polynomials; its left nullspace has dimension at least $n$ for dimension reasons, and one can build $P(z)$ and $Q(z)$ out of the entries of a $n\times 2nd$ matrix in this nullspace $[P_0\;P_1\;\dots\;P_d\;Q_0\;Q_1\;\dots\;Q_d]$). -As Gerry Myerson noted in the comments, $P=Q=0$ is ok if we do not impose any additional constraint. The proof above shows how to compute two polynomial such that $[P_0\;P_1\;\dots\;P_d\;Q_0\;Q_1\;\dots\;Q_d]$ is full-rank. We could ask for that as a normalizing condition. In the generic case, the left kernel of the Sylvester matrix has rank $n$, so with this constraint $P$ and $Q$ are well-defined up to left multiplication by a nonsingular matrix. - -1) is there a simple and efficient way to compute $P(z)$ and $Q(z)$ by exploiting the structure of the matrix? - -Currently my best shot would be Fourier transforms+a fast Cauchy-like solver like the GKO algorithm, for a total cost of $O(n^3d^2)$ and a rather cumbersome algorithm (I know about superfast solvers, but they seems to be unstable and convenient only for large input sizes). But this seems excessively complicated and costly. For the scalar case, one can block-triangularize the resultant matrix by running the Euclidean algorithm, but in the matrix case this does not work, as a nonzero leading coefficient need not be invertible. -Maybe there is an easy FFT-based solution (with cost $O(n^3d\log^{something}d)$) that I am missing? - -2) can we say more about the properties of $P(z)$ and $Q(z)$, for example that their determinants do not vanish for any $z$ whenever those of $A(z)$ and $B(z)$ don't? - -REPLY [2 votes]: In case anyone is interested, I have found a counterexample to (2) by delving into some old papers: it's in Gohberg, I.; Kaashoek, M. A.; Lerer, L.; Rodman, L. Common multiples and common divisors of matrix polynomials. II. Vandermonde and resultant matrices. Linear and Multilinear Algebra 12 (1982/83), no. 3, 159–203. -They show a pair of degree-1 matrix polynomials of size $n$ whose least common left (or right) multiple has degree $[n/2]+1$. If we take them as $A(z)$ and $B(z)$, then all nonsingular polynomials $C(z)$ and $S(z)$ satisfying the condition have degree at least $[n/2]$, instead of 1 as I hoped. -The paper seems to be an older version of a chapter in the Matrix polynomials book by Gohberg, Lancaster and Rodman, but the counterexample does not appear in the book.<|endoftext|> -TITLE: Modular Curves as Moduli Spaces of Elliptic Curves -QUESTION [8 upvotes]: Hi, -Is the modular curve defined as the quotient of the upper half-plane by an arithmetic group $ \Gamma $ always a moduli space of elliptic curves with extra structure? I know this is true for $ \Gamma_0(N), \Gamma_1(N), \Gamma(N) $, but I'm interested in some of other groups, particularly those of the form $ \Gamma_0(N|d) $ (notation from Conway and Norton). Are these curves moduli spaces, and if so for what classes of varieties? -Thanks! - -REPLY [11 votes]: For the groups $\Gamma$ in Conway-Norton, there is always a moduli problem of $\Gamma$-structures, but since the groups always contain $\Gamma_0(N)$ for some $N$, you won't be able to construct a universal family (because there is a $-1$ automorphism in the way). However, you will sometimes get a ``relatively representable'' problem in the sense of Katz-Mazur. -The upper half-plane quotients will be coarse spaces parametrizing objects of the following general form: You have a diagram of elliptic curves, with some isogenies of specified degrees between them, together with some data that tell you how much symmetry in the diagram you should remember. Since all of the groups normalize $\Gamma_0(N)$ for some $N$, the diagrams will typically involve cyclic isogenies of degree $N$ in some way, and the symmetrization will involve a subgroup of the finite quotient $N_{SL_2(\mathbb{R})}(\Gamma_0(N))/\Gamma_0(N)$. -The standard example is $\Gamma_0(p)^+$ for a prime $p$, which is generated by $\Gamma_0(p)$ as an index two subgroup, together with the Fricke involution $\tau \mapsto \frac{-1}{p\tau}$. The $\Gamma_0(p)$ quotient parametrizes diagrams $E \to E'$ of elliptic curves equipped with a degree $p$ isogeny between them. Taking the quotient of the moduli problem by the Fricke involution amounts to symmetrizing the diagram, so the $\Gamma_0(p)^+$ quotient parametrizes tuples $( \{ E_1, E_2 \}, E_1 \leftrightarrows E_2)$ of unordered pairs of elliptic curves, with dual degree $p$ isogenies between them. Equivalently, you can ask for a set of diagrams $\{E_1 \to E_2, E_2 \to E_1 \}$ where the maps are dual isogenies. -A less well-known example is the 3C group, which is an index 3 subgroup of $\Gamma_0(3|3)$, with Hauptmodul $\sqrt[3]{j(3\tau)} = q^{-1} + 248q^2 + 4124q^5 + \dots$. This group is labeled $\Gamma_0(3|3)$ in the Conway-Norton paper, because $\Gamma_0(3|3)$ is the eigengroup, namely the group that takes the Hauptmodul to constant multiples of itself. The 3C group contains $\Gamma_0(9)$ as a normal subgroup, with quotient $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. You can view the upper half-plane quotient as a parameter space of quadruples of elliptic curves, with a rather complicated system of cyclic 9-isogenies and correspondences that get symmetrized (more on this in the last paragraph). A more succinct expression follows from using the matrix $\binom{30}{01}$ to conjugate $\Gamma_0(3|3)$ to $\Gamma(1)$ and $\Gamma_0(9)$ to $\pm \Gamma(3)$. Then you're basically looking at a moduli problem that parametrizes elliptic curves $E$ equipped with an unordered octuple of symplectic isomorphisms $E[3] \cong (\mathbb{Z}/3\mathbb{Z})^2$ that form a torsor under the characteristic 2-Sylow subgroup $Q_8 \subset Sp_2(\mathbb{F}_3) \cong SL_2(\mathbb{Z})/\Gamma(3)$. -In general, you can encode moduli problems attached to arithmetic groups using the fact that congruence groups like $\Gamma(N)$ and $\Gamma_0(N)$ stabilize distinguished finite subcomplexes of the product of all $p$-adic Bruhat-Tits trees. Conway gives a explanation (that doesn't use the word "moduli") with pictures in his paper Understanding groups like $\Gamma_0(N)$. For example, when $N$ is a product of $k$ distinct primes, $\Gamma_0(N)$ stabilizes a $k$-cube. Given a finite stable subcomplex, there is a standard way to make a moduli problem out of it by assigning elliptic curves to the vertices, isogenies to the edges, such that the induced transformations on the Tate module behave as you would expect from traversing the product of buildings. To symmetrize, just enumerate orbits of the transformations you want, and demand a torsor structure. -In the case of the 3C group in the above paragraph, $\Gamma_0(9)$ pointwise stabilizes a subgraph of the 3-adic tree that is an X-shaped configuration spanned by 5 vertices. The edges coming out of the central vertex are in noncanonical bijection with points in $\mathbb{P}^1(\mathbb{F}_3)$, and to symmetrize, you can make an unordered 4-tuple of diagrams of 5 elliptic curves, related by the action of the subgroup $V_4 \subset PSL_2(\mathbb{F}_3)$ that preserves the cross-ratio.<|endoftext|> -TITLE: "incidental" intersections of a complete graph in the plane -QUESTION [10 upvotes]: Given a complete graph of n vertices (no three of which are no collinear) in the plane and straight edges, what is the maximal possible number of "incidental intersections" of edges, i.e., number of non-vertices at which two distinct edges intersect each other, not counting multiplicity? -This is a question that I pose to the students in my Mathematics for Elementary School Teachers as a way to understand mathematical conjecturing and proving -- and not always finding the solution. But it occurs to me that it might be handy to know whether the answer is actually known or not. - -REPLY [3 votes]: This problem (although phrased slightly differently) is #1.3.5 in Loren C. Larson's "Problem Solving Through Problems." -The rephrase of this is, "On a circle, n points are selected and the chords joining them in pairs are drawn. Assuming no three of the chords are concurrent (except at the endpoints), how many points of intersection are there?" -This is $\binom{n}{4}$, as any four of our points on the circle determine a (unique) intersection point, and any intersection point determines the boundary chords. -[This might've been better as a comment on Gerry's answer, but there's a whole family of similar problems to the question asker's in that chapter of the book (which are also excellent problems to give to a math club.)]<|endoftext|> -TITLE: What is known about the ultra-inverse limit? -QUESTION [5 upvotes]: Given a nonprincipal ultrafilter $\mu$ on $\mathbb{N}$ and a sequence of groups $G_i$, one can define its ultraproduct as: -$$ ^*\prod_{i\in \mathbb{N}}G_i:=\{(x_i)_{i \in \mathbb{N}}| x_i\in G_i\}/\sim$$, where $(x_i)_i\sim (y_i)_i$, iff $x_i=y_i$ $\mu$-almost everywhere. -Suppose you are given also group homomorphisms $f_{i+1}:G_{i+1}\rightarrow G_i$, then one could also consider something like an ultra- inverse limit: -$$\{[x_i]_i|f_i(x_i)=x_{i-1} \quad \mu-\mbox{almost everywhere}\}$$ -Has this been studied before? Is there a good source, where I could learn about this? - -REPLY [7 votes]: Either the even numbers or the odd numbers are $\mu$-large, so the condition degenerates to ultraproduct.<|endoftext|> -TITLE: Basic properties of Neron models -QUESTION [9 upvotes]: Let $R$ be a dvr with residue field $k$ and quotient field $K$. Define $S=Spec(R)$. Let -$A/K$ be an abelian variety. To my knowledge the Neron model of $A$ is a group scheme -${\cal N}/S$ with generic fibre $A$, which represents the functor -$$Y\mapsto Mor_K(Y\times_S Spec(K), A)$$ -on the category of smooth $S$-schemes. The morphism ${\cal N}\to S$ is smooth, in -particular it is flat and locally of finite type. -Question 1. Is it true that ${\cal N}\to S$ is of finite type? -I know that the special fibre ${\cal N}\times_S Spec(k)$ is in general not connected and -that the component group of the special fibre is an important invariant. -But what about the scheme ${\cal N}$ itself? -Question 2. Is it true that ${\cal N}$ is connected? -I strongly assumed that the answer to question 2 is "yes". (My reason to believe this: -Assume ${\cal N}$ is not connected. Then ${\cal N}=U\cup V$ for nonempty disjoint -open subsets $U$ and $V$ of ${\cal N}$. Then $A=U_K\cup V_K$ and $U_K$, $V_K$ are -disjoint open subsets of $A$. Furthermore $U\to S$, $V\to S$ are flat morphisms, hence -$U_K$ and $V_K$ are non-empty. This is a contradiction, because $A$ is connected.) -However I saw in the book of Bosch-Lutkebohmert-Raynaud examples of non-connected Neron -models. And I saw in Deligne's articles on Hodge theory the expression "connected -Neron model of $A$" (as opposed to "Neron model of $A$"). Hence I am very puzzled... -(I have to admit that I did not yet go through the construction of a -representing object of the functor above. That is probably the reason why -I cannot help myself at the moment.) - -REPLY [13 votes]: 1) Yes, that is part of Neron's theorem. (There also exist Neron models for semiabelian varieties, which are not of finite type in general.) -2) Yes, for the reason you give ($A$ is connected). When people say "connected component of the Neron model" they generally mean the open subgroup scheme which is in every fibre the connected component of the identity. -An alternative to BLR is the article "Neron models" by M. Artin in "Arithmetic Geometry" (ed. Cornell-Silverman). What you are asking is explained in the first section of that reference (look at 1.2 and 1.16).<|endoftext|> -TITLE: How can I find the average of two 2D curves? -QUESTION [10 upvotes]: I have a curve interpolation problem. -I have two closed curves that are defined on an X,Y plane. How can I define a 3rd curve that is the average of those two? Programmatically, I have a list of points for each curve, let's say N1 for curve 1 and N2 for curve 2, where N1 != N2 (most likely). -When I say 'average', initially, I would like the contribution of each curve to the final curve to be identical. Eventually, I would like to be able to weight the contributions of each curve (i.e., have my interpolated curve be 'closer' to one curve than another). -How can I go about doing this? -In a 1D case, I believe that the problem is somewhat easier, like I could solve it using some kind of projection (although my linear algebra is really rusty at this point). Is that intuition somewhat correct, and therefore can be extended to the 2D case? - -REPLY [2 votes]: Excellent question! The answer was not known last year. -The following recent paper may have the answer you need: -"Average curve of n smooth planar curves" by -Sati, Rossignac, Seidel, Wyvill, Musuvathy. -Computer-Aided Design. -Volume 70 Issue C, January 2016, -Pages 46-55. -http://dl.acm.org/citation.cfm?id=2831700 -It presents a fast, linear cost, construction that works very well when the curves are "compatible" (reasonably parallel). The paper also talks about how to use weighted averaging for blending and morphing between several curves. -The solution is independent of the order of the input curves and does not assume any given parameterization (like the timing that one of the comments mentions). - -Jarek<|endoftext|> -TITLE: Injectivity for bimodules and Hochschild cohomology -QUESTION [8 upvotes]: Let $A$ be a Banach algebra and let $X$ be an $A$-bimodule. Is there a notion of (relative) injectivity for $X$ which would imply that $\mathcal{H}^n(A,X)$ vanishes for all $n\ge 1$? Here $\mathcal{H}^n(A,X)$ denotes the continuous Hochschild cohomology of $A$ with coefficients in $X$ ($X$ can be assumed to be dual $A$-bimodule). -I expected there is such a notion, but after reading books by Helemskii, Runde and a few other sources on cohomology of Banach algebras I can't seem to find a general statement of this type, even though versions of projectivity and injectivity are discussed there. - -REPLY [4 votes]: A couple of people have encouraged me to post this as an answer, so here goes. I am currently without a copy of Helemskii's Pink Book so I can't give chapter-and-verse references as I would have liked. Everything that follows should be somewhere in there, although perhaps expressed slightly differently, and probably slightly better. Certainly what follows is too wordy, but I haven't had time to work out a condensed version. -To recap: Piotr is asking about the continuous Hochschild cohomology groups ${\mathcal H}^n(A,X)$ where $A$ is a Banach algebra and $X$ a Banach $A$-bimodule. To simplify the discussion slightly, I shall assume that $A$ has an identity element (which is indeed the case if $A$ is one of the usual convolution-type algebras associated to a discrete group) and that $X$ is unit-linked, i.e. that the identity of $A$ acts as the identity operator on $X$. -Conceptual/abstract POV (Helemskian) -As in the classical theory of Cartan-Eilenberg vintage, (continuous) Hochschild cohomology can be expressed in terms of relative Ext. One way to approach this, as Helemskii does, is to introduce the enveloping algebra $A^e$ of a unital Banach algebra $A$. -This has underlying Banach space $A\hat{\otimes} A$ (projective tensor product) and has multiplication defined by $(a\otimes b)\cdot (c\otimes d) = (ab\otimes dc)$. -(The definition is slightly different for the non-unital case, and the artificial dichotomy that arises in places in the Pink Book is something that vexes some of us. But I digress...) -The purpose of doing this is as follows: every Banach $A$-bimodule $X$ becomes a left Banach $A^e$-module via $(a\otimes b)\cdot x = axb$; and conversely, every left Banach $A^e$-module becomes a Banach $A$-bimodule via the same formula. Now, taking as read the definition of relative Ext that is given in the Pink Book, we have -$$ {\mathcal H}^n(A,X) \cong \operatorname{Ext}_{A^e}^n (A,X) $$ -This is an isomorphism of seminormed spaces for each $n$ (I guess it would be more precise to say an isomorphism of seminormed-space-valued $\delta$-functors or some such high-falutin' phrase) -Now, recall that if $B$ is a Banach algebra, then a left Banach $B$-module $X$ is said to be (relatively) $B$-injective if it satisfies the following: -whenever $N$ is a left Banach $B$-module and $M$ is a closed $B$-submodule of $N$ which is complemented as a Banach subspace, then each continuous linear $B$-module map $M\to X$ has a continuous linear extension to a $B$-module map $N\to X$. -Moreover, if $X$ is relatively $B$-injective then $\operatorname{Ext}_B^n(\cdot,X)=0$ for each $n\geq 1$. (The converse also holds, in fact.) Therefore: - -if $X$ is relatively $A^e$-injective, then ${\mathcal H}^n(A,X)=0$ for all $n\geq 1$. - -The get-your-hands-dirty approach (Johnsonite) -We continue to suppose that $A$ has an identity element. Now let $E$ be any Banach space and equip $V_E :={\mathcal L}(A\hat\otimes A, E)$ with the following natural $A$-bimodule structure: -$$ (b\cdot T \cdot a)(c\otimes d) = T(ac\otimes db) \quad\quad(T\in V_E). $$ -Claim: ${\mathcal H}^n(A,V_E)=0$ for all $n\geq 1$. -This is most easily proved by proving something stronger: -Exercise: Let $\delta: {\mathcal C}^n(A,V_E)\to {\mathcal C}^{n+1}(A,V_E)$ denote the Hochschild coboundary operator. Define $\sigma: {\mathcal C}^{n+1}(A,V_E) \to {\mathcal C}^n(A,V_E)$ by -$$ [\sigma\psi(a_1,\dots,a_n)]{(c\otimes d)} = [\psi(d,a_1,\dots, a_n)]{(c\otimes 1)}. $$ -Then $\delta\sigma(\psi)+\sigma\delta(\psi) =\psi$ for every $\psi\in\mathcal C^k(A,V_E)$. -We now observe the following: if $V$ is any Banach $A$-bimodule, and it can be written as $V\cong X\oplus Y$ where $X$ and $Y$ are closed $A$-sub-bimodules of $V$, then -$$ {\mathcal H}^n(A,V) \cong {\mathcal H}^n(A,X) \oplus {\mathcal H}^n(A,Y) \quad\hbox{for all $n$.} $$ -One can check this directly or appeal to the long exact sequence of Hochschild cohomology (which is a special case of the one for relative Ext). -Finally, for each Banach $A$-bimodule $X$ there is a canonical $A$-bimodule map $J:X\to V_X$ which is defined by $[J(x)](c\otimes d) = dxc$. Therefore: - -If there exists an $A$-bimodule map $P:V_X\to X$ such that $PJ$ is the identity, then ${\mathcal H}^n(A, X)=0$ for all $n\geq 1$. - -Clowns to the left of me, jokers to the right -As may be apparent to anyone who's read this far: the two conditions we have obtained on $X$, each of which implies that Hochschild cohomology with coefficients in $X$ vanishes, are one and the same condition. [The calculations in the second version actually show that $V_E$ is $A$-bi-injective - meaning the same as $A^e$-injective. This relied on $A$ having an identity element! Then, knowing that a complemented submodule of an injective module is injective, we see that the second condition implies $X$ is $A$-bi-injective.] The nice thing about the direct approach is that it gives one explicit formulas one can try even in settings where the coefficient module is not bi-injective (see for instance my first excuse for a paper ). Personally I think it is good to have both points of view. -It should lastly be noted that almost none of the above actually used analysis - everything is taken care of by working in a particular category with a particular tensor product. So what I have just written out is no more than was known at the time of Cartan-Eilenberg.<|endoftext|> -TITLE: Is there a list of all connected T_0-spaces with 5 points? -QUESTION [9 upvotes]: Is there some place (on the internet or elsewhere) where I can find the number and preferably a list of all (isomorphism classes of) finite connected $T_0$-spaces with, say, 5 points? -In know that a $T_0$-topology on a finite set is equivalent to a partial ordering, and wikipedia tells me that there are, up to isomorphism, 63 partially ordered sets with precisely 5 elements. However, I am only interested in connected spaces, and I'd love to have a list (most preferably in terms of Hasse diagrams). - -REPLY [11 votes]: Peter Jipsen has granted me permission to post static images of his posets. -4-element connected posets (10): - -5-element connected posets (44): - -Since the number (139) of distinct 5-element topological spaces (up to homeomorphism) is manageably small -and there don't seem to be any lists posted anywhere, I am posting one here. The first (second) row lists smallest closed (open) supersets of singletons and the rows that follow list all nontrivial open sets. -1 T_0 - a b c d e - ----- - a b c d e - ----- - a b c d e - ab ac ad ae bc bd be cd ce de - abc abd abe acd ace ade bcd bce bde cde - abcd abce abde acde bcde - -2 T_0 - ae b c d e - ----- - a b c d ae - ----- - a b c d - ab ac ad ae bc bd cd - abc abd abe acd ace ade bcd - abcd abce abde acde - -3 T_0 - ae be c d e - ----- - a b c d abe - ----- - a b c d - ab ac ad bc bd cd - abc abd abe acd bcd - abcd abce abde - -4 T_0 - ae be ce d e - ----- - a b c d abce - ----- - a b c d - ab ac ad bc bd cd - abc abd acd bcd - abcd abce - -5 connected T_0, poset (4,2) - ae be ce de e - ----- - a b c d abcde - ----- - a b c d - ab ac ad bc bd cd - abc abd acd bcd - abcd - -6 T_0 - ade b c d e - ----- - a b c ad ae - ----- - a b c - ab ac ad ae bc - abc abd abe acd ace ade - abcd abce abde acde - -7 T_0 - ad be c d e - ----- - a b c ad be - ----- - a b c - ab ac ad bc be - abc abd abe acd bce - abcd abce abde - -8 T_0 - ade be c d e - ----- - a b c ad abe - ----- - a b c - ab ac ad bc - abc abd abe acd - abcd abce abde - -9 T_0 - ade b c de e - ----- - a b c ad ade - ----- - a b c - ab ac ad bc - abc abd acd ade - abcd abde acde - -10 - a b c de de - ----- - a b c de de - ----- - a b c - ab ac bc de - abc ade bde cde - abde acde bcde - -11 T_0 - ad be ce d e - ----- - a b c ad bce - ----- - a b c - ab ac ad bc - abc abd acd bce - abcd abce - -12 connected T_0, poset (1,9) - ade be ce d e - ----- - a b c ad abce - ----- - a b c - ab ac ad bc - abc abd acd - abcd abce - -13 T_0 - ade be c de e - ----- - a b c ad abde - ----- - a b c - ab ac ad bc - abc abd acd - abcd abde - -14 connected T_0, poset (4,3) - ade be ce de e - ----- - a b c ad abcde - ----- - a b c - ab ac ad bc - abc abd acd - abcd - -15 T_0 - ade bde c d e - ----- - a b c abd abe - ----- - a b c - ab ac bc - abc abd abe - abcd abce abde - -16 connected T_0, poset (2,6) - ade bd ce d e - ----- - a b c abd ace - ----- - a b c - ab ac bc - abc abd ace - abcd abce - -17 connected T_0, poset (2,9) - ade bde ce d e - ----- - a b c abd abce - ----- - a b c - ab ac bc - abc abd - abcd abce - -18 T_0 - ade bde c de e - ----- - a b c abd abde - ----- - a b c - ab ac bc - abc abd - abcd abde - -19 - ade b c de de - ----- - a b c ade ade - ----- - a b c - ab ac bc - abc ade - abde acde - -20 connected T_0, poset (4,6) - ade bde ce de e - ----- - a b c abd abcde - ----- - a b c - ab ac bc - abc abd - abcd - -21 connected T_0, poset (3,6) - ade bde cde d e - ----- - a b c abcd abce - ----- - a b c - ab ac bc - abc - abcd abce - -22 connected T_0, poset (5,4) - ade bde cde de e - ----- - a b c abcd abcde - ----- - a b c - ab ac bc - abc - abcd - -23 - ade bde c de de - ----- - a b c abde abde - ----- - a b c - ab ac bc - abc - abde - -24 connected - ade bde cde de de - ----- - a b c abcde abcde - ----- - a b c - ab ac bc - abc - -25 T_0 - acde b c d e - ----- - a b ac ad ae - ----- - a b - ab ac ad ae - abc abd abe acd ace ade - abcd abce abde acde - -26 T_0 - acd be c d e - ----- - a b ac ad be - ----- - a b - ab ac ad be - abc abd abe acd - abcd abce abde - -27 connected T_0, poset (1,2) - acde be c d e - ----- - a b ac ad abe - ----- - a b - ab ac ad - abc abd abe acd - abcd abce abde - -28 T_0 - acde b ce d e - ----- - a b ac ad ace - ----- - a b - ab ac ad - abc abd acd ace - abcd abce acde - -29 connected T_0, poset (1,4) - ace bde c d e - ----- - a b ac bd abe - ----- - a b - ab ac bd - abc abd abe - abcd abce abde - -30 connected T_0, poset (2,1) - acde be ce d e - ----- - a b ac ad abce - ----- - a b - ab ac ad - abc abd acd - abcd abce - -31 T_0 - acde b ce de e - ----- - a b ac ad acde - ----- - a b - ab ac ad - abc abd acd - abcd acde - -32 T_0 - ace bd ce d e - ----- - a b ac bd ace - ----- - a b - ab ac bd - abc abd ace - abcd abce - -33 - ac b c de de - ----- - a b ac de de - ----- - a b - ab ac de - abc ade bde - abde acde - -34 connected T_0, poset (4,4) - acde be ce de e - ----- - a b ac ad abcde - ----- - a b - ab ac ad - abc abd acd - abcd - -35 connected T_0, poset (2,3) - ace bde ce d e - ----- - a b ac bd abce - ----- - a b - ab ac bd - abc abd - abcd abce - -36 connected T_0, poset (4,8) - ace bde ce de e - ----- - a b ac bd abcde - ----- - a b - ab ac bd - abc abd - abcd - -37 connected T_0, poset (1,5) - acde bde c d e - ----- - a b ac abd abe - ----- - a b - ab ac - abc abd abe - abcd abce abde - -38 T_0 - acde b cde d e - ----- - a b ac acd ace - ----- - a b - ab ac - abc acd ace - abcd abce acde - -39 - a b cde cde e - ----- - a b cd cd cde - ----- - a b - ab cd - acd bcd cde - abcd acde bcde - -40 connected T_0, poset (2,7) - acde bd ce d e - ----- - a b ac abd ace - ----- - a b - ab ac - abc abd ace - abcd abce - -41 - ac bc c de de - ----- - a b abc de de - ----- - a b - ab de - abc ade bde - abde - -42 connected T_0, poset (3,2) - acde bde ce d e - ----- - a b ac abd abce - ----- - a b - ab ac - abc abd - abcd abce - -43 connected T_0, poset (2,4) - acde bde c de e - ----- - a b ac abd abde - ----- - a b - ab ac - abc abd - abcd abde - -44 connected T_0, poset (3,1) - acde be cde d e - ----- - a b ac acd abce - ----- - a b - ab ac - abc acd - abcd abce - -45 T_0 - acde b cde de e - ----- - a b ac acd acde - ----- - a b - ab ac - abc acd - abcd acde - -46 - acde b c de de - ----- - a b ac ade ade - ----- - a b - ab ac - abc ade - abde acde - -47 - ae b cde cde e - ----- - a b cd cd acde - ----- - a b - ab cd - acd bcd - abcd acde - -48 connected T_0, poset (4,7) - acde bde ce de e - ----- - a b ac abd abcde - ----- - a b - ab ac - abc abd - abcd - -49 connected T_0, poset (4,9) - acde be cde de e - ----- - a b ac acd abcde - ----- - a b - ab ac - abc acd - abcd - -50 - ac bde c de de - ----- - a b ac bde bde - ----- - a b - ab ac - abc bde - abde - -51 connected - ae be cde cde e - ----- - a b cd cd abcde - ----- - a b - ab cd - acd bcd - abcd - -52 connected T_0, poset (3,7) - acde bde cde d e - ----- - a b ac abcd abce - ----- - a b - ab ac - abc - abcd abce - -53 connected T_0, poset (5,5) - acde bde cde de e - ----- - a b ac abcd abcde - ----- - a b - ab ac - abc - abcd - -54 connected - acde bde c de de - ----- - a b ac abde abde - ----- - a b - ab ac - abc - abde - -55 - acde b cde de de - ----- - a b ac acde acde - ----- - a b - ab ac - abc - acde - -56 connected - acde bde cde de de - ----- - a b ac abcde abcde - ----- - a b - ab ac - abc - -57 connected T_0, poset (1,7) - acde bcde c d e - ----- - a b abc abd abe - ----- - a b - ab - abc abd abe - abcd abce abde - -58 connected T_0, poset (3,4) - acde bcde ce d e - ----- - a b abc abd abce - ----- - a b - ab - abc abd - abcd abce - -59 connected T_0, poset (5,2) - acde bcde ce de e - ----- - a b abc abd abcde - ----- - a b - ab - abc abd - abcd - -60 connected - acde bc c de de - ----- - a b abc ade ade - ----- - a b - ab - abc ade - abde - -61 connected T_0, poset (3,9) - acde bcde cde d e - ----- - a b abc abcd abce - ----- - a b - ab - abc - abcd abce - -62 - acde b cde cde e - ----- - a b acd acd acde - ----- - a b - ab - acd - abcd acde - -63 - a b cde cde cde - ----- - a b cde cde cde - ----- - a b - ab - cde - acde bcde - -64 connected T_0, poset (5,7) - acde bcde cde de e - ----- - a b abc abcd abcde - ----- - a b - ab - abc - abcd - -65 connected - acde bcde c de de - ----- - a b abc abde abde - ----- - a b - ab - abc - abde - -66 connected - acde be cde cde e - ----- - a b acd acd abcde - ----- - a b - ab - acd - abcd - -67 connected - acde bcde cde de de - ----- - a b abc abcde abcde - ----- - a b - ab - abc - -68 connected - acde bcde cde cde e - ----- - a b abcd abcd abcde - ----- - a b - ab - abcd - -69 - acde b cde cde cde - ----- - a b acde acde acde - ----- - a b - ab - acde - -70 connected - acde bcde cde cde cde - ----- - a b abcde abcde abcde - ----- - a b - ab - -71 connected T_0, poset (1,1) - abcde b c d e - ----- - a ab ac ad ae - ----- - a - ab ac ad ae - abc abd abe acd ace ade - abcd abce abde acde - -72 connected T_0, poset (1,3) - abcde be c d e - ----- - a ab ac ad abe - ----- - a - ab ac ad - abc abd abe acd - abcd abce abde - -73 connected T_0, poset (2,2) - abcde be ce d e - ----- - a ab ac ad abce - ----- - a - ab ac ad - abc abd acd - abcd abce - -74 connected T_0, poset (4,5) - abcde be ce de e - ----- - a ab ac ad abcde - ----- - a - ab ac ad - abc abd acd - abcd - -75 - abc b c de de - ----- - a ab ac de de - ----- - a - ab ac de - abc ade - abde acde - -76 connected T_0, poset (1,6) - abcde bde c d e - ----- - a ab ac abd abe - ----- - a - ab ac - abc abd abe - abcd abce abde - -77 connected T_0, poset (2,8) - abcde bd ce d e - ----- - a ab ac abd ace - ----- - a - ab ac - abc abd ace - abcd abce - -78 connected T_0, poset (3,3) - abcde bde ce d e - ----- - a ab ac abd abce - ----- - a - ab ac - abc abd - abcd abce - -79 connected T_0, poset (2,5) - abcde bde c de e - ----- - a ab ac abd abde - ----- - a - ab ac - abc abd - abcd abde - -80 connected - abcde b c de de - ----- - a ab ac ade ade - ----- - a - ab ac - abc ade - abde acde - -81 - ab b cde cde e - ----- - a ab cd cd cde - ----- - a - ab cd - acd cde - abcd acde - -82 connected T_0, poset (5,1) - abcde bde ce de e - ----- - a ab ac abd abcde - ----- - a - ab ac - abc abd - abcd - -83 - abc bc c de de - ----- - a ab abc de de - ----- - a - ab de - abc ade - abde - -84 - a bc bc de de - ----- - a bc bc de de - ----- - a - bc de - abc ade - bcde - -85 connected T_0, poset (3,8) - abcde bde cde d e - ----- - a ab ac abcd abce - ----- - a - ab ac - abc - abcd abce - -86 connected - abe b cde cde e - ----- - a ab cd cd acde - ----- - a - ab cd - acd - abcd acde - -87 connected T_0, poset (5,6) - abcde bde cde de e - ----- - a ab ac abcd abcde - ----- - a - ab ac - abc - abcd - -88 connected - abcde bde c de de - ----- - a ab ac abde abde - ----- - a - ab ac - abc - abde - -89 connected - abe be cde cde e - ----- - a ab cd cd abcde - ----- - a - ab cd - acd - abcd - -90 connected - abcde bde cde de de - ----- - a ab ac abcde abcde - ----- - a - ab ac - abc - -91 connected T_0, poset (1,8) - abcde bcde c d e - ----- - a ab abc abd abe - ----- - a - ab - abc abd abe - abcd abce abde - -92 - a bcde bcde d e - ----- - a bc bc bcd bce - ----- - a - bc - abc bcd bce - abcd abce bcde - -93 connected T_0, poset (3,5) - abcde bcde ce d e - ----- - a ab abc abd abce - ----- - a - ab - abc abd - abcd abce - -94 connected - ae bcde bcde d e - ----- - a bc bc bcd abce - ----- - a - bc - abc bcd - abcd abce - -95 - a bcde bcde de e - ----- - a bc bc bcd bcde - ----- - a - bc - abc bcd - abcd bcde - -96 connected T_0, poset (5,3) - abcde bcde ce de e - ----- - a ab abc abd abcde - ----- - a - ab - abc abd - abcd - -97 connected - abcde bc c de de - ----- - a ab abc ade ade - ----- - a - ab - abc ade - abde - -98 connected - ae bcde bcde de e - ----- - a bc bc bcd abcde - ----- - a - bc - abc bcd - abcd - -99 - ade bc bc de de - ----- - a bc bc ade ade - ----- - a - bc - abc ade - -100 connected T_0, poset (4,1) - abcde bcde cde d e - ----- - a ab abc abcd abce - ----- - a - ab - abc - abcd abce - -101 connected - abcde b cde cde e - ----- - a ab acd acd acde - ----- - a - ab - acd - abcd acde - -102 connected - ade bcde bcde d e - ----- - a bc bc abcd abce - ----- - a - bc - abc - abcd abce - -103 connected T_0, poset (5,8) - abcde bcde cde de e - ----- - a ab abc abcd abcde - ----- - a - ab - abc - abcd - -104 connected - abcde bcde c de de - ----- - a ab abc abde abde - ----- - a - ab - abc - abde - -105 connected - abcde be cde cde e - ----- - a ab acd acd abcde - ----- - a - ab - acd - abcd - -106 - ab b cde cde cde - ----- - a ab cde cde cde - ----- - a - ab - cde - acde - -107 connected - ade bcde bcde de e - ----- - a bc bc abcd abcde - ----- - a - bc - abc - abcd - -108 - a bcde bcde de de - ----- - a bc bc bcde bcde - ----- - a - bc - abc - bcde - -109 connected - abcde bcde cde de de - ----- - a ab abc abcde abcde - ----- - a - ab - abc - -110 connected - ade bcde bcde de de - ----- - a bc bc abcde abcde - ----- - a - bc - abc - -111 connected - abcde bcde cde cde e - ----- - a ab abcd abcd abcde - ----- - a - ab - abcd - -112 connected - abcde b cde cde cde - ----- - a ab acde acde acde - ----- - a - ab - acde - -113 connected - abcde bcde cde cde cde - ----- - a ab abcde abcde abcde - ----- - a - ab - -114 connected - abcde bc bc de de - ----- - a abc abc ade ade - ----- - a - abc ade - -115 connected - abcde bcde bcde d e - ----- - a abc abc abcd abce - ----- - a - abc - abcd abce - -116 - a bcde bcde bcde e - ----- - a bcd bcd bcd bcde - ----- - a - bcd - abcd bcde - -117 connected - abcde bcde bcde de e - ----- - a abc abc abcd abcde - ----- - a - abc - abcd - -118 connected - ae bcde bcde bcde e - ----- - a bcd bcd bcd abcde - ----- - a - bcd - abcd - -119 connected - abcde bcde bcde de de - ----- - a abc abc abcde abcde - ----- - a - abc - -120 connected - abcde bcde bcde bcde e - ----- - a abcd abcd abcd abcde - ----- - a - abcd - -121 - a bcde bcde bcde bcde - ----- - a bcde bcde bcde bcde - ----- - a - bcde - -122 connected - abcde bcde bcde bcde bcde - ----- - a abcde abcde abcde abcde - ----- - a - -123 - abc abc c de de - ----- - ab ab abc de de - ----- - ab de - abc - abde - -124 connected - abe abe cde cde e - ----- - ab ab cd cd abcde - ----- - ab cd - abcd - -125 connected - abcde abcde c d e - ----- - ab ab abc abd abe - ----- - ab - abc abd abe - abcd abce abde - -126 connected - abcde abcde ce d e - ----- - ab ab abc abd abce - ----- - ab - abc abd - abcd abce - -127 connected - abcde abcde ce de e - ----- - ab ab abc abd abcde - ----- - ab - abc abd - abcd - -128 connected - abcde abcde cde d e - ----- - ab ab abc abcd abce - ----- - ab - abc - abcd abce - -129 connected - abcde abcde cde de e - ----- - ab ab abc abcd abcde - ----- - ab - abc - abcd - -130 connected - abcde abcde c de de - ----- - ab ab abc abde abde - ----- - ab - abc - abde - -131 connected - abcde abcde cde de de - ----- - ab ab abc abcde abcde - ----- - ab - abc - -132 - ab ab cde cde cde - ----- - ab ab cde cde cde - ----- - ab - cde - -133 connected - abcde abcde cde cde e - ----- - ab ab abcd abcd abcde - ----- - ab - abcd - -134 connected - abcde abcde cde cde cde - ----- - ab ab abcde abcde abcde - ----- - ab - -135 connected - abcde abcde abcde d e - ----- - abc abc abc abcd abce - ----- - abc - abcd abce - -136 connected - abcde abcde abcde de e - ----- - abc abc abc abcd abcde - ----- - abc - abcd - -137 connected - abcde abcde abcde de de - ----- - abc abc abc abcde abcde - ----- - abc - -138 connected - abcde abcde abcde abcde e - ----- - abcd abcd abcd abcd abcde - ----- - abcd - -139 connected - abcde abcde abcde abcde abcde - ----- - abcde abcde abcde abcde abcde - ----- - trivial topology<|endoftext|> -TITLE: Textbooks or notes on gradient flows in metric spaces -QUESTION [8 upvotes]: What is a good introduction in gradient flows in metric spaces? -I know the book Gradient flows: in metric spaces and in the space of probability measures by Luigi Ambrosio, Nicola Gigli and Giuseppe Savaré, but is too hard for an introduction (for me). -I'm looking for something with a similar content. - -REPLY [2 votes]: The book Topics in Optimal Transportation by Cédric Villani is not exactly on this topic but is very well written and contains a lot of related material good for background, motivation and applications. The book of Ambrosio, Gigli and Savaré is indeed pretty dry, but the results they established improved considerably on what was available in the literature. -The notes of Daneri and Savaré look good --- Savaré's presentations in a summer school this past June are available here.<|endoftext|> -TITLE: Hilbert Modular Newforms -QUESTION [8 upvotes]: Fix positive integers $k, N$ and let $\omega$ be a Dirichlet character mod $N$. -Let $f\in S_k(N,\phi)$ be a normalized newform (i.e. of weight $k$, level $N$ and character $\phi$) with fourier expansion $\sum_{n\geq 1} a(n)q^n$. In her paper 'Newforms and Functional Equations', Winnie Li showed that if $q$ is a prime dividing $N$ and $\phi$ is not a character mod $N/q$, then $|a(q)|=q^{\frac{k-1}{2}}$. In particular, $a(q)\neq 0$. This is Theorem 3.(ii) of the paper. I should also note that special cases of this were proven earlier Atkin-Lehner, Hecke and Ogg. -I am interested in the Hilbert modular version of this result. More specifically, let $\mathfrak{N}$ be an integral ideal of a totally real number field $K$ of degree $d$ over $\mathbb{Q}$, $\Phi$ be a Hecke character induced by a numerical character mod $\mathfrak{N}$ and $k\in (\mathbb{Z}_+)^d$. -Let $S_k(\mathfrak{N},\Phi)$ be the space of Hilbert modular cusp forms of weight $k$, level $\mathfrak{N}$ and character $\Phi$, viewed adelically to circumvent class number issues (I make no assumptions regarding the class number of $K$). Shimura, in 'The special values of the zeta functions associated to Hilbert modular forms' defined "Fourier coefficients" $C(\mathfrak{m},\textbf{f})$ for cusp forms $\textbf{f}\in S_k(\mathfrak{N},\Phi)$ (this on the bottom of page 649 of the paper). -Let $\textbf{f}\in S_k(\mathfrak{N},\Phi)$ be a normalized newform. I am interested in knowing when $C(\mathfrak{q},\textbf{f})=0$, where $\mathfrak{q}\mid\mathfrak{N}$. In their paper 'Twists of Hilbert Modular Forms', Tom Shemanske and Lynne Walling showed that if the numerical character inducing $\Phi$ is not defined mod $\mathfrak{N}/ \mathfrak{q}$, then this coefficient is nonzero whenever $\mathfrak{q}$ has degree 1 over the rationals or $\mathfrak{q}$ exactly divides $\mathfrak{N}$ (this is part 2 of Theorem 3.3 in the paper). In a remark directly after the Theorem, Tom and Lynne say that the reason for the restrictions on the prime $\mathfrak{q}$ is that this is what is needed to make Ogg's proof in the classical case go through. The 'truth', however, is that the $\mathfrak{q}$-th coefficient should always be nonzero in this case, without any restrictions on the degree of the prime $\mathfrak{q}$. A referee told Tom (who is my advisor) that the representation theory implies this stronger result. Tom lost the referee's report some time ago however, and doesn't know of any reference for this representation theoretic result. Does anyone know of a paper in which this result is proven without any restrictions on the prime $\mathfrak{q}$? -$\textbf{Question}:$ Let $\textbf{f}\in S_k(\mathfrak{N},\Phi)$ be a normalized newform, $\mathfrak{q}$ be a prime dividing $\mathfrak{N}$ and suppose that the numerical character inducing $\Phi$ is not defined mod $\mathfrak{N}\mathfrak{q}^{-1}$. Is it true that $C(\mathfrak{q},\textbf{f})\neq 0$? - -REPLY [4 votes]: I think I understand what Kevin is saying in the Theorem. Let $\mathfrak o$ be the ring of integers of $K$ and $\mathfrak p$ the maximal ideal. Define the following subgroups of -$G=GL(2, K)$: -$I= \begin{pmatrix}\mathfrak o^{\times} & \mathfrak o // - \mathfrak p & \mathfrak o \end{pmatrix}$ -$I_n=\begin{pmatrix} \mathfrak o^{\times} & \mathfrak o // \mathfrak p^n & 1+\mathfrak p^n\end{pmatrix}$ -$Z_n= \begin{pmatrix}1+\mathfrak p^n & 0 // 0 & 1+\mathfrak p^n\end{pmatrix}$ -Lemma. Let $\pi$ be a smooth representation of $I$ with a central character, such that $Z_n$ acts trivially, $Z_{n-1}$ does not act trivially and -the space of $I_n$-invariants is non-zero. Then the restriction of -$\pi$ to $I_n Z_0$ contains a one dimensional subrepresentation of the -form $\chi: \begin{pmatrix} a & b // c & d\end{pmatrix} \mapsto \chi_1(d)$, where -$\chi_1: \mathfrak o^{\times}\rightarrow \mathbb C^{\times}$ is a smooth character of conductor $\mathfrak p^n$. -Proof. Look at the action of the abelian group $Z_0$ on $\pi^{I_n}$. -The pair $(I_nZ_0, \chi)$ is a type for the Bernstein component, which contains the principal series representations that Kevin describes. In other words, if $\pi$ is an -irreducible smooth representation of $G$, then $Hom_{I_n Z_0}(\chi, \pi)\neq 0$ if and only if $\pi$ is a principal series rep with one character unramified, the other of conductor $\mathfrak p^n$. For this you could look either in the appendix by Heniart to: -http://www.math.u-psud.fr/~breuil/PUBLICATIONS/multiplicite.pdf -or in the book of Bushnell and Henniart. The main point being that the representation $\chi$ (as a representation of -$\begin{pmatrix} \mathfrak o^{\times} & 0 // 0 & \mathfrak o^{\times}\end{pmatrix}$) -shows up in the $U$-coinvariants of $\pi$, where $U$ is unipotent upper (lower?) triangular matrices.<|endoftext|> -TITLE: What is the lower bound for highly composite numbers? -QUESTION [10 upvotes]: if $x=d(n)$ is the number of divisors of $n$, what is the tightest lower-bound for $n$ only given $x$? -http://en.wikipedia.org/wiki/Highly_composite_number - -REPLY [2 votes]: It would be nice to have an inequality $n \ge f(x)$. If the poser wants numerical results, here are two: -The least number having exactly x divisors is given by OEIS sequence http://www.oeis.org/A005179. It is a pretty wild function. The nice paper by Grost is recommended. -The least number having x (or more) divisors is given by the OEIS sequence http://www.oeis.org/A061799.<|endoftext|> -TITLE: Intuition about the triangulation of a homotopy category K(A) -QUESTION [14 upvotes]: Let $\cal{A}$ be an additive category. Given a morphism of (cochain) complexes $f:X\rightarrow Y$ we can form the mapping cone $C_f$, which is the complex $X[1]\oplus Y$ with differential given by - $$\partial^n=\begin{pmatrix} -\partial_X^{n+1} & 0 \\\ f^n & \partial_Y^n\end{pmatrix}.$$ -We have a canonical sequence of complexes -$$ C_f[-1]\overset{\tilde{\beta}}{\longrightarrow} X\overset{f}{\longrightarrow}Y\overset{\alpha}{\longrightarrow}C_f\overset{\beta}{\longrightarrow}X[1] $$ -where $\alpha$ is injection, $\beta$ is projection, and $\tilde{\beta}$ is just $\beta[-1]$. -One can show that, up to homotopy, the morphisms $\alpha$ and $\beta$ only depend on the homotopy class of $f$ and so the above situation descends to the case of a morphism $f:X\rightarrow Y$ in the homotopy category of complexes $K(\cal{A})$. Triangles -$$ X\overset{f}{\longrightarrow}Y\overset{\alpha}{\longrightarrow}C_f\overset{\beta}{\longrightarrow}X[1] $$ -in $K(\cal{A})$ obtained in this way are called "standard triangles". We obtain the structure of a triangulated category on $K(\cal{A})$ by taking our distinguished triangles to be all triangles which are isomorphic (in $K(\cal{A}))$ to standard triangles. -I am trying to develop intuition and understand the motivation behind this triangulation on $K(\cal{A})$. I imagine that in some sense these triangles should capture the homotopy-theoretic information of morphisms of complexes, but at this point I do not have a clear understanding of the significance of these triangles. -I do have some understanding about how (in some vague sense) the sequence -$$ C_f[-1]\overset{\tilde{\beta}}{\longrightarrow} X\overset{f}{\longrightarrow}Y\overset{\alpha}{\longrightarrow}C_f\overset{\beta}{\longrightarrow}X[1] $$ -captures homotopy-theoretic information of the morphism $f:X\rightarrow Y$. For example, the map $\alpha:Y\rightarrow C_f$ can be thought of as a kind of "homotopy cokernel" of $f$ and the map $\tilde{\beta}:C_f[-1]\rightarrow X$ can be regarded as a kind of "homotopy kernel" of $f$. More precisely, a morphism $g:Y\rightarrow Z$ factors through $\alpha$ iff the composition $g\circ f$ is homotopic to 0. Indeed, you can show that there is a bijection between factorizations of $g$ through $\alpha$ and the homotopies from $g\circ f$ to $0$. In particular, there is a canonical homotopy of $\alpha \circ f$ to $0$. There is an analogous situation for $\tilde{\beta}$. One consequence of the above is that two morphisms of complexes $f_1,f_2:X\rightarrow Y$ are homotopic iff $f-g$ factors through the canonical map $X\rightarrow C_{id_X}$. -Because of such results I have some vague idea of how this business with the mapping cone of a morphism is well-entangled with questions of homotopy, but I would like a more precise understanding of the sense in which - -the standard triangle $ X\overset{f}{\longrightarrow}Y\overset{\alpha}{\longrightarrow}C_f\overset{\beta}{\longrightarrow}X[1] $ encodes the homotopy-theoretic information of $f:X\rightarrow Y$; -the standard triangulation on $K(\cal{A})$ encodes the homotopy information of the category of cochain complexes $C(\cal{A})$. - -Any comments which might help to pin-down the motivation for the standard triangulation on $K(\cal{A})$ will be thankfully received. - -REPLY [2 votes]: I am assuming this is not really news to anyone, but my take on triangules is that they provide a simple way to take "generalized quotients" of morphism. Here is what I mean by this: -If there is a surjective morphism, adding the kernel provides a short exact sequence and similarly, for an injective morphism, adding the cokernel does the same. -Now with the cone construction we can do the same for any morphism. The beauty is that pretty much anything that one can do with regular short exact sequences (and the long exact sequences they induce) can be done using distinguished triangles.<|endoftext|> -TITLE: Cobounded ⇒ cocompact? -QUESTION [12 upvotes]: Assume $\Gamma$ acts by isometries on a separable Hilbert space $H$, and - $$\operatorname{diam} H/\Gamma\le 1.$$ - Is it true that $H/\Gamma$ is compact? - - -Stupid example. Assume the action of $\Gamma$ on $H=\ell_2$ is generated by coordinate translations $x_n\mapsto x_n+\epsilon_n$. Then -$$\operatorname{diam} H/\Gamma=\tfrac12\cdot\sqrt{\sum_{n=1}^\infty\epsilon_n^2}.$$ -Thus, if $\operatorname{diam} H/\Gamma\le 1$ then $H/\Gamma$ is a quotient of Hilbert cube, and has to be compact. - -REPLY [4 votes]: The answer is "NO". To show this let us use the following: - -Lemma. Let $L$ be a lattice in $\mathbb R^q$ ($q$ is any positive integer). - Assume $$\operatorname{diam} \mathbb R^q/L>1000.$$ - Then there is a midpoint $m$ of two points in $L$ such that $|m-x|>1$ for any $x\in L$. - -Modulo Lemma one can construct an action of parallel translations the following way: -Let us construct inductively a sequence of lattices $L_q$ on $\mathbb R^q$ such that $\mathop{diam} \mathbb R^q/L_q<1000$ and such that $|x|>1$ for any $x\in L$. -Start with standard $L_1=\mathbb Z$ in $\mathbb R$. -To construct $L_{q}$ take -$$L_{q}'=L_{q-1}\times \mathbb Z\subset \mathbb R^{q-1}\times\mathbb R = \mathbb R^{q}.$$ -If $\mathop{diam} \mathbb R^q/L'_q < 1000$ set $L_q = L'_q$. -Othewise pass to the minimal lattice which contains $L'_q$ and the midpoint provided by the Lemma. -Applying this construction finitely many times you get a lattice $L_q$ with $\mathop{diam} \mathbb R^q/L_q<1000$. -Continue the process, we get lattice $L_\infty$ in $H$ which is a $1000$-net, its fundamental doamin contains a ball of radius 1; i.e. $H/L_\infty$ is not compact. -Proof of Lemma. -For $z\in\mathbb R^q$, denote by $\rho(z)$ the minimal distance to a point in $L$. -Take a point $z\in\mathbb R^q$ which maximize distance to $L$. -So $\rho(z)\ge 1000$. Then there is a couple of points $x,y\in L$ such that -$\angle xzy\ge\pi/2$ and $|x-z|=|x-z|=\rho(z)$. -Let $m$ be the midpoint for $x$ and $y$. -Then -$$|z-m|\le \frac{\rho(z)}{\sqrt{2}}$$ and therefore the distance from $m$ to any point of $L$ is at least $1000{\cdot}(1-\tfrac1{\sqrt{2}})>1$. $\square$<|endoftext|> -TITLE: What do I call type theory without Curry-Howard? -QUESTION [9 upvotes]: Is there a word I can say which will convey to type theorists that I am not thinking about types as propositions? -Background: as a category theorist, I am mostly interested in type theories as a language to express and prove things about the categories in which they naturally have semantics. E.g. typed lambda-calculus in cartesian closed categories, dependent type theory in locally cartesian closed categories, etc. Such categorical models can also be used when thinking of types as propositions, of course, but usually I am thinking of types (i.e. the objects in the categories in question) as more like sets (or groupoids or higher groupoids, sometimes). Sometimes the type theory even includes a separate notion of "proposition" which is interpreted by monomorphisms in the category, e.g. the geometric logic or "higher-order type theory" of toposes. -Of course, I can say "categorical type theory" to refer to the whole programme of semantics for type theories in categories. However, if I want to talk only about the type theory side of things, but I don't want people to start calling types "propositions" and terms "proofs," what can I say to convey the point of view I want to take? - -REPLY [2 votes]: "Term category of a hyperdoctrine." -The hyperdoctrine approach to categorical logic segregates the individuals (terms inhabiting types) in a category completely apart from the categories containing truth values and predicates (object-indexed truth values). In a hyperdoctrine nothing in the term category may be construed as a proposition -- those live elsewhere. -This contrasts with the approach of topoi, where there is an object of truth values $\Omega$ living in the same category as the type objects: propositions are morphisms into $\Omega$.<|endoftext|> -TITLE: Conditions for smooth dependence of the eigenvalues and eigenvectors of a matrix on a set of parameters -QUESTION [19 upvotes]: Let $A\in\mathcal M_n$ be an $n\times n$ real [symmetric] matrix which depends smoothly on a [finite] set of parameters, $A=A(\xi_1,\ldots,\xi_k)$. We can view it as a smooth function $A:\mathbb R^k\to\mathcal M_n$. - -1. What conditions should the matrix $A$ satisfy so that its eigenvalues - $\lambda_i(\xi_1,\ldots,\xi_k)$, - $i=1,\ldots,n$, depend smoothly on the - parameters $\xi_1,\ldots,\xi_k$? - -e.g. if the characteristic equation is $\lambda^3-\xi=0$, then the solution $\lambda_1=\sqrt[3] \xi$ is not derivable at $\xi=0$. - -2. What additional conditions should the matrix $A$ satisfy so that we can - choose a set of eigenvectors - $v_i(\xi_1,\ldots,\xi_k)$, - $i=1,\ldots,n$, which depend smoothly - on the parameters - $\xi_1,\ldots,\xi_k$? - -Update - important details - -The domain is simply connected -The rank of $A$ can change in the domain -The multiplicities of the eigenvalues can change in the domain, they can cross -The matrix $A$ is real symmetric -$n$ and $k$ are finite - -Update 2 - -A relaxation of the conditions of the problem: -For fixed $p=(\xi_{01},\ldots,\xi_{0k})$, can we find an open neighborhood of $p$ in the domain and a set of conditions ensuring the smoothness of the eigenvalues and the eigenvectors? - -REPLY [4 votes]: See the following papers by Armin Rainer (et al): -Perturbation theory for normal operators. arXiv:1111.4475 -Quasianalytic multiparameter perturbation of polynomials and normal matrices. -arXiv:0905.0837. -Denjoy-Carleman differentiable perturbation of polynomials and unbounded operators. Integral Equations and Operator Theory 71,3 (2011), 407-416. -arXiv:0910.0155<|endoftext|> -TITLE: Why are the following varieties symplectomorphic? -QUESTION [7 upvotes]: I saw a statement somewhere that for the Hirzebruch surfaces $F_n:=\mathbb{P}_{\mathbb{P}^1}(\mathcal{O}\oplus\mathcal{O}(n))$, $F_n$ and $F_m$ are symplectormorphic when $m$ and $n$ have the same parity. -My question is: Why is this true? -I can see that they are diffeomorphic by Freedman's Theorem: computing the intersection pairing on $\text{Pic}(F_n)=H^2(F_n, \mathbb{Z})$, which is an even form when n is even, and an odd form when n is odd. -But this result is deep and abstract, is there any easy way to construct a symplectomorphism? Certainly it is not given by polynomial maps, I'm wondering what it will be. - -REPLY [2 votes]: I think that $F(n)$ with a Kahler form such that the base $\mathbb{P}^1$ has area 2 and -the fibre has area 1, is symplectomorphic to $F(n+2)$ with the Kahler form such that base and fibre have area 1. -There is a "visual proof" of this which uses an idea of Leung and Symington (in an article called "Almost toric manifolds", on the arxiv). The operation is called "nodal trade", and is pictured below. -Figure A) is the image of the moment map of the toric variety $F(n)$. You can perturb the moment map to another Lagrangian fibration, so to replace a "corner" with a singular Lagrangian fibre (a pinched torus) over the red cross in figure B). Then you can slide this fibre along the dashed straight line until it hits the opposite edge and becomes another "corner". This gives figure C) which is the polytope of $F(n+2)$. -Leung and Symington explain these operations carefully in the above paper. -alt text http://www.freeimagehosting.net/uploads/a5a7b4ad8e.jpg<|endoftext|> -TITLE: positive elements in tensor products -QUESTION [10 upvotes]: Let $A \otimes B$ be the algebraic tensor of two $C^{\ast}$ -algebras, and an element x in $A\otimes B$ is positive if $x=yy^{\ast}$. Then is it always possible to write x in the form $x=\sum a_i\otimes b_i$, where $a_i$ and $b_i$ are positive elements? - -REPLY [18 votes]: I think the answer is no. The matrix -$$ -a=\begin{bmatrix} -1&0&0&1\\ -0&0&0&0\\ -0&0&0&0\\ -1&0&0&1 -\end{bmatrix} -$$ -is positive in $M_4(\mathbb{C})$. When we see this algebra as $M_2(\mathbb{C})\otimes M_2(\mathbb{C})$, it cannot be obtained as a sum of elementary tensors with positive entries. . -(ok, several hours later, here is the argument) -First, $a$ is positive because it is selfadjoint and $a=\left(\frac1{\sqrt2}a\right)^2$. Now, if we have a sum of elementary tensors in $M_2(\mathbb{C})\otimes M_2(\mathbb{C})$, it will look like -$$ -\sum_j\begin{bmatrix} -\alpha_j&\overline{\gamma_j}\\ \gamma_j&\beta_j\end{bmatrix} -\otimes -\begin{bmatrix}\alpha'_j&\overline{\gamma_j'}\\ \gamma_j'&\beta_j'\end{bmatrix} -=\begin{bmatrix} -\sum_j\alpha_j'\alpha_j& \sum_j \alpha_j'\overline{\gamma_j}& \sum_j\overline{\gamma_j'}\alpha_j&\sum_j\overline{\gamma_j'}\gamma_j\\ -\sum_j\alpha_j'\gamma_j& \sum_j \alpha_j'\beta_j&*&*\\ -*&*&*&*\\ -*&*&*&* - \end{bmatrix} -$$ -The assumption that each elementary tensor is made of the tensor of two positive matrices translates into $\alpha_j\geq0$, $\beta_j\geq0$, and $\alpha_j\beta_j\geq|\gamma_j|^2$ for all $j$ (and the "prime'' version too). Now if the matrix on the right is going to be our $a$ above, then the 2,2 entry forces the following: for each $j$, the product $\alpha_j'\beta_j=0$. If $\alpha_j'=0$, then $\gamma_j'=0$; and if $\beta_j=0$, then $\gamma_j=0$. That is, for each $j$, $\overline{\gamma_j'}\gamma_j=0$, and this forces the 1,4 entry to be $0$; but it is not zero in $a$.<|endoftext|> -TITLE: What's a good dense open of $\bar{M}_g,n(X,\beta)$? -QUESTION [6 upvotes]: The title says it all, what's a good dense open of $\bar{M}_g,n(X,\beta)$ which play the role of ${M}_g$ in $\bar{M}_g$? -My first (naive) guess is maps from a genus $g$ smooth curve to $X$ which represents the class $\beta$. But I'm a little bit concerned, is it dense for sure? Could it happen that in some cases one has singular curves only? Can a stable map from a singular curve always deform to a map from a smooth curve of genus $g$? - -REPLY [9 votes]: If $X$ is convex and $g = 0$, then you can take smooth curves with distinct marked point, this will be dense. However, in general the locus of smooth curves is not dense. An easy example is that of degree 1 unpointed maps from a curve of genus $g > 0$ to $\mathbb P^1$; such a curve must consist of a copy of $\mathbb P^1$ mapping isomorphically and some vertical components, hence it can not be smooth. -In general I would guess that there is no analogue of $M_g$. There are good philosophical reasons why it has to be so; the space $\overline{M}_g,n(X,\beta)$ is much too big, in general, and badly singular, it can't be expected to behave nicely in any way. The “good” part of $\overline{M}_g,n(X,\beta)$ is the virtual fundamental class; but that's just a class of cycles, it does not have open parts.<|endoftext|> -TITLE: Citing papers that are in a language that you do not read. -QUESTION [29 upvotes]: I have a basic question that others have definitely considered. -Often there are papers that originally appeared in a language that one might not understand (and I mean a natural language here). I would like to cite the original paper, because that is where the credit belongs. But on the other hand, doing so violates the golden-rule of read that paper that you cite! What should I do to overcome this dilemma? -So far, I have always cited the original, and if possible some other related work that has appeared in English---but sometimes, reviewers write back that I should not be citing papers written in a language different from English, which is what motivated me to ask this question. -Thanks for any useful advice. - -REPLY [4 votes]: Partly with the point of summarizing other answers/remarks: there are several points mixed together here. There is the issue of recognizing "prior art" (whether or not one's present work depends upon it). There is the issue of recognizing "necessary background" (often logically prior to the work in question). There is the issue of "competing, perhaps incomplete work" (to be civil, as well as being a mensch and informing readers about it). There is "compliance-necessitated reference", as in "icons" which, if not cited, will make the skeptical quasi-expert reader doubt the competence of the author. -Of course, there have been times in which otherwise-intelligent people did not realize they should write in English. (Sorry, part of my reason for making an "answer" was to publicly form that sentence. I did grow up in midwestern, mid-20th-century U.S., and was led to believe that people who didn't speak English were just being obstructionist jerks, since obviously English was the universal natural language... whew!) -Another perversity: more than one historically-famous mathematician has let slip to me in conversation that they take the viewpoint that not looking at someone else's paper relieves them from any obligation to cite it [even if it is prior work]. Not to my taste. -More perverse than pretending to ignore others' work is the actualy benefit their work may give you/one, even, or perhaps especially, if one hasn't assimilated it yet. As a historical example, it confused me for a decade or two that Siegel and Harish-Chandra (both at IAS) had apparently not communicated... ever?... so that the "holomorphic discrete series", visible in Siegel's and H. Braun's work in the late 1930's, were not visible... and, then, contrariwise, Shimura and other "modern" automorphic forms people seemed functionally oblivious to H-C's work 10-15 years earlier. -Must/should one personally certify anything cited? Well, sure, obviously, this would be desirable. Also, not feasible. Of course, if you're counting on the correctness of an obscure paper, you are on thin ice. If you're counting on correctness of an already-often-cited paper, then you are in better shape. No mystery here. Refereeing does not assure correctness, just makes it a tad more likely. If you claim to prove something scandalous, people will revisit all those innocent-seeming papers you cite. "Just the obvious." (But, contrary to the professional pose that ... oh, ... published papers are first-order predicate-logic ... correct? A needless conceit, of course, and we should not slip into it.) -The editorial pressures are corruptive. Yes. The professional competitive pressures are potentially corruptive, yes. But I think if we are honest with ourselves we can see what acknowledgements we should make. Prior work, even if we don't use it. Even if we are competitors. Give the reader (supposing they care!) a guide about how to arrive ... here. -Depending on how one understands the word, I think that, happily (to me) "honesty" is a good guide. -Oop, "was it in English?" What? Um... My own discussion, and all others, indicate that this cannot possibly be a legitimate issue. Write to the editor. But it is a "stimulating" question. :)<|endoftext|> -TITLE: index of a family of Dirac operators in $K^1$ -QUESTION [5 upvotes]: Suppose I have a family of Dirac operators over a compact base space B. From the paper of Atiyah and Singer about skew adjoint Fredholm operators we know that it has an index in $K^1(B)$. -Suppose furthermore I know "a lot" about these Dirac operators (like their spectrum, eigenspaces etc.) and $B$ is a simple space like e.g. a torus where I know everything about $K^1$ and the cohomology. -What methods are there to give an explicit description of the index in $K^1(B)$ or its image in the odd-dimensional cohomology? -Any suggestions or references welcome. - -REPLY [4 votes]: Two years ago my student Daniel Cibotaru wrote a dissertation entitled -Localization formulae in odd K-theory -in which he answers precisely this question in great generality. -More precisely, given a smooth family $(T_b)_{b\in B}$ of complex, Fredholm selfadjoint operators parametrized by a compact, connected, oriented smooth manifold $B$ he describes explicitely a (stratified) cycle in $B$ that is Poincare dual to the odd Chern character of this family. This cycle is non-homogeneous, i.e., it is a sum of cycles of various codimensions. The codimension $1$-part is the so called Maslov class or spectral flow class and it is more or less known. Daniel has a very nice description of the codimension $3$-part. For example if $B$ is a $3$-manifold, then the family defines a cohomology class in $H^3(B,\mathbb{Z})\cong \mathbb{Z}$ and Daniel explains how to compute this as a signed count of points in $B$. -More precisely, for a generic family $(T_b)_{b\in B}$, the locus of points $b$ where $\ker T_b\neq 0$ is an oriented surface $S\subset B$ and the family of vector spaces -$$ S\ni s\mapsto \ker T_s $$ -is a complex line bundle over $S$.The above integer is none other than the degree of this complex line bundle. -The higher codimension parts have explicit but rather complicated descriptions. -The philosophy of his dissertation can be easily described: he constructs a smooth model for the classifying space of $K^1$. This is an infinite dimensional Grassmanian-like object $\mathcal{X}$ equipped with a Schubert-like stratification by strata of finite codimensions. He then shows that the closures of the Schubert strata determine cohomology classes forming an integral basis of the cohomology of $\mathcal{X}$. The whole thing has a strong symplectic flavor.<|endoftext|> -TITLE: Structure of iterated $\mathbb{P}^1$-bundles -QUESTION [5 upvotes]: Let us call an algebraic variety $X$ an iterated $\mathbb{P}^1$-bundle if it is either a point or a locally trivial $\mathbb{P}^1$-bundle over $X'$, which is another iterated $\mathbb{P}^1$-bundle. -It is easy to construct examples of such varieties as toric varieties. Another (non-toric in general) examples of such varieties are the Bott-Samelson varieties. - -Is there a simple (combinatorial or other) description of these spaces? - -I would like this description to allow one to investigate some of the following: - -line bundles (ampleness, global generation etc.), -properties of the rank two vector bundle $\mathscr{E}$ on $X'$ such that $X = \mathbb{P}(\mathscr{E})$ (which exists and is determined up to multiplication by some line bundle on $X'$), -sections of the bundle $X\to X'$ (that is, line subbundles of $\mathscr{E}$), -construction of $\mathbb{P}^1$-bundles over $X$ (i.e., classes of rank two vector bundles), -degenerations over $\mathbb{A}^1$ to other iterated $\mathbb{P}^1$-bundles (in particular, their toric degenerations). - -REPLY [2 votes]: This is only a partial answer, but the paper -M. Willems, "K-théorie équivariante des tours de Bott", Duke Math. J. 132 (2006) -includes a combinatorial description of a degeneration of Bott-Samelson varieties to toric ones. (It's based on a construction of Grossberg and Karshon, and was also done algebraically by Pasquier.)<|endoftext|> -TITLE: Two notions of tangent vector for a Fréchet manifold -QUESTION [6 upvotes]: Let $X$ be a Frechet or Banach manifold. We can define tangent vectors by equivalence classes of smooth curves. But, we could also define them as derivations of germs of smooth functions. Do these two notions agree in this infinite dimensional setting? - -REPLY [6 votes]: To expand on Gjergji's answer a little: -According to Kriegl and Michor, the key for the identification is two properties on the model space: - -It be reflexive -It have the bornological approximation property - -Reflexivity is needed because, essentially, of what André mentions: linear maps in to a space are not always the same as linear maps out of linear maps out of the space! The derivational definition of tangent vectors is as operators on cotangent vectors (even if dressed up a derivations on germs) so it's a "dual of a dual" and you need reflexivity to get back where you started. -Approximation properties (of which there are many) are a way of saying that linear maps from one space to another aren't too messy, by which we mean that they behave a little like matrices. Matrices themselves correspond to things in $E \otimes E'$, but as we're in infinite dimensions then that's only finite rank operators. Still, as we're Grown Ups, we can handle approximations and so it's enough for us that any linear transformation can be approximated by a matrix. With the strong (or operator) topology, this only gets us compact operators, but with a weak topology we have a chance of getting the whole thing. The different types of approximation property correspond to different topologies on linear maps. -There are spaces (even Banach spaces) that don't have the requisite approximation property, but they're "not nice" spaces and also not ones that are usually encountered. All the "most common" spaces have the approximation property. -In particular, nuclear Frechet spaces are reflexive and have the approximation property, so - as Gjerji says - that's sufficient. However, Hilbert spaces are also reflexive and have the approximation property. -The last thing I have to say on this is that if your real question is "What's the right notion of tangent vector in infinite dimensions?" then you should look at the chart in Kriegl and Michor's book in section 33.21 (p351 in the current version online). From that chart, it's clear that the "right" definition is as equivalence classes of curves.<|endoftext|> -TITLE: Has a conjugation of SL2(Z) finite index in SL2(Z)? (Modular group) -QUESTION [7 upvotes]: Dear all, -I have a probably rather simple question: Suppose we have a Matrix $ M\in SL_2(\mathbb{Q}) $. Does the group $ M^{-1} SL_2(\mathbb{Z}) M \cap SL_2(\mathbb{Z})$ then always have finite index in $SL_2(\mathbb{Z})$? Why? Why not? -I really was not able to solve this problem! -All the best -Karl - -REPLY [7 votes]: There is a second proof which Tony Scholl hints at in the comments. This is probably secretly equivalent to the argument Greg writes up, but I find it easier to think about. -$SL_2(\mathbb{Z})$ is the subgroup of $SL_2(\mathbb{Q})$ preserving the lattice $L_1:=\mathbb{Z}^2$ inside $\mathbb{Q}^2$. Similarly, $M SL_2(\mathbb{Z}) M^{-1}$ is the gorup of matrices preserving $L_2 := M \cdot L_1$. So the group we are interested in is the group of matrices sending $L_1$ and $L_2$ to themselves. -Choose an integer $N$ such that $L_1 \cap L_2 \supseteq N L_1$ and $L_1 + L_2 \subseteq N^{-1} L_1$. Let $\Gamma$ be the subgroup of $SL_2(\mathbb{Z})$ which acts trivially on $L_1/ N^2 L_1$. The subgroup $\Gamma$ has finite index as it is the kernel of $SL_2(\mathbb{Z}) \to SL_2(\mathbb{Z}/N^2)$. -Now, $\Gamma$ stabilizes $N L_1$ and $N^{-1} L_1$, and acts trivially on $(N^{-1} L_1)/(N L_1)$. In particular, any lattice $K$ with $N^{-1} L_1 \supseteq K \supseteq N L_1$ will be taken to itself by $\Gamma$. We chose $N$ so that $L_2$ lies between $N^{-1} L_1$ and $N L_1$. So $\Gamma$ takes $L_2$ to itself, and we deduce that $\Gamma$ is contained in the group we are interested in. So the group we are interested in has index $\leq [SL_2(\mathbb{Z}) : \Gamma]$ in $SL_2(\mathbb{Z})$.<|endoftext|> -TITLE: an exercise about elliptic surface in Beauville's book -QUESTION [5 upvotes]: In Beauville's "Complex Algebraic Surfaces", given an elliptic surface $f : X \to C$ with a generic fiber $E$. Then either $\text{Alb}(X) \cong \text{Jac}(C)$ or there is an exact sequence of abelian varieties -$0 \to F \to \text{Alb}(X) \to \text{Jac}(C) \to 0$ -with $F$ being isogenous to $E$. -Assume that $X$ is a properly elliptic that is in the second case. Does anyone know an example where $F$ is isomorphic to $E$ (and $X$ is not a product), and an example where $F$ is not isomorphic to $E$? - -REPLY [8 votes]: Donu Arapura is right: -If you want $F$ isomorphic to $E$ you can do the following: -Consider the product $E \times P^1$. You find $F \simeq Alb(X) \simeq E$. -If you want $F$ and $E$ to be non isomorphic, one can do as follows. -Let $c$ be a 2-torsion point of $E$. Form the quotient $X$ of $E \times E$ by the involution defined by $(x, y) \mapsto (-x, y + c)$. -The first projection induced an elliptic fibration $X \to \mathbb P^1$. A generic fibre is $E$. -However, the albanese of $X$ is $E'$ = $E$/{0, $c$}. -The statement about the albanese follows by considering the elliptic fibration $X \to E'$ given by the second projection, and the statement of Beauville. -So let us continue this discussion with examples of Kodaira dimension one: -Take an elliptic curve $E$ and a curve $C$ with $g(C) \ge 2$. -Then $E \times C \to C$ is an example with $E$ and $F$ being isomorphic. -Now let $C$ be an hyperelliptic curve with involution $i \colon C \to C$. -Let $E$ be elliptic, and let $b \in E$ be a 2-torsion point. -We define $X$ to be the quotient of $E \times C$ by the involution given by $(x, y) \mapsto (x + b, i(y))$. -As before, the fibration $X \to \mathbb P^1$ gives an example for $E$ not isomorphic to $F$.<|endoftext|> -TITLE: Groups where every subgroup is a centralizer -QUESTION [6 upvotes]: Can one classify all finite groups $G$ with the property that every proper subgroup $H < G$ is the centralizer of some $g\in G$? - -REPLY [6 votes]: As Mark mentioned above, in such a group, every subgroup must be cyclic. Indeed, it is possible to show that the only such groups are either the finite abelian simple groups (all of whom clearly satisfy the required conditions), or groups of order $pq$, where $p$ and $q$ are distinct primes (Assume that $p\gneq q$. Then, given a subgroup of order $p$ or order $q$, we have that it must be the centralizer of any element of any element in it, as if it were anything larger, we would have that our group is abelian.) -To go in the other direction, let $G$ satisfy the required conditions. If $Z(G)\neq 1$, then we have that $Z(G)=C_{G}(x)$ for some $x\in G$. Then, $x\in Z(G)$, which implies that $C_{G}(x) = G$. Thus, $G$ is abelian, and as the centralizer of any element is now $G$, we have that $G$ is cyclic of prime order. -Now note that if $H$ is any proper subgroup, then $H=C_{G}(x)$ implies that $x\in H$, and that $Z(H)\neq 1$. Also note that the assumptions on our group descend to any subgroup. Thus every subgroup of $G$ is cyclic of prime order. Let $|G|=p_{1}...p_{n}$, with $p_{i}\lneq p_{i+1}$. All the sylows of $G$ are cyclic, and the transfer map into the one of smallest order, $p_{1}$, must be surjective, giving us a subgroup of order $p_{2}...p_{n}$. As all proper subgroups of $G$ are cyclic of prime order, we have that $|G|=pq$ (I'm sure there is an easier way to show the last bit).<|endoftext|> -TITLE: Lie's third theorem via differential graded algebras? -QUESTION [11 upvotes]: Dennis Sullivan, "Infinitesimal computations in topology", Publ. IHES: At the end of section 8, he writes, among other things, roughly the following. -Let $\mathfrak{g}$ be a (finite-dimensional, real) Lie algebra and let $\Lambda \mathfrak{g}^{\ast}$ be the Chevalley-Eilenberg complex (i.e. the -exterior algebra, -with the differential dual to the Lie bracket). He considers the spatial realization $\langle \Lambda \mathfrak{g}^{\ast}\rangle $ with -respect to the simplicial d.g.a. of $C^{\infty}$-forms on the standard simplices. This is a simplicial set, the $p$-simplices are -the d.g.a.-homomorphisms $\Lambda \mathfrak{g}^{\ast} \to \mathcal{A}(\Delta^p)$ to the de Rham forms on the simplex. -Let $G$ be the simply-connected Lie group with Lie algebra $\mathfrak{g}$. -"Theorem" (8.1)' (the quotation marks are due to Sullivan) -says that the fundamental group of $\langle \Lambda \mathfrak{g}^{\ast} \rangle $ is isomorphic to $G$. -Now the set of $p$-simplices of $\langle \Lambda \mathfrak{g}^{\ast} \rangle $ has a topology, induced from the $C^{\infty}$-topology) on the space of $\mathfrak{g}$-valued forms on the simplices and so -$\pi_1 (\langle \Lambda \mathfrak{g}^{\ast} \rangle )$ is a topological group. Sullivan also says that the above isomorphism is a homeomorphism. -It is not difficult to verify these assertions. Since it is rather clear how to describe the exponential map in this construction, we can also -recover the differentiable structure on $G$ from this method. -The upshot of this discussion is: once the existence of the simply-connected Lie group is known (this is Lie's third theorem, proven only decades after Lie by Cartan), it has the given abstract description. -It is well-known that Lie's third theorem is a pretty hard result (the standard proof goes via Ado's theorem). -It seems possible to reverse the logic of that argument and give a proof of Lie's 3rd theorem. -Given $\mathfrak{g}$, \emph{define} $G:=\pi_1 (\langle \Lambda \mathfrak{g}^{\ast} \rangle)$ -as a topological group. The exponential map $\mathfrak{g} \to G$ is given by the following formula: any $x \in \mathfrak{g}$ defines a constant -$1$-form on $\Delta^1$, hence a 1-simplex of $\pi_1 (\langle \Lambda \mathfrak{g}^{\ast} \rangle)$. That was the easy part; here are the nontrivial -parts: - -Show that $G$ is Hausdorff (probably difficult) -Put a smooth stucture on it, such that the exponential map is a local chart (probably the hardest part) -Once this is done, the simple connectivity of $G$ and the fact that $Lie (G)=\mathfrak{g}$ are probably both obvious. - -After all these preliminaries, I can ask my question: has this approach been written down properly? I am aware that a generalization -of this argument has been used -by Getzler http://arxiv.org/abs/math/0404003 and Henriques http://arxiv.org/abs/math/0603563, but in these papers, I do not find the details. It is of course also possible (maybe desirable) to banish all the fancy language from the discussion, leaving a definition of $G$ as the quotient of the space of $\mathfrak{g}$-valued 1-forms on the interval. - -REPLY [4 votes]: The details are here: -Marius Crainic, Rui Fernandes, Integrability of Lie brackets -http://arxiv.org/abs/math/0105033 -(To connect this to your question, notice that a morphism $T X \to \mathfrak{g}$ of Lie algebroids, which is the language they use, is dually the same as a morphism $\Omega^\bullet(X) \leftarrow CE(\mathfrak{g})$ of dg-algebras. For more see here)<|endoftext|> -TITLE: Role for generalized geometries in string theory -QUESTION [9 upvotes]: What role do generalized geometries (in terms of Dirac structures, for instance, symplectic, Poisson, complex, and generalized complex structures in the sense of Hitchin, Cavalcanti, and Gualtieri) play in string theory? -EDIT: More generally, what role to Dirac structures (subbundles of the generalized tangent bundle $TM \bigoplus T^*M$ which are maximally isotropic to the natural pairing and closed under the Courant bracket) play? - -REPLY [2 votes]: I would like to add that generalized geometry in the sense of Hitchin is also a good framework for the notion of brane (a very important notion in what physicists call M-theory) and also T-dualities one of which is mirror symmetry. Especially I have read an article basically saying that the category of generalized Calabi-Yau manifolds provide a good canditate for the Homological Mirror Symmetry conjecture but I don't remember the article anymore, sorry, maybe José Figueroa-O'Farrill does remember. See also the question: Mirror symmetries for generalized geometries ?. -EDIT: The article I was searching is: http://arxiv.org/abs/1106.1747.<|endoftext|> -TITLE: What is the L-function version of quadratic reciprocity? -QUESTION [21 upvotes]: Quadratic reciprocity theorems states that for two different odd prime p and q, -we have (p/q)(q/p)=(-1)^(p-1)(q-1)/4. -What is the statement of this theorem in L-function? - -REPLY [15 votes]: Let me try to say a few things from my point of view concerning Chandan Singh Dalawat's question about higher reciprocity. -The "Galois character" of the quadratic extension $K$ ramified only at $p$ is given by the -Kronecker symbol $\chi(q) = (p^*/q)$; the associated relation $\zeta(s) L(s,\chi) = \zeta_K(s)$ between zeta functions of the rationals, the Dirichlet L-function belonging to $\chi$, and the Dedekind zeta function of the field $K$ encodes the decomposition law of quadratic extensions (see the comment by Sam Derbyshire). -The quadratic reciprocity law in Euler's form states that this Kronecker symbol is a Dirichlet character with conductor $p$, i.e. that $\chi(q) = \chi(r)$ for positive primes -$q \equiv r \bmod p$. Thus it must be one of the $p-1$ Dirichlet characters modulo $p$, -and since $({\mathbb Z}/p{\mathbb Z})^\times$ is cyclic, is must be the quadratic Dirichlet -character modulo $p$. -This has an interpretation in terms of L-functions for Dirichlet characters; the zeta -relation $\prod L(s,\chi) = \zeta_L(s)$, where the product on the left hand side is -over all Dirichlet characters mod $p$ (suitably interpreted at the bad prime $p$), and -where $L$ is the field of $p$-th roots of unity, can be interpreted as the decomposition law for primes in cyclotomic extensions. The reason why there is a connection between these objects and the Kronecker character is, ultimately, the fact that $K \subset L$, -something that can be proved directly e.g. with Gauss sums, another tool for proving -reciprocity laws. In any case, the ultimate source of this proof is the realization of -the Kummer extension $K/{\mathbb Q}$ as a class field (inside the ray class field -$L/{\mathbb Q}$). -For getting the cubic reciprocity law, you can work with Kummer extensions over the field of cube roots of unity; comparing the decomposition laws for Kummer extensions and class fields gives you something like $(\pi/q)_3 = (q/\pi)_3$ for primary primes $\pi$ and rational primes $q$, which is strong enough to give you the full cubic reciprocity law with a couple of formal manipulations. Writing down the corresponding Artin and Dirichlet L-series is straightforward. -Something similar works for fourth powers, but then you run into trouble since you only get Eisenstein reciprocity out of this comparison of decomposition laws.<|endoftext|> -TITLE: What are D-branes, really? -QUESTION [36 upvotes]: In the past couple years, I've read many words pertaining to "D-branes" without feeling I have fully comprehended them. In broad terms, I think I get what they're about: They're supposed to serve as habitats for the ends of open strings and can be conceived of as submanifolds (of the target manifold in a sigma model), possibly augmented with a vector bundle, or a sheaf of [somethings], or maybe some other kind of label/data. (Corrections welcome.) -In the hopes of tightening my grasp on the concept, here are some of the questions that have been nagging me during my reading. - -What specifically is the definition of a D-brane, say in the context of a topological field theory? (Or what are the most promising provisional definitions?) What references are most accessible to a mathematical audience? -What picture should I have in my head when an author talks about "the moduli space of D-branes"? -What is the idea behind the "dynamics of D-branes" that researchers sometimes talk about? (Perhaps when I understand better how to think about these gadgets, it will be easier to conceive of how they should change over time.) -What goes into verifying (or at least asserting/conjecturing) that the elements of twisted K-theory classify "D-brane charges"? - -(Question reposted from here.) - -REPLY [21 votes]: There is an abstract algebraic formulation of QFT: this says that an $n$-dimensional QFT is a consistent assignment of spaces of states and of maps between them to $n$-dimensional cobordisms. -If one allows cobordisms with boundary here, one speaks of open-closed QFT. A D-brane in this context is the type of data assigned by the QFT to these boundaries. -There is also a geometric aspect to this: many abstractly defined QFTs are imagined to be "sigma-models". They are supposed to be induced by a process called "quantization" from a functional (the "action") on a space of maps $\Sigma \to X$ from a cobordism $\Sigma$ into a smooth manifold $X$ equipped with extra geometric data (such as metric, connections, etc.) -Under this correspondence, one may ask which abstract algebraic properties of the QFT derive from which geometric aspects of these "background structures". One finds that the data that the QFT assigns to boundaries comes from geometric data on $X$ that tends to look like submanifolds with their own geometric data on them (but may be considerably more general than that!). If so, this geometric data on $X$ is called a D-brane of the sigma-model. -There are many instances of this that are understood at the rough level at which quantum field theory was understood in the 20th century. One special case that is by now under fairly complete mathematical control and which serves as a good guide to the general concept of D-branes is what is called "2d rational CFT" . -There is a complete mathematical classification of 2d rational CFTs in their abstract algebraic form: they are given by special symmetric Frobenius algebra objects internal to a modular tensor category of representation of a vertex operator algebra. -Under this classification theorem, the boundary data = D-branes in the algebraic formulation can be proven to be precisely modules over this Frobenius algebra object. -In special nice cases, one understands where these come from geometrically. The notable example is the Wess-Zumino-Witten model, where the target space is a group manifold. Here one finds that in the simplest case the geometric data corresponding to these D-branes are submanifolds given by conjugacy classes, and carrying twisted vector bundles. More generally, though, the D-branes are given by cocycles in the twisted differential K-theory of the group. So the identification "D-brane = submanifold" is too naive, in general. The correct identification is: -geometric D-brane = geometric data on the sigma-model target space that induces boundary data of the corresponding algebraically defined worldvolume QFT. -For more see -http://ncatlab.org/nlab/show/brane<|endoftext|> -TITLE: Do maps have flows? -QUESTION [6 upvotes]: In A New Kind of Science: Open Problems and Projects(pg. 36). - -How can one extend recursive function definitions to continuous numbers? What is the continuous analog of the Ackermann function? The symbolic forms of the Ackermann function with a fixed first argument seem to have obvious interpretations for arbitrary real or complex values of the second argument. But is there a general way to extend these kinds of recursive definitions to continuous cases? Given a way to do this, how does it apply to recursive definitions like those on page 130? ... Stephen Wolfram - -The following is an example of a flow of a map from MO f(f(x))=exp(x)-1 and other functions “just in the middle” between linear and exponential. . -Consider $g(x)=e^x-1$. Then -$g^n(x)= x+\frac{1}{2!}n x^2+\frac{1}{3!} \left(\frac{3 n^2}{2}-\frac{n}{2}\right) - x^3+\frac{1}{4!} \left(3 n^3-\frac{5 - n^2}{2}+\frac{n}{2}\right) x^4 $ -$+\frac{1}{5!} - \left(\frac{15 n^4}{2}-\frac{65 n^3}{6}+5 - n^2-\frac{2 n}{3}\right) x^5 $ -$ +\frac{1}{6!} - \left(\frac{45 n^5}{2}-\frac{385 n^4}{8}+\frac{445 - n^3}{12}-\frac{91 n^2}{8}+\frac{11 n}{12}\right) - x^6 $ -$ +\frac{1}{7!}\left(\frac{315 n^6}{4}-\frac{1827 - n^5}{8}+\frac{6125 n^4}{24}-\frac{1043 - n^3}{8}+\frac{637 n^2}{24}-\frac{3 n}{4}\right) - x^7 + \cdots$ -Note that $g^0(x)=x, g^1(x)=e^x-1$ and that a symbolic mathematical program will also confirm that $g^m(g^n(x))=g^{m+n}(x) +O(x^8)$. -The half-iterate is also computed correctly, -$g^\frac{1}{2}(x)=x+\frac{x ^2}{4}+ \frac{x^3}{48} +\frac{x^5}{3840}-\frac{7 x^6}{92160} +\frac{x^7}{645120}$ -See MO What’s a natural candidate for an analytic function that interpolates the tower function? for more background. -Questions - -What evidence is there for believing that maps do not have flows? Is there anything known that would prevent proofs to establish existence, uniqueness, and convergence? References would be nice but an explanation would be better. -Consider $f(f(x))=g(x)$ where $g: \mathbb{R} \rightarrow \mathbb{R}$. Can $f: \mathbb{R} \rightarrow \mathbb{C}$ be an appropriate solution or must $f: \mathbb{R} \rightarrow \mathbb{R}$? -Likewise, is there any reason beyond aesthetics for believing that maps have flows? - -REPLY [2 votes]: Number 2 is answered in the affirmative if we take "functions" to mean "analytic functions". Take $\cos(x)$. It can be shown that there exists no function $f : \mathbb{R} \to \mathbb{R}$ such that $f(f(x)) = \cos(x)$. This follows by the fact: $\cos(x_0) = x_0$ where $x_0 = 0.739805...$, and $-1<\cos'(x_0) < 0$. Therefore if there was such an $f$, $\cos'(x_0) = f'(f(x_0))f'(x_0) = f'(x_0)f'(x_0)$ and $f'(x_0)$ cannot be real. -To show there does exists an $f: I \to I$, where $I$ is the immediate basin of attraction about $x_0$ and $\mathbb{R} \subset I$, is a little trickier. You'll have to track down a nice paper constructing it (it's a real hassle to construct); but it is possible. It would take about 10 pages of written work to get it out. Then $f:\mathbb{R} \to \mathbb{C}$. There exists exactly two such $f$ corresponding to $f'(x_0) = i\sqrt{|x_0|}$ or $f'(x_0) = -i\sqrt{|x_0|}$. -If you want a more trivial example just look at $-x$ which has a square root $ix$, but no real analytic square root.<|endoftext|> -TITLE: covering groups by infinitely many cosets -QUESTION [5 upvotes]: The classical Neumann lemma states that if a group is covered by finitely many cosets, then at least one of these cosets is the coset of a subgroup of finite index. (Actually, the lemma says more, namely that the group is covered by the cosets of subgroups of finite index.) -I wonder if there is an infinitary version of the lemma in the following sense: Suppose that $G$ is a group and $G=\bigcup_{i<\alpha}x_iH_i$, where the $H_i$ are subgroups and $\alpha$ is an ordinal less than some big cardinal $\kappa$. (Maybe even $\kappa$ is strongly inaccessible.) Is it necessarily the case that there is some $i$ for which the index of $H_i$ in $G$ is less than $\kappa$? - -REPLY [3 votes]: So it appears that my question is false. Here is a counterexample communicated to me by Ehud Hrushovski, although he attributes the counterexample to Macpherson and Neumann: Fix an uncountable cardinal $\kappa$ and let $G$ be the group of permutations of $\kappa$ with finite support. Let $G_n$ be the subgroup of $G$ consisting of those permutations which fix $n\in \omega$. Then $G=\bigcup G_n$ and each $G_n$ has index $\kappa$ in $G$.<|endoftext|> -TITLE: Understanding the unreducedness of a subscheme supported on fixed points -QUESTION [8 upvotes]: EDIT: Based on comments, I've decided to essentially rewrite this question. My apologies to those who commented below, whose comments seem a bit off topic when you try to match them with the current question. I've also posted a better-thought out follow-up here. -Lots of people like to study torus actions on varieties and manifolds, and in particular, like to study their fixed points. Most people I know just think about fixed points as a set, but they have a canonical scheme structure which carries more information. -As Bcnrd points out below, this subscheme of a scheme $X$ over a field $k$ is defined by looking at the functor on $k$-algebras defined abstractly by the $(\mathbb{G}_m)_A$-invariant points of $X(A)$. As Dave Anderson points out, if $X=\mathrm{Spec}(R)$, then this is simply the subscheme defined by the ideal generated by all elements of non-zero weight $I=R^{>0}R+R^{<0}R$. -For example, consider $R=\mathbb{C}[x,y,z]/(xy=z^n)$ where $x$ has weight 1, $y$ has weight -1 and $z$ has weight 0. In this case $I=(x,y)$, so $X^{\mathbb{G}_m}=\mathrm{Spec} \:\:\:\mathbb{C}[z]/(z^n)$. So, as you can see, the fixed point scheme doesn't have to be reduced, though if $X^{\mathbb{G}_m}$ is 0-dimensional, this can only happen if $X$ is not regular at the corresponding fixed point. -Now, imagine $X$ has a $\mathbb{G}_m$-equivariant resolution of singularities $\tilde X$, with a compatible $\mathbb{G}_m$ action, and $\tilde X^{\mathbb{G}_m}$ also 0 dimensional. - -Can I conclude anything about the length of $X^{\mathbb{G}_m}$ from knowing the length of $\tilde X^{\mathbb{G}_m}$ (which is just the number of fixed $k$-points by smoothness of $\tilde X$)? In a number of examples I'm looking at, these turn out to be equal, and I'm wondering how general a phenomenon this is. - -For example, the example $R=\mathbb{C}[x,y,z]/(xy=z^n)$ has a resolution with $n$ fixed points, and indeed, that's the length of $X^{\mathbb{G}_m}$. -Now, the examples I'm looking at have special features which may or may not be revelant, but I mention them in case they strike a chord. - -I'm looking at examples where $\tilde X$ is symplectic, and the $\mathbb{G}_m$-action is Hamiltonian. -Also in my examples, $\tilde X$ has a smooth $\mathbb{G}_m$-equivariant deformation $\tilde Y$, where the generic fiber is affine, and the fixed point scheme $\tilde Y^{\mathbb{G}_m}$ is flat and finite over the base. - -REPLY [3 votes]: Since a good picture for a length $n$ scheme set-theoretically supported at a point is some kind of limit of $n$ different reduced points getting closer together, let's try to understand what's going on that way. We can appeal to this - -Also in my examples, $\tilde{X}$ has a smooth $G_m$-equivariant deformation $\tilde{Y}$, where the generic fiber is affine, and the fixed point scheme $\tilde{Y}^{G_m}$ is flat and finite over the base. - -Probably in your examples this deformation (call it $\tilde{Y}_t$, where $t$ runs through the base space) descends to a deformation of $X$ (call it $Y_t$). ``Descends to'' means in particular that generically we have $Y_t = \tilde{Y}_t$. This is the case with the A_n resolution of C[x,y,z]/xy=z^n for example, and also for other cones associated with Springer theory. Now as $t \to 0$ the fixed points in $Y_t$ will collide at the singular fixed point in $X$, suggesting that the number of them is equal to the length. And as $t \to 0$ the fixed points of $\tilde{Y}_t$ will move bijectively to the fixed points of $\tilde{X}$. -I haven't worked out what general hypotheses make this argument legitimate--at least if $X$ is a complete intersection in a vector space on which the torus acts linearly, and your $\tilde{Y}$ is smooth over the base of the deformation, we seem to be all right. -Edit: Sorry, I think "complete intersection" is a distraction. Here's a formalization of the above argument. The situation is pretty special but there's not much you have to check. Say we have a commutative diagram - - f - Y~ ---> Y -g~| |g - v v - S~ ---> S - -that's Cartesian over a Zariski open subset of $S$, with $T$-actions on $\tilde{Y}$ and $Y$ compatible with $f$ and preserving the fibers of the vertical maps. Say that $\tilde{Y}$ and $Y$ are smooth, that $\tilde{g}$ is smooth and that $\tilde{g}\vert_{\tilde{Y}^T}$ is smooth and finite, that $g$ is flat and that $g\vert_{Y^T}$ is flat and finite. Finally, suppose that $X^T = Y^T \times_Y X$, $\tilde{X}^T = \tilde{Y}^T \times_{\tilde{Y}} \tilde{X}$. (Is this always true?) Then the length of $X^T$ = the length of $\tilde{X}^T$ because length is constant in flat families.<|endoftext|> -TITLE: Witten's topological twisting -QUESTION [13 upvotes]: I am reading the Witten's topological twisting for $N = 2$ Superconformal Field Theory(SCFT) http://arxiv.org/abs/hep-th/9112056 -In this paper Witten constructed 2 TQFTs i.e. A-model and B-model from an $N=2$ SCFT with a Kahler target manifold. My queries are the following : - -When we do the twist, we're actually changing the interpretation of the fermionic fields e.g. we're taking the field $\psi_{+}^{i}$ to be a section of $\Phi^{\star}(T^{0,1}X)$ instead of a section of $K^{\frac{1}{2}}$ tensor $\Phi^{\star}(T^{0,1}X)$.Mathematically, this seems OK. But on the physics side, are we changing any physics. -Are the topological A and B models still supersymmetric. Does the twisting preserves any supersymmetric. -And is this type of twisting always possible, for any SCFT. - -I am sorry for the question is not very clear, feel free to modify. - -REPLY [11 votes]: First, a historical note: the twisting procedure for $N=2$ SCFTs is due to Eguchi and Yang; although a twisting of sorts had already appeared in Witten's Topological Quantum Field Theory paper of 1988. -Let me give quick answers to your questions: - -Twisting per se does not change the physics: it merely allows one to identify a particular subsector of the theory. The topological theory is obtained by restriction to that subsector. -The topological A and B models are particular subsectors of the sigma model quantum field theory. Supersymmetries will generally not preserve those subsectors, hence the topological theories are no longer supersymmetric. -Any $N=2$ SCFT can be twisted in this way, but more generally one can twist also other conformal field theories. For example, one can twist string theories or also theories based on a Kazama algebra, such as the $G/G$ gauged Wess-Zumino-Witten model. More generally still, it used to be the case that all known two-dimensional topological conformal field theories are "cohomologically equivalent" to a twisted $N=2$ SCFT. (My information is probably out of date, since I last looked at this topic 15 years ago!)<|endoftext|> -TITLE: Propositional logic with categories -QUESTION [14 upvotes]: I have some vague sense that certain types of categories are related to certain types of logic. I've been meaning to learn more about this, so I thought I'd ask about the simplest case, propositional logic. In particular, I'd be interested in a statement of the completeness theorem for propositional logic using these ideas (something like a representation theorem for a certain class of categories). -Here's what I think is a start: a propositional calculus should, at least, behave like a bicartesian closed category $C$, possibly satisfying some extra conditions. The product should be $\vee$, the coproduct should be $\wedge$, and the exponential object should be $\rightarrow$. I think the terminal object is $\top$ and the initial object is $\bot$, and the exponential $p \rightarrow \bot$ should then be $\neg p$. Maybe this means the same thing as "Heyting algebra." -We're specifically interested in the free bicartesian closed category $C_P$ on a set $P$ of primitive propositions, and in functors from $C_P$ to the truth-value category $\mathbf{2} = \{ \bot \rightarrow \top \}$. If I've set up the definitions correctly, then a subset $S$ of $C_P$ semantically entails an object $t$ if and only if, for every functor $F : C_P \to \mathbf{2}$ such that $F(s) = \top \forall s \in S$, it is also true that $F(t) = \top$. -If I've set up the definitions correctly, functors like $F$ are defined by what they send to $\top$. If we want to characterize such sets syntactically, we write down some set of axioms and some set of inference rules telling us when objects sent to $\top$ let us construct other objects sent to $\top$, and then the completeness theorem tells us that these axioms and inference rules characterize the functors $F$. -But I'm clearly missing something important: at some point we introduce the law of the excluded middle, and then our Heyting algebras should collapse to Boolean algebras. But I'm not sure exactly when that point is. Can someone help me out? I suspect this is an exercise in a book on topos theory somewhere. - -REPLY [2 votes]: This paper by Abramsky relates Joyal's Lemma on the collapse of a cartesian closed category with a dualizing object to no cloning theorems and no deleting theorems in quantum mechanics.<|endoftext|> -TITLE: Newton equations, second order equation and (im)possible motions -QUESTION [15 upvotes]: I am am currently studying Newtonian mechanics from a conceptional and axiomatic point of view. Now, if I am not mistaken, one (but surely not all) statement of Newtons second law about nature is, that (almost ?) every motion of a "classical particle" (or "small body") can be described by a second order differential equation on $\mathbb{R}^3$ (or $\mathbb{R}^{3n}$ if one considers a system of $n$ particles). That is, if $I \ni t \mapsto x(t) \in \mathbb{R}^{3n}$ is a motion of $n$ particles in some environment, there is a smooth function $f \colon \mathbb{R}^{3n} \times \mathbb{R}^{3n} \times \mathbb{R} \to \mathbb{R}^{3n}$ (which describes the influence of the environment) such that $\ddot x(t) = f(x(t),\dot x(t), t)$ for all $t \in I$, thereby $I$ is an interval and $\dot x$ denotes the derivative of $x$. -Now my question is, if there is a good, mathematical sound intuition, which kinds motions are not allowed by Newtons second law because of the fact that it is a second order differential equation. -In other words: I want to analyze in detail, what Newtons seconds law tells us about nature. Especially I want to grasp, how the condition to be a second order differential equation gives restrictions to possible conceivable motions of particles. What would be allowed additionally if the equations were of $3$. or higher order? - -REPLY [2 votes]: This is maybe a bit of a simpleton's answer: -Motions of particles are described by second order equations because two particles starting from the same position and with the same velocity follow the same trajectory. -At least no experiment has ever shown evidence of the contrary.<|endoftext|> -TITLE: Commuting Matrices and the Weak Nullstellensatz -QUESTION [15 upvotes]: In the Wikipedia article on Hilbert's Nullstensatz, -http://en.wikipedia.org/wiki/Hilbert%27s_Nullstellensatz -the following application of the Weak Nullstensatz is mentioned: - -Commuting matrices -The fact that commuting matrices have a common eigenvector – and hence by induction stabilize a common flag and are simultaneously triangularizable – can be interpreted as a result of the weak Nullstellensatz, as follows: commuting matrices form a commutative algebra $K[A_1,\ldots,A_k]$ over $K[x_1,\ldots,x_k];$ the matrices satisfy various polynomials such as their minimal polynomials, which form a proper ideal (because they are not all zero, in which case the result is trivial); one might call this the '''characteristic ideal''', by analogy with the characteristic polynomial. -One then defines an eigenvector for a commutative algebra as a vector $v$ such that for all $x \in A$ one has $x(v) = \lambda(x)\cdot v$ for a linear functional $\lambda\colon A \to K.$ This simply linearizes the definition of an eigenvalue, and is the correct definition for a common eigenvector, as if $v$ is a common eigenvector, meaning $A_i(v)=\lambda_i v,$ then the functional is defined as $\textstyle{\lambda(c_0 I + c_1 A_1 + \cdots c_k A_k) := c_0 + \sum c_i \lambda_i}$ (treating scalars as multiples of the identity matrix $A_0 := I$, which has eigenvalue 1 for all vectors), and conversely an eigenvector for such a functional $\lambda$ is a common eigenvector. Geometrically, the eigenvalue corresponds to the point in affine $k$-space with coordinates $(\lambda_1,\ldots,\lambda_k)$ with respect to the basis given by $A_i.$ -Then the existence of an eigenvalue $\lambda$ is equivalent to the ideal generated by (the relations satisfied by) $A_i$ being non-empty, which exactly generalizes the usual proof of existence of an eigenvalue existing for a single matrix over an algebraically closed field by showing that the characteristic polynomial has a zero. - -I am somewhat struggling to make sense of that. The Weak Nullstellensatz says that I find a functional $\lambda \colon K[A_1, \dots, A_k] \to K$, and $\lambda(A_i)$ is an eigenvalue of $A_i$. But how do I conclude that a common eigenvector exists? - -REPLY [6 votes]: My "solution" is a bit strange, but I hope it is correct. -We regard $V$ as a module over $A$. $X = {\rm Spec}\ A$ is zero-dimensional. Then (as a sheaf on $X$) $V$ decomposes into a direct sum of sheaves over the finitely many points of $X$. This corresponds to a decomposition $V$ into subspaces corresponding to different eigenvalues in the sense you described. We must have a point $p$ such that the piece $V_p$ above it (which is the localization of $V$ at the maximal ideal $\mathfrak{m}$ of $A$ corresponding to $p$) is nonzero. Localizing at this ideal and restricting ourselves to the subspace $V_p$, we may assume that $A$ is local. If there would be no common eigenvector in $V_p$, then we would have $\mathfrak{m}V_p = V_p$, so by Nakayama's lemma $V_p = 0$, a contradiction. -darij: Does it deserve to be "awfully sophisticated" by your standards?<|endoftext|> -TITLE: Ergodic limits along subsets of $\mathbb{N}.$ -QUESTION [18 upvotes]: Let say that an infinite subsets $A$ of $\mathbb{N}$ is "nice w.r.to ergodic limits", if it can replace $\mathbb{N}$ in the individual ergodic theorem, that is, if it is such that the following statement is true: - -For any probability space - $(X,\Sigma,\mu),$ for any - measure-preserving transformation $T$ - on $X,$ for any $f\in L^1(X,\mu)$ the - ergodic means along $A,$ -$$M(f,T,A,t,x):=\big|\{j\in A\, : j\leq t\}\big|^{\,-1}\sum_{j\in A,\, j\leq t}f(T^{j}x)$$ -converge for a.e. $x\in X$ to the conditional expectation w.r.to the $T$ invariant $\sigma$-algebra, $\mathbb{E}(f|\Sigma_T),$ as $t \to +\infty.$ - -So $\mathbb{N}$ itself is nice in this sense, by Birkhoff's theorem; if $A$ is nice and $m$ is a positive integer, the set of translates $A+m$ is nice (the set of convergence with $T$ along $A+m$ coinciding with the $T^{\, m}$ pre-image of the set of convergence with $T$ along $A$). Also, a disjoint union of two nice sets is nice. -Is there any other structure on the family of these sets? What about e.g. the union of two of them? (at a glance it seems to me that something more can be said for the analogous cases of other ergodic theorems, e.g. for the $L^p$ convergence. -Looking at this very related question, and its answer, make me think that the situation may be non-trivial and interesting enough, so that it should have been studied. - -REPLY [10 votes]: These are called good universal sets. -Bourgain (1987) proved that sequences of the form $p(n)$, $n \in {\bf N}$, $p$ a non constant polynomial, are good. -He also proved (1988) that the set of primes is a good universal set for $L^p$ functions, $p> {(1+\sqrt{3})\over 2}$. This was later improved to $p>1$ by Wierdl, see its article for a short introduction to the problem http://www.springerlink.com/content/e027w4211n7784h1/fulltext.pdf . There is now an extensive litterature on the problem, following Bourgain's articles. -In another direction, note that a transformation T is mixing iff the ergodic theorem for T holds for all subsequences (see e.g. the book of Krengel, "ergodic theorems").<|endoftext|> -TITLE: sheaves and cosheaves -QUESTION [7 upvotes]: I am struggling hard to understand the pushforwards and pullbacks of cosheaves. Are they also cosheaves? And what are quasicoherent cosheaves? Is there anything like coquasicoherent cosheaves? Please tell me a good refernce on theses topics, if there is some. - -REPLY [10 votes]: I've written a preprint about (what I call) contraherent cosheaves of modules over the structure sheaf of rings of a scheme -- http://arxiv.org/abs/1209.2995 . These are a kind of dual creatures to quasi-coherent sheaves. The preprint will be updated and expanded (eventually).<|endoftext|> -TITLE: How to recognise that the polynomial method might work -QUESTION [67 upvotes]: A couple of days ago I was at a nice seminar given by Christian Reiher, during which he told us about a short proof of the following special case of a theorem of Olson. -Theorem. Let $(a_1,b_1),\dots,(a_n,b_n)$ be a sequence of points in ${\mathbb{Z}}_p^2$ with $n\geq 2p-1$. Then there is a non-empty subset $A\subset\{1,2,\dots,n\}$ such that $\sum_{i\in A}(a_i,b_i)=(0,0)$. -The short proof wasn't short in absolute terms, but was short if you were prepared to accept the following result of Noga Alon, known as the combinatorial nullstellensatz. -Theorem. Let F be a field and let P be a polynomial in n variables $x_1,\dots,x_n$ over F. Let $x_1^{t_1}\dots x_n^{t_n}$ be a monomial of maximal total degree $t_1+\dots+t_n$ that occurs in P with a non-zero coefficient, and let $S_1,\dots,S_n$ be subsets of F such that $|S_i|>t_i$ for every i. Then there exist $s_i\in S_i$ such that $P(s_1,\dots,s_n)\ne 0.$ -Once you have the combinatorial nullstellensatz, the special case of Olson's theorem (and I think the whole theorem) is reduced to a nice exercise: basically, once you sit down and think about it you quickly see that it makes sense to choose each $S_i$ to be {0,1}, and then a few simple tricks using Fermat's little theorem (the polynomial $1-x^{p-1}$ is zero if $x\ne 0$ and 1 otherwise) you can finish off quite easily. -This method is known as the polynomial method. My question is not how to apply the combinatorial nullstellensatz. It's how to recognise, when you see a problem, that the polynomial method might work. In this case, once you have that clue, it's easy to finish off. But how do you manage if there's nobody there to give you the clue? -I'm interested in this question in general: I've always found spotting that mathematical results can be used quite a difficult process -- somehow I have to do it for myself in one problem before I truly understand how to do it in other problems. And here's a good example where I have never spotted how to use the result. And had I been faced with the task of proving Olson's theorem, I don't think it would have occurred to me to use it. - -REPLY [10 votes]: I would like to add some more examples and references for the so called polynomial method that can help us recognise when it can be applied. -From what I understand so far, the polynomial method falls under these two big categories: -A. Construct an explicit polynomial (or a set of polynomials) that captures the given set. -B. Use interpolation arguments to get a polynomial whose degree (or some other property) is under control. -An important, and well known, subcategory of A is the dimension argument using polynomial spaces. -For example, if we want to show that there can be at most $n(n+1)/2$ equiangular lines in $\mathbb{R}^n$, then with each line we associate the polynomial $P_u = (u, x)^2 - \alpha^2(x, x)$ where $u$ is a fixed unit vector on the line and $arccos(\alpha)$ is the angle between every pair of lines. Then we show that $P_u(v) = (1 - \alpha) \delta_{u, v}$, proving that these polynomials are linearly independent. Now, all these polynomials are degree $2$ homogenous polynomials in $n$ variables, and hence the number is bounded above by the dimension of this spaces, $n(n+1)/2$. Another example that demonstrates this approach is the bound on $2$-distance sets in Euclidean spaces. For further details, and several examples I would refer to the manuscript "Linear Algebra Methods in Combinatorics" by Babai and Frankl. -Another subcategory is when you use the coefficients or degrees of the explicit polynomial. One of the simplest examples is perhaps the result by Blokhuis on nuclei of sets in $\mathbb{F}_q^2$. A nucleus of a set $S$ of points in $\mathbb{F}_q^2$ is a point $x$, not in $S$, such that every line through $x$ intersects $S$. Blokhuis showed that if $S$ has size $q + k$, with $k \leq q$, then all nuclei of $S$ are roots of a degree $k(q-1)$ polynomial (hint: an elementary symmetric polynomial), hence proving that the total number of nuclei is at most $k(q-1)$. -The Brouwer-Schrijver/Jamison bound on affine blocking sets also uses a property of polynomials over finite fields that, if a polynomial in $\mathbb{F}_q[t_1, \dots, t_n]$ vanishes on all points of $\mathbb{F}_q^n$ except one, then its degree must be at least $n(q-1)$. This is quite similar to the classical Chevalley-Warning theorem. And in fact, this can be proved via a dimension argument as well. -I think Combinatorial Nullstellensatz, and all its applications that I am aware of, fall under category A as well. Here, we construct a polynomial that is explicit enough so that we can determine the coefficient of its leading monomial. The answer by Gjergji Zaimi covers this approach nicely. If we want estimates on the total number of non-vanishing points, instead of just existence, then we can sometimes use the result by Alon-Füredi, as it was recently done by Clark, Forrow and Schmitt in this paper. -Category B seems to be somewhat more recent. It was used by Dvir in his proof of finite field Kakeya conjecture, and it has then found several applications. Terence Tao has written a nice survey on it which can be found here. Quoting Tao, - -Broadly speaking, the strategy is to capture (or at least partition) - the arbitrary sets of objects (viewed as points in some configuration - space) in the zero set of a polynomial whose degree (or other measure - of complexity) is under control; for instance, the degree may be - bounded by some function of the number of objects. One then uses tools - from algebraic geometry to understand the structure of this zero set, - and thence to control the original sets of object. - -Another important breakthrough using this approach was the result of Guth and Katz on Erdős distinct distances problem. -Can these two approaches (A and B) be combined? I would love to see an example of that. But this is what Terence Tao had to say about combining CN and Interpolation, - -roughly speaking, the idea is to start with a counterexample to the claimed extremal result, and then use this counterexample to design a polynomial vanishing on a large product set and which is explicit enough that one can compute a certain coefficient of the polynomial to be non-zero, thus contradicting the nullstellensatz. This should be contrasted with more recent applications of the polynomial method, in which interpolation theorems are used to produce the required polynomial. Unfortunately, the two methods cannot currently be easily combined, because the polynomials produced by interpolation methods are not explicit enough that individual coefficients can be easily computed, but it is conceivable that some useful unification of the two methods could appear in the future - -More References - -Aart Blokhuis, Polynomials in finite geometries and combinatorics. -A. A. Bruen, J. C. Fisher, The Jamison method in galois geometries. -Noga Alon, Discrete mathematics: methods and challenges. -Gyula Károlyi, The polynomial method in additive combinatorics. -Simeon Ball, Polynomials in Finite Geometries and The Polynomial Method in Galois Geometries. -Terence Tao and Van Vu, Chapter 9 in Additive Combinatorics. -Zeev Dvir, Incidence Theorems and Their Applications. -Terence Tao, Algebraic combinatorial geometry: the polynomial method in arithmetic combinatorics, incidence combinatorics, and number theory. -Peter Sziklai, Polynomials in finite geometries and Applications of Polynomials over Finite Fields. -Larry Guth, Polynomial Method in Combinatorics.<|endoftext|> -TITLE: Do convex and decreasing functions preserve the semimartingale property? -QUESTION [18 upvotes]: Some time ago I spent a lot of effort trying to show that the semimartingale property is preserved by certain functions. Specifically, that a convex function of a semimartingale and decreasing function of time is itself a semimartingale. This was needed for a result which I was trying to prove (more details below) and eventually managed to work around this issue, but it was not easy. For twice continuously differentiable functions this is an immediate consequence of Ito's lemma, but this cannot be applied in the general case. After failing at this task, I also spent a considerable amount of time trying to construct a counterexample, also with no success. So, my question is as follows. - -1) Let $f\colon\mathbb{R}^+\times\mathbb{R}\to\mathbb{R}$ be such that $f(t,x)$ is convex in $x$ and continuous and decreasing in $t$. Then, for any semimartingale $X$, is $f(t,X_t)$ necessarily a semimartingale? - -Actually, it can be assumed here that $X$ is both continuous and a martingale which, with some work, would imply the general case. -As it turns out, this can be phrased purely as a real-analysis question. - -2) Let $f\colon\mathbb{R}^+\times\mathbb{R}\to\mathbb{R}$ be such that $f(t,x)$ is convex in $x$ and decreasing in $t$. Can we write $f=g-h$ where $g(t,x)$ and $h(t,x)$ are both convex in $x$ and increasing in $t$? - -Stated like this, maybe someone with a good knowledge of convex functions would be able to answer the question one way or the other. -For $f(t,x)$ convex in $x$ and increasing in $t$ then approximation by smooth functions and applying Ito's lemma allows us to express $f(t,X_t)$ as the sum of a stochastic integral and an increasing process -$$ -f(t,X_t)=\int_0^t\frac{\partial f}{\partial x}(s,X_s)\,dX_s+V_t,\qquad{\rm(*)} -$$ -so it is a semimartingale. If, instead, $f$ is decreasing in $t$, then an affirmative answer to question 2 will reduce it to the easier case where it is increasing in $t$, also giving a positive answer to the first question. -Explaining why question 1 implies 2 is a bit trickier. If 2 was false, then it would be possible to construct martingales $X$ such that the decomposition (*) holds where the variation of $V$ explodes at some positive time. -This problem arose while I was trying to prove the following: is a continuous martingale uniquely determined by its one dimensional marginals? For arbitrary continuous martingales this is false, but is known to be true for diffusions $dX_t=\sigma(t,X_t)\,dW$ for Brownian motion $W$ and smooth parameter $\sigma$. The idea is to back out $\sigma$ from the Kolmogorov forward equation. This is well-known in finance as the local volatility model. However, I was trying to show rather more than this. All continuous and strong Markov martingales are uniquely determined by the one dimensional marginals. I was able to prove this, and the relation between the marginals and joint distributions of the martingale has many nice properties (I wrote a paper on this, submitted to the arXiv, but not published as I am still working on changes asked for by the referees). The method was to reformulate the Kolmogorov backward equation in terms of the marginals. This does use Ito's lemma, requiring twice differentiability, but can be circumvented with a bit of integration by parts as long as $f(t,X_t)$ is a semimartingale for the kinds of functions mentioned above. The question above arose from trying, and failing, to prove this. Without an answer to this question, the problem becomes much much harder, as many of the techniques from stochastic calculus can no longer be applied (and, approximating by semimartingales didn't seem to help either). The work around was very involved, part of which I published as a standalone paper here and the rest forms most of a paper I submitted to the arXiv here. Adding together those papers, it comes to maybe 50 pages of maths and a lot of effort to work around the question above. - -Update: I posted more details on the various equivalent forms of this question here. I also posted a possible counterexample, without proof but with some numerical evidence suggesting that it may indeed be a counterexample. - -REPLY [4 votes]: I still have no idea what the answer to this question is. However, it is possible to attack the problem in several different ways, and there are various different (but logically equivalent) ways of stating it. I'll post some of these as an answer now, as it seems rather long to fit in the original statement of the question. Maybe one of the reformulations below will help lead to a resolution of the problem. -This answer is already very long, and I've been trying to shorten it as much as I can. I can't see any way of giving proper proofs of all the statements below without making it a lot longer. So, I'll only give very few details of he proofs here. I will, however, list each of the equivalent formulations H1-H6 below in the logical order, so that each statement can be proven to be equivalent to the preceding one without too much work. Let's start from statement 2 of the original question, which I will refer to as Hypothesis (H1) for the purposes of this answer. - -Hypothesis (H1): Let $f\colon\mathbb{R}^+\times\mathbb{R}\to\mathbb{R}$ be such that $f(t,x)$ is convex in $x$ and decreasing in $t$. Then, $f=g-h$ where $g(t,x)$ and $h(t,x)$ are both convex in $x$ and increasing in $t$. - -The decomposition in (H1) exists if and only if it exists locally. Letting -$I=[0,1]$ denote the unit interval, we get the following equivalent statement. - -Hypothesis (H2): Let $f\colon I^2\to\mathbb{R}$ be such that $f(t,x)$ is convex and Lipschitz continuous in $x$ and decreasing in $t$. Then, $f=g-h$ where $g(t,x)$ and $h(t,x)$ are convex in $x$ and increasing in $t$. - -The truth of this statement remains unchanged if it is restricted to functions ƒ which are zero on the three edges $I\times\{0\}$, $I\times\{1\}$, $\{0\}\times I$ of the unit square. I'll use $D$ to denote the set of such functions satisfying the conditions of (H2). Then, whenever the decomposition in (H2) exists, it is always possible to choose $g$, $h$ to be zero on $I\times\{0\}$, $I\times\{1\}$ and $\{1\}\times I$. From now on, whenever the decomposition $f=g-h$ in (H2) is referred to, it will be assumed that $g$, $h$ are chosen to satisfy these conditions. -We can strengthen (H2) by also placing a uniform bound on the terms $g$, $h$ in the decomposition. Here, $\lVert g\rVert$ denotes the supremum norm and $f_x$ denotes the partial derivative. - -Hypothesis (H3): There is a constant $K\gt0$ such that, for all $f\in D$, the decomposition $f=g-h$ as in H2 exists and can be chosen such that $\lVert g\rVert,\lVert h\rVert\le K\lVert f_x\rVert$. - -Statement (H3) is particularly convenient because of the following: to prove that (H3) holds, it is enough to look at a dense subset of functions in $D$. Taking limits of the decompositions would then extend to the result to all of $D$. So, it is enough to concentrate on, say, smooth functions or piecewise linear functions. -Next, it is useful to choose the decomposition in (H2) to minimize $\lVert g\rVert$ and $\lVert h\rVert$. - -Lemma 1: Suppose that $f\in D$ and that the decomposition $f=g-h$ as in (H2) exists. Then, there is a unique maximal choice for $g$, $h$. That is, if $f=g_1-h_1$ is any other such decomposition then $g\ge g_1$ and $h\ge h_1$. - -I'll refer to the decomposition in Lemma 1 as the optimal decomposition. As it not clear that any such decomposition should exist, I'll briefly sketch the argument now. The idea is to discretize time, using a partition of the unit interval $0=t_0\lt t_1\lt\cdots\lt t_r=1$. Denote the convex hull of a function $u\colon I\to\mathbb{R}$ by $v=H(u)$, which is the maximum convex function $v\colon I\to\mathbb{R}$ bounding $v$ from below, -$$ -\begin{align} -v(x) &= \sup\left\{w(x)\colon w\le u{\rm\ is\ convex}\right\}\\ -&=\inf\left\{\left((b-x)u(a)+(x-a)u(b)\right)/(b-a)\colon a\le x\le b\right\}. -\end{align} -$$ -The optimal decomposition in discrete time can be constructed as functions $h_k\colon I\to\mathbb{R}$, starting at the final time $k=r$ and working backwards to $k=0$, -$$ -h_r(x) = 0,\ h_{k-1}={\rm H}\left(h_k+f(t_k,\cdot)-f(t_{k-1},\cdot)\right). -$$ -Interpolate this to be piecewise constant in time, defining $h(t,x)=h_k(x)$ for $t_{k-1}\lt t\le t_k$. Then, $h(t,x)$ and $g\equiv f+h$ are convex in $x$ and increasing in time $t$, restricting to times in the partition. - -Lemma 2: Suppose that $f\in D$ and let $0=t_{n,0}\lt t_{n,1}\lt\cdots\lt t_{n,r_n}=1$ be a sequence of partitions of the unit interval. For each $n$ let $h^n(t,x)$ be the function corresponding to the partition and piecewise constant in $t$, as constructed above. We also suppose that the partitions have mesh going to zero and eventually include all times at which $f$ is discontinuous. Then one of the following holds. - -$f$ decomposes as in (H2) and $h^n(t,x)\to h(t,x)$ pointwise on $I^2$, where $f=g-h$ is the optimal decomposition. -$f$ does not have a decomposition as in (H2) and $h^n(0,x)\to-\infty$ for all $0\lt x\lt1$. - - -The idea is that, if $h^n$ has any limit point $h$, then $h(t,x)$ and $g=f+h$ will be convex in $x$ and increasing in $t$, giving the decomposition required by (H2). Furthermore, by construction, if $f=g^\prime-h^\prime$ is any other such decomposition, then $h^n\ge h^\prime$ at times in the partition, so $h\ge h^\prime$. This shows that the decomposition is optimal and, as the optimal decomposition is unique, all limit limit points of $h^n$ are the same, so $h^n\to h$. The only alternative is that $h^n$ has no limit points, in which case the second statement of the Lemma holds. Using this construction of the optimal decomposition, (H3) can be shown to be equivalent to the following. - -Hypothesis (H4): There is a constant $K\gt0$ such that, for all smooth functions $f,g\colon I^2\to\mathbb{R}$ with $\lVert f\rVert$, $\lVert g\rVert$, $\lVert f_x\rVert$, $\lVert g_x\rVert$ bounded by 1 and $f(t,x)$, $g(t,x)$ convex in $x$ and respectively decreasing and increasing in $t$, then, - $$\int_0^1\int_0^1 f_{xx}g_t\,dxdt \le K.$$ - -As this statement is quite different from the preceding ones, I'd better give some explanation now. The idea is to use integration by parts, -$$ -\begin{align} -\int_0^1\int_0^1(f_{xx}g_t+g_{xx}f_t)\,dxdt &= \left[\int_0^1(f_xg_t+g_xf_t)\,dt\right]_{x=0}^1-\left[\int_0^1f_xg_x\,dx\right]_{t=0}^1\\ -&\le 6(\Vert f_x\Vert \Vert g\Vert + \Vert g_x\Vert \Vert f\Vert). -\end{align} -$$ -If $f$ and $g$ are increasing in time then the terms on the left hand side are both positive, so we get bounds for the integrals of $f_{xx}g_t$ and $g_{xx}f_t$ individually. Hypothesis (H3) extends this to the case where $f$ is decreasing in time, implying (H4). -Conversely, suppose that (H4) holds. Letting $f=g-h$ be the decomposition computed along a partition as described above, we can use the fact that the convex hull $v=H(u)$ of a function $u$ satisfies $v_{xx}(u-v)=0$ to get the equality $(h_{k-1})_{xx}(h_k-h_{k-1}+f(t_k,\cdot)-f(t_{k-1},\cdot))=0$. This leads to the following inequalities, -$$ -\begin{align} -\frac12\Vert h\Vert^2&\le\frac12\int_0^1 h_x(0,x)^2\,dx\le\sum_{k=1}^r \int_0^1(h_{k-1})_{xx}(h_{k-1}-h_k)\,dx\\ -&=-\sum_{k=1}^r\int_0^1 (h_{k-1})_{xx}(f_k-f_{k-1})\,dx. -\end{align} -$$ -Hypothesis (H4) can be used to bound the final term, showing that $h$ cannot diverge as the mesh of the partition goes to zero, so we get convergence to an optimal decomposition satisfying a bound as in (H3). -The hypothesis can now be formulated as a statement about martingales. - -Hypothesis (H5): There is a constant $K\gt0$ such that, for all $f\in D$ and martingales $0\le X_t\le1$, $f(t,X_t)$ decomposes as $M+V$ where $M$ is a martingale and $V$ has variation - $$ -\mathbb{E}\left[\int_0^1\,\vert dV\vert\right]\le K\Vert f_x\Vert. -$$ - -The idea is that $g(t, x)\equiv\mathbb{E}[(X_t-x)_+]$ is convex in $x$ and increasing in $t$. -In the case where $f$, $g$ are smooth and $X$ is a continuous martingale, -Ito's formula can be used to split $f(t,X_t)$ into a martingale term plus the sum of the increasing process $\frac12\int f_{xx}(t,X_t)\,d[X]_t$ and the decreasing process $\int f_t(t,X_t)\,dt$, which have expectations $\iint f_{xx}g_t\,dxdt$ and $\iint f_tg_{xx}\,dxdt$ respectively. -Finally, by adding a randomly occurring jump term to any semimartingale, and with a bit more work, it is possible to reduce this to the martingale case. This gives the statement asked in the original question. - -Hypothesis (H6): Let $\colon\mathbb{R}^2\to\mathbb{R}$ be such that $f(t,x)$ is convex in $x$ and continuous and decreasing in $t$. Then, for any semimartingale $X$, $f(t,X_t)$ is a semimartingale.<|endoftext|> -TITLE: Quantum Group Calculations in Mathematica -QUESTION [8 upvotes]: I'm trying to learn how to do algebraic manipulations in Mathematica but not finding the help very helpful. I'm going to ask about a specific quantum group example related to a previous question of mine. For $SU_q(N)$, how would I use Mathematica to show that -$$ -S(u^1_2)u^3_1 = q^{-1}u^3_1S(u^1_2). -$$ -I am, of course, assuming that such a thing can be done in Mathematica. If I am wrong in this assumption, could someone please direct me a package that can do this calculation? Gap, Magma? - -REPLY [7 votes]: I just found the link to the QuantumGroups Mathematica package by Scott Morrison mentioned by Noah: -http://katlas.org/wiki/QuantumGroups%60<|endoftext|> -TITLE: Blowups of Cohen-Macaulay varieties -QUESTION [12 upvotes]: Suppose that $X$ is a Cohen-Macaulay normal scheme/variety and $\pi : Y \to X$ is a proper birational map with $Y$ normal. -Question: Is $Y$ also Cohen-Macaulay? Are there common conditions which imply it is? -If $Y$ is not normal I know of several ways to show that the answer to the first question is no. -There are obvious spectral sequences but I don't see how to deduce what I want from them, perhaps I'm being dumb (or maybe there is an obvious example). - -REPLY [11 votes]: An example was given in Section 3 of this paper by Cutkosky: "A new characterization of rational surface singularities" (The scheme $Z$ in the last page, which is a blow up of some $m$-primary ideal of a regular local ring of dimension $3$, is normal but not Cohen-Macaulay). -The algebraic side of this example has been studied quite a bit, so perhaps more explicit examples are known. I am not an expert here, but you can check out a paper by Huckaba-Huneke here, or papers by Vasconcelos (he has a book called "Arithmetic of Blow-up algebras" which discussed, among other things, Serre's condition $(S_n)$ on Rees algebras), and the references there.<|endoftext|> -TITLE: Monodromy with algebraic fundamental groups -QUESTION [12 upvotes]: Topological setting -Say we have a fiber bundle: $p: X \rightarrow B$. Let $s: B \rightarrow X$ be a section. Then $\pi_1(B,b)$ acts on $\pi_1(F_b,s(b))$ (where $F_b:=p^{-1}(b)$) by: Let $\gamma \in \pi_1(B,b)$ and $\delta \in \pi_1(F_b,s(b))$. Deform $\delta$ above $\gamma$ in such a way that above $\gamma(t)$, $\delta(t)$ will be a loop in $p^{-1}(\gamma(t))$ with base point $s(\gamma(t))$. At $t=1$ we will arrive at some new loop in $\pi_1(F_b,s(b))$. This defines the action of $\gamma$ on $\delta$. -Question -Is there a way to define this algebraically? Meaning: -Let $p:X \rightarrow B$ be a map of integral schemes (varieties if you prefer?). Assume it is flat and with isomorphic fibers. Let $s:B \rightarrow X$ be a section. Can we then define, similarly, a monodromy action of $\pi_1(B,b)$ ($b$ a geo. point) on $\pi_1(X,s(b))$? -And maybe my setting is off. What is the right setting for this (algebraically), and how would you define this action? - -REPLY [2 votes]: Here is a way to change the question to make it easier to answer. -Let $p:X \to B$ be a topological fiber bundle with fiber $F$, such that $F$, $X$, and $B$ are all connected and tame (say, CW complexes). Then there is a homotopy exact sequence -$$\pi_2(B) \to \pi_1(F) \to \pi_1(X) \to \pi_1(B) \to 1.$$ -Let $N$ be the quotient of $\pi_1(F)$ by the image of $\pi_2(X)$. Then $N$ is also a subgroup of $\pi_1(X)$, and -$$\pi_1(X)/N \cong \pi_1(B).$$ -So $\pi_1(B)$ acts on $N$ because a quotient group always acts by conjugation on the kernel. -If $X$ has a global section as a bundle over $B$, this implies that the term $\pi_2(B) \to \pi_1(F)$ is trivial (because a sphere in $\pi_2(B)$ then lifts to a disk mapped to $X$ whose boundary is a constant loop in $F$). If this term is trivial, whether or not $X$ has a global section, then $\pi_1(F) = N$ and so $\pi_1(B)$ acts on $\pi_1(F)$. -Now, suppose that $p:X \to B$ is a map of algebraic varieties and $F$ is some fiber over a closed point. Then the étale fundamental group $\tilde{\pi}_1$ is a functor on the relevant category. So you at least get a sequence -$$\tilde{\pi}_1(F) \to \tilde{\pi}_1(X) \to \tilde{\pi}_1(B) \to 1.$$ -Then maybe the composition of the first two maps is trivial because $F$ is sent to a closed point. For a general $p$ (taking simple examples from topology), this sequence is not an exact sequence. But, if there are suitable conditions on $p$ for it to be analogous to a fiber bundle, then maybe it should be exact. Note that the étale fundamental group $\tilde{\pi}_1(V)$ is the profinite completion of the analytic fundamental group $\pi_1(V)$ when $V$ is a variety over $\mathbb{C}$. On the negative side, profinite completion is not an exact functor on groups. (But what about for fundamental groups of complex algebraic varieties?) -If it is exact, then I'm not sure whether the kernel in $\tilde{\pi}_1(F)$ can be naturally related to anything like an étale $\tilde{\pi}_2(B)$, but maybe so.<|endoftext|> -TITLE: Comparison between Hamiltonian Floer cohomology and Lagrangian Floer cohomology of the diagonal -QUESTION [15 upvotes]: Let X be a compact symplectic manifold with a form $\omega$. And $X \times X$ is equipped with the symplectic form $(\omega,-\omega)$. The diagonal $\Delta:X \mapsto X \times X$ is a Lagrangian submanifold. So, in this question, Hochschild (co)homology of Fukaya categories and (quantum) (co)homology. Tim Perutz says "PSS is a canonical ring isomorphism from QH∗(X) to the Hamiltonian Floer cohomology of X, and the latter can be compared straightforwardly to the Lagrangian Floer cohomology of the diagonal." I have no doubt that this second assertion is straightforward, since I have consulted a couple of references and no one spells this out. But, I don't quite see it. I believe what I am missing is the relationship between holomorphic strips in $X\times X$ and holomorphic cylinders in X. (Edit: I would also like to understand the comparison of the product structures too) -Edit: here is a rough geometric idea which might have something to do with the truth. I want to assume that my Hamiltonian is time independent and that all orbits are actually fixed points. Given a map of a strip into $X\times X$ the two projections give us two strips into X. The idea is to glue the strips together to form a cylinder which is a map into X. Of course, this doesn't take into account issues of compactifications and so on... Anyways, if someone would be happy to spell it out I would appreciate it. - -REPLY [9 votes]: Michael's answer describes the sort of thing I had in mind when I brusquely dismissed this as "straightforward" in the earlier, big-picture-style answer that Daniel quoted. (Michael, thanks for explaining.) -This isomorphism intertwines the product structures: think of cutting a pair of pants into two identical pieces. -I'd like to add a cautionary note. Before you can compare Hamiltonian Floer homology to the Lagrangian Floer self-homology of the diagonal, you have to define both groups. The difficulties in doing so are not identical. In the Lagrangian case, disc-bubbling potentially obstructs the construction. -For particularly simple manifolds - monotone ones, where $c_1$ is a positive multiple of $[\omega]$ when evaluated on spherical homology classes - both groups have reasonably straightforward definitions. This is a setting where the isomorphism really works. -For Calabi-Yau manifolds, the definition of Hamiltonian Floer homology is still not too complicated (it involves understanding transversality of evaluation on moduli spaces of somewhere-injective holomorphic spheres). But Lagrangian Floer homology of a Maslov-index-zero Lagrangian in a Calabi-Yau is hard to construct. So far as I know, one generally has to work rationally and to compute obstructions in the sense of Fukaya-Oh-Ohta-Ono. Presumably, in the case of the diagonal these obstructions vanish. -So there is a definite reason for preferring the Hamiltonian Floer homology model when trying to formulate things precisely: it bypasses the obstructions, and (for Calabi-Yau manifolds) the virtual transversality theory.<|endoftext|> -TITLE: Technical phd after 33 -QUESTION [6 upvotes]: Is it a good idea to get a math/stats/theoretical cs phd after 33 in the hope of becoming a professor? I am an electrical engineer who did his phd in EE at a low ranked school. I have been working in industry with a lost soul. I did enjoy my research and published quite a few papers. I have taken courses till Algebraic number theory and audited lectures on etale cohomology. I am thinking of a decent school like UCLA or UC Berkeley to pursue a second doctorate. - -REPLY [11 votes]: In my opinion, people are not giving the OP enough credit for already having a PhD in a related field. This means s/he has a better grasp of what it means to get a PhD than 99% of the general population, has already shown considerable technical skill, the drive to complete a task which requires years of hard work, and so forth. -So far as I know, by far the most common reason for someone to drop out of a doctoral program in math (or any discipline) is that they just didn't really know what they were getting into: they had no way of properly gauging the scope and amount of work involved. -I am currently involved in graduate admissions at UGA. I have only been doing this for a little while, so my ideas may change, but at the moment if I saw a candidate who had a decent undergraduate background, test scores within the normal range of our successful applicants (e.g., greater than 50% on the math subject GRE) and already had a PhD, I would be tempted to put them towards the top of the list. Thinking it through as a hypothetical admissions decision, my only major concern would be that the candidate's undergraduate background may not be as fresh or relevant as that of our other strong applicants. Thus I would recommend taking a few math classes at the advanced undergraduate level. With good performance on those, I see no reason not to count the candidate's previous experience as an advantage, rather than his/her age as a disadvantage.<|endoftext|> -TITLE: Grothendieck topologies, Mayer-Vietoris, and points -QUESTION [13 upvotes]: I am trying to think about certain problems in the theory of motives without having a proper background in Grothendieck topologies and the like, hoping to teach myself the related techniques in the process. Here is a rather specific question that I've stumbled upon; I would appreciate any references and/or explanations of the relevant issues. -Consider a topology "of reasonable size". What I have in mind is the Zariski, Nisnevich, or etale site of an algebraic variety. Let us call the objects of this topology "open sets". My understanding is that there is also some notion of "points" of a Grothendieck topology, and that for the three topologies mentioned above these are the spectra of the local rings of the points of the scheme, the spectra of Henselizations of these local rings, and the spectra of the strict Henselizations of these local rings, respectively. Please correct me if this is wrong. -I think one can define a "cohomology theory" on a site as a sequence of contravariant functors, indexed by nonnegative integers, from the category of open sets to, say, abelian groups, such that whenever an open set is covered by two other ones, there is a cohomological Mayer-Vietoris sequence. Given a cohomology theory, one can also define its value on any point of the site simply by passing to the inductive limit. -Suppose that I have a morphism of cohomology theories on, say, the Nisnevich site of an algebraic variety. Assume that this morphism is an isomorphism at all points. Does it follow that it is an isomorphism of cohomology theories (i.e., an isomorphism on all open sets)? - -REPLY [3 votes]: You definition of a cohomology theory is rather strange since it does not seem to depend on the choice of the site (it corresponds to Zariski topology). -For the correct definition of cohomology theory (I don't know it by heart:)), the answer to your question is probably 'yes'. The corresponding condition is 'topos has enough points' -(for example, see http://webcache.googleusercontent.com/search?q=cache:htiGfZlZqc0J:ncatlab.org/nlab/show/point%2Bof%2Ba%2Btopos+site+has+enough+points+sheaf&cd=1&hl=en&ct=clnk). -It is fullfiled for the topologies your mentioned. Moreover, this condition was mentioned explicitly in some papers of Voevodsky or Suslin.<|endoftext|> -TITLE: Is there a characterization of free groups in terms of the unitary dual? -QUESTION [13 upvotes]: If $G$ is a countable discrete group, I'm curious if it is possible to decide whether $G$ is a free group only by looking at properties of $Rep(G)$, the collection of (equivalence classes of) strongly continuous unitary representations of G on separable Hilbert space. -If this question is too crude, consider the following refinement: -If $G$ is a countable discrete group, is it possible to decide whether $G$ is a free group only by looking at properties of $Rep(G)$ viewed as the unitary dual of $G$, the topological space of (unitary equivalence classes of) unitary representations of $G$ on separable Hilbert space, equipped with Fell's topology? - -REPLY [4 votes]: There is a very strong form of residual finiteness that holds for free groups, property FD, introduced by Lubotzky and Shalom, which says that finite representations are dense in the unitary dual (with respect to the Fell topology). So property FD is certainly a necessary condition for the group to be free. -If $S\subset G$ is a generating set for $G$, then we may detect whether $S$ generates $G$ freely from the unitary dual. Let $F_S$ be the free group generated by $S$. Then the map $F_S\to G$ induces a bijection of unitary duals $\tilde{G}\to\tilde{F_S}$ if and only if $S$ generates $G$ freely. This is because free groups are residually finite, so if $w\in F_S$ were a relator $G$, then there would be a finite group $H$ and a homomorphism $\varphi:F_S\to H$ such that $\varphi(w)\neq 1$. So the finite unitary representation of $F_S$ corrsponding to $\varphi$ would not be in the image of $\tilde{G}\to \tilde{F_S}$.<|endoftext|> -TITLE: Embedding Theorem for big line bundles -QUESTION [5 upvotes]: Kodaira embedding theorem says that a positive line bundle is ample, i.e. high tensor powers are holomorphically embeddable into complex projective space of high dimension. -However, ampleness is not stable under blow-ups. Usually a replacement is to consider big line bundles, which is stable under blow-ups. -Is there an embedding theorem for big line bundles? One cannot hope that an embedding of high tensor powers of a big line bundle to be everywhere injective, but can we have injectivity almost everywhere? -Here's the precise question. Let $X$ be a complex compact manifold. Let $L$ be an ample line bundle over $X$. Let $f:Y\to X$ be a blow up, or series of blow ups. Is there some condition for the pullback $f^*L$ to be an embedding outside the exceptional locus of $f$? - -REPLY [3 votes]: Colin, -Francesco has already mentioned the "embedding theorem for big line bundles" that you were probably looking for, but I would like to add that this is pretty much the definition of big. -An alternative point one could make is that according to Kodaira's Lemma "big=ample+effective", so a high power of a big divisor is very ample on the complement of an effective divisor. -As for your last question, if $f:Y\to X$ is any proper morphism and $\mathscr L$ is an ample line bundle on $X$, then if $g$ denotes the morphism(!) defined by the global sections of very high powers of $f^*\mathscr L$, then it follows that $f$ factors through $g$. In particular, then $g$ is an isomorphism to its image (hence an embedding) wherever $f$ is (if $f$ is not birational, then this "wherever" is the empty set).<|endoftext|> -TITLE: Can an etale (phi, Gamma) module be an extension of non-etale ones? -QUESTION [10 upvotes]: This question is about p-adic representations of $\mathrm{Gal}(\overline{\mathbb{Q}}_p / \mathbb{Q}_p)$ and $(\varphi, \Gamma)$-modules. By theorems of Fontaine, Cherbonnier-Colmez and Kedlaya, the category of p-adic representations of $\mathrm{Gal}(\overline{\mathbb{Q}}_p / \mathbb{Q}_p)$ is equivalent to each of the following three categories: - -etale $(\varphi, \Gamma)$-modules over Fontaine's ring $\mathbb{B}_{\mathbb{Q}_p}$ -etale $(\varphi, \Gamma)$-modules over the subring $\mathbb{B}^{\dagger}_{\mathbb{Q}_p}$ -slope zero $(\varphi, \Gamma)$-modules over the Robba ring $\mathcal{R}$ (also known as $\mathbb{B}^{\dagger}_{\mathrm{rig}, \mathbb{Q}_p}$). - -It's well known that slope 0 $(\varphi, \Gamma)$-modules over the Robba ring can sometimes be written as extensions of other Robba-ring $(\varphi, \Gamma)$-modules which are not themselves of slope 0. (Indeed there is the whole rich theory of trianguline representations, whose Robba-ring $(\varphi, \Gamma)$-modules are built up entirely from rank 1 pieces.) -My question: does this happen for either of the other two categories of $(\varphi, \Gamma)$-modules? Can one have a short exact sequence of $(\varphi, \Gamma)$-modules over $\mathbb{B}_{\mathbb{Q}_p}$ or $\mathbb{B}^{\dagger}_{\mathbb{Q}_p}$ where the middle term is etale but the two end terms are not? - -REPLY [9 votes]: In the first two cases, the slopes of $\varphi$-modules are given by the "standard" Dieudonné-Manin decomposition. In particular, subobjects of étale objects are étale. -For more info, see (for example) chapter 4.5 of Kedlaya's "Slope Filtrations Revisited".<|endoftext|> -TITLE: Automorphic forms on GL(3) -QUESTION [10 upvotes]: We know that the classical Maass forms on GL(3) are depicted, for instance, in D.Goldfeld's book. I wonder that if there exists "holomorphic" automorphic forms on GL(3) as an analogue of GL(2) case. If those forms exist, where can I find the materials which concretely tell the story of them? -Thanks in advance. - -REPLY [3 votes]: Slightly old question, but: There are Maass forms on GL(3) whose minimal K-type is non-spherical. There is a paper of Miyazaki on the (g,K) module structure of GL(3), and I'm working on a more complete description.<|endoftext|> -TITLE: Ramanujan and algebraic number theory -QUESTION [25 upvotes]: One out of the almost endless supply of identities discovered by Ramanujan -is the following: -$$ \sqrt[3]{\sqrt[3]{2}-1} = \sqrt[3]{\frac19} - \sqrt[3]{\frac29} + \sqrt[3]{\frac49}, $$ -which has the following interpretation in algebraic number theory: the fundamental unit -$\sqrt[3]{2}-1$ of the pure cubic number field $K = {\mathbb Q}(\sqrt[3]{2})$ becomes a cube in the extension $L = K(\sqrt[3]{3})$. -Are there more examples of this kind in Ramanujan's work? - -REPLY [13 votes]: $$(7 \sqrt[3]{20} - 19)^{1/6} = \ \sqrt[3]{\frac{5}{3}} -- \sqrt[3]{\frac{2}{3}},$$ -$$\left( \frac{3 + 2 \sqrt[4]{5}}{3 - 2 \sqrt[4]{5}} -\right)^{1/4}= \ \ \frac{\sqrt[4]{5} + 1}{\sqrt[4]{5} - 1},$$ -$$\left(\sqrt[5]{\frac{1}{5}} + \sqrt[5]{\frac{4}{5}}\right)^{1/2} -= \ \ (1 + \sqrt[5]{2} + \sqrt[5]{8})^{1/5} = \ \ -\sqrt[5]{\frac{16}{125}} + \sqrt[5]{\frac{8}{125}} + \sqrt[5]{\frac{2}{125}} - \sqrt[5]{\frac{1}{125}},$$ -and so on. Many of these were submitted by Ramanujan as problems to the -Journal of the Indian Mathematical Society. See the following link: -http://www.math.uiuc.edu/~berndt/jims.ps for more precise references. -Quote: "although Ramanujan never used the term unit, and -probably did not formally know what a unit was, -he evidently realized their fundamental properties. -He then recognized that taking certain powers of units -often led to elegant identities."<|endoftext|> -TITLE: Examples of prime numbers in nature -QUESTION [16 upvotes]: Finding primes in signals is seen as a sign of some kind of intelligence - see e.g. the role of primes in the search for extraterrestrial life (see e.g. here). This is because there are relatively few examples of numbers that appear in nature because they are prime. -One example of the use of prime numbers in nature is as an evolutionary strategy used by cicadas of the genus Magicicada (see e.g. here or here: [1]) -My question: -Do you know of any other instances where prime numbers occur in nature? Could you please also give a source/link - and perhaps some background. Thank you. - -[1] Goles, E., Schulz, O. and M. Markus (2001). "Prime number selection of cycles in a predator-prey model", Complexity 6(4): 33-38 - -Edit: Obviously many people misunderstood me. I didn't mean the occurrence of prime numbers just by coincidence - but because they are prime. The cicadas example - although being controversial - at least hints at some kind of evolutionary strategy. - -REPLY [8 votes]: I think this question is better asked as "are there examples of prime SEQUENCES in nature?" -The fact that a starfish has 5 points, or that a sunflower has 23 petals or whatever, proves very little about primes in nature. It could be complete coincidence that the number happens to be prime. -What if starfish are found to have either 3, 5, or 7 points? Does that imply a prime sequence? Maybe its just something about odd numbers > 1. -The Cicada example definitely implies a prime sequence, not just a single number which happens to be prime.<|endoftext|> -TITLE: Topological dimension versus cohomological dimension -QUESTION [10 upvotes]: This should be really well known but I don't seem to find a statement about it nor a question in MO answering this. -Consider a Compact Hausdorff topological space $X$. The cohomological dimension of $X$ is the first natural number where the cohomology vanishes. The topological dimension is as defined in wikipedia (see also Hurewicz Wallman's book). -For CW complexes, it seems easy to prove that finiteness of topological dimension implies finiteness of the other. However, for general spaces I don´t figure out how to prove (or construct a counterexample) to -Q) Does finite topological dimension of $X$ implies finite cohomological dimension? -Maybe there are well known inequalities also (for example, it seems that for spaces homotopy equivalent to CW complexes one should have that the topological dimension is bigger than or equal to the cohomological one). Information about this kind of inequalities would be welcome. - -REPLY [16 votes]: First of all, "cohomological dimension" means two different things. The cohomological dimension of $X$ could be the largest $n$ such that $H^n(X)$ does not vanish. This definition is appropriate for defining the cohomological dimension of a group $G$ to be the cohomological dimension of its classifying space $BG$. It could also mean the supremum of $n$ for all closed subsets $C \subseteq X$ such that $H^n(X,C) \ne 0$. This second definition is more appropriate for geometry. By this second definition, an $n$-ball is $n$-dimensional, because you can let $C$ be its boundary. In this second, more geometric approach, it's standard and better to use Čech cohomology. So I should really write $\check{H}^n(X,C)$, but I won't bother. Also, you can ask for the cohomological dimension with various coefficients, and you might get different answers. -There is a very interesting review by Dranishnikov of cohomological dimension theory of compact metric spaces. It includes his famous construction of a compact metric space whose covering dimension is $\infty$ and whose cohomological dimension (over $\mathbb{Z}$) is 3. -Theorem: (elementary) If $X$ is any topological space and $A$ is an abelian group, then $\dim_A X \le \dim X$, where the right side is the covering dimension. -As Torsten Ekedahl points out, this inequality follows quickly from the definition of Čech cohomology and the fairly similar definition of covering dimension. In both cases you use the nerve of an open covering. In Čech cohomology, you take the limit of the cohomology of the nerve. In covering dimension, you take the lim inf of the dimension of the nerve, QED. -Theorem: (elementary) (1) $\dim_A X \le \dim_{\mathbb{Z}} X$; (2) For every $A$, $\dim_A X = 0$ iff $\dim X = 0$; (3) $\dim_\mathbb{Z} X = 1$ iff $\dim X = 1$; (4) For every $A$ and every $n$-dimensional compact simplicial complex $K$, $\dim_A K = n$. -I would guess that (4) also applies to compact CW complexes. -Theorem: (Alexandroff) If $X$ is a compact metric space and if its covering dimension $\dim X$ is finite, then $\dim X = \dim_{\mathbb{Z}} X$. -Theorem: (Pontryagin) For each prime $p$, there is a "surface" $X$ whose $\mathbb{Z}/p$-dimension is 2, and whose $\mathbb{Z}/q$-dimension is 1 for any prime $q \ne p$. -In light of Alexandroff's theorem, if you want cohomological dimension and covering dimension to differ for a compact metric space, the covering dimension has to be infinite. Whether this was possible was a long-standing open problem; Dranishnikov found the first example. He also describes an example of Dydak and Walsh of a compact metric space $X$ such that $\dim_{\mathbb{Z}} X = 2$ but $\dim_{\mathbb{Z}} X \times X = 3$.<|endoftext|> -TITLE: Scheme theoretic closure of a locallly closed subscheme -QUESTION [6 upvotes]: In the book "The Geometry of Schemes" of Eisenbud and Harris, page 26, it is said that the scheme theoretic closure of a closed subscheme Z of an open subscheme U is the closed subscheme of X defined by the sheaf of ideals consisting of regular functions whose restrictions to U vanish on Z. -I cannot verify this assertion when the open immersion of U in X is not quasi-compact (I mean I cannot prove that this sheaf of ideals is quasi-coherent). -Am I missing something here ? - -REPLY [2 votes]: It seems indeed that example 2.10 in the Stacks project morphisms of schemes chapter provides a counter example, where the sheaf of ideals of regular functions whose restrictions to U vanish on Z is not quasi-coherent, because if it were part (3) of lemma 4.3 would be fulfilled. Thank you again for this hint, and if I am not mistaken, I have answered my own question, and an errata should be added for this book on page 26, where the open immersion of U in X should be supposed quasi-compact.<|endoftext|> -TITLE: Small complete categories in a Grothendieck topos -QUESTION [22 upvotes]: It is a classical theorem of Freyd that if a small category is complete (has all small limits—in fact, having small products suffices), then it is a preorder (has at most one morphism between any two objects). The proof of this theorem (which can be found here or in CWM) is non-constructive, i.e. it uses the Law of Excluded Middle. Therefore, it can potentially fail in the internal logic of an elementary topos. And in fact, it does fail in the effective topos, and more generally in realizability topoi, where there do exist small complete categories that are not preorders. -However, I have heard it said that Freyd's theorem cannot fail in a Grothendieck topos; i.e. that a small complete category in a Grothendieck topos must still be a preorder—despite the fact that the internal logic is still in general intuitionistic, so that Freyd's proof cannot work. Can someone explain why this is, or (even better) give a reference containing a proof? - -REPLY [12 votes]: Hi. I mentioned that I had thought about this on nForum a while back - sorry I didn't get back to you sooner. The following sketch of a proof is mainly due to Colin McLarty. -Two features which distinguish a Grothendieck topos from a more general topos are - -That it has a geometric morphism to Sets, namely the global sections functor. -That it has an object of generators (i.e. there is an object G such that if $f,g: A \to B$ are not equal then there exists an arrow $h: G \to A$ with $fh \neq gh$) - -Let $C$ be a small complete category object in a Grothendieck topos $T$ which is not a preorder. Then $C^G$ is also a small complete category in this topos essentially because exponentials commute. The global sections functor applied to $C^G$ gives a small complete category in the category of sets which is not a preorder (the property of being a small complete category is preserved by geometric morphisms, and the special property of G allows the property of being "not a preorder" to carry through), which is a contradiction. -It is a little easier to think about in the case of sheaves on some topological space. There a small complete category object which is not a preorder would have to fail to be a preorder on some open set, and the sections on that open set would be a small complete category which is not a preorder. $G$ takes the place of this open set above. -If you have any questions about this let me know. In particular I can write out all of the adjunctions showing various properties are preserved, but I don't want to get too nitty gritty if it isn't useful to you. -Kind regards, -Steven Gubkin<|endoftext|> -TITLE: Under what hypotheses are schematic fixed points of a flat deformation themselves flat? -QUESTION [5 upvotes]: This is something of a follow-up question to this one; I hope people won't think this is a duplicate. At least in my head, it seems like a distinct enough question to merit a fresh start. -All my schemes will be finite type over an algebraically closed field $k$. Let $X\to S$ be a flat family of affine schemes over smooth affine base. Let's say for now that each fiber and the whole family have rational singularities, and thus are Cohen-Macaulay. Assume, furthermore, that $X$ has an action of the group scheme $T=(\mathbb{G}_m)_S$; this is the same data as a grading on $k[X]$ such that $k[S]$ has degree 0. -Now, we can take the schematic fixed points $X^T$ of this family, which is a subscheme of $X$ whose points over any ring are invariant points of $X$. Concretely, this is the vanishing set of the ideal generated by all functions of non-zero degree. - -Must the morphism $X^T\to S$ be flat? If not, are there stricter hypotheses than I gave above which would assure it is? - -For example, consider the family $$X=\mathrm{Spec}[x,y,z,a_0,\dots, a_{n-1}]/(xy=z^n+a_{n-1}z^{n-1}+\cdots + a_0)$$ where $S=\mathrm{Spec}[a_0,\dots, a_{n-1}]$ with $x$ having degree 1, $y$ degree $-1$ and $z,a_i$ having degree 0. In this case $$X^T=\mathrm{Spec}[z,a_0,\dots, a_{n-1}]/(z^n+a_{n-1}z^{n-1}+\cdots + a_0=0),$$ which is, of course, flat over $S$, even though the number of closed points in a fiber (the number of roots of $z^n+a_{n-1}z^{n-1}+\cdots + a_0$) varies from $n$ to $1$. - -REPLY [8 votes]: Here is a counterexample. Let $\mathbb G_{\rm m}$ act on $\mathbb A^2$ by $t\cdot(x,y) = (tx,t^{-1}y)$, and let $f\colon \mathbb A^2 \to \mathbb A^1$ be defined by $f(x,y) = xy$. -I am positive that when $X$ is smooth over $Y$, the fixed point scheme is also smooth; but I doubt that one can say much more, in general. -[Edit] Here is a variant. Let $\mathbb G_{\rm m}$ act on $\mathbb A^4$ by $t\cdot(x,y, z, w) = (tx,t^{-1}y,tz,t^{-1}w)$, and let $f\colon \mathbb A^4 \to \mathbb A^1$ be defined by $f(x,y) = xy + zw$.<|endoftext|> -TITLE: Baker-Campbell-Hausdorff formula: prime divisors of denominators -QUESTION [10 upvotes]: Consider the Baker-Campbell-Hausdorff formula (Wikipedia page): -$$Z(X,Y) := X + Y + \frac{1}{2}[X,Y] + \frac{1}{12}[X,[X,Y]] - \frac{1}{12}[Y,[X,Y]] + \dots$$ -Many sources, including the Wikipedia page, have an explicit expression for the terms, so the question I'm asking should be answerable just from that expression. -For a prime p and a natural number n, denote by $f(p,n)$ the largest k such that, if we truncate the formula to terms that involve products of length at most $n$, then one or more of the denominators is divisible by $p^k$. Note that this truncation is valid when we are working in a situation of nilpotency class $n$. -It's pretty easy to see that $f(p,n) = 0$ for $n < p$, and is nonzero for $n \ge p$. -My question: Is there a direct explicit expression for $f(p,n)$ (or a sandwiching of it between two fairly close expressions)? For instance, inspection of the first few terms suggests that $f(2,n) = n - 1$, but I'm not sure how to derive this from the general expression. -Analogue: In the power series for the exponential $e^x$, the analogue to $f(p,n)$ is the sum $[n/p] + [n/p^2] + [n/p^3] + \dots$ where $[]$ denotes the greatest integer function. -UPDATE: Chapter 3 of the Springer Lecture Notes by Klass, Leedham-Green, and Plaskett (access online if you have an a university subscription) contains some estimates. However: (i) I'm not sure all the numerical calculations there are correct, since they don't agree with others I have seen, (ii) the authors aren't concerned about the precise growth of $f(p,n)$ -- they only care that it grows slowly enough that the series converges under certain conditions. - -REPLY [8 votes]: If the homogeneous component $Z_n(X,Y)$ of $Z(X,Y)$ of degree $n$ is represented in the Lyndon basis $\mathcal{L}_n$, or in any basis $\mathcal{B}_n$ whose transformation -matrix $T_{\mathcal{L_n}\to\mathcal{B_n}}$ has determinant $\pm 1$, then an explicit formula for the exponent $f(p,n)$ of the highest power of $p$ that divides the least common multiple of the denominators of the coefficients of $Z_n(X,Y)$ is given by -$$ -f(p,n)=\frac{n-s_p(n)}{p-1}+\lfloor\log_p(s_p(n))\rfloor. -$$ -Here, $s_p(n)=\alpha_0+\ldots+\alpha_r$ denotes the sum of the digits in the $p$-adic expansion $n=\alpha_0+\alpha_1p+\ldots+\alpha_r p^r$, $0\leq\alpha_i -TITLE: Is the box product of morphisms associative? -QUESTION [5 upvotes]: Suppose $(C,\otimes)$ is a symmetric monoidal finitely-cocomplete category such that $\otimes$ preserves colimits. Given two morphisms $a:A_1\to A_2$ and $b:B_1\to B_2$, define $a\Box b$ to be the induced "lower right corner map" $A_1\otimes B_2\coprod_{A_1\otimes B_1} A_2\otimes B_1\to A_2\otimes B_2$. -Is $\Box$ associative? Does it give a monoidal product on $Arr(C)$? -I'm having trouble figuring out what the diagram for this should look like if it's true (although a quick "arithmetic check" shows that it should be). -If it's not associative for some stupid reason that I've overlooked in the general case presented here, is it true in the case that $C$ is a presheaf category and $\otimes$ is the cartesian product? - -REPLY [2 votes]: I accepted Todd's answer because it was very helpful, but I'm going to write this answer because I figured out a very useful heuristic way to think about this (although I think that actually proving the equivalence wouldn't be very tough (or worthwhile)). -There is a very nice geometric description that makes the combinatorics of the problem much much clearer: Imagine that all of the maps involved are just the inclusion $0\to \mathbf{R}$. Then the box product of $n$ of these maps is the inclusion $X\hookrightarrow \mathbf{R}^n$, where $X$ is the union of all codimension 1 hyperplanes spanned by the axes. We can associate a commutative diagram with this description, which is generated by the different ways to glue over different maps, and this exactly represents the general problem. The reason why embedding in $\mathbf{R}^n$ in this way is so effective is that any two of the relevant objects are glued along their intersection. The embedding encodes the explicit combinatorial information so you don't need to worry about it. -In particular, without this picture (or a similar one) in mind, it will very quickly get extremely difficult to deal with box products of anything more than a few maps, since there are already tons of ways to glue together just three planes passing through the origin if you do it piece-by-piece. This is because using "pushout notation" as above, we not only have to keep track of the pieces we're gluing together, but also what we're gluing them along. If you don't understand the geometric picture, you're essentially proving it by brute force by drawing every single possible pushout path (see Todd's answer; indeed if there were more than three maps, the commutative diagram would be four dimensional) to the required map. -I hope people learn from my mistakes!<|endoftext|> -TITLE: Existence of a smooth function with nowhere converging Taylor series at every point -QUESTION [20 upvotes]: By Borel's theorem, for any sequence of real numbers $a_n,$ there is a $C^{\infty}$-function -$f:\mathbb{R}\to\mathbb{R}$ whose Taylor series at 0 is $\sum a_nx^n.$ In particular, there are $C^{\infty}$-functions whose Taylor series at a point has 0 radius of convergence. -Motivated by this I have the following question. Is there a $C^{\infty}$-function -$f:\mathbb{R}\to\mathbb{R}$ whose Taylor series has 0 radius of convergence at every -point in $\mathbb{R}?$ I realize that this might sound like a homework problem, but, well, -it's not. - -REPLY [3 votes]: Good. I thought something like this might work. -One more question before I get off this subject, which might be harder (or easier!): what about the other extreme? Is there a smooth function on an interval in $\mathbb R$, not analytic on any subinterval, whose Taylor series at every point has positive radius of convergence? The Fabius function might be an example, but this is questionable since the $n$'th derivative has maximum $2^{\sigma(n)}$, where $\sigma(n)=\frac{n(n+1)}{2}$, which is not quite good enough using a crude estimate for radius of convergence if there are points where many derivatives are close to the maximum.<|endoftext|> -TITLE: Complex Analysis applications toward Number Theory -QUESTION [16 upvotes]: I'm an undergrad who is taking a Complex Analysis Course mainly for its applications in number theory. -So I would like to ask some guidelines about which theorems/concepts should I focus on in order to develop a narrower path for self study. -In addition, it would be helpful to know if there is a book that does a good job showing off how the -complex analysis machinery can be used effectively in number theory, -or at least one with a good amount of well-developed examples in order to provide a wide background of the tools that complex analysis gives in number theory. - -REPLY [23 votes]: This question is like asking how abstract algebra is useful in number theory: lots of it is used in certain areas of the subject so there's no tidy answer. You probably won't be using Morera's theorem directly in number theory, but most of single-variable complex analysis is needed if you want to understand basic ideas in analytic number theory. A few topics you should pay attention to are: the residue theorem, the argument principle, the maximum modulus principle, infinite product factorizations (esp. the Hadamard factorization theorem), the Fourier transform and Fourier inversion, the Gamma function (know its poles and their residues), and elliptic functions. Basically pay attention to the whole course! There really isn't a whole lot in a first course on complex variables where one can say "that you should ignore if you are interested in number theory". -If you want to be careful and not just wave your hands, you need to know conditions that guarantee the convergence of series and products of analytic functions (and that the limit is analytic), the existence of a logarithm of an analytic function (it's not the composite of the three letters "log" and your function), that let you reorder terms in series and products, that justify termwise integration, and of course the workhorse of analysis: how to make good estimates.<|endoftext|> -TITLE: Monstrous Moonshine -QUESTION [8 upvotes]: Wikipedia claims that the group of units of Z24 (1,5,7,11,13,17,19,23), which all have order 2, and are isomorphic to (Z/2Z)^3 have an important connection to Monstrous Moonshine theory, however, I cannot find any other reference besides Wikipedia that claims this --- It was recommended on sci.math that I pose this question here. -Perhaps it's a mistake? And he meant that the primes of the Monster, which continue to 71, are what are considered in Moonshine. -Paul Hjelmstad, B.M, B.A. -[Edit (PLC): Here is the relevant passage from wikipedia:] - -24 is the highest number $n$ with the property that every element of the group of units $(\mathbb{Z}/n\mathbb{Z})^{\times}$ of the commutative ring $\mathbb{Z}/n\mathbb{Z}$, apart from the identity element, has order $2$; thus the multiplicative group $(\mathbb{Z}/24\mathbb{Z})^{\times} = \{1,5,7,11,13,17,19,23\}$ is isomorphic to the additive group $(\mathbb{Z}/2\mathbb{Z})^3$. This fact plays a role in monstrous moonshine. - -REPLY [4 votes]: $\newcommand{\Q}{\mathbf Q} \newcommand{\Z}{\mathbf Z}$ -I don't know about monstrous moonshine, but $(\Z/24\Z)^\times$ is the group of automorphisms of the maximal elementary abelian $2$-extension $\Q_2\left(\root2\of{\Q_2^\times}\right)=\Q_2(\root2\of5, \root2\of3, \root2\of2)=\Q_2(\zeta_{24})$ of $\Q_2$. See for example Lemma 8 of Lecture 19 of my course on Local arithmetic.<|endoftext|> -TITLE: Symmetric subspace of linear operators -QUESTION [6 upvotes]: This is a question that stemmed from fooling around with unitary t-designs. -Let -\begin{equation} - \mathbb{V} = \mathrm{span} \{\; U^{\otimes t}\; |\; U \in \mathrm{U}(d)\} -\end{equation} -Where $\mathrm{U}(d)$ is the unitary group acting on $\mathbb{C}^d$. -What can we say about $\dim \mathbb{V}$? What tools can I use to figure out the dimension or good lower bounds for it? -I've read (p. 162) about the analogous question for $\mathbb{U} = \mathrm{span}\{|\psi\rangle^{\otimes t}\;|\; |\psi\rangle \in \mathbb{C}^d\}$. In particular, we can show that $\dim \;\mathbb{U} = {d + t - 1 \choose t - 1}$. To use that approach, we need to change $\mathrm{U}(d)$. -If we change $\mathrm{U}(d)$ to $\mathbb{C}^{d\times d}$ (i.e. all the matrices on $\mathbb{C}^d$) then is it obvious that the resulting space is the same as $\mathbb{V}$? If they are the same, then the question becomes easier, since then we have the space of all linear operators that are invariant under permutations of the $t$ registers and we can just use the formulate for $\dim \mathbb{U}$ thinking of the operators as row vectors. -(I am not sure of the tags for this, so feel free to change them) - -REPLY [6 votes]: Here is another proof of the same result (it is probably the same as the one given by darij grinberg but I do not understand all the words. It has perhaps more his place as a comment, but I do not know how to write comments). In the case $t=2$, it is Claim 1 in Lemma 1.3.2 of http://arxiv.org/pdf/1002.4595v2 -Denote by $P : M_d(\mathbb C)^{\otimes t} \to M_d(\mathbb C)^{\otimes t}$ the linear map such that $P(A_1 \otimes \dots \otimes A_t) = 1/t!\sum A_{\sigma(1)} \otimes \dots \otimes A_{\sigma(t)}$, where the sum os over all permutations of $\{1,\dots ,t\}$. $P$ is the natural projection on the symmetric tensors. -Let $(H_k)$ be the assertion "$\mathbb V$ contains the space $F_k$ spanned by $P(h_1 \otimes h_2 \otimes \dots \otimes h_t)$ for any hermitian matrices $h_1,...,h_t$ with $h_{k+1} = h_{k+2} = \dots h_t=1$" ($1$ is the identity matrix). -By polarity, $(H_k)$ is equivalent to the assertion "$\mathbb V$ contains $P(h_1 \otimes h_2 \otimes \dots \otimes h_t)$ for any hermitian matrices $h_1,...,h_t$ with $h_1 = h_2 = \dots =h_k$ and $h_{k+1} = h_{k+2} = \dots h_t=1$". -Prove $(H_k)$ by induction on $k$. Take indeed a symmetric matrix $h$, and consider $f(x) = \exp(ixh)^{\otimes t}$ and develop $f$ in power series: $f(x) = \sum_n x^n a_n$. Then $a_n$ belongs to $\mathbb V$ for all $n$. For example $a_1 = i t P(h\otimes 1 \ \dots \otimes 1)$ which proves $(H_1)$. -One moment of thinking gives you that you can write $a_k=i^k C(t,k) P(h\otimes \dots h \otimes 1 \dots \otimes 1) + $something in $F_{k-1}$, where $C(t,k)$ is the binomial coefficient and there are $k$ $h$'s and $(t-k)$ $1$'s in $h\otimes \dots h \otimes 1 \dots \otimes 1$. This allows to proceed by induction. -Thus for $k=t$, and using that $M_d(\mathbb C)$ is the linear span of the hermitian matrices, you get that $\mathbb V$ is the whole space of symmetric tensors.<|endoftext|> -TITLE: Is there a measure zero set which isn't meagre? -QUESTION [58 upvotes]: A subset of ℝ is meagre if it is a countable union of nowhere dense subsets (a set is nowhere dense if every open interval contains an open subinterval that misses the set). -Any countable set is meagre. The Cantor set is nowhere dense, so it's meagre. A countable union of meagre sets is meagre (e.g. all rational translates of the Cantor set). -There can also be meagre sets of positive measure, like "fat Cantor sets". To form a fat Cantor set, you start with a closed interval, then remove some open interval from the middle of it, then remove some open intervals from the remaining intervals, and so on. The result is nowhere dense because you removed open intervals all over the place. If the sizes of the intervals you remove get small fast, then the result has positive measure. -So does meagreness have any connection at all to measure? Specifically, are all measure zero sets meagre? - -REPLY [26 votes]: There's already been some good answers to this. However, this is something that I have also thought about recently, because I happen to have come across several meagre sets of full Lebesgue measure in some of my answers to other questions. In fact, in my experience on MO, meagre sets with full Lebesgue measure actually seems to be more the rule than the exception. So, I'll add these to the list. - -In this math.SE question and this MO question, David Speyer was trying to find the set of θ such that $\sum_{n=1}^\infty \sin(n^r\theta)/n$ converges (r > 1 an integer). He was concerned about the θ = 1 case but, from my answer on MO and David's answer on math.SE it can be seen that it converges for almost every θ but, at the same time, it only converges for θ in a meagre set. -Along a similar line, this MO question was asking for which θ the asymptotic bound $\sum_{n=1}^N{\rm sign}(\sin(n\pi\theta))=O(N^x)$ holds. For 1/2 < x < 1 my answer shows that it holds for almost every θ but, at the same time, it only holds for θ in a meagre set. -This question asks if there are 2x2 matrices C such that Tr(Cn) is dense in the reals as n runs through the positive integers. Bjorn Poonen shows that the answer is yes. In fact, his proof is easily modified to show that Tr(Cn) fails to be dense only on a meagre set. However, my answer shows that |Tr(Cn)| is either bounded or tends to infinity (so, not dense) for almost every C. -The above examples really just come down to the following point. The set of real numbers with finite irrationality measure (i.e., non-Liouville numbers) is meagre. However, almost every real number has irrationality measure 2. - -Along similar lines, the set of normal numbers is meagre and has full Lebesgue measure (see also, Andreas' answer). The set of real numbers whose continued fraction quotients have geometric mean converging to Khinchin's constant is meagre with full Lebesgue measure. The set of real numbers whose continued fraction quotients occur according to the Gauss-Kuzmin distribution is meagre with full Lebesgue measure. And so on...<|endoftext|> -TITLE: An analytic proof of the De Franchis theorem -QUESTION [9 upvotes]: The De Franchis theorem in its simplest form states that given two compact Riemann surfaces $\Sigma_{g_1},\Sigma_{g_2}$ where $g_1,g_2 > 1$, there are only finitely many non-constant holomorphic mappings from $\Sigma_{g_1}$ into $\Sigma_{g_2}$. -The complete proof can be found in P.Samuel's book Lectures on Old and New Results on Algebraic Curves. This proof uses quite a bit of Algebraic Geometry, in which I have very little background. -Does anyone know how to prove this result using purely complex analytic ideas? - -REPLY [2 votes]: We break the proof up into two steps: -1. The set of holomorphic maps is compact in the compact open topology. This is an easy consequence of the hyperbolicity of the Riemann surfaces - -The space of holomorphic maps is discrete. Suppose that there is a one parameter family (f_t) of maps between the Riemann surfaces. The derviative (\frac{\partial f}{\partial t}\0 is then a holomorphic section of the pull-back of the tangent bundle of the range over the domain. By hyperbolicity there are no non-zero sections. - -You can find arguments of this type in -M. Kalka, B. Shiffman and B. Wong, Finiteness and Rigidity Theorems for Holomorphic Mappings, - Michigan Math. J. 28 (1981), 289-295.<|endoftext|> -TITLE: Analysis of a quadratic diophantine equation -QUESTION [11 upvotes]: Hi! This is my first post on Math Overflow. I have two equations: $a(3a-1) + b(3b-1) = c(3c-1)$ and $a(3a-1) - b(3b-1) = d(3d-1)$. I'm trying to find properties of $a$ and $b$ that lead to solutions, where $a, b, c, d \in \mathbb{N}$. I'm having trouble applying any of the techniques in my abstract algebra book, as they mostly only apply to linear Diophantine equations. -So far, I only really have managed to deduce the following things: -$2b(3b-1) + d(3d-1) = a(3a-1) +b(3b-1)$ -$2b(3b-1) = c(3c-1) - d(3d-1)$ -$2b(3b-1) = (c-d)(3(c+d)-1)$ -Any ideas on where to go from here would be greatly appreciated. Thanks! - -REPLY [4 votes]: For this system one can find a general rational parametrization and -then N&S conditions for integer solutions. -Adding the pair: -$x^2 + y^2 = z^2 + 1$ -$x^2 - y^2 = t^2 - 1$ -gives: -$2 x^2 = z^2 + t^2$ -which has a general rational parametrization (GPR): -$(z + t)/2 = (v^2 - 1) x / (v^2 + 1)$ -$(z - t)/2 = 2 v x / (v^2 + 1)$ -Adding these gives an expression for z and plugging this back -in the first of the original pair then gives: -$(y/x)^2 - (1/x)^2 = 4 v (v^2 - 1)/(v^2 + 1)^2$ ( $= 4 R$ say) -which has GPR: -$y/x, 1/x = (L^2 + R)/L, (L^2 - R)/L$ -and replacing $u := L (v^2 + 1)$ (to give an obvious simplification) -yields a general rational solution of the original pair as: -$D = u^2 - v (v^2 - 1)$ -$D x = u (v^2 + 1)$ -$D y = u^2 + v (v^2 - 1)$ -$D z = u (v^2 + 2 v - 1)$ -$D t = u (v^2 - 2 v - 1)$ -Homogenizing these by taking $u, v = a/c, b/c$ with $(a, b, c) = 1$, -we now investigate how to specialize this to integer solutions. -First, $y$ is an integer iff $a^2 c - b (b^2 - c^2)$ divides $2 a^2 c$. -Equivalently, there is an integer $L$ such that: -$b L (b^2 - c^2) = (L - 2) a^2 c$ -Then two cases arise, depending on the parity of L. -Case 1 L odd -We show that this is impossible (given the other constraints of -the problem). -If $L$ is odd then $(L, L - 2) = 1$ and thus for some integer $n$ -we have: -$a^2 c, b(b^2 - c^2) = L n, (L - 2) n$ -Multiplying the equations for $z$ and $t$ by $2 a b$, and plugging -the above pair into the result gives: -$2 a b z = a^2 (L - 2) + 2 b^2 L$ -$2 a b t = a^2 (L - 2) - 2 b^2 L$ -So letting $a, b = A e, B e$ with $(A, B) = 1$ implies the following, -in which $2 L / A$ and $(L - 2) / B$ are integers: -$2 z = A (L - 2) / B + B (2 L / A)$ -$2 t = A (L - 2) / B - B (2 L / A)$ -For z, t to be integers we require $A (L - 2) / B$ and $B (2 L / A)$ -to both odd or both even. -If they are both odd then A and $2 L / A$ must be both odd, which -is impossible. -If they are both even then A even implies B odd and thus $2 L / A$ -even, and A odd implies $(L - 2) / B$ even. So in either case this -implies L even, contrary to hypothesis. -So that leaves us with .. -Case 2 L even -Denoting $m := L / 2$ for convenience, we must how have for some -integer $n$ : -$a^2 c, b(b^2 - c^2) = m n, (m - 1) n$ [*] -which, as in Case 1, implies: -$2 z = A (m - 1) / B + B (2 m / A)$ -$2 t = A (m - 1) / B - B (2 m / A)$ -Again $A (m - 1) / B$ and $B (2 m / A)$ must be either both odd -or both even.. -Both odd leads to the same contradiction as Case 1 as it requires -$A$ and $2 m / A$ both odd. -So they must be both even, which is the case iff $A \equiv m \mod(2)$ -(provided that when $m$ is odd, $(m - 1) / B$ is even, in other -words $B$ does not divide out the power of 2 dividing $m - 1$). -Furthermore from the form of $z$, $t$, as $f \pm g$, they have -the same parity. So adding and subtracting the original pair -implies that $x, y$ are integers iff $z, t$ are integers. -Note that the above isn't an explicit integer solution. All I -have done is reduce the problem to the pair [*], to which I -have a draft solution that needs checking. But if anyone else -wishes to nip in first with a solution to these then obviously -feel free!<|endoftext|> -TITLE: impact of Poincaré duality on functional equation -QUESTION [7 upvotes]: Given a variety $X/\mathbf{F}_q$ and a sheaf $\mathcal{F}$ on it, what is the relation of $L(X,\mathcal{F},T)$ and $L(X,D(\mathcal{F}),T)$? - -REPLY [3 votes]: Perhaps I am missing something here. But my quick reaction is that surely they are the same. Assuming that $D$ is the full Grothendieck-Verdier duality functor, then $H^i(\overline{X},\mathcal{F})^*\simeq H^{2n-i}(\overline{X},D(\mathcal{F}))$ (Here base change from $\mathbb{F}_q$ t$D(\mathcal{F})$o its algebraic closure commutes with duality). -Then since the characteristic polynomial of an endomorphism of a vector space is the same as that of its transpose, and n is even, nothing changes. -One (possibly esoteric) way of thinking about the (cohomological) L-function, is as action induced by $1-t\sigma$ (where $\sigma$ is Frobenius) on the determinant line (tensored with $\mathbb{Q}_l(t)$, strictly speaking) of the perfect complex $R\pi_!(\mathcal{F})$ of $\mathbb{Q}_l$ vector spaces. Then the determinant of the dual is the dual of the determinant -- but these are both scalars acting on dual lines and so are the same. -PS Perhaps what you really want is the proof of the functional equation along these lines. The key point is to not only replace $\mathcal{F}$ by $D(\mathcal{F})$ but also to replace $t$ by $1/(q^nt)$.<|endoftext|> -TITLE: Is every finitely generated group colimit of residually finite groups -QUESTION [7 upvotes]: I was listening to a talk about ultraproducts and one result there suggested, that every finitely generated group can be written as a colimit of residually finite groups (over a directed system). -As I don't see a trivial proof, I expect it to be false (it is a statement about all groups). But I don't see a counterexample. - -REPLY [12 votes]: The answer is: This does not hold in general. -If the group in question is finitely generated, then the maps into the colimit will eventually be surjective. If the group in question is also finitely presented, then eventually, all relations will hold in the groups in the colimit diagram. Hence, you can split the surjection and see the colimit group as a subgroup of one of the groups in the diagram. -Hence, your assertion would imply that every finitely presented group is residually finite (since this property passes to subgroups), which of course is wrong. Higman's group is the classical counterexample: -$$H:= \langle a,b,c,d \mid ba = a^2b, cb=b^2c, dc=c^2d, ad=d^2a \rangle.$$ -It has no finite quotients and is finitely presented.<|endoftext|> -TITLE: Decomposition of an algebraic group in an affine and a proper part -QUESTION [6 upvotes]: Let $K$ be a perfect field. In what follows, an algebraic group $G/K$ is by definition a group scheme of finite type over $K$. -The following seems to be well-known: -Theorem: Let $G/K$ be a connected smooth algebraic group. Then there is a connected smooth affine normal closed subgroup $N$ of $G$, an abelian variety $A/K$ and a homomorphism $G\to A$ with kernel $N$ such that the sequence -$$0\to N\to G\to A\to 0$$ -is exact for the fppf-topology (say). -Can someone give me a proper reference or a hint why this is true? (Checking the literature I find on the one hand plenty of references treating affine algebraic groups, and on the other hand references containing the theory of abelian varieties, but I was surprised not to find a reference containing a proof of this Theorem about the "mixed case".) - -REPLY [12 votes]: This question is answered in the Wikipedia page for algebraic groups. The article says that this is a difficult result of Chevalley, and it has a link to a modern write-up by Brian Conrad of Chevalley's result. So that is surely a proper reference. No particular hint leaps out at me for "why" it is true, but I can say something about what the proof is really saying. The subgroup $N$ appears as the common kernel of all algebraic homomorphisms from $G$ to all abelian varieties. So it is a functorial construction and $N$ is actually a characteristic subgroup, not just a normal subgroup. Relatively early on it is shown that there are no non-trivial algebraic homomorphisms from an affine group to an abelian variety, in fact not even any non-trivial algebraic morphisms that don't have to be homomorphisms. -It is also relatively quick to show that $G/N$ is an abelian variety. The really hard part is to show that the kernel is affine. You might as well let $G = N$ and you might as well let $K$ be algebraically closed. The hard theorem is that if $G$ does not have any non-trivial homomorphisms to an abelian variety, then it is affine. It is important to remember that $G$ is affine if and only if it is linear, i.e., an algebraic subgroup of $\text{GL}(V)$ for some finite-dimensional vector space $V$. This vector space $V$ is the most difficult construction of the paper.<|endoftext|> -TITLE: Do we know the Chow groups of spheres? -QUESTION [10 upvotes]: Let $k$ be a field (of char. not $2$) and $X_k=\text{Spec} (k[x_1,\cdots,x_n]/(x_1^2+\cdots +x_n^2-1))$. Do we know the Chow groups $A_i (X_k)$? I could not find any references, even for $X_{\mathbb R}$. -What (I think) I know: the K-groups were computed by Swan, so we know the total Chow group up to torsions. In codimension $1$ (i.e., class groups) I am fairly certain the answers are known. - -REPLY [6 votes]: The book -The Algebraic and Geometric Theory of Quadratic Forms. by R. Elman, N. Karpenko and A. Merkurjev (American Mathematical Society Colloquium Publications, 56., American Mathematical Society, Providence, RI, 2008. 435 pp.) -contains a lot of information. I was just reading -B. Totaro, The automorphism group of an affine quadric, Math. Proc. Cambridge Philos. Soc. (2007) vol. 143 (1) pp. 1-8 -and he is referring to this book for information on Chow groups of spheres.<|endoftext|> -TITLE: Is this true that algebraic spaces etale and surjective over a scheme is a scheme ? -QUESTION [6 upvotes]: Let X be an algebraic space such that there exists an etale surjective map f: X -> Y, where Y -is a scheme. Is it true that this implies X is also a scheme? - -REPLY [8 votes]: The point must be to avoid separatedness hypotheses on $f$. (D. Knutson proved algebraic spaces locally quasi-finite and separated over schemes are schemes; he may have had noetherian hypotheses, in which case those hypotheses are removed in the appendix to Champs Algebriques.) -Anyway, in the absence of separatedness the answer is negative. In the Introduction to Knutson's book there's a procedure beginning with smooth algebraic variety $Y$ of dimension $> 1$ over a field and "replacing" a smooth hypersurface in $Y$ with an irreducible etale double cover while remaining smooth and irreducible. This resulting $X$ is an algebraic space etale over $Y$, and the rank-jumping of $f:X \rightarrow Y$ is really weird: it jumps up along a closed set rather than down, contrary to Zariski's Main Theorem (in the formulation of EGA) for locally quasi-finite flat maps of schemes. Hence, $X$ is not a scheme (though the failure to be a scheme can surely be seen in a zillion other ways). -The preceding does actually have an "application". It yields a geometrically irreducible smooth algebraic space $X$ over $\mathbf{Q}$ of dimension 2 such that $X_{\mathbf{C}}$ admits an analytification in the sense of complex-analytic spaces but $X_k$ does not admit an analytification in the sense of rigid-analytic spaces or Berkovich $k$-analytic spaces for any non-archimedean field $k$ of characteristic 0. (This is explained in detail in Example 3.1.1 of my paper "Non-archimedean analytification of algebraic spaces" with M. Temkin.)<|endoftext|> -TITLE: Lie algebras to classify Lie groups -QUESTION [7 upvotes]: What does the classification of Complex Semi-simple Lie algebras buy us in terms of classifying Lie groups? Certainly it classifies complex semi-simple lie groups but can we get any better? I know we can take compact real forms of the semi-simple algebras and there are several theorems about topological similarities for Lie groups with the same Lie algebra. How far can we take this? What is the biggest class of Lie groups we can rope in by this method? - -REPLY [4 votes]: Here is an answer from a different point of view : every connected Lie group is almost a semidirect product of a semisimple Lie group (Levi factor) and a connected solvable normal Lie subgroup (its solvalble radical). See Onishchik and Vinberg's Lie Groups and Lie Algebras III for a more precise statement. In that sense, the classification of real Lie groups is reduced to that of semisimple and solvable ones. Although somewhat crude, this classification helps for example in the study of unitary representations. (See papers by M. Duflo on the orbit method.) -Another subtle issue with the classification of semisimple Lie groups is that over -$\mathbb C$, the simply connected group corresponding to a semisimple Lie algebra is always linear, whereas over $\mathbb R$, this is not the case. For example, the simply connected Lie group corresponding to $\mathfrak{sl}_2(\mathbb C)$ is $\mathrm{SL}_2(\mathbb C)$, but the simply connected Lie group -corresponding to $\mathfrak{sl}_2(\mathbb R)$ is an infinite-sheeted covering of $\mathrm{SL}_2(\mathbb R)$. In other words, $\pi_1(\mathrm{SL}_2(\mathbb C))=\{e\}$ whereas -$\pi_1(\mathrm{SL}_2(\mathbb R))=\mathbb Z$. (Here $\pi_1$ means the first fundamental group.)<|endoftext|> -TITLE: Is the Fourier transform of 1/(1-log(1-x^2)) (supported in [-1,1]) integrable? -QUESTION [13 upvotes]: This question was suggested when trying to find an explicit example of a continuous function with compact support in $\mathbb{R}$ whose Fourier transform is not integrable. The existence of such a function was proved by an abstract argument in this MO question, but no explicit example was given. It is clear that such a function cannot be differentiable everywhere. -The functions $(1-x^2)^{\alpha}\_{+}$ for $0<\alpha<1$ look like reasonable candidates, since they have a singularity at $\xi=\pm1$ and converge as $\alpha\to0$ to the characteristic function of $[-1,1]$, whose Fourier transform is not integrable. However $\hat f_{\alpha}(\xi)=O(|\xi|^{-(1+\alpha)})$ as $|\xi|\to\infty$, and hence is integrable. -For any $\alpha,\beta>0$, the function -$$g_{\alpha,\beta}(x)=(1-\beta\log(1-x^2))^{-\alpha}\hbox{ if }|x|<1,\quad 0 \hbox{ if } |x|\ge1$$ -is more singular than any $f_{\alpha}$. So my questions are: - -Is $\hat g_{\alpha,\beta}$ is integrable? -What is the asymptotic behaviour of $\hat g_{\alpha,\beta}$ at infinity? - -I suspect, based on numerical calculations, that the answer to the first question is no, at least for sufficiently small $\alpha$ and $\beta$. - -REPLY [14 votes]: If one performs a smooth dyadic decomposition of $g_{\alpha,\beta}$ around the singularities $x = \pm 1$ (i.e. using smooth partitions of unity to decompose $g_{\alpha,\beta}$ into pieces that are localised in the region $1-|x| \sim 2^{-n}$ for $n \geq 0$), and then takes the Fourier transform of these pieces, one soon arrives at the conclusion that $\hat g_{\alpha,\beta}$ decays like $\frac{1}{|\xi| \log^\alpha |\xi|}$ (times something like $\sin \xi$) as $\xi \to \infty$ (and one can then back up this intuition with stationary phase, or use some localised form of the Plancherel theorem for an $L^2$ averaged result, which should be enough for the application at hand). So absolute integrability should fail for $\alpha \leq 1$.<|endoftext|> -TITLE: Is the Rado graph a Cayley graph? If so, what is the group like? (And other questions...) -QUESTION [29 upvotes]: The countable random graph, also known as the Rado graph, is characterized as the unique countable graph in which every two disjoint finite sets $A$ and $B$ of vertices admit a vertex $p$ connected to every vertex in $A$ and to none in $B$. This is a kind of saturation property, and any two countable graphs exhibiting it are isomorphic by a fun back-and-forth construction; indeed, any finite partial isomorphism extends in this situation to a full isomorphism. The same argument shows that every countable graph has many copies inside the Rado graph. -The Rado graph can also be described as the almost-sure result of constructing an edge between any two distinct integers with independent probability $p\in(0,1)$, essentially because the probability of fulfilling the property for any two $A$ and $B$ is one, since there are infinitely many independent chances to get the pattern correct. (Note, in this question, graphs have no trivial loops or parallel edges.) -My questions are: - -Is the Rado graph a Cayley graph? That is, can one orient and label the edges with generators in such a way that the Rado graph is the resulting Cayley graph of the corresponding group presentation? (I expect that the answer is yes, by building up a simply transitive group action and employing Sabidussi's theorem, but I would be interested to see an elegant account of this idea if it works.) -If so, is the resulting group unique? What is its nature? -For example, can one make the Rado graph a Cayley graph of a group with torsion? without torsion? -Does the group exhibit a probabilistic account, as the graph does? -If the Rado graph is a Cayley graph, what universal properties does the resulting group enjoy similar to the attractive universal properties of the graph? For example, every countable Cayley graph finds a copy in the Rado graph; can we always find a copy whose Cayley labeling extends to a Cayley labeling of the full Rado graph? - -It seems that the ideas of the random graph extend in several ways to the context of directed graphs. For example, one could ask that for every disjoint triple of sets $A$, $B$ and $C$, there is a node $p$ pointing at all nodes in $A$, pointed at by all nodes in $B$ and with no edges between $p$ and nodes in $C$. This saturation property again characterizes the directed graph up to isomorphism by a back-and-forth construction. I suppose this graph would also be the almost-sure result of building the Rado graph probabalistically, but also orienting the edges with independent probability $p$. - -Are any of the questions above affected by this change? That is, in this version of the questions, the edge orientations are already given, but satisfy the saturation property. -And what of the uncountable analogues of the random graph? For example, if the Continuum Hypothesis holds, then there is a (unique) graph of size $\aleph_1$ satisfying the saturation property with respect to all countable disjoint subsets $A$, $B$, that is, having a vertex connected to all vertices in $A$ and to none in $B$. Is this graph a Cayley graph, and so on? - -REPLY [10 votes]: The answer that I started to write is a combination of Gjergji's answer and David's answer. Let $G$ be a countable group which is not a finite union of translates of non-principal square root sets and a finite set. Then as David says, you get infinitely many independent chances to find $p$. As in Joel's original set-up, you can also use biased coin tosses to make the edges. -I'm left wondering when a countably infinite group is a finite union of translates of non-principal square root sets and a finite set. I guess that $C_2^{\infty} \times A$ is an example, if $A$ is a finite group which is not an elementary 2-group. Certainly no elementary $p$-group (or, additively, no vector space over $\mathbb{Z}/p$) is an example, and neither is $\mathbb{Z}^n$. And I guess that $\text{SL}(n,\mathbb{Z})$ isn't an example either. - -There are two different kinds of directed graphs, graphs with and without 2-cycles. The question posed considers the kind without 2-cycles, but I don't see a problem defining either kind of Rado graph on the same class of groups. For that matter, you could have various kinds of colored Rado graphs, where some of the edge colors are directed and others are undirected. (So, allowing 2-cycles amounts to having two undirected colors and one directed color.) -I don't think that it really says a whole lot about the group. It says that the Rado graph has such a huge automorphism group that many groups act freely transitively on it. It looks like the Cameron-Johnson condition on the group $G$ is actually not only sufficient for a freely transitive action, but also necessary, at least to obtain the $n$-colored Rado graph for every finite $n$ as a Cayley graph of $G$. Still, the groups $G$ that are impossible are essentially impossible for a local reason. -As for the uncountable analogues of the random graph, I might guess that the same constructions still work sometimes, but I would ask someone like Joel Hamkins. :-)<|endoftext|> -TITLE: Transitive shifts with multiple fully supported MMEs -QUESTION [6 upvotes]: This is a sequel to my earlier question, where I asked for an example of a shift space that is mixing but not intrinsically ergodic -- that is, it has multiple measures of maximal entropy (MMEs). Steve Huntsman's answer referred me to a paper by Haydn that gives such an example: however, in that example the two MMEs are supported on disjoint compact subsets of the shift space, and so in some sense the shift can be viewed as two intrinsically ergodic shifts that have been "glued together". -Is there an example of a transitive shift with multiple MMEs that are all fully supported? More precisely, does anybody know of a transitive shift space $X\subset \{0,1,\dots,p-1\}^\mathbb{Z}$ for which there are two distinct ergodic measures $\mu_1, \mu_2$ such that - -$h_{\mu_1}(\sigma) = h_{\mu_2}(\sigma) = h_\mathrm{top}(X,\sigma)$; -$\mu_i(U)>0$ for every open set $U\subset X$ and $i=1,2$? - -REPLY [3 votes]: Poking through the DGS book mentioned in Ian's answer I came along a reference to a paper that turns out to be exactly what I wanted when I asked this question originally, so I'll post it here for the sake of closure and because it's a nice example. -The paper is by Wolfgang Krieger: On the uniqueness of the equilibrium state, Mathematical Systems Theory 8 (2), 1974, p. 97-104. -The example is the Dyck shift, which is easiest to understand in terms of brackets. The alphabet of the shift is a collection of $2n$ symbols that come in $n$ pairs; each pair has a left element and a right element. So with $n=2$ we can write the four symbols as ( ) [ ]. The shift space $X$ comprises all sequences on these symbols in which the brackets are "opened and closed in the right order". So for example, ( ) [ ] is a legal word, as is ( ( ( ) [ ] [, but ( [ [ ) is illegal because the ( bracket cannot be closed before the [ brackets are. -Sticking with $n=2$, let $B_-\subset X$ be the set of all sequences in which every left bracket has a corresponding right bracket, and $B_+$ be the set of all sequences in which every right bracket has a corresponding left bracket. One can show that every shift-invariant measure has $\mu(B_- \cup B_+) = 1$ by partitioning the complement into a countable collection of disjoint sets indexed by the location of the first/last left/right bracket with no partner. -Define a map $\pi_+\colon B_+ \to \{0,1,2\}^\mathbb{Z}$ by sending ( to 0, [ to 1, and both ) and ] to 2. Then $\pi_+$ is an isomorphism between the two shift spaces because every right bracket has a corresponding left bracket, and hence its identity as ) or ] is uniquely determined by the rules of the shift. Similarly, the analogous map $\pi_- \colon B_- \to \{0,1,2\}^\mathbb{Z}$ is an isomorphism. -Because every ergodic invariant measure on $X$ is supported on either $B_-$ or $B_+$, we conclude that $h(X) = \log 3$ and that there are exactly two ergodic measures of maximal entropy $\mu_{\pm} = \nu \circ \pi_{\pm}$, where $\nu$ is the $(\frac 13, \frac 13, \frac 13)$-Bernoulli measure on the full 3-shift. Each of these measures gives positive measure to every open set in $X$, and each is of positive entropy -- indeed, each is Bernoulli, which is part of what makes this answer so satisfying to me. -Note that for larger values of $n$ the same argument shows that $h(X) = \log(n+1)$.<|endoftext|> -TITLE: Birkhoff's theorem about doubly stochastic matrices -QUESTION [6 upvotes]: Birkhoff's theorem states: -The set of $n \times n$ doubly stochastic matrices is a convex set whose extreme points are the permutation matrices -This theorem seems to be commonly attributed to Birkhoff (perhaps also von Neumann). But I recall listening to a talk by Harold Kuhn, where he said that this theorem should actually be attributed to some $P$ where $P \in \{$Jacobi, Dénes Kőnig, Jenő Egerváry, Somebody else?$\}$. -Question: Does anybody know whom Kuhn might have meant, and to whom this theorem should really be attributed? -I would be very happy to learn the connection (also, yes, am embarrassed that despite listening carefully during the talk, I have still forgotten!) - -REPLY [4 votes]: From Schneider's "The Birkhoff-Egervary-Konig theorem for matrices over lattice ordered abelian groups" after the following theorem is presented - -Let $G$ be a lattice ordered abelian group. Every generalized doubly stochastic matrix with elements from $G$ is the sum of generalized permutation matrices. - -it is remarked that the theorem for $G=\mathbb{Z}$ was obtained by Konig in 1916 and for $G=\mathbb{R}$ by Birkhoff in 1946 however in 1931 "Egervary proved a result for integral matrices which is more general than Konig's theorem. He observed that by continuity considerations his theorem may be shown to hold for real matrices. Thus in this way one obtains a result which contains Birkhoff's theorem." There are many references in the paper.<|endoftext|> -TITLE: Examples of categorification -QUESTION [31 upvotes]: What is your favorite example of categorification? - -REPLY [4 votes]: The canonical example in my mind is: -Sets ~> vector spaces ~> linear categories -This is not so trivial -- it is relevant to the topic of extended TQFTs.<|endoftext|> -TITLE: Sheaves without global sections -QUESTION [15 upvotes]: The line bundle $O(-1)$ on a projective space or $O(-\rho)$ on a flag variety has a property that all its cohomology vanish. Is there a story behind such sheaves? -Here are more precise questions. Let $X$ be a smooth complex projective surface (say, a nice one like Del Pezzo or K3). Does there always exist a coherent locally free sheaf $M$ whose derived global sections vanish? Can one describe all such sheaves? Is there a coarse moduli space of such sheaves? - -REPLY [6 votes]: On a K3-surface if a line bundle $\mathcal L$ has vanishing cohomology, then -$\mathcal L^2=-4$ by the Riemann-Roch formula. Conversely, if $\mathcal L^2=-4$ -and neither $\mathcal L$ nor its inverse has global sections, then also the -first cohomology vanishes. -If $X$ is a projective K3-surface with $\mathrm{Pic}(X)=\mathbb Z$, then -$\mathcal L^2\geq0$ for all line bundles and hence no line bundle has all its -cohomology vanishing. On the other hand there are (plenty of) K3-surfaces with -an ample line bundle $\mathcal M$ and a line bundle $\mathcal L$ with $\mathcal -M\cdot\mathcal L=0$ and $\mathcal L^2=-4$. Then neither $\mathcal L$ nor its -inverse has global sections (as otherwise we would get a strictly effective -divisor with zero intersection with $\mathcal M$). Thus for line bundles at -least the problem restricted to K3-surfaces depends very much on the surface. -As for the idea of consider the cokernel $B$ of $\mathcal O_X\rightarrow -F_\ast\mathcal O_X$. We have that the cohomology of $B$ vanishes if $X$ is -ordinary. The general complete intersection is ordinary (by a theorem of -Deligne) so there are plenty of those. Raynaud has also shown that if $X$ is a -curve then $B\bigotimes \mathcal L$ has vanishing cohomology for the general -line bundle $\mathcal L$ of degree $0$.<|endoftext|> -TITLE: Would a supersymmetric theory of von Neumann algebras be useful? -QUESTION [7 upvotes]: While looking over the first chapter of -1) Quantum Fields and Strings: A Course For Mathematicians (P. Deligne, P. Etingof, D.S. Freed, L. Jeffrey, D. Kazhdan, J. Morgan, D.R. Morrison and E. Witten, eds.,), 2 vols., American Mathematical Society, Providence, 1999. -I wondered whether there would be any use to developing a theory of super-von Neumann algebras, mimicking the usual theory. Not knowing whether or not this would be a sterile or trivial exercise, I never tried. -I have always wondered, though: - -Would there be any benefit in developing a theory of super-von Neumann algebras, and if so what would the benefit likely be? Particularly, could such a theory tell us anything useful about ordinary von Neumann algebras we otherwise couldn't easily obtain? - -Perhaps this is a trivial question, but I'm curious if anyone with broader knowledge can shed some light on this. -Of course, the dream is that looking at something like this would miraculously unveil something cool like a canonical time-evolution on $II_{1}$-factors. -(This is another candidate for the 'dumb question' tag!) - -REPLY [6 votes]: Maybe a dump answer. There is the notion of $\mathbb{Z}_2$-graded von Neumann algebra acting on a $\mathbb{Z}_2$ graded Hilbert space. There is also the notion of a graded commutant etc. and it is used in the context fermionic nets, CAR algebra, supersymmetric conformal nets etc. -So it looks for me more like "trivial exercise".<|endoftext|> -TITLE: Can a single DVR witness all specializations on a variety? -QUESTION [14 upvotes]: If $X$ is a noetherian scheme with points $x$ and $\xi$ so that $x$ is in the closure of $\{\xi\}$, then there exists a discrete valuation ring $V$ and a map $Spec(V)\to X$ sending the generic point to $\xi$ and the closed point to $x$. That is, any closure relation among point of $X$ can be witnessed by a map from a DVR. This follows from the commutative algebra fact that any noetherian local ring is dominated by a DVR. (Applied to the local ring of $x$ in $\overline{\{\xi\}}$) -I'd like to know if I can rearrange the quantifiers: can I choose a single $V$ that will witness any closure relation among points of $X$? The answer is NO, because the characteristics of residue fields of points of $X$ can jump around. For example, if $X=Spec(\mathbb Z)$ and the residue field of $V$ has characteristic $p$, there's no way $V$ can witness the specialization $(q)\in\overline{\{(0)\}}$ for a prime $q$ different from $p$. If I remove this problem by requiring $X$ to be of finite type over a field, I feel like the answer might be yes. - -If $X$ is a scheme of finite type (may as well be a variety) over a field $k$, does there exist a DVR $V$ so that for any closure relation $x\in\overline{\{\xi\}}$ among points of $X$, there is a map $Spec(V)\to X$ sending the generic point to $\xi$ and the closed point to $x$? - -Remark 1: I think it's worth pointing out one way to construct these witnessing maps from DVRs. The local ring at the generic point of a divisor on a smooth variety is a DVR, so if $\overline{\{\xi\}}$ is smooth and $\overline{\{x\}}$ is a divisor, it's easy. Otherwise, you can (in characteristic zero) resolve the singularities of $\overline{\{\xi\}}$ and blow up a point that maps to $x$. Now you have a smooth variety whose generic point maps to $\xi$ with a divisor whose generic point maps to $x$, so the local ring at the generic point of the divisor is a DVR that witnesses the specialization. Perhaps reasoning like this can somehow be used to construct $V$. -Remark 2: If $V$ exists, it's huge. For example, if $X=\mathbb A^2$, then the generic point of $X$ specializes to the generic point of any curve in $\mathbb A^2$, so the residue field of $V$ will have to contain the function fields of curves of arbitrarily large genus. Maybe there's a small way to accomplish this, but I don't mind if $V$ ends up being a ginormous ultraproduct or something. - -REPLY [5 votes]: I think you already answered the question yourself and so did Konstantin Ardakov. But this question was still marked as open... -Suppose that the ground field is the field Q of rational numbers. Then let V be the power series ring C[[t]] where C is the field of complex numbers. This is an example. -To prove it, as you say, it suffices to prove that any local domain A which is a localization of a smooth algebra over Q, can be dominated by C[[t]]. Choose a regular system of parameters x_1, ..., x_d in the maximal ideal of A. Consider the quotient B of A(t_1, ..., t_d) by x_i - t_i x_1 where t_1, ..., t_d are variables and where A(t_1, ..., t_d) is the localization of A[t_1, ..., t_d] at the prime ideal m_A(A[t_1, ..., t_d]). Then B is a regular local ring of dimension 1 and A injects into B. The completion of B is K[[t]] where K is a finitely generated extension of Q. Hence B, and therefore A, is dominated by C[[t]]. -An Italian geometer might replace the use of C by a universal overfield.<|endoftext|> -TITLE: Decomposing the plane into intervals -QUESTION [10 upvotes]: I posted this on Stack Exchange and got a lot of interest, but no answer. -A recent Missouri State problem stated that it is easy to decompose the plane into half-open intervals and asked us to do so with intervals pointing in every direction. That got me trying to decompose the plane into closed or open intervals. The best I could do was to make a square with two sides missing (which you can do out of either type) and form a checkerboard with the white squares missing the top and bottom and the black squares missing the left and right. That gets the whole plane except the lattice points. This seems like it must be a standard problem, but I couldn't find it on the web. Question: So can the plane be decomposed into unit open intervals? closed intervals? - -REPLY [7 votes]: Conway and Croft show it can be done for closed intervals and cannot -be done for open intervals in the paper: -Covering a sphere with congruent great-circle arcs. -Proc. Cambridge Philos. Soc. 60 1964 787–800.<|endoftext|> -TITLE: Is every nonnegatively curved plane conformal to the complex plane? -QUESTION [9 upvotes]: Is it true that any complete metric of nonnegative (Gauss) curvature on $\mathbb R^2$ is conformally equivalent to the standard Euclidean metric on $\mathbb R^2$? -Remarks: locally any Riemannian metric on a surface is conformally equivalent -to the standard $\mathbb R^2$, due to existence of isothermal coordinates. Thus a Riemannian metric defines a conformal structure on the surface, and the uniformization theorem says that a simply-connected open surface is conformally equivalent to the complex plane or the unit disk. Thus the question is whether a complete nonnegatively curved plane is conformally equivalent to the complex plane. - -REPLY [7 votes]: Cheng-Yau proved that: A complete Riemannian manifold with non-negative Ricci curvature -and at most quadratic growth for volumes of balls as the radius goes to infinity is -parabolic. -EDIT (by Igor Belegradek). Various criteria for parabolicity are found in the survey of Grigoryan. In particular, on page 177 it is mentioned that a a complete Riemannian manifold with at most quadratic volume growth is parabolic. For complete open nonnegatively curved surfaces the volume growth is at most quadratic by Bishop-Gromov. On the other hand, there are parabolic complete manifolds with arbitrary fast volume growth (see page 180 of the same survey). Finally, the very first proof of parabolicity of complete nonnegatively curved plane seems to be due to Blanc-Fiala (1941); the reference is in Huber's paper mentioned in my comment to Anton's answer.<|endoftext|> -TITLE: Approximation by locally Lipschitz functions -QUESTION [9 upvotes]: Could you tell me what is the name and/or reference for the following theorem: - -Let $M$ be a metric space. Then any continuous function $f:M\to\mathbb R$ can be a be uniformly approximated by a locally Lipschitz functions. - -REPLY [7 votes]: oh what a good question it is ! i think this question is from the following paper : -see : https://projecteuclid.org/download/pdf_1/euclid.rae/1230939175 -and your question is the same with theorem 2 in this paper ! -is it helpful ? thank you !<|endoftext|> -TITLE: Looking for reference on Serre's talk "linear rep and number of points mod p" -QUESTION [10 upvotes]: Actually I am not sure this is a legitimate question on MO. In April and June of this year Serre gave two talks on the same title "linear representations and the number of points mod p", one in ETH Number theory Days Zurich, another during Prof. Gross's birthday conference in Boston. Unfortunately I was in neither of those, nor could I find another reference about this talk online. -In the proof of Weil conjecture for curves, say an elliptic curve $E$ over $\mathbf{Q}$ has good reduction mod $p$, then the characteristic polynomial of the Frobenius operator on the Tate module for $E$ mod $p$ will essentially give us the number of points on the curve in $\mathbf{F}_{p^n}$. So I would say this is an obvious example of relations between two-dimensional representation and number of points. But I wonder if Serre has more. E.g. (tangentially related) Here Mazur was interested in the Chebyshev bias (which is usually a quite analytic phenomenon) among the number of points corresponding to different $p$ (which, by the way, has few references also), and I'd like to know if the representation side could shed some light on this. -Of course since I didn't attend the talk, Serre could be talking about totally different things. I would greatly appreciate it if anyone attended the talk/have seen such notes/heard about this circle of ideas share some comments on this. Thanks! - -REPLY [10 votes]: You might be interested in Serre's recent book -Lectures on $N_X(p)$, AK Peters, Taylor and Francis, New York, 2011, 163 pages, -a version of which is available on his website at the Collège de France. -Serre deals with the following basic question : Let $X$ be a scheme of finite type over $\mathbf{Z}$. What can you say about the number $N_X(p)$ of points of $X$ over the finite field $\mathbf{F}_p$, as $p$ varies (over the primes) ? For example, it is not entirely trivial (Theorem 1.1) to show that $X$ is empty if and only if $N_X(p)=0$ for all sufficiently large $p$. -Another aspect of the question is the relationship between the analytic space $X(\mathbf{C})$ and the sequence $N_X(p)$. For example, can you recover the dimension of $X(\mathbf{C})$, or the number of irreducible components of $X(\mathbf{C})$ from the sequence $N_X(p)$ ? See Theorem 1.2 for this. -Later (Thoerem 6.15) he discusses what conclusion can be drawn if $|N_X(p)-N_Y(p)|<2$ for a set of $p$ of density $1$.<|endoftext|> -TITLE: Quantum probability experiment? -QUESTION [5 upvotes]: I am looking for an example (or definition) of a quantum probability experiment (if there is such a thing). Ideally it should have these properties: - -Be purely mathematical; no mention of physics or other empirical sciences; -in the example, all variables should be replaced by constants that are as small or simple as possible without collapsing back into classical probability; -it should state what are the analogues of the ingredients of a classical probability experiment: sample space $\Omega$, outcome $\omega\in\Omega$, event $E\subseteq\Omega$, probability $\mathbb P(E)$, random variable $X:\Omega\rightarrow V$ where $V=\mathbb R$ or something else. -show explicitly how $\sigma$-additivity or finite additivity $\mathbb P(\sqcup_i A_i)=\sum_i \mathbb P(A_i)$ fails, or is replaced by some other rule, as the case may be. Use of measure theory is welcome. -be short enough to fit in an MO answer. - -I looked at Greg Kuperberg's draft article here and answer here, and while I had trouble extracting what I am describing above, it made me hopeful that such a thing might be possible. - -REPLY [10 votes]: There is an old example due to Kochen and Specker of sort-of this point, which later was replaced by a much better example due to Bell. (Actually, the Kochen-Specker construction is related to an earlier, stronger result of Andrew Gleason.) Bell's example, after some tidying up, was actually tested in a physical experiment by Alain Aspect, and by many other people since. In the Kochen-Specker thought experiment, finite additivity does not directly fail. Rather, you can exhibit a set of Booleans which cannot generate a traditional Boolean algebra that is consistent with the quantum interpretation. In other words, you cannot have "underlying determinism" that is consistent with natural embeddings of classical probability into quantum probability. -Consider the quantum probability space $M_3 = M_3(\mathbb{C})$ of $3 \times 3$ matrices. For each line $L \in \mathbb{C}^3$, there is a corresponding Boolean event given as an operator by the orthonormal projection onto $L$. When Boolean events commute, they generate a commutative algebra, and thus a classical probability space, and one has traditional additivity. So if you have an orthonormal frame of lines, you get a copy of the Boolean algebra of events on three points. However, these Boolean algebras are inconsistent, even just using lines in $\mathbb{R}^3$. There exists a collection of 31 lines in $\mathbb{R}^3$ that do not admit a boolean function $f$ such that exactly one of $f(L_1)$, $f(L_2)$, and $f(L_3)$ is true for every orthogonal frame $\{L_1,L_2,L_3\}$. (Following Gleason, Kochen and Specker found 117 lines, but this was simplified to 31 by Conway and Kochen.) The set of lines is complicated, but there is a simpler construction of Peres in $\mathbb{R}^4$, which then establishes the contradiction for $M_4$. Namely, you should take the 12 diagonals of the regular 24-cell and the 12 diagonals of the dual 24-cell. Then it is an exercise to check that these 24 lines have the same inconsistency with respect to orthonormal frames. -At another level, the contradiction that you seek is not possible without an extra structure on classical or quantum probability. Classical probability makes a category whose objects are Boolean algebras or $L^\infty$ algebras, and whose morphisms are, say, stochastic maps. (Actually you want contravariant stochastic maps. An map from algebra $A$ to algebra $B$ yields a map of states from $B$ to $A$, and stochastic maps are usually defined on states.) Quantum probability does also, where the objects are von Neumann algebras and the morphisms are quantum stochastic maps, by definition completely positive, normal, unital maps. However, any such category embeds in the classical probability category or even the category Set. This formal result has been considered important by people who want to make quantum probability look classical. (E.g., it is essentially Bohm's point.) It means that if you have a single quantum system, you can always build a countability additive classical probability model "behind" it. - -However, classical probability is also a tensor category to model joint systems, and so is quantum probability. Bell's theorem is that there does not exist a tensor embedding (speaking loosely; I'd have to think about how to rigorously define what it excludes) of quantum probability into classical probability. In physics terminology, a non-tensor-preserving construction such as that of Bohm is called "non-local". -Here is a description of an optimized form of Bell's construction called the CHSH inequality. Let $a_1, a_2 \in A$ and $b_1, b_2 \in B$ be four Booleans or, more conveniently, four $\pm 1$-valued Bernoulli random variables. Then their correlation matrix $E[a_j b_k]$ in $A \otimes B$ satisfies this elementary inequality in classical probability: -$$E[a_1 b_1] + E[a_1 b_2] + E[a_2 b_1] - E[a_2 b_2] \le 2.$$ -To prove this inequality, consider that each term is $\pm 1$ if you plug in four values of the variables; if the first three term are 1 then all four are equal and the last term is $-1$. (You can easily make them unbiased so that $E[a_j b_k]$ really is the correlation matrix.) However, if you take $A \cong M_2$ (the qubit system) and $B \cong M_2$, then it is easy to find four such random variables and a state on $A \otimes B$ such that instead, -$$E[a_1 b_1] + E[a_1 b_2] + E[a_2 b_1] - E[a_2 b_2] = 2\sqrt{2}.$$ -(This is the maximum possible value in quantum probability.) This is a classical impossible set of correlations. You can repeatedly interrogate system $A$ by randomly measuring either $a_1$ or $a_2$, and the same for $B$. The correlation makes you think that $A$ and $B$ are in communication with each other, even when it cannot be true. This is what was actually demonstrated in Aspect's experiment. His experiment directly violated the CHSH inequality, in a protocol in which there wasn't enough time for light to travel from $A$ to $B$. -Here are a few more comments about the violation demonstrated in Aspect-type experiments. What is actually measured is -$$E = E[(-1)^{(j+1)(k+1)} a_j b_k],$$ -where $j$ and $k$ are also random variables taking values in $\{1,2\}$. If $A$ and $B$ (or Alice and Bob) could communicate, then they could easily make $E > \frac12$, or even as close to 1 as they please, because they would both know the pair $(j,k)$ and they could pick a favorable pair of answers. But in the experiment, they are separated, and Alice is only told $j$ and Bob is only told $k$. In this case, they must separately provide $a_j$ and $b_k$, and the CHSH inequality applies if $A$ and $B$ are classical. -This is a somewhat formal and blasé summary of the illusion of telepathy in quantum probability that was first constructed by Bell. I have a livelier description of the same formulas in a colloquium talk that I gave at Berkeley.<|endoftext|> -TITLE: A Distinct parts/Odd parts identity for standard Young tableaux -QUESTION [9 upvotes]: Let $\lambda$ denote a partition of size $n$. Let -$$d_{\lambda}= \text{number of distinct parts of } \lambda $$ -$$o_{\lambda}= \text{number of odd parts of } \lambda $$ -$$f_{\lambda}= \text{number of standard Young tableau of shape } \lambda $$ -Given an involution $\pi \in S_{n}$, whose insertion tableau has shape $\lambda$, it is well known (via the Robinson-Schensted correspondence, and neatly outlined in Sagan's book on the Symmetric Group) that : -$$ o_{\lambda^{t}}= \text{number of fixed points in the involution } \pi $$ -$$ \sum_{\lambda \vdash n} f_{\lambda}= \text{number of involutions in } S_{n} $$ -In the aforementioned formulae, $\lambda^{t}$ refers to the conjugate of the partition $\lambda$. -Now, some computations I have carried out for Kronecker products of two irreducible characters of $S_{n}$ revealed the following identity in a special case: -$$\sum_{\lambda \vdash n}d_{\lambda}f_{\lambda}=\sum_{\lambda \vdash n}o_{\lambda}f_{\lambda}$$ -Note that the right hand side actually counts the total number of fixed points in all involutions in $S_{n}$. I did manage to prove the above result in general, but I am hoping someone could guide me to a proof which is bijective, i.e say uses the RS correspondence to establish the left hand side equals the the total number of fixed points in all involutions in $S_{n}$. -Also, I'd like it if I could be directed to where this and/or similar sums appeared.(as an exercise in a book, or in some paper). -Thanks! -Edit: I had a look at Sagan, which I did not have handy last night and made a minor change in saying the number of fixed points in an involution $\pi \in S_{n}$ is the number of odd columns in the insertion tableau of $\pi$. -Edit(10/27): -I thought I should put down the idea that I had. But since I am not sure if this should count as an answer, I am putting it in the body of the question. -Note that -$$\sum_{\lambda \vdash n}d_{\lambda}f_{\lambda}=\sum_{\lambda \vdash n+1}f_{\lambda}-\sum_{\lambda \vdash n}f_{\lambda}$$ -So all that remains to be shown is the nice fact that the total number of fixed points in all the involutions of $S_{n}$ is the difference between the number of involutions in $S_{n+1}$ and the number of involutions in $S_{n}$. - -REPLY [2 votes]: "all that remains to be shown is the nice fact that the total number of fixed points in all the involutions of $S_n$ is the difference between the number of involutions in $S_{n+1}$ and the number of involutions in $S_n$." -And this is straightforward: every involution in $S_n$ can be extended to an involution in $S_{n + 1}$ either by replacing a fixed point $i$ with a cycle $(i, n + 1)$ or by adding $n + 1$ as a new fixed point. -(It is unclear to me whether you had already seen that this fact has such a simple bijective proof; it's also not clear to me whether this satisfies your desire for a completely bijective proof.)<|endoftext|> -TITLE: kernel of G(Z/p^2 Z)->G(Z/pZ) is the lie algebra of G over Z/pZ? -QUESTION [5 upvotes]: Let $G$ be an affine algebraic group defined over $\mathbf Z$. The kernel of the natural homomorphism $G(\mathbf Z/p^2\mathbf Z)\to G(\mathbf Z/p\mathbf Z)$, if abelian, is a group which comes along with the conjugation action of $G(\mathbf Z/p\mathbf Z)$. -In the case where $G$ is a classical group, this kernel is isomorphic (as a set with $G(\mathbf Z/p\mathbf Z)$-action) to the Lie algebra $\mathfrak g(\mathbf Z/p\mathbf Z)$ of $G(\mathbf Z/p\mathbf Z)$ (which comes with the adjoint action). It seems that this should be the case in general. -Does anyone know of a reference for this kind of thing? - -REPLY [8 votes]: Take a look at Waterhouse's book - Introduction to affine group schemes. I think Theorem 12.2 is what you're looking for.<|endoftext|> -TITLE: Levy's isoperimetric inequality for sphere -QUESTION [11 upvotes]: Let me recall subj: -If $s>0$, $A$ and $B$ are two subsets of $\mathbb{S}^{n}$, $|A|=|B|$ ($|\cdot|$ stands for the Lebesgue measure on the sphere) and $B$ is a cup $B=\{ (x_1,x_2,\dots,x_n)\in \mathbb{S}^n, x_n\leq t \}$ (for some $t\in [-1,1]$), then $|A_s|\geq |B_s|$, where $A_s$ means $s$-neighborhood of the set. -It leads to measure concentration inequalities for the sphere and so has numerous applications. So I guess that Levy's initial proof was simplified, maybe not once. What is the easiest proof of the inequality and where to read it? - -REPLY [2 votes]: A different symmetrization-based proof is given in this review article by Schechtman (pp. 7-8); see the previous page for references.<|endoftext|> -TITLE: Which elements in SL2(Q) are conjugated to an element in SL2(Z) -QUESTION [11 upvotes]: Dear all, -once again my question is all about $SL_2(\mathbb{Z})$ and $SL_2(\mathbb{Q})$ ! Which elements in $M \in SL_2(\mathbb{Q})$ can you write in the following form: -$M= NBN^{-1}$ -with $N \in GL_2(\mathbb{Q})$ and $B \in SL_2(\mathbb{Z})$? -It is surely nor all of $SL_2(\mathbb{Q})$ (look at traces), but I do not have any guess which matrices I get! -Thank you very much again! -Karl - -REPLY [17 votes]: You can do this if and only if the trace of $M$ is an integer. -By the theory of the rational canonical form if matrices -$A$ and $B$ over $\mathbb{Q}$ have the same characteristic -polynomial and neither has a repeated eigenvalue they are -conjugate by a matrix over $\mathbb{Q}$. This almost does it, -save for some fiddling about when the eignvalue of $M$ is repeated.<|endoftext|> -TITLE: Expectation of first positive value in random walk -QUESTION [5 upvotes]: Let $p$ be a parameter in $]0,1[$. Let $(X_k)_{k\geq 0}$ be an independent, identically distributed sequence of random variables, such that each $X_k$ takes values only in -$\lbrace -1, \frac{1-p}{p} \rbrace$ and $P(X_k=-1)=1-p$ (so that $X_k$ has mean $0$). -Let $S_n=X_1+X_2+ \ldots +X_n$ for $n\geq 1$ and let $N$ denote the smallest -integer such that $S_{N} > 0$ (it is well known that $N$ exists almost surely). -What is the expectation of $S_{N}$ ? -If $p$ is of the form $1-\frac{1}{k}$ where $k$ is an integer, it is easily seen -that $S_{N}$ is constant and equal to $\frac{1}{k-1}$. -Update 10/26/2010: In general, $S_N$ can only take a finite number of values, so the expectation is finite, as noted in the comments below. It seems -that the distribution of $S_N$ should be computable using some simple algebra, -but I was unable to do this. - The finite-set of values property allows one however to compute $E(S_N)$ to a reasonable -acurracy for a given $p$. For $p=\frac{1}{3}$, the expectation is larger than 1 -and does not seem to be rational. - -REPLY [2 votes]: (EDITED, to comply with quite accurate objections by Louigi and Byron) -Assume that with full probability $X_k$ is either $-1$ or a random positive integer (this includes the setting of the question when $p=1/(k+1)$ with $k$ a positive integer but note that $X_k$ may take more than one positive integer values). Then, Wiener-Hopf factorization formula becomes simple enough to compute the distribution of $S_N$. -More precisely, let $N$ denote the first time $n\ge1$ such that $S_n>0$ (as in the OP's post) and let $M$ denote the first time $n\ge1$ such that $S_n\le 0$ (note the "lower than or equal to"). In the centered and bounded case the OP is interested in, $N$ and $M$ are both almost surely finite and Wiener-Hopf formula reads -$$ -(1-E(e^{iuS_N}))(1-E(e^{iuS_M}))=1-E(e^{iuX}), -$$ -for every real number $u$ and every $X$ distributed as the steps $X_k$. Here, $S_M=-1$ on $[X_1=-1]$ and $S_M=0$ on $[X_1>0]$. This yields -$$ -q(1-e^{-iu})E(e^{iuS_N})=E(e^{iuX};X>0)-p, -$$ -with $q=P[X=-1]$ and $p=1-q=P[X>0]$. This provides the full distribution of $S_N$ and, differentiating both sides at $u=0$, the expectation of $S_N$. The end result is -$$ -E(S_N)=E(X+X^2;X>0)/(2q). -$$ -If $X=-1$ or $X=k$ with $k$ a positive integer, then $[X > 0]=[X=k]$ and $p=1/(k+1)$, and one sees that $S_N$ is uniformly distributed on the integers from $1$ to $k$ and that $E(S_N)=(k+1)/2$.<|endoftext|> -TITLE: Compute Lie algebra cohomology -QUESTION [15 upvotes]: Is there a computer algebra system that is able to compute the Lie algebra cohomology in a given representation? What if the Lie algebra is finite dimensional? -In my case I would like to be able to compute the cohomology in the following situation: -let $\mathfrak{g}\subset \mathfrak{h}$ be an inclusion of finite dimensional complex Lie albebras, I'd like to compute the cohomology of $Hom(\mathfrak{g}, \mathfrak{h}/\mathfrak{g})$. - -REPLY [17 votes]: I have looked at this question a few years ago (with some more recent sporadic gilmpses), so I am definitely not uptodate. Here it goes, anyway, what I have learned back then: - -GAP. -GAP is a wonderful tool, but I would not call its cohomology computation capabilities efficient. As -far I understand, it implements a straightfoward approach by constructing appropriate matrices and solve a linear algebra problem. One sees easily how the matrix size grows prohibitively with the dimension of an algebra. Seems to be not suitable for anything beyond "toy" problems. -Mathematica. -Mathematica-based package "SuperLie" for computations in Lie (super)algebras, including cohomology, written by Pavel Grozman and used by Dimitry Leites and his collaborators in their recent papers (see arXiv): http://www.equaonline.com/math/SuperLie/ . Seems to have a very steep learning curve but looks quite impressive. Seems to outperform GAP, but how far - I don't know. -C. -There is program (constantly evolving as far I understand) for computations of cohomology of Lie (super)algebras by Vladimir Kornyak (see arXiv). He is doing things beyond straightforward linear algebra approach - for example, he tries to split the cochain complex to smaller subcomplexes and perform reduction modulo an appropriate prime (in the case of zero characteristic). Seems to be comparable with "SuperLie" (as far as cohomology is concerned). It is written in plain C and does not have overheads of computer algebra systems. Unfortunately, Kornyak does not disclose (at least publically) sources or even binaries. -REDUCE. -N. v.d Hijligenberg and G. Post, Computation by computer of Lie superalgebra homology and cohomology, Acta Appl. Math. 41 (1995), 123-134 http://dx.doi.org/10.1007/BF00996108 - haven't looked at it. -LiE -J. Silhan, Algorithmic computations of Lie algebras cohomologies, Proceedings of the 22nd Winter School ``Geometry and Physics'', Rend. Circ. Mat. Palermo Suppl. No. 71 (2003), 191-197 http://dml.cz/handle/10338.dmlcz/701718 . A more specific program, implemented in LiE, for computation of cohomology of parabolic subalgebras of classical Lie algebras and related, basing on celebrated Kostant's work. Haven't looked at it thoroughly. -Magma. -I never bothered with this commercial package which seems to be comparable with GAP. -? -D.V. Reshetnikov, Computation of cohomology groups of the Lie algebras of type $B_n$ and $C_n$, Russian Math. (Izv. VUZ) 53 (2009), N8, 58-59 http://dx.doi.org/10.3103/S1066369X0908009X . Here the author reports (quite uselessly, I should admit, as no further details are given) about a program for computation of Lie algebra cohomology developed by him. - -I personally think that there is a big unexplored area here - one should use heavily sparsity of occuring matrices (currently none of the programs described above seems to use it). Which kind of sparsity it is, is not clear apriori, and, moreover, I suspect that for different algebras and modules one will have different kinds of sparsity. This makes an interesting connection with methods and tricks from numerical linear algebra. -Again, take all this with a grain of salt, as things may have changed since I looked at them.<|endoftext|> -TITLE: Subfactor summer reading list -QUESTION [10 upvotes]: Many people I talk to lament the nonexistence of a coherent source for learning the theory of subfactors. -Could someone suggest a nice (ordered) list of books/papers to work through to obtain a suitable background in this theory, assuming the audience is comprised of mathematicians familiar with the basics of von Neumann algebra theory? - -REPLY [2 votes]: I also think a good starting point is Jones and Sunders. -If you are interested in type III subfactors, the literature seems to me a bit rare, but there are some lecture notes: -Kosaki - "Type III factors and index theory" -In algebraic quantum field theory one studies "nets" which associates to certain space time regions (under certain conditions) a type III factor. Subfactors play an important role. There is e.g. the paper: -Longo, Rehren - "Nets of Subfactors"<|endoftext|> -TITLE: What's a mathematician to do? -QUESTION [292 upvotes]: I have to apologize because this is not the normal sort of question for this site, but there have been times in the past where MO was remarkably helpful and kind to undergrads with similar types of question and since it is worrying me increasingly as of late I feel that I must ask it. -My question is: what can one (such as myself) contribute to mathematics? -I find that mathematics is made by people like Gauss and Euler - while it may be possible to learn their work and understand it, nothing new is created by doing this. One can rewrite their books in modern language and notation or guide others to learn it too but I never believed this was the significant part of a mathematician work; which would be the creation of original mathematics. It seems entirely plausible that, with all the tremendously clever people working so hard on mathematics, there is nothing left for someone such as myself (who would be the first to admit they do not have any special talent in the field) to do. Perhaps my value would be to act more like cannon fodder? Since just sending in enough men in will surely break through some barrier. -Anyway I don't want to ramble too much but I really would like to find answers to this question - whether they come from experiences or peoples biographies or anywhere. -Thank you. - -REPLY [33 votes]: It's 12:40am New Year's Day, so maybe not the best time to be writing MO responses, but I loved reading everyone's encouraging answers to this question. (BTW: @Lubin: Taking algebra, with a dollop of algebraic number theory, from you was inspiring.) Anyway, my answer to the OP would be to learn lots and to work hard on problems that interest you. Don't worry so much about whether you're proving breakthrough theorems, just try as hard as you can to understand the parts of mathematics that interest you the most. (By "understand", of course, I mean get into the guts, figure out what's really going on, and prove as much as you can.) Also don't worry that you won't solve every problem (or even a majority of the problems) on which you work, and don't worry that you won't ever feel you fully understand everything about a problem; that's why there's always more to investigate. Then, after a decade or two, feel free to look back, and I think you'll find that you have made a contribution to humanity's knowledge of mathematics. -And even when you're doing research, it's hard (at least, I've found it hard) to decide on the significance of what you've done. I think the difficulty is that after working hard on a problem for a year or two and making enough progress to write a paper, one understands the problem so well that everything that one can prove seems trivial, while everything that's left undone seems hopeless. So maybe my saying "wait a decade or two" is a bit excessive, but it's definitely worth waiting a couple years before you decide on the quality of the work you've done. -Happy New Year to one and all at MO.<|endoftext|> -TITLE: Is the cut locus of a generic point in a hyperbolic manifold a generic polyhedron? -QUESTION [12 upvotes]: Let $p\in M$ be a point in a closed riemannian manifold $M$. Recall that the cut locus of $p$ is the subset of $M$ consisting of all points that are connected to $p$ by at least 2 distance-minimizing geodesics. -I will start with a general question: - -Is it true that for generic metrics on $M$ and generic points $p$, the cut locus of $p$ is a $(n-1)$-dimensional polyhedron with generic singularities? - -By "generic singularities" I mean that $M$ is a simple polyhedron. See for instance this paper of Alexander and Bishop. -This property is certainly not satisfied for some important specific metrics: for instance, if $M$ is a round sphere the cut locus is a point, no matter where $p$ is. If $M$ is a flat torus, we get a generic polyhedron for generic flat metrics. What about hyperbolic manifolds? So, this is my question: - -Let $M^n$ be a hyperbolic $n$-manifold. Is the cut locus of a generic point a $(n-1)$-polyhedron with generic singularities? - -Of course I am mostly interested in the case $n=3$. In dimension $n=2$ one may also pick a generic hyperbolic metric. - Edit: In dimension 1, a simple polyhedron is a graph with vertices of valence 2 or 3. In dimension 2, it is a polyhedron such that the link of a point is either a circle, a circle with a diameter, or a circle with three radii. -In general, a $n$-dimensional compact polyhedron is simple if every point has a neighborhood which is the cone over the $(k-1)$-skeleton of the $(k+1)$-simplex, times a $(n-k)$-disc. - -REPLY [7 votes]: The question you pose is stated as an open question (in the 3-dimensional hyperbolic case) in the following paper: -Díaz, Raquel; Ushijima, Akira -On the properness of some algebraic equations appearing in Fuchsian groups. -Topology Proc. 33 (2009), 81–106. -Quoting from the review on mathscinet: -[the paper] takes its motivation from the fact that, apparently, the statement about the genericity of Dirichlet fundamental polyhedra is open for $\mathbb{H}^3$. (According to the authors, the paper of T. Jørgensen and A. Marden [in Holomorphic functions and moduli, Vol. II (Berkeley, CA, 1986), 69--85, Springer, New York, 1988] has a gap which the present authors have so far been unable to fix.)<|endoftext|> -TITLE: How to show that a space has the homotopy type of wedge of spheres ? -QUESTION [5 upvotes]: Let me try and put the question in context. I am studying certain subsets of the tangent bundle of a sphere. I also have a regular CW complex which is a deformation retract of such a subset. Hence I have a description of the cells and the information that tells me which cell is in the boundary of which other cell. Fortunately this cell complex is a homotopy colimit of a diagram of spaces. As a result I can compute the cohomology groups (but not the product). -All my examples concerning $S^1$ and $S^2$ show that these subsets have the homotopy type of wedge of copies of $S^1$ and $S^2$ respectively. Hence I am trying to prove that this is the case in all dimensions. In this process the only thing I was able to prove that there is a retraction from these subsets to the underlying sphere. -So I would like to know about various methods to show that a space is a wedge of spheres. -I understand that this question might sound vague and the information too little. - -REPLY [4 votes]: To follow up a bit on Mikael's answer, the notion of non-pure shellability is probably more relevant to your situation. Shellable simplicial complexes are wedges of spheres of equal dimension, but non-purity allows different dimensional spheres. You should look at papers by Michelle Wachs and Anders Bjorner if you're interested. However, this will require finding a simplicial decomposition of your space, which may be a challenge. -Added: Since this is now the accepted answer, I figure I should give the precise references. Both papers are on JSTOR (follow the links). -Björner, Anders; Wachs, Michelle L. -Shellable nonpure complexes and posets. I. -Trans. Amer. Math. Soc. 348 (1996), no. 4, 1299–1327. -http://www.jstor.org/stable/i311403 -Björner, Anders; Wachs, Michelle L. -Shellable nonpure complexes and posets. II. -Trans. Amer. Math. Soc. 349 (1997), no. 10, 3945–3975. -http://www.jstor.org/stable/i311413<|endoftext|> -TITLE: Is arbitrary union of closed balls in $\mathbb{R}^n$ Lebesgue measurable? -QUESTION [26 upvotes]: Is an arbitrary union of non-trivial closed balls in the Euclidean space $\mathbb{R}^n$ Lebesgue measurable? If so, is it a Borel set? -@George -I still have two questions concerning your sketch of proof. -First, how can you guarantee each of the open balls in the countable union has radius greater than or equal to 1? -Second, I don't know how to use convexity to prove $\mu (B') \leq (1+\epsilon)^{N}\mu(B)$ - -REPLY [22 votes]: No, in dimension $N>1$, it does not have to be Borel measurable. E.g., in 2 dimensions, consider, a non Borel measurable subset of the reals $S$, and let $A$ be the union of closed unit balls centered at points $(x,0)$ for all $x\in S$. The intersection of $A$ with $\mathbb{R}\times \{1\}$ is the non-Borel set $S \times \{1\}$, so $A$ is not Borel. -On the other hand, for $N=1$, any union of non-trivial closed intervals is Borel-measurable. If $A$ is such a union and $B$ is the union of the open interiors, then it can be seen that $A$ is just the union of $B$ with (at most countably many) endpoints of connected components of $B$. - -Lebesgue measurability does hold, however. Faisal posted a link for this as I was typing my answer, but I think its still worth giving a brief sketch of the proof I was starting to type (Edit: added more detail, as requested). - -Reduce the problem to that of balls with at least some positive radius $r$ and within some bounded region. To do this, suppose that $S$ is the set of closed balls and $S_r$ denotes the balls of radius at least $r$ and with center no further than $r$ from the origin. Then, -$$ -\cup S=\bigcup_{n=1}^\infty\left(\cup S_{1/n}\right). -$$ -As the measurable sets are closed under countable unions, it is enough to show that $\cup S_r$ is Lebesgue measurable for each $r>0$. So, we can assume that all balls are of radius at least $r$ and are within some bounded distance of the origin. -Let $A$ be the union of the closed balls, and $B\subseteq A$ be the union of their interiors. This is open so, by second countability, is a union of countably many open balls of radius at least $r$. Also, $A$ lies between $B$ and its closure $\bar B$. -Show that the boundary $\bar B\setminus B$ of $B$ has zero measure. If we scale up the radius of each of the countable sequence of open balls used to obtain $B$ by a factor $1+\epsilon$ to get the new set $B^\prime$ then $\mu(B^\prime)\le(1+\epsilon)^N\mu(B)$. Showing this is the tricky part, but it does follow from convexity of the balls: If the balls have radius $r_k$ and centres $x_k$, then consider the sets -$$ -B_t=\bigcup_{k=1}^\infty B(r_k,tx_k) -$$ -for real $t$, so that $B_1=B$. The function $t\mapsto\mu(B_t)$ is increasing in $t\ge0$*. Also, $B^\prime= (1+\epsilon)B_{1/(1+\epsilon)}$ giving, -$$ -\mu(B^\prime)=(1+\epsilon)^{N}\mu(B_{1/(1+\epsilon)})\le(1+\epsilon)^{N}\mu(B) -$$ -as claimed. As $\bar B\subseteq B^\prime$ we get $\mu(\bar B\setminus B)\le((1+\epsilon)^N-1)\mu(B)$ which can be made as small as we like by choosing ε small. - -* Edit: in my initial response, I was thinking that this answer is enough to prove that $\mu(B_t)$ is increasing in $t$. However, as Mizar points out in the comments below, this is not clear. Actually, I don't think we can reduce it to that case. However, the result is still true, by the Kneser-Poulson conjecture. This states that if the centres of set of unit balls in Euclidean space are all moved apart, then the measure of their union increases. Although only a conjecture, it has been proved for continuous motions, which applies in our case. Also, expressing each ball of radius greater than some arbitrarily small $r > 0$ as a union of balls of radius $r$, then it still applies in our case for balls of non equal radii. - -Edit: having seen Faisal's explanation, the proof I outline here is completely different to his. The result Faisal quotes is a bit more general as it applies to convex sets with nonempty interior, rather than just balls. However, the proof given above also works for symmetric convex sets with nonempty interiors. As every convex set with nonempty interior is a union of (translates of) symmetric ones, this implies the same result - -REPLY [19 votes]: A google search reveals that an arbitrary union of (nondegenerate) convex sets is Lebesgue measurable: see -Balcerzak and Kharazishvili. On uncountable unions and intersections of measurable sets. -Georgian Math. J. 6 (1999), no. 3, 201–212. -Edit: As requested, here's a summary of the proof. -The authors prove that an arbitrary union of (closed, nondegenerate) $n$-simplices $\{ S_t \}_{t \in T}$ in $\mathbb{R}^n$ is Lebesgue measurable. First a preliminary definition: - -A bounded set $X \subset \mathbb{R}^n$ is said to be $\alpha$-regular, for $\alpha$ a positive real number, if $\lambda(X) \geq \alpha \lambda(V(X))$, where $V(X)$ is a closed ball with minimal diameter for which the inclusion $X \subset V(X)$ holds. - -Observe that an $n$-simplex is $\alpha$-regular for some $\alpha \in (0,1]$. Thus -$$ \bigcup_{t \in T} S_t = \bigcup_{m=1}^\infty \ \bigcup \{ S_t \colon S_t \text{ is } \textstyle{\frac{1}{m}}\text{-regular} \}. $$ -So in order to show that $\cup_t S_t$ is Lebesgue measurable, it suffices to show that $X_m = \cup \{ S_t \colon S_t \text{ is } \frac{1}{m}\text{-regular} \}$ is Lebesgue measurable for all $m \in \mathbb{Z}_{>0}$. Towards this end, given $x \in S_t$ and $c \in (0,1)$, let $S_t(x,c)$ denote the image of $S_t$ under the map $y \mapsto x + c(y-x)$. Then -$$ \mathcal{F}_m = \{ S_t(x,c) \colon S_t \text{ is } \textstyle{\frac{1}{m}}\text{-regular}, x \in S_t, c \in (0,1) \} $$ -is a Vitali covering of $X_m$. The Vitali covering theorem now takes us home: the countable subcollection $\mathcal{F}_m^\ast \subset \mathcal{F}_m$ produced by the theorem has a Lebesgue measurable union $\cup \mathcal{F}_m^\ast$, which also satisfies -$$ \bigcup \mathcal{F}_m^\ast \subset X_m \quad\text{and}\quad \lambda(X_m \backslash \bigcup \mathcal{F}_m^\ast) = 0. $$<|endoftext|> -TITLE: The free group $F_2$ has index 12 in SL(2,$\mathbb{Z}$) -QUESTION [20 upvotes]: Is there someone who can give me some hints/references to the proof of this fact? - -REPLY [4 votes]: May be the discussion is out of date, but I like to present a simple way to show that the subgroup S generated by -$$ -\left(\begin{array}{cc} 1 & 2\cr - 0 & 1 \end{array}\right) -$$ and -$$ -\left(\begin{array}{cc} 1 & 0\cr - 2 & 1 \end{array}\right) -$$ has index 12 in $SL(2,Z)$. -First of all Mark Sapir noted that by Kargopolov S is a group of matrices -$$ -\left(\begin{array}{cc} 1+4k_1 & 2n_1\cr - 2n_2 & 1+4k_2 \end{array}\right) -$$ -Lemma. Let $G$ be a group, -$H$ be a subgroup of $G$, $N$ be a normal subgroup of $G$, that is subgroup of $H$. -Then ind(G:H)=ind(G/N:H/N). -Proof. Indeed, let $x\not\in H$. Then $xN\cap HN\subseteq xH\cap H=\emptyset$. So the natural -homomorphism $G\to G/N$ sends different classes to different classes. -Now consider the natural homomorphism $\phi:SL(2,Z)\to SL(2,Z_4)$. -$G=SL(2,Z)$, $H=S$ and Ker$(\phi)=N$ satisfy the Lemma. -So, ind$(SL(2,Z):S)$=ind$(SL(2,Z_4):\phi(S))=(2*4^2+2*2*2+4*2):4=12$<|endoftext|> -TITLE: Realizing a homology by a smooth immersion -QUESTION [7 upvotes]: An alternative title is: When can I homotope a continuous map to a smooth immersion? -I have a simple topology problem but it's outside my area of expertise and I worry may be rather subtle. Any help would be appreciated. -The set-up is the following: -Let $M$ be some (closed say) $n$ dimensional manifold and suppose that $\Sigma_1$ and $\Sigma_2$ are two closed submanifolds of $M$ of dimension $k$. Note that $\Sigma_1$ and $\Sigma_2$ are allowed to intersect (in my situation they are also embedded but I don't believe this is effects anything). Suppose in addition that $\Sigma_1$ and $\Sigma_2$ are homologous. If $k\leq n-2$ I would like a compact manifold with boundary $\Gamma$ with $\partial \Gamma=\gamma_1\cup \gamma_2$ and a smooth immersion $F:\Gamma\to M$ so that $F(\gamma_i)=\Sigma_i$. In other, words the homology between $\Sigma_1$ and $\Sigma_2$ can be realized by a smooth immersion. -I believe by approximation arguments one can always get a smooth such $F$ without restriction on $k$ but it need not be an immersion (especially if $k=n-1$). My gut is that when you have $k\leq n-2$ since the dimension of the image of $F$ is codimension one you have enough room to perturb it to be an immersion. That is that $F$ is homotopic rel boundary to our desired immersion. -Unfortunately, I don't know enough to formalize this and all my intuition comes from considering curves and domains in $\mathbb{R}^3$ so I'm afraid there may be obstructions in general. -References would be greatly appreciated. -Thanks! -Edit: -As suspected, the question is somewhat subtle . To make it tractable lets assume that $M$ is a $C^\infty$ domain in $\mathbb{R}^3$ (so is a fairly simple three-manifold with boundary) and that the $\Sigma_i$ are curves. This is where my intuition says that there should be such a smooth immersion. - -REPLY [4 votes]: The answer is known in the special case you mentioned (curves in a 3-manifold) -Actually it is true more generally: Namely -if k = n-2 and your k-dimensional oriented submanifolds $\Sigma_1$ and $\Sigma_2$ represent the same INTEGER homology class in the ambient ORIENTED n-manifold M, and they are disjoint and embedded, -then there is an embedded k+1 dimensional submanifold in the cylinder $M \times I$ with boundary $\Sigma_1$ in $M\times {0}$ and $\Sigma_2$ in $M \times {1}.$ -This a trivial consequence of the fact, that -$K(Z,2) = MSO(2).$ -The homotopy classes [M, K(Z,2)] give the 2-dimensional cohomology group, hence also the k-th homology group of M. -The homotopy classes [M, MSO(2)] give the codimension 2 embedded submanifolds up to an embedded cobordism (by the Thom construction). QED. -In general the answer is negative, as it was mentioned by others.<|endoftext|> -TITLE: Some questions about Invexity -QUESTION [7 upvotes]: Recently, I am looking into a non-convex optimization problem whose points satisfying KKT conditions can be obtained. Then the problem becomes how to decide whether the KKT conditions are sufficient for the global optimality of the solution. It is said on Wikipedia(link text) that for "invex function" the KKT is sufficient for optimality. By doing some searching, I was led to a concept called "invexity" and the book "Invexity and Optimization" -( http://www.springerlink.com/content/978-3-540-78561-3#section=155972&page=1). -But, why this "invexity" research, which was first proposed in 1980's, is only confined in a small group of people. And the book is cited for only 4 times. -In my current understanding, invexity is a generalization of convexity, and has some very good properties as in convexity, which should be very attractive and should have drawn lots of people's attention. -Is it because this concept is not interesting, not useful? Or it is being studied under other names? - -REPLY [7 votes]: I second Czenek's recommendation. Moreover, it could be helpful to know this cute little characterization of invexity (due to Craven and Grower, see also here): -Theorem: A differentiable function $f$ is invex if and only if every stationary point is a global minimum. -Proof: Let $f$ be invex and $u$ stationary. Then $\nabla f(u) = 0$ and hence $f(x)-f(u)\geq 0$ for all $x$. Conversely, let every stationary point be globally optimal. Then -$$ -\eta(x,u) = \begin{cases}0 & \text{if}\ \nabla f(u) = 0\\\\ -\frac{f(x)-f(u)}{\|\nabla f(u)\|^2}\nabla f(u) & \text{else}\end{cases} -$$ -shows that $f$ is invex. -Hence, invexity is just another way to say the every stationary point is globally optimal. Since the latter property is fulfilled by quite "arbitrary" functions, one may conclude that invexity does not impose too many structure.<|endoftext|> -TITLE: Applications of Grothendieck-Riemann-Roch? -QUESTION [52 upvotes]: I am currently trying to learn a bit about Grothendieck-Riemann-Roch... -To try to get a better feeling for it, I am looking for examples of nice applications of GRR applied to a proper morphism $X \to Y$ where $Y$ is not a point. I already I know of a fair number of nice applications of HRR, i.e. GRR when $Y$ is a point. I've read through some of the relevant sections of Fulton's Intersection Theory book, but I've only found applications of HRR there, though it's very possible that I overlooked something. -I am also interested in seeing worked-out, explicit, concrete examples, with explicit Chow/cohomology classes. -Thanks much! - -REPLY [3 votes]: Following the answer of Sasha, I can give you here an explicit example of computations using GRR. -Suppose you have a smooth projective surface $S$ and a $(-2)$-curve $i : C \hookrightarrow S$, that is $C^2=-2$, $C\simeq \mathbb{P}^1$. Then for any integer $a\in\mathbb{Z}$ the object $i_\ast O_C(a)$ is a spherical object in $D^b(S)$, and thus you can consider the spherical twist (in the sense of Seidel and Thomas) $T:= T_{\mathcal{O}_C(a)}$. -Now the action $T^H$ of $T$ on the cohomology $H^{\ast} (S,\mathbb{Q})$ (following Huybrechts, "Fourier-Mukai Transforms in Algebraic Geometry") is easily computable once you know the chern character $ch(i_*\mathcal{O}_C(a))$. -I do these computations in cohomology: -By GRR for $i : C \hookrightarrow S$ you have -$$Todd(S) ch(i_\ast O_C(a)) = i_\ast (ch(O_C(a)) Todd(C)) = i_\ast([C]+(a+1)[x])$$ -for $[x]$ the (cohomology) class of a point in $C\simeq \mathbb{P}^1$. -You can get rid of $Todd(S)$ using that $K_S\cdot C=0$, and you obtain -$$ch(i_\ast O_C(a)) = [C]+(a+1)[x].$$<|endoftext|> -TITLE: Non-dominant polynomial maps in the plane -QUESTION [14 upvotes]: Let $P(x,y), Q(x,y)$ be polynomials of two variables over an algebraically closed field $k$. Suppose that the map $(x,y) \mapsto (P(x,y),Q(x,y))$ is not a dominant map from $k^2$ to $k^2$. Does this mean that one has $P(x,y) = R(T(x,y))$ and $Q(x,y) = S(T(x,y))$ for some polynomials $R,S,T$? -For fixed $x$, it seems to me that the curve $\{ (P(x,y), Q(x,y)): y \in k \}$ either degenerates to a point or is a curve of genus zero, which ought to kill off the problem, but I don't have enough of an understanding of the genus zero plane curves to do this. -A closely related question: if $P(x,y)-c$ is reducible for all $c \in k$ (or maybe all but finitely many $c$), does this mean that $P(x,y) = R(T(x,y))$ for some polynomials $R, T$, with $T$ having strictly smaller degree than $P$? I played around a bit with some Galois theory but was only able to convince myself that this question was more or less equivalent to the original question. - -REPLY [4 votes]: My questions are now answered, but just for sake of completeness I wanted to record my own "low tech" approach to the questions which seem to solve the problem in the characteristic zero case, when there are no Frobenius issues and all finite extensions involved are automatically separable. (But my algebraic geometry/Galois theory is so lousy that there may be some mistakes here.) It may be that the other answers are basically a modernised version of these arguments that don't rely on separability. -My idea was to use the two questions to solve each other by induction on the degree of the polynomials involved. If one has a positive answer to the second question, one can deduce the first as follows. We can take $P$ to have smaller or equal degree to $Q$. There are two cases, depending on whether $P(x,y)-c$ is generically reducible or generically irreducible. (We can of course assume $P$ non-constant.) In the former case, one can use the second question to replace $P$ by a polynomial $T$ of lower degree and use the induction hypothesis. In the latter case, if $(x,y) \to (P(x,y),Q(x,y))$ is not dominant, then its image is in a curve, and so for generic $c$, the locus of $P(x,y)-c$ must be contained in a finite union of loci of $Q(x,y)-c'$ for some $c'$ depending on $c$. But $P(x,y)-c$ is irreducible, so by the nullstellensatz this shows that $Q(x,y)-c'$ is some multiple of $P(x,y)-c$ for some $c'$, and we can again use the induction hypothesis to finish up. -Now we use the first question to solve the second. If $P(x,y)-c$ is reducible, then for generic $c$ I think one can use abstract nonsense to factor $P(x,y)-c$ in $K(x,y)$ where $K$ is some finite extension of $k(c)$. Using the primitive element theorem (here is where I need separability and thus characteristic zero), this means (I think) that $P(x,y)-f(z)$ is reducible in $k(x,y,z)$ for some non-trivial polynomial $f$. (It may be that $f$ is merely a rational function rather than a polynomial, but it does not seem to really affect the argument either way.) But, by viewing $P(x,y)-f(z)$ as a polynomial in $z$ with coefficients in $k(x,y)$, we see that all factors of $P(x,y)-f(z)$ (again viewed as polynomials of $z$ with coefficients in $k(x,y)$) must have coefficients that lie in the algebraic closure of $k(P)$. Pick a non-constant coefficient $Q(x,y)$ of this type, then $(x,y) \mapsto (P(x,y),Q(x,y))$ is not dominant, and $Q$ has degree strictly less than $P$, and then the first question gives the desired representation $P=R(T)$. -Combining the two implications with an induction on degree seems to give the claim.<|endoftext|> -TITLE: Logarithmic differentials -QUESTION [7 upvotes]: I have a general and a more special question. I begin with the general one: If $X/k$ is an -algebraic scheme over a field $k$ and $D$ is a divisor with normal crossings on it, then there is the so-called sheaf of log-differentials $\Omega_{X/k}(\log(D))$ on it. Is there a good reference for the very basic properties of this kind of differetials? In particular I am interested in behavior under pullback and analogues to the two exact sequences one has for usual differentials. -Now I come to the special question. Let $k=\mathbb F_q$ and $B/k$ a smooth projective curve. Let $F: B\to B$ be the Frobenius morphism. ($F$ is the identity on the underlying topological space of $B$ and the map ${\cal O}_B(U)\to {\cal O}_B(U)$ induced by $F$ is $x\mapsto x^q$ for every open $U\subset B$.) Let $S=\sum a_P P$ be an effective divisor on $B$. Is it true that $F^* \Omega_{X/k}(\log(S))=\Omega_{X/k}(\log(F^* S))$ (rather then -$F^* \Omega_{X/k}(\log(S))=\Omega_{X/k}(\log(S))$)? (Here $F^* S=\sum a_P^q P$.) -My interest in these matters arise from an attempt to understand the paper "Purely inseparable points on curves of higher genus" http://www.mathjournals.org/mrl/1997-004-005/1997-004-005-004.pdf -I find the results in this paper very interesting. For example they have been applied towards full Mordell-Lang in positive characteristic, and I have other applications in mind. -But I am very puzzled. In the proof of the Theorem in this paper, I do not understand why $\Omega_B((F^n)^{-1} S)=\Omega_B(S)$ and why $\deg(P^*\omega)$ should be bounded. Also I do not see a reason why the separable map $g$ occurring later in the proof should be non-constant. -Remark: In this paper the ground field $k$ seems to be an issue that should be discussed a bit. The paper starts with "Let $k$ be a field of characteristic $p>0$". I think the Corollary is false as it stands, at least in the case where $k$ is algebraically closed with $trdeg(k/{\mathbb F}_p)\ge 1$. (In that case, if $K=k(X)$ and $C/K$ is a smooth projective curve of genus $\ge 2$ which is defined over $k$ will have $|C(k)|=\infty$, -$|C(K)|=\infty$ and $|C(K^{\frac{1}{p^\infty}})|=\infty$, but of course $C$ needs not be birational to a curve defined over a finite field.) Also there seem to be counterexamples to the Theorem itself, if $k$ is a general field of positive characteristic. These counterexamples disappear, if we assume $k$ finitely generated over its finite prime field, and I think the whole paper was meant to address that case. So at the moment I try to understand everything in the case where $k$ is finitely generated of positive characteristic. But until now I failed even in the easiest case where $k$ is finite ... -EDIT: Maybe I should state here the Claim, whose proof I finally want to understand. -Claim: Let $k$ be a finitely generated field of positive characteristic. Let $K/k$ be a finitely generated field extension with $trdeg(K/k)=1$. Let $C/K$ be a smooth projective curve over $k$ of genus $\ge 2$. Assume that $C$ is not birational over $\overline{K}$ to a curve which is defined over a finite field. Then $C(K^{\frac{1}{p^\infty}})$ is finite. - -REPLY [4 votes]: For a reference on logaritmic sheaves, see Esnault-Viehweg "Lectures on vanishing theorems", in particular the first six Chapters. -The Frobenius morphism is also considered in the same book, in Chapters 8 and 9. -But in order to answer your question, no deep results are needed. In fact, over a curve $B$ the sheaf $\Omega_B(\log S)$ is invertible and isomorphic to $\Omega_B(S)$. This because, in general, there is a short exact sequence -$0 \to T_B(-S) \to T_B(- \log S) \to T_S \to 0$, -and $T_S$ is zero for a zero-dimensional scheme. -On the other hand, by [Hartshorne, II prop. 6.9] for any divisor $D$ on $B$ we have -$\deg(F^*D) = q \cdot \deg(D)$. -Then -$\deg (F^*\Omega_B(\log S))=q \cdot \deg(K_B)+q \cdot \deg(S)$, -$\deg (\Omega_B(F^*\log S))=\deg(K_B)+q \cdot \deg(S)$, -$\deg (\Omega_B(\log S))=\deg(K_B)+\deg(S)$, -hence the invertible sheaf $F^*\Omega_B(\log S)$ is in general different from both $\Omega_B(F^*\log S)$ and $\Omega_B(\log S)$. -The fact that $\Omega_B((F^n)^{-1}S))=\Omega_B(S)$ comes simply from the fact that $F$ is ramified everywhere with ramification index $q$, so the set-theoretic preimage of $S$ is nothing but $S$.<|endoftext|> -TITLE: Hanging a ball with string -QUESTION [24 upvotes]: What is the shortest length of string that suffices to hang - a unit-radius ball $B$? - -This question is related to an earlier MO question, but I think different. - - -Assume that the ball is frictionless. -Perhaps $B$ is a billiard ball, and the string is nylon thread. - - -Start under the restriction that you are not permitted to cut -the string. - - -But you may tie knots, whose total length $\epsilon$ is negligible -(or we can take the infimum of all lengths). - - -Let $h$ be the distance from the topmost hanging point to the north -pole of $B$. - - -Here is a possible solution of length $L = 3\pi + h$. - - - -The green portion is $h$. -The blue forms one loop from north pole $N$ through south pole $S$ -and back up to $N$, at which point it is tied and then -descends again to tie at $S$. I am a bit uncertain if this $120^\circ$ partitioning could -be maintained without friction. -Just one great circle of length $2\pi$ would leave hemispheres exposed, allowing the -ball to fall out under the slightest perturbation. -Variations are obtained by altering the assumptions above. -Suppose there is friction $\mu$. Perhaps $B$ is a tennis ball, -and the string is twine. -Maybe then a type of spiral shorter than $3\pi$ would work? -Allowing cutting of the string may help. -Maybe then one could fashion a bird's nest into which $B$ -nestles, achieving a length closer to $L = 2\pi + h$? -Tying a knot above the north pole of $B$ could conceivably help, -in which case the length of $h$ might play a role. -Any ideas would be welcomed, including sharpenings of the problem specification. -I am especially interested to hear of a provably optimal solution -under any variation. -Addendum (28Oct10). Here is a depiction of Scott's suggestion in the comments: - - -Addendum (10Feb11). Martin Demaine at MIT contacted me to inform me that this -question was asked and answered long ago: H.T. Croft wrote a paper, "A Net to Hold a Sphere," -J. London Math. Soc., 39, (1964) pp.1-4. (PDF here.) He credits the problem and solution -to A.S. Besicovitch in a paper from 1957, same title, Math. Gaz. XLI, pp. 106-7. Here is Croft's first sentence: - -Besicovitch [1] has shown: if a net of inextensible string encloses a sphere of unit radius - in such a way that the sphere cannot slip out, then the length of the string is strictly greater - than $3 \pi$, and this is false with any greater constant replacing $3 \pi$. - -This result accords with the answers below, by Scott and drvitek. - -REPLY [7 votes]: There is another stable solution of total length $3\pi + h + \epsilon$ for any $\epsilon > 0$. We take a circle of constant latitude $\delta < 0$ (sufficiently small) and then connect this circle to the north pole via two diametrically opposed strings. This is then clearly stable. Furthermore this is equivalent to your solution (plus Scott Carnahan's epsilon-modification). -However, there is in fact a stable solution of total string-length $2\pi+h+\epsilon$. We simply take the almost-equatorial circle in the last solution and drag it to the south pole, so that we have a small circle there along with two diametrically opposed support strings connecting it to the north pole. This solution is (barely) stable, although physically it is not exactly easy to implement (a very small but non-negligible disturbance will suffice to remove the ball). -The reason for the two answers: this was my thought process in action. -EDIT: Please disregard the above; both of these solutions are unstable. -In fact Scott's example is part of a general class of solutions of total length $3\pi+h+\epsilon.$ Take any two diametrically opposite points, and draw a small spherical triangle containing one of the points. Connect all six possible edges between the four points by geodesics, and finally rotate the arrangement so that one of the strings passes through the north pole. -Here is a short proof that any stable configuration must have at least four points where three or more strings meet. If there are three or less points, there is a hemisphere $H$ which contains all of the points. Take the complement $H^C$ of this hemisphere; we can remove any strings in $H^C$ because they cannot be geodesics. As $H^C$ doesn't contain the north pole, the sphere can fall out. -Note that if any strings are not geodesics between junctions, we can ignore the strings. -Here is a short proof (that is not quite rigorous) that could provide a lower bound. Suppose there is not a loop (that is, a set of points connected by geodesics) that is completely contained in the southern hemisphere. Then we may drag any points in the southern hemisphere into the northern hemisphere. (This statement is the part I can't make completely rigorous.) So the sphere must be unstable. Now if we have a loop in the southern hemisphere, there must be at least two strings meeting at the north pole, as otherwise we could simply slide all of the string off one side of the sphere. -So we must have a loop in the southern hemisphere and (not necessarily direct) connections from at least two of these points to the north pole. I can't figure out how to work a good lower bound from here.<|endoftext|> -TITLE: When is $G$ isomorphic to $G \times G$? -QUESTION [16 upvotes]: Is there a finitely generated nontrivial group $G$ such that $G \cong G \times G$? -Here are some properties which such a group $G$ has to satisfy: - -$G$ is not abelian (otherwise $G$ is a noetherian $\mathbb{Z}$-module, and the composition of the first projection $G \times G \to G$ with an isomorphism $G \cong G \times G$ will be bijective, i.e. $G$ is trivial). -$G$ is perfect (apply the first observation to $G/G'$) - -REPLY [14 votes]: As a geometric group theorist, one would of course relax the question by allowing passage to finite index subgroup, i.e. one would ask for groups such that G and GxG have finite index subgroups, which are isomorphic. One then calls G an GxG commensurable, and for this weaker property there are a lot of interesting examples. My favourite one right now is the Grigorchuk group. But even commensurability to GxG is a very restrictive property: It implies, for instance, that if G is infinite, then it has infinite asymptotic dimension. I just stumbled over this result in the thesis of J. Smith. The proof is almost trivial: Since G is coarsely equivalent to GxG, it is coarsely equivalent to G^n for all n. Now Z embeds -quasi-isometrically into G (since G is infinite), and hence Z^n embeds coarsely into G^n (hence G), so asdim G is at least asdim Z^n for all n, and we conclude. This is in particular the case if G and GxG are isomorphic. The upshot is, that for a group of finite asymptotic dimension one cannot have G=GxG, not even up to finite index.<|endoftext|> -TITLE: When is it possible to construct a joint law from its two-dimensional marginals? -QUESTION [6 upvotes]: My question is much more specific than the title: - -Given a symmetric - distribution $\Xi$ on $\mathbb R^2$, when is it possible to construct a sequence $\xi_1,\xi_2,\dots$ of random variables such that the joint distribution of any two of them is $\Xi$? - -For example, if the pdf of $\Xi$ is decomposable: $p_\Xi(x,y) = p(x) p(y)$, then one can just take a sequence of independent r.v.'s. -(To construct certain counterexample related to fractional Brownian motion) I am particularly interested in the pdf $p_\Xi(x,y) = \frac{(a+1)(a+2)}2 |x-y|^{a}1_{[0,1]}(x)1_{[0,1]}(y)$, $a\in(-1,0)$. - -REPLY [10 votes]: I recently came upon this question in the context of distributions taking values in a finite set, but since yours take values in the compact interval $[0,1]$ I don't think much will go wrong applying the answer to your setting. -Certainly a sufficient condition is that you can construct an exchangeable sequence $\chi_1,\chi_2,\ldots$ for which the marginal of $\chi_1$ and $\chi_2$ is $\Xi$, or equivalently that $\Xi$ is a mixture of i.i.d. distributions per de Finetti's theorem. It turns out this is also necessary. -To see this suppose that $\Xi$ satisfies the condition you give on its two-variable marginals. Then for any finite $n$ there is a finite sequence of random variables $\xi_1,\ldots,\xi_n$ (just take the first $n$ variables of your given sequence) all of whose two-variable marginals are $\Xi$. We can construct a new sequence of random variables $\chi_1,\ldots,\chi_n$ by randomly permuting the $\xi_1,\ldots,\xi_n$. By linearity the marginal of any two of these will still be $\Xi$. -But the distribution of the $\chi_1,\ldots,\chi_n$ is invariant under arbitrary permutations, by symmetry. Using a diagonalization and compactness argument, we get from the existence of such a sequence $\chi_1,\ldots,\chi_n$ for all finite $n$ the existence of an exchangeable sequence $\chi_1,\chi_2,\ldots$ whose two-variable marginal distribution is $\Xi$.<|endoftext|> -TITLE: Extremely messy proofs -QUESTION [142 upvotes]: Currently in my undergraduate courses I am being taught how to set up various machinery using slick, short proofs and then how to apply that machinery. What I am not being taught, largely, is what came before these slick, short proofs. What did mathematicians do before so-and-so proved such-and-such lemma? Where, in other words, are the tedious, long proofs that we can look to as examples of the horrible mess we are escaping? What insights helped mathematicians escape those messes? -Right now I am particularly interested in examples from measure theory. What did people do before, for example, Dynkin's lemma or Caratheodory's extension theorem? Or were these tools available from near the start? -An answer should include both some indication of how tedious and long the old approach was and how much slicker and shorter the modern approach is. Ideally, it should also discuss how the transition between the two happened. -(If you prefer the old approach to the modern approach, for example for pedagogical reasons, that would also be interesting to hear about.) - -REPLY [7 votes]: The Quadratic Equation formula: Al-Khawarizmi (c 800 AD) did not have negative numbers, nor zeros, and also did not possess the needed algebraic notation. Therefore he had to devote 6 chapters of his treatise on algebra ("Hisab al-jabr w'al-muqabala") to different types of quadratics, and the rules for solving each one of them. -See "Quadratic, cubic and quartic equations", MacTutor History of Mathematics archive.<|endoftext|> -TITLE: Properties of a non-sofic group -QUESTION [23 upvotes]: This question is, essentially, a comment of Mark Sapir. I think it deserves to be a question. -A countable, discrete group $\Gamma$ is sofic if for every $\epsilon>0$ and finite subset $F$ of $\Gamma$ there exists an $(\epsilon,F)$-almost action of $\Gamma$. See, for example Theorem 3.5 of the nice survey of Pestov http://arxiv.org/PS_cache/arxiv/pdf/0804/0804.3968v8.pdf. -Gromov asked whether all countable discrete groups are sofic. It is now widely believed that there should be a counterexample to this. -Since most groups are sofic, it would be useful to have a collection of properties that would imply that a group is not sofic...so one can then construct a beast having such properties. - -What are some abstract properties of $\Gamma$ that would imply $\Gamma$ is not sofic? - -An open question of Nate Brown asks whether all one-relator groups are sofic. I'd be interested to know what properties of a one-relator group $\Gamma$ would imply that $\Gamma$ is not sofic. - -REPLY [5 votes]: Sofic groups fulfill determinant conjecture. -This implies in particular that there exists a natural constant $c$ such that given a matrix $M$ over the integral group ring of a given sofic group $G$, we have that -$$ -|tr_{vN} \exp(-cM) - \dim_{vN}\ker M| < \frac{1}{3}. -$$ -This can be used to show that some problems about the group are decidable. Suppose a group $G$ is torsion-free, has decidable word problem, fulfills Atiyah conjecture, and is sofic. Then there is an algortihm which decides whether a given matrix $M$ over the integral group ring has non-trivial kernel, as an operator on $[l^2(G)]^{\dim M}$. -Indeed, given $M$ it's easy to bound its $l^2$ norm and based on this to decide how many terms in $tr_{vN}\exp(-cM)$ have to be computed in order to be less than $\frac{1}{6}$ from the actual value of $tr_{vN} \exp(-cM)$. Call this approximation $a$ (it can be computed since the word problem is decidable). Now, because $G$ is torsion free and fulfills Atiyah conjecture, we know that $\dim_{vN}\ker M$ is an integer, and it's equal to $0$ iff $M$ has trivial kernel. So $M$ kas trivial kernel if and only if $a<\frac{1}{2}$ -Similar algorithm works if a group has bounded torsion, since $\frac{1}{3}$ in the first equation can be exchanged with any postivie real number. I seem to have read that there exist Tarski monsters with decidable word problem. That means that in principle :-) one could try to show that there's no such algorithm for these Tarski monsters and arrive at the conclusion that either these monsters are non-sofic or they don't fulfill Atiyah conjecture.<|endoftext|> -TITLE: Algebraic topology for nonlinear compact operators -QUESTION [7 upvotes]: There are analogues of certain basic notions in algebraic topology in the theory of Banach spaces. For example, the Brouwer fixed point theorem generalizes to the Schauder fixed point theorem, and the idea of the degree of a map generalizes to the Leray-Schauder degree. In both cases, you must restrict yourself to considering (nonlinear) compact operators: operators that take bounded sets to relatively compact sets. -How far can this analogy be extended? How much of algebraic topology can be made to work in this setting? Suppose we consider arbitrary bounded subsets (not necessarily convex) of a Banach space as our class of spaces, and then compact operators between those. (If that's the wrong choice, feel free to correct it.) Can you define some sort of infinite-dimensional homology group so that the degree is an element of it? is there an analogue of the Lefschetz fixed point theorem? Or does the analogy break down if you try to go beyond defining a degree? - -REPLY [3 votes]: Critical groups capture more information than Leray Schauder degree. -Instead of a number, critical groups are a sequence of homological groups, which depends on local behaviour in the variational setting. -For more details, see the book infinite dimensional Morse theory by Kung Ching Chang.<|endoftext|> -TITLE: Algorithm to find an acyclic matching on a poset -QUESTION [5 upvotes]: Are there algorithms / theorems to find an acyclic matching on the Hasse diagram of a poset. -I am particularly interested in the face poset of a regular CW complex. -Also, how to decide if the given acyclic matching is perfect (impossible to add one more vertex). -Is there some structure on the set of all acyclic matchings ? - -REPLY [2 votes]: Let me mention the random approach by Bruno Benedetti and Frank Lutz: http://arxiv.org/abs/1303.6422<|endoftext|> -TITLE: P-adic L functions -QUESTION [22 upvotes]: My question is how should one think of p-adic L functions? I know they have been constructed classically by interpolating values of complex L-functions. Recently I have seen people think about them in terms of Euler systems. But we know only a few Euler systems and there are lot of p-adic L functions. -In case of elliptic curves(at least over $\mathbb{Q}$) complex L-functions give information about the Galois representations. Should the p-adic L-function give some information about some p-adic Galois representation? It seems to be the case in case of cyclotomic fields where we think of the cyclotomic character as a 1-dimensional representation. -I apologize in advance if my questions are vague. I am just starting to learn about the subject. - -REPLY [25 votes]: There are three way to obtain $p$-adic L-functions. The big dream is that one can do all of them for a large class of $p$-adic Galois representations $V$. To study them one starts best to look at the cases $\mathbb{Q}(1)$ for the classical Kubota-Leopoldt $p$-adic $L$-functions or the Tate-module of an elliptic curve etc. -Let $K_{\infty}=\mathbb{Q}(\mu_{p^{\infty}})$ be the union of all cyclotomic fields of roots of unity of $p$-power order. Let $G$ be its Galois group, which is isomorphic to $\mathbb{Z}_p^{\times}$. - -Attached to $V$ there is a complex $L$-function and there are conjectures saying that certain values are algebraic and satisfy to certain congruences modulo powers of $p$, e.g. Kummer congruences. So in some cases, one can show the algebraicity and the congruences. So the values fit together to a $p$-adic analytic function. But the better way of presenting the $p$-adic $L$-function is by constructing a measure on the Galois group $G$ with values in $\mathbb{C}_p$. One can then evaluate the $p$-adic $L$-function on characters of the group $G$. This way the $p$-adic $L$-function resembles a lot its complex counterpart as they are described in Tate's thesis. See Lang's Cyclotomic Fields or Washington or Mazur-Tate-Teitelbaum for instance. -On the algebraic side, we have a Selmer group or a class group that we watch growing in the tower $K_{\infty}/\mathbb{Q}$. The characteristic series of the dual of this Selmer group as a $\Lambda$-module is a sort of a generating function for this growth. Like zeta-functions for varieties over finite fields. These characteristic series are in fact power-series, but they are defined up to a unit (as they are generators of some ideal). Greenberg's paper give a good introduction to this side. -The Euler system (if we are lucky to be in one of the few cases where we have one) is a system of norm-compatible cohomology classes. In particular they give an element in $H^1(K_n, V)$ for each intermediate field $K_n$. But there should be an element over sufficiently many abelian extensions. The norm-compatibility is involves a factor that looks like an Euler factor of the complex $L$-function. There is a general map, called the Coleman map or the logarithme élargi or whatever, from the inverse limit of the $H^1(K_{n,p}, V)$ to a ring of power-series. The image of the Euler system under this map should be the analytically defined $p$-adic $L$-function. Typically one shows that they satisfy the same interpolation property. - -In some sense the Euler system is the bridge between the analytic and the algebraic world. Under the Coleman map it links to the analytic side. In the other direction, one can form derivative classes out of the cohomology classes. These derived classes can be analysed locally and they can be used to bound the Selmer group and hence the characteristic series. That is how one can prove the main conjecture in some cases in one direction. Probably a good place to start is Coates-Sujatha. -The $p$-adic $L$-function of an elliptic curve is conjectured to satisfy a $p$-adic Birch and Swinnerton-Dyer formula. (Mazur-Tate-Teitelbaum and Bernardi-Perrin-Riou in the supersingular case). On the algebraic side instead, we know almost that the characteristic series satisfies this formula. The order of vanishing is known to be at least as large as the rank and if they agree then the leading term has the desired shape involving the Tate-Shafarevich group; of course only up to a $p$-adic unit. -In the geometric case, say an elliptic curve over a function field $K$ of a curve over a finite field $k$, the complex and the $p$-adic function are the same ($p\neq\text{char}(k)$), since they are both just a polynomial with integer coefficients. Tate's Bourbaki talk on BSD shows how one can use the tower $K_{\infty} = \bar{k} \cdot K$ to prove a good deal about BSD. Iwasawa theory tries to mimic this. -So I believe that $p$-adic $L$-functions are just as nice and interesting as their complex counterparts. Even if they seem more mysterious and the definition is less straight forward, we sometimes know more about them. Now I stop otherwise I am going to write a book about it here...<|endoftext|> -TITLE: Online introduction to Lattice Theory? -QUESTION [12 upvotes]: Apart from J. B Nation's (revised) Notes on Lattice Theory, is there any other (mostly introductory) material on Lattices available online? - -REPLY [2 votes]: Old thread, but who knows this might be useful to someone. A video lecture on Lattices: -https://www.youtube.com/watch?v=qPtGlrb_sXg<|endoftext|> -TITLE: What are some open problems in toric varieties? -QUESTION [17 upvotes]: In light of the nice responses to this question, I wonder what are some open problems in -the area of toric geometry? In particular, - -What are some open problems relating to the algebraic combinatorics of toric varieties? - -and - -What are some open problems relating to the algebraic geometry of toric varieties? - -REPLY [12 votes]: My favourite is Oda's Strong Factorization Conjecture: - -Can a proper, birational map between smooth toric varieties be factored as a composition of a sequence of smooth toric blow-ups followed by a sequence smooth toric blow-downs? - -Note that if you are allowed to intermingle the blow-ups and blow-downs (the weak version) it has been proved. In fact, it was proved for general varieties in characteristic 0 using the toric case: -Torification and Factorization of Birational Maps. Abramovich, Karu, Matsuki, Wlodarczyk. -A conjectural algorithm for computing toric strong factorizations can be found in the following arXiv article: -On Oda's Strong Factorization Conjecture. Da Silva, Karu.<|endoftext|> -TITLE: Growth of Coefficients of cusp forms -QUESTION [6 upvotes]: Hi -I am curious about the growth of coefficients of cusp forms. I am aware of Ramanujan-Petersson conjecture/theorem in general terms but I was hoping for a more detailed description of precise statements on the growth of all the cusp form coefficients and whether these were tight. Also, for the Fourier coefficients of Hasse-Weil L-series of elliptic curves (meaning what can we say about these without making use to modularity). Any good references would be very useful also. Thanks! - -REPLY [10 votes]: Let me focus on how to deduce a bound for all coefficients $a_n$ assuming the Deligne bound on $a_p$ (this is a standard argument). -Let $f$ be a newform of weight $k$, level $N$ and Nebentypus character $\psi$ modulo $N$. Assume the Deligne bound $|a_p| \leq 2p^{(k-1)/2}$, and let us prove the inequality $|a_n| \leq d(n) n^{(k-1)/2}$ for every $n \geq 1$ (here $d(n)$ is the number of positive divisors of $n$). -Since both sides are multiplicative in $n$, it suffices to consider the case $n=p^m$, where $p$ is prime. Put $u_m=a_{p^m}$. By looking at the Euler factor of $f$ at $p$, we know the following formal identity : -$\sum_{m \geq 0} u_m X^m = \frac{1}{1-a_p X +\psi(p) p^{k-1} X^2}$. -Now there are two main cases : -1) $p$ divides $N$. Then $\psi(p)=0$ and $u_m = a_p^m$ for every $m \geq 0$. It can be shown, using purely analytical arguments, that $|a_p| \leq p^{(k-1)/2}$ (see Theorem 3 in Li's article "Newforms and functional equations", Math. Ann., 1975). Thus we get $|u_m| \leq p^{m(k-1)/2}$ which gives the desired inequality (and in fact, a stronger one). -2) $p$ doesn't divide $N$. Put $1-a_p X +\psi(p) p^{k-1} X^2 = (1-\alpha X)(1-\beta X)$. In the case $k \geq 2$, the Weil conjectures proved by Deligne imply that $|\alpha|=|\beta|=p^{(k-1)/2}$ (this still holds if $k=1$, by a theorem of Deligne and Serre). There are two subcases : -2a) $\alpha \neq \beta$. Solving the linear recurrence relation satisfied by $u_m$ gives -$u_m = \frac{\alpha^{m+1}-\beta^{m+1}}{\alpha-\beta} = \alpha^m+\alpha^{m-1} \beta + \ldots + \beta^m$. -In particular $|u_m| \leq (m+1) p^{m(k-1)/2}$ which is what we want. -2b) $\alpha = \beta = \frac{a_p}{2}$. In this case $u_m = (m+1) \alpha^m$ for every $m \geq 0$, and the inequality also follows. -Remarks : It is conjectured that case 2b never happens if $k \geq 2$ (semi-simplicity of crystalline Frobenius). Note also that in case 2a, we can write $u_m = \beta^m (1+\lambda+ \ldots + \lambda^m)$, where $\lambda := \alpha/\beta$ satisfies $|\lambda|=1$ and $\lambda \neq 1$. Thus in fact we get a bound of the form $|u_m| \leq C \cdot p^{m(k-1)/2}$, where the constant $C$ depends only on the argument of $\alpha/\beta$. Finally, note that we also have the strict bound $|a_p|<2p^{(k-1)/2}$ in case 2a. -EDIT : case 2b can indeed happen in the case $k=1$ (thanks to David for pointing this out).<|endoftext|> -TITLE: Nonseparable example in dimension theory? -QUESTION [17 upvotes]: Could you give me an example of a complete metric space with covering dimension $> n$ all of which closed separable subsets have covering dimension $\le n$? - -The question closely related to this one. - -REPLY [5 votes]: There are completely metrizable spaces $X$ with $\dim(X)=1$ and $\mathrm{ind}(X)=0$, where $\dim$ denotes the covering dimension and $\mathrm{ind}$ the small inductive dimension. The first such example is due to P. Roy (see "nonequality of dimensions for metric spaces", TAMS 1968), but there are many others. My favourite is due to E.K. van Douwen (see "The small inductive dimension can be raised by the adjunction of a single point", Indag.Math. 1973). -Note that if $Y$ is a separable (not necessarily closed) subspace of such an $X$ then $\dim(Y)=0$ since $\dim$ and $\mathrm{ind}$ coincide for separable metrizable spaces. So any such $X$ satisfies what you want for $n=0$.<|endoftext|> -TITLE: Who is Kirszbraun? -QUESTION [17 upvotes]: Kirszbraun's theorem is one of my favorite theorems in mathematics. -I always wanted to know something about Kirszbraun, or at least to see his picture. -Do you have any information about him? -(I know only trivial things, like where he was publishing and that he was working in Warsaw.) -P.S. Thanks to Lukasz Grabowski, I fould the following entry about Kirszbraun in Polish Biographical Dictionary. - -REPLY [7 votes]: It is very, VERY likely that Kirszbraun published only one paper. Namely, if you read carefully that short entry in Polish (written by E.Marczewski = Szpilrajn), you'll notice that in the penultimate sentence it says that upon completing his studies he got a job as an actuary ("aktuariusz", the person who calculates insurance costs) in an insurance company named "Przyszlosc"; therefore, it is quite reasonable to assume that he didn't continue any form of a(n academic) career as a research mathematician. On the other hand, a footnote an the title page of its Fundamenta paper clearly indicates that the paper is an abridged and improved version of his "Magister" (= Master) thesis, defended back in 1930 and prepared in the preceding 4-year period. So, it is rather safe to jump to the conclusion that Über die zusammenziehenden und Lipschitzschen Transformationen is his only publication, and that he didn't pursued any further mathematical research.<|endoftext|> -TITLE: Examples where the analogy between number theory and geometry fails -QUESTION [25 upvotes]: The analogy between $O_K$ ($K$ a number field) and affine curves over a field has been very fruitful. It also knows many variations: the field over which the curve is defined may have positive or zero characterstic; it may be algebraically closed or not; it may be viewed locally (by various notions of "locally"); it may be viewed through the "field with one element" (if I understand that program) and so forth. -Often when I've dealt with this analogy the case is that the geometric analog of a question is easier to deal with than the arithmetic one, and is strongly suggestive for the veracity of the arithmetic statement. -My question is: what are some examples of where this analogy fails? For example, when something holds in the geometric case, and it is tempting to conjecture it's true in the arithmetic case, but it turns out to be false. If you can attach an opinion for as to why the analogy doesn't go through in your example that would be extra nice, but not necessary. - -REPLY [2 votes]: If $A/K$ is an abelian variety, $v$ is a place of $K$, $h$ is the global height function and $\lambda_v$ is the local height function at $v$, then comparing $h(P)$ and $\lambda_v(P)$ for $P \in A(K)$ varies a lot depending on the situation. $\lambda_v(P) = O(1)$ if $K$ is a function field of characteristic zero, $\lambda_v(P) = O(h(P)^{1/2})$ (usually) in positive characteristic and this cannot be improved, and $\lambda_v(P) = O(\log(h(P)))$ conjecturally for number fields (and is definitely not $O(1)$).<|endoftext|> -TITLE: Is there a 7-regular graph on 50 vertices with girth 5? What about 57-regular on 3250 vertices? -QUESTION [23 upvotes]: The following problem is homework of a sort -- but homework I can't do! -The following problem is in Problem 1.F in Van Lint and Wilson: - -Let $G$ be a graph where every vertex - has degree $d$. Suppose that $G$ has - no loops, multiple edges, $3$-cycles - or $4$-cycles. Then $G$ has at least - $d^2+1$ vertices. When can equality - occur? - -I assigned the lower bound early on in my graph theory course. Solutions for $d=2$ and $d=3$ are easy to find. Then, last week, when I covered eigenvalue methods, I had people use them to show that there were no solutions for $d=4$, $5$, $6$, $8$, $9$ or $10$. (Problem 2 here.) I can go beyond this and show that the only possible values are $d \in \{ 2,3,7,57 \}$, and I wrote this up in a handout for my students. -Does anyone know if the last two exist? I'd like to tell my class the complete story. - -REPLY [12 votes]: Additional random facts. -The Peterson Graph can be obtained by identifying the antipodal points of a dodecahedron and it has $S_5$ as its automorphism group (order 120 of course). -There are a number of geometric constructions of the Hoffman-Singleton Graph (the 25 points and 25 non-vertical lines of an affine plane over $Z_5$ are used in one , the 15 points and 35 lines of projective 3-space over $Z_2$ in another). The automorphism group has order 252000. -A Moore graph of degree 57, if it exists, would have a trivial automorphism group: would have to have a small automorphism group. -edit See the comment below from Chris Godsil -Aschbacher, M. "The Non-Existence of Rank Three Permutation Group of Degree 3250 and Subdegree 57." J. Algebra 19, 538-540, 1971. -Here is a good reference from 2010: Search for properties of the missing Moore graph which shows among other things that if such a graph exists then it has automorphism group of order at most 375. -later Since we have new interest I'll add some beautiful well known facts. - -The triangle graph $T_5$ is the line graph of $K_5$ and is regular of degree 6 with 10 vertices. So $S_5$ acts on it and that is the full automorphism group. As mentioned by N. Elkies, the Peterson graph is the complement of $T_5$. $T_5$ has five maximal cliques $K_4$ corresponding to the 5 vertices. These become the five totally disconnected 4-vertex induced sub-graphs (independent sets) mentioned by R. Bell. If we fix one such independent 4-set, connect one new vertex with each of the six pairs and then connect each of these to the one pair disjoint from it, we get the Peterson Graph. So this is the points and edges of a tetrahedron. -In a Moore graph of order 7 the largest independent sets have 15 vertices. The incidence between these and the other 35 is the same as that between the points and blocks of a certain resolvable Steiner triple system and (equivalently) that between the 15 points and 35 lines of PG(3,2). These descriptions leave some edges unspecified, but: -Consider the 35 triples from $\{a,b,c,d,e,f,g\}$ as labels for 35 vertices and connect each to the four with labels disjoint from its own. There are 30 heptads being choices of seven triples no two disjoint (so forming a Fano plane). $S_7$ is transitive on these but $A_7$ has two orbits of size 15. If we use one such orbit to label 15 more vertices and make the obvious connections, we get the Moore graph of order 7. -The Peterson graph has a nice description in terms of the 4 points and 6 edges of a tetrahedron or PG(3,1) if we abuse notation. The Moore graph of order 7 has a nice description in terms of the 15 points and 35 lines of PG(3,2). Now, PG(3,7) has 400 points and 2850 lines and if there is a Moore graph of order 57 (warning! warning! Many would conjecture that there is none!) then it has 400+2850 vertices of which at most 400 could be independent... The fact that a large automorphism group has been ruled out makes this an unpromising approach, but who knows?<|endoftext|> -TITLE: When are infinitely many points in the orbit of a polynomial integers? -QUESTION [6 upvotes]: This question is inspired by a riddle in math.stackexchange. -Let $P$ be a polynomial, and $O = \{P^{(n)}(0) : n \geq 0\}$ be its orbit under zero (viewed as a set). Suppose that $O$ contains infinitely many integers. Is it true that for some $n$, $P^{(n)}$ is a polynomial with integral coefficients? -We can ask the same question replacing integers with rationals. -EDIT: Nick and David gave simple counterexamples for the first question. -Still open: - -In the setting of the original question, is it true that some composition power of $P$ takes integers to integers? -The original question with rationals. - -REPLY [5 votes]: A slight variant that turns out to be non-elemenatry is to replace the polynomial with a rational function. This leads to: -Theorem: Let $R(x)\in\mathbf{Q}(x)$ be a rational function of degree at least 2, let $\alpha\in\mathbf{Q}$ be an initial value, and suppose that the orbit $O_R(\alpha)=\{R^{(n)}(\alpha) : n\ge0\}$ contains infinitely many integers. Then the second iterate $R^{(2)}(x)$ of $R$ is a polynomial. -For specific $R$ there are often easy proofs, but in general the proof seems to require some non-trivial result on Diophantine approximation such as Thue's theorem. There's an exposition of the proof in The Arithmetic of Dynamical Systems (Springer 2007), Section 3.7. See Section 3.8 for a stronger result saying roughly that as $n$ gets large, then the numerator and denominator of $R^{(n)}(\alpha)$ have about the same number of digits. (There's one extra technical condition for this last result.)<|endoftext|> -TITLE: Degree of balls in finitely-generated subgroups of SL_2(C) -QUESTION [8 upvotes]: Let $T$ be a finite symmetric set generating a Zariski dense subset of an algebraic group $G$ (specifically, $PSL_2(\mathbb{C})$ or its subgroups). Is there an $\alpha>0$ such that the set $T^{\leq n}$ of words of length at most $n$ is not in any codimension-1 subvariety of degree $n^{\alpha}$? -"Escape from subvariety" arguments seem to prove similar results that are not polynomial. - -REPLY [4 votes]: In the case of $\text{PSL}(2,\mathbb{C})$, the set of words of length $O(n^3)$ is not contained in a subvariety of degree $n$. A polynomial of degree $n$ is a linear combination of matrix entries of the irreps of highest weight $\le n$. Let $A_n$ be the direct sum of the corresponding matrix algebras; its dimension is a sum of consecutive squares which is then $O(n^3)$. The generating set $S$ yields a set of operators in $A_n$, which then yields an algebra filtration of $A_n$. Since $S$ generates a Zariski dense subgroup, some term of this algebra filtration is eventually all of $A_n$. On the other hand, the filtration is generated by the term of degree 1, i.e., one can write -$$A_n^{(k+1)} = A_n^{(k)}A_n^{(1)},$$ -taking all linear combinations of all products of pairs on the right side. So the dimensions of the terms of the filtration have to keep going up by at least one until the filtration terminates, which then gives you the $O(n^3)$ bound. -There is a similar argument for any linear algebraic group, except that the algebra $A_n$ is different for each one. -This is a really interesting question. Although this argument does give you the bound that you wanted, I have the feeling that the bound is not optimal.<|endoftext|> -TITLE: Folner sequences of amenable groups of exponential growth -QUESTION [15 upvotes]: Let $G$ be an amenable group of exponential growth and let $S$ be a finite symmetric generating set. For each $k$, let $B_{k}$ be the closed ball of radius $k$ about the identity element in the corresponding Cayley graph of $G$ and let $b_{k} = |B_{k}|$. If $\lim b_{k+1}/b_{k}$ exists, then $\lim b_{k+1}/b_{k} = \lim b_{k}^{1/k} > 1$ and this easily implies that no subsequence of the $B_{k}$ forms a Folner sequence for $G$. But is this also true for those amenable groups of exponential growth for which $\lim b_{k+1}/b_{k}$ does not exist? - -REPLY [2 votes]: I'm fairly sure [from personal communications] this question is (widely) open in general [and a negative answer would come as a surprise to many]. But the only groups I know of where it is known that $\lim \frac{b_{k+1}}{b_k}$ do not exist are the ones described here (as linked by Andreas Thom). Now, for these groups it is clear the answer is yes. -It turns out that, for the generating sets given there, these groups have "pinched" exponential growth. So as a consolation (or perhaps a tiny step of progress to answer the question) here is a partial answer for groups (and generating sets) that have pinched exponential growth. -Remark: for examples of such pairs and as well as groups where one generating set has pinched exponential growth while the other does not, see at the end of this answer. -Notation: Let $B(n)$ be the ball of radius $n$ and $b_n = |B(n)|$. Let $S(n) = B(n) \setminus B(n-1)$ (with $S(0) = \lbrace e_G \rbrace$) be the spheres and $s_n = |S(n)|$. -Definition: Say a pair $(G,S)$ (where $S$ is a finite generating set of the group $G$) has pinched exponential growth if there are constants $K1$. -Let $g = \lim_n b_n^{1/n} = \inf_n b_n^{1/n}$. Now, note there are infinitely many $n_i$ so that, for $n \leq n_i$, $s_{n_i} \geq s_n$ (otherwise $s_n$ is bounded and $b_n$ grows at most linearly). Clearly, $s_{n_i} \leq b_{n_i} \leq (n_i+1) s_{n_i}$, so -$$ -g = \liminf b_{n_i}^{1/n_i} \geq \liminf_i s_{n_i}^{1/n_i}. -$$ -But $b_n \leq \sum_{j=0}^n s_j$ so -$$g = \liminf_i b_{n_i}^{1/n_i} \leq \liminf (n_i+1)^{1/n_i} s_{n_i}^{1/n_i} = \liminf |S(n_i)|^{1/n_i}. -$$ -Thus far we have -$$g = \liminf |S(n_i)|^{1/n_i}.$$ -However, $n \mapsto s_n$ is also sub-multiplicative, i.e. $s_{n+m} \leq s_n s_m$. -Hence -$$g = \lim_n s_n^{1/n} = \inf_n s_n^{1/n}.$$ -The equality with the $\inf$ implies $s_n/g^n \geq 1$. -Next note that pinched exponential growth is equivalent to $\limsup b_n/g^n = L < \infty$ (for some $L$). Putting this together -$$ -\begin{array}{rll} -\displaystyle \liminf \frac{s_{n+1}}{b_n} -& \displaystyle = g \liminf \frac{s_{n+1}}{g^{n+1}} \cdot \frac{g^n}{b_n} \\ -& \displaystyle \geq g \liminf \frac{s_{n+1}}{g^{n+1}} \bigg( \limsup \frac{b_n}{g^n} \bigg)^{-1} -& \displaystyle \geq g /L -\end{array} -$$ -This implies that -$$ -\frac{ |B(n+1)|}{|B(n)|} \geq 1 + \frac{g}{L} -$$ -as desired. -Three remarks on pinched exponential growth: -1- the property depends on the generating set. If $(G,S)$ has pinched exponential growth. Then $G \times \mathbb{Z}$ with "summed" generating set $\lbrace (s,0) \mid s \in S \rbrace \cup \lbrace (e_G,\pm 1) \rbrace$ has also pinched exponential growth. With the "product" generating set $\lbrace (s,\epsilon) \mid s \in S , \epsilon = \pm 1 \rbrace$ it does not have pinched growth. Also, $G \times G$ with the product generating set has pinched exponential growth, but with the summed generating set it does not. -2- with the "usual" generating sets, solvable Baumslag-Solitar and some lamplighters groups have pinched exponential growth (this can be seen on their growth series). For the growth series of $BS(1,n)$ see Collins, Edjvet and Gill, Growth series of the group $\langle x,y \mid x^{-1}y x=y^\ell \rangle$, Arch. Math. 62:1--11, 1994. For the growth series of lamplighters of the form $(\oplus_{i \in \mathbb{Z}} H) \rtimes \mathbb{Z}$) where the growth series of $H$ is known (i.e. $H$ is finite, or Abelian, or $BS(1,n)$, or a lamplighter or ...) see Johnson, Rational growth of wreath products, in "Groups St Andrews 1989" volume 2 309--315. -3- If the growth series is rational then having pinched exponential growth is the same as having exactly one root on the convergence radius. Groups with two roots on their convergence radius are known to exist, but I don't know if there is a group (with rational growth series and) 3 roots on the convergence radius.<|endoftext|> -TITLE: How many minors I need to check to conclude all minors will vanish ? -QUESTION [16 upvotes]: Given a $m \times n$ matrix $n>m$, I was trying to check if all its $m \times m$ minor vanish. -I remember hearing that one really does not need to check all possible minors in order to conclude that all of them would vanish. -If such a result is true, how many minors will do the job and which ones ? -I am wondering if it is even possible to calculate the value of all minors based on the value of a nicely chosen "generating subset" ? - -Edit:- The question which I had asked does not have an affirmative answer as explained -by Steven Sam. But matrix minors do satisfy some relationships see the answer by Sheikraisinrollbank below. If someone can modify the question to a more appropriate one (in light of Steven Sam and Sheikraisinrollbank answers ) please feel free to do so. -I have often come across a situation (more so at present than ever before) where in order to answer a problem in my subject area I am led to questions which are totally different areas about which I have absolutely no familiarity. Most often these are quite basic and I would suppose well known to any one who works in those areas. It is natural that a person who is not familiar with a given field will end up asking for "a result of the following kind" rather than a precise question. For a person who is knowledgeable I understand the question may be irritating or look ill posed but mind you the hapless fellow is not a graduate student in the given field and please do not judge him accordingly. I think its desirable that if someone knows how to reformulate the question to something so that it becomes well posed or meaningful it should be done. Why not edit the question to something so that it becomes a valid well posed question, to something which is obviously much more interesting than which was originally posed ? - -REPLY [12 votes]: To counterbalance Steven Sam's answer some (b/c the OP's intuition is correct in a sense): -It's true that the right way to check that all m by m minors are zero in practice is Gaussian elimination. However, while the minors may be linearly independent, they satisfy quadratic relations ("Plucker relations", see for instance the wikipedia article on Grassmannians) that allow you to deduce some things. In the simplest non-trivial case of 2 by 4 matrices, writing $m_{ij}$ for the $(i,j)$th $2$ by $2$ minor one has $$m_{12}m_{34}-m_{13}m_{24}+m_{14}m_{23}=0.$$ This might have some theoretical value for the OP's situation that Gaussian elimination does not. For instance, in this case it allows one to deduce that if $m_{12}$ and $m_{13}$ are both zero then either $m_{14}$ or $m_{23}$ is zero. But it's hard to know if this helps without knowing somewhat more about the motivating problem.<|endoftext|> -TITLE: Counting roots: multidimensional Sturm's theorem -QUESTION [13 upvotes]: Sturm's theorem gives a way to compute the number of roots of a one-variable polynomial in an interval [a,b]. Is there a generalization to boxes in higher dimensions? Namely, let $P_1,\dotsc,P_n\in \mathbb{R}[X_1,\dotsc,X_n]$ be a collection of $n$ polynomials such that there are only finitely many roots of $P_1=P_2=\dotsb=P_n=0$. I want to be able to compute the number of roots in $[a,b]^n$. I do not care if the roots are counted with or without multiplicity. -I would also be interested in upper bounds on the number of roots that is similar to Descartes' rule of signs. The only work in this connection that I managed to find is by Itenberg and Roy, who postulated a conjectural extension of Descartes' rule of signs, which however later was shown to be false. - -REPLY [2 votes]: The Hermite method for real root counting generalizes to the multivariate case if your system of polynomial inequalities has only a finite number of COMPLEX roots. this was shown by Becker and Wörmann and independently by Pedersen, see: -http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.40.9539<|endoftext|> -TITLE: Are there non-reflexive vector spaces isomorphic to their bi-dual? -QUESTION [32 upvotes]: Let $V$ be an infinite dimensional topological vector space and consider the natural application $\iota\colon V\to V^{**}$. The space $V$ is said to be reflexive if $\iota$ is an isomorphism. -Are there examples where $\iota$ fails to be an isomorphism but $V$ and $V^{**}$ are nevertheless isomorphic? -Can one find an example where $V$ is a Banach space and the isomorphism is actually an isometry? - -REPLY [4 votes]: James' construction can be iterated, in order to produce a countable family of pairwise non-isomorphic Banach spaces with the same property. -Look at the following paper: -Marek Wójtowicz - "Finitely Nonreflexive Banach Spaces" -Proceedings of the American Mathematical Society -Vol. 106, No. 4 (Aug., 1989), pp. 961-965. -EDIT. The family $Z_n$ of Banach spaces constructed by Wójtowicz has the following properties: - -$Z_n$ is isomorphic to $Z_n^{**}$; -$Z_n$ is $n$-reflexive; -if $n < m$ then $Z_n$ is not isomorphic in $Z_m$ and $1$-complemented in $Z_m$.<|endoftext|> -TITLE: Partial orders arising from $l$-spaces -QUESTION [6 upvotes]: Let $X$ be a $l$-space, i.e. a locally compact totally disconnected hausdorff space, which is not compact. Then $P = \{K : K \subseteq X \text{ compact-open}\}$ is a basis for the topology. Regard $P$ as a partial order with respect to "$\subseteq$". -Question: Which partial orders are isomorphic to partial orders which arise from $l$-spaces as above? Note that they have finite infima and suprema and a smallest element, but not a maximal element. But I doubt that this is already the whole characterization. - -REPLY [3 votes]: In - -M. H. Stone. Applications of the Theory of Boolean Rings to General Topology. Transactions of the American Mathematical Society Vol. 41, No. 3 (May, 1937), pp. 375-481 - -Marshall Stone proved (see Theorem 4) that there is a duality between locally compact totally disconnected Hausdorff spaces and Boolean rings (possibly without unit). Boolean rings are in turn equivalent to generalized Boolean algebras. The duality assigns to each space the generalized Boolean algebra of the compact open subsets. So the answer to your question seems to be that such posets are precisely the generalized Boolean algebras.<|endoftext|> -TITLE: Intuition for rational functions -QUESTION [6 upvotes]: I asked this on mathematics stack exchange and did not receive answer . I hope it is good manners to ask here. Thank you very much. -Let $X$ be integral scheme and $\mathcal K$ sheaf of rationnal functions on $X$. For any -point $y\in X$ different of generic point we know that fiber of $\mathcal K$ (defined as usual as $\mathcal K _y / \mathcal m_y \mathcal K_y$) is zero. I'll be very gratefull if you explain intuitively why this is so, in language of restriction of $\mathcal K$ to reduced subscheme $Y=\overline{\{y \} }$. I have difficulty because many rationnal functions on $X$ can be restricted to nonzero rationnal functions on $Y$ . How is that compatible with fiber of $\mathcal K$ equals zero at $y$? - -REPLY [10 votes]: The non-classical aspect of this setup is that you're using a quasi-coherent sheaf that is not coherent, and beyond the coherent case one cannot expect information about a fiber (e.g., vanishing, 6 generators, etc.) to "propogate" to information in a neighborhoood (which would be the spirit behind the choice of word "coherent", I suppose). Computing the fiber of the field of all rational functions at a non-generic point likely has no classical counterpart, much as in number theory one doesn't ever try to reduce $\mathbf{Q}$ modulo 5, only $\mathbf{Z}_{(5)}$ or its subrings.<|endoftext|> -TITLE: How do I approach Optimal Control? -QUESTION [16 upvotes]: Other than learning basic calculus, I don't really have an advanced background. I was curious to learn about Optimal Control (the theory that involves, bang-bang, Potryagin's Maximum Principle etc.) but any article that I start off with, mentions the following: "Consider a control system of the form..." and then goes on to defining partial differential equations. In short, I am lost. -Can someone suggest me a path I should take to learn more about Optimal Control from the very basics? - -REPLY [2 votes]: If you have a background in differential geometry, you will probably like the two books of Jurdjevic, Geometric Control Theory, and Optimal Control and Geometry: Integrable Systems, both from Cambridge Univesity Press.<|endoftext|> -TITLE: Which manifolds are homeomorphic to simplicial complexes? -QUESTION [92 upvotes]: This question is only motivated by curiousity; I don't know a lot about manifold topology. -Suppose $M$ is a compact topological manifold of dimension $n$. I'll assume $n$ is large, say $n\geq 4$. The question is: Does there exist a simplicial complex which is homeomorphic to $M$? -What I think I know is: - -If $M$ has a piecewise linear (PL) structure, then it is triangulable, i.e., homeomorphic to a simplicial complex. -There is a well-developed technology ("Kirby-Siebenmann invariant") which tells you whether or not a topological manifold admits a PL-structure. -There are exotic triangulations of manifolds which don't come from a PL structure. I think the usual example of this is to take a homology sphere $S$ (a manifold with the homology of a sphere, but not maybe not homeomorphic to a sphere), triangulate it, then suspend it a bunch of times. The resulting space $M$ is supposed to be homeomorphic to a sphere (so is a manifold). It visibly comes equipped with a triangulation coming from that of $S$, but has simplices whose link is not homemorphic to a sphere; so this triangulation can't come from a PL structure on $M$. - -This leaves open the possibility that there are topological manifolds which do not admit any PL-structure but are still homeomorphic to some simplicial complex. Is this possible? -In other words, what's the difference (if any) between "triangulable" and "admits a PL structure"? -This wikipedia page on 4-manifolds claims that the E8-manifold is a topological manifold which is not homeomorphic to any simplicial complex; but the only evidence given is the fact that its Kirby-Siebenmann invariant is non trivial, i.e., it doesn't admit a PL structure. - -REPLY [8 votes]: Regarding Charles Rezk's second question: -This leaves open the possibility that there are topological manifolds which do not admit any PL-structure but are still homeomorphic to some simplicial complex. Is this possible? -For dimension 4, it follows from the Poincare conjecture that a 4-manifold is triangulable iff smoothable (which is also equivalent to having PL structure for dimension <8). See Problem 3 of http://www.maths.ed.ac.uk/~aar/haupt/sandro.pdf. Also see this presentation. -For dimension >4, Springer Online Reference Works claims that "the imbedding $PL \subset TRI$ is also irreversible in the same strong sense (there exist polyhedral manifolds of dimension $\geq 5$ that are homotopy inequivalent to any PL-manifold)", but gives no examples. In this presentation it is stated that "All oriented closed 5-manifolds triangulable", so I think among them there may be some with nontrivial KS invariant and hence cannot bear PL structure. -In addition, This book (p.168, Theorem 18.4) seems to contain a result that strengthens the one mentioned in Paul's answer. -Added: This paper (22.5. Example) explicitly gives an example of "A topological manifold which is homeomorphic to a polyhedron but does not admit any PL structure".<|endoftext|> -TITLE: What is the order of a in (Z/nZ)*? -QUESTION [6 upvotes]: I was recently thinking about efficient algorithms for modular exponentiation. This led me to the (more interesting, in my opinion) question: - -Let $1 < a < n$ be an integer relatively prime to $n$. What is the order of ${\overline{a}}$ in $\mathbb{Z}/n\mathbb{Z}^*$ (the multiplicative group of $\mathbb{Z}/n\mathbb{Z}$)? - -I did some Google searching, but all I could find were the obvious facts that the order should divide the order of the group $\phi(n)$ and the exponent of the group $\lambda(n)$ (see Carmichael function). I asked several people if anything more could be said, but the answers were generally: "Some people study this. It is really hard." However, I couldn't find any other references. - -Is this a question that has been seriously considered? If so, what is known and does anyone have any good references? - -I am happy to suppose that we know a priori the prime factorization of both $a$ and $n$. Even given this information, is there something precise that can be said? -Because this is a (potentially) open problem, it is possible that it should be a community wiki page, I am not entirely certain what the policy is there. If so, someone please wiki-hammer this, as I have not the power! It might also be deserving of the open-problem tag? -Edit: I do in fact have the power to make community wiki posts (which I discovered by checking the faq) just not to edit someone else's. Still, I would prefer that this be a "real" question unless that is inappropriate. - -REPLY [7 votes]: You seem to have been given some misinformation so I'll answer this question although I think it is elementary. You want to find the order of $a$ modulo $n$. The prime factorization of $a$ is largely irrelevant, the prime factorization of $n$ is crucial since otherwise you don't know the order of the group. Conversely, knowing the order of $a$ for many $a$'s will allow you to factor $n$. I'll assume you can factor $n$. -If $n$ is prime, then the group is cyclic, so any factor of $n-1$ is the order of some element. There isn't much more that can be said, you can't eyeball the order except in some obvious cases such as $a=\pm 1$. If you know a factorization of $n-1$, then you can run through the divisors of $n-1$ to find the order. If you don't know the factorization of $n-1$ then brute force is basically all you can do. -If $n$ is the power of a prime $p$, then if you can compute the order modulo $p$ (say $d$), it is easy to compute it modulo $n$ by finding the highest power of $p$ dividing $a^d-1$. This is an exercise which most number theory textbooks do when discussing primitive roots modulo prime powers. -In general, you get the order modulo $n$ by factoring $n$, and using the Chinese remainder theorem to reduce to the above cases.<|endoftext|> -TITLE: What's coherent about coherent sheaves? -QUESTION [19 upvotes]: In a recent answer to a recent question, BCnrd wrote - -[...] beyond the coherent case one cannot expect information about a fiber (e.g., vanishing, 6 generators, etc.) to "propogate" to information in a neighborhoood (which would be the spirit behind the choice of word "coherent", I suppose). [...] - -Is that what motivates the adjective coherent? Is this documented somewhere? - -REPLY [18 votes]: Looking at the paper of MALATIAN -"Faisceaux analytiques: étude du faisceau des rélations entre p fonctions holomorphes", -Séminaire Henry Cartan, tome 4 (1951-52), exp. n.15, p. 1-10 -one finds the -Definition 3 - -"On dit qu'un sous-faisceau analytique $\mathcal{F}$ de $\mathcal{O}_E^q$ is $cohérent$ au point $x \in E$, s'il existe un voisinage ouvert $U$ de $x$ et un système fini d'elements $u_i \in \mathcal{O}_U^q$ jouissant de la propriété suivante: pour tout $y \in U$, le sous-module de $\mathcal{O}_U^q$ engendré par les $u_i$ est précisement $\mathcal{F}_y$. -On dit qu'un faisceau $\mathcal{F}$ est cohérent (tout court) s'il est coherent en tout point de $E$." - -And, in the following page: - -"...En d'autre termes, cette condition exprime que le faisceau $induit$ par $\mathcal{F}$ sur l'ouvert $U$ est "engendré" par un sous-module de $\mathcal{O}_U^q$." - -Reading this, it seems that the original definition given by Cartan in its seminar is somehow related to the "coherent behaviour" of $\mathcal{F}$ as a subsheaf of $\mathcal{O}_U^q$, in terms of generation of the stalks. -EDIT. -However, this is not the whole story. Loking at the introduction of the book of Grauert-Remmert, as Brian suggests, it appears that the word "coherent" was actually introduced by Cartan some years before, in the middle of the '40; in fact, he investigated the so-called "coherent systems of punctual modules" when studyng the Cousin's problem. But he does not mention this previous work in his Seminar, when he introduces coherent analytic sheaves. -Grauert-Remmert write that - -"coherence is, in a vague sense, a local principle of analytic continuation". - -And Cartan himself, in its collected works, says - -"En gros on peut dire que, pour en $A$-faisceaux $\mathcal{F}$ cohérent en un point $a$ de $A$, la connaissance du module $\mathcal{F}_a$ détermine les modules $\mathcal{F}_x$ attachées aux points $x$ suffisamment voisins de $a$."<|endoftext|> -TITLE: `Naturally occuring' $K(\pi, n)$ spaces, for $n \geq 2$. -QUESTION [42 upvotes]: [edited!] Given a group $\pi$ and an integer $n>1$, what are examples of Eilenberg-Maclane spaces $K(\pi, n)$ that can be constructed as "known" manifolds? (or if not a manifold, say some space people had a pre-existing desire to study before $K(\pi,n)$ spaces were identified as being of interest) - -Constructing $K({\bf Z}, 2)$ as ${\bf CP}^{\infty}$ is the only example I know - but there must be more out there. -I'm interested in concrete examples (like the one above) that could, e.g., be given in a Topics grad course for topology students. They seem to be scarse, so it would be nice to know what was known. -Note: I've excluded $n=1$ because most people know examples (or can figure them out) in this case. - -REPLY [32 votes]: If $M$ is a hyperfinite type $I\!I\!I_1$ factor, then (at least conjecturally), its group of outer automorphisms is a $K(\mathbb Z,3)$. -This is based on the following three properties of that von Neumann algebra: -• The group of unitary central elements of $M$ is a circle, and thus a $K(\mathbb Z,1)$. -• The group of unitaries in $M$ is contractible. -• The automorphism group of $M$ is contractible (conjectural). -To see that $Out(M)\cong K(\mathbb Z,3)$, apply the long exact sequence of homotopy groups to the following two fiber sequences: -$$ -U(Z(M)) \to U(M) \to Inn(M) -$$ -$$ -Inn(M) \to Aut(M) \to Out(M) -$$ -As a consequence, we also get that $BOut(M)\cong K(\mathbb Z,4)$. -I recommend my talk "A K(ℤ,4) in nature" (MSRI, April 2014), for an explanation of how to realize $Out(M)$ as the automorphism group of a naturally occurring mathemtical object.<|endoftext|> -TITLE: Localizing an arbitrary additive category -QUESTION [11 upvotes]: Under which conditions localizing an additive category by some class S of morphisms yields and additive category? It seems easy to define certain addition on morphisms if we fix their representatives as zig-zags (i.e. compositions of 'old' morphisms with inverses of morphisms in S; here I use the fact that 'my' S is closed with respect to direct sums of morphisms), but I am not sure at all that this addition will not depend on the choice on representatives. Is there any reasonable condition that will ensure this? I definitely do not want to restrict myself to abelian or triangulated categories. -It seems that in the situations I am interested in, any morphism is a composition of the embedding of a direct summand, an inverse of a morphism from S, and an 'old' morphism (i.e. it is 'almost a fraction'). The Ore conditions are not fulfilled (in general, probably); yet some weakening of them could hold. -I would be deeply grateful for any associations here! -My examples are: -For an additive (pseudo-abelian) category B consider some full triangulated (thick) subcategory D of $K^b(B)$; then my S for B is the set of morphisms in B that yield objects of D (if considered as complexes of length 1). -In particular, S is always closed with respect to compositions and direct sums of morphisms. -In fact, I am interested in all aspects of this setup! - -REPLY [3 votes]: Here is the statement of Exercise 8.4 p. 202 of -[KS] Categories and Sheaves by Kashiwara and Schapira: - -(a) Let $\mathcal C$ be an additive category and $\mathcal S$ a right multiplicative system. Prove that the localization $\mathcal C_{\mathcal S}$ is an additive category and $Q:\mathcal C\to\mathcal C_{\mathcal S}$ is an additive functor. - -I'll describe what I think is the argument Kashiwara and Schapira have in mind. -Statement (a) follows from - -(b) there is a pre-additive category on $\mathcal C_{\mathcal S}$ making $Q$ additive, -(c) $Q$ is essentially surjective. - -Statement (b) is easy, and is treated in a very detailed way at the beginning of the following text of Dragan Miličić: -http://www.math.utah.edu/~milicic/Eprints/dercat.pdf -As (b) is obvious, it only remains to explain why (b) and (c) imply (a). The key is Lemma 8.2.3 (ii) p. 169 of [KS], which says: - -(d) Let $\mathcal C$ be a pre-additive category; let $X,X_1,$ and $X_2$ be objects of $\mathcal C$; and, for $a=1,2$, let $X_a\xrightarrow{i_a}X\xrightarrow{p_a}X_a$ be morphisms satisfying - $$ -p_a\circ i_b=\delta_{ab}\ \operatorname{id}_{X_a},\quad i_1\circ p_1+i_2\circ p_2=\operatorname{id}_X. -$$ - Then $X$ is a product of $X_1$ and $X_2$ by $p_1,p_2$ and a coproduct of $X_1$ and $X_2$ by $i_1,i_2$. - -Proof. For any $Y$ in $\mathcal C$ we have -$$ -\operatorname{Hom}_\mathcal C(Y,p_a)\circ\operatorname{Hom}_\mathcal C(Y,i_b)=\delta_{ab}\ \operatorname{id}_{\operatorname{Hom}_\mathcal C(Y,X_a)}, -$$ -$$ -\operatorname{Hom}_{\mathcal C}(Y,i_1)\circ\operatorname{Hom}_{\mathcal C}(Y,p_1)+\operatorname{Hom}_{\mathcal C}(Y,i_2)\circ\operatorname{Hom}_{\mathcal C}(Y,p_2)=\operatorname{id}_{\operatorname{Hom}_{\mathcal C}(Y,X)}. -$$ -This implies that $\operatorname{Hom}_{\mathcal C}(Y,X)$ is a product of $\operatorname{Hom}_{\mathcal C}(Y,X_1)$ and $\operatorname{Hom}_{\mathcal C}(Y,X_2)$ by $\operatorname{Hom}_{\mathcal C}(Y,p_1),\operatorname{Hom}_{\mathcal C}(Y,p_2)$, and thus, $Y$ being arbitrary, that $X$ is a product of $X_1$ and $X_2$ by $p_1,p_2$, and we conclude by applying this observation to the opposite category. q.e.d. -(d) implies: - -(e) Let $F:\mathcal C\to\mathcal C'$ be an additive functor of pre-additive categories; let $X,X_1,$ and $X_2$ be objects of $\mathcal C$; and, for $a=1,2$, let $X_a\xrightarrow{i_a}X\xrightarrow{p_a}X_a$ be morphisms such that $X$ is a product of $X_1$ and $X_2$ by $p_1,p_2$ and a coproduct of $X_1$ and $X_2$ by $i_1,i_2$. Then $F(X)$ is a product of $F(X_1)$ and $F(X_2)$ by $F(p_1),F(p_2)$ and a coproduct of $F(X_1)$ and $F(X_2)$ by $F(i_1),F(i_2)$. - -In view of (e), (b) and (c) imply (a).<|endoftext|> -TITLE: Why are polynomials easier to handle with than integers? -QUESTION [17 upvotes]: This may seems to be an elementary question, but I found no answers on MO nor google. -I have always heard "polynomials are easier to handle with than integers". For example: - -When $n$ is quite large, maybe 200 or more, it's relatively easier to factorize a polynomial $f$ of degeree $n$ than to factorize an integer with $n$ bytes. -When multiplying large integers, we see them as polynomials,use techniques such as FFT,intepolations to multiply polynomials,and then back to integers. -3.The zeta functions of $F[x]$ and $\mathbb{Z}$, and the former are easier to study than the latter. - -Of course there are other examples, but because of my shortage of knowledge, I can only lise these above. -So my question is (as in the titile): Why are polynomials easier to handle with than integers? I ask this because contrary to our intuitives, polynomials are "more complex" objects than integers. - -REPLY [5 votes]: In the Euclidean division of polynomials, quotient and remainder are uniquely defined, and compatible with addition. This fails for integers. -There is no integer analogue of constant polynomials. I guess that the search for the "field with one element" is in a sense motivated by this.<|endoftext|> -TITLE: Does homeomorphic and isomorphic always imply homeomorphically isomorphic? -QUESTION [51 upvotes]: Let $(G,\cdot,T)$ and $(H,\star,S)$ be topological groups such that - -$(G,T)$ is homeomorphic to $(H,S)$ and $(G,\cdot)$ is isomorphic to $(H,\star)$. -Does it follow that $(G,\cdot,T)$ and $(H,\star,S)$ are isomorphic as topological groups? - -If no, what if they are both Hausdorff? What if they are both Hausdorff and two-sided complete? - -REPLY [16 votes]: Any two $2$-dimensional rational vector spaces in $\mathbb R$ are isomorphic as groups and also order-isomorphic (I believe), hence homeomorphic when given the relative topology from $\mathbb R$. But any isomorphism as topological groups would have to be given by multiplication by a real number. So for example $\mathbb Q+\mathbb Q\pi$ and $\mathbb Q+\mathbb Q\sqrt 5$ are a counterexample. -EDIT: Or the same thing with $\mathbb Z$ instead of $\mathbb Q$.<|endoftext|> -TITLE: Computing `$\pi_1 S^1$` using groupoids -QUESTION [9 upvotes]: I believe it is possible to compute $\pi_1 S^1$ by applying the groupoid version of the Seifert-Van Kampen Theorem (in the version presented in May's Concise Course) to a covering of the circle by three arcs. Is there an account like this somewhere in the literature? Ideally I'd like a discussion that a student familiar with May's book would be able to read. (May doesn't take a 2-categorical approach to groupoids, and so he does not discuss the fact that a diagram of groupoids that is a point-wise equivalence induces an equivalence of colimits. This is rather important for computations.) -Edit: this last statement is false in general! I was thinking of homotopy colimits. The relevant (correct) fact appears in Ronnie Brown's book: retracts of pushouts are pushouts. This is the means by which one compares the Van Kampen theorem for the full fundamental groupoid - as in May's book - with the Van Kampen theorem for the fundamental groupoid on a set of basepoints.) - -REPLY [8 votes]: A rather belated comment on these! I like the comparison between the circle $S^1$ as obtained from the unit interval $[0,1]$ by identifying $0$ and $1$ in the category of spaces, and the group of integers $\mathbb Z$ as obtained from the groupoid $\mathcal I$, which has objects $0$ and $1$ and exactly one arrow $\iota:0 \to 1$, by identifying $0$ and $1$ in the category of groupoids. -I got hold of the idea in the 1960s from writing the first edition of this book that all of 1-dimensional homotopy theory was better expressed in terms of groupoids rather than groups This led to the question: are groupoids useful in higher homotopy theory? Is the 1-dimensional case a ``one-off''? or not? -I liked the more exciting prospect, but it took 9 years to get with Philip Higgins in 1974 a good definition in dimension 2, namely the homotopy double groupoid of a pair of spaces, and a 2-dimensional van Kampen theorem. -June 2, 2018 More background to this area is in the paper published this year: -Modelling and Computing Homotopy Types:I -One of the points made is that the usual base point approach seems to assume all interesting spaces are path connected; yet the circle is a counterexample since it is nicely represented as a union of non connected spaces. The algebra of groupoids nicely copes with the more general case; one main point of my 1967 paper on the groupoid version of the van Kampen theorem was to obtain by such a method the fundamental group of the circle! -Note also that if the theorem is proved by verification of the universal property, then the proof for $\pi_1(X,C)$ is no more difficult than the proof for $\pi_1(X,c)$; the case of many sets is given in this paper.<|endoftext|> -TITLE: Elliptic curves over finite fields -QUESTION [6 upvotes]: I have basic questions about elliptic curves over finite fields. - -Where to find general references? Hartshorne for instance restricts to algebraically closed ground fields. -Over an arbitrary field $K$, is the right definition of an elliptic curve a smooth proper -curve of genus 1 with a choice of $K$-rational point? -What is known about the structure of the group of $K$-rational points when $K$ is finite? In particular, how much does it depend on the curve? -Are there simple examples where you can explicitly see all of the $K$-rational points, $K$ finite? - -REPLY [2 votes]: I completely agree with the earlier answers. Just two remarks... - -For question 3, If $K$ is finite of cardinality $q$, then $E(K)$ is isomorphic to $\mathbf Z/n\mathbf Z\times\mathbf Z/m\mathbf Z$, where $n$ divides gcd$(q-1,m)$. -Concerning your last question, here is a simple example where you can explicitly 'see' all the $K$-rational points without direct computations; I hope it may interest you. Consider the elliptic curve -$$E:Y^2=X^3+1$$ -defined over the finite field $K=\mathbf F_p$, where $p=3n+2$ is a prime number. Then, there is a bijection $\varphi:K\to E(K)-\lbrace O\rbrace$ (where $O$ is the point at infinity) given by -$$\varphi(t)=\left((t^2-1)^{2n+1},t\right).$$<|endoftext|> -TITLE: Is there a ruler and compass construction of the common perpendicular of two geodesics in H^3? -QUESTION [9 upvotes]: Assume we have two geodesics in the Poincaré ball model of $\mathbb{H}^3$, viewed as arcs intersecting the boundary of and contained in the Euclidean unit sphere in $\mathbb{R}^3$. Is there a ruler and compass construction in $\mathbb{R}^3$ to construct their common perpendicular geodesic? -It seems that there is in 2 dimensions, see here. However the obvious generalisation to 3 dimensions would give the common perpendicular to geodesic planes containing the original two geodesics, rather than between the geodesics themselves. - -REPLY [4 votes]: I remembered that the isometries for the upper half space model of $\mathbf H^3$ are given by the action of $PSL_2\mathbf C$ on the points at infinity (the $xy$-plane as the complex numbers). So, given four geodesic endpoints, move one to $0$ with a pure translation. We now have three complex numbers $A,B,C,$ where $A$ is the other endpoint of the geodesic with an endpoint at $0.$ Define a complex number $\gamma$ that solves -$$ ABC \gamma^2 + 2 BC \gamma + (B+C-A) =0, $$ -where you might as well pick $\gamma = 0$ if the constant term $B+C-A=0.$ -Next, apply the linear fractional transformation -$$ h(z) = \frac{z}{\gamma z + 1} $$ -to the plane. -The result is $$ h(A) = h(B) + h(C) $$ -That is, the midpoint of $0$ and $h(A)$ is the same as the midpoint of $h(B)$ and $h(C).$ So the common orthogonal geodesic is the vertical ray through $h(A)/2$ and allowing the third coordinate to vary. Then map everything back to your originals. -Note that there is no answer if two of your original geodesic endpoints coincide. In that case, they lie in a common flat, and lines asymptotic at infinity do not share a perpendicular. -See if I can do the link the right way this time, Milnor's survey on the first 150 years of hyperbolic geometry can be downloaded.<|endoftext|> -TITLE: Vanishing of higher homotopy groups of finite complexes -QUESTION [10 upvotes]: Throughout, let $X$ be a connected finite CW-complex. If the universal covering of $X$ is contractible, then $\pi_n(X)=0$ for all $n \geq 2$. In this case $X$ is a model for $B\pi_1(X)$. -I am wondering whether this is the only reason why higher homotopy groups vanish above a certain degree. More precisely: - -Question: Let $k \geq 3$ be an integer. Can it happen that $\pi_n(X) = 0$ for all $n \geq k$ and $\pi_{k-1}(X)\neq 0$? - -REPLY [14 votes]: No, it cannot happen. In a paper by McGibbon and Neisendorfer, it is proven that if X is a 1-connected space and its mod-p-homology is non-zero in some degree, but zero in all higher degrees, then the $\pi_n X$ contain a subgroup of order p for infinitely many n. This can be applied to the universal cover of your X.<|endoftext|> -TITLE: Finiteness of higher homotopy groups of finite complexes -QUESTION [6 upvotes]: Throughout, let $X$ be a connected finite CW-complex. - -Question: If $X$ is of dimension $n$. Is there some integer $n'$ (maybe depending only on $n$), such that all homotopy groups $\pi_k(X)$ for $k \geq n'$ are finite? - -For the spheres $S^n$, $n'=2n+1$ works by Freudenthal's Suspension Theorem and Serre's result that the stable homotopy groups in that range are finite. More generally, if $\pi_1(X)=0$, then the Milnor-Moore theorem relates the rational homotopy groups to the rational homology of the loop space of $X$ and I believe that this can be used to get a similar conclusion. But what if $\pi_1(X) \neq 0$? -EDIT: Igor Belegradek (besides answering the question) pointed out that what I stated in the last three lines is not correct. - -REPLY [18 votes]: The answer is no in a very strong way even for simply-connected complexes. In rational homotopy theory there is a famous dichotomy between elliptic and hyperbolic spaces: a simply-connected finite complex is either elliptic or hyperbolic. Elliptic means that all but finitely many homotopy groups are finite. Hyperbolic means that the sum of ranks of first $k$ homotopy groups grows exponentially with $k$. In some sense most spaces are hyperbolic. If I remember correctly, $m$-fold connected sum of $S^2\times S^2$ with itself is hyperbolic if $m>1$. You can read more of this in the book "Rational homotopy theory" by Felix-Halperin-Thomas.<|endoftext|> -TITLE: Arguments against large cardinals -QUESTION [60 upvotes]: I started to learn about large cardinals a while ago, and I read that the existence, and even the consistency of the existence of an inaccessible cardinal, i.e. a limit cardinal which is additionally regular, is unprovable in ZFC. Nevertheless large cardinals were studied extensively in the last century and (apart from attempts that went too far as the Reinhardt-Cardinals) nobody ever found a contradiction to ZFC. As a consequence it seems to me that set theorists today don't consider the possibility of the non-existence of a large cardinal. Therefore my questions: -Why is it so unreasonable to think that the existence of large cardinals contradicts ZFC? Are there any mathematicians that do believe that large cardinals don't exist? And what are their arguments? -EDIT: I want to say thank you for all your very interesting answers and comments. It will take some time for me to fully understand them though, however I feel like I have already learned a lot due to this discussion. Thank you! - -REPLY [11 votes]: Large cardinals offer a detailed coherent picture — with a single principle, that of symmetry, reaching even (essentially) the strongest large cardinals. They continually offer new results — without inconsistency. For weaker large cardinals (up through Woodin cardinals) we have canonical well-understood fine-structural models, with a variety of intuitively true (or otherwise well-analyzed) seemingly unrelated combinatorial principles about real numbers and other sets implying their consistency. And while we are not there yet, we expect many more such results at the level of supercompact cardinals. -However, per the question title, the rest of the answer will be arguments against large cardinals, which are important both because of genuine doubts (especially about certain things) and for the insight the arguments offer. Many of the arguments also apply to ZFC, but become stronger as the large cardinal strength increases. -The arguments separate into: -• Arguments for inconsistency of large cardinals. -• Arguments for non-existence of large cardinals (even if they are consistent). -• Arguments for not accepting large cardinals as axioms. -Moreover regarding truth/consistency, we can approximately separate mere doubts from affirmative arguments for falsehood/inconsistency, with most arguments being the former. -Arguments against acceptance as axioms: -• Doubts about large cardinal axioms (see the other paragraphs). -• The importance for the foundations of mathematics to be free of unnecessary doubt. -• The unfortunate general disinterest in foundations. Most mathematicians do not even know exactly what ZFC is. (This is relevant here because 'acceptance' has a sociological component.) -• Related to the above, the limited need to go beyond ZFC for results most mathematicians care about. -• Knowing that something is provable in a weak theory (or just ZFC) often gives us important information that goes beyond knowing that it is true. -• One can still use large cardinal axioms by including them as assumptions. -Doubts about consistency: -• Doubt about metaphysical existence of infinite sets, or if they exist, about intuitive reasoning about such sets. Many believe that the observable physical universe is finite and/or that the operation of physical laws is computable, and that humans do not have privileged access to truth. Also, some facts that are obviously true about finite sets fail for the infinite. -• Lack of inner model and core model theory beyond Woodin cardinals. While (as set theorist John Steel put) "all roads lead to projective determinacy", we do not yet know whether all roads lead to analogs of supercompact cardinals. -  - Core model induction has not yet reached a Woodin limit of Woodin cardinals (but there is progress). Core models (including core model induction) are our main method of showing that natural plausible combinatorial propositions at low levels of $V$ imply consistency of large cardinals. -  - Existence of canonical inner models for a stationary set of Woodin cardinals is still open as of 2018. -  - While under plausible conjectures, current definitions of fine-structural models reach subcompact cardinals (and slightly further), supercompact cardinals show a qualitively different and anti-core-model behavior, such as consistency of indestructibility. -• Lack of sufficient heuristic 'proofs' of consistency of ZFC: -  - It is hard to find compelling arguments for consistency of ZFC that make no mention of infinity (other than the argument that consequences of ZFC have been well-understood with no inconsistency found; also, "hard" need not mean "impossible"). -  - Lack of enough natural true arithmetic statements that are known to imply consistency of ZFC. -  - Lack of reasonable ordinal notation systems that have been proved to reach ZFC (or even just $\mathrm{Z}_2$). Ordinal notation systems (and to a lesser extent, canonical inner models) give us a well-understood picture of how a theory operates, which (among many other benefits) counteracts doubts about consistency. -Affirmative historical arguments for inconsistency: -• Historically, paradoxes (including Russell paradox) were used to argue against infinity (including consistency of infinity) but these arguments have receded, as ZFC has no shown sign of being inconsistent through such a paradox. -• Kunen's inconsistency was used to argue against superficially similar axioms: Since existence of nontrivial elementary $j:V_{λ+2}→V_{λ+2}$ is inconsistent, why should using $V_{λ+1}$ be different? However, the current understanding is that there is no inconsistency without the axiom of choice (but see below), and unlike $V_{λ+2}$, choice does not cause problems for $V_{λ+1}$ and (starting from a stronger axiom for ZF) we can get choice in a generic extension. -• More recently, Hugh Woodin has argued (see his work on the HOD dichotomy) that Reinhardt cardinals for ZF are inconsistent. Specifically, for cardinals such as measurable (or even strong), there are canonical ordinal definable inner models that use the restrictions of the extenders from the embeddings on $V$ (and get the same large cardinals from these extenders), but he was able to show that modulo consistency of a strengthening of ZF + Reinhardt cardinal (that is weaker than Berkeley cardinals), this need not hold for supercompact cardinals. Personally I think that this failure is why supercompacts have been so useful for consistency of combinatorial propositions — unlike weaker cardinals, the embeddings (in a sense) capture a large part of $V$, and thus cannot be generally included into canonical inner models as is — but there is still much that we do not understand. -In between consistency and existence, we have various degrees of soundness. Properties inconsistent with the axiom of choice can still hold at those levels. Thus, for example, near the top of the known hierarchy (and thus, with the highest doubts about consistency), we have the $Π^V_2$ statement "For every cardinal $κ$, there is a model of 'ZF + Berkeley cardinal' closed under $κ$ sequences." The truthfulness of such statements is generally regarded as closely related to consistency. However, this belief assumes a fundamental soundness of the large cardinal hierarchy, which has strong analytical and empirical support, but can also be argued against (for example, see the arguments for $V=L$). -Doubts on existence of large cardinals: -• Formalists and adherents to various related philosophies believe that infinite (or failing that, uncountable) sets do not actually exist. What may exist are finite quasimodels (or countable models) of various set axioms, but they hold that there is no single preferred truth predicate for $(V,∈)$. The lack of consensus on CH is often cited to support this. Various large cardinal axioms may hold in some but not other theories/quasimodels/models, and the use of ZFC is a useful common convention. For them, existence of large cardinals morphs into goodness of using ZFC + A (for various A) for mathematical work. -• Consistency of large cardinals is fundamentally different from existence. To fully accept inaccessible cardinals, a set platonist would need to know that they metaphysically exist, or for a symmetry platonist (a type of truth-value platonist), that we have enough symmetry that truth values of statements at the level of inaccessible cardinals are unambiguous. -• The replacement axiom schema (and choice) may intuitively follow from "For every ordinal $α$, a process that can be always be continued can be iterated $α$ times", but this reasoning is insufficient to reach inaccessible cardinals. -• By Kunen's inconsistency, some otherwise natural large cardinal axioms are inconsistent with choice. Perhaps we will discover a stronger natural true combinatorial principle that refutes more large cardinal axioms. -• A canonical theory for a given expressiveness level may require consistency strength that on its face corresponds to a higher level. For example, a true reasonably complete axiomatization of second order arithmetic requires projective determinacy, and projective determinacy can be ascertained by just studying real numbers (disclaimer: not everyone agrees with this), which in turn gives consistency of Woodin cardinals. In this manner, doubt about consistency of Woodin cardinals may translate into doubt about existence of the set or class of all real numbers (and quantification over it). We do not know how much strength is required for third order arithmetic and higher expressiveness levels, and a breakdown would argue for non-existence of higher levels. -• Large cardinals can be destroyed by forcing. For example, we can destroy all measurables by Prikry forcing (making them cofinality $ω$ without changing other cofinalities), and it is arguably unclear why the ground model rather than the extension is a better analog to $V$. Also, by adding clubs, starting with a model of GCH, I think we can destroy all Mahlo cardinals while preserving GCH and cofinalities; and no forcing can resurrect Mahloness. -Affirmative arguments for non-existence: -• Predicativist argument for $V=L$. Predicativists hold that certain objects are (essentially) constructed in stages, and that quantification is fully permitted only over the previously constructed stages. Most predicativists start with only natural numbers, but if one also accepts arbitrary ordinals, the result is the constructible universe $L$ — each stage of $L$ is added predicatively. One would then object to zero sharp analogously to the traditional predicativist objection to zero hyperjump (Kleene's $\mathcal{O}$). Also, some mathematicians simply like $L$ as a canonical tidy universe that satisfies ZFC (and is $Σ^1_2$-correct). -• Against measurable cardinals: -  - The nonrigidity (nontrivial elementary embedding $V→M$) arguably qualitatively alters the behavior of $V$ and is contrary to how $V$ behaves at lower levels. -  - Measurable cardinals cause a few counterexamples in topology. -  - Under $HOD(ℝ)⊨\mathrm{AD}$, $ω_1^V$ is the least measurable in HOD (assuming the proof of lack of lower measurables applies here). The absence of lower measurable cardinals, combined with the fact that every ordinal definable subset of $ω_1$ is constructible from a real, and that the measure on $ω_1$ is 'caused' by an external structure (the club filter on $ω_1$) (along with the general largeness of measurables and existence of Prikry forcing) suggest a possibility that measurable cardinals do not occur in the cumulative hierarchy on their own, but reflect a restriction on $\mathcal{P}(κ)$. -(However, on the balance, I find that evidence supports existence of measurables.)<|endoftext|> -TITLE: Is the analysis as taught in universities in fact the analysis of definable numbers? -QUESTION [109 upvotes]: Ten years ago, when I studied in university, I had no idea about definable numbers, but I came to this concept myself. My thoughts were as follows: - -All numbers are divided into two classes: those which can be unambiguously defined by a limited set of their properties (definable) and those such that for any limited set of their properties there is at least one other number which also satisfies all these properties (undefinable). -It is evident that since the number of properties is countable, the set of definable numbers is countable. So the set of undefinable numbers forms a continuum. -It is impossible to give an example of an undefinable number and one researcher cannot communicate an undefinable number to the other. Whatever number of properties he communicates there is always another number which satisfies all these properties so the researchers cannot be confident whether they are speaking about the same number. -However there are probability based algorithms which give an undefinable number as a limit, for example, by throwing dice and writing consecutive numbers after the decimal point. - -But the main question that bothered me was that the analysis course we received heavily relied on constructs such as "let $a$ be a number such that...", "for each $s$ in the interval..." etc. These seemed to heavily exploit the properties of definable numbers and as such one can expect the theorems of analysis to be correct only on the set of definable numbers. Even the definitions of arithmetic operations over reals assumed the numbers are definable. Unfortunately one cannot take an undefinable number to bring a counter-example just because there is no example of undefinable number. How can we know that all of those theorems of analysis are true for the whole continuum and not just for a countable subset? - -REPLY [16 votes]: You can also talk about arithmetically definable real numbers: those for which the Dedekind cut of rationals is of the form: -$$\{m/n: \forall x_1 \exists x_2 \ldots \forall x_{k-1} \exists x_k\, -p(m,n,x_1,\ldots,x_k)=0\},$$ -where the $x$'s range over integers, and $p$ is a polynomial with integer coefficients. -Then on this definition of definability: $e$ and $\pi$ and all the familiar reals are definable. Only countably many numbers are definable. There must be other real numbers which are undefinable. And it all makes sense, and is even provably consistent, in ordinary set theory. -A standard reference for this way of thinking is the system $ACA_0$ in Simpson's Subsystems of Second-Order Arithmetic. -The cost of this metamathematical simplicity is a small change to the mathematics: any definable bounded sequence of reals has a definable least upper bound, but an uncountable definable set of reals may not. Feferman's notes on Predicative Foundations of Analysis show how to develop standard analysis on this basis. If we changed mathematics as taught in universities to be based on predicative analysis, few undergraduates or people outside the math department would notice much difference.<|endoftext|> -TITLE: How should one think about non-Hausdorff topologies? -QUESTION [65 upvotes]: In most basic courses on general topology, one studies mainly Hausdorff spaces and finds that they fit quite well with our geometric intuition and generally, things work "as they should" (sequences/nets have unique limits, compact sets are closed, etc.). Most topological spaces encountered in undergraduate studies are indeed Hausdorff, often even normed or metrizable. However, at some point one finds that non-Hausdorff spaces do come up in practice, e.g. the Zariski topology in algebraic geometry, the Fell topology in representation theory, the hull-kernel topology in the theory of C*-algebras, etc. -My question is: how should one think about (and work with) these topologies? I find it very difficult to think of such topological spaces as geometric objects, due to the lack of the intuitive Hausdorff axiom (and its natural consequences). With Hausdorff spaces, I often have some clear, geometric picture in my head of what I'm trying to prove and this picture gives good intuition to the problem at hand. With non-Hausdorff spaces, this geometric picture is not always helpful and in fact relying on it may lead to false results. This makes it difficult (for me, at least) to work with such topologies. -As this question is somewhat ambiguous, I guess I should make it a community wiki. -EDIT: Thanks for the replies! I got many good answers. It is unfortunate that I can accept just one. - -REPLY [5 votes]: Some rather strong non-Hausdorff topologies arise quite naturally in topological dynamics: orbit spaces of group actions, leaf spaces of foliations, etc. These spaces are not often studied in their own right, because the Hausdorff space of which they are the quotient is usually sitting right there in front of our eyes and is quite adequate for all purposes. -However, in studying $Out(F_n)$, the outer automorphism group of a free group $F_n$ of rank $n$, a certain non-Hausdorff orbit space plays a central role in the theory that was first developed in the papers by Bestvina, Feighn, and Handel on the Tits alternative for $Out(F_n)$. The group $F_n$ may be identified with the fundamental group of some finite graph $G$, and with the deck transformation group of the universal cover $\tilde G$. Its Gromov boundary $\partial F_n$ is a Cantor set identified with the space of ends of $\tilde G$. The "space of lines upstairs", denoted $\tilde{\mathcal B}(F_n)$, is the space of all 2-point subsets $\partial F_n$ equipped with the Hausdorff topology, and may be identified with the "lines in $\tilde G$"---bi-infinite edge paths in $\tilde G$ without backtracking---equipped with the topology having a basis element for each finite path $\tilde\alpha$ in $\tilde G$ consisting of all lines containing $\tilde\alpha$. The left action of $F_n$ on itself, or its deck transformation action on $\tilde G$, extends continuously to an action on $\partial F_n$ and induces an action on $\tilde{\mathcal B}(F_n)$ whose non-Hausdorff orbit space $\mathcal B(F_n)$ is the "space of lines (downstairs)"; it may be identified with the set of bi-infinite edge paths in $G$ without backtracking (ignoring parameterization) equipped with the topology having a basis element for each finite path $\alpha$ in $G$ consisting of all lines having $\alpha$ as a subpath. -Here is some intuition used in understanding $\tilde{\mathcal B}(F_n)$. In analogy with Thurston's unstable geodesic laminations, an outer automorphism $\phi \in Out(F_n)$ can have an "attracting lamination", a subset $\Lambda \subset \tilde{\mathcal B}(F_n)$ which is the closure of a single point $\ell$ such that for some open set $U \subset \tilde{\mathcal B}(F_n)$, the forward iterates $\phi^i(U)$ form a neighborhood basis for $\ell$, plus additional properties on $\ell$ ruling out some simple pathological possibilities. This point $\ell$ is thought of as a "generic leaf" of $\Lambda$, and the properties of $\ell$ are often used as stand-ins for properties of $\Lambda$ as a whole. As a simple example: in the graph $G$, given an edge $E$, the line in $G$ realizing a given generic leaf $\ell$ crosses $E$ if and only if every generic leaf of $\Lambda$ crosses $E$, in which case one might wish say that $\Lambda$ itself crosses $E$. But it is important to draw the distinction between such and edge $E$ and some other edge $E'$ which is crossed by some nongeneric leaves of $\Lambda$ but not by any generic leaves. It sometimes happen that $\Lambda$ is "minimal" and so every leaf is generic and one's intuition is more easily developed; but the very real possibility of nongeneric leaves is important to understand as well. -The non-Hausdorff space $\mathcal B(F_n)$ can indeed be described as a certain quotient of a certain Hausdorff space, but that Hausdorff space depends on rather unnatural choices: $\mathcal B(F_n)$ is a quotient of the space of parameterized bi-infinite edge paths in $G$ with the compact open topology. The unnaturality of this construction lies in the unnaturality of the choice of the graph $G$, which varies over the Culler-Vogtmann "outer space" of $F_n$. The graph $G$ is not well-defined up to homeomorphism, although it is well-defined up to homotopy equivalence. This unnaturality is perhaps one factor that led to the emergence of the non-Hausdorff space $\mathcal{B}(F_n)$ as the central object, rather than the space of bi-infinite bi-infinite edge paths in some choice of $G$. -This situation is somewhat similar to what happens in the study of the mapping class group of a closed oriented surface $S$ of genus at least $2$. As one varies over all possible hyperbolic structures on $S$, the quotient space of the geodesic foliation of the projective line bundle of $S$ has an invariant definition independent of the choice of hyperbolic structure, namely it is the non-Hausdorff quotient space $\partial^2 \pi_1 S / \pi_1 S$. That "invariance" has never been particularly inviting to dynamicists, who would rather think about the geodesic foliation itself (or maybe the geodesic flow on the unit tangent bundle). But when Thurston developed his theory of geodesic laminations on $S$, he often took pains to point out that the space of geodesic laminations was well-defined independent of the somewhat unnatural choice of hyperbolic structure.<|endoftext|> -TITLE: What are Central Limit Theorems and why are they called so? -QUESTION [10 upvotes]: I know two opinions: -1) "Central" means "very important" (as it was central problem in probability for many decades), and CLT is a statement about Gaussian limit distribution. If the limit distribution of fluctuations is not Gaussian, we should not call such statement CLT. -2) "Central" comes from "fluctuations around centre (=average)", and any theorem about limit distribution of such fluctuations is called CLT. -Which is correct? - -REPLY [6 votes]: There's a nice little book on this subject: The Life and Times of the Central Limit Theorem. The book goes over the history of the theorem from its embryonic form to its more or less final form 200 years later. It also gives precise statements of some of the modern variations of the theorem and indicates directions of research.<|endoftext|> -TITLE: What is the image of the half/full twist in the Hecke algebra, in the Kazhdan-Lusztig basis? What is the corresponding complex of Soergel bimodules? -QUESTION [6 upvotes]: Let $B_n$ be the braid group on $n$ strands. It has generators $\tau_i$ for $i = 1,\ldots,n-1$ which exchange the $i$th and $(i+1)$st strands, and which satisfy the relations -$\tau_i \tau_j = \tau_j \tau_i$ for $|i-j| \ne 1$ and $\tau_i \tau_{i+1} \tau_i = \tau_{i+1} \tau_i \tau_{i+1}$. -$B_n$ maps to the Hecke algebra $H_n$, which has the same generators and the additional relation $(\tau_i-1)(\tau_i + q) = 0$. Composing this map with the Jones-Ocneanu trace yields the HOMFLY polynomial of the closure of a given braid. -In the braid group, there is the half-twist $\Delta$. It exchanges the $i$-th and $(n-i)$th strands by rotating the whole braid counterclockwise. Its square, the full twist, generates the center of the braid group. -One particularly natural choice of basis for the Hecke algebra is the Kazhdan-Lusztig canonical basis. - - -What is the image of $\Delta^k$ in $H_n$, in the Kazhdan-Lusztig basis? - - -This is a decategorified version of the question I am really interested in: - - -What is the complex of Soergel bimodules corresponding to $\Delta^k$ ? - - -A related question, in case anyone happens to know: - - -What is the HOMFLY homology of a torus knot? - -REPLY [4 votes]: As Geordie says, this is really hard, and will get much harder as n increases (I'd guess $n=3$ is about an order of magnitude harder than $n=2$, and that trend continues). -The relevant papers on this are Stosic:Homological thickness and stability of torus knots and Rozansky:An infinite torus braid yields a categorified Jones-Wenzl projector. Jake Rasmussen had some similar results, but it looks like he never wrote them up (I guess fatherhood has been distracting for him). In fact, you might try emailing Marko; it's possible that the thick/Schur categorification story can actually help here and make things a bit more concrete. I wouldn't hold your breath, though. -Perhaps the most famous fact about $\Delta^k$ is that for the correct power of $q$, the limit $\lim_{k\to\infty}q^?\Delta^k$ is actually the Jones-Wenzl projector. Lev's paper is about how this holds in the categorified picture as well, though he's working in the Temperley-Lieb case, not full Soergel bimodules.<|endoftext|> -TITLE: How many simply connected subsets of an n-by-m grid? -QUESTION [8 upvotes]: Given an n-by-m square grid graph, how many ways are there to choose a subset of the vertices which is simply connected? Here, a subset of vertices is simply connected if the vertices, together with any edges or interior faces connecting them amongst themselves, form a contractible subregion of the grid. More formally, we can naturally embed the grid graph into the plane. Then I want to count subsets of vertices such that the union of the dual 2-cells forms a simply connected region in the plane. - -Let me try to be a little bit clearer this time. Let's work directly with the dual, since that is easier to visualize. Hence, my question is: - -Consider a grid of square tiles of dimensions n-by-m, with each of the nm tiles distinctly labeled. How many distinct (labeled) simply connected subsets of tiles are there as a function of n and m? - -Because the tiles are labeled, rotation or translation to get the same polyomino isn't allowed. I'm trying to count all subsets. Commenter JBL points out the sequence for m=n at Sloane's, which also links to a lot of work by Artem M. Karavaev on this problem. - -REPLY [4 votes]: $$ $$ -edit 3 : The original answer given apllies again -Now that the question is clarified and specified labeled square tiles (although now changing what $n$ and $m$ denote from the number of vertices along each dimension to now denoting the number of tiles' edges along each dimension) arranged in a $m \times n$ rectangular pattern, and asks about the number of simply connected subsets of tiles. -The number of simply-connected subset of tiles for an $m \times n$ rectangular array of labeled square tiles is equal to the number of different labeled graphs induced by rooted trees of size greater than or equal to $0$ nodes for the 2-dimensional lattice graph of size $m \times n$. The numbers given below, in the old version of the answers, assumes that $m$ and $n$ represented the number of vertices and $m-1$ and $n-1$ represented the number of faces or tiles in the dual-graph of the $m \times n$ lattice graph originally specified. -[edit 2] : old problem with this question as was previously posed -It's possible to use the same vertex-set to define multiple polyominos, if rotation is not allowed, because a vertex-set alone is not sufficient to specify a polyomino shape and which edges need to be considered. - -Thus the question as posed, requesting vertex sets, is not rigorously defined enough to admit a solution. - -For $m=4, n=3$, the vertex set {$1,2, \cdots, 12$} consisting of all $12$ of the vertices can define two polyominos which are distinct if rotation is not allowed: -1---2 3---4 1---2---3---4 -| | | | | | -5 6---7 8 5 6---7 8 -| | | | | | -9--10--11--12 9--10 11--12 - -Thus, the vertex set alone is not sufficient to distinguish different valid polyominos. Vertices and edges must be specified; or alternatively for the dual the faces must be specified (as described in the older part of this answer, below). -Here's my example of a 16-vertex set for $m=4, n=4$ which generates two different polyominos which remain distinct even under rotation. - 1---2 3---4 1---2---3---4 - | | | | | | - 5 6---7 8 5 6---7 8 - | | | | | | - 9 10--11 12 9 10 11 12 - | | | | | | | | -13--14 15--16 13--14 15--16 - -old answer is still below -If by your question you mean how many different subgraphs of an $m \times n$ lattice graph consist of a single connected component, rather than multiple components, then the answer is going to be different from Pietro Majer's approach. -In that case, for $m=n=2$, let's label each vertex in the lattice graph as $v_{x,y}$ with $1 \le x \le m$ and $1 \le y \le n$. Would you consider the selections $S_1=${$v_{1,1}, v_{1,2}, v_{2,1}$} and $S_2=${$v_{2,2}, v_{1,2}, v_{2,1}$} to be different selections of vertices or the same, because the single face adjoining these $2$ of the $4$ ways to select $3$ of the $4$ vertices in this case is considered part of the selection. If so, then would selecting a single edge on a face be considered as selecting the face of the polyomino? -If not, then there's another approach to the solution. Given an $m \times n$ lattice graph, with vertices labeled by their $x,y$ coordinates: - -call the total set of vertices $S$ -a subset $S_i \in S$ consists of $0$ to $m\cdot n$ vertices, and each subset can (for labeling's sake) be labeled with an integer $0 \le i \le 2^{mn}$. -define the distance of one subset $S_a$ to another subset $S_b$ to be the minimal Manhattan distance from the elements of subset $S_a$ and subset $S_b$ -define a subset $S_a \in S$ as meeting your condition if the minimal distance of each vertex in that subset from the other elements of that subset is $1$ - -$S_a = ${$ v_1, v_2, ... , v_k $} -$\forall v_j \in S_a, \textrm{distance}(v_j, S_a - v_j)=1$ -where $S_a - v_j$ represents the set generated by starting with $S_a$ and removing element $v_j$ - -If instead of that approach, you are really talking about polyominos or the faces of the $m \times n$ lattice: -You can generate the dual of this graph as the $(m-1)\times(n-1)$ lattice and count the number of single component subgraphs in that. -In regard to JBL's comment to the question, I think you're right that Sequence A140517 at http://oeis.org/A140517 is the right sequence for an $(n-1) \times (n-1)$ grid. I don't think the question's original poster has clarified the question well enough since the question still asks for "vertex subsets", rather than saying "faces" on the 2-d lattice. I have a different approach for a different answer below, where you can take the dual of the $m \times n$ lattice and get an $(m-1) \times (n-1)$ lattice with each vertex in this graph representing a face in the original lattice graph. Then, the count of the single component subgraphs of this dual is the answer. -The original poster needs to clarify the question and give some example answers to $m=2, n=2$. Does the solution for $m=2, n=2$ count only the single face generated by all $4$ vertices as the only solution, do the different sets of vertices taken three at a time count as separate solutions? Would taking two vertices that are distance $1$ apart count as a solution at all? The question really needs to be clarified.<|endoftext|> -TITLE: How does Berger-Moerdijk's relative Boardman-Vogt work? -QUESTION [9 upvotes]: In "The Boardman-Vogt resolution of operads in monoidal model categories," the authors construct factorizations of sufficiently nice operad maps $P\to Q$ into a cofibration followed by a weak equivalence $P\rightarrowtail W(H,Q)_P\stackrel{\sim}{\to}Q$. This construction depends on a choice of interval $H$. When $P$ is the unit operad $I$, this is called $W(H,Q)$ and is a cofibrant replacement for $Q$. My question is about the relative construction when $P\ne I$. -The factorization $W(H,Q)\_P$ is constructed as a sequential colimit of pushouts. For simplicity, concentrate everything in arity $1$ so that we can ignore $\Sigma_n$ actions; then these pushouts are of the form -$H(T)\otimes\underline{Q}(T)\gets {(H\otimes Q)}^-\_P(T)\to W_{k-1}(H,Q)_P.$ -The pushout of this diagram (taken as $T$ ranges over all appropriate trees) is $W_{k}(H,Q)_P$. It is my understanding that the middle entry of this pushout is the colimit over a cube with the final corner omitted, and that the left entry is that final corner. The map to the right factor, which is constructed inductively, - -informally puts edge-lengths to 0, whenever the vertices of the edge are both labelled by elements of P, and then applies the corresponding attaching map of the absolute W-construction. - -The middle entry of the pushout is somewhat involved to describe explicitly; it occurs on p833--834 and relies on p819--821 in the published version of the paper. Let's pick an example,and I'll do a calculation, and hopefully someone can point out the mistake I'm making. -Let's look at the bivalent tree with four vertices $(v_1,v_2,v_3,v_4)$ and three internal edges $(e_{12}, e_{23}, e_{34})$. I believe that one part of the diagram making up the middle entry is: -$I_{12}\otimes H_{23}\otimes I_{34}\otimes Q_1\otimes I_2\otimes I_3\otimes Q_4$ -and that this maps along the edges of the cube into both -$I_{12}\otimes H_{23}\otimes I_{34}\otimes Q_1\otimes I_2\otimes Q_3\otimes Q_4$ -and -$I_{12}\otimes H_{23}\otimes I_{34}\otimes Q_1\otimes P_2\otimes P_3\otimes Q_4$ -In the first case, the attaching map to $W_{k-1}(H,Q)\_P$ collapses the edge $e_{34}$, getting the vertex label $Q_3\otimes Q_4$ since the edge between them is labeled by $I$ and also forgets $v_2$ because it is labeled by $I$, using the product on $H$ to get the label on the resulting edge. So we get a tree with vertices $(v_1, v_{34})$ and edges $(e_{134})$. These are labeled by $Q$, $Q$, and $H$. -In the second case, the attaching map first collapses the edge $e_{23}$ because it is between two elements of $P$, composing $P_2\otimes P_3\to P$, and then collapses the other two edges because they are of length $0$, getting $Q_1\otimes P\otimes Q_4\to Q$. In this case, we get a tree with only one vertex and no internal edges. -In order for the attaching map to be well-defined on the colimit defining the middle entry, if we begin in $I_{12}\otimes H_{23}\otimes I_{34}\otimes Q_1\otimes I_2\otimes I_3\otimes Q_4$ then we must get the same eventual answer in $W(H,Q)_P$ whichever one of these procedures we choose. Then, shifting for concreteness' sake to, say, the topological category and the Boardman-Vogt interval, we get that the formal composition on the two vertex tree $q_1, q_2$ with edge length $r$ is equivalent to the composition $q_1\circ q_2$ on the one-vertex tree. This is a problem because it means that the entire structure collapses to $Q$, which is not in general sufficiently cofibrant. - -REPLY [8 votes]: It turns out that the construction as written in the paper is incorrect, but there is a fix known to the authors. Instead of collapsing $P\otimes H\otimes P$ to $P$, one only collapses $I\otimes P\otimes H\otimes P\otimes I$ (where the $I$ factors are included in $H$ with length 1, not 0). This gets rid of the problem above and the rest of the construction checks out. Many thanks to Gijs Heuts for communicating this to me.<|endoftext|> -TITLE: Is the norm of a $0-1$ matrix (almost) attained on a $0-1$ vector? -QUESTION [19 upvotes]: I'd like to state explicitly a problem which was somehow hidden in my three-week-old post: - -Does there exist an absolute constant $c>0$ with the property that for any matrix $M\in{\mathcal M}_{m\times n}(\{0,1\})$ (zero-one matrices with $m$ rows and $n$ columns), there is a non-zero vector $x\in\{0,1\}^n$ such that $\|Mx\|/\|x\|\ge c\|M\|$? - -(Here $\|\cdot\|$ denotes both the Euclidean norms in ${\mathbb R}^m$ and ${\mathbb R}^n$ and the induced operator norm.) - -I can prove the conclusion with $c\sim 1/\sqrt{\log n}$ even in the case $M\in{\mathcal M}_{m\times n}({\mathbb R})$, and an example due to Greg Kuperberg shows that this is, essentially, best possible. The question is, can one make an improvement under the assumption that all elements of $M$ are restricted to the values $0$ and $1$? - -REPLY [11 votes]: As I have figured out recently, the answer is no. The full proof is somewhat technical and I cannot supply full details within the framework of an MO post, but here is the idea behind the construction. -Start with a symmetric matrix $A\in{\mathcal M}_{n\times n}(\{0,1\})$ such that the Perron-Frobenius eigenvalue of $A$ is much larger than the absolute value of any other eigenvalue, and the corresponding eigenvector is "reasonably simple". Now take a high tensor power of $A$. We get a symmetric zero-one matrix $M$ which inherits the spectral gap of the original matrix $A$, and hence the norm $\|Mx\|$ is controlled by the projection of $x$ onto the Perron-Frobenius eigenvector, say $v$, of $M$. Being a tensor power of the Perron-Frobenius eigenvector of the original matrix $A$, the vector $v$ can be analyzed, and with some effort can be shown to be "oblique" in the sense that it is "not aligned" with any zero-one vector. Hence, if $x$ is a zero-one vector, then the projection of $x$ onto $v$, and therefore the norm $\|Mx\|$, are small. -A precise result I was able to prove along these lines is as follows: there exist matrices $M\in{\mathcal M}_{n\times n}(\{0,1\})$ of arbitrarily large order $n$ such that for any non-zero vector $x\in\{0,1\}^n$, we have - $$ \|Mx\| \ll \left( \frac{\log\log n}{\log n} \right)^{1/8} \|M\|\|x\|. $$<|endoftext|> -TITLE: What does the Riemann-Hurwitz formula tell us on the Picard variety -QUESTION [6 upvotes]: Let $f:X\longrightarrow Y$ be a finite separable morphism of smooth projective integral curves over an algebraically closed field. -Then we have a linear equivalence of Weil divisors on $X$: $$ K_X=f^\ast K_Y + R.$$ Here $$R=\sum \textrm{length} (\Omega_{X/Y})_p [p]$$ is the ramification divisor on $X$. This is the Riemann-Hurwitz theorem. -We have a short exact sequence $$ 0 \longrightarrow \textrm{Pic}^0(X) \longrightarrow \textrm{Pic}(X) \longrightarrow \mathbf{Z} \longrightarrow 0,$$ where $\textrm{Pic}(X)\longrightarrow \mathbf{Z}$ is the degree map. -We know what the Riemann-Hurwitz theorem tells us on the degree part of $\textrm{Pic}(X)$. It gives us the topological data $g(X)$ in terms of the degree of $R$, the genus of $Y$ and the degree of $f$. -But what does it tell us on $\textrm{Pic}^0(X)$? - -REPLY [9 votes]: The answer is quite classical when $f \colon X \to Y$ is an unramified double cover. -In this case Riemann - Hurwitz formula gives -$g(X)-1 = 2g(Y)-2$. -Consider the following three natural maps: -$f^* \colon J(Y) \to J(X)$, -$Nm \colon \textrm{Pic}^0(X) \to \textrm{Pic}^0(Y), \quad Nm(\sum a_ip_i):= \sum a_if(p_i)$ -$\tau \colon J(X) \to J(X)$, -where $f^*$ is induced by the pull-back of $0$-cycles, $Nm$ is the norm map and $\tau$ is the involution induced by the double cover $f$. -Then - -$\textrm{Ker} \; f^*=\langle L \rangle$, where $L$ is a point of order $2$ in $J(Y)$; -the connected component of $Nm^{-1}(0)$ containing the identity coincides with the image of $I-\tau$. It is an Abelian subvariety of $\textrm{Pic}^0(X)$ of dimension $g(Y)-1$, that is denoted by $\textrm{Prym}(X, \tau)$. - -Moreover, under the identification of $\textrm{Pic}^0(X)$ with $J(X)$, the principal polarization of $J(X)$ restricts to twice a principal polarization on $\textrm{Prym}(X, \tau)$. -The geometry of Prym varieties is very rich. In particular, Riemann-Hurwitz identity -$K_X =f^*K_Y$ -induces subtle relations between the Theta divisor $\Theta$ of $X$ and the Theta divisor $\widetilde{\Theta}$ of $\textrm{Prym}(X, \tau)$. -You can look at [Arbarello-Cornalba-Griffiths-Harris, Geometry of algebraic curves, Appendix C] or at [Birkenhake-Lange, Complex Abelian Varieties, Chapter 12] for further details. -In the general case, it is possible to define the so-called generalized Prym varieties, at least where $f \colon X \to Y$ is a tame Galois branched cover. Look for instance at the paper of MERINDOL -"Varietés de Prym d'un revetement galoisien [Prym varieties of a Galois covering]" -Journal Reine Angew. Math. 461 (1995), 49-61.<|endoftext|> -TITLE: Dual of bounded uniformly continuous functions -QUESTION [25 upvotes]: Let $(X,d)$ be a metric space, and let $C_u(X)$ be the Banach space of bounded uniformly continuous functions on $X$ (with the uniform norm). How can I characterize its dual space $C_u(X)^*$? -I would guess it can be described as some space of measures. I would even be interested in the case $X=\mathbb{R}$. -Obviously if $X$ is compact this is just the signed (or complex) Radon measures on the Borel $\sigma$-algebra of $X$. If $d$ is a discrete metric then we have all finitely additive measures on $X$. But more generally I do not know. -Edit: If $C_b(X)$ is the Banach space of all bounded continuous functions on $X$, we of course have $C_u(X) \subset C_b(X)$ as a closed subspace, and we know that $C_b(X)^*$ can be identified with the space of finite, regular, finitely additive signed Borel measures on $X$. Certainly each such measure gives us a continuous linear functional on $C_u(X)$, so we have a map $C_b(X)^* \to C_u(X)^*$ which is just the restriction map, but it is not injective. Conversely, by Hahn-Banach each bounded linear functional on $C_u(X)$ extends to a bounded linear functional on $C_b(X)$, but not in a canonical way. -Also, it is clear that in general $C_u(X)^*$ contains more than just the countably additive measures, since e.g. if $X=\mathbb{R}$ it contains some Banach limits. So we have all the countably additive finite Radon measures, but not all the finitely additive finite regular measures. This is why I would imagine that $C_u(X)^*$ consists of all finitely additive measures satisfying some condition that is more than "regular" but less than "countably additive". But I have no idea what it might be. -As mentioned in comments, I am happy to know about nontrivial special cases: $X$ locally compact, $X$ complete and separable, etc. - -REPLY [7 votes]: This is a belated reply to your query, more a comment but too long for that and, anyway, I don't have that option. -Firstly, there is a concept of a compactification which plays for metric spaces the same role as the Stone-Cech compactification for completely regular spaces (it works even for uniform spaces). It is called the Samuel compactification and the measures you are looking for are the Radon measures on this compact set (which contains the original space in a canonical way). In functional analytic terms it is the spectrum of the Banach algebra you are using---the bounded, uniformly continuous functions on the metric space (again, more generally, uniform space---in fact, I think that the category of uniform spaces is the more natural one for your question). -I kind of suspect that you are looking for ways to get nice spaces of measures (or spaces of nice measures) as duals of spaces of bounded continuous or uniformly continuous functions. If so, this has been investigated in some detail. In the case of continuous functions this is accomplished by the so-called strict topology (introduced originally by Buck for locally compact spaces, then generalised to completely regular spaces by several mathematicians in the 70's). -Pachl (in Pac. J. Math. paper "Measures as functionals on uniformly continuous functions"---available online) did something similar for measures on uniform spaces---he called them uniform measures. In particular, he proved a deep theorem on compactness in this space---a theorem that is not as well known as it should be. -A unified approach to these subjects can be found in the book "Saks Spaces and Applications to Functional Analysis".<|endoftext|> -TITLE: Why are Tamagawa numbers equal to Pic/Sha? -QUESTION [23 upvotes]: For a connected algebraic group $G$ over a global field $K$ with adeles $A$, the Tamagawa number of $G$ is the volume of $G(A)/G(K)$. It is conjectured (and often known) to be rational, namely the quotient of the order of the Picard group divided by the order of the Tate-Shafarevich set. -Is there some simple explanation for the appearance of these integers? For example, is there some natural object that turns out to have volume 1 that divides into equal pieces, indexed by TS, and then copies indexed by Pic reassemble to form $G(A)/G(K)$? You can throw in bundles where the fibers have volume 1, if necessary. - -REPLY [3 votes]: I found the Sha part of the answer in Voskresenskii's book. -I'm looking at a quotient and I want cohomology to appear. -Cohomology is the obstruction to invariants being exact, so I -should lift this to a quotient of Galois modules. -$G(\bar K)$ has a Galois action, as does $G(A_K\otimes \bar K)$. -Their quotient $Q$ has fixed points $Q(K)=(G(A_L)/G(L))^{Gal(L/K)}$, -where $L$ is a splitting field. -The long exact sequence in cohomology -$$0\to G(K)\to G(A_K)\to Q(K)\to H^1(K;G)\to -H^1(K;G(A_{-}))=\prod H^1(K_\nu;G)$$ -shows that $Q(K)$ differs from $G(A_K)/G(K)$ by exactly the -Tate-Shafarevich group. -In the torus case, one studies the failure of multiplicativity of -the Tamagawa number in extensions $T'\to T\to T''$. -The sheafified quotient $Q_{T'}(K)$ naturally appears as the (maximum possible) kernel of -$T(A)/T(K)\to T''(A)/T''(K)$. -Thus it is better to look at $Q_{T'}(K)\to Q_T(K)\to Q_{T''}(K)$. -This is still not right exact, so $H^1(K;Q_T)$ enters, which -Nakayama duality says is Pontrjagin dual to $Pic(T)$. Perhaps this is saying that we should form a $Pic(G)^*$ gerbe over $Q_G$? -I still have some hope for a geometric explanation for the Picard group.<|endoftext|> -TITLE: What is the depth of the "provability heirarchy"? -QUESTION [7 upvotes]: I am not a logician or set theorist, so hopefully this makes sense. Let $T$ be a theory which is expressive enough to make statements like "Statement $A$ has a proof in $T$"; for example, $T$ might be capable of expressing elementary arithmetic and proving certain basic arithmetic facts. -Let $\Pi^0(T)$ consist of those statements $A$ in $T$ such that either $A$ or $\neg A$ admits a proof in $T$. In general, for $i\in \mathbb{N}$ let $\Pi^i(T)$ consist of those statements $A$ such that the statement "Con(T) implies that there does not exist a proof of $A$, or a proof of $\neg A$, in $T$" is in $\Pi^{i-1}(T)$. Intuitively, $\Pi^1$ should consist of those statements $A$ for which one can prove in $T$ that neither $A$ nor $\neg A$ admits a proof in $T$, $\Pi^2$ should consist of those statements $A$ such that one can prove in $T$ that one cannot tell whether or not they admit a proof, but where one knows this fact, and so on. -Godel's first incompleteness theorem implies that $\Pi^0(T)$ does not contain every sentence of $T$; in particular, the proof amounts to constructing a sentence in $\Pi^1(T)$. -Now I'm sure that if this heirarchy is not nonsense for some reason that I'm missing, then it must be well-studied. So here goes: - -1) If this set-up is studied, what is it called? -2) In standard theories, e.g. ZFC, is every sentence contained in $\bigcup_{i\in \mathbb{N}} \Pi^i(T)$? It seems to me that any ZFC proof that this is not the case must be non-constructive, and so would be pretty interesting. -3) Are there any standard theories $T$ of finite (known) depth in this heirarchy, i.e. every sentence is contained in $\Pi^i(T)$ for some fixed $i>0$? If so, can this be proven in $T$? -4) One can also define this heirarchy in a relative setting; e.g. one can take two theories $T\subset S$ and ask about $S$-proofs of statements about the existence of proofs in $T$. Is this studied? Are answers to the above questions known in these cases? - -Motivation: I recently watched this talk by Voevodsky, and ever since I've been wondering about how much we actually know about what we can prove. One might hope that even if we can't prove every "true" statement, then we can at least always prove that a statement does not admit a proof in our theory, if that is indeed the case. That seems unlikely, and amounts to the claim that all statements in our theory are in $\Pi^1(T)$, but I think that it is the best situation one can hope for (at least naively) given Godel's theorems. So, is the situation hopeless? Give it to me straight, doc. - -REPLY [8 votes]: I believe the concept you are looking for is that of "iterated consistency extension." A very nice treatment is given by Torkel Franzén in his book Inexhaustibility: a non-exhaustive treatment. See also this related question and this blog article by Mike O'Connor.<|endoftext|> -TITLE: Fourier dimension of the sum of sets -QUESTION [7 upvotes]: This question came up when my supervisors, my colleague, and I were considering arithmetic progressions in sets of fractional dimension. In particular, we were interested in "extracting" Salem sets from other sets, and we came across the following question: -We define the Fourier dimension of $E \subseteq \mathbb{R}$, $\mathrm{dim}_F(E)$, as the supremum of $\beta \in [0,1]$ such that for some probability measure $\mu$ supported on $E$, -$ -|\widehat{\mu}(\xi)| \leq C|\xi|^{-\beta/2}. -$ -Suppose $E_1,E_2$ are two subsets of $\mathbb{R}$. What is the relationship, if any, between $\mathrm{dim}_F(E_1 + E_2)$ and $\mathrm{dim}_F(E_1)$, $\mathrm{dim}_F(E_2)$? -A quick calculation shows -$ -\mathrm{dim}_F(E_1 + E_2) \geq \min(\mathrm{dim}_F(E_1) + \mathrm{dim}_F(E_2),1), -$ -but can we do any better than this? Or can strict inequality hold? -For some motivation, we may look to Hausdorff dimension instead (it is known that $\mathrm{dim}_F(E) \leq \mathrm{dim}_H(E)$; a set $E$ such that $\mathrm{dim}_F(E) = \mathrm{dim}_H(E)$ is called a Salem set). I believe Falconer gave an example of sets $E_1$ and $E_2$ such that $\mathrm{dim}_H(E_1) = 0 = \mathrm{dim}_H(E_2)$, yet $\mathrm{dim}_H(E_1+E_2) = 1$. - -REPLY [3 votes]: It is possible that $\dim_F(E_1)=\dim_F(E_2)=0$ yet $\dim_F(E_1+E_2)=1$, so there is no inequality in the opposite direction. -In fact, Falconer's example of sets $E_1, E_2$ such that $\dim_H(E_1)=\dim_H(E_2)=0$ but $\dim_H(E_1+E_2)=1$ already works. In Falconer's example, not only $E_1+E_2$ has dimension $1$, but in fact $E_1+E_2$ is an interval. Hence $\dim_F(E_1)=\dim_F(E_2)=0$ (since $\dim_F$ is bounded above by $\dim_H$) but $\dim_F(E_1+E_2)=1$. -There are also examples which are not forced by the Hausdorff dimensions of the sets. Indeed, let $C$ be the ternary Cantor set. It is a classical result of Kahane and Salem that if $\mu$ is any measure supported on $C$, then $\widehat{\mu}(\xi)\nrightarrow 0$ as $|\xi|\to\infty$; in particular, $\dim_F(C)=0$. Clearly, the same is true for any dilate $t C$ with $t\neq 0$. -On the other hand, it is well known that $C+C$ equals the interval $[0,2]$. Indeed, since $C$ and also $tC$ have Newhouse thickness equal to $1$, Newhouse's gap lemma (the endpoint version), guarantees that $C+tC$ has nonempty interior for all $t\neq 0$, and therefore also Fourier dimension $1$ (note that I'm not claiming that the indicator function of $C+tC$ has Fourier decay, just that there is a measure supported on (an interval in) $C+tC$ that does). -Hence, for all $t$ we have $\dim_F(C)=\dim_F(tC)=0$ but $\dim_F(C+ tC)=1$. This shows that the opposite inequality fails in a rather dramatic function: it does not even hold typically in the sense of Marstrand's Theorem. -Morally speaking, there is no reason why $$\dim_F(E_1+E_2)=\min(1,\dim_F(E_1)+\dim_F(E_2))$$ should hold in general. Leaving Hausdorff dimension considerations aside, if $E_1$ or $E_2$ are not Salem, this tells us that there are some resonances in the construction of the sets at a set of frequencies (possibly very sparse). These special frequencies will in general be lost in the sum $E_1+E_2$ (unless $E_1$ and $E_2$ also resonate to each other in some strong form), so one would expect that $\dim_F(E_1+E_2) > \dim_F(E_1)+\dim_F(E_2)$. However, I suspect it is not trivial at all to give specific examples where $\dim_H(E_1+E_2)<1$ (because proving Salemness or even some good decay of Fourier coefficients is usually hard). -On the other hand, equality $\dim_F(E_1+E_2)=\min(1,\dim_F(E_1)+\dim_F(E_2))$ can certainly hold. Indeed, it is easy to see this is always the case when $E_1$ and $E_2$ are Salem and additionally one of them has coinciding Hausdorff and upper box-counting dimension (which is the case for all known constructions of Salem sets). Indeed, denoting upper box-counting dimension by $\dim_B$, it is well known that $\dim_H(A+B)\le \dim_H(A)+\dim_B(B)$, so in this case $\dim_H(E_1+E_2)\le \dim_H(E_1)+\dim_H(E_2)$, and it follows from salem-ness and $\dim_F\le \dim_H$ that -$$ -\dim_F(E_1+E_2)\le \dim_F(E_1)+\dim_F(E_2). -$$<|endoftext|> -TITLE: Character table for the affine group of Z/p^nZ -QUESTION [6 upvotes]: Initial caveat: the following question could probably be answered by Google, MathSciNet or my library, if I could find the right search terms or book... but I've not had any luck today, so I hope someone can point me to a reference. -(The question is related to some of my older questions concerning characters of finite groups. All representations/characters are over the complex field.) -I am trying to estimate a certain invariant associated to finite groups, and recently thought that a useful toy example to play with would be -$$ G = \left\{ -\left( \begin{matrix} a & b \\ 0 & 1 \end{matrix} \right) \;\mid\; - a,b\in {\mathbb Z}/p^n{\mathbb Z}, p \nmid a - \right\} $$ -where $p$ is a prime and $n\geq 1$. I guess this might be called the "affine group" of the ring $R={\mathbb Z}/p^n{\mathbb Z}$? -Now the invariant can be calculated pretty easily (modulo tedious sums) once we know the character table of $G$, but this means more than knowing the degrees of the irreducible characters; I need to know the values they take on the various conjugacy classes inside $G$. -For $n=1$ this does not take long to do directly and can also be found as an example in various introductory-level textbooks on representation theory. However, for $n=2$ the best I could find was a section in a paper of several authors, where they just work out the character table by hand after first finding the characters via induction ($G$ is a semi-direct product arising from the action of the group of units in $R$ on the additive group of $R$). Now since I want to continue to higher $n$, I seem to be faced with three options: -1) Slog through the computation myself (which is probably good for my mathematical soul, but takes time & brainpower I need to spend on other things) -2) Learn how to ask a computer to do this (see previous parenthetical remark) -3) Find a reference which just gives the table. -So before embarking on 1), I thought I'd ask here. Most sources I could find from a crude skim online and in my library only discussed linear groups over finite fields; but I'm hoping that the construction here is sufficiently natural that it might have been treated already and written up somewhere. - -REPLY [6 votes]: The groups you are interested in are sometimes called false Tate extensions in number theorists' jargon. They are Galois groups of the Galois closures of extensions of $\mathbb{Q}$ obtained by adjoining the $p^n$-th roots of a $p$-th power free element. The irreducible representations are very explicitly described in Vladimir Dokchitser's PhD thesis, beginning of chapter 3. -Alternatively, the characters, together with their values, are easy to compute using a procedure described in Serre's Linear representation book, part II, Section 8.2. He explains how to obtain all the irreducible characters of any group that is a semi-direct product when the normal subgroup is abelian. Note that the character of an induced representation is easy to compute in terms of the original character and the coset-action. -And of course one more remark: for any given $p$ and $n$, MAGMA will just give you the character table. -Edit: Since you seem unsure, how to ask a computer for the character table, here is MAGMA code as an example. You will easily adopt it to any other package that handles character tables: -p:=3; n:=2; -Z:=Integers(); -gl:=GL(2,quo); -pr:=PrimitiveRoot(p^n); - -G:=sub; -CharacterTable(G);<|endoftext|> -TITLE: Asymptotics of Product of consecutive primes -QUESTION [5 upvotes]: I am looking for the asymptotic growth of product of consecutive primes. Is there anything that is known about this growth? - -REPLY [5 votes]: You can also prove that $$\displaystyle\lim_n\left(\prod_1^n p_i\right)^{1/p_n} = e$$ -(where $p_i$ is the $i$-th prime number and $e$ is Euler's exponential number)<|endoftext|> -TITLE: Is there any formal foundation to ultrafinitism? -QUESTION [84 upvotes]: Ultrafinitism is (I believe) a philosophy of mathematics that is not only constructive, but does not admit the existence of arbitrarily large natural numbers. According to wikipedia, it has been primarily studied by Alexander Esenin-Volpin. On his opinions page, Doron Zeilberger has often expressed similar opinions. -Wikipedia also says that Troelstra said in 1988 that there were no satisfactory foundations for ultrafinitism. Is this still true? Even if so, are there any aspects of ultrafinitism that you can get your hands on coming from a purely classical perspective? -Edit: Neel Krishnaswami in his answer gave a link to a paper by Vladimir Sazonov (On Feasible Numbers) that seems to go a ways towards giving a formal foundation to ultrafinitism. -First, Sazonov references a result of Parikh's which says that Peano Arithmetic can be consistently extended with a set variable $F$ and axioms $0\in F$, $1\in F$, $F$ is closed under $+$ and $\times$, and $N\notin F$, where $N$ is an exponential tower of $2^{1000}$ twos. -Then, he gives his own theory, wherein there is no cut rule and proofs that are too long are disallowed, and shows that the axiom $\forall x\ \log \log x < 10$ is consistent. - -REPLY [2 votes]: One can construct a consisten logic which explicitly limits computational complexity of valid statements, and thus is ultrafinitist: https://arxiv.org/abs/2106.13309<|endoftext|> -TITLE: Is there much theory of superalgebras acting on manifolds by alternating polyvector fields? -QUESTION [10 upvotes]: Usual story: vector fields on $M$, with their Lie bracket, form a Lie algebra. We can consider "actions" of some other Lie algebra ${\mathfrak g}$ on $M$ by looking at Lie homomorphisms ${\mathfrak g}\to Vec(M)$. -Now soup up $Vec(M)$ to alternating multivector fields (i.e. sections of $\Lambda^\bullet TM$), and extend the Lie bracket to the Schouten bracket, making it into a graded Lie superalgebra. (Careful: for the grading to work right, a section of $\Lambda^k TM$ must have degree $k-1$.) -My question: -Has there been much theory developed of graded Lie superalgebras acting on (non-super) manifolds? -For example, if ${\mathfrak g}$ is generated by a single generator $\pi$ of degree $1$, with $[\pi,\pi]=0$ (which one would ordinarily only expect in even degree), then ${\mathfrak g}$-manifolds are Poisson manifolds. - -REPLY [4 votes]: I am aware of one context in which something similar to this has been studied. I apologise in advance for "tooting my own horn" since this is based on a paper I co-authored in 2008 and some other work of mine. -The context is that of the geometric construction of Lie superalgebras, which arises quite naturally in the study of supersymmetric supergravity background. No Physics is necessary to understand the ideas, though -- so do read on! -The geometric data depends on the supergravity theory in question, but it will invariably include a lorentzian spin manifold $(M,g)$ and a connection $D$ on the spinor bundle. (There may be other fields and the spinor bundle might be twisted by a line bundle with its own connection, but such details are peripheral to the main story.) Spinor fields in the kernel of $D$ (subject, perhaps, to additional algebraic conditions) are called (supergravity) Killing spinors and they form a vector space we will denote $\mathfrak{k}_1$. -The symmetric square of the spinor bundle can be decomposed in terms of exterior powers of the tangent bundle and hence one can "square" Killing spinors to obtain polyvector fields satisfying some differential conditions. For example, in so-called Freund-Rubin backgrounds, these polyvectors are dual to so-called special Killing forms. -Let $\mathfrak{k}_0$ denote the image of the Killing spinors under the squaring map. Then in many cases, on the super-vector space $\mathfrak{k} = \mathfrak{k}_0 \oplus \mathfrak{k}_1$ there is a Lie superalgebra structure, called the Killing superalgebra of the background, where the $\mathfrak{k}_1 \otimes \mathfrak{k}_1 \to \mathfrak{k}_0$ component of the Lie bracket is precisely the squaring map of spinor fields. And in some cases, but not in general, the $\mathfrak{k}_0 \otimes \mathfrak{k}_0 \to \mathfrak{k}_0$ component of the bracket is induced from the Schouten bracket of polyvectors. The polyvectors which appear have odd degree, whence they are in the even subspace of the Lie superalgebra. -A similar construction works also in riemannian geometry and is related to a geometric construction of certain exceptional Lie superalgebras such as $F_4$ and $E_8$. Although in those constructions only vector fields were considered in $\mathfrak{k}_0$ and it would be interesting if there were an extension involving higher-rank polyvector fields.<|endoftext|> -TITLE: BRST cohomology -QUESTION [25 upvotes]: I am reading some work on Mirror Symmetry from Physics perspective,the physicists seem to use some aspects of BRST quantization and BRST cohomology. What is BRST Quantization and BRST cohomology, in realm of Mathematics. More precisely, -What are the BRST complexes, how do we get this cohomology, what is the relation (if any) of this cohomology theory to say de Rham or Cech cohomology. Where do the mathematicians use this theory. What is it in context of $N =2 $ Superconformal field Theory, most relevant for Mirror Symmetry. -I appreciate both physics and mathematical ideas. - -REPLY [24 votes]: Stepping back from the 'oid and Theo's fascinating answer, I thought I would try to complement it -- after all, the title of my PhD thesis (PDF file) did contain the phrase BRST cohomology. It also allows me to expand on my answer to this earlier MO question. -For the impatient let me start with a slogan: - -BRST is symplectic reduction - -Of course, like all slogans this is an oversimplification, but I hope that it gets the main point across. -Let's consider for definiteness a finite-dimensional symplectic manifold $(M,\omega)$ and a connect Lie group $G$ acting on $M$ via symplectomorphisms. Let us assume, for simplicity, that for every $X \in \mathfrak{g}$, the Lie algebra of $G$, the fundamental vector field on $M$ is hamiltonian and moreover that the map $\mathfrak{g} \to C^\infty(M)$ is a Lie algebra homomorphism, where the Lie algebra structure on $C^\infty(M)$ is given by the Poisson bracket. In other words, the action of $G$ gives rise to an equivariant moment mapping $\Phi: M \to \mathfrak{g}^*$. -Symplectic reduction is a procedure by which we go from $(M,\omega)$ to a lower-dimensional symplectic manifold $(\tilde M, \tilde \omega)$ and it consists of two steps: - -we restrict to the submanifold $M_0 = \Phi^{-1}(0)$, i.e., the zero-momentum submanifold, and -we descend to the quotient $\tilde M = M_0/G$ with the induced symplectic structure $\tilde \omega$ defined by -$$ \pi^* \tilde \omega = \imath^*\omega $$ -where $\pi: M_0 \to \tilde M$ and $\imath: M_0 \to M$ are the natural projection and embedding, respectively. - -(This assumes for simplicity that $0$ is a regular value of the moment map and that $G$ acts on $M_0$ in such a way that the quotient is smooth.) -In the simplest case, BRST cohomology is a cohomology theory which yields $C^\infty(\tilde M)$ from $C^\infty(M)$. It is the cohomology of the total differential of a double complex consisting horizontally of the Koszul complex which defines a resolution of $C^\infty(M_0)$ in terms of free $C^\infty(M)$-modules, and vertically of the Chevalley--Eilenberg complex computing the cohomology of the Lie algebra $\mathfrak{g}$ with coefficients in the $C^\infty(M)$-modules in the Koszul resolution. The main theorem is the isomorphism -$$H^0_{\mathrm{BRST}}(C^\infty(M)) \cong C^\infty(\tilde M)$$ -as Poisson algebras. -This admits a number of generalisations: to not-necessaarily finite-dimensional Poisson manifolds and where there is no group action but just a coisotropic distribution; what in the Physics literature goes by the name of first-class constraints, a nomenclature due to Dirac. -The real power of BRST cohomology lies in the fact that the BRST complex is a complex of Poisson algebras and the differential is an inner Poisson derivation. This means that it is amenable to quantisation and that is the use to which it is put in the literature mentioned by the OP. -The problem is to quantise a constrained hamiltonian system $(M,\omega)$ whose reduced "phase space" is the symplectic quotient $(\tilde M, \tilde\omega)$. This means that the physical dynamics are those in $(\tilde M, \tilde \omega)$ and hence that is the classical system one ought to be quantising. Alas, this is usually difficult: firstly, quantisation is intrinsically difficult but also $(\tilde M, \tilde \omega)$ is usually geometrically complicated. BRST gives a way to do this by quantising the usually simpler system $(M,\omega)$ and then take the quantum BRST cohomology: -$$ -\matrix{C^\infty(M) & \stackrel{\mathrm{quant}}{\longrightarrow} & \mathcal{H}\cr -\downarrow & & \downarrow\cr -C^\infty(\tilde M) & \stackrel{\mathrm{quant}}{\longrightarrow} & \tilde{\mathcal{H}} \cr} -$$ -where the horizontal arrows are quantisations and the vertical arrows are (zeroth degree) BRST cohomology: classical in the left and quantum in the right. -The virtues of the "north-east" path is that we get to quantise a simpler system and that symmetries are usually preserved. In the Physics literature this goes by the name of "covariant quantisation". This is particularly useful in the quantisation of the bosonic and NSR strings and their brethren.<|endoftext|> -TITLE: Can the unit complex 1-dimensional disc be embedded isometrically into complex euclidean space? -QUESTION [12 upvotes]: Let $D$ denote the unit complex 1-dimensional disc, together with the hyperbolic metric $h_D=\frac{4dz\wedge d\bar{z}}{(1-|z|^2)^2}$of curvature $-1$. By Nash's embedding theorem, we can always embed the disc $D$ real-analytically and isometrically into real Euclidean space ${\mathbb{R}}^n$ for some large $n$. (I think $n=5=2\dim_{\mathbb{R}}D+1$ is sufficient.) -I'm wondering whether we have a complex-analytic embeddeding of the disc $D$ into complex Euclidean space. Consider complex Euclidean space ${\mathbb{C}}^n$ with the standard Euclidean metric $h_{\mathbb{C}^n}=dz_1\wedge d\bar{z_1}+\cdots+dz_n\wedge d\bar{z_n}$. Does there exist a holomorphic map $f:D\to {\mathbb{C}}^n$ for some large $n$ that is at the same time an isometry, i.e. $f^*h_{\mathbb{C}^n}=h_D$? -If we write the map $f$ in terms of coordinates $f=(f_1,\ldots, f_n)$, this question has the an equivalent formulation solely in terms of holomorphic functions. This question is asking whether there are holomorphic functions $f_i:D\to {\mathbb{C}}$ on the disc, such that $\frac{4}{(1-|z|^2)^2}=f_1(z)\overline{f_1(z)}+\cdots + f_n(z)\overline{f_n(z)}$ for all $z\in D$. - -REPLY [13 votes]: The answer is 'no', there is no holomorphic curve in $\mathbb{C}^n$ (for any $n$) such that the induced metric has constant negative curvature. To my knowledge, this was first proved by E. Calabi many many years ago, essentially using the structure equations for holomorphic curves in $\mathbb{C}^n$. The proof is easy, but it involves knowing something about the structure equations, and I don't want to try to explain that here. A reasonable source for this is Blaine Lawson's Lectures on Minimal Submanifolds, Chapter IV (see Theorem 11). -There is a more general fact, namely that the only minimal surface (let alone holomorphic curve) in $\mathbb{C}^n$ that has constant Gaussian curvature is a flat plane. This is due to M. Pinl, Minimalfl\"achen fester Gau{\ss}chen Kr\"ummung, Math. Ann. 136 (1958), 34-40.<|endoftext|> -TITLE: Furstenberg's Conjecture on 2-3-invariant continuous probability measures on the circle -QUESTION [29 upvotes]: Hillel Furstenberg conjectured that the only $2$-$3$-invariant probability measure on the circle without atoms is the Lebesgue measure. More precisely: - -Question: (Furstenberg) Let $\mu$ be a continuous probability measure on the circle such that - $$\int_{S^1} f(z) d\mu = \int_{S^1} f(z^2) d\mu = \int_{S^1} f(z^3) d\mu, \quad \forall f \in C(S^1).$$ Is $\mu$ the Lebesgue measure? - -The assumption implies that the Fourier coefficients $\hat\mu(n)$ satisfy $$\hat\mu(n) = \hat \mu(2^k3^ln), \quad \forall k,l \in \mathbb N, n \in \mathbb Z.$$ Furstenberg's question is known to have an affirmative answer if one makes additional assumptions on the entropy of the measure. -The basic strategy is usually to show that a non-vanishing (non-trivial) Fourier coefficient implies the existence of an atom. The standard tool to construct atoms in a measure on the circle is Wiener's Lemma, which says that as soon as there exists $\delta>0$ such that the set -$\lbrace n \in \mathbb Z \mid |\hat\mu(n)| \geq \delta \rbrace$ has positive density in $\mathbb Z$, $\mu$ has an atom. More precisely, the following identity holds: -$$\sum_{x \in S^1} \mu(\lbrace x \rbrace)^2 = \lim_{n \to \infty} \frac1{2n+1} \sum_{k=-n}^n |\hat \mu(k)|^2.$$ -Clearly, $$|\lbrace 2^k3^l \mid k,l \in \mathbb N \rbrace \cap [-n,n] | \sim (\log n)^2 $$ so that Wiener's Lemma can not be applied directly. My question is basically, whether this problem can be overcome for any other subsemigroup of $\mathbb N$. - -Question: Is there any subsemigroup $S \subset \mathbb N$ of zero density known, such that every $S$-invariant continuous probability measure on $S^1$ is the Lebesgue measure? What about the subsemigroup generated by $2,3$ and $5$? - -REPLY [21 votes]: Manfred Einsiedler and Alexander Fish have a paper (arxiv.org/abs/0804.3586) showing that a multiplicative subsemigroup of $\mathbb N$ which is not too sparse satisfies the desired measure classification. The semigroups they consider are still somewhat large, and in particular not contained in finitely generated semigroups. -One can check that the set of perfect squares $E$ satisfies measure classification as follows: since $\frac{1}{N}\sum_{n=1}^N \exp(2\pi i n^2 \alpha)\to 0$ for every irrational $\alpha,$ one can conclude as in Wiener's lemma that an atomless measure $\mu$ on $S^1$ satisfies $\lim_{N\to \infty} \frac{1}{N}\sum_{n=1}^N|\widehat{\mu}(n^2)|=0.$ As in the original post, it follows that an atomless $E$-invariant measure is Lebesgue.<|endoftext|> -TITLE: Group cohomology and cohomology in non-abelian categories -QUESTION [5 upvotes]: One defines the $H^n(G,M)$ where $M$ is a $\mathbb{Z}[G]$ module as $Ext^n_{\mathbb{Z}[G]}(\mathbb{Z},M)$ where $\mathbb{Z}$ is viewed as a trivial $\mathbb{Z}[G]$-module. -Is this part of a general pattern for how to define cohomology for non-abelian categories? -In groups we see that we switch to the abelian category of $\mathbb{Z}[G]$-modules. If this idea is indeed extended for general non-abelian categories, what abelian category do we switch to? - -REPLY [7 votes]: Yes, this generalizes to Hochschild cohomology and André-Quillen cohomology. -Given a category $\mathcal{C}$ with finite limits and an object $X$, you can form the category of Beck modules $\operatorname{Ab}(\mathcal{C} / X)$, abelian group objects in the slice category over $X$. When $\mathcal{C}$ is the category of groups, a Beck module over $G$ is precisely a split extension of $G$ with abelian kernel, so Beck modules can be identified with $\mathbb{Z}[G]$-modules. -In nice cases, the category $\operatorname{Ab}(\mathcal{C} / X)$ is abelian, and the forgetful functor $\operatorname{Ab}(\mathcal{C} / X) \to \mathcal{C} / X$ has a left adjoint, called abelianization $\operatorname{Ab}_X$. The Hochschild cohomology of a Beck module $M$ is defined to be $HH^i(X; M) = \operatorname{Ext}^i(\operatorname{Ab}_X X, M)$. In the groups case, $\operatorname{Ab}_G G$ is the augmentation ideal, so Hochschild cohomology is just group cohomology shifted by 1. -Hochschild cohomology itself is only an approximation to André-Quillen cohomology, which is somehow a more homotopically correct notion (I don't know enough about any of this to explain what this means, exactly). In the case of groups, these two cohomology theories coincide. I suggest having a look at Martin Frankland's thesis, which gives a good overview of these things. It works out explicit examples (groups, abelian groups, associative algebras, commutative algebras) very nicely in Appendix A.<|endoftext|> -TITLE: What recent discoveries have amateur mathematicians made? -QUESTION [146 upvotes]: E.T. Bell called Fermat the Prince of Amateurs. One hundred years ago Ramanujan amazed the mathematical world. In between were many important amateurs and mathematicians off the beaten path, but what about the last one hundred years? Is it still possible for an amateur to make a significant contribution to mathematics? Can anyone cite examples of important works done by amateur mathematicians in the last one hundred years? -For a definition of amateur: - -I think that to make the term "amateur" meaningful, it should mean someone who has had no formal instruction in mathematics past undergraduate school and does not maintain any sort of professional connection with mathematicians in the research world. – Harry Gindi - -REPLY [2 votes]: The dentist Leon Bankoff, who died in 1997, published several mathematical papers. He has an Erdos number of 1. He has Wikipedia page. Here is a link to an interview of him by G.L. Alexanderson. -http://math.fau.edu/yiu/AEG2013/BankoffCMJ.pdf<|endoftext|> -TITLE: Rooks on a lifeline -QUESTION [7 upvotes]: The short version of this question is: - -If $G$ is a graph whose nodes are associated with squares of a chessboard, such that no two nodes in the same row or column of the board are adjacent, we want to associate rooks with the vertices of $G$, such that at most one rook appears in each row and column of the chessboard under the constraint that the vertices containing the rooks induce a connected subgraph of $G$ (thus, the rooks are connected to each other with a lifeline, or lifegraph if you want to be specific). -A maximal configuration of rooks is such that no rooks can be added to the chessboard without violating the constraint that each column/row contain only one rook. -Question: A back-tracking depth first search will find all maximal configutions. Will a back-tracking breadth-first search do the same? - -Let's consider an $m \times n$ chessboard that will be inhabited by rooks. As is usual with chess problems in graph theory, each square is represented by a vertex: let $v_{s,t}$ represent the square at row $s$ and column $t$ of the board. Now, let $G$ be an arbitrary graph on the set of vertices $V$ that comprise the squares of the chessboard. -We want to fill the chessboard with rooks (said another way: we want to associate rooks with vertices), such that: - -Every row and every column contain at most one rook (more formally, if a rook is associated with a vertex $v_{i,j}$, then no rook will be associated with $v_{i,s}$ for $s \in \{1,\ldots,n\}$) or $v_{t,j}$ for $t \in \{1,\ldots,m\}$), -The vertices with rooks must induce a connected subgraph of $G$ (hence the reference to a "lifeline" that must connect all rooks). - -An example of a chessboard with a single rook (indicated by the black vertex) is shown here - -The gray squares show the area covered by the rook - no other rook can be placed on any of the gray squares. Note that I have neglected to color the squares themselves black and white as they should appear on a real chessboard. Edges are indicated by red lines. -Let's consider how we might extend the neighborhood of the black node - call it $v$. There are three vertices adjacent to $v$, but we can only place rooks on at most two of them at a time without violating the constraint that only a single rook cover a given row/column. In fact, there are two ways of placing the rooks, as shown here: - -and here (broken link). -A maximal configuration is a valid placement of rooks (according to our constraints) that cannot be extended by the addition of another rook without violating the constraints. -We can enumerate all maximal configurations by performing a back-tracking depth-first search from each node (i.e. each square on the chessboard). With a depth-first search, we only add one rook to the chessboard at a time. - -Suppose that we perform a back-tracking breadth-first search instead. At each step, we add as many rooks the board as possible. Of course there possibly are many different ways of adding as many rooks as possible at each step. This is exactly what is done in the two images above: the maximum number of rooks are added in both possible configurations. -Will this strategy also enumerate all maximal configurations? - -REPLY [6 votes]: The answer is that you have gotten too greedy! I give my squares coordinates in R^2. Place vertices on the points (z,z) for integers z with absolute value less than 100 and also at (-2,2), (1,2) and (-1,-2). Place edges between (z,z) and (z+1,z+1) for all z and also place edges between ((-2,2) and (0,0)) , ((-1,-2) and (0,0)) , ((-1,-2) and (1,1)), ((2,1) and (0,0)), ((2,1) and (-1,-1)). So 3 non diagonal vertices and 5 non diagonal edges. -Obviously the best we can do is place rooks on the diagonal from (-100,-100) to (100,100). -Now if you start your search at a vertex on the diagonal at or left of (-1,-1) (resp. at or right of (1,1)) then you won't get to the other side of the diagonal since at (-1,-1) (resp (1,1)) you take the edge to (2,1) (resp (-1,-2)) which cuts you off from (2,2) (resp (-2,-2)). That also explains why you can't start at (2,1) or (-1,-2). If you start at (-2,2) you are cut off from both diagonals as you can't cross (-2,-2) and (2,2). But if you start at the origin, you take the edge to (2,2) which shows you can't start at the origin either. -Thus you got too greedy. Close though, and interesting.<|endoftext|> -TITLE: How do you describe vector bundles on elliptic curves? -QUESTION [18 upvotes]: Throughout "curve" means smooth projective curve over an algebraically closed field. -Motivation and Background -I read somewhere that Atiyah has classified vector bundles on elliptic curves. My understanding is that the story is roughly: every vector bundles breaks up as a direct some of indecomposable vector bundles. The indecomposable vector bundles are further divided by their degree and rank. The set $Ind(d,r)$ of isomorphism classes of indecomposable vector bundles of a fixed degree $d$ and rank $r$ has the structure of variety isomorphic to the Jacobian of the curve; namely the curve itself. In fact $Ind(d,r) \cong Ind(0, gcd(d,r) )$ so its enough to consider the degree $0$ vector bundles. -Now for $V,V'$ vector bundles on any curve $C$ its true that $\deg V \otimes V' = \deg V\cdot rk(V') + \deg V'\cdot rk(V)$. So in the case of an elliptic curve the set $Ind(0,r)$ is a torsor for $Pic^0(C)$. Additionally, there is a unique isomorphism class $V_r \in Ind(0,r)$ characterized by the fact that $h^0(V_r) \ne 0$ (in fact $h^0(V_r)=1$). -The Question -Since curves of low genus are usually "simple enough" to easily describe explicitly, I would like to see how explicit I can be with a description of the $V_r$, or at least $V_2$. So my question is simply, given an explicit elliptic curve in $\mathbb{P}^2$, say $zy^2 - x(x-z)(x+z)$can you reasonably show how to construct $V_2$; i.e. give cocyles for it $\phi_{ij} \colon U_{ij} \to GL_2(k)$, or produce a graded module for it? -Initial Thoughts -$V_2$ should correspond to the unique nontrivial extension in $Ext^1(\mathcal{O}, \mathcal{O}) \cong H^1(\mathcal{O}) \cong k$.On a curve $C$, $0 \to \mathcal{O}_C \to K(C) \to K(C)/\mathcal{O}_C \to 0$ is a flasque resolution of the structure sheaf and an element in $H^1(\mathcal{O}_C)$ corresponds to a map $\alpha \colon \mathcal{O}_C \to K(C)/\mathcal{O}_C$. Then the desired extension should be the pullback of $0 \to \mathcal{O}_C \to K(C) \to K(C)/\mathcal{O}_C \to 0$ via $\alpha$. -The trouble with this is that I can't seem to pin down what $\alpha$ is. I now that, via Serre duality, it corresponds to a global section of $H^0(\omega_C)$ which I can explicitly describe as a differential on the curve. Using the example I mentioned above, on the affine patch where $z \ne 0$, it is: $\frac{dx}{2y} = -\frac{dy}{3x^2-1}$ but I guess I don't understand Serre duality well enough to determine $\alpha$ from this. Also I'm not even sure if this will lead to a reasonable way of getting at the cocylces of $V_2$. And this is so close to "just using the definitions" I have to imagine there might be a better way to go about this... - -REPLY [29 votes]: A description via cocycle usually is not convenient to work with. From my point of view, a description as an extension is much more useful. But if you want something else, I would advice the following. Since your curve $C$ is given as a double covering of $P^1$, that is as a relative spectrum of the sheaf of algebras $A = O \oplus O(-2)$ on $P^1$ (with $O$ summand being generated by the unit and with multiplication $O(-2)\otimes O(-2) \to O$ given by the ramification divisor $D$), the category of coherent sheaves on $C$ is identified with the category of sheaves on $P^1$ with a structure of a module over this algebra. -Spelling this out, a sheaf on $C$ is a sheaf $F$ on $P^1$ with a morphism $\phi:F(-2) \to F$ such that the composition $\phi\circ\phi(-2):F(-4) \to F(-2) \to F$ coincides with the morphism given by $D$. The structure sheaf $O_C$ corresponds to $A$. Hence the extension of $O_C$ with $O_C$ corresponds to an extension of $A$ with $A$, that is we have an exact sequence -$$ -0 \to O \oplus O(-2) \to F \to O \oplus O(-2) \to 0. -$$ -Since $Ext^1_A(A,A) = Ext^1_O(O,A) = H^1(A) = H^1(O \oplus O(-2))$, we see that the nontrivial extension corresponds to the extension of $O$ in the right term by $O(-2)$ in the left term. Thus $F = O \oplus O(-1)^2 \oplus O(-2)$ and the above sequence can be rewritten as -$$ -0 \to O \oplus O(-2) \to O \oplus O(-1) \oplus O(-1) \oplus O(-2) \to O \oplus O(-2) \to 0 -$$ -The maps are given by matrices $\left[\begin{smallmatrix}1 & 0\cr 0 & x\cr 0 & y\cr 0 & 0 \end{smallmatrix}\right]$ and $\left[\begin{smallmatrix} 0 & y & -x & 0 \cr 0 & 0 & 0 & 1 \end{smallmatrix}\right]$. The map $\phi:F(-2) \to F$ providing $F$ with a structure of an $A$-module is given by the matrix $\left[\begin{smallmatrix} 0 & g & h & -f \cr x & xy & -x^2 & h \cr y & y^2 & -xy & -g \cr 0 & y & -x & 0 \end{smallmatrix}\right]$. Here $f$ is the polynomial in $x$ and $y$ of degree $4$ such that $div(f) = D$, and $g,h$ are polynomials such that $f = gx +hy$.<|endoftext|> -TITLE: Commutator subgroup does not consist only of commutators? -QUESTION [25 upvotes]: Let $G$ be a group, $G'=[G, G]$. - -"Note that it is not necessarily true that the commutator subgroup - $G'$ of $G$ consists entirely of - commutators $[x, y], x, y \in G$ (see [107] for some finite group examples)." - -Quoted from http://www.math.ucdavis.edu/~kapovich/EPR/ggt.pdf page 8. -Anybody can provide the examples? I can't find the book [107]. - -REPLY [4 votes]: Sorry for adding one more answer, but here's a simple argument based on using that if the center is large enough then the commutator map has too small image to cover the derived subgroup, and even showing that, in suitable (varying) finite groups, the commutator length can be unbounded. -Let $p$ be an odd prime. Consider the free 2-step-nilpotent Lie algebra over $\mathbf{Z}/p\mathbf{Z}$ on $n$ generators. It has dimension $n+n(n-1)/2$, and its center has dimension $n(n-1)/2$ and coincides with the derived subalgebra. -The law $x\ast y=x+y+\frac12[x,y]$ defines a group law (this is the Baker-Campbell-Hausdorff formula), defining a group $G$ of order $p^{n+n(n-1)/2}$ (and exponent $p$). -The center $Z$ of $G$ has order $p^{n(n-1)/2}$ and coincides with the derived subgroup. The commutator map $G\times G\to [G,G]$ factors through $G/Z\times G/Z$, and hence its image contains at most $p^{2n}=|G/Z|^2$ elements. As soon as $2n -TITLE: Geometry and Integrability in Other Bundles -QUESTION [5 upvotes]: Background: Suppose $E=TM$ is the tangent bundle to some differentiable manifold $M^n$. If we specify some subbundle $D\subset TM$ (distribution of $k$-planes) then there are two natural situations that arise. We may have that at some point $p$ there is an immersed submanifold $N^k\subset M^n$ passing through $p$ such that for all $q\in N$, $T_q N=D_q$. If this is true then we say $D$ is a integrable at $p$. Otherwise it is non-intregrable. -Frobenius Integrability states that every involutive distribution is completely integrable, i.e. $M$ is foliated by integral manifolds. One way of phrasing involutivity is to require that any lie bracket of vector fields lying in $D$ stay in $D$. There is another version, which is the one that I prefer, using differential forms. Note that any $k$-distribution is cut out locally by $n-k$ independent 1-forms $\theta_1,\ldots,\theta_{n-k}$. Call $\Theta$ the ideal generated by these, then we say $\Theta$ is involutive if $d\Theta\subset \Theta$,i.e. it is a differential ideal. -Situation: This last version generalizes to arbitrary vector bundles (with connection) easily. Suppose $E$ is a rank $N>n$ vector bundle over $M^n$. Then I can specify any sub-vector-bundle $D\subset E$ by -$$D_x:=\lbrace v\in E_x | \theta_1(v)=\ldots=\theta_{n-k}(v)=0\rbrace $$ -for some collection of 1-forms, i.e. sections of the dual bundle $E^*$. I can extend the above definition of involutive distribution to this subbundle in the obvious way. Let me take this as a definition of integrable subbundle. -Question: If the geometric concept associated to an integrable subbundle of $TM$ is a foliation, what is the geometric concept associated to an integrable subbundle of $E$? -I am only beginning to dig into exterior differential systems, so any well articulated answer/exposition is appreciated. -EDIT: The differential one gets from a connection actually just lands in vector valued forms, i.e. the Twisted de Rham complex. So the complex one would like to get (which looks like a Koszul complex possibly) is not obtained in a canonical way with a connection. Other differential operators would be needed. - -REPLY [5 votes]: As it was already pointed out, on a bare vector bundle there is no intrinsic notion of "integrability". However, things change when you pass to a Lie algebroid: In this case the vector bundle $E$ is equipped with two extra things: a Lie bracket for the smooth secitons $\Gamma^\infty(E)$ and an "anchor" map $\rho\colon E \longrightarrow TM$ which is a bundle map such that it yields a Leibniz rule for the two Lie brackets involveld: the one on $E$ and the canonical one on $TM$. -The tangent bundle itself is clearly an example for a Lie algebroid and so is a bundle of Lie algebras etc. Yet another example comes from Poisson geometry: any Poisson tensor on $M$ makes $T^*$ a Lie algebroid. -Now there are many notions of integrabilty associated to this structure: first, the image of the anchor map $\rho$ is a involutive distribution inside $TM$, now of course the rank may vary. Nevertheless there is a good notion of integrability (Stefan-Sussman). But you can also ask yourself whether a Lie algebroid "integrates" to something called a Lie groupoid in a similar way as a Lie algebra integrates to its simply-connected connected Lie group. This is a by far more complicated question which has ultimately been solved by Crainic and Fernandes only a couple of years ago. -Yet another instance where we encounter "integrability" for a general vector bundle is the generalized geometry: here you start with a vector bundle $E$ which is equipped with the extra structure of what is called a Courant algebroid. This is a maximally indefinite fibre metric, a bracket for the sections, and an achor map. Now the axioms are rather funny (weird) but $E = TM \oplus T^*M$ provides an example. Inside a Courant algebroid you can ask for maximally isotropic subbundles $L \subseteq E$ which are closed under the bracket ("integrable"!) These are called Dirac structures and play an important role in many recent applications in geometric mechanics of constraint systems, generalized geometries a la Hitchin, Poisson geometry etc. -The literature is waste and a little googling in the arXive will provide more than you want :)<|endoftext|> -TITLE: Loop spaces and infinite braids -QUESTION [16 upvotes]: The Artin braid groups $B_n$ and the symmetric groups $S_n$ are closely related by the maps $1 \to P_n \to B_n \to S_n \to 1$. The infinite symmetric group has interesting interactions with homotopy theory, due to a result of Barratt-Priddy(-Segal(-Quillen(-others))) that "identifies" the sphere spectrum $QS^0$ with $S_{\infty}$. - -In light of the short exact sequence above, are there any Barrat-Priddy-esque results on the infinite braid group $B_{\infty}?$ Is there even a loop space structure on $B_{\infty}?$ -Ditto for the pure braids $P_{\infty}$, the kernal of $B_{\infty} \to S_{\infty}$. - -REPLY [17 votes]: Yes, $B \beta_\infty$ is homology equivalent to $\Omega^2_0 S^2$, the zero component of the double loop space of $S^2$. The map $B_\infty \to S_\infty$ induces the obvious stablisation map $\Omega^2_0 S^2 \to Q_0S^0$.<|endoftext|> -TITLE: is there any way to bound the number of CM points by height functions? -QUESTION [9 upvotes]: It is known that if $X$ is a curve over a number field $F$ equipped with a flat regular model over $O_F$ the integer ring, one can define, using a suitable ample line bundle with an Hermitian metric, the notion of height for points in $X$. In particular, there are only finitely many algebraic points with height lower than a given constant. -What happens when one studies the heights of CM points on the compactified modular curve, with respect to the ample line bundle defining modular forms of large weight (plus suitable Hermitian metric)? Is it possible to study the distribution of CM cycles through height functions? Does one have similar results as in the previous example, say bounding the number of CM points by heights? -This is also motivated by the Bogomolov conjecture. Roughly speaking, for $A$ an abelian variety over a number field, and $Z\subset A$ an irreducible closed subvariety that is not a torsion subvariety (abelian subvariety translated by a torsion point), then the "number" (or some other numerical invariants) of torsion subvarieties in $Z$ is bounded in terms of suitable constants in heights, degrees, that only depends on $Z$. It might be of interest to work out an analogue for modular curves and perhaps other moduli spaces like Shimura varieties. -Excuse me for the vague statements above and thanks in advance for comments and references. - -REPLY [2 votes]: For your titular question, Beats me. Personally I'm not aware of anyone who's studied the distribution of CM points with respect to height in the way you describe. -What I have seen papers that study the distribution of CM points with respect to things other than height and papers that look at the height of CM points (as well as papers that say quite a lot about the Faltings height of a CM abelian variety). -Here are a few different papers of note on those topics: -Equidistribution of CM points: -http://www.math.ucla.edu/~wdduke/preprints/modud.pdf - Gives an amazing asymptotic result on the trace of a CM $j$-invariant (i.e. a result on $X(1)$). -http://www.math.columbia.edu/~szhang/papers/ZhangIMRN.pdf - Gives an analogue on more general modular curves and quaternionic Shimura Curves as well as a connection to the Andre-Oort conjecture -Also good for an overview is the "Equidistribution in Number Theory" volume edited by Granville and Rudnick http://www.springer.com/mathematics/numbers/book/978-1-4020-5403-7 -Heights of CM points/cycles: -http://www.math.columbia.edu/~szhang/papers/HCMI.pdf - Heights of CM Points I--The Gross-Zagier formula -http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.32.4204 - This attempts to identify affine elliptic modular curves by the presence of CM points of large enough height(logarithmic height on $\overline{\mathbf{Q}}$). -I'd like to emphasize that this does not pretend to be a complete reference list and I don't pretend to know much about the theory of heights. Rather this is a smattering of exciting ideas in the area.<|endoftext|> -TITLE: A property of continuous maps with respect to compact subsets -QUESTION [6 upvotes]: I'm interested in continuous maps between topological spaces $f:X\to Y$ such that for any compact subset $L$ of $Y$ contained in $f(X)$, there is a compact subset $K$ of $X$ such that $L$ is contained in $f(K)$. -Proper maps satisfy this, but there are examples of continuous maps which don't, for example with discrete spaces, taking a non-stationary convergent sequence extended at infinity. -I would like to know if there are characterizations for those topological spaces which have enough compact subsets in the sense that: each real-valued continuous function satisfy this property. - -REPLY [3 votes]: Such maps are called compact-covering maps, and are a somewhat well-known and well studied object. They came up naturally in many contexts, and if you look for that keyword in mathscinet, you will find many matches. As Bruno mentions, it is very easy to construct examples of maps in very ordinary settings that fail to be CC.<|endoftext|> -TITLE: Does there exist a shot in ideal pocket billiards? -QUESTION [33 upvotes]: Assume you have one shot with the cue ball in pocket billiards (a.k.a. pool), with -the game idealized in that no spin is placed on the cue ball in -the initial shot, all collisions between billiard balls, and balls with the cushions, -are perfectly elastic, and there is arbitrarily small felt-ball rolling friction so that -reflections may be assumed to continue as long as necessary. - -Q1. Is there always some direction in which to shoot the cue ball so that - some ball goes into a pocket (is sunk), and although other balls also may - be sunk, the cue ball itself is not (so there is no scratch)? - - - - -My sense is that the answer should be 'Yes' because irrationally angled -paths will eventually visit pockets. But the presence of up to 15 balls, -and the requirement that the cue -ball not scratch, seems to leave open the possibility that there -is some initial configuration of the balls that thwarts every possible shot. -It might be necessary to assume that the cue ball is initially not -touching any other ball? -Two other variations: - -Q2. Is there always a shot to sink a particular ball $B^*$? -Q3. Is there always a shot to sink a particular ball $B^*$ and no other ball? - -The latter might be too -strong, as all six pockets could conceivably be "protected" by non-$B^*$ balls. - -REPLY [11 votes]: How about the following? It suggests the answer to Q1 is no. -The 2,15,6,9 and cueball are in a horizontal row and touching (or very nearly touching). The 14,cueball and 4 are in a vertical row and (very nearly) touching. -When the cueball is hit, one or both of the following will happen (depending on which direction the cueball is hit). - -The cueball moves horizontally to the right. -The 4 moves vertically downwards, hitting the 8-ball, which travels down the table, off the bottom cushion and back up the table, off the left cushion and hits the 15 head on and slightly from the left (the angle in my illustration is probably not accurate though). The 8 and 15 will come back down the table, while the cueball is knocked horizontally to the right along with the 9 and, possibly, the 6. - -In any case, the white should go in the corner pocket off the 3. - -[Edit: The layout has been modified since the initial revision, but it is still the same idea.] -We need to be careful here and check that something unintended doesn't happen, like the 8,15 or 4 going in a pocket before the white. This seems unlikely and, in any case, we can put extra balls on the table to block their path if necessary. -There is still one problem though. If the cueball is hit at a specific angle, the white will start moving horizontally towards the 3, and the 9 will follow it shortly afterwards. What happens if, just after the cueball deflects off the 3, the 9 catches up and hits it? Maybe this is enough to deflect it away from the pocket? I'm not sure about this. Maybe the setup can be simplified a bit. -[The pool table above was copied from Wikimedia commons] - -Here's a simpler example. The white ball is hemmed in by the top-right corner pocket. All you can do is to either hit it straight into the pocket, or against the 7 or 14. If the 7 is hit, it rolls along the top cushion into the 3, which bounces off the left cushion into the 12, which rolls towards towards the top right corner, pocketing the white. -Similarly, hitting the 14 will knock the 9 towards the bottom cushion and back up into the 8, which again pockets the white.<|endoftext|> -TITLE: Cardinality of classes -QUESTION [23 upvotes]: I am trying to define an embedding whose range includes classes. Is there a coherent way of assigning "cardinality" to proper classes? - -REPLY [20 votes]: Erin, there is no need to do this. I do not know of any practical reasons for doing it. And, of course, "cardinality" has to be properly interpreted to make some sense of the word. -In extensions of set theory where classes are allowed (not just formally as in ZFC, but as actual objects as in MK or GB), sometimes it is suggested to add an axiom (due to Von Neumann, I believe) stating that any two classes are in bijection with one another. Under this axiom, the "cardinality" of a proper class would be ORD, the class of all ordinals. (By the way, by class forcing, given any proper class, one can add a bijection between the class and ORD without adding sets, so this assumption bears no implications for set theory proper.) -Without assuming Von Neumann's axiom, or the axiom of choice, I know of no sensible way of making sense of this notion, as now we could have some proper classes that are "thinner" than others, or even incomparable. Of course, we could study models where this happens (for example, work in ZF, assume there is a strong inaccessible $\kappa$, and consider $V_\kappa$ as the universe of sets, and $Def(V_\kappa)$ in Gödel's sense (or even $V_{\kappa+1}$) as the collection of classes).<|endoftext|> -TITLE: Upper bound for size of subsets of a finite group that contains a sum-full set -QUESTION [7 upvotes]: Problem -I'm looking for an upper bound for the number $k(G)$ of a finite group $G$, defined as follow: - -Let $\mathcal{F}_k$ be the family of subsets of $G$ with size $k$, and we - define $k(G)$ be the minimum $k$ such that every subset $X \in \mathcal F_k$ - contains a non-empty sum-full set $S$, which is a set satisfies - $$ S \subseteq S+S := \{ x+y \mid x,y \in S \}. $$ - -Note that the inequality $k(G) \leq |G|$ holds trivially since there is only one -subset in $\mathcal F_{|G|}$ which is $G$ itself, and $G$ is a semigroup indeed. -Are there any papers or references about this number $k(G)$? Does it have a name? I'm interesting in particularly upper bounds of $k(G)$, but any related results are fine. - -Motivation -The restricted Davenport number $\hat{D}(G)$ of a group $G$, is defined as the smallest number $d$ such that given a subset $A \in \mathcal F_d$, there exists a zero-sum non-empty subset $S \subseteq A$, that is, -$$ \sum_{x \in S} x = 0, $$ -where $0$ is the identity in $G$. -In the paper "On a conjecture of Erdos and Heilbronn", Szemeredi has proved: -$$\hat{D}(G) = O(\sqrt{|G|}). $$ -Hamidoune and Zemor set a precise bound $\sqrt{2}$ on the constant of the big-O notation. -I'm trying to provide a link between $\hat{D}(G)$ and the number $k(G)$; it seems to me that the size of sum-full sets in $G$ may related to the zero-sum problem. I'll provide the justification in another post, which is highly related. - -REPLY [4 votes]: First of all the $k(G)$ cannot be smaller than the size of any proper subgroup of $G$, because if $H$ is a proper subgroup, $gH$ is a coset, $g\not\in H$, then $gHgH$ does not intersect $gH$ (if $ghgh'=gh''$, then $g\in H$). For example, if $G$ is Abelian, $|G|$ is not prime (i.e. $G$ is not cyclic of prime order), then $k(G)>\sqrt{|G|}$: look at the size of a maximal proper subgroup.<|endoftext|> -TITLE: Difference between turnstile and implication -QUESTION [9 upvotes]: Does anyone know the difference between proving that -|- phi ------------------- -|- ( psi -> phi ) - -and proving that -|- phi -> ( psi -> phi) ? - -REPLY [3 votes]: This is not an answer to the question intended by OP, but rather a philosophical meta-answer and might be interesting to some readers. -The use of sign $\vdash$ is completely different from $\rightarrow$. You can not prove -"$\varphi \vdash \psi$", since it is not a proposition. It is a speech-act, it is invented by Frege and has been discussed extensively in philosophy of mathematics and philosophy of language. This speech-act is called assertion and is composed of two separate parts "|" and "-". It is an act of judgment. For more details see this SEP article. This distinction between a proposition and a judgment is very important and essential for Martin-Lof's type theory and philosophy. -In fact, from this viewpoint, $\varphi \vdash \psi$ is a common formal misuse because the assumptions cannot be before the speech-act. For this reason Martin-Lof prefers to write the assumptions of the assertion after the proposition in his type theories. -For difference between $\vdash$ and $\rightarrow$ see this article.<|endoftext|> -TITLE: Most memorable titles -QUESTION [172 upvotes]: Given the vast number of new papers / preprints that hit the internet everyday, one factor that may help papers stand out for a broader, though possibly more casual, audience is their title. This view was my motivation for asking this question almost 7 years ago (wow!), and it remains equally true today (those who subscribe to arXiv feeds, MO feeds, etc., may agree). - -I was wondering if the MO-users would be willing to share their wisdom with me on what makes the title of a paper memorable for them; or perhaps just cite an example of title they find memorable? -This advice would be very helpful in helping me (and perhaps others) in designing better, more informative titles (not only for papers, but also for example, for MO questions). -One title that I find memorable is: - -Nineteen dubious ways to compute the exponential of a matrix, by Moler and van Loan. - - -The response to this question has been quite huge. So, what have I learned from it? A few things at least. Here is my summary of the obvious: Amongst the various "memorable" titles reported, some of the following are true: - -A title can be memorable, attractive, or even both (to oversimplify a bit); -A title becomes truly memorable if the accompanying paper had memorable substance -A title can be attractive even without having memorable material. -To reach the broadest audience, attractive titles are good, though mathematicians might sometimes feel irritated by needlessly cute titles -Titles that are bold, are usually short, have an element of surprise, but do not depart too much from the truth seems to be more attractive in general. -5.101 Mathematical succinctness might appeal to some people---but is perhaps not that memorable for me---so perhaps such titles are attractive, but maybe not memorable. -If you are a bigshot, you can get away with pretty much any title! - -REPLY [3 votes]: On the more applied side of things, I'm quite fond of the following (sub)title: -Estimating the number of unseen species: A bird in the hand is worth $\log n$ in the bush<|endoftext|> -TITLE: compact-open topology -QUESTION [9 upvotes]: Is there a natural reason for defining the compact-open topology on the set of continuous functions between two locally compact spaces. For example "to make ... functions continuous". Or in another way of asking this, is there an adjoint functor of the functor, say F, which assigns the topological space $F(X,Y):=Hom_C(X,Y)$ (with the compact-open topology on it) to the couple X,Y. - -REPLY [20 votes]: I always liked the following reason: -Let's call a topology on a space "admissible" if the evaluation function $e: Hom(X,Y) \times X \rightarrow Y$ is continuous. Then the compact-open topology is coarser than any other admissible topology. In particular, in any case where the compact-open topology is admissible, it is the smallest possible topology that does this. -EDIT: See comments for some references. I don't claim any originality here :)<|endoftext|> -TITLE: Is there an analogue of mathscinet for physics? -QUESTION [10 upvotes]: I've been looking recently at some papers in physics, from journals that are not listed in mathscinet. Is there is a similar database for physics, with reviews and citation links? I'd like to see where the papers I'm currently looking at have been referenced, in order to follow the subject forward in time. - -REPLY [10 votes]: In addition to freely available Google Scholar and SPIRES, and subscription-based Web of Science and Scopus, there is a free NASA Astrophysics Data Systems database which, contrary to its title, appears to have broader scope than SPIRES, at least as far as mathematical physics is concerned; it provides abstracts (but not reviews), citations, and, for some older papers, their full-text scanned versions. Now it is sort of integrated with arXiv.org: when looking at the abstract of any arXiv preprint, you see the link to its citations and references at NASA ADS under References & Citations. This database has, inter alia, a specialized physics and geophysics search engine.<|endoftext|> -TITLE: Notation: Exponent of a group -QUESTION [6 upvotes]: The exponent of a group $G$ is the least positive $n$ such that $g^n = e$ for all $g \in G$. This is obviously a sensible name for the concept. -A notational awkwardness arises however when the group $G$ is abelian and written additively. I find it grating to refer to the least positive $n$ such that $\forall g \in G$ $ng = e$ as the exponent because there is nothing going on that even looks like exponentiation. -Is there an alternate terminology that can be used in this situation? - -REPLY [2 votes]: First, I appologise since this should rather be a comment (but I cannot yet comment). -Yet, in the process of reading this question and some of the linked material I got confused, and since I very frequently use the term exponent for abelian groups denoted additively, and know various people who do likewise, I am quite interested in this, too. - -I looked at the linked content of Fuchs's book but I failed to locate the part that Mark Sapir is referring to. (I have no access to the book at the moment, and thus would be greatful for a precise quote of the relevant part.) -The only thing I found online is that 'periodic' is used a synonym for 'torsion', but no mention of period of a group. -Also in Lang I was unable to find the mentioned usage, it seems Dylan Moreland made the same experience. Indeed, it seems period of an element is used in Lang also when the notation is multiplicative. - -Sorry, for this abuse of 'answer', but I would be grateful for clarification and did not know how to express this except via and 'answer'.<|endoftext|> -TITLE: What is a good example of a complete but not model-complete theory, and why? -QUESTION [10 upvotes]: The standard examples of complete but not model-complete theories seem to be: -- Dense linear orders with endpoints. -- The full theory $\mathrm{Th}(\mathcal{M})$ of $\mathcal{M}$, where $\mathcal{M} = (\mathbb{N}, >)$ is the structure of natural numbers equipped with the relation $>$ (and nothing else, i.e. no addition etc). -Can anyone explain or give a reference to show why any of these two theories are not model-complete, or give another example altogether of a complete but not model complete theory (with explanation)? - -REPLY [3 votes]: A theorem of Chang asserts that any theory which is axiomatized by for-all there exists sentences and is categorical in some infinite power is model complete. -Lindstrom gave an example that completeness does not suffice.<|endoftext|> -TITLE: On Bourbaki's characterization of projectives... -QUESTION [5 upvotes]: In Bourbaki's Commutative Algebra we have the following theorem: -II.5.2 Let $A$ ba a ring and $P$ an $A$-module. TFAE: -(i) $P$ is a f.g. projective module\ -(ii) $P$ is a finitely presented module and, for every maximal ideal $m$ of $A$, $P_m$ is a free $A_m$-module.\ -(iii) $P$ is a f.g. module, for all $p \in$ Spec($A$), the $_p$-module $P_p$ is free and, if we denote its rank by $r_p$, the function $p \mapsto r_p$ is locally constant in the topological space Spec($A$). -There is more to this theorem but I have stated what I need for my question. Namely, unless I am missing something, for (iii) implies (i) don't we need the extra condition that $A$ is a reduced ring? I see their proof and cannot find the problem but Eisenbud suggests that there exists a counterexample in problem 20.13 in Commutative Algebra with a View Toward Algebraic Geometry. I must be missing something! - -REPLY [4 votes]: Dear fishibones, an important point is that at the beginning of Chapter II, Bourbaki states that all the rings he will consider are commutative. This implies that if a free module has finite dimension, this dimension is unambiguously defined. This property is called the Invariant Basis Number property (IBN) and may fail for non-commutative rings (Counterexamples can be found in, say, Lam's Lectures on Modules and Rings, Springer GTM 189.) This unambiguously defined dimension of a finitely generated free module is what Bourbaki calls its rank. -Let me emphasize that Bourbaki only uses "rank" in the above sense i.e. for finitely generated free modules. If $M$ is such a module of rank $r$ over the ring $A$, then for any prime ideal $\frak {p}$$ \in A$ the modules -$M_{\frak p}$ over $A_{\frak {p}}$ and $M_{\frak {p}} \otimes_{A_{\frak p}} \kappa (p)$ over $\kappa (p)$ -also are free of rank $r$. With this definition and use of rank, Bourbaki's Théorème 1 that you mention is perfectly correct (bien sûr!) -There is no contradiction with Eisenbud's exercise: he does not assume that his module $M$ has all its localizations $M_{\frak {p}}$ free over $R_{\frak {p}}$, whereas Bourbaki does. -Eisenbud only assumes that the dimension of the $\kappa ({\frak p})$- vector space $M_{\frak {p}} \otimes_{R_{\frak p}} \kappa (p)$ is locally constant, while Bourbaki does not even mention this dimension when $M_{\frak {p}}$ is not $R_{\frak {p}}$-free.<|endoftext|> -TITLE: How is Fredkin and Toffoli's Conservative Logic related to Linear Logic? -QUESTION [14 upvotes]: In the answers to this question, Timothy Gowers asks: - -I've been interested in this question for some time. I haven't put any serious thought into it, so all I can offer is a further question rather than an answer. (I'm interested in the answers that have already been given though.) My question is this. Is there a system of logic that will allow us to prove only statements that have physical meaning? - -One answer to this question is given by Fredkin and Toffoli's conservative logic, which is an attempt to give a system of digital logic consistent with various abstract physical principles such as conservation laws, reversibility, and one-to-one composition (ie, no unbounded fanout, since signal strength degrades when signals are split). However, to a proof theorist, the constraints they describe sound hauntingly similar to the language used to motivate linear logic. Furthemore, the circuit diagrams they draw look like string diagrams in monoidal categories, which are models of linear logic. -So my own question is, how is conservative logic related to linear logic? - -REPLY [2 votes]: The question seems to be groping for a fancy, specific answer when, in my view, the most important connection is relatively basic and general. -In mathematics, as you say, you have symmetric monoidal categories. A word in a symmetric monoidal category has a set of inputs and a set of outputs and a certain type of labelled graph in between them. Symmetric monoidal categories are used in many situations. They were inspired by multilinear algebra with tensors; I gather that they are also important in linear logic. -In theoretical computer science, there is the notion of a circuit, and the closely related notion of a straight-line program. A circuit is also a labelled acyclic graph with inputs and outputs. Computer scientists began with boolean circuits, but these days there are a lot of study of arithmetic circuits over any ring, as well as quantum circuits composed of unitary gates. A precursor to quantum circuits is the clever definition of reversible and conservative circuits of Toffoli and Fredkin. In the paper cited, they have in mind a dynamical interpretation of graphs that don't have to be acyclic. However, the case of what they did that has had the most influence is finite, acyclic graphs. In this case, their essential insight is that you can have a perfectly good model of boolean circuits based on permutations of $\{0,1\}^n$ rather than functions from $\{0,1\}^a$ to $\{0,1\}^b$. -So the point is that the math concept of monoidal categories, which are broadly important, is basically the same as the CS concept of circuits, which are also broadly important. Every monoidal category, maybe together with some distinguished generators, gives you a new type of circuit, so that you can then ask circuit complexity questions. A particular monoidal category may or may not yield interesting circuit questions, and two different monoidal categories might yield equivalent circuit questions. Nonetheless, there are many interesting different cases. -Actually, not-necessarily-acyclic graphs can also be interpreted as words in symmetric monoidal categories that are also "closed" or "pivotal". There are several ways to interpret tensor words of this type as circuits; one of these ways (for some types of words) is as dynamical circuits. -Linear logic and conservative logic are both called "logic", and they both use monoidal categories. (Also, as Peter Shor mentioned, linear logic is partly inspired by quantum probability, while conservative logic is used in quantum computation.) Other than that, they don't look particularly related to me.<|endoftext|> -TITLE: M24 moonshine for K3 -QUESTION [19 upvotes]: There are recent papers suggesting that the elliptic genus of K3 exhibits moonshine for the Mathieu group $M_{24}$ (http://arXiv.org/pdf/1004.0956). Does anyone know of constructions of $M_{24}$ analogous to the FLM construction of the monster as the automorphism group of a holomorphic $c=24$ CFT (aka VOA)? In particular, the monster has $2^{1+24}. \cdot O/Z_2$ as the centralizer of an involution and the Conway group acts as automorphisms of the 24-dimensional Leech lattice. $M_{24}$ has $2^{1+6}:L_3(2)$ as the centralizer of an involution and $L_3(2)$ (with an additional $Z_2$) is the automorphism group of a 6-dimensional lattice with 42 vectors of norm 4 (not unimodular obviously). String theory on K3 gives rise to a $c=6$ CFT (not holomorphic). There are obvious differences between the two situations, but enough parallels to make me suspect a connection, hence the question. - -REPLY [4 votes]: This is not an answer, but perhaps someone can build off it. I suppose you want something different from the $A_1^{24}$ lattice CFT construction mentioned in the paper that you cited. -I wouldn't be surprised if one could apply a technique along the lines of John Duncan's constructions of vertex superalgebras with actions of larger sporadic groups. For example, you might try to tensor 12 free fermions together to get a $c=6$ superalgebra, then take an orbifold by an involution (but I have no idea if that would work). -An alternative method of construction is by codes. You can get a $c=12$ VOA with an $M_{24}$ action using Golay code construction on $L(1/2,0)^{\otimes 24}$ (see e.g., Miyamoto's paper), but it sounds like this precise construction might not be what you want.<|endoftext|> -TITLE: Zeros of a combination of exponentials -QUESTION [7 upvotes]: Is there any known result about the necessary and sufficient conditions for the existence of zeros for a function $f(x)=\sum_{n=1}^{N} a_n e^{b_n x}$, where $a_n,b_n \in \mathbb{R}\, \forall n=1,2,\cdots,N$, $a_1,a_N >0$, $b_1 < b_2 < \cdots < b_N $ and $x \in \mathbb{R}$? -It is known (see "Problem and Theorems in Analysis II" by Polya and Szego) that using a generalization of Descartes' rule of signs it possible to say that, named with $Z$ the number of changes of sign in the sequence of the $a_n$ and with $Z_0$ the number of zeros of $f(x)$, $Z-Z_0 \geq 0$ is an even integer. -The number $Z-Z_0$ should be even since for $x \rightarrow -\infty$, the dominant therm of $f(x)$ is $a_1 e^{b_1 x}>0$ and for $x \rightarrow +\infty$ the dominant term is $a_N e^{b_N x}>0$. -This gives an upper limit for the number of zeros, but there is any way to say "$f(x)$ should have at least $M$ zeros", with $0 < M \leq Z$? -Thanks in advance, -Nico - -REPLY [3 votes]: Note that we can assume wlog that $b_n\geq 0.$ In the case they are rationals, writing $b_n=p_n/q$, with $p_n\in\mathbb{N},\\ $ $q\in\mathbb{N}_+,\\ $ and $t:=e^{x/q},\\ $ puts everything into the case of positive roots of a real polinomial, with not more, nor less generality. The book by Pólya and Szegő has a section on the location and number of positive roots of a polynomial; in any case, whatever you can say for it can clearly be translated for your exponential equation. Then, the case of real $b_n$ can certainly be treated by approximation.<|endoftext|> -TITLE: About isogenies of abelian varieties -QUESTION [6 upvotes]: Why it is true that, over an algebraically closed field, any abelian variety is isogenous to a principally polarized abelian variety? - -REPLY [9 votes]: This is to fill some prerequisites to BCnrd's comment-answer. -First of all, there are several definitions of a polarization on an abelian variety, and the most "coordinate-free" one is that it is a homomorphism $\lambda:A\to A^t = Pic^0(A)$ given by some (non-unique) ample divisor $D$, so that $\lambda(a) = \mathcal O_A( T^*_a D - D)$, where $T_a:A\to A$ is the translation by $a\in A$. A polarization is principal if $\lambda$ is an isomorphism. -Then the basic steps, all requiring proof, are: - -Every abelian variety (over a field) has a polarization. -$\lambda$ is surjective, and $K(\lambda)=\ker\lambda$ is a finite group subscheme of $A$ of length $d^2$. The degree of $\lambda$ is defined to be $d$. Thus, a polarization is principal iff $d=1$. -$K(\lambda)$ comes with a perfect skew symmetric Weil pairing $b: K(\lambda)\times K(\lambda) \to \mathbb G_m$ with the values in the multiplicative group. "Perfect pairing" means that it defines an iso from $K(\lambda)$ to its Cartier dual. -If $H\subset K(\lambda)$ is a subgroup which is isotropic w.r.t. this pairing (i.e. $b$ restricted to $H\times H$ is trivial), then $\lambda$ descends to a polarization on $B=A/H$ of degree $d/|H|$. - -So then it is enough to find a nontrivial isotropic subgroup $H$, replace $A$ by $A/H$, and continue by induction, until you reach $d=1$. -In char 0, this is trivial: just pick any cyclic subgroup in $K(\lambda)$. Since $b$ is skew-symmetric, it is automatically isotropic. -In char p, you need to get into the classification of finite group schemes, and learn the difference between $\mathbb Z/p\mathbb Z$, $\mu_p$, and $\alpha_p$. Once you learn that (Introduction to affine group schemes by Waterhouse is a good source), then you are perhaps ready to read BCnrd's comment-answer.<|endoftext|> -TITLE: Eisenstein mod p Hilbert modular forms -QUESTION [5 upvotes]: I have a question regarding non-cuspidal Hilbert modular forms. If one starts with a non-parallel weight for example, it is easy to prove that there are no Eisenstein series of any level, or as is generally stated, all forms are cuspidal. My question is what happens with mod p Hilbert modular forms? Are there (non-zero) non-cuspidal mod p Hilbert modular forms of non-parallel weight? (say at least when one or all the weights are greater than 1). -For classical modular forms, if the weight is greater than 1, the mod p modular forms are exactly the reduction of global modular forms, so the naive answer would be that there are none, but I am not too familiar with mod p Hilbert modular forms... - -REPLY [4 votes]: The partial Hasse invariants $h_1,\ldots,h_d$ are mod $p$ Hilbert modular forms of non-parallel weight whose $q$-expansion at each cusp is equal to 1. The forms $h_1-1,\ldots,h_d-1$ generate the kernel of the $q$-expansion map over $\mathbb{F}_p$. Technically, these forms are of weight $(0,\ldots,0,p,-1,0,\ldots,0)$ or $(0,\ldots,0,p-1,0,\ldots,0)$, at least when $p$ is unramified, but you can always multiply them by some large parallel weight form to get something of everywhere positive weight. As you remarked, they have no characteristic 0 lift on account being non-cuspidal and having non-parallel weight. -If you want a more detailed account of these guys, and mod $p$ Hilbert modular forms in general, I recommend Goren's Lectures on Hilbert Modular Varieties and Modular Forms, especially chapter 5, and Andreatta-Goren's Hilbert Modular Forms: mod p and p-adic aspects, available on Goren's website.<|endoftext|> -TITLE: Simple curves on non-orientable surfaces. -QUESTION [18 upvotes]: Given an element in the (first) homology group of a surface, I would like to know if it can be represented as a simple closed curve. For orientable surfaces, this is well-known, but I wasn't able to find a reference for non-orientable surfaces. -For orientable surfaces, the sphere with $g$ handles has homology $\mathbb{Z}^{2g}$, and an element $(a_1, \dots, a_{2g}) \in \mathbb{Z}^{2g}$ can be represented by a simple closed curve if and only if $gcd(a_1, \dots, a_{2g})=1$. This is classical and actually not too hard to prove. -For non-orientable surfaces, the sphere with $k$ crosscaps has homology $\mathbb{Z}_2 \times \mathbb{Z}^{k-1}$. -Let $(a, b_1, \dots, b_{k-1}) \in \mathbb{Z}_2 \times \mathbb{Z}^{k-1}$. If $k$ is odd and $gcd(b_1, \dots, b_{k-1})=1$, then we can represent this element as a simple curve, regardless of the value of $a$ by the orientable case. But there are other homology classes not of this form which can be represented by simple curves. For example, I think $(0,2,0) \in \mathbb{Z}_2 \times \mathbb{Z}^2$ can be represented by a simple curve on the sphere with 3 crosscaps. Incidentally, while I am at it, I'd also like to know why it is common to use $\mathbb{Z}_2$-homology when working with non-orientable surfaces, as opposed to $\mathbb{Z}$-homology (which is what I am using). - -REPLY [6 votes]: (I found this MO thread while searching for a reference similar to the one requested by the OP, so I'm adding the following reference in case it is helpful to other future visitors) -Representing codimension one homology classes on closed nonorientable manifolds by submanifolds: William H. Meeks, III. Illinois Journal of Mathematics, Volume 23, Number 2, June 1979 -In dimension two the theorem in this paper implies that an integer homology class on a connected closed nonorientable surface can be represented by an embedded circle if and only if the class is primitive or twice a primitive class. -(A class in $H_1(M,Z)$ is called primitive if the induced class in $H_1(M, Z)/\text{Torsion}$ is the zero class or is not a nontrivial multiple of any other class.) -In a further paper (Representing homology classes by embedded circles on a compact surface) Meeks appears to be claiming that the same result also holds for nonorientable surfaces with boundary, but the referenced paper is listed as a preprint and does not appear to have been actually been published (it doesn't, for instance, appear on the list of publications in his CV)<|endoftext|> -TITLE: Monotone functions are differentiable a.e. and Hilbert's Fifth Problem: what's the connection? -QUESTION [23 upvotes]: In Andrew Gleason's interview for More Mathematical People, there is the following exchange concerning Gleason's work on Hilbert's fifth problem on whether every locally Euclidean topological group is a Lie group (page 92). - -MP: Is there some "human" story you can tell us about the breakthrough when it came? -Gleason: Yes, there's a really remarkable story about that. Sometime -- I can't tell you the exact date but let's say around 1949 -- I was doing other things too, and one of the things that I found very interesting and very curious and which I really felt I should try to understand better was a very famous theorem to the effect that a monotonic function is almost everywhere differentiable. It's a rather remarkable and very difficult theorem -- it's not easy to prove. A very very hard theorem of analysis and a really surprising theorem. Well, at the time I was sort of speculating about this theorem, but it wasn't for at least two years that I suddenly realized that that would solve the problem I was dealing with! Knowing that, in connection with some other stuff I had been working on, really put the whole thing together. It was a realization that although this theorem had been on my mind for maybe two years, I had never recognized that it was crucial to the arguments that I was trying to work through in the Hilbert problem. I hadn't realized it. Then suddenly it just came to me. -MP: It just came to you? -Gleason: That's right. It just came to me that I could use this technique, this theorem, in connection with these curves in Hilbert space that I was dealing with -- and get the answer! ... - -I've never studied Hilbert's Fifth Problem or its solutions, but I've always been curious what Gleason meant by this connection. Can anyone shed some light on this? - -REPLY [23 votes]: Well, I cannot say for certain, but I did know Gleason well (he was my thesis advisor, and we wrote a paper together after that) and I have written an essay about Gleason's work on the Fifth Problem (in the Gleason Memorial article in the AMS Notices --- http://www.ams.org/notices/200910/rtx091001236p.pdf ) and based on that I think I can make a reasonable guess about what he had in mind. Recall that what Gleason actually proved was that a locally compact group without arbitrarily small subgroups is a Lie group (then Montomery and Zippin proved that a locally Euclidean group did not have small subgroups). A key idea in Gleason's proof was the construction of a unique one-parameter subgroup through any point sufficiently close to the identity (i.e., essentially, constructing the exponential map) and this in turn depended on showing the existence of a unique square root for elements near the identity (see his paper "Square roots in locally Euclidean groups"). I believe that it is the step going from square roots to one-parameter subgroups that used ideas from the monotonic implies differentiable a.e. theorem. The "these curves" that he mentions in that More Mathematical People article, can only be the one-parameter subgroups. For more details, see my Notices article above, particularly the section called "Following in Gleason’s Footsteps".<|endoftext|> -TITLE: Complex Projective Space Spin and Dirac: Part II -QUESTION [6 upvotes]: (1) Let $M$ be a complex manifold of real dimension $2n$, and denote the line bundle of complex $(N,0)$-forms by $\Omega^{(N,0)}(M)$. When $M = CP^N$, the line bundles are indexed by the integers, and so, $\Omega^{(N,0)}(CP^N)$ must correspond to a integer. What is this integer? In the $N=1$ case, the corresponding integer is $2$. This suggests, that in general, $\Omega^{(N,0)}(CP^N)$ corresponds to $2N$. Is this true? -(2) For the anti-canonical spin$^c$ structure of $CP^N$, the spinor bundle is isomorphic to -$$ -S := (\Omega^{(0,0)}(CP^N) + \cdots + -\Omega^{(0,N)}(CP^N)) \otimes S_N, -$$ -where $S_k$ is the square root of $\Omega^{(N,0)}(CP^N)$ (square root wrt tensoring as multiplication). What does the square root mean when when line bundle integer is odd? In the $N=2$ case, this is seen to reduce to $\cal{E}_{-1} \otimes \cal{E}_1$, where $\cal{E}_p$ is the line bundle corresponding to the integer $p$. Is this the $Z2$ grading on the spinor bundle? If so, what does this look like in higher dimensions? -(3) Finally, for a given spin connection $\nabla$, to define a Dirac operator we would need a Clifford representation, ie a map -$$ -C:(\Omega^{(1,0)} \oplus \Omega^{(0,1)}) \otimes S \to S. -$$ -For $N=2$, this should be given by a map -$$ -C: (\Omega^{(1,0)} \oplus \Omega^{(0,1)}) \otimes (\cal{E}_{-1} \oplus \cal{E}_1) \to \cal{E}_1 \oplus \cal{E}_1. -$$ -What is this rep? What does it look like for higher order $N$? -Note: the first subindex in the second $\cal{E} \oplus \cal{E}$ above should be $-1$, I'm having tex problems when I try to write it as such though. - -REPLY [3 votes]: Here is a representation theoretical method for the computation of the (infinitesimal character of) the canonical bundle of any compact homogeneous Kahler manifold which include Grassmanians and complex projective spaces as particular cases. The computation for the complex projective space will be given explicitely. These spaces are coset spaces of a compact Lie group by an isotropy group which is a centralizer of a torus. The important step is to identify the generators of the torus in the Dynkin diagram of the compact Lie group. In the case of the $CP^N$, the torus is one dimensional and its generator corresponds to the simple root $\aplha_1$ or $\alpha_N$ of $A_N$. The other roots correspond to Grassmannians. -A nice explanation of this correspondence is given by John Baez here. -This method is sometimes called painted Dynkin diagrams is decribed in the following references by: by Alexeevsky and Bordemann, Forger and Romer). -Now the infinitesimal character corresponding to the canonical bundle is the sum of all roots of $A_N$ containing the generator of the torus which will be taken here as $\alpha_1$. In $A_N$ the roots are sums of continuous strings of simple roots, thus the roots containing $\alpha_1$ are: $\alpha_1, \alpha_1+\alpha_2, \alpha_1+\alpha_2+\alpha_3,. . .$, Their sum is therefore $\lambda = N\alpha_1+(N-1)\alpha_2+(N-2)\alpha_3,. . .$. The infinitesimal characters of the basic line bundles correspond to the fundamental weights, thus we need to transform the result to the weight basis. Using: -$\alpha_1 = 2 w_1 - w_2, \alpha_i = -w_{i-1} + 2w_{i} + w_{i+1}$, for $i = 2, ..., N-1, \alpha_N = -w_{N-1}+ 2w_{N}$, we get: $\lambda = (N+1) \w_1$. The generalization to cases where the torus is multidimensional such as flag manifolds is obviousperformed by just duplication of the above procedure for evey torus generator -2) When N+1 is odd, tyen $CP^N$ does not admit a spin structure, however, one can tensor with the square root of a "line bundle" of odd infinitesimal character to obtain a $spin^c$ structure. Of course, neither the "square root of the canonical bundle" nor -the square root of the line bundle exist separately, only their product is a line bundle over $CP^N$. -3) If we choose to work with a spin connection induced from the Levi-Civita connection, -then the Dirac oprtaor '$D: \wedge(T^{0,1}) \otimes S\rightarrow \wedge(T^{0,1}) \otimes S$ is just the sum of the twisted Dolbeault operator and its adjoint as given in page 82 of Friedrich's book.<|endoftext|> -TITLE: Is there any sense in which Dirichlet density is "optimal?" -QUESTION [25 upvotes]: A philosopher asked me an interesting math question today! We know that there are sets S of integers which don't have a "natural" or "naive" density -- that is, the quantity (1/n)|S intersect [1..n]| doesn't approach a limit. The "analytic" or "Dirichlet" density exists whenever the naive density does, and is equal to it -- but sometimes is well-defined even when the naive density is not, for instance when S is the "Benford set" of integers whose first digit is 1. (See this related MO question.) -Anyway, the philosopher asked: why stop at Dirichlet density? Is there a sequence of probability measures p_1, p_2, p_3, .... on the natural numbers such that - -Whenever S has a Dirichlet density, the limit of p_i(S) exists and is equal to the Dirichlet density, but -there are subsets of the natural numbers such that p_i(S) approaches a limit but S has no Dirichlet density? - -(Possibly clarifying addition: to obtain naive density, one can take p_i to be the measure assigning probability 1/i to each integer in [1..i] and 0 to integers greater than i. To get Dirichlet density, take p_i to be the measure with p_i(n) = 1/(n^{1+1/i}zeta(1+1/i)). In either case, the corresponding density of S is the limit in i of p_i(S), whenever this limit exists. So what I have in mind is densities which can be thought of as limits of probability measures, though perhaps there are reasons to have yet more general entities in mind?) -If so, are there any examples which are interesting or which are used for anything in practice? - -REPLY [15 votes]: The best way to impress a philosopher is to tell him/her about ultrafilters. A (non-principal) filter on $\mathbb N$ is a set of infinite subsets of $\mathbb N$ closed under intersections and taking super-sets. A maximal filter (under inclusion) is called an ultrafilter. There are plenty of those but nobody saw them since their existence depends on the axiom of choice. For every filter $\omega$ one can define the concept of convergence of sequences of real numbers: a sequence $b_n$ converges to $b$ if for every $\epsilon$ the set of $i$'s such that $|b-b_i|\le \epsilon$ is in $\omega$. If $\omega$ is an ultrafilter, then every bounded sequence of real numbers has unique $\omega$-limit. It is not true if $\omega$ is not an ultrafilter. The smallest interesting filter (called Fréchet filter) consists of all sets with finite complements. The limit corresponding to that filter is the ordinary limit studied in Calculus. You can start from the Fréchet filter and add sets to it to produce bigger and bigger filters. Each filter gives you a Dirichlet-like density. If $\omega$ is an ultrafilter, then all sets will have density (between 0 and 1). Otherwise there will always be sets without density assigned.<|endoftext|> -TITLE: Pair consisting of a compact manifold and Morse function -QUESTION [5 upvotes]: Consider the following situation: -Let $M$ and $M'$ be two closed manifolds and suppose $f:M\to \mathbb{R}$ and $f':M'\to \mathbb{R}$ are smooth morse functions on $M$ and $M'$ respectively. We say the pair $(M,f)$ and $(M',f')$ are equivalent if there is a smooth diffeomorphism $\phi:M\to M'$ so that $f'\circ \phi=f$ and consider equivalence classes $[(M,f)]$. -My questions are: -1) Has this been studied at all? and if so -2) Is there any sort classification result for (oriented) surfaces $M$? -What I'm looking for in 2) is that any equivalence class $[(M^2, f)]$ contains some geometrically nice example...for instance an immersed surface in $\mathbb{R}^3$ with the morse function given by restriction of the $z$ coordinate function. I'm mostly interested in examples where $f$ has as few critical points as the topology of $M$ allows. -I apologize if this is trivial as it came up while I was playing around with an idea for a different problem and (like most aspects of topology) is sadly not something I'm very familiar with. -If it helps, the example I have in my head is to take a genus-2 surface in $\mathbb{R}^3$ that looks like a figure 8 and morse function given by restricting the $z$ coordinate and contrast it with the genus-2 surface in $\mathbb{R}^3$ that looks like $\infty$ (i.e. the first one on its side) with Morse function given by restricting the $z$ coordinate. These two pairs shouldn't be equivalent as the number of components of level curves of the first morse function is at most 2 while the second has level curves of the morse function with 3 components (I apologize for the crummy graphics). This is in spite of both morse functions having the same number of critical points (six). I was wondering how many other examples of morse functions there were with 6 critical points that weren't equivalent to these two. -Edit: -I felt I should add...heuristically what I am interested in is to what extent can one use a Morse function to say whether the handles (in a surface) are "next to each other" or "on top of one another". - -REPLY [3 votes]: You might have a look at Hatcher and Thurston's paper "A presentation for the mapping class group of a closed orientable surface". They use Morse functions on the surface to facilitate the derivation of their presentation. On page 223 and following, they discuss how to form a graph from a Morse function on a surface, which is the leaf space of the level sets of the Morse function. This graph, together with its map to $\mathbb{R}$, uniquely determines the Morse function on the surface (up to graph isomorphism preserving the function). In particular, one need only know the values of the Morse function at the vertices of the graph (corresponding to the critical points of the Morse function on the surface) since it is monotonic on each of the edges. The space of such functions will therefore be parameterized by a subset of a vector space, satisfying certain inequalities. As they indicate on p. 224, the surface with the Morse function may be recovered from the graph (with its Morse function) by embedding it in $\mathbb{R}^3$ and taking the boundary of a regular neighborhood (after a small perturbation).<|endoftext|> -TITLE: Non-representable functor, representable on locally Noetherian schemes? -QUESTION [9 upvotes]: What is an example of a functor $F : \mathbb{C}\text{-Sch.} \to \text{Sets}$ with the property that -the restriction of $F$ to locally Noetherian $\mathbb{C}$-schemes can be represented by a locally Noetherian $\mathbb{C}$-scheme, but that scheme does not represent $F$. -I'd be particularly nice to see a "real-world" example (though this might be stretching the notion -of "real-world"). - -REPLY [18 votes]: Define $F(X) = {\rm{Hom}}_{\mathbf{C}}(X,{\rm{Spec}}(R/tR))$ where $R$ is the valuation ring of an algebraic closure of $\mathbf{C}((t))$. Note that every element of the maximal ideal of $R/tR$ is nilpotent yet also an $N$th power for arbitrarily large $N$. For any noetherian $\mathbf{C}$-algebra $A$, every $\mathbf{C}$-algebra map $R/tR \rightarrow A$ carries -the maximal ideal into the nilradical of $A$. But the nilradical of $A$ has all elements -with vanishing $n$th power for some uniform $n$ (depending on $A$) since $A$ is noetherian, so in fact the maximal ideal of $R/tR$ is killed by any such map. In other words, the restriction of $F$ to the full subcategory of locally noetherian objects is represented by ${\rm{Spec}}(\mathbf{C})$. -This is a perfectly "real world" example, since valuation rings of algebraic closures of complete discretely-valued fields come up all the time in number theory and rigid-analytic geometry.<|endoftext|> -TITLE: Quantum cohomology of partial flag manifolds -QUESTION [17 upvotes]: Is there a place in the literature where the quantum differential -equation (or even just quantum cohomology algebra) -of partial flag manifolds $G/P$ is computed for arbitrary semi-simple $G$ and -arbitrary parabolic $P$? I actually think that I know one way -to formulate (and prove) the answer but -I was sure that this was well-known and to my surprise I couldn't find the reference -for the general case (the case when $P$ is a Borel subgroup is well-known and there is -a lot of literature for other parabolics in the case when $G$ is a classical group but again -I couldn't find a treatment of the general case). -For the quantum cohomology algebra many papers mention a result of Peterson (which I think coincides with what I want when one takes the appropriate limit going from quantum $D$-module to quantum cohomology algebra) which describes it, but I was unable to find a published proof of this result. Is it written anywhere? - -REPLY [5 votes]: Among many other nice results, the paper "Totally Positive Toeplitz Matrices and Quantum Cohomology of Partial Flag Varieties" by Konstanze Rietsch contains a proof of Peterson's result. It's available at arXiv:math/0112024. The result appears as Theorem 4.2. -I believe Peterson's theorem says that if one takes the opposite Schubert cell $B_{-} w_P B/B$ and intersects that with what is now called the Peterson variety, then the coordinate ring of that space is the quantum cohomology of $G/P$. -Section 2 of Harada and Tymoczko's paper "A positive Monk formula in the S^1-equivariant cohomology of type A Peterson varieties" has a concise description of the Peterson variety. This paper is available on the arxiv at arXiv:0908.3517.<|endoftext|> -TITLE: Which model of computation is "the best"? -QUESTION [11 upvotes]: In 1937 Turing described a Turing machine. Since then many models of computation have been decribed in attempt to find a model which is like a real computer but still simple enough to design and analyse algorithms. -As a result, we have dozen of algorithms for, e.g., SORT-problem for different models of computation. Unfortunately, we even cannot be sure that an implementation of an algorithm with running time $O(n)$ in a word RAM with bit-vector operations allowed will run faster than an an implementation of an algorithm with running time $O(n \cdot \log{n})$ in a word RAM (I am talking about "good" implementations only, of course). -So, I want to understand which of existing models is "the best" for designing algorithms and I am looking for an up-to-date and detailed survey on models of computation, which gives pros and cons of models and their closeness to reality. - -REPLY [2 votes]: Since you're interested in running times this is probably not what you're after, but IMHO the best model of real-life computation is a machine consisting of communicating finite-state automata or of communicating Turing machines. -Consider a parallel machine consisting of n communicating Turing machines. Each Turing machine has a tape, consisting of squares, and a head which can read from and write on any given square. Also, any Turing machine, in addition to its usual states, has available to it n-1 special states where it can read, at that instant, what another of the machine's heads happen to be reading. (e.g. the nth Turing machine has n-1 states, and if it were in special state i, it reads what the ith Turing machine is reading at that moment). The -parallel machine obviously can be programmed to include all normal deterministic computation (just program the first TM to do something), but with reasonable assumptions it also includes non-deterministic (or if you prefer, non-computable) behaviour. -For instance, consider a parallel machine with just two machines. And consider this program. One machine shifts between a square reading a 0 and a 1, while the other machine shifts to the end of its input, and then it reads whatever the first machine's head happens to be reading. It then outputs this digit. Obviously what the parallel machine outputs depends on the speeds with which the respective machines operate. It is therefore non-deterministic. That is, the same program will behave differently depending on which 2-TM parallel machine you run it on, because the speed at which a TM processes a step determines the behaviour. -Or look it this way. Consider two computers attached to a printer. It is non-deterministic who will print on the printer first, a feature which cannot be modeled correctly by a single Turing machine.<|endoftext|> -TITLE: Odd-bit primes ratio -QUESTION [18 upvotes]: Say that a number is an odd-bit number if -the count of 1-bits in its binary representation is odd. -Define an even-bit number analogously. -Thus $541 = 1000011101_2$ is an odd-bit number, -and $523 = 1000001011_2$ is an even-bit number. - -Are there, asymptotically, as many odd-bit primes as even-bit primes? - -For the first ten primes, we have -$$ -\lbrace 10, 11, 101, 111, 1011, 1101, 10001, 10011, 10111, 11101 -\rbrace -$$ -with 1-bits -$$ -\lbrace 1, 2, 2, 3, 3, 3, 2, 3, 4, 4 \rbrace -$$ -and so ratio of #odd to $n$ is $5/10=0.5$ at the 10-th prime. -Here is a plot of this ratio up to $10^5$: - - - - -(Vertical axis is mislabeled: It is #odd/$n$.) - - -I would expect the #odd/$n$ ratio to approach $\frac{1}{2}$, except perhaps the fact that primes ($>2$) are -odd might bias the ratio. The above plot does not suggest convergence -by the 100,000-th prime (1,299,709). -Pardon the naïveness of my question. -Addendum: Extended the computation to the $10^6$-th prime (15,485,863), where it still -remains 1.5% above $\frac{1}{2}$: - -REPLY [24 votes]: Yes. This was proven in -C. Mauduit and J. Rivat, Sur un problème de Gelfond: la somme des chiffres des nombres premiers, Ann. Math. -I found this by searching for "evil prime" and "odious prime" in the OEIS. More precisely, they prove the Gelfond conjecture: - -Let $s_q(p)$ denote the sum of the digits of $p$ in base $q$. For $m, q$ with $\gcd(m, q-1) = 1$ there exists $\sigma_{q,m} > 0$ such that for every $a \in \mathbb{Z}$ we have -$$| \{ p \le x : s_q(p) \equiv a \bmod m \} | = \frac{1}{m} \pi(x) + O_{q,m}(x^{1 - \sigma_{q,m}})$$ -where $p$ is prime and $\pi(x)$ the usual prime counting function. - -REPLY [21 votes]: Yes, the ratio approaches 1/2. This was proven in -C. Mauduit et J. Rivat, Sur un probléme de Gelfond: la somme des chiffres des nombres premiers. -See Three topics in additive prime number theory for exposition. Also, the poorly-named sequences in Sloane: A027697 and A027699.<|endoftext|> -TITLE: When are non-quasi-coherent sheaves used? -QUESTION [6 upvotes]: Non-quasi-coherent sheaves of $\mathcal O_X$ modules on a scheme seem like a wild concept to me; are they actually used for something? - -REPLY [5 votes]: One can think the adeles on a curve (or higher adeles on other spaces) as a sheaf of $\mathcal O$-algebras. That is, consider the sheaf $B(U)=\prod_{x\in U}\mathcal O_x$, where $\mathcal O_x$ is the completion of $\mathcal O$. Then the sheaf $A=B\otimes K$, where $K$ is the sheaf of rational functions, has the adeles as global sections. There is a short exact sequence $\mathcal O\to K\times B\to A$. One can tensor a quasicoherent sheaf with this to obtain a resolution to compute cohomology. Indeed, Weil introduced the adeles (after the earlier ideles) specifically to prove Riemann-Roch. I'm not sure when this was reinterpreted in terms of sheaves, which were only introduced later.<|endoftext|> -TITLE: Does it ever make sense NOT to go to the most prestigious grad school you can get into? -QUESTION [8 upvotes]: I'm a senior undergrad at a top-ish(say, top 15) math school. I'm a solid, not stellar, student. This year I'm taking the qualifying exam grad courses in algebra and analysis and have been taken aback by the "pressure cooker" atmosphere among grad students here. That is, even moreso than in the undergraduate program. -If I'm self driven, could going to a "less prestigious" school afford me more space(I mean in a psychological sense) to produce a more solid contribution to math? By "less prestigious", I mean a school "ranked" significantly lower than the range of schools that I could comfortably get into. For me, "less prestigious" would be ranked around 40-60 on, say, USNews or NRC. -My reasoning is that at such a school, I would be more able to learn the fundamentals at my own pace, as opposed to a pace dictated to me by the program. I know I want to do math, and I think my learning style may be better suited to going at my own pace. Thoughts? - -REPLY [12 votes]: The pace issue is a misconception of yours. You will have to learn the graduate level mathematics as quickly as is necessary to begin your dissertation work if you are going to a PhD program. Essentially, you have to catch up to a century's worth of mathematics in a year and a half. So deal with it. -But to answer the original question, what will be essential to your survival in grad school and beyond is your relationship with your PhD advisor. I have seen students from not top programs land better jobs because their advisors took the time and care to help their students, bring them to conferences, introduce them to others working in the field, help them select post-doc positions, etc. -If you know the area in which you want to work, then you should select grad school based on that. If you are not sure, then you should select among the best to which you are admitted, and select an advisor based upon her area of expertise and her ability to get grad students into good schools. Also check out the social network among the grad students. Is it solid? Will others help you in your studies. -Ultimately, what makes your dissertation great is your own creativity and your depth of thought, but all else being equal, a good advisor is as important as the school that she teaches at. - -REPLY [4 votes]: It's true the atmosphere of a school make a big difference. Without knowing anything more, I would suggest applying to a variety of schools, then visit the schools you get into. This will give you a much better sense of whether or not the school is a good fit for you than just the ranking. I think after visiting, you should be able to narrow your choice down to a couple schools you like, and at that point, probably either would be a good choice. -Incidentally, I wrote some tips on visiting grad schools after I visited quite a few. You might also check out the career advice section on Terry Tao's blog, though I'm not sure there's anything specific about choosing a grad school.<|endoftext|> -TITLE: Can 'network' and 'graph' be used interchangably? -QUESTION [5 upvotes]: This question is with regards to terminology. I am writing a journal paper that describes the software I develop at my workplace call Navigator. The software is used for visualization of networks/graphs. At work, we commonly interchange the use of the words 'networks' and 'graphs' to refer to the same concept. But for writing the journal paper, I would like to know if they can be safely interchanged, or if they refer to slightly different concepts? -The software is used for modeling arbitrary entities as nodes and arbitrary connections between them as edges between the nodes. The graphs are not directed. No flow is implied, but some idea of flow can be added by attaching numerical/textual attributes to the edges. - -REPLY [2 votes]: I worked at a telecommunications equipment company for several years, and although there was no hard-and-fast distinction between "graph" and "network," I would say that we tended to use "graph" for an abstract mathematical object and "network" for something whose nodes had some kind of physical existence. -It sounds to me that your software deals with entities that are on the abstract end of the spectrum, so I would tend to call them graphs. "Network" may carry a connotation that your software is used for designing specific kinds of physical networks, such as telephone networks, road networks, etc.<|endoftext|> -TITLE: "Linear algebra" over Z/nZ - reference please! -QUESTION [7 upvotes]: Let A be a matrix with entries in Z/nZ. (n is not assumed to be prime.) Then the size of the row span is the size of the column span. All computations are mod n, so both these numbers are finite. -I believe you can prove this using Smith normal form: both the size of the row span and the size of the column span will be the same after passing to the Smith normal form (lift the matrix entries to Z to compute the Smith normal form). When A is in Smith normal form, these quantities can be easily computed. -I would like to find a reference for this fact, and would be grateful to anyone who could provide one. - -REPLY [4 votes]: Your idea of using Smith normal form leads directly to a solution: But you need to verify that for every matrix $M$ with entries in $\mathbb{Z}/n\mathbb{Z}$ there are invertible matrices $A$ and $B$ with entries in $\mathbb{Z}/n\mathbb{Z}$ such that $AMB$ is in Smith normal form. It is essential that $A$ and $B$ be invertible as matrices over $\mathbb{Z}/n\mathbb{Z}$, otherwise the row and column spaces of $AMB$ won't necessarily be the same size as those of $M$. -A ring with the property that every matrix is equivalent to one in Smith normal form is called an elementary divisor ring. In the book "Matrices over Commutative Rings" by William Brown (Marcel Dekker 1993) it is shown (Theorem 15.8 and 15.9) that every principal ideal ring is an elementary divisor ring. It is simple to check that the rings $\mathbb{Z}/n\mathbb{Z}$ are principal ideal rings. So there's the reference you asked for. If you can't find the book by Brown the article by Kaplansky mentioned in the comments has the same material in a more general setting.<|endoftext|> -TITLE: Rank 2 vector bundle on a product of elliptic curves -QUESTION [7 upvotes]: Let $E$, $F$ be two complex elliptic curves, and $A=E \times F$. Let us denote by -$\pi_E \colon A \to E, \quad \pi_F \colon A \to F$ -the natural projections. For all $p \in F$ let us write $E_p$ instead of $\pi_F^*(p)$. -Now let us fix $p \in F$ and consider the unique indecomposable rank $2$ vector bundle $\mathcal{F}$ on $A$ defined by the extension -$0 \to \mathcal{O}_A \to \mathcal{F} \to \mathcal{O}_A(-E_p) \to 0$ -My main question is the following: - -What is the restriction of $\mathcal{F}$ to a fibre of type $E_x$, for $x \in F$ general? - -Related to this, another question is - -What are $\pi_{F*} \mathcal{F}$ and $R^1 \pi_{F*} \mathcal{F}$? - -Observe that the restriction of $\mathcal{F}$ to $E_x$ is given by an extension -$0 \to \mathcal{O}_{E_x} \to \mathcal{F}|_{E_x} \to \mathcal{O}_{E_x} \to 0$, -so by Atiyah's classification of vector bundles over an elliptic curve we have that $\mathcal{F}|_{E_x}$ is either $\mathcal{O}_{E_x} \oplus \mathcal{O}_{E_x}$ or the unique non-trivial extension of $\mathcal{O}_{E_x}$ with itself. -But I cannot decide what happens generically. -Motivation. I met this problem when I started the investigation of some triple covers $f \colon X \to A$ such that -$f_*\mathcal{O}_X=\mathcal{O}_A \oplus \mathcal{F}^{\vee}$. -It is worth noticing that -the vector bundle $\mathcal{F}$ is the easiest example of indecomposable vector bundle on $A$ which is not simple (in fact, $\textrm{End}(\mathcal{F})=\mathbb{C} \oplus \mathbb{C}$). - -REPLY [4 votes]: It's the nontrivial extension. Here is why: -Let $\pi=\pi_F$ and consider -$$ -0\to \pi_*\mathscr O_A \to \pi_*\mathscr F \to \pi_*\mathscr O_A(-E_p)\to R^1\pi_*\mathscr O_A \to \dots -$$ -Now -1) $\pi_*\mathscr O_A\simeq \mathscr O_F$ and $\pi_*\mathscr O_A(-E_p)\simeq\mathscr O_F(-p)$ -2) $R^1\pi_*\mathscr O_A\simeq R^1\pi_*\omega_A$ is torsion-free by Kollár's theorem (I guess it is also isomorphic to $\omega_F\simeq \mathscr O_F$ so you don't even need Kollár here), so -3) if $\pi_*\mathscr F\to \pi_*\mathscr O_A(-E_p)$ is not surjective, then it is zero. -4) $\pi_*\mathscr F$ is torsion free and hence locally free. -Suppose that the restriction of your sequence to any fiber is a trivial extension. Then $\pi_*\mathscr F$ has to have rank $2$ and hence by 1) and 3) the map $\pi_*\mathscr F\to \pi_*\mathscr O_A(-E_p)\simeq \mathscr O_F(-p)$ is surjective. -Next, take the short exact sequence -$$ -0\to \mathscr O_F\to \pi_*\mathscr F \to \mathscr O_F(-p) \to 0. -$$ -This is necessarily the trivial extension, but also by construction it is the push-forward of your original sequence. -Then the pull-back of this sequence maps functorially to the original sequence and you get an isomorphism on the sides, so you get an isomorphism in the middle as well. That implies that $\mathcal F$ is a trivial extension, which is an obvious contradiction. -Now as a side result we get that $\mathscr O_F\simeq \pi_*\mathscr O_A\simeq \pi_*\mathscr F$ from the beginning of the long exact sequence and -$$\begin{multline} -0\to \pi_*\mathscr O_A(-E_p)\simeq\mathscr O_F(-p)\to R^1\pi_*\mathscr O_A\simeq\mathscr O_F\to R^1\pi_*\mathscr F \to \\ \to R^1\pi_*\mathscr O_A(-E_p)\simeq \mathscr O_F(-p)\otimes R^1\pi_*\omega_A\simeq \mathscr O_F(-p)\to 0 -\end{multline}$$ -so $\pi_*\mathscr F\simeq \mathscr O_F$ and $R^1\pi_*\mathscr F\simeq \mathscr O_p\oplus \mathscr O_F(-p)$. -EDIT: added the last exact sequence and corrected conclusion according to Francesco's comment.<|endoftext|> -TITLE: Fibonacci = Leonardo Pisano? -QUESTION [6 upvotes]: Leonardo of Pisa is best known as Fibonacci; various stories found in books and on the web claim that the name Fibonacci was invented by Edouard Lucas or Guillaume Libri in the 19th century, and that it means "son of Bonacci" (Leonardo's father was apparently called -Guglielmo Bonaccio). Heinz -Lueneburg found out that the name Fibonacci had been used already by John Leslie in 1820. -Are there any facts known as to who, why and when invented the name Fibonacci? -Or wasn't it invented at all? - -REPLY [7 votes]: From The Fabulous Fibonacci Numbers by A.S. Posamentier and I. Lehmann (Prometheus Books, New York (2007), pp. 17-18): - -Leonardo Pisano - or Leonardo of Pisa, Fibonacci - his name as recorded in history, is derived from the Latin "filius Bonacci," or a son of Bonacci, but it may have been more likely derived - from "de filius Bonacci," or family of Bonacci. He was born to Guglielmo (William) Bonacci and his wife in the port city of Pisa, Italy, around 1175, shortly after the start of construction of a famous bell tower, the Leaning Tower of Pisa. - -The authors also indicate in a footnote that - -it is unclear who first used the name Fibonacci; however, it seems to be attributed to Giovanni Gabriello Grimaldi (1757-1837) at around 1790 or to Pietro Cossali (1748-1815).<|endoftext|> -TITLE: Still Difficult After All These Years -QUESTION [72 upvotes]: I think we all secretly hope that in the long run mathematics becomes easier, in that with advances of perspective, today's difficult results will seem easier to future mathematicians. If I were cryogenically frozen today, and thawed out in one hundred years, I would like to believe that by 2110 the Langlands program would be reduced to a 10-page pamphlet (with complete proofs) that I could read over breakfast. -Is this belief plausible? Are there results from a hundred years ago that have not appreciably simplified over the years? From the point of view of a modern mathematician, what is the hardest theorem proven a hundred years ago (or so)? -The hardest theorem I can think of is the Riemann Mapping Theorem, which was first proposed by Riemann in 1852 and (according to Wikipedia) first rigorously proven by Caratheodory in 1912. Are there harder ones? - -REPLY [71 votes]: Difficulty is not additive, and measuring the difficulty of proving a single result is not a good measure of the difficulty of understanding the body of work in a given field as a whole. -Suppose for instance that 100 years ago, there were ten important theorems in (say) complex analysis, each of which took 30 pages of elementary arguments to prove, with not much in common between these separate arguments. (These numbers are totally made up for the purposes of this discussion.) Nowadays, thanks to advances in understanding the "big picture", we can now describe the core theory of complex analysis in, say, 40 pages, but then each of the ten important theorems become one-page consequences of this theory. By doing so, we have actually made the total amount of pages required to prove each theorem longer (41 pages, instead of 30 pages); but the net amount of pages needed to comprehend the subject as a whole has shrunk dramatically (from 300 pages to 50). This is generally a worthwhile tradeoff (although knowing the "low tech" elementary proofs is still useful to round out one's understanding of the subject). -There are very slick and short proofs now of, say, the prime number theorem, but actually this is not the best measure of how well we understand such a result, and more importantly how it fits in with the rest of its field. The fact that we can incorporate the prime number theorem into a much more general story of L-functions, number fields, Euler products, etc. which then ties in with many other parts of number theory is a much stronger sign that we understand number theory as a whole.<|endoftext|> -TITLE: $S^n \to S^m \to B$ bundle: possible? -QUESTION [12 upvotes]: Sphere bundles and bundles over spheres are everywhere and are excellent things to get one's hands dirty with. - -(1a) But when can we have a bundle $S^n \to S^m \to B?$ It seems like requiring the total space of a sphere bundle to be a sphere is pretty restrictive. -(1b) Does the answer to (1a) depend on a choice of category (PL, TOP, etc)? - -There's an $S^1-$ bundle over $CP^1$ with total space $S^3$, but that's the only example I can find. - -(2) Do people know examples other than the one above? - -REPLY [8 votes]: This question was studied in a paper by Browder who proves the following. -Thereom. Consider any Serre fibration of $F\to S^n\to B$ where $F$, $B$ are connected polyhedra. Then $F$ is homotopy equivalent to $S^1$, $S^3$, or $S^7$. If $F=S^1$, then $B$ is homotopy equivalent to $CP^k$ with $2k+1=n$. If $F=S^7$, then $B$ is homotopy equivalent to $S^8$. -One expects that if $F= S^3$, then $B$ is homotopic to a quaternionic projective space of a suitable dimension but Browder mentions (if I am reading it correctly) that this is not true, and there are other examples. - Caution: there might be something wrong with the above theorem because it rules out existence of $S^7$ Hopf bundles over the octonian projective plane. Is there such a bundle? Hatcher's comment above says there isn't, am I reading it right? Could someone clarify the situation? I care because Browder's theorem was used in some geometric problems involving boundary at infinity of nonnegatively curved manifolds.<|endoftext|> -TITLE: Principal L-functions on GL(n) -QUESTION [5 upvotes]: What does the principal L-functions on GL(n), $n \geq 3, n \in \mathbb{Z}$, look like? -Where can I find materials about principal L-functions on GL(n)? - -REPLY [16 votes]: There are several ways to attack standard L-functions (I prefer "standard" over "principal" because it is associated to the standard representation of GL(n) on an n-dimensional space). -I'm going to assume that by "look like" you mean the formula for the local factors as a function of the local data. At primes where the local representation is spherical (which happens for all but finitely many primes), the local factor is -$$1\over(1-\alpha_1q_v^{-s})\ldots(1-\alpha_nq_v^{-s})$$ -where the $\alpha_i$ are the Satake parameters for the local representation, which are essentially of the form $q_v^{-s_i}$, where the $s_i$ are the exponents of the character of the unramified principal series into which the (spherical) local representation injects (by the Borel-Matsumoto theorem), up to some normalization issues. I'm not going to touch non-spherical or archimedean primes. -On to the different methods (in what follows, all cusp forms are assumed to generate irreducible representations): -1) Godement-Jacquet: In the spirit of Tate's thesis, take a cusp form f on G=GL(n) (and f' in the dual representation) and a Schwartz-Bruhat function $\Phi$ on M(n) and integrate -$$\int_{Z_{\bf A}G_k\backslash G_{\bf A}} \langle g\cdot f,f'\rangle \Phi(g)|det(g)|^s dg$$ -where the brackets are the inner product pairing. This converges for Re(s) sufficiently large. -The uniqueness of the inner product makes this factor over primes into local integrals of similar shape. You then prove that the local integrals give the correct factor for supercuspidal representations, and then you prove an induction formula to show that the integral gives the correct factor for all representations parabolically induced from supercuspidal. (It may be possible to calculate the factors directly for unramified principal series, but I don't remember.) And then you have to deal with the archimedean primes. This proves that the integral produces the L-function for judicious choice of f, f', $\Phi$. -To prove analytic continuation and functional equation, you invoke a version of Poisson summation for M(n) and proceed as in Tate's thesis. (You end up with lots of weird terms that you show are zero by invoking cuspidality of f.) -2) "Doubling method" Rankin-Selberg integrals: Take a cusp form f on G=PGL(n), a cusp form f' in the dual representation, and a specific Eisenstein series on PGL(n$^2$) and integrate -$$\int_{{C_{\bf A}G_k\times G_k\backslash G_{\bf A}\times G_{\bf A}}}E((g_1,g_2))f(g_1)f'(g_2) dg_1dg_2$$ -where C is the center of PGL(n$^2$). The embedding of $G\times G$ in PGL(n$^2$) is defined by the natural bi-regular action of $G\times G$ on M(n). -The Eisenstein series gives the integral a meromorphic continuation and functional equation. Unwind it and out pops the Godement-Jacquet integral. -When dealing with Eisenstein series, you usually have to do some extra work to get that there are only finitely many poles. However, if I recall correctly (and I'm not sure that I do), the Eisenstein series here is the mirabolic one, which you can prove has nice properties (essentially using the Godement-Jacquet arguments). -Added: To extend to GL(n), you have to add in an extra term to deal with the fact that the center of $G\times G$ is not compact after you mod out by the center of GL(n$^2$), so the naive integral would diverge for trivial reasons. (And you can't just mod out by more in the integral because the Eisenstein series is not constant on the center of $G\times G$, even though it has trivial central character) -3) Cogdell's Eulerian integrals: Take f on GL(n) and f' on GL(n'), with n>n'. Integrate -$$\int_{Z_{\bf A}GL_{n'}(k)\backslash GL_{n'}({\bf A})}Pf(g)f'(g)|det(g)|^{s-1/2}dg$$ -where P is a projection operation on f. P is brilliantly designed to let this integral unwind completely into what is essentially a Mellin transform of Whittaker functions (which then factors over primes by the uniqueness of Whittaker models). More work shows that this is the L-function of the product. -Note that the decay of the cusp forms implies that the integral is entire. A little observation gives the functional equation. -4) Langlands-Shahidi method: (edited) The constant term of an Eisenstein series on GL(n+n') attached to cusp forms on GL(n) and GL(n') contains the L-function for the product (in your case take n'=1). In fact, it is -$${L_S(s,f\otimes f')\over L_S(1+s,f\otimes f')}\cdot A_S(s,f\otimes f')$$ -where $A_S$ represents bad-prime factors, which generally you have to work to control (when it is even possible). The meromorphic continuation of the Eisenstein series is equivalent to that of the constant term, so you get that this quotient of L-functions is meromorphic. -When n'=1, you have the mirabolic Eisenstein series, so you know that the only poles in the constant term come from the zeroes of $L(1+s,f\otimes f')$, giving you, up to the $A_S$, the analytic continuation of the L-function you care about. -I never studied the method, so I hope everything I said was correct. -References: -1) Godement-Jacquet (Zeta functions of simple algebras), Jacquet has two survey articles (in PSPM 26 and 33) -2) Gelbart, Piatetski-Shapiro, and Rallis (Explicit construction of automorphic L-functions) -3) Cogdell's website has his expository articles on his integral representation -4) Shahidi has a new book out on the method, and he has expository articles here and there.<|endoftext|> -TITLE: Human checkable proof of the Four Color Theorem? -QUESTION [17 upvotes]: Four Color Theorem is equivalent to the statement: "Every cubic planar bridgeless graphs is 3-edge colorable". There is computer assisted proof given by Appel and Haken. Dick Lipton in of his beautiful blogs posed the following open problem: - -Are there non-computer based proofs of the Four Color Theorem? - -Surprisingly, While I was reading this paper, -Anshelevich and Karagiozova, Terminal backup, 3D matching, and covering cubic graphs, the authors state that Cahit proved that "every 2-connected cubic planar graph is edge-3-colorable" which is equivalent to the Four Color Theorem (I. Cahit, Spiral Chains: The Proofs of Tait's and Tutte's Three-Edge-Coloring Conjectures. arXiv preprint, math CO/0507127 v1, July 6, 2005). - -Does Cahit's proof resolve the open problem in Lipton's blog by providing non-computer based proof for the Four Color Theorem? - -Cross posted on math.stackexchange.com as Human checkable proof of the Four Color Theorem? - -REPLY [29 votes]: This is too long for a comment, so I am placing it here. -In this article of the Notices of the AMS, Gonthier describes a full formal proof of the four-color theorem, which makes explicit every logical step of the proof. -Although this formal proof has been checked by the Coq proof system, it would seem to be a category error to view this proof as a computer-based proof of the same kind as Appel and Haken's. The situation with Gonthier's proof is that we essentially have a full written text constituting a verified formal proof of the four-color theorem in first order logic. -And that is a state of certainty that most theorems in mathematics, including many of the classical results, have not yet attained.<|endoftext|> -TITLE: Norm of commutators (bis) -QUESTION [6 upvotes]: This question is slightly related to a popular one with the same title (see here). -Let $k$ be a field with characteristic zero. It is known (see Exercise 310) that a matrix $A\in M_n(k)$ is nilpotent if and only if it is a commutator of its own: there exists a $B$ such that $A=AB-BA$. Of course, $B$ is not unique. - -Consider the complex case ($k=\mathbb C$). Endow $M_n(\mathbb C)$ with your beloved norm, preferably either the operator norm $\|\cdot\|_2$ or the Schur--Frobenius--Hilbert--Schmidt norm $\|\cdot\|_F$. If $A$ is nilpotent, what is the smallest value of $\|B\|$, where $B$ is a factor in $A=AB-BA$ ? What is the smallest constant $\mu(n)$ such that for every $n\times n$ nilpotent $A$, there exists such a $B$ with $\|B\|\le\mu(n)$ ? Actually, is there such a finite $\mu(n)$ ? - -Edit. When ${\rm rk}A=1$, that is $A=xy^*$ with $y^*x=0$, one can always take $B$ such that $\|B\|_2=\frac12$ or $\|B\|_F=\sqrt2/2$. Just take $B$ diagonal in a unitary basis $\{\frac{x}{\|x\|_2},\frac{y}{\|y\|_2},\ldots\}$, with eigenvalues $-\frac12,\frac12,0,\ldots,0$. - -REPLY [6 votes]: Consider the matrix $$\left\[ \begin{array}{ccc} 0 & 1 & k \\\ 0 & 0 & 1\\\ 0 & 0 & 0\end{array}\right\].$$ Then the norm of $B$ depends on $k$ (just solve the system of linear equations $AB-BA=A$). So the answer to your question seems to be "no".<|endoftext|> -TITLE: What is a square root of a line bundle? -QUESTION [25 upvotes]: If ${L}$ is a line bundle over a complex manifold, what does the square root line bundle $L^{\frac{1}{2}}$ mean? -After some google, I got to know that there are certain conditions for the existence of square root line bundle. In particular,I've following questions : - -What is the square root of a line bundle, what are the conditions for its existence. -More importantly, how to think of the square root line bundle. Intuitively, It seems that a square root bundle $L^{\frac{1}{2}}$ is a line bundle s.t. the tensor bundle obtained by taking the tensor product of $L^{\frac{1}{2}}$ with itself gives the line bundle $L$.(correct if I am wrong). - -I am particularly interested in square root of the Canonical Line bundle over a Riemann Sphere and the relation of square root bundle to Spinors in QFT. -Please provide some references to look at. - -REPLY [25 votes]: A comment which I am surprised has not been mentioned yet. -There is a short exact sequence of sheaves on X -$$ -0 \to \{\pm 1\} \to {\cal O}_X^\times \to {\cal O}_X^\times \to 0 -$$ -where the second map sends a nonvanishing holomorphic function to its square. You get a long exact sequence of cohomology groups, which in the relevant range is -$$ -H^0(X, {\cal O}_X^\times) \to H^1(X, \{\pm 1\}) \to Pic(X) \stackrel{2}{\to} Pic(X) \to H^2(X, \{\pm 1\}) -$$ -where map on the Picard group sends a line bundle $\cal L$ to its square ${\cal L}^{\otimes 2}$. One therefore gets an obstruction to extracting a square root of the line bundle in the cohomology group $H^2(X, \{\pm 1\})$, and an obstruction to uniqueness which is the aforementioned 2-torsion line bundle. -Chasing the homological algebra explicitly, in Mike Skirvin's answer he mentions taking the square root of the transition functions $g_{\alpha \beta}$ defining the line bundle. Generally, if you do this it will not work out - you must make nonunique choices of $h_{\alpha \beta}$ with $h_{\alpha \beta}^2 = g_{\alpha \beta}$, and you may not get a cocycle. The coboundary $\delta h$ will be a cycle, but because you know its square is $\delta g \equiv 1$ you find that $\delta h$ is made up of locally constant functions taking only the values $\pm 1$. This is a representing cocycle for the obstruction, and asking whether it is a coboundary is equivalent to asking whether one can fix the signs on the $h_{\alpha \beta}$ to make them match up and give you a line bundle.<|endoftext|> -TITLE: When I can safely assume that a function is a Laplace transform of other function? -QUESTION [18 upvotes]: If I have a function and I want to represent it as being the Laplace transform of another, that is, I want to be sure that there is $\hat{f}(s)$ such that my function $f(x)$ can be written as: -$$f(x) = \int ds \hat{f}(s) \exp(-sx)$$ -what conditions should I impose over $f(x)$? -In other words, what are the conditions for the Fourier–Mellin–Bromwich integral -$$\hat{f}(s) = \frac{1}{2\pi i} \int_{\gamma - i\infty}^{\gamma + i\infty} f(x) \exp(sx) dx$$ -to exist? - -REPLY [8 votes]: This kind of question is very interesting, and I too would like to know answers. -Sorry to self-publicise; I hope it's not regarded as impolite, but since I have also considered this exact kind of question, it's quickest just to refer to my own paper (and the references I give in there): Laplace Transform Representations and Paley–Wiener Theorems for Functions on Vertical Strips by Zen Harper, Documenta Math. 15 (2010) 235-254. -Given an analytic function on a vertical strip, I try to find conditions which guarantee it can be represented by a bilateral Laplace transform (in various senses). I definitely don't claim to have any kind of complete answer, but it's the best I know of (by definition! If I knew any better answers, I would have written them in my paper!). -It seems like there are still many open questions about this.<|endoftext|> -TITLE: Counterexample for the Open Mapping Theorem -QUESTION [28 upvotes]: I would like to ask a counterexample for the classical theorem in functional analysis: the open mapping theorem in the case that $Y$ is Banach, but $X$ is not Banach to show that the completeness of X is crucial. -In details, find a continuous linear mapping $T:X \to Y$ such that $T(X)=Y$ and $Y$ is Banach but $T$ is not open. -If we can construct this, we could get an interesting example: there exists a bijective linear (contiuous) mapping between two normed space $X$ and $Y$, and only one of them is Banach. -The counterexamples for the case when $Y$ is not Banach is simple, but I didn't come up if I need $X$ is not Banach and $Y$ is Banach. -Thanks! - -REPLY [15 votes]: Minh, this is not a new answer since you have already a satisfying number of them, but it's a riddle for you, since you are interested in this topic (too long to be posted as a comment). -"Theorem". All Banach norms on a real vector space $X$ are equivalent. -"Proof". (sketch). Let $\|\cdot\|_1$ and $\|\cdot\|_2$ two Banach norms on $X$. Consider $\|\cdot\|_3:=\|\cdot\|_1+ \|\cdot\|_2. $ Prove that it is actually a norm. Prove that a sequence converge to $x\in X$ w.r.to $\|\cdot\|_3$ if and only if it converges to $x$ both w.r.to $\|\cdot\|_1$ and w.r.to $\|\cdot\|_2$. Prove that a sequence is Cauchy wrto $\|\cdot\|_3$ if and only if it is Cauchy both w.r.to $\|\cdot\|_1$ and w.r.to $\|\cdot\|_2$. Deduce that $\|\cdot\|_3$ is complete. Apply the OMT tho the identity from $(X, \|\cdot\|_3)$ to $(X, \|\cdot\|_1)$ and from $(X, \|\cdot\|_3)$ to $(X, \|\cdot\|_2)$, and deduce that the three norms are equivalent. -However, if all Banach norms on $X$ are equivalent, all linear forms are continuous, and in infinite dimension there are non-continuous linear forms.<|endoftext|> -TITLE: Invertible matrices satisfying $[x,y,y]=x$ -QUESTION [28 upvotes]: I have been thinking about this question for quite some time but now this question by Denis Serre revived some hope. - Question. Let $x,y$ be invertible matrices (say, over $\mathbb C$) and $[x,y,y]=x$ where $[a,b]=a^{-1}b^{-1}ab$, $[a,b,c]=[[a,b],c]$. Does it follow that some power of $x$ is unipotent? -The motivation is this. Consider the one-relator group $\langle x,y \mid [x,y,y]=x\rangle$. It is hyperbolic (proved by A. Minasyan) and residually finite (that is proved in my paper with A. Borisov). If the answer to the above question is "yes", then that group would be non-linear which would provide an explicit example of non-linear hyperbolic group. - Update 1. Can $x$ in the above be a diagonal matrix and not a root of 1? - Update 2. The group is residually finite, so it has many representations by matrices such that $x, y$ have finite orders (hence their powers are unipotents). - Update 3. The group has presentation as an ascending HNN extension of the free group: $\langle a,b,t \mid a^t=ab, b^t=ba\rangle$. So it is related to the Morse-Thue map. Properties of that map may have something to do with the question. See two quasi-motivations of the question as my comments . - -REPLY [13 votes]: The answer is "No". Indeed, consider the 1-related group $G=\langle x,y \mid [x,y,y]=x\rangle$ That group has a presentation $\langle a,b,t \mid a^t=ab, b^t=ba\rangle$ (easy to check). Thus it is an ascending HNN extension of the free group. The group $G$ is hyperbolic (proved by Minasyan using the Bestvina-Feighn combination theorem). By a theorem of Hagen and Wise, the group (and almost every other hyperbolic ascending HNN extensions of a free group) acts geometrically on a finite dimensional CAT(0)-cube complex. By a result of Ian Agol then the group $G$ is virtually special and hence linear, i.e., there exists an injective homomorphism $\phi$ from $G$ to a special linear group (over $\mathbb{Z}$). Since $G$ is hyperbolic, it does not contain nilpotent non-Abelian subgroups. Thus the pair of integer matrices $(\phi(a), \phi(b))$ is an example showing that the answer is "no".<|endoftext|> -TITLE: The non-convergence of f(f(x))=exp(x)-1 and labeled rooted trees -QUESTION [7 upvotes]: This question is closely related to MO f(f(x))=exp(x)-1 and other functions “just in the middle” between linear and exponential. Consider $e^{e^x-1}$, this is the generating function of the Bell numbers. A more general way to look at Bell numbers is as rooted trees, hierarchies of height 2. Given $g(x)=e^x-1$, $g^n(x), n \in \mathbb{N}$ is the generating function of hierarchies of height n. See page 107 - 110 of Analytic Combinatorics. The ECS should have the integer sequences associated with hierarchies of different heights. Also see OEIS - - Integer sequence height OEIS - {1,1/2,1/8,0,1/32,-7/128,1/128,159/256} 1/2 A052122 - {0,1,1,1,1,1,1,1,1} 1 - {1,2,5,15,52,203,877,4140} 2 A000110 - {1,3,12,60,358,2471,19302,167894} 3 A000258 - {1,4,22,154,1304,12915,146115,1855570} 4 A000307 - {1,-1,2,-6,24,-120,720,-5040} -1 A000142 - {1,-2,7,-35,228,-1834,17582,-195866} -2 A003713 - -Several solutions for $f(f(x))=e^x-1$ have been proposed on MO, but the work of I.N. Baker is cited as proving that $f(x)$ has no convergent solution, "even in an ϵ-ball around 0." I am currently trying to read the original German, to understand Baker's proof. -Question 1 Could someone summarize Baker's proof? It is frequently referred to and an explanation in English would be wonderful. -Question 2 Formal power series can contain useful information, even if the are divergent. It seems that divergent series are not treated with quite the contempt they used to be. I believe on the Tetration Forum that someone raised the possibility of $f(x)$ being Borel summable. What are the potential options for "rehabilitating" a series that is not nicely convergent. -Question 3 If $g(x)=e^x-1$, $g^n(x), n \in \mathbb{N}$ is the generating function of hierarchies of height n, doesn't $g(x)=e^x-1$, $g^n(x), n \in \mathbb{R}$ consists of labeled rooted trees of fractional height? So shouldn't $f(x)=g^\frac{1}{2}(x)$ be the generating function for labeled rooted trees of height $\frac{1}{2}$? -Doesn't the divergence of $f(x)=g^\frac{1}{2}(x)$ imply that a label rooted tree of height $\frac{1}{2}$ have infinitely many leaves, that the width of the tree is infinite. Can't be use the fact that we are working with a labeled rooted tree to constrain the width of the tree from becoming infinite? - -REPLY [2 votes]: This is an attempt to summarize some work related to question 2 which I do not fully understand myself. I am summarizing Sections 1.1 and 1.2 of Dudko's thesis, whose exposition is excellent (including work of earlier authors which he describes). -Set $F(z) = e^z-1$, so $F(z) = z+z^2/2+z^3/6+O(z^4)$. This discussion will apply to functional square roots of any $F(z)$ of the form $z+c z^2+O(z^3)$ for $c \neq 0$. Set $f(w) = 1/F(1/w)$, so $f(w) = w - 1/2 + w/12 + O(w^2)$. We will attempt to find a composition square root $f^{\langle 1/2 \rangle}(w)$ for $w$, the change of coordinates $w \mapsto 1/z$ will then change it into a compositional square root for $F$. -Suppose that we had an invertible holomorphic function $\alpha$ obeying -$$\alpha(f(w)) = \alpha(w)-1/2. \quad (\ast)$$ -For now, I'll be sloppy about on what region $\alpha$ is defined; this will eventually be a crucial issue. Such an $\alpha$ is called a Fatou coordinate. -Then we could define fractional compositions $f^{\langle s \rangle}$ by $f^{\langle s \rangle}(w) = \alpha^{-1}(\alpha(w)-s/2)$ and we would clearly have $f^{\langle s \rangle} \circ f^{\langle t \rangle} = f^{\langle s+t \rangle}$ and $f^{\langle 1 \rangle}=f$. -There is a unique formal power series solution -$$\alpha(w) = w+\frac{1}{6} \log w + \sum_{n \geq 1} c_n w^{-n}$$ -to $(\ast)$. -Dudko shows (Theorem 37) that, for any $\delta>0$ there is an $R>0$ such that the sum $\sum c_n z^{-n}$ is Borel summable on a region of the form $U_+ = \{ r e^{i \theta}: r > R, \theta \in (-\pi+\delta, \pi - \delta) \}$ and is separately Borel summable on a region of the form $U_- = \{ r e^{i \theta}: r > R, \theta \in (\delta, 2 \pi - \delta) \}$. Here the integral defining the first Borel sum is along the positive real axis, and the integral for the second is on the negative real axis. However, the two Borel summations have different values! I'm not sure how to translate that Borel summability into the Borel summability you are looking for, but it seems in the same neighborhood.<|endoftext|> -TITLE: Essentially one random metric on $\mathbb{S}^2$? -QUESTION [17 upvotes]: I heard it claimed that there is, in some sense, only -one random metric on $\mathbb{S}^2$. -I would appreciate any pointer to literature that explicates -this intriguing claim. -So far my own searches have not struck a suitable source. -I don't know enough about the topic to even define -what constitutes a random metric in the context of this claim, -so I cannot ask a sharper question. -I seek references to learn more. Thanks! -Addendum. -Reading the literature kindly suggested by jc, I believe that the source is the work -of -Jean-Francois Le Gall, and in particular, his paper -"The topological structure of scaling limits of large planar maps" -Invent. Math. 169 (2007), no. 3, 621--670 -arXiv:math/0607567v2 math.PR. -He shows that a random quadrangulation converges in the Gromov-Hausdorff metric -to a limiting metric space. -Here is a quote from Le Gall's lectures at Clay Inst. on the topic: - -This limiting random metric space, which is called the Brownian map, - can be viewed as a "Brownian surface" in the same sense as - Brownian motion is the limit of rescaled discrete paths. - The Brownian map is almost surely homeomorphic to the two-dimensional sphere, although it has Hausdorff dimension 4. - -REPLY [17 votes]: I'm not sure if this is what is being referred to, but there is a notion of random planar maps which are random triangulations of the sphere and the "scaling limit" of large triangulations is expected to yield a random metric space with the topology of $S^2$ (called the "Brownian Map") though this has not yet been completely proved. Various funny results are known though, such as that $S^2$ with this random metric has Hausdorff dimension 4. -The webpage of Le Gall contains lecture notes around these ideas. See particularly this recent introduction. Olivier Bernardi also has pretty slides for two lectures on the topic "Scaling Limit of Random Planar Maps". - - -update of April 10, 2011: Grégory Miermont has posted a preprint claiming a proof of the result I was vaguely describing above for quadrangulations, rather than triangulations (see the final section of his preprint for thoughts on "universality"). According to the introduction, Jean-François Le Gall has also proven a similar result with a different method recently as well. The abstract: - -We prove that uniform random quadrangulations of the sphere with $n$ faces, endowed with the usual graph distance and renormalized by $n^{-1/4}$, converge as $n\to\infty$ in distribution for the Gromov-Hausdorff topology to a limiting metric space. We validate a conjecture by Le Gall, by showing that the limit is (up to a scale constant) the so-called {\em Brownian map}, which was introduced by Marckert & Mokkadem and Le Gall as the most natural candidate for the scaling limit of many models of random plane maps. The proof relies strongly on the concept of {\em geodesic stars} in the map, which are configurations made of several geodesics that only share a common endpoint and do not meet elsewhere.<|endoftext|> -TITLE: Algebraic Curves and Phase Diagrams of Physical Systems -QUESTION [10 upvotes]: Lots of low degree curves arise naturally as the phase spaces of physical systems (that is, the curve parameterized by $(q,p)$ where $q$ is a generalized position variable and $p$ is a generalized momentum variable (such that $p=\dot{q}$, etc). -For instance, if degree is 1, then we can construct the curve as the phase space of a particle moving with constant velocity, and it is parameterized by $(x,0)$, so is given by the equation $y=0$, up to a choice of coordinates. -For $d=2$, a conic, we can use, for instance, a spring, whose position is $\cos(x)$, and so $(\cos x,\sin x)$ parameterizes a circle (ellipses are also fairly easy to see how to do this way, and parabolas are linearly accelerated particles, though I don't see hyperbolas immediately) -For $d=3$, it is well known that a simple pendulum has phase space an elliptic curve with roots the initial height, the length of the pendulum, and minus the length. -Are there other nice examples like this? Is there a natural physical system that realizes hyperbolas? Does every real elliptic curve arise in this way? How about quartic or higher curves? I assume there are physical systems that work, as every ODE is modeling SOME system, but are there well-known examples that arise naturally in physics? - -REPLY [2 votes]: I have a nice example concerning the electric field in two dimensions, which I'm not sure if physicists know about. One can easily phrase it in terms of ODE. -The electric field lines of a system of point charges in two dimensions form a pencil of (real) algebraic curves, the degree of which equals the sum of (the absolute values of) the charges (which have to be integers to begin with, and the system only has finite number of charges). You get a lot of singular ones in particular, and irreducible singular ones. -I'd be happy to see a reference of it. It's so natural that one would expect that the 19th century physicists/mathematicians know about them.<|endoftext|> -TITLE: Do these properties characterize differentiation? -QUESTION [34 upvotes]: Let $L: C^\infty(\mathbb{R}) \to C^\infty(\mathbb{R})$ be a linear operator which satisfies: -$L(1) = 0$ -$L(x) = 1$ -$L(f \cdot g) = f \cdot L(g) + g \cdot L(f)$ -Is $L$ necessarily the derivative? Maybe if I throw in some kind of continuity assumption on $L$? If it helps you can throw the "chain rule" into the list of properties. -I can see that $L$ must send any polynomial function to it's derivative. I want to say "just approximate any function by polynomials, and pass to a limit", but I see two complications: First $\mathbb{R}$ is not compact, so such an approximation scheme is not likely to fly. Maybe convolution with smooth cutoff functions could help me here. Even if I could rig up something I am concerned that if polynomials $p_n$ converge to $f$, I may not have $p_n'$ converging to $f'$. My Analysis skills are really not too hot so I would like some help. -I am interested in this question because it is a slight variant of a characterization given here: -Why do we teach calculus students the derivative as a limit? -I am not sure whether or not those properties characterize the derivative, and they are closely related to mine. -If these properties do not characterize the derivative operator, I would like to see another operator which satisfies these properties. Can you really write one down or do you need the axiom of choice? I feel that any counterexample would have to be very weird. - -REPLY [34 votes]: Here is a slightly more general take on this question. First notice that your condition $L(1) = 0$ is redundant. That is because if you take $f = g = 1$ in your third condition, you get $L(1) = L(1) + L(1)$. It is a theorem that goes back as far as I know to Chevalley (see around page 76 in his Theory of Lie Groups) that if $M$ is a $C^\infty$ manifold, then any linear map $L$ of $C^\infty(M)$ to itself that satisfies the derivation condition (your third condition) is a smooth vector field. This means that in local coordinate $(x_1,\ldots,x_n)$ it has the form $L(f) = \sum_i h_i {\partial f\over \partial x_i}$ where the $h_i$ are smooth functions. Moreover, if we compute $L(x_j)$ using this formula we see that $h_j$ is just $L(x_j)$. So in particular if a linear map $L$ of $C^\infty (R^n)$ to itself satisfies the derivation condition and $L(x_i) = \delta_{ij}$, then $L = {\partial \over \partial x_j}$.<|endoftext|> -TITLE: local structure of free $\mathbb{R}$ actions -QUESTION [8 upvotes]: Assume the topological group $\mathbb{R}$ acts properly on a space $X$. Does then the projection map $p:X\rightarrow \mathbb{R}\backslash X$ have local sections ? -(for every $\mathbb{R}x\in \mathbb{R}\backslash X$, there is a open neighbourhood $U \subset \mathbb{R}\backslash X$) and a section of $p|_{p^{-1}(U)}:p^{-1}(U)\rightarrow U$). -Are there any nice conditions for $X$, that imply the existence of local sections ? - -REPLY [10 votes]: Such a theorem is proved for completely regular spaces $X$ in my article: -On the Existence of Slices for Actions of Non-Compact Lie Groups, Richard S. Palais, The Annals of Mathematics, Second Series, Vol. 73, No. 2 (Mar., 1961), pp. 295-323 . -Actually, that paper considers more general groups than just $\mathbb{R}$ (any locally compact group) and for actions a little more general than proper (what I call Cartan G-Spaces). The paper is available from JSTOR. (Note that what I prove under these circumstances is the existence of a slice. But since for a proper action the isotropy group at any point is compact, for the case of $\mathbb{R}$ this means that all isotropy groups are trivial, so the action is automatically free, and a slice is a local section.)<|endoftext|> -TITLE: $A_5$-extension of number fields unramified everywhere -QUESTION [32 upvotes]: So I was having tea with a colleague immensely more talented than myself and we were discussing his teaching algebraic number theory. He told me that he had given a few examples of abelian and solvable extensions unramified everywhere for his students to play with and that he had find this easy to construct with class field theory in the back of his head. But then he asked me if I knew how to construct an extension of number fields with Galois group $A_{5}$ and unramified everywhere. All I could say at the time (and now) is: - -There are Hilbert modular forms unramified everywhere. -There are Hilbert modular forms whose residual $G_{{F}_{v}}$-representation mod $p$ is trivial for all $v|p$. -There are Hilbert modular forms whose residual $G_{F}$-representation mod $p$ has image $A_{5}$ inside $\operatorname{GL}_{2}(\mathbb F_{p})$. - -Suppose there is a Hilbert modular form satisfying all three conditions. Then the Galois extension through which its residual $G_{F}$-representation factors would have Galois group $A_{5}$ and would be unramified everywhere. -Can this be made to work? -Regardless of the validity of this circle of idea, can you construct an extension of number fields unramified everywhere and with Galois group $A_{5}$? - -REPLY [9 votes]: Although the question was only about unramified $\mathfrak{A}_5$-extensions, and has been completely answered, it might not be superfluous to mention the following paper which I happened to come across today: -MR0819826 (87e:11122) -Elstrodt, J.(D-MUNS); Grunewald, F.(D-BONN); Mennicke, J.(D-BLF) -On unramified $A_m$-extensions of quadratic number fields. -Glasgow Math. J. 27 (1985), 31–37. -An explicit description is given of unramified extensions $S/k$ with Galois group equal to the alternating group $A_n$, where $k$ is a quadratic number field. The authors prove that if $f(x)\in {\bf Z}[x]$ is a monic, irreducible polynomial of degree $n$ with square-free discriminant and Galois group $S_n$, then $S/k$ is an unramified $A_n$-extension. Here $S$ denotes the splitting field for $f(x)$ over ${\bf Q}$ and $k={\bf Q}(\sqrt{\Delta})$, where $\Delta$ is the determinant of $f$. The proof involves a series of calculations which show that $S/k$ has relative different 1. -In the final section, 84 examples of unramified $A_5$-extensions of quadratic fields are given. In 15 of the cases the quadratic field is real and in 69 cases it is imaginary. This list contains an example (with real quadratic field) due to E. Artin, which was mentioned by S. Lang [Algebraic number theory, see p. 121, Addison-Wesley, Reading, Mass., 1970]. -Reviewed by Charles J. Parry -Addendum (2011/03/30) Kedlaya's preprint mentioned by Speyer is now available on the arXiv. A corollary is that for each $n\geq3$, infinitely many quadratic -number fields admit everywhere unramified degree-$n$ extensions whose normal -closures have Galois group $\mathfrak{A}_n$. -Addendum Kedlaya's paper has now appeared in the Proceedings of the AMS 140 (2012), 3025--3033<|endoftext|> -TITLE: Finding the Boundary Faces of the Zonohedron -QUESTION [5 upvotes]: A zonotope is a linear combination of m vectors with coefficients in [0,1]: $Z = \{ \sum \lambda_i v_i : 0 \leq \lambda _i \leq 1 \}$. The fancy way is to say it's the Minkowski sum of line segments in Rn. -One step in a certain geometric construction is to find the boundary faces of a zonohedron (3D). The boundary points all have λ=0 or 1 but that requires finding 2m points. It's also not clear which ones are within the polygon and which ones are corners. It might be possible to do it faster inductively. -In two dimensions you can arrange the set $\{ v_i, -v_i: i = 1 \dots m\}$ in a circle and add them in clockwise order. In 3D, I might arrange the vectors in a sphere, but then I'm not sure in which order to add the vectors. I read somewhere, this is like integrating a discrete version of the Gauss map. - -REPLY [6 votes]: If you are looking for facets rather than vertices, the answer is easier to formulate in terms of flats of the corresponding matroid. Take the (n-1)-dim flats - each corresponds to 2 facets (one on each side). For vertices, the answer is easier to formulate in terms of oriented matroids. For more on zonotopes and connections to matroids, read here: - -A. Björner, M. Las Vergnas, B. Sturmfels, N. White, G. Ziegler, Oriented Matroids. Cambridge. -Gunter M. Ziegler, Lectures on Polytopes, Springer.<|endoftext|> -TITLE: If $k[S]$ is noetherian, is S finitely generated? -QUESTION [5 upvotes]: Let $S$ be a semigroup. If $S$ is abelian, then it follows that the semigroup algebra $k[S]$ is finitely generated if and only if $S$ is. -What if we relax the condition on $k[S]$, so that $k[S]$ is only noetherian. Does it in this case follow that $S$ is finitely generated? - -REPLY [9 votes]: It is an open problem (or was, last time I checked!) whether the noetherianity of $k[S]$ implies finite generation of $S$, when $S$ is not abelian. -This is discussed in chapter 5 of Noetherian semigroup algebras by Eric Jespers and Jan Okniński, along with various cases where we know that $S$ is finitely generated. They prove, for example, that this is so if $k[S]$ satisfies a polynomial identity, and this gives the case in which $S$ is abelian as a corollary.<|endoftext|> -TITLE: Wikipedia's definition of 'locally free sheaf' -QUESTION [13 upvotes]: Let $R$ be a, say, noetherian ring and $M$ an $R$-module. The Wikipedia article on 'locally free sheaf' tells me that the following two statements are equivalent: - -The module $M$ is locally free (Edit: this means there is an open cover $\{U_i\}$ of $Spec R$ such that every $\tilde{M}_{|U_i}$ is free as an ${\mathcal{O}_{Spec R}}{|U_i}$-module.) -$M_p$ is a free $R_p$-module for every prime ideal $p$ of $R$. - -I see that these two things are equivalent if $M$ is finitely generated but I cannot see this in general, even if $R$ is noetherian. Am I missing something or is there a mistake on Wikipedia? -If the latter case is true, has anybody an example of a (non-finitely generated) $R$-module $M$ over a noetherian $R$ such that $M_p=(R_p)^{n_p}$ for every prime ideal $p$ of $R$ and such that $M$ is not locally free? - -REPLY [26 votes]: Dear roger123, let $R$ be a commutative ring and $M$ an $R$-module ( which I do not suppose finitely generated). In order to minimize the risk of misunderstandings, allow me to introduce the following terminology: -Locfree The module $M$ is locally free if for every $P \in Spec (R)$ there is an element $f \in Spec(R)$ such that $f \notin P$ and that $M_f$ is a free $R_f$ - module. -Punctfree The module $M$ is punctually free if for every $P \in Spec (R)$ the $R_{P} $ - module $M_P$ is free. -Fact 1 Every locally free module is punctually free. Clear. -Fact 2 Despite Wikipedia's claim, it is false that a punctually free module is locally free. -Fact 3 However if the punctually free $R$- module $M$ is also finitely presented, then it is indeed locally free. -Fact 4 A finitely generated module is locally free if and only it is projective. -Fact 5 A projective module over a local ring is free.This was proved by Kaplansky and is remarkable in that, let me repeat it, the module $M$ is not supposed to be finitely generated. -A family of counterexamples to support Fact 2 Let R be a Von Neumann regular ring. This means that every $r\in R$ can be written $r=r^2s$ for some $s\in R$. For example, every Boolean ring is Von Neumann regular. Take a non-principal ideal $I \subset R$. Then the $R$- module $R/I$ is finitely generated (by one generator: the class of 1 !), all its localizations are free but it is not locally free because it is not projective (cf. Fact 4) .The standard way of manufacturing that kind of examples is to take for R an infinite product of fields -$\prod \limits_{j \in J}K_j$ and for $I$ the set of families $(a_j)_{j\in J}$ with $a_j =0$ except for finitely many $j$ 's. -Final irony In the above section on counterexamples I claimed that the $R$ - module $R/I$ is not projective.This is because in all generality a quotient $R/I$ of a ring $R$ by an ideal $I$ can only be $R$ - projective if $I$ is principal . And I learned this fact in...Wikipedia !<|endoftext|> -TITLE: Expected inverse determinant with independent rows -QUESTION [5 upvotes]: Let $a_1,a_2,\dots,a_n$ be independent identically distributed random vectors in $\mathbb R^n$. I need a bound for $E[|\det A|^{-1}]$, where $A$ is the matrix composed out of these vectors. -More specifically, these vectors take their values on a curve. -And more generally, I will be happy even if there is an estimate for a "non-square" determinant, precisely, for $E[G^{-1/2}]$, where $G$ is the Grammian determinant of $n$ iid vector in $\mathbb R^m$ and $m>n$. - -Update: yes, I need an upper bound. -Precisely, I have something like $a_i = (f_1(\xi_i),f_2(\xi_i),\dots,f_n(\xi_i))$, where $f_j(x) = |x|^{-\alpha_i}\sin(\beta_i x)$ with $\alpha_i\in(1,2)$ and look for an estimate in terms of $|\alpha_i-\alpha_j|$ and $|\beta_i-\beta_j|$. -I'm flexible with the choice of distribution for $\xi$, but it should be the same for all sets of $\alpha$'s and $\beta$'s. - -REPLY [3 votes]: Perhaps my paper "How good is Hadamard's inequality for determinants", Canad. Math. Bull. 27 (1984) 260-264 may help. -A rough statement of the main theorem is the following. Assume that we have a probability distribution on the set of all real $n \times n$ matrices $A$ such that: (1) the density at $A$ only depends on the lengths of the columns of $A$; and (2) the probability that $\det A$ is nonzero is 1. -Define the Hadamard ratio $h(A)$ to be $|det A|$ divided by the product of the lengths of the columns. -Then for nearly all $A$ the value of $h(A)^2$ is close to $n!/n^n$. (The theorem gives more precise results.)<|endoftext|> -TITLE: Amenability of groups -QUESTION [21 upvotes]: Let $G$ be non-amenable finitely generated group. -1) Is it true that there exists a sequence $S(n)$ of sets which generate $G$ and such that -$\frac{1}{|S(n)|}||\sum_{g\in S(n)} \lambda(g)||\rightarrow 0$ when $n\rightarrow \infty$. -2) The same as (1), but $S(n)$ is finite subset of $G$. -Here $\lambda:G\rightarrow B(l^2(G))$ is left regular representation of $G$. -Also 1) is reformulation of 2). -Edit: here are some discussions on the question. - -REPLY [9 votes]: It was proved by Nagnibeda-Smirnova and myself (http://arxiv.org/abs/1206.2183) that if a group $\Gamma$ contains an infinite normal subgroup $N$ such that $\Gamma/N$ is not amenable then the question above for $\Gamma$ is true. This gave many examples of groups which are non-amenable and does not contain $\mathbb{F}_2$ as a subgroup, such as Burnside and Golod-Shafarevych groups. As an easy consequence: if $\Gamma$ is not amenable then for $\Gamma\times \mathbb{Z}$ the question is true. -The question in the complete generality was finally solved by Andreas Thom (http://arxiv.org/abs/1306.1767) -There are applications to percolation theory (due to Nagnibeda-Pak) and to operator space analog of von Neumann's conjecture (due to Pisier) listed in the citations above. -To my current knowledge (please correct me here), the question itself should be contributed either to Gilles Pisier or to Nagnibeda-Pak.<|endoftext|> -TITLE: Galois Groups of a family of polynomials -QUESTION [26 upvotes]: I've stumbled across the family of polynomials -$ f_p(x) = x^{p-1} + 2 x^{p-2} + \cdots + (p-1) x + p $, -where $p$ is an odd prime. It's not too hard to show that $f_p(x)$ is irreducible over $\mathbb{Q}$ -- look at the Newton polygon of $f_p(x+1)$ over $\mathbb{Q}_p$ and you see that it factors as the product of an irreducible polynomial of degree $p-2$ and a linear. Since $f_p(x)$ has no real roots (look at the derivative of $f_p(x) (x-1)^2$) it must be irreducible over $\mathbb{Q}$. It's also not hard to see that the only primes dividing the discriminant are $2, p$ and primes dividing $p+1$. I would expect that the Galois group of a randomish polynomial would be the full symmetric group. Indeed, according to Magma this is true for $f_p(x)$ for $p=3,5, \dots, 61$ with the exception of $p=7,17$. So my question is -- are these the only exceptions? -Added later: I've had Magma find the Galois group for primes through 101 and it found another exception: $p=97$. So my initial guess was wrong. -Another addition: If one looks at odd $p$ (not just prime) for $p < 100$ there is another exception, 49. Also 241 is not an exception (misread magma's output). -The ideas in the following two papers may be of help: -"On the Galois Groups of the exponential Taylor polynomials" by Robert Coleman, in L'Enseignement Mathematique, v 33 (1987) pp 183-189 -and -"On the Galois Group of generalized Laguerre polynomials" by Farshid Hajir, J. Th´eor. Nombres Bordeaux 17 (2005), no. 2, 517–525 (also available on the author's web page). - -REPLY [12 votes]: Let $\alpha$ be a root of a polynomial -$f(x) \in \mathbf{Q}[x]$ of degree $n$, let $K = \mathbf{Q}(\alpha)$, -$L$ be the Galois closure of $K$, and -$G = \mathrm{Gal}(L/\mathbf{Q}) \subset S_n$. -How does one prove that a permutation group contains $A_n$? -Following Jordan, the usual method is to show that it -is sufficiently highly transitive. Also following Jordan, -to do this it suffices to construct subgroups of $G$ which -act faithfully and transitively on $n-k$ points and trivially -on the other $k$ points (for $k$ large, $\ge 6$ using CFSG), and -to show that $G$ is primitive. (The standard method for doing this is -to find $l$-cycles for a prime $l$.) -In the context of a Galois group, the most obvious place -to look for "elements" is to consider the decomposition -groups $D$ at places of $\mathbf{Q}$. -If $l$ is unramified in $K$, this corresponds to looking -at a Frobenius element (conjugacy class). In practice -(as far as a computation goes) this is quite useful, -but theoretically it is not so great unless there is a -prime $l$ for which the factorization is particularly clean. -This leaves the places which ramify in $L$. -For example, if $v = \infty$, one is considering -the action of complex conjugation; if there are exactly -two complex roots then $c$ is a $2$-cycle, and from Jordan's -theorem (easy in this case) we see that if $G$ is primitive -then $G$ is $S_n$. -The proposed method (following Coleman et. al.) for proving that -$G$ contains $A_n$ is somewhat misguided, I think. The -key point about the polynomial -$\sum_{k=0}^{n} x^k/k!$ is that the corresponding field is ramified at many primes, -and the decomposition groups at these primes give the -requisite elements. Conversely, the polynomial considered in this -problem corresponds to a field with somewhat limited ramification -- as has been noted, the only primes which ramify divide -$p(p+1)$. -It can be hard to compute Galois groups of random families of polynomials in general. I do not know if this is true in the present case, but given the lack of motivation I won't spend any more time thinking about it than the last hour or two, and instead give some partial results. However, the methods -given here may well apply more generally. -Let $n = p - 1$. -CLAIM: Suppose that $p+1$ is exactly divisible by a prime $l > 3$. -Then $G$ contains $A_{n}$. (This applies to a set $p$ -of relative density one inside the primes.) -STEP I: Factorization of $p$; $G$ is primitive. -Let $f(x) = x^{p-1} + 2 x^{p-2} + \ldots + p$. -Note that -$$(x-1)^2 f(x) = x(x^{p} - 1) - p(x-1) = x^{p+1} - 1 - (p+1)(x-1).$$ -We deduce that -$f(x) \equiv x(x-1)^{p-2} \mod p$, and that -$$p = \mathfrak{p} \mathfrak{q}^{p-2}$$ -for primes $\mathfrak{p}$ and $\mathfrak{q}$ in -the ring of integers $O_K$ of $K$ both of norm $p$. -(To show this one needs to check that $[O_K:\mathbf{Z}[\alpha]]$ -is co-prime to $p$ - one can do this by considering the Newton -Polygon of $f(x+1)$.) -Let $D \subset G$ be a decomposition -group at $p$. This corresponds -to choosing a simultaneous embedding of the roots -of $f(x)$ into an algebraic closure of the $p$-adic numbers. -We see that we may write -$f(x) = a(x) b(x)$ as polynomials over the $p$-adic numbers (which -I can't latex at this point for some reason), - where $a(x) \equiv x \mod p$ -has degree one and $b(x) \equiv (x-1)^{p-2}$ is -irreducible of degree $p-2$ and -corresponds to a totally ramified extension. -Clearly $D$ acts transitively on the $p-2 = n-1$ roots of $b(x)$ and fixes -the roots of $a(x)$. Since $D \subset G \cap S_{n-1}$, we -see that $G \cap S_{n-1}$ is transitive in $S_{n-1}$ and -so $G$ is $2$-transitive (and hence primitive). -Step II: Factorization of $l$: -Let $l$ be a prime dividing $p+1$. We -assume that $l \ge 5$ and $l$ exactly divides $p+1$. -We see that -$$f(x) \equiv (x-1)^{l-2} \prod_{i=1}^{k-1} (x-\zeta^i)^{l}$$ -where $\zeta$ is a $k$th root of unity and $kl=p+1$. -This suggests that: -$$l = \mathfrak{p}^{l-2} \prod_{i=1}^{k} \mathfrak{q}^l.$$ -This also follows from a Newton polygon argument applied -to $f(x - \zeta^i)$. (Warning, this uses that $l$ exactly divides $p+1$.) -Step III: Some basic facts about local extensions: -Lemma 1. Suppose the ramification degree of $E/\mathbb{Q}_l$ -is $l^m$. Then the ramification degree of the Galois -closure of $E$ is only divisible by primes dividing $l(l^m-1)$. -Proof. Kummer Theory. -Lemma 2. Suppose that $h(x) \in \mathbf{Q}_l[x]$ -is an irreducible polynomial of degree $k$ with $(k,l) = 1$, -such that the -corresponding field $E/\mathbf{Q}_l$ is totally ramified. -If $F$ is the splitting field of $h(x)$, then -$\mathrm{Gal}(F/\mathbf{Q}_l) \subset S_k$ contains a $k$-cycle. -Proof: From a classification of tamely ramified extensions, there -exists an unramified extension $A$ such that $[EA:A] = [E:\mathbf{Q}_l]$ -and $EA/A$ is cyclic and Galois. It follows that -$\mathrm{Gal}(EA/A)$ acts transitively and faithfully on the roots -of $h(x)$, and is thus generated by a $k$-cycle. -Step IV: $G$ contains an $l-2$-cycle. -Consider the decomposition group $D$ at $l$. -The orbits of $D$ correspond to the factorization of $l$ in $O_K$. -On the factors corresponding to primes of the form -$\mathfrak{q}^p_i$, the image of $D$ factors through a group -whose inertia has degree divisible only by primes dividing -$l(l-1)$, by Lemma 1. On the other hand, on -the factor corresponding to $\mathfrak{p}^{l-2}$, the image of -inertia contains an $l-2$ cycle, by Lemma 2. - Since $(l(l-1),l-2) = 1$, we see -that $D \subset G$ contains an $l-2$ cycle. -Step V: Jordan's Theorem. -Since $G$ is primitive, and $G$ contains a subgroup -that acts transitively and faithfully on $l-2$ points -(and trivially on all other points), we deduce -(from the standard proof of Jordan's theorem) -that $G$ is $n-(l-2)+1 = n+3-l$ transitive. This is at least $6$ -(since $n+2$ is at least $2l$) -and so $G$ contains $A_n$ (by CFSG). -STEP VI: (for you, dear reader) -Find the analogous argument when $p+1$ is exactly divisible -by $l^k$ for some $k \ge 2$ --- try to construct a cycle -of degree $l^k - 2$, although be careful as it will no -longer be the case (as it was above) that -$[O_K:\mathbf{Z}[\alpha]]$ was co-prime to $l$. -This still leaves $p-1$ either a power of $2$ or a power -of $2$ times $3$, which might be annoying --- one would -have to think hard about the structure of the decomposition group at $2$ in those cases.<|endoftext|> -TITLE: What is known about the "moduli space of morphisms" $X \to Y$? -QUESTION [8 upvotes]: Let $X$ and $Y$ be projective varieties. I am assuming that there is some construction of a "moduli space" parametrizing the morphisms $X \to Y$. For instance, if one identifies such morphisms with their graphs in $X \times Y$, one can at least hope that these graphs would correspond to a locally closed subscheme of the Chow or Hilbert schemes for $X \times Y$, although this may not be the best way to construct such a space. In the case of $X = Y = \mathbb{P}^n$, this space should probably be a disjoint union of $PGL_n$ with $\mathbb{P}^n$ [Edit:This is not correct; I know a theorem that the image must be either a point or all $\mathbb{P}^n$, but grossly misapplied it by neglecting maps of degree > 1]. (I am requiring the varieties to be projective because, for instance, if $Y$ is the affine line, then the morphisms $X$ to $Y$ would naturally correspond to the global sections of $X$, which for general $X$ is too big to fit in one nice scheme, being infinite-dimensional. Although I suppose it might work if we confined ourselves to e.g. morphisms of a fixed degree, and we are already doing something of this sort in looking at Chow or Hilbert schemes, so perhaps my concern here is needless.) -Is there a standard space that represents these morphisms, and if so, what is known about it? Is there a good reference for this? - -REPLY [6 votes]: There is a good non-stacky intro to this in the first chapter of Rational curves on algebraic varieties by János Kollár. You can also find some very nice applications there, such as Mori's Bend & Break.<|endoftext|> -TITLE: What are the $p$-adic representations of $\hat{\mathbb{Z}}$ ? -QUESTION [12 upvotes]: A continuous representation $\hat{\mathbb{Z}} \rightarrow GL_n(\mathbb{Q}_p)$ is determined by the image of $1$. But the image of $1$ does not always defines such a representation (consider for example the representation which sends $1$ on $p$ from $\mathbb{Z}$ to $GL_1(\mathbb{Q}_p)$). So my question is : what are the conditions on the image of $1$ ? -For example if $n=1$, then I know that $1$ must be sent on an element of $\mathbb{Z}_p^\times$, but I don't know if the converse is true. -EDIT: Correction about the example. - -REPLY [19 votes]: I am going to write a community wiki answer here which people can vote up. -(See this meta thread concerning the Mathoverflow user, -which bumps questions with no voted-up answer.) -Main result: A homomorphism $f: \mathbb Z \to GL_n(\mathbb Q_p)$ extends continuously to -$\hat{\mathbb Z}$ if and only if the image of $f$ can be conjugated into $GL_n(\mathbb Z_p)$. -Proof: -If $f:\hat{\mathbb Z} \to GL_n(\mathbb Q_p)$ is continuous, the image is compact, hence contained -in a maximal compact subgroup, which can be conjugated into $GL_n(\mathbb Z_p)$. -Conversely, if $f:\mathbb Z \to GL_n(\mathbb Q_p)$ lands in a compact subgroup, -then the closure of the image is compact, hence profinite (any compact subgroup of $GL_n(\mathbb Q_p)$ is profinite), and hence $f$ extends to $\hat{\mathbb Z}$ -(since $\hat{\mathbb Z}$ is precisely the profinite completion of $\mathbb Z$). -QED -As noted in the comments, to tell if a matrix (e.g. $f(1)$) can be conjugated into -$GL_n(\mathbb Z_p)$, one simply has to look at the characteristic polynomial, -and ask that all the coefficients lie in $\mathbb Z_p$, with the constant term being -a unit. Thus to apply the theorem in practice, one simply computes the characteristic polynomial of $f(1)$ and see if its satisfies these conditions. -EDIT: Now actually made community wiki; sorry about that --- I thought I had already clicked the CW box, -but obviously not. (The point is that the above argument is just a rephrasing of what is in the comments.)<|endoftext|> -TITLE: Simplest proof of dimension of solution space for linear ODEs -QUESTION [7 upvotes]: Given a general n-th degree linear ODE, what's the easiest way to prove that there are precisely n linearly independent solutions? - -REPLY [7 votes]: I don't know of an easy/easiest way to prove it for a general $n$th degree linear ODE, but it is worth pointing out that in the constant coefficient case you can get this from elementary linear algebra. The idea is that if $N$ is a positive integer and you have complex numbers $c_1, \dots, c_N$, then the solutions to the differential equation -$$ -\sum_{n=0}^N c_k y^{(k)} = 0 -$$ -(here $y^{(k)}$ denotes the $k$th derivative of $y$, interpreted as $y$ when $k=0$) are precisely the elements of the kernel of the operator -$$ -T = \sum_{n=0}^N c_k D^k -$$ -where $D$ is differentation, regarded as an operator on a vector space $V$ of functions (there is some freedom in what particular space you choose here; say the set of all infinitely differentiable functions $\mathbb{R} \to \mathbb{C}$). From the fundamental theorem of algebra, you know there are complex numbers $\omega, \omega_1, \dots, \omega_N$ with the property that the polynomial $\sum_{n=0}^N c_k z^k$ factors as $\omega \prod_{n=1}^n (z - \omega_n)$; it follows that your operator $T$ also factors, in the algebra of operators on $V$, as -$$ -T = \omega \prod_{n=1}^N (D - \omega_n I), -$$ -where $I$ denotes the identity operator on $V$. -The point is that each of the operators $D - \omega_n I$ has a one-dimensional kernel by basic calculus. (For any $k$, the function $f(t) = \exp(kt)$ is a solution to $y' = k y$, and if $g$ is any other, the quotient rule for derivatives shows that $(g/f)' = 0$. So by a standard argument involving the mean value theorem, $g/f$ is constant; so $\{f\}$ is a basis for $D - kI$.) -And it is a basic linear algebra fact that a product of $n$ operators with one-dimensional kernel, can have kernel of dimension at most $n$. (Follows from the more general assertion that if $S_1: V \to V$ and $S_2: V \to V$ are any operators, the dimension of the kernel of $S_1 S_2$ is at most the dimension of the kernel of $S_1$ plus the dimension of the kernel of $S_2$. This very easy consequence of the rank-nullity theorem--- and does not require $V$ to be finite dimensional.) -Why is the kernel of $T$ exactly $n$-dimensional? Well, just write down $n$ linearly independent elements in it, as they do in textbooks. (Of course, if you have the better sort of textbook, the entire argument just given is in there.) -For non-constant coefficients, factoring the corresponding differential operator is no longer the way you want to approach this. But for a lot of ODE, you can still get reasonably elementary theorems about the dimension of the kernel of the operator by applying some kind of transform (e.g. the Laplace transform) and getting in a position where it is just algebra again.<|endoftext|> -TITLE: CW complexes and paracompactness -QUESTION [23 upvotes]: It seems like when we assume "niceness" in homotopy theory we assume that $X$ has the homotopy type of a CW complex, and in fiber bundle theory we assume that $X$ is paracompact. How do these two interact? Is any space with the homotopy type of a CW complex paracompact? (In particular, is $I^I$ paracompact?) -(CW complexes are always paracompact and Hausdorff. According to Milnor (http://www.jstor.org/stable/1993204) a paracompact space that is "equi locally convex" will have the homotopy type of a CW complex. Also according to that paper, if $X$ has the homotopy type of a CW complex and $K$ is actually a finite complex then $X^K$ has the homotopy type of a CW complex.) - -REPLY [8 votes]: For the fact that CW complexes are paracompact, I think that the proof in this book is better than I have seen elsewhere: - - -\bib{frpi:cst}{book}{ - author={Fritsch, Rudolf}, - author={Piccinini, Renzo~A.}, - title={Cellular structures in topology}, - series={Cambridge studies in advanced mathematics}, - publisher={Cambridge University Press}, - date={1990}, - volume={19}, -}<|endoftext|> -TITLE: orbits in locally compact group -QUESTION [7 upvotes]: As everyone knows if $x\in S^1$, then the set $\{ x^n \}$ is either finite or dense. Under which condition is true for any other locally compact group, i.e if $G$ is a locally compact group, and $x\in G$ is a non-trivial element, then the set $x^n$ is either finite or dense. - -REPLY [8 votes]: It is easy to produce groups $G$ in which the set $\{x^n\}$ is finite for every $x\in G$. E.g, let $F$ be any finite group; then $G=F^{\mathbb{N}}$ is a compact, totally disconnected group with this property. So in the OP we may assume that $G$ contains some element generating an infinite cyclic dense subgroup. -I claim that, in this case, $G$ is indeed isomorphic to $S^1$. First, $G$ is locally compact abelian. Here is a result I found in the book by W. Rudin, "Fourier analysis on groups" (Wiley, 1962). Say that a locally compact abelian group is {\it monothetic} if it contains a dense cyclic subgroup. Theorem 2.3.2: every monothetic group is either compact or isomorphic to $\mathbb{Z}$ (as a topological group). Coming back to the OP, our group $G$ is monothetic. By the theorem just quoted, $G$ must be compact (as $\mathbb{Z}$ clearly does not satisfy the assumptions of the OP). -Let $\hat{G}$ be the dual of $G$, a discrete, infinite abelian group. By Pontrygin duality, it is enough to prove that $\hat{G}$ is infinite cyclic. Dualizing the assumptions in the OP, we see that every homomorphism $\hat{G}\rightarrow S^1$ either has finite image, or is injective. In particular, $\hat{G}$ is just infinite (infinite, but every proper quotient is finite). It is not very difficult to see that a just infinite, abelian group is infinite cyclic (see McCarthy, Donald, Infinite groups whose proper quotient groups are finite. I. Comm. Pure Appl. Math. 21 1968 545–562).<|endoftext|> -TITLE: For which $b$ it is possible that $S^n$ can have a Lorentz metric? Why? -QUESTION [10 upvotes]: It is possible that on a sphere $S^n$ there is a natural Riemannian metric in $R^(n+1)$. But it is not always possible for pseudo Riemann metric since the sum of two symmetric matrix which are not positive definite but may have rank different from the two matrix. So I wonder what is the sufficient and necessary condition for the dimension of a sphere which can be endowed with a Lorentz metric. - -REPLY [19 votes]: I would just like to point out that a slightly more general version of the statement in Moroianu's answer holds: -A compact orientable manifold $M$ admits a Lorentzian metric if and only if $\chi(M)=0$. -Clearly, simply-connected manifolds are orientable, hence the above statement is more general. But, e.g., tori clearly admit product Lorentzian metrics. -For the proof of the above statement, recall $M$ admits a semi-Riemannian metric of index $k$ if and only if the tangent bundle $TM$ admits a sub-bundle, or distribution, of rank $k$. Thus, $M$ admits a Lorentzian metric if and only if it admits a line bundle (which is also equivalent to $M$ admitting a vector field that never vanishes). The obstruction to the existence of a line bundle on $M$ is given by the Euler class of $M$ (see e.g. Davis and Kirk "Lecture notes in Algebraic Topology"). By the Gauss-Bonnet-Chern Theorem, the Euler characteristic is the Euler class evaluated on the fundamental class of $M$, so that, for compact orientable manifolds, the obstruction is precisely $\chi(M)$.<|endoftext|> -TITLE: third stable homotopy group of spheres via geometry? -QUESTION [68 upvotes]: It is ''well-known'' that the third stable homotopy group of spheres is cyclic of order $24$. It is also ''well-known'' that the quaternionic Hopf map $\nu:S^7 \to S^4$, an $S^3$-bundle, suspends to a generator of $\pi_8 (S^5)=\pi_{3}^{st}$. It is even better known that the complex Hopf map $\eta:S^3 \to S^2$ suspends to a generator of $\pi_4 (S^3) = \pi_{1}^{st} = Z/2$. For this, there is a reasonably elementary argument, see e.g. Bredon, Topology and Geometry, page 465 f: - -By the long exact sequence, $\pi_3 (S^2)=Z$, generated by $\eta$. -By Freudenthal, $\pi_3 (S^2) \to \pi_4 (S^3) = \pi_{1}^{st}$ is surjective. -Because $Sq^2: H^2(CP^2;F_2) \to H^4(CP^2;F_2)$ is nonzero, the order of $\eta$ in $\pi_{1}^{st}$ is at least $2$ (the relation between these things is that $\eta$ is the attaching map for the $4$-cell of $CP^2$). -By a direct construction, $2\eta$ is stably nullhomotopic. Essentially, $\eta g = r \eta$, where $r,g$ are the complex conjugations on $S^2=CP^1$ and $S^3 \subset C^2$. $g$ is homotopic to the identity, $\eta=r\eta$. The degree of $r$ is $-1$, so after suspension (but not before), composition with $r$ becomes taking the additive inverse. Therefore $\eta=-\eta$ in the stable stem. - -My question is whether one can mimick substantial parts of this argument for $\nu$. Here is what I already know and what not: - -There is a short exact sequence $0 \to Z \to \pi_7 (S^4) \to \pi_6 (S^3) \to 0$ that can be split by the Hopf invariant. Thus $\nu$ generates a free summand. -is the same argument as for $\eta$. -using the Steenrod operations mod $2$ and mod $3$ on $HP^2$, I can see that the order of $\nu$ in $\pi_{3}^{st}$ is at least $6$. -this is a complete mystery to me and certainly to others-:)). How can I bring $24$ in via geometry? How do I relate the quaternions and $24$? What one sees immediately is that one has to be careful when talking about conjugations in the quaternionic setting, in order to avoid proving the false result ''$2 \nu=0 \in \pi_{3}^{st}$''. - -I know that this result goes back to Serre, but I cannot find a detailed computation in his papers and it seems that the calculation using the Postnikov-tower and the Serre spectral sequence is a bit lengthy. There are three other approaches I know but they are much less elementary: -Adams spectral sequence, J-homomorphism (enough to show that the order of $\nu$ is $24$), framed bordism (supported by things like Rochlin's theorem and Hirzebruch's signature formula). -Any idea? P.S.: if there is a similar argument for the octonionic Hopf fibration $S^{15} \to S^8$ (the stable order is 240), that would be really great. - -REPLY [14 votes]: There is a "half-geometric" observation that since -$$ -\pi_3^s\cong H_3(\tilde A_N),\ N\gg0 -$$ -where $\tilde A_N$ is the universal central extension of the $N$th alternating group, elements of $\pi_3^s$ might be described by certain 3-cycles (hopefully of geometric origin). -Seemingly this geometry is easier to discern after the embedding $\tilde A_N\hookrightarrow\mathrm{St}_N(\mathbb Z)$ into the Steinberg group of the integers. -(For a ring $R$, the group $\mathrm{St}_N(R)$ is the universal central extension of $\mathrm E_N(R)$ - the group which for most decent rings is the same as $\mathrm{SL}_N(R)$; one has $K_3(R)\cong H_3(\mathrm{St}(R))$ which in good cases stabilizes after some $\mathrm{St}_N(R)$.) -In "The generalized Grassmann invariant" (late 70ies) K. Igusa introduced technique of pictures to construct a homomorphism -$$ -\chi:K_3(\mathbb Z[\pi])\to H_0(\pi;\mathbb F_2[\pi]); -$$ -in his own words, "The definition of $\chi$ comes from very intuitive geometric considerations, but unfortunately the algebraic analogue is rather clumsy." -Anyway geometry is still there as this map is strongly related to pseudoisotopy of compact manifolds (with $\pi_1=\pi$ and $\pi_2=0$). The image of $\pi_3^s(B\pi_+)\to K_3(\mathbb Z[\pi])$ is in the kernel of $\chi$, and for trivial $\pi$ all this enabled him to detect an element of order 48 in $K_3(\mathbb Z)$; $\pi_3^s$ is a subgroup of index 2 there. His picture representing a generator of $K_3(\mathbb Z)$ - -has been haunting me for decades. It shows something like "$1/2$ of the generator of $\pi_3^s$ living outside $\tilde A_N\hookrightarrow\mathrm{St}_N(\mathbb Z)$" and there surely must be something simpler which represents the generator of $\pi_3^s$ inside $\tilde A_N$ itself. This should not be that difficult as $\tilde A_N$ comes equipped with a very explicit and nice embedding into $\mathrm{Spin}(N)$...<|endoftext|> -TITLE: Lipschitz properties of minima/minimizers of convex functions of two variables -QUESTION [5 upvotes]: Suppose I have a function $f(x,y)$ from $\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ that is convex in both $x$ and $y$. Set -$g(y) = \min_{x} f(x,y)$ -What I would like is for $g(y)$ to be Lipschitz: -$|g(y) - g(y')| \le c \cdot \| y - y' \|$ -Unfortunately, $f(x,y)$ may have a very poor Lipschitz constant for general $x$. Are there general conditions on $f$ for which the minima are Lipschitz? -Alternatively, when can we say the minimizer $x^{\ast}(y) = \arg \min_x f(x,y)$ is Lipschitz in $y$? -I've tried looking in a few convex optimization books for answers, but no luck. - -REPLY [5 votes]: I encountered the same problem three years ago and found some relevant literature. Here are a few. See also the refs therein. -Lipschitz Behavior of Solutions to Convex Minimization Problems. -Jean-Pierre Aubin, -Mathematics of Operations Research, Vol. 9, No. 1. (Feb., 1984), pp. 87-111. -Lipschitz continuity of solutions of linear inequalities, programs and complementarity problems. -O. L. MANGASARIAN and T.-H. SHIAU. -SIAM J. CONTROL AND OPTIMIZATION, 25(3), 1987. -Lipschitz Continuity of Solutions of Variational Inequalities with a Parametric -Polyhedral Constraint. -N. D. Yen, -Mathematics of Operations Research, Vol. 20, No. 3. (Aug., 1995), pp. 695-708. -On Lipschitzian Stability of Optimal Solutions of Parametrized Semi-Infinite -Programs. -Alexander Shapiro, -Mathematics of Operations Research, Vol. 19, No. 3. (Aug., 1994), pp. 743-752. -SHARP LIPSCHITZ CONSTANTS FOR BASIC OPTIMAL SOLUTIONS AND BASIC FEASIBLE SOLUTIONS OF LINEAR PROGRAMS. -Wu Li, -SIAM J. CONTROL AND OPTIMIZATION -Vol. 32, No. I, pp. 140-153, January 1994<|endoftext|> -TITLE: Are there more connected or disconnected graphs on $n$ vertices? -QUESTION [11 upvotes]: Suppose we are talking about graphs with $n$ labeled vertices. Which graphs are more common: connected or disconnected? - -REPLY [3 votes]: I like Jonah Ostroff short and sweet proof, but the key to it lies in the fact that there is not a bijection between the set $S_1$ of connected graphs and the set $S_2$ of disconnected graphs over $n$ labeled vertices for $n \ge 4$, as follows: - -the complement of each disconnected graph is a connected graph (which Ostroff points out) -the complement of a connected graph can also be a connected graph -thus the cardinality of the set of connected graphs must be larger than the cardinality of the disconnected graphs, because while there is a one-to-one mapping of each disconnected graph onto a connected graph, there exist connected graphs which do not map to a disconnected graph - -For example, for $n=4$: -Take the $12$ possible un-drected Hamiltonian paths of length $4$ on a graph over four labeled vertices. -The complement of each of these paths is also a hamiltonian path. -Since we know that the complement of a disconnected graph is obviously connected for $n>3$, then the number of connected graphs is at least equal to the number of disconnected graphs. Hoewever, since for $n>3$, the complements of at least some of the connected graphs are also connected graphs, that means that there must be more connected graphs than there are unconnected graphs. -The $12$ Hamiltonian paths are those connected graphs over $4$ vertices whose complements are also connect: thus the remaining $2^6 - 12 = 52$ graphs are divided into pairs of complement graphs which are connected and disconnected, -yielding a total of $26$ disconnected graphs, and $26+12=38$ connected graphs -over the set of $64$ labeled graphs over $4$ labeled vertices. -The path graphs of length $n$ on the set of $n$ vertices are the canonical example of connected graphs whose complements are also connected graphs (for $n>3$).<|endoftext|> -TITLE: Mirror symmetries for generalized geometries ? -QUESTION [7 upvotes]: For Calabi-Yau three-folds we have $\mathcal{mirror \ symmetry}$: a map that associates most Calabi-Yau three-folds $M$ another Calabi-Yau three-fold $W$ such that $ h^{1,1}(M) = h^{2,1}(W)$ and $ h^{1,1}(W) = h^{2,1}(M)$ where $h^{i,j}$ are the Hodge numbers of the Calabi-Yau. In string theory such a duality leads to the conjecture that the type IIA superstring theory compactified on $M$ is equvilalent to the type IIB compactified on $W$. -$ \textbf{Question} :$ Are there extensions of mirror symmetry applied to generalized geometries (in the sense of Hitchin, Cavalcanti, and Gualtieri)? If so, what is the state of the art of this topic/question? - -REPLY [3 votes]: Mirror symmetry is at the most fundamental level an isomorphism of N=(2,2)-supersymmetric conformal field theories attached to different geometric data, which acts on the supersymmetries as a prescribed outer automorphism (switching A- and B-twists). Calabi-Yaus give rise to such SCFTs, hence one can ask for two CYs to be mirror -- though this is not a map in general, rather a correspondence that can then be pinned down more precisely in terms of large volume limits and SYZ fibrations etc. Mirror symmetry also has many rougher manifestations, such as an isomorphism of topological field theories of A- and B-type attached to geometric data, which now no longer need to be CY. In any case the point is that the generalized geometries you mention are precisely appropriate backgrounds to define such (2,2) SCFTs (see eg articles of Kapustin from around 04), so from the beginning they have a mirror symmetry question (look for pairs giving rise to the same SCFT up to this outer automorphism).<|endoftext|> -TITLE: Betti number and harmonic forms -QUESTION [11 upvotes]: On a compact, boundaryless, Riemannian manifold, the dimension of the space of harmonic k-forms is equal to the k-th Betti number. Is this correct (by Hodge theory)? For example, on the surface of a unit sphere $S^2$ and any equivalent topology, the 0-th Betti number is 1, which is consistent with the fact that any harmonic function on a compact, boundless, Riemannian manifold is a constant. More importantly to me, what about the 1st Betti number? I think the only harmonic 1-form on the topological class of $S^2$ is zero --- you can't comb every hair on a sphere towards the same direction --- right? -Basically, it is easy to prove both results (1st betti number = 0 and the only harmonic 1-form is zero) on the ideal case of $S^2$. But if I want to go further, some topological tool has to be involved? -I am a PDE guy, so if there is any PDE-based proof to the above question, it'd be highly appreciated! - -REPLY [8 votes]: If I understood your question correctly, it is: how can I compute the dimension of the space of harmonic forms? -There is one class of Riemann manifolds where it is possible to write down the harmonic forms explcitly, without topology. These are the symmetric spaces (for example spheres, projective spaces, compact Lie groups, tori, Grassmann manifolds, flag manifolds). The space of harmonic forms on the symmetric space $M=G/H$ is the same as the invariant forms, which is $\Lambda^* (\mathfrak{g}/\mathfrak{h})^G$. And you can compute this algebraically. I guess that each explicit computation of harmonic forms, including your example on $S^2$, is based on that principle (maybe for Riemann surfaces and nonsymmetric homogeneous spaces there are exceptions to this rule). -As pointed out by Gunnar, the dimension of the space of harmonic forms on $M$ does not depend on the metric. So you can conclude that there is no harmonic 1-form on $S^2$, in any wild metric. -Of course, most manifolds do not carry a symmetric metric. Here, if you want to compute the dimension of the space of harmonic forms, it might be easiest to compute first the cohomology and then appeal to the Hodge theorem.<|endoftext|> -TITLE: Reference request: The first cohomology of SL(2,Z) with coefficients in homogeneous polynomials -QUESTION [21 upvotes]: Let $H_k$ be the vector space of degree $k$ homogeneous polynomials in two variables.I'm looking for a reference for the fact that $H^1(SL(2,\mathbb Z);H_k)=M^0(k+2)\oplus\overline{M^0(k+2)}\oplus E_{k+2}$, where - -$M^0(k+2)$ is the space of cuspidal modular forms of weight $k+2$. -$\overline{M^0(k+2)}$ is its conjugate. -$E_k$ is $1$-dimensional if $k\geq 4$ is even, and is zero otherwise. - -I know how to get the $M^0(k+2)\oplus\overline{M^0(k+2)}$ piece. By the Eichler-Shimura isomorphism, this is the same as the cuspidal cohomology $H^1_{cusp}(SL(2,\mathbb Z);H_k)$, spanned by cocycles that vanish on the matrix $T$=[1 1 \ 0 1]. Serge Lang's book on Modular Forms has a detailed explanation of how to get this. I'd really like to know how to get the $E_{k+2}$ piece, which as far as I understand, corresponds somehow to the Eisenstein series, which is the "extra" modular form of weight $k+2$ that is not cuspidal. -It's not too hard to see that $H^1(SL(2,\mathbb Z);H_k)=H^1(PSL(2,\mathbb Z);H_k)$, with $PSL(2,\mathbb Z)$ having the nice presentation $\langle S,T\,|\, (ST)^3=S^2=I\rangle$, so one should be able to represent the $E_{k+2}$ cocycle by specifying its values on $S,T$. -So, to summarize, I'd be satisfied with any proof that $H^1(SL(2,\mathbb Z);H_k)=M^0(k+2)\oplus\overline{M^0(k+2)}\oplus E_{k+2}$, but I'd be happiest with an answer that lets me get my hands on the $E_{k+2}$ cocycle as explicitly as possible, perhaps even telling me its values on $S$ and $T$. Any comments that would help clarify this are appreciated. -Edit: I think I have a better understanding of what the "extra" Eisentstein cocycle is, based on Kevin's comments. It seems that the cuspidal cocycles vanish on $T$, whereas the Eisenstein cocycle vanishes on $S$, although I don't see how to show, for example, that there is only one dimension's worth of cocycles vanishing on $S$, up to coboundaries. (Edit: I'm not entirely sure about this.) -Edit: Shimura's Introduction to the Theory of Automorphic Forms only covers the cuspidal part of the above isomorphism. - -REPLY [17 votes]: The result you're looking for is contained in the following article : -Haberland, Klaus. Perioden von Modulformen einer Variabler and Gruppencohomologie I (German) [Periods of modular forms of one variable and group cohomology I], Math. Nachr. 112 (1983), 245-282. -Let $S_k$ (resp. $M_k$) be the space of holomorphic cusp forms (resp. holomorphic modular forms) for $\Gamma = SL_2(\mathbf{Z})$. Let $\Gamma_{\infty}$ be the stabilizer of $\infty$ in $\Gamma$. Let $V_k$ be the space of polynomials of degree $\leq k-2$ with complex coefficients. Haberland proves an exact sequence -\begin{equation} -(*) \qquad 0 \to S_k \oplus \overline{S_k} \to H^1(\Gamma,V_k) \to H^1(\Gamma_\infty,V_k) \to 0. -\end{equation} -Let $T = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \in \Gamma_{\infty}$. There is a natural map $V_{k+1} \to H^1(\Gamma_\infty,V_k)$ sending a polynomial $P$ to the cocycle $c_P$ determined by $c_P(T) = P(X+1)-P(X)$. It is easy to check that this map induces an isomorphism $\psi : V_{k+1}/V_k \cong H^1(\Gamma_\infty,V_k)$, so that the latter space is one-dimensional. -The "Eisenstein cocycle" you're looking for is a natural map $\delta : M_k \to H^1(\Gamma,V_k)$ which Haberland constructs the following way (actually I learnt this construction and many other properties of $\delta$ during Zagier's 2002-2003 lectures at the Collège de France). -Let $f \in M_k$. Let $\widetilde{f}$ be an Eichler integral of $f$, that is any holomorphic function on $\mathcal{H}$ such that -\begin{equation} -\left(\frac{1}{2\pi i} \frac{d}{dz}\right)^{k-1} \widetilde{f}(z) = f(z). -\end{equation} -Note that $\widetilde{f}$ is unique up to adding some element of $V_k$. Since we integrate $k-1$ times, the function $\widetilde{f}$ should be thought of as a function of "weight" $k-2\cdot (k-1) = 2-k$ (of course this isn't true in the strict sense). Let us make this more precise. -For any $n \in \mathbf{Z}$, let $|_n$ denote the weight $n$ action of $SL_2(\mathbf{R})$ on the space of complex-valued functions on $\mathcal{H}$ (so that any $f \in M_k$ is a fixed vector of the weight $k$ action of $\Gamma$). Note also the weight $2-k$ action gives the usual action of $\Gamma$ on $V_k$. The crucial fact is that we have -\begin{equation} -\widetilde{f} |_{2-k} (\gamma-1) \in V_k \qquad (\gamma \in \Gamma). -\end{equation} -This can be proved using Bol's identity -\begin{equation} -\left(\frac{d}{dz} \right)^{k-1} (F |_{2-k} g) = \left(\frac{d^{k-1} F}{dz^{k-1}} \right) |_k g -\end{equation} -which holds for any holomorphic function $F$ on $\mathcal{H}$ and any $g \in SL_2(\mathbf{R})$. -Since $\gamma \mapsto \widetilde{f} |_{2-k} (\gamma-1)$ is obviously a coboundary in the space of functions on $\mathcal{H}$, it defines a cocycle in the space $V_k$. Therefore we get $\delta(f) \in H^1(\Gamma,V_k)$ and this element doesn't depend on the choice of $\widetilde{f}$. Thus we have constructed $\delta : M_k \to H^1(\Gamma,V_k)$. -It is not difficult to check that if $f =\sum_{n \geq 0} a_n e^{2i\pi nz}$ then the image of $\delta(f)$ in $H^1(\Gamma_\infty,V_k)$ is the image of the polynomial $\frac{a_0 \cdot (2\pi i)^{k-1}}{(k-1)!} \cdot X^{k-1} \in V_{k+1}$ under the isomorphism $\psi$ above. In particular $\delta$ is injective, and the exact sequence $(*)$ gives the isomorphism you want. -Note that there is a distinguished choice of $\widetilde{f}$, namely -\begin{equation} -\widetilde{f} = \sum_{n \geq 1} \frac{a_n}{n^{k-1}} e^{2i\pi nz} + \frac{a_0 \cdot (2\pi i)^{k-1}}{(k-1)!} z^{k-1}. -\end{equation} -Let $c_f \in Z^1(\Gamma,V_k)$ be the cocycle associated to this choice of $\widetilde{f}$. Let us compute the value of $c_f$ on $T$ and $S= \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. First as explained above, we have -\begin{equation} -c_f(T)=\frac{a_0 \cdot (2\pi i)^{k-1}}{(k-1)!} ((X+1)^{k-1}-X^{k-1}). -\end{equation} -To compute $c_f(S)$, Haberland uses the natural integral representation of $\widetilde{f}$ in terms of $f-a_0$, and gets -\begin{equation} -c_f(S) = \frac{(2\pi i)^{k-1}}{(k-2)!} \int_0^{\infty} \left(f(z)-\frac{a_0}{z^k}-a_0 \right) (z-X)^{k-2} dz -\end{equation} -(there is a similar but more complicated formula for $c_f(\gamma)$ for any $\gamma \in \Gamma$, see below). Then $c_f(S)$ can be expressed in terms of the special values of $L(f,s) := \sum_{n=1}^\infty a_n/n^s$ at integers $s = 1,\ldots,k-1$. It is then a good exercise to compute $c_f(S)$ when $f$ is the Eisenstein series $E_k$, in terms of Bernoulli numbers and of $\zeta(k-1)$ (this is Satz 3 in Haberland's article, Kapitel 1). -Please tell me if something isn't clear in my explanation. -EDIT : I found the following expression for $c_f(\gamma)$ where $\gamma \in \Gamma$. It is quite complicated (maybe it could be somewhat simplified) : -\begin{equation} -\begin{aligned} -\frac{(k-2)!}{(2\pi i)^{k-1}} c_f(\gamma) &= \int_{z_0}^{\infty} (f(z)-a_0)(z-X)^{k-2} dz + \int_{\gamma^{-1} \infty}^{z_0} \left(f(z) -\frac{a_0}{(cz+d)^k} \right) (z-X)^{k-2} dz \\ -& + \frac{a_0}{k-1} \left((X-z_0)^{k-1}-(X-\gamma z_0)^{k-1} |_{2-k} \gamma + X^{k-1} |_{2-k} (\gamma-1) \right) -\end{aligned} -\end{equation} -where $z_0 \in \mathcal{H}$ is arbitrary and $\gamma= \begin{pmatrix} a & b \\ c & d \end{pmatrix}$.<|endoftext|> -TITLE: Rigidity of triangle comparison in Alexandrov spaces -QUESTION [9 upvotes]: For $CAT(\kappa)$ spaces $X$ we have following rigidity result: if equality holds in any of the comparison distances between a triangle $\Delta$ in $X$ and the corresponding comparison triangle $\tilde\Delta$ in the model space $M_\kappa$ of constant curvature $\kappa$, then the convex hull of $\Delta$ is isometric to the convex hull of $\tilde\Delta$. -In Alexandrov spaces the picture is different. We do not have such a rigidity anymore. A counterexample can be found by considering two copies of a spherical triangle glued by their boundaries. -The rigidity result that we can obtain is following (I think at least, but I do not know any reference): under the assumption of equality in any of the comparison distances we can embed isometrically the convex hull of the comparison triangle $\tilde\Delta$ in $X$, such that two sides coincide with the corresponding two sides of $\Delta$, but it may happen that the third side does not coincide anymore. An example of this can be seen in the counterexample above. -So my question is following. In the example given above the space is singular. Does anybody know an non singular example (i.e. manifold of sectional curvature $\geq \kappa$), where we do not have the same rigidity as in the $CAT(\kappa)$ case? -Remark: We assume for our triangle $\Delta$ that its perimeter is $<2\pi/\kappa$ and each side has length $<\pi/\kappa$. -Edit: Ok, I will be more sprecific with my question: -Is there a Riemannian manifold $M$ with sectional curvature $\geq \kappa$ and a -a triangle $(x,y,z)$ in $M$ and a point $p$ in the side $yz$ -such that if $(\tilde x, \tilde y, \tilde z)$ is a comparison triangle in $M^2_\kappa$, and $\tilde p$ the corresponding point in $\tilde y \tilde z$, -we have the equality $d(x,p)=d(\tilde x,\tilde p)$ but the triangle $(x,y,z)$ (that is, the 1-dimensional object) cannot be filled with a triangle of constant sectional curvature $\kappa$ (that is, the 2-dimensional object). Such an example is easy to construct if we admit singular Alexandrov spaces, but I do not know any manifold examples. - -REPLY [7 votes]: The question is not stated precisely. -So I'm free to say anything :) - -If you are interested in "non-uniqueness" then the anser is "NO". In any such triangle $[x y z]$ there are at least two distinct geodesics between directions of $[x y]$ and $[x z]$; -thus the space of directions at $x$ can not be a sphere. -If you are interested in "the third side is wrong", the answer is "YES". Take $x=(0,0,0)$, $y=(1,0,1)$ and $z=(0,1,-1)$ in $\mathbb R^2\times [-1,1]$. -Then glue $\mathbb R^2\times \{1\}$ to it-self along reflection $(u,v)\mapsto (-u,v)$ -and -glue $\mathbb R^2\times \{-1\}$ to it-self along reflection $(u,v)\mapsto (u,-v)$. -You get a singular Alexandrov space with triangle $[x y z]$ where the side $[y z]$ might be wrong for some filling of the hinge at $x$. BUT this triangle admins a flat filling. -It is easy to smooth this spaces near both singular lines to obtain Riemannian manifold with the same property. -I know one example of a triangle in singular Alexandrov 3-space such that all distances between points on sides are the same as the corresponding distances in the model triangle, but it can not be filled with a flat triangle. This is a bit tricky to construct. I can not make this example to be Riemannian (and I feel that it is impossible). - -P.S. The answer to the "specific" question is "NO". -Take a 2-dimensional nonnegatively curved manifold $M$ with a pair of points $p,q$ such that there are two minimizing geodesics from $p$ to $q$. One can choose a triangle with vertexes $x=(0,p)$, $y=(1,q)$ and $z=(-1,q)$ in the product $\mathbb R\times M$ which does not admit flat filling. -On the other hand any such triangle satisfies your condition for midpoint of $[y z]$.<|endoftext|> -TITLE: Why is "abelian" infrequently capitalized? -QUESTION [39 upvotes]: Posted with input from meta for improvement. I usually read, e.g. "Gaussian integers" and "Riemannian metrics", and occasionally "euclidean" or "cartesian" or even "lorentzian space", but the latter examples are relatively uncommon. I have also seen, e.g. "artinian" (and it has been commented that "algorithm" might be interpreted in a similar context, though the "al-" prefix seems to me to militate against it). I can't imagine that the relative standing of the individual has much effect, because, hey: Gaussian. Perhaps algebraists have collectively decided that spaces with some structure are fair game for the lower-case treatment? -Scott Carnahan pointed out that some people apparently subscribe to consistent conventions, e.g. "don't capitalize when the name is modified to form an adjective. These seem somewhat inconsistent with current practice, though." As Harry Gindi put it: "I think that if you can say X is (term) without the noun following it, it is generally left uncapitalized. So, notice, for instance, 'the ring X is noetherian', 'the ring X is artinian', 'the group X is abelian', 'the square F is cartesian', 'the square F is cocartesian', etc." -So, to take Deane Yang's formulation (note already the discrepancies in capitalization!): "Is there really a rational explanation why it's 'abelian', 'noetherian', and 'artinian' but 'Euclidean', 'Riemannian', and 'Lorentzian'?" - -REPLY [6 votes]: A simple rule for this is that, if the adjective has a precise mathematical meaning, it should be lower-cased: for example, abelian (= commutative) group, euclidean metric, or gaussian (= normal) distribution. If the meaning is vaguer, referring to the method, style, or approach generally associated with the originator, then capitalizing is appropriate: for example, Bayesian statistics. If the word does not carry an adjectival ending—e.g. Hilbert (space)—it should always be capitalized.<|endoftext|> -TITLE: Reference request: number theory of Z[1/p] -QUESTION [9 upvotes]: Can anyone suggest a good place to read up on the number theoretic properties of and techniques for $\mathbb{Z}[1/p]$, (that is, rational numbers with only powers of a prime $p$ in the denominator)? -I find myself struggling to answer some of the more basic questions about this ring, especially whether or not it is a Euclidean domain, and if so, what the associated Euclidean function/algorithm is. -Thanks in advance! - -REPLY [5 votes]: The general fact here is that any localization of a Euclidean domain is again a Euclidean domain. I will restrict myself to the case where the Euclidean norm on $R$ is multiplicative, i.e., -satisfies $|xy| = |x| |y|$ (as does the absolute value on $\mathbb{Z}$, of course), and in this case I will define an explicit Euclidean norm on the localized ring in terms of the given norm -and the (let's say saturated, WLOG) multiplicative subset $S$. -For a ring $R$, I write $R^{\bullet}$ for $R \setminus \{0\}$. -Since $R$ is Euclidean, it is a UFD, so to give a function $|\ |: R \setminus \{0\} \rightarrow \mathbb{Z}^{> 0}$ such that $|1| = 1$, $|xy| = |x| |y|$ and $x \in R^{\times} \iff |x| = 1$, it is enough to send every principal prime ideal $(\pi)$ to some integer $n_{\pi} > 1$. (This holds because the multiplicative monoid of principal nonzero $R$-ideals is the free commutative monoid on the principal prime ideals.) Then the norm of an arbitrary nonzero element of $R$ is defined by the uniqueness of factorization into principal prime ideals. -The multiplicative group $R_S^{\bullet}$ of a localization $R_S$ is the free commutative monoid on the principal prime ideals $(\pi)$ such that $(\pi) \cap S = \emptyset$. One can view this naturally as a submonoid of $R^{\bullet}$ and therefore define an induced norm $| \ |_S$. In other words, if $x \in R^{\bullet}$, write $x = s_x x'$ where $s_x \in S$ and $x'$ is prime to $S$. then, for any $s \in S$, -$|\frac{x}{s}|_S = |x|_S = |s_x x'|_S = |x'|_S = |x'|$. -Note that for all $x \in R$, we have $|x|_S \leq |x|$. -Let us now show that if $R$ is Euclidean under $| \ |$, $R_S$ is Euclidean under $|\ |_S$: -for $A \in R_S$ and $B \in R_S^{\bullet}$, we must find $Q \in R_S$ such that $|A-QB|_S < |B|_S$. There exist $a,b \in R$ and $s \in S$ such that $A = \frac{a}{s}$, $B = \frac{b}{s}$. Then, since $s \in R_S^{\times}$, $|a-Qb|_S = |\frac{a}{s} - Q \frac{b}{s}|_S = |A - QB|_S$ and $|b|_S = |\frac{b}{s}|_S = |B|_S$, so without loss of generality -we may take $s = 1$. -As above, write $b = s_b b'$, and choose $q \in R$ such that $|a-qb'| < |b'|$. Put $Q = \frac{q}{s_b}$. Then -$|a - Q b|_S = |a- \frac{q}{s_b} b|_S = |a-q b'|_S \leq |a-qb'| < |b'| = |b'|_S = |b|_S.$ -For your particular question $R = \mathbb{Z}$, the Euclidean norm is the usual absolute value, and $S = \{2^a \ | \ a \in \mathbb{Z}^+\}$.<|endoftext|> -TITLE: Aut(G) = $C_3$, G = ? -QUESTION [10 upvotes]: Is there a group G such that Aut(G) = $C_3$? What if we replace 3 with a prime number p? - -REPLY [15 votes]: The original question has already been answered, but Jared Weinstein asked in a comment about what happens if we don't assume the axiom of choice. I've convinced myself that it's consistent with ZF to have a vector space over $\mathbb{F}_2$ with automorphism group $C_3$. In case any set theorists (other than me) are looking at this question, here's the model I have in mind. (It's a permutation model, using atoms, but the Jech-Sochor theorem suffices to convert it into a ZF-model.) Start with the full universe $V$ built from a countable set $A$ of atoms (and satisfying AC). In $V$, give $A$ the structure of a $\mathbb{F}_4$-vector space, obviously of dimension $\aleph_0$. (The relevance of the 4-element field $\mathbb{F}_4$ is that the two elements that are not in the 2-element subfield are cube roots of 1, so multiplication by either of them gives an automorphism of order 3.) Let $G$ be the group of automorphisms of this vector space, and let $M$ be the Fraenkel-Mostowski-Specker permutation submodel of $V$ determined by the group $G$ with finite subsets of $A$ as supports. In $M$, $A$ is an $\mathbb{F}_4$-vector-space. Multiplication by the elements of $\mathbb{F}_4\setminus\mathbb{F}_2$ gives a $C_3$-action on the underlying abelian group. Fairly easy calculations (admittedly not yet written down) convince me that this abelian group has no automorphisms in $M$ beyond this copy of $C_3$.<|endoftext|> -TITLE: Evil Fourier Coefficients -QUESTION [15 upvotes]: Let $f:[0,1]\to[0,1]$ be the classical devil's staircase. -Has anybody ever computed (or studied) the fourier coefficient of $f(x)$? -Related question: is the fourier series of $f(x)-x$ normally convergent (with respect to uniform norm)? - -REPLY [23 votes]: The Fourier transform of the derivative $\mu$ of the Devil staircase is explicitely stated on the wikipedia page of the Cantor distribution, in the table at the right, -under the heading "cf" (characteristic function). Its value is -$$ \int_0^1 e^{itx} d\mu(x) = e^{it/2}\ \ \prod_{k=1}^\infty \cos(t/3^k)$$ -Just multiply by $-1/it$, add $1/it$, and you get the Fourier transform of the Devil staircase. -A word on the proof. The Cantor distribution is the weak limit of the functions obtained by summing the indicator functions of the 2^n intervals generating the Cantor set at the nth step -(after renormalization). The Fourier transform of these sums can be computed explicitely. Then let n goes to infinity.<|endoftext|> -TITLE: Resources for teaching arithmetic to calculus students -QUESTION [19 upvotes]: Every time we teach calculus we discover that a significant portion of our students never understood arithmetic. I don't mean that they can't multiply numbers, but rather that they don't know intuitively that a car going 15 miles an hour goes 1 mile in 60/15=4 minutes (i.e. that division is the arithmetic operation corresponding to this problem). -It would be entirely inappropriate to teach them as if they were 10 years old, even if we had 3 months to teach them arithmetic. -Usually these are fairly intelligent individuals considering that they managed to get through high school mathematics well enough to get into a good college or university despite this handicap, and this deficiency in their background is not their fault in any way. They are likely to be able to pick up arithmetic quite quickly, and figure out from that why they have been a little befuddled through all of high school math. -I would hope this problem has been studied and ways to help these students have been proposed. -I am looking for references either to resources for these students or resources for instructors trying to help these students in the context of a calculus (or precalculus) class. - -REPLY [20 votes]: I have TA'ed a "Mathematics for Future Elementary School Teachers" course. The point of the course is to develop a deep understanding of elementary school math (read: An actual understanding, rather than a knowledge of how to do computations). The book we used was Sybilla Beckmann's "Mathematics for Elementary Teachers". -At the end of the course, most students could really explain why 2/3 of 4/5 of a cup of milk was 8/15 of a cup of milk, and could draw a picture which showed why it was true. Ditto for the addition of fractions, and the algorithms for addition, multiplication, and division. I had many students who were flabbergasted that no one had ever shown them why these things were true before. Of course, I didn't actually show them: Sybilla's book is geared toward activities which help students to discover why these things work on their own or in small groups. The role of the teacher is to direct and clarify. -The reason that this course works, though, is because the students (at least initially) think that they are only learning how to explain these things to elementary school students. You never come right out and say "You do not understand addition, and I am going to show you". So it is a unique circumstance. Even then there are many students who resist the course because they feel like they don't have to put in any work to understand such "basic concepts". A lot of these students turn around when they realize that they do not really understand, and see that they are doing poorly on examinations and homework. Some of them do not ever feel comfortable enough to face their ignorance, and these people generally do not do so well in the course. A teacher must be humble enough to realize when they do not understand something, so it is a good thing that this course is a requirement for future teachers. -If you are serious about starting a course focused on elementary school math at the college level, which I think is a GREAT idea, I would use Beckmann's book. It is really fantastic. If you want more info, like an actual plan for a quarter's worth of work, I could email one to you.<|endoftext|> -TITLE: Roots of polynomial-like equations with irrational powers -QUESTION [5 upvotes]: Is there any known theory for equations like $a_1 x^{y_1} + a_2 x^{y_2} +..a_nx^{y_n}=0$ where the $y_i$'s are arbitrary irrationals? Can you say anything about the disposition of roots? - -REPLY [2 votes]: Note that these are simply a special case (generalization? depending on terminology) of exponential polynomials. -Or rather, to be more accurate, they can be treated as such. Let $t=\ln x$, and then your monomials become $a_i e^{ty_i}$. The change of variables does introduce some technical issues, obviously (in dimension $n$, you need to do it separately for each of the $2^n$ quadrants, and then possibly iterate on the coordinate zero hyperplanes). -Over the complexes, as was pointed out by Gerry, there is potentially infinitely many roots. Over the reals, the number of roots is still finite, and some of the elementary results like Descartes's rule of signs do translate fairly well to that setting. (The fact that for ordinary polynomials, the derivative is either 0 or has fewer roots is not so crucial in many applications). -In several variables, the theory of fewnomials developed by Khovanskii gives some (pessimistic) upper bounds on the number of real roots (it also gives some restrictions on the patterns of the complex roots). -None of this is overly difficult, but I don't see the point of going into more details without a more specific question.<|endoftext|> -TITLE: Kahler structure on flag manifolds -QUESTION [8 upvotes]: Does every complex flag manifold have a natural Kähler structure? If so, what is it? - -REPLY [8 votes]: Every flag manifold $M=G^{\mathbb{C}}/P=G/C(S)$ where $P$ is a parabolic subgroup and $C(S)=P\cap G$ is the centralizer of a torus $S\subset G$, admits a finite number of -invariant Kähler structures. In particular the complex presentation $G^{\mathbb{C}}/P$ -gives rise to an finite number of invariant complex structures (i.e. integrable almost complex structures commuting with the isotropy representation of $M$). Any such complex structure is determined by an invariant ordering $R_{M}^{+}$ on the set of complementary roots -$R_{M}=R\backslash R_{K}$ of $M$ and explicitly is given by -$$ -J_{o}E_{\pm \alpha}=\pm i E_{\pm\alpha}, \quad a\in R_{M}^{+} -$$ -where $E_{\alpha}$ are root vectors with respect a Weyl basis of $\frak{g}^{\mathbb{C}}$. -On the other hand, the real presentation $G/C(S)$ makes $M$ a (homogeneous) Kähler manifold, -as a (co)-adjoint orbit of an element $w\in\frak{g}$ in the Lie algebra $\frak{g}$ of the compact connected (semi)simple Lie group. Flag manifolds exhaust all compact homogeneous Kähler manifolds corresponding to a compact connected semi-simple Lie group. -To be more specific, $M$ admits a finite number of Kähler structures which are parametrized by the well-known $\frak{t}$-chambers (connected components of the set of regular elements of $\frak{t}$) where -$$ -{\frak{t}} =( H\in{\frak{h}} : (H, \Pi_{0})=0 ) - $$ -is a real form of the center ${\frak{z}}$ of the isotropy subgroup $K=C(S)$. -Here $\frak{h}$ is the Cartan subalgebra corresponding to a maximal torus $T$ of $G$ which contains $S$, and $\Pi_{0}\subset\Pi$ is the subgroup of simple roots which define (the semi-simple part of) the complexification $\frak{k}^{\mathbb{C}}$ (note that $K=C(S)=P\cap G$ is a reductive Lie group). We have -$$ -{\frak{z}}^{\mathbb{C}}={\frak{t}}\oplus i {\frak{t}}, \ \ \ {\frak{k}}^{\mathbb{C}}={\frak{z}}^{\mathbb{C}}\oplus{\frak{k}}_{ss}^{\mathbb{C}} -$$ -where ${\frak{z}}^{\mathbb{C}}$ is the complexification of the center ${\frak{z}}$ and ${\frak{k}}_{ss}^{\mathbb{C}}$ is the semi-simple part of the reductive complex Lie subalgebra ${\frak{k}}^{\mathbb{C}}$ -In particular, there exists a natural 1-1 correspondence between elements from ${\frak{t}}$ and -closed invariant 2-forms on $M$. Symplectic 2-forms (non-degenerate) correspond to regular elements $t$ of ${\frak{t}}$. -Note that the corresponding symplectic form corresponding to a regular element $t_{0}$ is the Kirillov-Kostant-Souriau 2-form in the (co)-adjoint orbit $Ad(G)t_{0}$, that is -$$ -\omega_{t_{0}}(X, Y)=B(t_{0}, [X, Y]), \ \ X, Y\in T_{t_{0}}M. -$$ -For more details see: D. Alekseevsky: Flag manifolds (11. Yugoslav Geometrical seminar, Divcibare, 10-17 October 1993, 3-35. -This article is a very good review on the geometry of flag manifolds.<|endoftext|> -TITLE: Partitioning a Rectangle into Congruent Isosceles Triangles -QUESTION [15 upvotes]: Is it possible to partition any rectangle into congruent isosceles triangles? - -REPLY [24 votes]: No. Note that the acute angle of your triangle must divide $\pi/2$ (look at a corner), so there are countably many such triangles (up to similarity), and hence you get only a countable set of possible ratios of sides.<|endoftext|> -TITLE: An anecdote by R. Schmidt -QUESTION [27 upvotes]: Did anybody here ever read those lines by R. Schmidt (?) where he talked about the terseness of articles in group theory in the days prior to the conclusion of the classification of the finite simple groups? -The complete story might have appeared in the Notices of the AMS or a similar publication or in an online recollection of group-theoretic anecdotes; I think that I read it 14 or 15 years ago. All I can recall is that, at some point of his write-up, R. Schmidt (?) brought up that famed Group Theory Year in UC (1960-1961) and a hilarious exploit by J. H. Conway and David Wales from the epoch. -I really hope that somebody out there has read it recently and can tell me where I can find it again. -Thanks. -Added (Nov 8/2010) The said Conway incident went something like this: "one day he (Professor Freese, as you can read in the first reply to the thread) found Conway and Wales working with some relations and generators. Conway and Wales said they wanted to see if it was a group what they had there. Professor Freese could not help but ask -How could it not be?, -to which, they immediately replied -It could be infinite!" -Added (Nov 19/2010) I'm sure that in that elusive note one could also find the following story: -"A typical journal page(?) in those days would look something like this -Theorem. All groups are finite. -Proof. Deny." -I also remember that I didn't really get the joke at the time, but that's another story... -Does any of you remember reading something like this once? C'mon fellas! I know I could not have been the only one in here to catch it back in the day. - -REPLY [46 votes]: As William DeMeo says, I was the one who told this story. I was a grad student at CalTech from 1968-72. One quarter John Conway was visiting and working with David Wales, searching for finite simple groups. CalTech had an afternoon tea everyday. Conway and Wales had a group presentation on the blackboard and I asked what the problem was. They said they were trying to determine if it was a group. After a minute I got up the nerve to ask them how could it not be a group. They both turned to me and in unison said "It could be infinite!" -I believe this search did construct a simple group of order 145926144000 that Rudvalis had suggested might exist. It is sometimes called the Rudvalis-Conway-Wales group.<|endoftext|> -TITLE: Non-Hölder continuous devil's staircases -QUESTION [8 upvotes]: Let $f:[0,1]\to[0,1]$ be a devil's staircase in the usual sense. (That is, $f$ is continuous, non-decreasing, $f'=0$ on a set of full Lebesgue measure.) We also require the complement to the set where $f'$ vanishes to have Hausdorff dimension zero. -Question. Is it true that $f$ is not Hölder continuous? -(This looks plausible, since $f$ has `very little room' where it can grow so it has to grow very fast - at least, at some points.) - -REPLY [9 votes]: Let $K$ be the bad set. -Assume $f$ is Hölder continuous with exponent $\alpha$. -Since Hausdorff dimension of $K$ is zero, -given $\epsilon>0$ we can cover $K$ by open intervals $\left]a_i,b_i\right[$ -with length $\ell_i=b_i-a_i$ has such that -$$\sum_n\ell_n^\alpha<\epsilon\ \ \ \ \ (*)$$ -and $\ell_n<\epsilon$ for any $n$. -Set $v_i=f(b_i)-f(a_i)$. -Since $f$ is Hölder continuous, -$$v_i < C{\cdot}\ell_i^\alpha.\ \ \ \ \ (**)$$ -But clearly -$$\sum v_i=1$$ -which contradicts $( * )$ and $( * * )$. -Did I miss something? - -REPLY [7 votes]: If a map $f$ is Lipschitz, then it is a standard result in dimension theory that $\dim_H f(Z) \leq \dim_H Z$ for all $Z$. More generally, if a map $f$ is Hölder with exponent $\alpha \in (0,1]$, then one has the inequality -$$ -\dim_H f(Z) \leq \frac 1\alpha \dim_H Z. \qquad \qquad (*) -$$ -The proof of this uses the same sorts of calculations as in Anton's answer. Now the function $f$ that you describe has the property that there is a Cantor set $C$ such that $f$ is locally constant on the complement of $C$. The complement of $C$ is open, and hence is a countable union of open intervals, so its image under $f$ is a countable set. -In particular, this implies that $f(C)$ is the entire interval $[0,1]$ with an at most countable set removed. Thus $\dim_H f(C) = 1$, and if $f$ were Hölder continuous with exponent $\alpha>0$, the inequality in (*) would give -$$ -1 = \dim_H f(C) \leq \frac 1\alpha \dim_H Z = 0, -$$ -a contradiction.<|endoftext|> -TITLE: Why is every factor a tensor product of a $\sigma$-finite factor and a factor of type I? -QUESTION [7 upvotes]: Here I am not assuming the factor is represented on a separable Hilbert space. This is quoted on page 370 of Takesaki II, then a bit later on page 381, and I haven't been able to find a proof prior to this point in the book or in Takesaki I. - -REPLY [6 votes]: Note that "$\sigma$-finite" is a tricky notion. For example, any ${\rm\ II}_1$ factor, even non-separable, is $\sigma$-finite, because the finite trace is a faithful state. -In any case, the argument one needs is that of Makoto Yamashita, with a few clarifications. Note that for type I and type II factors, the assertion is trivial: for type I, the $\sigma$-finite factor can be taken to be $\mathbb{C}$. For type ${\rm\ II}_1$, it is always $\sigma$-finite. And any type ${\rm\ II}_\infty$ is the tensor of a ${\rm\ II}_1$ with a ${\rm\ I}_\infty$, so again the assertion holds. -This leaves us then with a factor of type III. The question is why does there exist a projection $p$ with $pMp$ $\sigma$-finite. It is well-known that any von Neumann algebra has a faithful semifinite normal weight. From this one can deduce that there has to exist a projection $p$ where the weight is finite. And then one deduces that, restricted to $pMp$, the weight is a faithful normal state. As $M$ is type III, one can construct a family of pairwise orthogonal projections $\{p_j\}_{j\in J}$ with $p_j$ equivalent to $p$ for all $j$. This equivalences can be used to construct a system of "matrix units", from where the isomorphism -$$ -M\simeq(pMp)\otimes B(\ell^2(J)) -$$ -follows.<|endoftext|> -TITLE: Sturm chain analogue for exponential polynomials? -QUESTION [7 upvotes]: I'm going to define an exponential polynomial of degree $k$ as a function $f$ of the form -$f(x) = \sum_{i=1}^k c_ie^{\alpha_ix}$ ($\alpha_i$s real). -My first question is: is there an algorithm for counting the number of real roots of such an expression, with complexity depending only on the degree $k$? -I strongly suspect that the answer to this question is yes, and that the answer is known (seeing as Tarski's exponential function problem is all but solved), but I can't find it described anywhere. -My second question is: can somebody tell me what this algorithm is? Or give me a hint? -I vaguely remember reading somewhere that there was a known method analogous to the method of Sturm chains for polynomials... But I haven't been able to figure out what it should be, nor have I been able to find where I read that claim. My best guess is that we can get rid of terms of such an expression by first dividing $f$ by an exponential $e^{\alpha x}$ to make a term constant, differentiating, and then multiplying by that same exponential. If we call this operation $D_{\alpha}$, we get -$D_{\alpha}f(x) = \sum_{i=1}^k c_i(\alpha_i-\alpha)e^{\alpha_ix}$. -The nice thing is that $D_{\alpha}f$ acts analogously to the derivative of $f$, i.e. between any consecutive zeroes of $f$ there is a zero of $D_{\alpha}f$. The problem is that I can't think of a good analogue to the division algorithm for exponential polynomials (maybe we don't need one?). -Edit: When I say that Tarski's exponential function problem is "all but solved," I mean that all that is missing from the full solution is a proof of Schanuel's conjecture. I'm not saying that Schanuel's conjecture is easy, but given this result it seems to me that we should be able to describe some sort of explicit algorithm for deciding problems like this one, although the proof of correctness of such an algorithm might require us to assume Schanuel's conjecture holds. - -REPLY [3 votes]: Wilkie's proof of o-minimality of exponentiation tells you (without needing to assume Schanuel's Conjecture) that for each $k$ there is a bound $M_k$ such that any exponential polynomial $\sum_{i=1}^k c_ie^{\alpha_i x}$ has at most $M_k$ real zeros [assuming that the $c_i$ and $\alpha_i$ are real.] -Perhaps Khovanskii's work gives an explicit bound on the $M_k$.<|endoftext|> -TITLE: Spectral sequences: opening the black box slowly with an example -QUESTION [117 upvotes]: My friend and I are attempting to learn about spectral sequences at the moment, and we've noticed a common theme in books about spectral sequences: no one seems to like talking about differentials. -While there are a few notable examples of this (for example, the transgression), it seems that by and large one is supposed to use the spectral sequence like one uses a long exact sequence of a pair- hope that you don't have to think too much about what that boundary map does. -So, after looking at some of the classical applications of the Serre spectral sequence in cohomology, we decided to open up the black box, and work through the construction of the spectral sequence associated to a filtration. And now that we've done that, and seen the definition of the differential given there... we want some examples. -To be more specific, we were looking for an example of a filtration of a complex that is both nontrivial (i.e. its spectral sequence doesn't collapse at the $E^2$ page or anything silly like that) but still computable (i.e. we can actually, with enough patience, write down what all the differentials are on all the pages). -Notice that this is different than the question here: Simple examples for the use of spectral sequences, though quite similar. We are looking for things that don't collapse, but specifically for the purpose of explicit computation (none of the answers there admit explicit computation of differentials except in trivial cases, I think). -For the moment I'm going to leave this not community wikified, since I think the request for an answer is specific and non-subjective enough that a person who gives a good answer deserves higher reputation for it. If anyone with the power to thinks otherwise, then feel free to hit it with the hammer. - -REPLY [2 votes]: In another direction, you can consider the Hodge spectral sequence of complex manifolds: for compact Kähler manifolds it degenerates ar $E_1$ but examples can be provided for which the Hodge spectral sequence degenerates at an arbitrary page and is fairly explicit. You can look at this paper: http://www.mathematik.uni-marburg.de/~rollenske/papers/froelicher.pdf<|endoftext|> -TITLE: Integer polynomial (of degree >1) all of whose values are square-free -QUESTION [8 upvotes]: Is there an integer polynomial $ A \in {\mathbb Z} [ X ]$ of degree $d\geq 2$ such that for any integer $n\in {\mathbb Z}$ , $ A(n) $ is a square-free integer? - -REPLY [13 votes]: No. WLOG $A$ is irreducible. Pick a sufficiently large prime $p$ dividing $A(k)$ for some $k$ (there are infinitely many such primes, for example by the argument here). In particular pick $p$ large enough so that it is relatively prime to the coefficients of $A$ and to $d$, and so that it does not divide the discriminant of $A$. Then the congruence $A(x) \equiv 0 \bmod p^2$ has a solution by Hensel's lemma.<|endoftext|> -TITLE: Projective variety with no syzygies but not isomorphic to projective space -QUESTION [16 upvotes]: Let $A:={\mathbb{C}}[x_0,\ldots,x_n]=\oplus_{d=0}^{\infty} {\mathbb{C}}[x_0,\ldots, x_n]_d$ be the graded complex algebra of polynomials in $n+1$ variables, graded by degree. -Suppose $L$ is a line bundle over a projective manifold $X$ such that the ring of plurisections of $L$, i.e. $\oplus_{d=0}^{\infty} H^0(X,L^d)$ is isomorphic to $A$ as graded complex algebras. -Does that imply that $X$ is necessarily complex projective space ${\mathbb{CP}}^n$ and $L$ is the tautological line bundle? - -REPLY [6 votes]: Here is an example for Torsten's Addendum: a line bundle with the required properties which is not semi-ample (no multiple is globally generated): -Take an $\pi:X\to \mathbb P^n$ as in Torsten's example, that is, with $\pi_*\mathcal O_X\simeq \mathcal O_{\mathbb P^n}$. Also assume that there is an exceptional divisor $E$, so let's just say it is a birational morphism contracting the divisor $E$ to a point. Now take $\mathcal L=\pi^*\mathcal O_{\mathbb P^n}(1)$ and consider the short exact sequence: -$$ -0\to \mathcal L^{\otimes n} \to (\mathcal L(E))^{\otimes n} \to \mathcal O_{nE}(nE) \to 0 -$$ -The sheaf on the right has no global sections, so the other two have the same $H^0$. Therefore -$$ -H^0(X,(\mathcal L(E))^{\otimes n}) = H^0(\mathbb P^n, \mathcal O_{\mathbb P^n}(n)) -$$ -Finally $(\mathcal L(E))^{\otimes n}$ cannot be globally generated, because -$H^0(X,\mathcal L^{\otimes n})=H^0(X,(\mathcal L(E))^{\otimes n})$, so - ${\rm supp}\,E$ is always in the base locus (or in other words, if it had a section whose zero section did not contain ${\rm supp}\,E$, then its restriction to $nE$ would give a nonzero section of $\mathcal O_{nE}(nE)$). - -And here is what you can assert based on your condition: -Naive statement: Essentially the only reason your desired claim can fail is via Torsten's example. -More precise statement: Let $k$ be a field, $S:=k[x_0,…,x_n]$ and suppose $\mathcal L$ is a line bundle on a projective variety $X$ (over $k$) such that $R(X,{\mathcal L})=\oplus_{d=0}^\infty H^0(X,\mathcal L^d)\simeq S$ as graded $k$-algebras. Then there exists a diagram -$$ -X \overset\sigma\longleftarrow \widetilde X \overset\phi\longrightarrow \mathbb P^n -$$ -such that $\sigma$ is a projective birational morphism and $\phi$ is surjective. Furthermore, $\sigma^*\mathcal L\supseteq \phi^*\mathcal O_{\mathbb P^n}(1)$. Finally if $n=\dim X$, then $\phi$ is also birational and hence $X$ is a rational variety. -Proof: -By the assumption ${\rm Proj}\, R(X,\mathcal L)\simeq \mathbb P^n$ and there exists a dominant rational map $\psi: X\dashrightarrow \mathbb P^n$. Let -$$ -X \overset\sigma\longleftarrow \widetilde X \overset\phi\longrightarrow \mathbb P^n -$$ -be the resolution of indeterminacies of $\psi$ via blow-ups on $X$. Then $\sigma$ is a projective birational morphism and hence $\widetilde X$ is also projective and $\sigma^*\mathcal L\supseteq \phi^*\mathcal O_{\mathbb P^n}(1)$. It follows that the induced morphism $\phi$ has to be surjective (since $\widetilde X$ is projective and the morphism is defined everywhere). Since this morphism is given by global sections of sufficiently high powers of a line bundle (i.e., $\phi^*\mathcal O_{\mathbb P^n}(m)$ for some $m>0$), it follows that it has connected fibers and $\phi_*\mathcal O_{\widetilde X}\simeq \mathcal O_{\mathbb P^n}$ and hence if $n=\dim X$, then $\phi$ is birational.<|endoftext|> -TITLE: What is the probability that every pair of students is at some point in the same classroom? -QUESTION [10 upvotes]: A cohort in a school consists of 75 students who study for 6 years. Each year, the students are randomly distributed into 3 classrooms of 25 students each. What is the probability that, after 6 years, each student has at some point been in a classroom with every other student? -More generally: Starting with an edgeless (undirected) graph on $cn$ vertices, let a round consist of first randomly partitioning the vertices into $c$ disjoint sets of $n$ vertices each, then adding an edge between every pair of nonadjacent vertices that lie in the same set. What is the probability that, after $y$ rounds, the result is a complete graph? -I asked this question on math.stackexchange but received no fully useful response (see here, where I've also posted an answer with further discussion and generalization and partial "solutions"). I'd especially like to know about tools for the exact answer, but approximations or bounds would also be interesting. -The particular case above was posed by a friend, who teaches in a school with those values of $c$, $n$, and $y$. In that particular case the answer is easily seen to be "Don't hold your breath, pal." - -REPLY [13 votes]: A Poisson approximation should be good here, for appropriate ranges of the parameters. Consider the events $A_{ij}=$"$i$ and $j$ are never put in the same class". -Broadly speaking, if you have a large number of these events, and each individual one has low probability, and any pair of events are close to independent, and there are no anomalous higher-order dependencies (here in particular I am being completely vague) then you expect the total number of events that occur to be approximately Poisson (with mean given by the total expected number of events). -Here $P(A_{ij})=\Big(\frac{(c-1)n}{cn-1}\Big)^y$ and the total number of events is -$\left(\begin{smallmatrix} cn \\ 2 \end{smallmatrix}\right)$. -If the first of those terms is small and the product is neither too small nor too large, then one would expect the distribution of the total number of events to occur to be approximately Poisson with mean -$\left(\Big(\frac{(c-1)n}{cn-1}\Big)^y\left(\begin{smallmatrix} cn \\ 2 \end{smallmatrix}\right)\right)$ -and hence the probability that none of them occur should be roughly -$\exp\left(-\Big(\frac{(c-1)n}{cn-1}\Big)^y\left(\begin{smallmatrix} cn \\ 2 \end{smallmatrix}\right)\right)$. -I tried a small simulation for c=3, n=25 and y=20, for which the approximation gives a probability of 0.33571. All students met each other on 33676 of 100000 trials, which is encouraging enough for the accuracy of the approximation in this case. -(there are conditions on the dependency between the events which guarantee that this sort of approximation is (asymptotically) reliable. For example, if the events are all positively correlated in an appropriate sense. But that doesn't seem to apply here. So for the moment what is written above is merely heuristic. But maybe someone else will see how to be more precise).<|endoftext|> -TITLE: Ideal Class Number -QUESTION [11 upvotes]: As far as I know there are two proofs of the finiteness of the ideal class group of a number field. One is due to Minkowski using the "geometry of numbers" and another one is due Chevalley using "ideles". -My question is divided into two parts: -1: Is there any other proof? -2: Second question needs some preliminary background. Let $K$ be a number field and suppose $\mathcal{O}_K$ is its ring of integers. The group $SL_2(\mathcal{O}_K)$ acts on $\mathbb{P}^1(K)$, and one can show the ideal class number is equal to the number of orbits of this action. So proving the finiteness of orbits implies the finiteness of the ideal class number. Is there any proof for this? - -REPLY [4 votes]: In the usual proof of the class number formula, i.e. the computation of the residue of the -Dedekind zeta function at $s = 1$, it is used that the class number is finite and that the unit group has the right number of generators. But the proof essentially works also if you do not use this fact, and in the end you get both results for free - the price you have to pay is a presentation which is a bit messier than usual because you have to allow for the possibility that h is infinite and that you have too few units. It is a good exercise to go through the proof in the real quadratic case, though. -Edit. The situation is not as simple as I thought it is. The problem is the following: by counting lattice points you can easily prove that the number of ideals with norm $< X$ in any given ideal class $C$ is equal to $cX + O(\sqrt{x})$. I would have thought that the existence of infinitely many ideal classes quickly produces nonsense, but this is wrong. In fact, the constants in the O-term may depend on the ideal classes. The usual proof of the finiteness of the ideal class group shows that there is a finite constant $c'$ such that the error term is less than $c' \sqrt{X}$. The problem is what to do without this information. -It follows, if I am right, quite easily that if $h$ is not finite, then the number of ideals with norm $< X$ is not of the form $O(x)$, i.e. grows fast than $cX$ for any constant $c$. In the quadratic case, using the fact that the number of ideals with norm $m$ can be expressed in terms of Legendre symbols implies that the number of ideals with norm $< X$ is $L(1,\chi) X + O(\sqrt{X})$, and now we get a contradiction plus a proof that $L(1, \chi)$ does not vanish (which in turn implies the fact that the Dedekind zeta function of the quadratic field has a pole of order $1$ in $s = 1$, if you know that it is analytic). -I have not yet seen what to do for general number fields.<|endoftext|> -TITLE: Tight bound for sum of entries of the inverse of a nonnegative matrix -QUESTION [6 upvotes]: While playing around with certain non-negative matrices, I got stuck at the following question. -Let $A$ be a strictly positive-definite $n \times n$ matrix ($n \ge 3$), with ones on the diagonal, and all other entries in the range $[0,1]$. How should I go about proving a tight bound on the sum of the entries of $A^{-1}$. In symbols, how should I go about trying to compute the smallest number $\gamma(n)$ such that $$1^TA^{-1}1 \le \gamma(n).$$ -Any pointers to related work, or some possible ways to attack the problem will be very useful. Of course, if you think this question is not well-formulated, please help me improve it! -Remarks: -a. Notice that for the special case where $A$ is the identity matrix, we have equality, and $\gamma(n)=n.$ -b. If the vector of all ones, i.e., $1$, happens to be an eigenvector of $A$, then also we have an instant answer. -More background -The reason I reached the above bounding problem was that I was looking at the matrix -$$\begin{bmatrix} A & 1\\\\ -1^T & n\end{bmatrix},$$ -and trying to prove that it is positive semidefinite. So, either I could show that $A-11^T/n \succeq 0$, or $1^TA^{-1}1 \le n$. Now, since the bound with $n$ might not always hold, I started searching for a $\gamma(n)$ that holds. Perhaps, I need to further cleanup my question? - -REPLY [7 votes]: When $n\ge3$, there does not exist such a bound. To see this, take $n=3$ and the following matrix -$$A=\begin{pmatrix} -1 & a & 0 \\\\ a & 1 & a \\\\ 0 & a & 1 \end{pmatrix}.$$ -If $a< 1/\sqrt2$, it is positive definite with non-negative entries. However -$$1^TA^{-1}1=\frac{3-4a}{1-2a^2}$$ -is not bounded as $a\rightarrow1/\sqrt2$.<|endoftext|> -TITLE: Non-algebraizable Formal Scheme? -QUESTION [15 upvotes]: What is an example of a formal scheme that is not algebraizable? -Recall that, if $X$ is a locally noetherian scheme and $Z$ is a closed subset (of the underlying topological space), then one can form the formal completion of $X$ along $Z$ which is sometimes denoted $X_{/Z}$. This is a formal scheme whose underlying topological space is $Z$. -What is a formal scheme that is not of this form? -Update: Emerton and Francesco Polizzi suggested several examples that arise in the study of deformations of varieties with trivial canonical bundle. It'd be nice to see some more elementary, explicit examples as well. -Update 2: In comments, Francesco Polizzi mentioned that further examples can be found in [Hironaka-Matsumura, "Formal functions and formal embeddings" J. math. soc. Japan 20, Theorem 5.3.3 ] and [Hartshorne, Ample subvarieties of algebraic varieties, p. 205]. -This is too long to fit into comments: -@FP: Thanks! I'm not sure I quite follow the argument for non-algebraizability in the book. Sernesi states that, if $X \to \text{Spec}(\bar{A})$ is an algebrization, then $X$ would admit a non-trivial line bundle "since $X$ is of finite type over an integral scheme." Furthermore, he states that this line bundle can be chosen to "correspond to a Cartier divisor whose support does not contain $X_{s}$ [the special fiber] and has nonempty intersection with $X_{s}$." (note: The notation $X$, $X_s$ is different in the text.) -It is not clear to me why such a line bundle exists: $\mathbb{A}^n$ is a finite type scheme over an integral scheme that has no non-trivial line bundles. -I understand how this shows that there is no algebraization by a $\bar{A}$-projective scheme, but why is there no algebraization by an arbitrary scheme? -I was a little nervous about the argument (Raynaud has an example of a family of Abelian varieties over a nodal curve with non-projective total space), but my concern was needless. -Here is one argument. Let $X_0/\mathbb{C}$ be an algebraic $K3$-surface. We assume algebraizability and derive a contradiction. The statement about existence of non-algebraic deformations is very strong: In fact, there exists a 1st order deformation $f_1 \colon X_1 \to \text{Spec}(\mathbb{C}[t]/(t^2))$ with the property that the restriction of any line bundles $L_1$ on $X_1$ to $X_0$ is numerically trivial. We use this deformation to derive a contradiction. -By definition, there exists a morphism $f_1 \colon \text{Spec}(\mathbb{C}[t]/(t^{2})) \to \text{Spec}(\mathbb{C}[[x_1, \dots, x_{20}]])$ with property that the versal deformation restricts to $X_1$. Now factor this morphism as $\text{Spec}(\mathbb{C}[t]/(t^{2})) \to \text{Spec}(\mathbb{C}[[t]]) \to \text{Spec}(\mathbb{C}[[x_1, \dots, x_{20}]])$ (by lifting the images of $x_1, \dots, x_{20}$ under $f_1^{*}$). -If $X_{t} \to \text{Spec}(\mathbb{C}[[t]])$ is the restriction of the versal deformation, then the generic fiber is an algebraic $K3$-surface, hence admits an ample line bundle. The total space $X_{t}$ of the family is regular, so it is possible to extend this line bundle to a line bundle $L_{t}$ on $X_{t}$. But then the restriction of $L_{t}$ to the special fiber is not numerically trivial (by flatness); however, no such line bundle can even lift to 1st order. Contradiction. - -REPLY [11 votes]: Bhargav's example is really an example of a non-algebraic formal subscheme of the affine plane. Such examples are ubiquitous in foliation theory : a differential equation and, more -generally, a foliation on a (smooth) algebraic variety -has local leaves which are smooth formal schemes. This follows from -the formal Frobenius theorem (in positive characteristic, the foliation needs to have p-curvature zero). Sometimes, these leaves are the formal completions of an algebraic -subvariety, but often not. However, these leaves are isomorphic, at formal schemes, -to the formal completion at the origin of an affine space; from the intrinsic -point of view, they thus are algebraizable. -The theorems of Hironaka, Matsumura, Hartshorne to which Francesco Polizzi -refers are in the same spirit, -but concern formal subschemes along an algebraic subvariety. -They don't apply to formal subschemes based at a point. -Actually, Arakelov geometry allows to establish analogs of these theorems and -algebraize some formal subschemes based at a point (eg leaves of a foliation). -See papers of Bost (Pub. Math IHES, vol. 93, 2001), and of Bost and myself -(Manin Festschrift, 2010). - -REPLY [5 votes]: I find it more or less illusory to ask for non-algebraizable formal schemes which would not -fit into the scope of deformation theory. Indeed, a formal scheme $\hat X$ over $C[[t]]$, say, -is nothing but a family of schemes $(X_n)$, where $X_n$ is a scheme over $C[t]/(t^{n+1})$ -together with isomorphisms of $X_n$ with $X_{n+1}\otimes C[t]/(t^{n+1})$. -On the other hand, I wonder whether classical examples of non-algebraic analytic spaces, -or algebraic spaces, could be constructed in the category of formal schemes, but I have no precise answer to give.<|endoftext|> -TITLE: When do 3D random walks return to their origin? -QUESTION [25 upvotes]: The probability of a random walk returning to its origin is 1 in two dimensions (2D) but only 34% in three dimensions: This is Pólya's theorem. I have learned that in 2D the condition of returning to the origin holds even for step-size distributions with finite variance, and -as Byron Schmuland -kindly explained in this Math.SE posting, even for distributions with infinite variance, recurrence depends upon the details of the step-length tail distribution. But this is all in 2D. -My question is: - -Are there conditions on the step-size and step-direction distributions in three dimensions (3D) that ensure - that the walk will return to the origin with probability 1? - -Of course I exclude here a step-direction distribution that squashes 3D → 2D. -But perhaps partial dimensional compression suffices? - (source) -(3D random-walk image credit to http://logo.twentygototen.org/.) - -REPLY [2 votes]: The answer to your question appears to be contained in a 1928 Theorem of Polya, on simple random works on a lattice $\mathbb Z^d$. As a soft-reference, see page 15 of this short paper, using only very basic mathematics (Stirling's formula, etc.). The crux of the solution is to show that in $3$ dimensions, the average number of returns to the origin is finite, and so a.e you eventually get lost for good!<|endoftext|> -TITLE: endomorphism of factor: can it be idempotent up to congugacy? -QUESTION [7 upvotes]: Let $M$ be a factor, and let $\phi:M\to M$ be an irreducible endomorphism -("irreducible" means that the relative commutant of $\phi(M)$ in $M$ is trivial). -Let's also assume that $\phi$ is not invertible. -Is it possible to have $\phi\circ \phi$ conjugate to $\phi$? -In other words, is it possible to have an endomorphism $\phi$, and a unitary $u\in M$, such that $$\phi(\phi(x))=u\phi(x)u^*,\quad\forall x\in M.$$ -If this is possible, I would like to see an example. - -Note: an answer to the above question would also settle this question. - -REPLY [7 votes]: This is not possible. If it were, then using the notation above, given any $x \in \phi(M)$, we would have $x u^* = u^* \phi(x)$, and $\phi(x) u = u x$. Hence, for any $x \in \phi(M)$ we have -$$ -x u^* \phi(u^*) u^2 = u^* \phi(x u^*) u^2 -$$ -$$ -= u^* \phi(u^*) \phi \circ \phi (x) u^2 = u^* \phi(u^*) u^2 x. -$$ ` -Hence $u^* \phi(u^*)u^2 \in \phi(M)' \cap M = \mathbb C$ and so $\phi(u^*) \in \mathbb C \cdot u^*$. Then, for any $y \in M$ we would have that -$$ -\phi \circ \phi (y) = u \phi(y) u^* -$$ -$$ -= \phi(u y u^*). -$$ -Since $\phi$ is injective we then have $\phi(y) = u y u^*$, and hence $\phi$ is invertible. -If you don't require that $\phi(M)$ be irreducible then this is possible.<|endoftext|> -TITLE: Quantum PCP Theorem -QUESTION [9 upvotes]: Although I think I know the answers to these, I'd just like to collect them all in one place. -What is the quantum PCP theorem, what implications does its proof have for simulation of Hamiltonians and is following Irit Dinur's reproof of the classical version the best/only current mode of attack (and if so why?) What is the sort of math/physics/theoretical CS background needed to approach this problem? - -REPLY [3 votes]: One more comment: -The quantum PCP conjecture implies that calculating the energy of the Gibbs state of a quantum system in constant temperature is QMA-Complete. This answers the question: - -what implications does its proof have for simulation of Hamiltonians? - -Given a Hamiltonian $H$, the Gibbs state is defined by: -$$\rho_{\mathrm{gibbs}} = \frac{1}{Z}\exp(-H/T)$$ -where the normalization factor $Z=\operatorname{Tr}(\exp(-H/T))$. -It is known that the Gibbs state minimizes the free energy (H - TS): -$$\rho_{\mathrm{gibbs}}=\operatorname{argmin}_{\rho \succeq 0, \operatorname{Tr}(\rho)=1} \operatorname{Tr}(H\rho) + T \cdot \operatorname{Tr}(\rho \ln \rho)$$ -Let $\rho_{\min}$ be one of the pure ground states of $H$, with energy $E_0$. Since $\rho_{\min}$ is a pure state, $Tr(\rho \ln \rho)=0$. -Therefore, -$$\operatorname{Tr}(H\rho_{\mathrm{gibbs}}) + T \cdot Tr(\rho_{\mathrm{gibbs}} \ln \rho_{\mathrm{gibbs}}) \leq \operatorname{Tr}(H \rho_{min}) = E_0.$$ -Since $\operatorname{Tr}(\rho \ln \rho) \leq n \ln d$, where $n$ is the number of qudits (of dimension $d$ in the systems, -$$\operatorname{Tr}(H\rho_{gibbs}) \leq E_0 + T n \ln d .$$ -By choosing $T = O\left(\frac{\epsilon}{\ln d}\right)$ one gets an additive approximation of $\epsilon n \leq \epsilon m$ to the ground energy, which is $QMA$-hard by the quantum PCP conjecture ($m$ is the number of terms in the Hamiltonian, therefore it is at least linear in the number of qubits in the system). -I believe this result is a folklore, and clearly also holds in the classical setting. I learned about it from the following paper: -Brandão, F. G., & Harrow, A. W. (2013, June). Product-state approximations to quantum ground states. In Proceedings of the 45th annual ACM symposium on Symposium on theory of computing (pp. 871-880). ACM.<|endoftext|> -TITLE: Total space of the line bundle $\mathcal{O}(1)$ over $\mathbb{P}^n$ -QUESTION [22 upvotes]: It is well known that total space of the tautological line bundle $\mathcal{O}(-1)$ over projective space $\mathbb{P}^n$ is closed subvariety of $\mathbb{P}^n\times\mathbb{A}^{n+1}$. My question is how to realize total space of $\mathcal{O}(1)$ over $\mathbb{P}^n$ in such manner, i.e. I need an embedding of $Tot(\mathcal{O}(1))$ in simple variety and defining equations. -Thanks. - -REPLY [11 votes]: For me the best description of $Tot(O(1))$ is the tautological --- as the relative spectrum of the sheaf of algebras $A = O \oplus O(-1) \oplus O(-2) \oplus \dots$ on $P^n$: -$$ -Tot(O(1)) = Spec_{P^n}(A). -$$ -This allows to work with $Tot(O(1))$ more effectively than any other description. For example, a coherent sheaf on $Tot(O(1))$ can be represented by a quasicoherent sheaf $F$ on $P^n$ together with a morphism $F(-1) \to F$ inducing on $F$ a structure of an $A$-module.<|endoftext|> -TITLE: Embedding of Baumslag-Solitar group into a certain group -QUESTION [9 upvotes]: Let $G$ be a group -generated by $a_0, a_1, a_2$ with relations: -$a_0 a_1 a_0^{-1}=a_1^4$ -$a_1 a_2 a_1^{-1}=a_2^4$ -$a_2 a_0 a_2^{-1}=a_0^4$ -I am wondering if $BS(1,4)=\langle a,b:bab^{-1}=a^4\rangle$ is embedded into G via $a\mapsto a_1$, $b\mapsto a_0$ -Remark: the group is constructed in analogy to Higman group - -REPLY [3 votes]: A few minutes of running GAP shows that the group has order $6751269$. Hence, as the final edit of the accepted answer points out, the group is finite, and no Baumslag-Solitar group embeds in it.<|endoftext|> -TITLE: Bounds on $\|P^{k+1} - P^k\|$ for $n$ by $n$ stochastic matrix $P$ with trace $n-1$ and integer $k\gg n$ -QUESTION [10 upvotes]: The problem: -We have a $n$-state Markov chain with arbitrary initial distribution and transition matrix $P$ that is arbitrary except that we know that $P$ has trace $n-1$. Of course $P$ is also a stochastic matrix. -Let $b(n,k)$ denote the suprememum, over all such matrices $P$ and initial distributions, of the total variation distance between the distribution of this chain after $k$ and $k+1$ time steps. I am primarily interested in bounds for fixed $n$ as $k$ goes to infinity. In the remaining description of partial results I will therefore focus on the following question: what is the largest $\alpha \ge 0$ such that $b(n,k)=O(k^{-\alpha})$, where big-oh hides a function of $n$. -For fixed $n$ all matrix norms are equivalent, so it suffices to bound some matrix norm of $P^{k+1} - P^k$ by $O(k^{-\alpha})$. In particular using the vector $\ell_p$ norm $|\cdot|_p$ and its induced matrix norm $\|\cdot\|_p$ it suffices to bound $|(P^{p+1}-P^k)v|_p$ for all vectors $v$ with $|v|_p \le 1$. -I'm using terminology primarily from Horn and Johnson's Matrix Analysis book. -Partial results and discussion: -One can use Markov chain couplings to show that $\|P^{k+1}-P^k\|_1 = O(1/\sqrt{k})$. (See Lemma 10 in http://www.cs.brown.edu/~ws/papers/regret.pdf .) -All eigenvalues of a stochastic matrix are at most 1 in absolute value and the trace of a matrix is equal to the sum of eigenvalues, so we conclude that all eigenvalues of $P$ have real part between 0 and 1. -Consider the special case of $P$ symmetric. Then $P$ is symmetric then it is orthogonally diagonalizable with real eigenvalues. For any $0 \le \lambda \le 1$ it is easy to show that $|\lambda^{k+1} - \lambda^k| = O(1/k)$, so therefore for symmetric $P$ we have $\|(P^{p+1}-P^k)\|_2 = O(1/k)$. -Consider the special case of circulant matrices, i.e. matrices $P$ where $P_{ij}$ is a function of $i - j \mod n$. The trace restriction implies that the diagonal entries of $P$ are $(n-1)/n$. By Gersgorin disks it follows that all eigenvalues are within $1/n$ of $1-1/n$ in the complex plane. If my calculations are right the maximum over that circle of $|\lambda^{k+1}-\lambda^k|$ is $\Theta(1/\sqrt{nk})$ for $k \gg n$. Finally the eigenvectors for circulant matrices are orthogonal, so we conclude that $\|(P^{p+1}-P^k)\|_2 = \Theta(1/\sqrt{nk})$. -It turns out that circulant matrices are diagonalized by discrete Fourier transform matrices, i.e eigenvector $v_i$ satisfies $(v_i)_j = \omega^{ij}$ where $\omega=e^{2 \pi i / n}$ is a root of unity. This allows us to sharpen the above bounds a bit. One can show that any eigenvalue of $P$ is of the form $1 - 1/n + (1/n)\cdot\sum_i \alpha_i \omega^i$ where $\sum_i \alpha_i = 1$, $\alpha_i \ge 0$ (convex combination), and $\alpha_0 = 1 - 1/n$. For $k \gg n$ one can show that $|\lambda^{k+1}-\lambda^k|$ is $O(n/k)$ for any $\lambda$ of this form. -Given the above results I have a strong suspicion that for general stochastic $P$ with trace $n-1$ we have $\|P^{k+1}-P^k\| = O(1/k)$. Of course you might say that a better conjecture would be limited to matrices $P$ that are unitarily diagonalizable (a.k.a. normal), but I haven't found examples where non-normality makes things worse so I'm conjecturing $O(1/k)$ holds for non-normal matrices as well. One simple example of a non-normal matrix $P$ with $\|P^{k+1}-P^k\|=\Theta(1/k)$ is $P=\left(\begin{array}{cc}1 - \varepsilon & 0 \\ \varepsilon & 1\end{array}\right)$. (This has trace strictly greater than $n-1$, but that's fine as the trace can be corrected be extended by adding two dummy states without significantly changing the problem.) -Extending the above eigenvalue-based proofs to the general case runs into a huge roadblock: eigenvectors are not in general orthogonal. This is a problem because the initial distribution vector may in general have unboundedly large coefficients when translated into the eigenvector basis. -The mixing time of these Markov chains can be unbounded, so techniques for bounding the mixing time are not obviously helpful. -Edit: the following example shows that the eigenvectors can have an arbitrarily small angle. -$$P=\left(\begin{array}{ccc}1-\varepsilon & 0 & 0 \\ \varepsilon & 1-\delta & 0 \\ 0 & \delta & 1\end{array}\right)$$ -Its (right) eigenvectors are $(\delta-\varepsilon,\varepsilon,-\delta)^T$ , $(0,1,-1)^T$ and $(0,0,1)^T$. As $\delta$ approaches $\varepsilon$ the first and second eigenvectors become arbitrarily close to parallel. - -REPLY [10 votes]: OK, I think I have a full answer at this point, so let me post it. - -Step 1. (algebra). - -If $P$ is an $n\times n$ stochastic matrix and $\lambda$ is an eigenvalue of $P$ with $|\lambda|=1$, then $\lambda^k=1$ for some $k\le n$ and $1,\lambda,\lambda^2,\dots\,\lambda^{k-1}$ are eigenvalues of $P$. -Indeed, let $x$ be an eigenvector. WLOG, $\lambda\ne 1$. Let $S=\{j:|x_j|=\max_k|x_k|\}$. Then if $i\in S$ and $p_{ij}>0$, then $j\in S$. Now, if $i\in S$ and $p_{ij}>0$, then $x_j=\lambda x_i$. Thus, if we have some entry in $x$, we also have $\lambda^k$ times this entry for every $k$, but the number of different entries is at most $n$. Moreover, we can assume that one of the entries is $1$ and split the indices in $S$ into groups $S_m$ by the rule $j\in S_m$ iff $x_j$ is in the half-open counterclockwise arc from $\lambda^m$ to $\lambda^{m+1}$ so that $i\in S_m$, $p_{ij}>0$ imply $j\in S_{m+1}$. From here we immediately see that $P-\lambda^qI$ is not invertible for every $q$ (the $S$-block annihilates the vector $y_j=\lambda^{qm}$ for $j\in S_m$ and the full determinant has the determinant of the $S$-block as a factor. Thus, all powers of $\lambda$ are eigenvalues. - -Step 2. (compactness argument). - -Consider a convergent sequence $P_k$ of $n\times n$ stochastic matrices with the limit $P$. Assume that $P_k$ have eigenvalues $\lambda_k$ and $\lambda_k$ are not contained in any Stolz angle. Then we may assume that $\lambda_k\to\lambda$, $|\lambda|=1$. Clearly, $\lambda$ is an eigenvalue of $P$. If $\lambda\ne 1$, then $P$ has several eigenvalues summing to $0$ (powers of $\lambda$), so $\operatorname{Tr} P\le n-2$, which makes it not a limit point of your set. But If $\lambda=1$, it is even worse, because, if $\lambda_k$ approach $1$ tangentially, then $\lambda_k^{m_k}$ can tend to any point on the unit circle but they are also eigenvalues of $n\times n$ stochastic matrices (powers of $P_k$) and so are their limits. Thus, we have some fixed (but depending on $n$) Stolz angle, containing all the eigenvalues of your matrices. - -Step 3. (harmonic analysis) - -Let $f(m)=\sum_{k=1}^n c_k\lambda_k^m$ for $m\ge 0$ and $0$ for $m<0$ where $\lambda_k$ lie in some fixed Stolz angle $A$. Then -$$ -Vf=\sum_{m\in\mathbb Z}|f(m+1)-f(m)|\le C(A,n)\max_m |f(m)| -$$ -Proof: -We begin with a - -Complex analysis lemma - -Let $F(z)=\sum_{k=1}^n c_k e^{\mu_k}z$ where $\mu_k\in\mathbb C$, $|\mu_k|\le 1$. Then $F$ has at most $C(n)$ zeroes in the unit disk. -Proof: Let $m$ be the maximum of $|f|$ over the unit disk. Then the first $n$ derivatives at the origin are bounded by $C(n)m$. But $\Phi(t)= F(zt)$ ($|z|=1$) satisfies an $n$-th order differential equation $\Phi^{(n)}=\sum_{k=0}^{n-1}b_k\Phi^{(k)}$ with coefficients $b_k$ obtained by expansion of the polynomial $\prod_{k=1}^n (x-z\mu_k)$, which are bounded by $2^n$, say. The standard ODE theory implies that $\Phi$ is bounded by $C'(n)m$ on $[-2,2]$, so the ratio of the maximum of $F$ over the disk of radius $2$ and over the unit disk is bounded, which is enough to control the number of zeroes in the unit disk (each Blaschke factor moves it up fixed number of times). Rescaling and covering, we conclude that if $|\mu_k|<\mu$, then there may be only $C(n,K)$ zeroes in the disk of radius $K/mu$. -Now, - -Induction - -If $n=1$, the claim is obvious: the maximum is just $c_1$ and $|\lambda_1^k-\lambda_1^{k+1}|\le C(A)(|\lambda_1|^k-|\lambda_1|^{k+1})$ -Let $n>1$. Write $\lambda_k=e^{-\mu_k}$ with $|\mu_k|\le C(A)\operatorname{Re\mu_k}$. Let $\mu=\max|\mu_k|=\mu_n$. Note that $f$ is the trace of $F(z)=\sum_{k=1}^n c_ke^{-\mu_k z}$ on integers. The derivative of the real or imaginary part of $F(t)$ can have only $C(2n,K)$ zeroes on $[0,K/\mu]$, so the real and the imaginary parts have a bounded number of intervals of monotonicity there whence $f$ has variation dominated by its maximum on $[0,K/\mu]$. Now, choose $K=K(A)$ so that $\gamma=\lambda_n^N$ is less than $1/2$ in absolute value where $N\approx K/\mu$. The function $g(m)=f(m+N)-\gamma f(m)$ for $m\ge 0$ and $0$ for $m<0$ is bounded by $2\max|f|$ and has one term less. Thus, by the induction assumption, $Vg\le C(n)\max|f|$. -To recover $f$ from $g$, note that $f(m)-\gamma f(m-N)=G(m)$ where $G(m)=g(m-N)$ for $m\ge N$ and $G(m)=f(m)$ for $m\le N$. Note that $VG$ is still under control because we have bounded both $Vg$ and the part of $Vf$ corresponding to the interval $[0,N]$. Now it remains to iterate this recurrence to get $f(m)=G(m)+\gamma G(m-N)+\gamma^2 G(m-2N)+\dots$ and to use the shift invariance of and the triangle inequality for the total variation. - -Step 4. (the end) - -Each entry of the matrix $P^k$ is of this form (assuming that the eigenvalues are distinct, which is a dense case). Thus, the total variation of each entry is bounded by some $C(n)$ depending on $n$ only. This is equivalent to $\sum_k\|P^k-P^{k+1}\|\le C(n)$ but the sequence of norms ($\ell^\infty$) is non-increasing, so it is $O_n(k^{-1})$. -Feel free to ask questions :). I suspect this all is written in some obscure textbooks but to do the literature search now is beyond my abilities.<|endoftext|> -TITLE: Can regular expressions be made unambiguous? -QUESTION [15 upvotes]: When investigating regular languages, regular expressions are obviously a useful characterisation, not least because they are amenable to nice inductions. On the other hand ambiguity can get in the way of some proofs. -Every regular language is recognized by an unambiguous context-free grammar (take a deterministic automaton which recognises it, and make a production $R \rightarrow tS$ for every edge $R \stackrel{t}{\rightarrow} S$ in the DFA, and $R \rightarrow \epsilon$ for every accepting state $R$). -On the other hand, the natural "grammar" for a regular language is its regular expression. Can these be made unambiguous? -To be precise, let's define a parse for a regular expression (this is I think a natural definition, but not one I've seen named before). - -$x$ is an $x$-parse of $x$, if $x$ is a symbol or $x=\varepsilon$ -$(y, 0)$ is an $R\cup R'$-parse of $x$, if $y$ is an $R$-parse of $x$ -Similarly, $(y,1)$ is an $R\cup R'$-parse of $x$, if $y$ is an $R'$-parse of $x$ -$(y_1, y_2)$ is an $RR'$-parse for $x_1x_2$, if $y_i$ is an $R$-parse for $x_i$ for $i=1,2$ -$[]$ is an $R^*$-parse for $\varepsilon$ -$[y_1, y_2, \dots, y_n]$ is an $R^*$-parse of $x_1x_2\cdots x_n$, if $y_i$ is an $R$-parse for $x_i$ for $1 \le i \le n$ - -In short, the parses of a string tell us how a regular expression matches a string if it does. -A regular expression $R$ is unambiguous if, for every $x \in L(R)$, there is only one $R$-parse of $x$. - -Given a regular expression, is there an unambiguous regular expression which matches the same language? - -REPLY [3 votes]: Regarding your question about disambiguation without exponential increase: if you're willing to move to generalized regular expressions (regular expressions with the complement and intersection operators as primitives, rather than exponential-blowup-inducing derived operations), then yes: assuming longest-match-convention on the Kleene star the only source of ambiguity will be the alternation (vertical bar) construct, and you can disambiguate every instance of this by simply turning "a|b" into "a|(b&~a)" ("a or b-and-not-a"). -Note that adding intersection and complementation does not alter the class of languages which may be defined. The languages which can be defined are still exactly the regular languages. -I think this suggests that you probably can't disambiguate in general without exponential blowup because disambiguation is "pretty similar to" (vague term, I know) intersection with the complement. -Here's a nice paper which presents many of the results needed above.<|endoftext|> -TITLE: What is the p-adic valuation of a number? -QUESTION [14 upvotes]: There seem to be two conflicting definitions for p-adic valuation in the literature. -Firstly, for any non-zero integer n, we have $\nu=\nu_p(n)$ is the greatest non-negative integer such that $p^\nu$ divides $n$. Secondly, we have $|n|_p$ which is defined as $1/p^\nu$. [These definitions can be extended to the rationals.] -$\nu$ is defined as the p-adic valuation in Khrennikov, Nilson, P-adic deterministic and random dynamics (for example) and $|\cdot|_p$ is defined as the p-adic valuation in Khrennikov, P-adic and group valued probabilities, in Harmonic, wavelet and p-adic analysis (for example). - -Question: Is there a preferred definition for p-adic valuation? - -REPLY [43 votes]: I will explain what's going on. We call $\lvert x\rvert_p$ the $p$-adic absolute value of $x$ and $v_p(x)$ the $p$-adic valuation of $x$. The distinction that is made by the two terms "absolute value" and "valuation" is completely standard… in English. However, Khrennikov is originally from Russia and in Russian there is one term for both concepts (нормирование = normirovanie, with stress — for English speakers I am not making this up — on the second syllable). There is a term "absolute value" in Russian, but it is not an abstract concept; it refers only to the usual absolute value on the real or complex numbers (and quaternions?). This is perhaps why Khrennikov is using the term "valuation" incorrectly to refer to an absolute value function. -(I'm giving a course in Moscow this semester and I found this point frustrating when I was preparing my initial lectures. In different books I found the same word used for an absolute value and for a valuation and couldn't find the term that exclusively means absolute value. Eventually I determined there isn't one; you just know by context what meaning is intended. Native speakers are welcome to correct me here.) -UPDATE (3 years later): I learned from a student in St. Petersburg that the mathematicians there use separate terms for an absolute value $\lvert\cdot\rvert$ on a field and its corresponding valuation $v$: they call $\lvert x\rvert$ the norm (норма) of $x$ and $v(x)$ the exponent (показатель) of $x$. -UPDATE (11 years later): Consistent with Laurent's answer, the Russian Wikipedia page for absolute value (find the English one and then click on Русский) refers to an absolute value as нормирование and a valuation as экспоненциальное нормиорование, where the first word is eksponentsialnoe = exponential. - -REPLY [7 votes]: I would say it's incorrect to call $|\cdot|_p$ a valuation; it is an absolute value or better yet a norm, but certainly not a valuation. Note also that for any $0 < \alpha < 1$, the formula $|n|=\alpha^{\nu_p(n)}$ defines a norm and it's customary but by no means obligatory to take $\alpha = 1/p$. -See http://en.wikipedia.org/wiki/Valuation_%28algebra%29 for the definition of a valuation and the remark that "Some authors use the term exponential valuation rather than "valuation". In this case the term "valuation" means "absolute value"." - -REPLY [6 votes]: The conflict is just that some people use the words valuation and absolute value interchangeably. The term "p-adic valuation", used correctly, refers to $\nu$, though perhaps in some areas of math the prevailing choice is the other way around.<|endoftext|> -TITLE: Riemannian metric induced by a metric -QUESTION [12 upvotes]: Let $M$ be a smooth manifold, $\rho(p, q)$ — a differentiable metric on $M$. Can we construct Riemannian metric $g(X,Y)$ on $TM$ that induces $\rho(p, q)$? Under what conditions? -I'm sure this question has been dealt with, I just didn't find it in the quick survey of literature :) - -REPLY [17 votes]: Here is a closely related question that may have been what the OP was driving at. Suppose that you ONLY know a Riemannian manifold as a metric space---that is you know the point set and the distance between any two points, but you do not know the metric tensor or even the differentiable structure. Can you nevertheless reconstruct these from the distance function. The answer is that you can. See: -http://www.ams.org/proc/1957-008-04/S0002-9939-1957-0088000-X/S0002-9939-1957-0088000-X.pdf<|endoftext|> -TITLE: What does the word "symplectic" mean? -QUESTION [25 upvotes]: I know the definition of symplectic structure, symplectic group, and so on. But what does the word "symplectic" itself mean? -Meta question: I have many other mathematical words whose etymologies are obscure to me. Is it OK for me to ask one question per such word? - -REPLY [3 votes]: From Pietro Majer's comments I learn that "symplectic" is the past participle of a classic Greek verb which means "to embrace". -Consequently I would just remark then how surprising is the effectiveness of this adjective to reflect the pervasiveness of the ideas from symplectic geometry in modern mathematics, which interconnect many different subjects. -Infact, from the introduction to the paper "Symplectic Geometry" by A.Weinstein, I quote: - -I think it is not unfair to say that symplectic geometry is of interest today, not so much as a theory in itself, but rather because of a series of remarkable "transforms" which connect it with various areas of analysis.<|endoftext|> -TITLE: Can a non-Borel set be a standard Borel space? -QUESTION [9 upvotes]: Recall that a standard Borel space is a measurable space $(X,\mathcal{M})$ (i.e. a set with a $\sigma$-algebra) such that there exists a 1-1 bimeasurable map $\phi$ from $(X,\mathcal{M})$ to $[0,1]$ (the latter equipped with its Borel $\sigma$-algebra $\mathcal{B}_{[0,1]}$). It is known that any Borel subset of a complete separable metric space is a standard Borel space. -Suppose now that $E$ is a non-Borel subset of $[0,1]$ (or any other complete separable metric space), such as the Vitali set. We can equip $E$ with the $\sigma$-algebra $\mathcal{M}$ induced by its inclusion into $[0,1]$, namely $$\mathcal{M} = \{ B \cap E : B \in \mathcal{B}_{[0,1]}\}.$$ $\mathcal{M}$ is also the Borel $\sigma$-algebra generated by the subspace topology on $E$, which is also the metric topology on $E$. -Is it possible that $(E,\mathcal{M})$ is a standard Borel space? I would think not. Clearly the inclusion map $E \hookrightarrow [0,1]$ is not bimeasurable (though it is measurable), but it's less clear that no other injection could be bimeasurable. -Wikipedia gives an example using a set $E$ of outer measure 1 and inner measure 0, and shows indirectly that it cannot be standard Borel by equipping it with the probability measure $P(B \cap E)=m(B)$, noting that the inclusion map $X : E \to [0,1]$ is a uniformly distributed random variable, and observing that $X$ does not admit a regular conditional distribution given itself. However, it is not so clear how to extend this to other non-Borel sets (particularly those with outer measure 0), and in any case it would be nice to have a direct proof if possible. -Thanks! - -REPLY [6 votes]: Strictly speaking, a standard Borel space can also be finite or countable. -Keeping in mind this minor point, -A subset of $[0,1]$ (endowed with the restriction of the Borel $\sigma$-algebra) is a standard Borel space if and only if it is a Borel subset of $[0,1]$. -This comes from the fact that the image of a Borel subset by an injective Borel map between two standard Borel spaces is a Borel set. -This result is proven for example in Dudley "Real analysis and Probability" or Cohn "measure theory". -Now for the proof. Let $\phi$ be the isomorphism between $([0,1],{\cal B})$ and $(E,{\cal M})$. The space E is a subset of $[0,1]$, so we get a Borel injection from $[0,1]$ to $[0,1]$ whose image is precisely $E$. Hence $E$ is a Borel subset of $[0,1]$.<|endoftext|> -TITLE: What polynomials biject from $\mathbb{N}^{2}$ to $\mathbb{N}$? -QUESTION [10 upvotes]: Perhaps there are none with integral coefficients; so let us admit rational coefficients. The map $(x, y) \mapsto x + \frac{1}{2}(x + y)(x + y + 1)$ is well known, and swapping $x$ and $y$ in the formula yields another, so we have two for starters. - -REPLY [6 votes]: Describing such bijections is an open problem. Maximal result (there is no other bijections among polynomials of degree not higher than 4) are contained in - -John S. Lew, Arnold L. Rosenberg, - Polynomial indexing of integer lattice-points I. General concepts and quadratic polynomials, J. Number Theory 10 (1978) pp 192-214, doi:10.1016/0022-314X(78)90035-5. -Polynomial indexing of integer lattice-points II. Nonexistence results for higher-degree polynomials, J. Number Theory 10 (1978) pp 215-243, - doi:10.1016/0022-314X(78)90036-7<|endoftext|> -TITLE: Asymptotics for forbidden subwords -QUESTION [5 upvotes]: Fix an alphabet $A$ and consider words of length $n$ over $A$. Fix a set $B$ of $k$ forbidden subwords (subword is not necessarily connected, i.e. $abb$ is a subword of $abcb$). Can anything be said about the asymptotics of number of permissible words (i.e. words that don't containt any word from $B$ as a subword)? (a particular case - what if $n=k^{1+\epsilon}$ and we let $k \rightarrow \infty$?) - -REPLY [3 votes]: If you fix $B$ then the situation is described by a DFA (deterministic finite automaton), i.e. the set of permissible words is a regular language, and so has a rational generating function; therefore, the number of permissible words grows either exponentially or polynomially. -Re your general question, if you take $B = \{ a : a \in A \}$ (or better, $B$ consists of the empty word) then there are no permissible words. On the other hand, if all the words in $B$ have size greater than $n$, then all words are permissible. So $n = k^{1+\epsilon}$ is not really meaningful. -Maybe you're worried that the last example (all words in $B$ are bigger than $n$) is cheating. You can take $A = \{a,b,c,d\}$ and $B = \{a^kb^{m-k} : 0 \leq k \leq m \}$. The set $B$ is reduced (i.e. no word is a subword of any other word), and yet the number of permissible words is exponential; we can construct such sets $B$ with arbitrary size. -It seems reasonable (see Bill's comment below) to assume that the set of words under the subword relation is a wqo (well-quasi-ordering), and so there is no infinite reduced $B$. Therefore we can't ask whether there's an infinite reduced $B$ which allows exponential growth; if $B$ need not be reduced, take $B = \{a^m : m \geq 1\}$. -Edited to explain the acronyms re Bill's comment.<|endoftext|> -TITLE: Pseudonyms of famous mathematicians -QUESTION [91 upvotes]: Many mathematicians know that Lewis Carroll was quite a good mathematician, who wrote about logic (paradoxes) and determinants. He found an expansion formula, which bears his real name (Charles Lutwidge) Dodgson. Needless to say, L. Carroll was his pseudonym, used in literature. -Another (alive) mathematician writes under his real name and under a pseudonym (John B. Goode). (That person, by the way, is Bruno Poizat: it's no secret, even MathSciNet knows it.) - -What other mathematicians (say dead ones) had a pseudonym, either within their mathematical activity, or in a parallel career ? - -Of course, don't count people who changed name at some moment of their life because of marriage, persecution, conversion, and so on. - -Edit. -The answers and comments suggest that there are at least four categories of pseudonyms, which don't exhaust all situations. - -Professional mathematicians, who did something outside of mathematics under a pseudonym (F. Hausdorff - Paul Mongré, E. Temple Bell - John Taine), -People doing mathematics under a pseudonym, and something else under their real name (Sophie Germain - M. Le Blanc, W. S. Gosset - Student)), -Professional mathematicians writing mathematics under both their real name and a pseudonym (B. Poizat - John B. Goode), -Collaborative pseudonyms (Bourbaki, Blanche Descartes) - -REPLY [2 votes]: Křesomysl Blizzard https://zbmath.org/authors/blizzard.kresomysl was a pseudonym of the collective of Walter Schachermayer, Erik G. F. Thomas and Heinrich von Weizsäcker for the publication -Blizzard, Křesomysl, A Krein-Milman set without the integral representation property., Frolík, Zdeněk (ed.), Abstracta. 8th winter school on abstract analysis. Abstracts of papers presented at the winter school, WSAA 8, Moravská Bouda, Czech Republic, 1980. Prague: Czechoslovak Academy of Sciences. 39-42 (1980). ZBL1437.46007. -Note that the DML CZ https://dml.cz/handle/10338.dmlcz/701173 did only partially resolve the identity as E.G.F. Thomas with the remark "Author corrected, K. Blizzard did not participated at the school"; the full identity is given in [20] of Schachermeyer's homepage https://www.mat.univie.ac.at/~schachermayer/pubs/index.php.<|endoftext|> -TITLE: Homotopy theory for spanning trees of a graph -QUESTION [13 upvotes]: I am studying a paper of L. Lovász, ``A homology theory for spanning trees of a graph,'' but professor Babai has told me that Lovász later realized that this work is better framed in the language of homotopy theory. -Does anyone know if Lovász (or anyone else) published a homotopy-based treatment of the results in the above paper? - -REPLY [7 votes]: I think that this has something to do with what's called A-homotopy theory. In fact, the following two papers on A-homotopy theory reference Lovasz's paper, and from looking at the first paper below it seems like it's exactly the sort of thing you're looking for. -MR1808443 (2001k:57029) Barcelo, Hélène; Kramer, Xenia; Laubenbacher, Reinhard; Weaver, Christopher Foundations of a connectivity theory for simplicial complexes. Adv. in Appl. Math. 26 (2001), no. 2, 97–128. (Reviewer: Andrew Vince), 57Q05 (05B35 05C10 55P99 55Q05) -MR2163440 (2006f:52017) Barcelo, Hélène; Laubenbacher, Reinhard Perspectives on $A$-homotopy theory and its applications. Discrete Math. 298 (2005), no. 1-3, 39–61. (Reviewer: Jean-Louis Cathelineau), 52B40 (05B35 05E25 37F20 52C35 55R80 57Q05)<|endoftext|> -TITLE: Planar layouts of bipartite graphs -QUESTION [7 upvotes]: Instances of SAT induce a bipartite graph between clauses vertices and variable vertices, and for planar 3SAT, the resulting bipartite graph is planar. -It would be very convenient if there was a planar layout that had all the variable vertices in one line and all the clause vertices in a straight line. This can't be done because such a graph would be outerplanar, and $K_{2,3}$ isn't. -But maybe a weaker layout is possible. - -Is it possible to lay out any planar - bipartite graph $G = (A \cup B, E)$ - such that - -All vertices of $B$ are on a straight line -A can be partitioned into $A_1 \cup A_2$ such that all vertices of $A_1$ - are on a parallel straight line to the - left of $B$, and all vertices of $A_2$ - are on a parellel straight line to the - right of $B$. - - -This seems to relate to track drawings of planar graphs. - -REPLY [4 votes]: Any planar graph can be drawn with curves for the edges and its vertices in any position in the plane. -But with straight line segment edges, it's not always possible, even for graphs in which every vertex in A has degree exactly two, and even if you relax the straight-line requirement for A and only require that the vertices in B be on a straight line. For, these graphs are exactly the graphs formed by subdividing every edge of an arbitrary planar graph G. And a drawing of this type, for a graph formed from G in this way, is exactly a two-page book embedding of G. But a planar graph G has a two-page book embedding only if edges can be added to it to make it Hamiltonian. So if you start with a graph G that is maximal planar and non-Hamiltonian, such as the Goldner–Harary graph, and subdivide every edge, you will get a planar bipartite graph that cannot be drawn in the way you request. -As an aside, relaxing the requirement that A be drawn on two lines parallel to the line through B does allow some additional graphs to be drawn, even though the above argument shows that it doesn't allow them all. For instance, Louigi has shown that the cube has no drawing on three parallel lines, but it does have one where B is on a straight line and A is on two sides of it:<|endoftext|> -TITLE: How should one think about sheafification and the difference between a sheaf and a presheaf -QUESTION [38 upvotes]: The first time I got in touch with the abstract notion of a sheaf on a topological space $X$, I thought of it as something which assigns to an open set $U$ of $X$ something like the ring of continuous functions $\hom(U,\mathbb{R})$. People said that sections of a sheaf $F$, i.e. elements of $F(?)$, are something which allow 'glueing' like in the example: if two functions $f:U\to\mathbb{R}$ and $g:V\to\mathbb{R}$ coincide on the intersection $U\cap V$ there is an unique function $U\cup V\to\mathbb{R}$ restricting to $f$ and $g$. So a sheaf consists of 'glueable' objects. -A presheaf, say of rings, on a topological space $X$ is a functor $F:Op(X)^{op}\to Rng$ where $Op(X)$ denotes the category of open sets of $X$. One may generalize all this using the terms 'site' and 'topos' but let's consider this easy situation. A sheaf is a presheaf fulfilling an extra condition, so there is an inclusion of categories -$$ - Pre(X)\leftarrow Shv(X):i. -$$ -Please excuse the awful notation but this inclusion functor admits a left-adjoint, the sheafification functor -$$ -f:Pre(X)\leftrightarrow Shv(X):i. -$$ -Since I got in touch with schemes, I think of a presheaf or a sheaf as of a space. There is a notation of a 'stalk' $F_x\in Rng$ at a point $x$ of $X$. I think of a stalk as the point $x$ of the space $F$. The inclusion functor and the sheafification functor both respect the stalks. -For example the sheafification of a constant presheaf is a locally constant sheaf. -My first question is: - -How shoud I really think about sheafification? - -A presheaf of sets is the canonical co-completion of a category: You take a (small) category $S$ which does not allow glueing (=has not all colimits) and then $Pre(S)$ has all colimits. $S$ is fully and faithfully embedded into $Pre(S)$ with the Yoneda embedding $Y:S\to Pre(S)$. This functor does not respect colimits, so, loosely speaking, the way of glueing is not respected in this transition. Maybe, considering sheaves instead of presheaves is a way of repairing this failure. -My second question is: - -With respect to the interpretation above, what really makes the difference between a presheaf and a sheaf and how should I visualize that difference, if I think of a presheaf as if it is a space? - -Thank you. - -REPLY [2 votes]: It's worth thinking of sheafification as a "localization" process. For example if you take the pre-sheaf on $X$ defined for each $U$ to be the set of constant functions $U\to Y$, then the "sheafification" is the set of "locally constant" functions on each $U$. -Or let $F(U)$ be the set of bounded functions $U\to Y$ for some metric space $Y$. Then the sheafification of $F$ gives for each $U$ the locally bounded functions on $U$. -If $F(U)=\{\star\}$ if the closure of $U$ is compact, and $F(U)=\emptyset$ if the closure of $U$ is not compact, then the sheafification of $F$ is $F(U)=\{\star\}$ if $U$ is locally compact. -Indeed, practically any time you talk about a "local" version of a property, you are implicitly dealing with a sheafification.<|endoftext|> -TITLE: Is it possible to recover the degree of a field extension from a list of elements and the ground field? -QUESTION [6 upvotes]: I'm interested to know if there is anything known about recovering the degree of a field extension, $E/k$, given $E=k(\alpha_1,\ldots, \alpha_n)$ (here I'm assuming that the extension is of finite degree. Obviously there are some silly examples anyone could eyeball, like $\sqrt[m]{d}$ for $n=1$ and $\alpha_1=d\in k$ or the case $E=\mathbb{F}_q$ and $k=\mathbb{F}_p$ where we can just divide orders (intermediate fields are clearly equally trivial). Where, to give the necessary bit of care, we assume this is a nontrivial extension. If $E/k$ is Galois and we can appeal to other bits of theory, we might also get the degree by calculation of the Galois group('s order). Is there anything known about more general extensions? It is conceivable given that $E/k$ is an extension of algebraic number fields that the theory of ideals might give an insight, especially given the (IMO) rather fascinating fact that $\mathcal{O}_E$ is finitely generated as a $\mathbb{Z}$-module is an equivalent statement, and using the machinery of algebraic number theory, or some other extra structure, but I'm principally concerned with the more general theory if any exists or if just anything is known about this problem. - -REPLY [4 votes]: To put this one to rest, I will answer the more precise question that, after much prodding, we got Adam to formulate in the comments. I am merely paraphrasing a comment of Qiaochu. -If you are given the $\alpha_i$ as roots of irreducible polynomials, then the degree is not a function of the $\alpha_i$. Of course, when you adjoin only one root, the degree of the extension is just the degree of the minimal polynomial. But as soon as you adjoin two roots, you cannot recover the degree. Linear independence over the base field doesn't help either: let $k=\mathbb{Q}$, let $\alpha_1$, $\alpha_2$ be two distinct roots of $x^9-2$. Then they can generate -$\mathbb{Q}(\sqrt[9]{2},\mu_3)$ or $\mathbb{Q}(\sqrt[9]{2},\mu_9)$ for some 9-th roots $\sqrt[9]{2}$ of 2, depending on whether $\alpha_1$ and $\alpha_2$ differ by a 3-rd or by a 9-th root of unity. Accordingly, the degree will be either 18 or 54. In either case, the roots will be linearly independent over $\mathbb{Q}$, so they satisfy your conditions. -Adjoining roots of distinct polynomials won't help either, since you can just take some other element of the top field whose minimal polynomial has some roots over the bottom field. Now, if instead you adjoin roots of polynomials, whose splitting fields are disjoint over the base field, then the degree is just the product of the degrees of the minimal polynomials.<|endoftext|> -TITLE: What abstract nonsense is necessary to say the word "submersion"? -QUESTION [15 upvotes]: This question is closely related to these two, but the former doesn't go far enough and the latter didn't attract much attention, and anyway I want to ask the question slightly differently. -Recall that in the category of Topological spaces or in the category of Manifolds, a submersion is a (not necessarily surjective!) map $f: X \to Y$ so that for each point $x\in X$, there exists open neighborhood $f(x) \in U \subseteq Y$ and a map $g: U \to X$ splitting $f$, i.e. $f \circ g = \operatorname{id}_U$. This definition does not generalize well to other categories: it requires at least that "points" know a lot about the objects, and that we know what are "open neighborhoods". - -My question is: How much extra "abstract nonsense" structure do I need to put on a category for it to have a good theory of submersions? - -On the one hand, the surjective submersions of manifolds are all regular epimorphisms (does this characterize the surjective submersions?), and so I could imagine defining "submersion" to mean a map that factors as a regular epi and a regular mono (I think that the regular monos in manifolds are the open embeddings?). Then it seems that I don't need any extra structure, but I have not checked that this conditions characterizes submersions. -On the other hand, (surjective?) submersions form a Grothendieck pretopology, and hence determine a Grothendieck topology. Conversely, I would have assumed that a Grothendieck topology (which is extra structure on a category) determines which maps are submersions, although I am sufficiently new to this that I don't have a proposal for such a definition. - -REPLY [2 votes]: Given two manifolds $M$and $N$ and a differentiable map $f:M\to N$, pull back the tangent bundle of $N$. The derivative arrow $Df: TM \to f^*TN$ is a morphism of vector bundles over $M$ and a regular epimorphism iff $f$ is a submersion. So the extra structure we need is something like the tangent bundle on every object of the category.<|endoftext|> -TITLE: Certain double covers of cubic surfaces -QUESTION [5 upvotes]: Let $S$ be a smooth cubic surface in $\mathbb{P}^3$. I would like to understand that variety $V$ that parametrizes lines $\ell$ such that $\ell \cdot S=3P$ with $P \in S$. At any point $P \in S$, let $\Pi_P$ be the tangent plane, and let $\Gamma_P=\Pi_P \cap S$. Generically, $\Gamma_P$ is a plane cubic with a node at $P$ and therefore two tangents at $P$. Each of these satisfies $\ell \cdot S=3P$. This leads us to the fact that $V \dashrightarrow S$ is a double cover. I wanted to know if $V$ has a name, and also what can we say about it. - -REPLY [7 votes]: For any surface of degree at least three we can consider more generally a similar construction of the set of lines $\ell$ with $\ell \cdot S \ge 3$. This is called the asymptotic double cover of $S$. See -McCrory, C., Shifrin, T. Cusps of the projective Gauss map, J. Differential Geom. 19 (1984), 257–276. -or my paper: -Surfaces in $P^3$ over finite fields, in Topics in Algebraic and Noncommutative Geometry: Proceedings in Memory of Ruth Michler, C. Melles et al. eds., Contemporary Math. 324 (2003) 219-226.<|endoftext|> -TITLE: Best lower bound for off-diagonal Ramsey numbers -QUESTION [5 upvotes]: What are the current best lower bounds for off-diagonal Ramsey numbers $R(k,l)$ with $l$ of order unity and asking for asymptotic behavior for large $k$, such as $R(k,4)$, $R(k,5)$, and so on? (please include any log factors, too!) Other than the more complicated arguments of Kim for $R(k,3)$, are all the other best lower bounds from the Lovasz local lemma? - -REPLY [8 votes]: The best bounds I know of are due to Tom Bohman for $R(k,4)$ and Bohman and Peter Keevash for $R(k,5)$ and beyond. Both rely on using the differential equations method to analyze the following process: Start with the empty graph, and at each step add an edge uniformly at random among all edges which do not create a $K_t$. The bounds they achieve are -$$R(k,t) \geq c_t \left( \frac{k}{\log k} \right)^{\frac{t+1}{2}} (\log k)^{\frac{1}{t-2}}$$ -The final term in this product corresponds to the improvement over the bounds obtained using the Local Lemma. For $t=3$ it matches Kim's bound up to a constant factor.<|endoftext|> -TITLE: What condition on a "bibundle between categories" generalizes "right-principal bibundle between groupoids"? -QUESTION [6 upvotes]: My question is long on background and motivation, and almost but not quite answered over at the nLab. I'll write up a bunch before asking my question (feel free to skip to the end or look at the title), so that (i) those who know more than me can see exactly what I do and do not understand (ii) those that know less than me might learn something (iii) to clarify my own thoughts. -Background -Let $\mathcal S$ be a reasonably nice category: for example, I want it to at least have either all finite limits, or I want it to have a good theory of submersions; and I need some extra conditions (see comments), that I haven't fully thought through, but they should be satisfied in various "geometrical" categories like (your favorite convenient category of) Topological Spaces or Manifolds. A span in $\mathcal S$ is a diagram $X \leftarrow M \rightarrow Y$, and if I can, I will require that the right map be a submersion. There is a two-category $\operatorname{Span}(\mathcal S)$ constructed in the usual way: 1-morphisms are spans, and 2-morphisms are maps of spans that cover the identity morphisms on the bases. -A category object in $\mathcal S$ is a span $X = \{X_0 \leftarrow X_1 \rightarrow X_0\}$ and 2-morphisms $$ \{ X_0 \leftarrow X_0 \rightarrow X_0 \} \overset i \to \{X_0 \leftarrow X_1 \rightarrow X_0\}$$ $$ \{X_0 \leftarrow (X_1 \underset{X_0}\times X_1) \rightarrow X_0\} \overset m \to \{X_0 \leftarrow X_1 \rightarrow X_0\}$$ in $\operatorname{Span}(\mathcal S)$ making $X$ into an algebra object. (The domain of $m$ is the 1-composition of $X$ with itself in $\operatorname{Span}(\mathcal S)$, and the domain of $i$ is the 1-identity span on $X_0$.) Writing $\circ$ for the 1-composition in $\operatorname{Span}(\mathcal S)$, the requirement is that the various maps $X\circ X \circ X \to X$ formed from $m,i$ all agree. -A functor in $\mathcal S$ $\{X_1 \rightrightarrows X_0\} \to \{Y_1 \rightrightarrows Y_0\}$ is a pair of maps $X_1 \to Y_1$ and $X_0 \to Y_0$ making some diagrams commute. But there tend not to be enough functors when $\mathcal S$ does not satisfy the axiom of choice. For example, if $\mathcal S$ is the category of manifolds, then certain categories (the groupoids, which I will define in a moment) are supposed to present stacks, but the functor that associates to each manifold $M$ the groupoid of functors $\{M \rightrightarrows M\} \to \{X_1 \rightrightarrows X_0\}$ into some groupoid $X$ does not satisfy the right descent axioms. -Instead, the usual thing to do is to define the notion of "right-principal bibundles". Let $X = \{X_1 \rightrightarrows X_0\}$ and $Y = \{Y_1 \rightrightarrows Y_0\}$ be categories. An $X,Y$-bibundle is a span $B = \{X_0 \leftarrow B_1 \rightarrow Y_0\}$ and 2-morphisms $X \circ B \to B$ and $B \circ Y \to B$, such that all the various 2-morphisms $X \circ X \circ B \circ Y \circ Y \to B$ agree (as above, $\circ$ is the 1-composition in $\operatorname{Span}(\mathcal S)$). The "tensor product over $Y$" gives a "composition" of bibundles which is associative up to a canonical associator that satisfies a pentagon. -Given a span $B = \{X_0 \leftarrow B_1 \rightarrow Y_0\}$, there is another span -$$ X_0 \leftarrow (B_1 \underset{X_0 \times Y_0}\times B_1) \rightarrow Y_0 $$ -and a "diagonal" map from $B$ to this other span. Let $B$ be an $X,Y$-bibundle. Using the diagonal map, one can build a map -$$ B_1 \underset{Y_0}\times Y_1 \to B_1 \underset{X_0}\times B_1$$ -which is actually a map of objects over $X_0 \times Y_0 \times Y_0$. On (generalized) elements, this map is $(b,y) \mapsto (b,by)$. The bibundle $B$ is right-principal if this map is an isomorphism. A category $X$ is a groupoid if it is right-principal as an $X,X$-bibundle. If $X,Y$ are groupoids, a functor $f: X \to Y$ determines an $X,Y$-bibundle where the middle object is $X_0 \underset {Y_0} \times Y_1$, and it is right-principal. -Then the point is that the 2-category whose objects are groupoids in $\mathcal S$ and whose one-morphisms are right-principal bibundles embeds as a full sub-2-category into the category of "stacks", which I will not define. For a precise version of this story, see for example C. Blohmann, Stacky Lie groups, 2007. -Question -Any functor of categories in $\mathcal S$ determines a bibundle, but if the categories are not groupoids, then the bibundle is not (usually) right-principal; for example, the identity bibundle is not. I do want the bibundles of groupoids to be "morphisms" of categories, so I don't want to just take functors. On the other hand, as observed in Op. cit., if we don't demand some sort of "right-principality" condition, then the 2-category which allows all bibundles as 1-morphisms has neither products nor a terminal object. Hence my question is: - -What is the correct notion of "bibundle" for (internal) categories that generalizes "right-principal bibundle of groupoids"? - -The answer to my question is almost "anafunctor". An anafunctor seems to have a bit more structure than a bibundle, since it is an object not in $\operatorname{Span}(\mathcal S)$ but in $\operatorname{Span}(\text{categories in }\mathcal S)$. (Conversely, at least when $\mathcal S$ is the category of sets, bibundles wihout any conditions are the same as profunctors.) If I understood better how to go between anafunctors and bibundles, I would probably be happy. - -REPLY [3 votes]: Hi Theo. This is something I have been working on lately. In Makkai's original paper on anafunctors he defines a condition on an anafunctor which makes it saturated. His motivations are logical, in that he wants the 'image' of a point in the domain category (pullback and pushdown along the span) to be closed under isomorphism. This is satisfied for his examples of (co)limit anafunctors, or an anafunctor arising from a universal property. In the case that he talks about saturated anafunctors between groupoids, this is the same as a right principal bibundle (in Set). When we replace groupoids by categories, then he defines a saturated anafunctor to be an anafunctor such that the underlying span between the cores is saturated. Thus the 'cover' (in Makkai's case, a surjection of sets) is a right principal bibundle for the underlying groupoids. -This is done so that the canonical 2-functor $Cat \to Cat_{sat.ana}$, where we do not assume choice, sends fully faithful, essentially surjective functors to equivalences in the bicategory $Cat_{ana}$. Actually I'm fudging here, because Makkai defines $Cat_{sat.ana}$ as a an anabicategory - a category weakly enriched over $Cat_{ana}$. -Note also that Street at one point defines in his Oberwolfach descent notes a definition of a 'torsor for a category' which is just the same as Makkai's definition. He correctly notes that there are non-invertible maps between such 'torsors', unlike the group(oid) case. -So to cut a long story short, it is possible to define a saturated anafunctor for internal groupoids and hence categories. (email me if you would like some notes on this) -But! This is not the same as a torsor for an internal category, as one could extract or otherwise from various places, e.g. Moerdijk's or Johnstone. See this answer. The two definitions are aiming at very different things. For example, a saturated anafunctor with codomain a topological monoid (as a one-object category) is trivial, but a torsor for the same is not. In general internal saturated anafunctors are about inverting weak equivalences between internal categories (fully faithful and essentially 'surjective': i.e. some map admitting local sections) but internal torsors are about characterising maps between topoi and classifying topoi and stuff (you can tell I know less about the latter).<|endoftext|> -TITLE: Bolyai's construction -QUESTION [6 upvotes]: Here is Bolyai's construction (in Klein model), which I learned recently from answer of Will Jagy to this question. -It is a Compass-and-straightedge construction of asymptotically parallel line in absolute geometry. - -Do you know an elementary proof showing that Bolyai's construction really does what it suppose to? -"Elementary" means without calculations and without referring to the models. -Comments: - -I know a simple proof in Klein model (thanks to A. Akopyan). (It is easy to guess from the picture above. Draw one more vertical line and note that cross ratios for two quadruples of points on two asymptotically parallel lines is the same.) -If there is a simple argument, I would include it in the course on Foundation of geometry which I'm teaching at the moment. -At the moment I do not know a simple argument in the Poincaré disk model. - -REPLY [4 votes]: There is a proof in Introduction To Non Euclidean Geometry by Harold E. Wolfe -It is an elementary proof without calculations or referring a models. -You can get a copy here: -http://www.archive.org/details/introductiontono031680mbp<|endoftext|> -TITLE: When can we detect forcing? -QUESTION [13 upvotes]: First, a rather broad question: has there been any work on what, given a model $M$ of set theory, we can say about those models of set theory $N$ and posets $\mathbb{P}$ such that $\mathbb{P}\in N$ and $M=N[G]$ for some $G$ $\mathbb{P}$-generic over $N$? -Second, a more specific question. Let a poset $\mathbb{P}$ be $detectable$ if there is some sentence $\phi_\mathbb{P}$ in the language of set theory + a constant denoting $\mathbb{P}$ such that for all $M$ with $\mathbb{P}\in M$, we have $(M, \mathbb{P})\models\phi\iff M=N[G]$ for some model $N$ with $\mathbb{P}\in N$ and $G$ $\mathbb{P}$-generic over $N$. What posets are detectable? [Answered in the comments by Amit Kumar Gupta.] -Finally, an incredibly general question. Let $M$ be a model of $ZFC$, $\mathcal{C}$ a class of posets in $M$. Say $\mathcal{C}$ is $consistent$ if there is some elementary extension $N$ of $M$ such that, for all $\mathbb{P}\in \mathcal{C}$, there is some $N_\mathbb{P}\models ZFC$ with $\mathbb{P}\in N_\mathbb{P}$ and some $G$ $\mathbb{P}$-generic over $N_\mathbb{P}$ such that $N=N_\mathbb{P}[G]$. What are the consistent classes like? Can we say anything interesting about them? -I'm not sure if these questions are meaningful, or - even assuming they are - if they are interesting. Basically, what I'm interested in is the notion of inverse forcing - similar in an aesthetic sense, at least to me, to inverse Galois theory - and I haven't run into anything along these lines yet. - -REPLY [5 votes]: Here is a nice observation that is not mentioned in the previous answers: -Even if we know that a model is a forcing extension, we may still not know either the forcing or the original model. Of course, this is clear if $V=M[G]$ where $G$ is generic over $M$ for some poset ${\mathbb P}$ that admits some ${\mathbb Q}$ as a factor, so $G=H*I$ for some $H$ generic over $M$ for ${\mathbb Q}$ and $V=(M[H])[I]$ for $I$ generic over $M[H]$ for the poset that is usually denoted by ${\mathbb Q}/G$. But in fact, $V$ could be generic over different, ${\subseteq}$-incomparable models for very different posets. -Here is an example (probably due to Woodin?): Assume a tiny amount of determinacy ($\Sigma^1_2$-determinacy is an overkill). It follows that, for any real $x$, ${\mathcal P}(\omega)\cap L[x]$ is countable, so there is a real $w$ such that $x\le_T w$ and $L[w]$ is a forcing extension of $L[x]$ by Cohen forcing. Also, there is a $w'$ with $x\le_T w'$ and $L[w']$ a forcing extension of $L[x]$ by random forcing. The amount of determinacy we are assuming allows us to invoke Martin's cone theorem to conclude that (from a mild large cardinal hypothesis): - -There is a real $x$ such that for any real $y$, if $x\le_T y$, then $L[y]$ is both a forcing extension of some model $L[z]$ by Cohen forcing, and a random extension of some model $L[z']$ by random forcing. - -(I know a few years ago a student of Kunen was looking at a similar problem, but I did not keep up with their results.)<|endoftext|> -TITLE: English translation of Lambert's Theorie der Parallellinien? -QUESTION [6 upvotes]: Does anyone know if there is an available (published or unpublished) English translation of Johann Lambert's Theorie der Parallellinien? I was able to find it online in German by way of the bibliography of Jeremy Gray's Worlds out of Nothing at this link, but I have been unable to find it in English. Any help would be appreciated. - -REPLY [7 votes]: Vincenzo de Risi of Max Planck Institute (https://www.mpiwg-berlin.mpg.de/en/users/vderisi) is preparing an English edition of this paper to be published by Springer.<|endoftext|> -TITLE: Higher Composition Law -QUESTION [7 upvotes]: Prof M.Bhargava's work on "Higher Composition Law" which solved some outstanding conjectures on number theory seems to be very interesting topic. I have seen his papers but, in spite of the titles, it is not easy to understand (Of course in my point of view, for sure there are many people who can understand it easily). -Do you know any lecture note or expository paper which explains more details and some explicit example? especially his work on composition law for binary quadratic form. -Thanks - -REPLY [9 votes]: I have lecture notes that I'd like to turn into a book one day. I have not yet had time to adapt to my new TeX system, and the drawings done via ps-tricks do not yet come out as planned. In addition, there are gaps and mistakes that I have not yet had time to fill and correct (the chapter on composition should be essentially correct and complete, however). In any case, the present set of notes, for the time being, can be found -here. Comments and -corrections are welcome. - -REPLY [5 votes]: Try the Bourbaki talk given by Karim Belabas a couple of years ago : -Paramétrisation de structures algébriques et densité de discriminants [d'après Bhargava] -Astérisque, Vol. 299 (2005), Exp. No. 935, pp. 267-299, Séminaire Bourbaki. Vol. 2003/2004,MR 2167210.<|endoftext|> -TITLE: What facts in commutative algebra fail miserably for simplicial commutative rings, even up to homotopy? -QUESTION [29 upvotes]: Simplicial commutative rings are very easy to describe. They're just commutative monoids in the monoidal category of simplicial abelian groups. However, I just realized that a priori, it's not clear that even some of the simplest facts we can prove for ordinary commutative rings (in particular those that depend integrally on the axiom of choice, or even those that depend on the law of the excluded middle) will hold for simplicial commutative rings. However, we have at least one saving grace. That is, the interesting parts of simplicial commutative algebra come from considering things up to homotopy. -So, for example, as far as it makes sense, can we prove that every simplicial ideal of a simplicial commutative ring is weakly equivalent to one contained in a maximal simplicial ideal? Perhaps a better way to state this would be something like, "every noncontractible simplicial commutative ring admits at least one surjective map to a simplicial commutative ring that's weakly equivalent to a simplicial field", or some variation on where the homotopy equivalence appears. Given that the axiom of choice does not necessarily hold in $sSet$, it doesn't seem reasonable to think that the ordinary theorem will hold. -Is there a version of the Hilbert basis theorem that holds up to isomorphism? How about weak equivalence? -What other well-known theorems will fail, even up to homotopy? - -REPLY [8 votes]: Maybe it is too late to answer this, but here is a classical "epic fail" which I'm sure you've already seen in disguise: -Even for the constant ring $\mathbb{Z}$, thought of as a simplicial ring, short exact sequences of free modules of finite rank (i.e. free abelian group of finite rank on each degree) don't split. -Simplicial free abelian groups are just complexes of abelian groups. Short exact sequence of complexes produces a long exact sequence, and if the boundary map is non-zero then it certainly won't split. -I don't have a more conceptual way to "explain" this, but for me it is like such failure are caused by some non-abelian input, (e.g. those coming from a topological space or homotopy groups.)<|endoftext|> -TITLE: Examples of sequences whose asymptotics can't be described by elementary functions -QUESTION [25 upvotes]: It is somewhat miraculous to me that even very complicated sequences $a_n$ which arise in various areas of mathematics have the property that there exists an elementary function $f(n)$ such that $a_n = \Theta(f(n))$ or, even better, $a_n \sim f(n)$. Examples include - -Stirling's approximation $n! \sim \sqrt{2\pi n} \left( \frac{n}{e} \right)^n$ (and its various implications), -The asymptotics of the partition function $p_n \sim \frac{1}{4n \sqrt{3}} \exp \left( \pi \sqrt{ \frac{2n}{3} } \right)$, -The prime number theorem $\pi(n) \sim \frac{n}{\log n}$, -The asymptotics of the off-diagonal Ramsey numbers $R(3, n) = \Theta \left( \frac{n^2}{\log n} \right)$. - -What are examples of sequences $a_n$ which occur "in nature" and which provably don't have this property (either the weak or strong version)? Simpler examples preferred. -(I guess I should mention that I am not interested in sequences which don't have this property for computability reasons, e.g. the busy beaver function. I am more interested in, for example, natural examples of sequences with "half-exponential" asymptotic growth.) - -REPLY [3 votes]: Suppose F is a field of finite characteristic and u,v,w lie in A=F[x,y]. Let a_n be the dimension of the vector space A/(u^n,v^n,w^n). Suppose a_1 is >0 and finite. Then as n grows, -(a_n)/(n^2), though bounded above and bounded away from 0 generally has an oscillating "fractal-like" behavior.<|endoftext|> -TITLE: affine morphism -QUESTION [7 upvotes]: Let $f:X\to Y$ be a morphism of varieties over a field $k,$ such that $X(\overline{k})\to Y(\overline{k})$ is bijective. Is $f$ necessarily an affine morphism? - -REPLY [9 votes]: No. Here is an example: -Let $g:\mathbb A^2\to Y$ be a morphism that glues two closed points $P$ and $Q$ together and otherwise it is an isomorphism. Now let $X=\mathbb A^2\setminus \{P\}$ and $f$ the restriction of $g$ to $X$. -If you add to the conditions that $f$ is projective, then the statement is true, because then $f$ is finite and hence affine. -EDIT: -A minute ago there was another question asking how to define $Y$. It has now disappeared, but perhaps it is still interesting to include some references. -1) See this MO question and the discussion. -2) See this paper of Karl Schwede. Especially, Theorem 3.4 in general and Corollaries 3.6 and 3.9 in particular. -3) Try to construct it directly. If you get stuck, look at Karl's paper and try to carry out the computation in this special case. It might be a worthy exercise if you have never done anything like this before.<|endoftext|> -TITLE: Convergence of an empirical distribution w.r.t. the Hellinger distance -QUESTION [10 upvotes]: Let $P$ be a probability distribution on a finite set $\mathcal{X}$ and let $X_1, X_2, \ldots, X_n$ be drawn i.i.d. according to $P$. Define the empirical distribution: -$\hat{P_n}(x) = \frac{1}{n} \sum_{i=1}^{n} 1_{X_i = x}$ -Let $d_H(P,Q)$ be the Hellinger distance: -$d_H(P,Q) = \left( \frac{1}{2} \sum_{x \in \mathcal{X}} ( \sqrt{P(x)} - \sqrt{Q(x)} )^2 \right)^{1/2}$ -Is there a nice expression for the expected distance between $\hat{P_n}$ and $P$? That is, is there some formula like -$\mathbb{E}[ d_H(P,Q) ] = C \frac{1}{n} - O(\frac{1}{n^2})$ -where $C$ can be written out explicitly? Or if the rate of convergence is slower than $1/n$, can we get the exact rate of convergence? -For context, if we consider the KL-divergence or $L_1$ distance then we can get explicit expressions for the first term in the rate of convergence of $\hat{P_n}$ to $P$. Can we do the same for the Hellinger distance? -It would be interesting to know this for densities as well, but maybe the discrete problem is easier. - -REPLY [5 votes]: it is possible to show that $\mathrm {E}d(P,\hat{P_n})\sim \frac{C}{\sqrt{n}}$ and specify the value of $C$. -let -$$D_n^2 =\sum_{x \in \mathcal{X}} \left( \sqrt{P(x)} - \sqrt{\hat{P_n}(x)} \right)^2 = 2d^2(P,\hat{P_n}). $$ -$4nD_n^2$ is known in statistics [for reasons unclear to me] as the freeman-tukey goodness-of-fit [gof] statistic for testing the null hypothesis that $X\sim P$. like the better known pearson chi-squared gof statistic, it also has [under the null hypothesis] an asymptotic chi-squared distribution with $k-1$ df. here $k=|\mathcal{X}|$. -the statistic $D_n^2$ seems to have been first considered by Matusita in On the Estimation by the Minimum Distance Method. in Decision Rules, Based on the Distance, for Problems of Fit, Two Samples, and Estimation, Matusita develops some asymptotic [and other] properties of $D_n^2$, including the fact that under the null hypothesis, as $n\to\infty$, -$$\kern-1.9in (1)\kern1.9in 4nD_n^2\ \buildrel{\mathcal L}\over{\to}\ \chi^2_{k-1}.$$ -it is also shown there that -$$\kern-.88in (2)\kern.88in 4nD_n^2\ \le\ \mathbb{X}^2_n\ :=\ n\sum_{x \in \mathcal{X}} \frac{\left({\hat P}(x)-P(x)\right)^2}{P(x)}. $$ -$\mathbb{X}^2_n$ is, of course, the pearson chi-squared gof statistic, and it is well-known that under the null hypothesis $X\sim P$, as $n\to\infty$, -$$\kern-2in (3)\kern2in \mathbb{X}^2_n\ \buildrel{\mathcal L}\over{\to}\ \chi^2_{k-1}.$$ -it is also easily seen that for all $n\ge 1,\ \mathrm {E} \mathbb{X}^2_n\ =\ k-1$. -together with (3) [and non-negativity], this entails that $\mathbb{X}^2_n$ is uniformly integrable. in view of (2), so is $4nD_n^2$, so it follows from (1) that -$$\mathrm {E}4nD_n^2\to \mathrm {E}\chi^2_{k-1}\ =\ k-1\ \mathrm{as}\ n\to\infty$$ -and -$$\mathrm {E}2\sqrt{n}D_n\to \mathrm {E}\chi_{k-1}\ \mathrm{as}\ n\to\infty.$$ -[for more details on connections between convergence in law, uniform integrability and convergence of expectations, see Billingsley 1st ed, p32 theorem 5.4 or Billingsley 2nd ed pp31-32 theorems 3.4 and 3.5.]<|endoftext|> -TITLE: Techniques for lower-bounding angle between two eigenvectors of a matrix -QUESTION [5 upvotes]: Are there any techniques for lower-bounding the angle between eigenvectors of a matrix? Or a lower bound on the related quantity of the condition number of the matrix of eigenvectors? In particular I'm looking for bounds that depend on the difference in the corresponding eigenvalues, with larger angles when the eigenvalues are more separated. -For symmetric matrices (and more generally normal matrices) the angles are of course all right angles. I'm looking for techniques that apply to non-normal matrices. -(The particular class of matrices that I care about is stochastic matrices with trace $n-1$ as described in my previous question Bounds on $||P^{k+1} - P^k||$ for $n$ by $n$ stochastic matrix $P$ with trace $n-1$ and integer $k>>n$. .) - -REPLY [4 votes]: Here's a generalization of the result that eigenvectors of a real symmetric matrix with distinct eigenvalues are orthogonal: -Let $x$ and $y$ be unit eigenvalues of a real matrix $A$ with eigenvalues $\lambda$ and $\mu$, respectively. Then -$$ -\lambda\langle x,y\rangle -=\langle Ax,y\rangle -=\langle x,A^\mathrm{T}y\rangle -=\langle x,Ay\rangle-\langle x,(A-A^\mathrm{T})y\rangle -=\overline{\mu}\langle x,y\rangle-\langle x,(A-A^\mathrm{T})y\rangle. -$$ -Rearranging and applying standard inequalities then yields -$$ -|\langle x,y\rangle| -\leq\frac{\|A-A^\mathrm{T}\|_2}{|\lambda-\overline{\mu}|}. -$$ -The numerator of this fraction makes intuitive sense, since any correlation between eigenvectors must come from the skew-symmetric part of the matrix. Furthermore, the denominator provides the relationship you suggested in your question, especially when the eigenvalues are real.<|endoftext|> -TITLE: deformation and obstruction of a pair (X, D) -QUESTION [9 upvotes]: Let $X$ be a smooth variety and $D \subset X$ be an effective, reduced, irreducible divisor. My question is the following. -1.What is the first order deformation and obstrution for the pair $(X, D)$? -2.In particular, if $D$ is a singular divisor and has a global smoothing such that the line bundle $\mathcal{O}(E)\vert_E$ also extends to a general smoothing, then under what conditions can we deform the pair so that the induced deformation of the pair gives a smoothing of $D$. - -REPLY [5 votes]: Let me expand a little bit my previous comment. For the sake of simplicity, I will assume first that $D$ is smooth. -Then one has a short exact sequence -$0 \to T_X(- \log D) \to T_X \to N_{D|X} \to 0 \quad (*)$, -where $T_X(- \log D)$ is the sheaf of tangent vectors on $X$ which are tangent to $D$. The group $H^1(X, T_X(\log D)$ classifies the first-order deformations of the pair $(X, D)$, and the obstructions lie in $H^2(X, T_X(-\log D))$. -Taking cohomology in $(*)$, we obtain -$H^0(D, N_{X|D}) \stackrel{\alpha}{\to} H^1(X, T_X(-\log D)) \to H^1(X, T_X) \stackrel{\beta}{\to} H^1(D, N_{D|X}) \to H^2(X, T_X(-\log D)).$ -It is well known that the group $H^0(D, N_{D|X})$ classifies first-order embedded deformations of $D$ in $X$, with $X$ fixed. Therefore the map $\alpha$ naturally associates to every such a deformation the corresponding deformation of the pair $(X, D)$. -Moreover, we can interpret the map $\beta$ as the obstruction to lifting an abstract first-order deformation of $X$ to a deformation of the pair $(X, D)$. -For instance, let $X$ be a smooth projective surface and $D$ a nonsingular rational curve with -$D^2=-1$. Then $H^1(D, N_{D|X})=0$, so every abstract deformation of $X$ lifts to a deformation of the pair $(X, D)$. -If instead $D^2=-2$, there are in general abstract deformations of $X$ which do not come from deformations of the pair. For example, the general deformation of a Kummer surface (which contain 16 $(-2)$-curves) is a $K3$ surface without $(-2)$-curves. -When $D$ is singular, the same arguments hold with $N_{D|X}$ replaced by the so-called equisingular normal sheaf $N'_{D|X}$. See Sernesi's book [Deformation of algebraic schemes, Chapter 3] for further details.<|endoftext|> -TITLE: Why does the (S2) property of a ring correspond to the Hartogs phenomenon? -QUESTION [24 upvotes]: Hartogs Theorem says every function whose undefined locus is of codim 2 can be extend to the whole domain. I saw people saying this corresponds to the (S2) property of a ring. But I can't see why this is true. Can anybody explain this or give a heuristic argument? - -REPLY [11 votes]: Here's a somewhat more elementary argument that (S2) implies the Hartogs condition. More precisely, I will show that if $X$ is an (S2) noetherian scheme, then any rational function defined outside a closed subset of codimension two can be extended to the whole domain. (This extension is unique by definition of a rational function.) -Assume, by way of contradiction, that $X$ is an (S2) noetherian scheme and $f$ is a rational function on $X$ that is defined outside a closed set of codimension at least two, but cannot be extended to the whole domain. Let $\mathscr{I}$ be the ideal of denominators of $f$; in other words, over an open affine $\operatorname{Spec} A$, -$$I = \{g \in A \mid g f \in A\}.$$ -This is well-defined as a sheaf since the ideal of denominators is preserved under flat pullback (and in particular, localization); see this mathoverflow question. -If $g \in A$ is a nonzerodivisor, then $g \in I$ if and only if $f = a / g$ for some $a \in A$, hence the name "ideal of denominators." One can check that the closed subscheme $Z \subset X$ corresponding to $\mathscr{I}$ is, set-theoretically, the "indeterminacy locus of $f$": the smallest closed subset such that $f$ is defined over $X \smallsetminus Z$. By hypothesis, $f$ can be defined outside a closed subset of codimension two, so $\operatorname{codim} Z \geq 2$. Equivalently, whenever $W$ is an irreducible component of $Z$, then the local ring $\mathscr{O}_{X,W}$ has dimension at least two. Since $X$ is assumed to be (S2), every maximal regular sequence in $\mathscr{O}_{X,W}$ has length at least two. -Since $W$ is an irreducible component of the subscheme corresponding to $\mathscr{I}$, it follows that the radical of $\mathscr{I}_W \subset \mathscr{O}_{X,W}$ is precisely the maximal ideal $\mathfrak{p}$. (Algebraically, $\mathfrak{p}$ is a minimal prime over $I$, and corresponds to the generic point of $W$.) Let $g,h \in \mathfrak{p}$ form a regular sequence (which exists since $X$ is (S2)). Replacing $g$ and $h$ by appropriate powers, we may assume that they are both contained in $\mathscr{I}_W$. By definition of regular sequence, $g$ is a nonzerodivisor. Since $h,g$ is a also a regular sequence, $h$ is a nonzerodivisor. Thus, $g$ and $h$, being nonzerodivisors that lie in the ideal of denominators, are in fact denominators of $f$: there exist $a, b \in \mathscr{O}_X,W$ such that -$$\frac{a}{g} = \frac{b}{h} = f$$ -$$ah = bg.$$ -Since $g,h$ is a regular sequence, $h$ is a nonzerodivisor mod $g$. When we mod out by $g$, the equation above becomes $ah \equiv 0$, which would imply $a \equiv 0 \pmod{g}$. In other words, $a \in (g)$. But since $f = a/g$, this would imply that $f \in \mathscr{O}_{X,W}$, a contradiction since $f$ cannot be extended over $W$.<|endoftext|> -TITLE: Does $\pi_1$ have a right adjoint? -QUESTION [15 upvotes]: Eilenberg and Mac Lane showed that given a group $G$ there exists a pointed topological space $X_G$ such that $\pi(X_G,\bullet)\cong G$. It is obviously a way to "invert direction" to the functor $\pi_1\colon \mathbf{Top}^\bullet\to \mathbf{Grp}$ to a functor $\mathcal K\colon \mathbf{Grp}\to \mathbf{Top}^\bullet$ such that $\pi_1(\mathcal K(G),\bullet)\cong G$ (almost by definition). This is equivalent to say that there exists a natural transformation (equivalence, in this case) between $\pi_1\circ\mathcal K$ and $\mathbf{1}_{\mathbf{Grp}}$, which turns out to resemble some sort of counity. -It would be wonderful if I could define an adjunction between the two categories in exam, given by the two functors. Everytime I try to think about some sort of unity to this hypotetical adjunction I poorly fail: considering the vast literature in the field of algebraic topology, I believe in only two possible cases. The first, nothing interesting arises from this adjunction. The second, there is no sort of adjunction. -The key point, quite trivial, to answer is: it is well known that an adjunction is uniquely determined by one among unity and counity, provided the one is universal. But $\boldsymbol\varepsilon\colon \pi_1\circ\mathcal K\to \mathbf{1}_{\mathbf{Grp}}$ is an equivalence: can I conclude that it is universal? - -REPLY [2 votes]: This question was answered a long time ago, but let me mention a couple of relevant papers which invest on Fosco's intuition and partially confirms it. - -Paul C. Kainen, Weak adjoint functors. P.C. Math Z (1971) 122: 1. - -The main point of the paper is the observation that $\mathsf{K}( -, n)$ is a weak adjoint for $\pi_n$. The paper insists many subtleties, like (non-)preservation of limits and (non-)functoriality of the weak adjoints. Very soon Kainen changed research topic, but it is worthy to mention also the paper below. - -Paul C. Kainen, Universal coefficient theorems for generalized homology and stable cohomotopy, Pacific J. Math. 37 (1971) 397– 407.<|endoftext|> -TITLE: Why Lagrangian cobordism? -QUESTION [13 upvotes]: There are a good number of quantum topology papers in which a TQFT-like set-up is constructed as a functor to the category of vector spaces from some category of cobordisms which satisfy some "Lagrangian" condition. For example, Cheptea-Habiro-Massuyeau consider a category whose morphisms are cobordisms $M$ between closed oriented surfaces $F_+$ and $F_-$, where we choose Lagrangian subgroups $A_{\pm}$ of $H_1(F_\pm)$ correspondingly, and where we require that $H_1(M)=m_-(A_-)+m_+H_1(F_+)$ and that $m_+(A_+)\subseteq m_-(A_-)$ in $H_1(M)$ (the $m_\pm$ are inclusion maps). Similar conditions are imposed in many other papers. -I have never understood why such conditions are imposed. One half-thought I have is that it is related to Wall's result that the kernel of the inclusion of $H_1(\partial M)$ in $H_1(M)$ is a Lagrangian subgroup of $H_1(\partial M)$. Another half-idea is that it might be some weak 2-framing condition for the cobordism or something. - -What is the conceptual explanation for this Lagrangian condition and its variants? Does it have anything to do with framing? (or orientation?) Why do "symplectic" and "Lagrangian" have anything to do with TQFT? (is it all just Wall's result in some guise?) - -Every time I see a Lagrangian condition I feel very stupid for not knowing what it's doing there. - -REPLY [6 votes]: A historical reason for the importance of things Lagrangian in a TQFTy setting comes from Floer Homology and a conjecture of Atiyah that ``the two'' Floer theories are the same for 3-manifolds. Floer defined one homology (monopole Floer homology) based on the Chern Simons action functional, with the generators of its chain complex (the analogues of critical points in Morse Theory) being flat connections. He built another homology theory (the more standard one, today) to deal with the Arnol'd conjecture in Hamiltonian mechanics. If you take a Heegegard splitting of the 3-manifold, then, associated to the dividing surface you have the moduli space of flat connections, which is a symplectic manifold. You can then start to set up a Hamiltian Floer theory on this moduli space. Those flat connections over the surface which extend into the three-manifold on one side of the surface define a Lagrangian submanifold of the moduli space. Those flat connections which extend to the other side define another Lagrangian submanifold. The symplectic Floer business on the moduli space, set up properly, gives you a count the intersections of these two Lagrangian submanifolds, and hence of the set flat connections which extend to the whole 3-manifold. Atiyah conjectured these two Floer theories are the same. -I suggest as an entry the Math Reviews article, See MR1283871, by Donaldson on the proof of a special case of this conjecture by Deitmar Salamon and Stamatis Dostoglou. -A philosophical reason is Weinstein's "symplectic creed": everything is Lagrangian. -The classical version of a quantum state is a Lagrangian submanifold. So if one's -TQFT yields a "semi-classical" Hilbert space, say a symplectic vector space, then the states (wavefunctions) of the "Hilbert space" be Lagrangian subspaces<|endoftext|> -TITLE: Descent of closed subschemes over finite fields -QUESTION [10 upvotes]: Let $p$ be a prime number, and $k$ a finite field with $q=p^n$ elements. Let $X$ be a scheme over $k$ and denote by $X'$ its base change to an algebraic closure $\bar k$ of $k$. Denote by $F_X:X\rightarrow X$ the Frobenius endomorphism of $X$ relative to $k$, and by $F'_X$ its base change to $X'$. -Let $f:Z\rightarrow X'$ be a closed subscheme, over $\bar k$, that is invariant under $F'_X$, meaning that there exists a $k'$-morphism $G:Z\rightarrow Z$ so that $f\circ G=F'_X\circ f$. Is it then true that $Z$ admits a descent to $k$? - -REPLY [10 votes]: If $Z$ is reduced, then the answer is yes, and the proof is rather elementary. -First, we can suppose $X$ is affine. Indeed, cover $X$ by affine open subsets $X_i$'s. Then $Z\cap X'_i$ is also stable by $F'_X$ because both $X'_i$ and $Z$ are stable by $F'_X$. If we can show that $Z\cap X'_i$ is defined over $k$, then $Z$ is clearly defined over $k$ too. -So suppose $X=\mathrm{Spec} A$ is affine. The closed subscheme $Z$ is defined over some finite extension $K/k$ of degree $r$. Let $I$ be the defining ideal of $Z$ in $X_K$. We have to show that $I$ is stable by $\mathrm{Gal}(K/k)$. Consider the generator $\sigma: \lambda \mapsto \lambda^{q^{r-1}}$ of $\mathrm{Gal}(K/k)$. Let $b=\sum_i a_i\otimes \alpha_i\in I$ with $a_i\in A$ and $\alpha_i\in K$. Denote by $(F_{X})_K$ the relative Frobenius $F_X$ of $X$ extended to $K$. Then -$$\sigma(b)^q=(\sum_{i} a_i\otimes \alpha_{i}^{q^{r-1}})^q=\sum_{i} a_i^q\otimes \alpha_{i}=(F_{X})_K(b)\in I$$ -Therefore $\sigma(b)\in \sqrt{I}=I$ and we are done.<|endoftext|> -TITLE: Z/48 and Moonshine Beyond the Monster -QUESTION [7 upvotes]: I am interested in pursuing an understanding of K-theory. Primarily, the -$K_3(\mathbb{Z})$ algebraic K-group over ring of integers of an algebraic number field and its relationship to the $\mathbb{Z}/48$ ring of integers modulo 48. -This is (of course), again, from Terry Gannon's "Moonshine Beyond the Monster" -where he talks about many amazing coincidences with the number 24, the -Riemann Zeta Function $\Sigma_{n=1}^\infty (1/n)^{-1} = -1/12$, Apery's constant, where -$\Sigma_{n=1}^\infty (1/n)^2 = \pi^2/6$ (which he states are both synonomous in their relationship to $K_3(\mathbb{Z})\leftrightarrow \mathbb{Z}/48$....) -A little harder to discern is the (possible) relationship of the Bimonster, -$(M \times M) \rtimes \mathbb{Z}/2 \to M \wr 2$, and the Incidence Graph of the M-13 pseudogroup -with 13 points and lines, (the 13 point, 13 line projective plane, where here, -the coincidence would appear to be the number 26, which is the dimension of Bosonic String Theory (2 + 24 dimensions, the quantum harmonic oscillator on a 2-dimbrane, which relates to -1/12 above per John Baez "My Favorite Number is 24")). It's tempting to see the resemblance of 24 relating to the Monster, -and 48 to the BiMonster, but that seems to obvious. Finally, is there any -relevance in bringing in the M12-Mathieu group here, being so close to the -M13-pseudogroup? I apologize ahead of time if this last paragraph is "shooting -the moon" but hopefully my first two paragraphs are well-stated questions. - -REPLY [15 votes]: I am not sure if this helps you since it doesn't contain any speculations on connections with finite simple groups, but the torsion in $K_{2n+1}(\mathcal{O}_F)$ is now fairly well understood for arbitrary $n$ and for rings of integers $\mathcal{O}$ of arbitrary number fields, thanks to Suslin, Voevodsky, Rost, et al. The torsion subgroups are essentially certain Tate modules, not only as groups but in fact as Galois modules. In particular, the way 48 appears in the description of $K_3(\mathbb{Z})$ is that it happens to be twice the biggest $n$, such that the exponent of $(\mathbb{Z}/n\mathbb{Z})^\times$ divides 2, and this observation fits into a bigger picture. -For a sketch of this bigger picture, see e.g. the article by Charles Weibel "Algebraic K-Theory of Rings of Integers in Local and Global Fields" in the Handbook of K-theory. He also gives references for the proofs of the various known results. -Edit: I should also mention that the connection with the Riemann zeta function (or, in the case of number fields, with the Dedekind zeta function of the number field), is - at least conjecturally - not a coincidence either. The special values of Dedekind zeta functions are conjecturally described by sizes of torsion subgroups and volumes of free parts of higher $K$-groups of the rings of integers. This is the Lichtenbaum conjecture. That in turn is also part of a bigger conjectural picture, the Bloch-Kato conjecture on Tamagawa numbers. In short: the connection that you are asking about is a special case of a large conjectural framework of global-local principles, which all take their cue from the analytic class number formula, but are completely out of reach at present.<|endoftext|> -TITLE: Undecidability in Conway's Game of Life -QUESTION [12 upvotes]: I strongly believe that - given the rules of Conway's Game of Life and an initial configuration - it is not decidable by a Turing Machine whether a given pattern will emerge, let alone as a stable pattern, be it static, moving, and/or rotating. - -How can this be proven? - -I guess, this kind of uncomputability would go far beyond the "simple" unpredictability of non-linear systems. - -REPLY [5 votes]: This is proved (or, really, a sketch of the proof is given) in the second volume of the extraordinary [Elwyn R. Berlekamp, John H. Conway, and Richard K. Guy, Winning Ways for your Mathematical Plays. Academic Press, 1982]<|endoftext|> -TITLE: a question on Folner sets -QUESTION [12 upvotes]: Does this property characterizes amenability or there are examples of non-amenable groups satisfying it? -Let $G$ be finitely generated group. -Property: -There exists $C<1$ such that for every $S\subset G$ - finite set, there exists $F \subset G$ - finite, such that -$|sF \Delta F|\leq C\cdot |F|$ for every $s\in S$ -Currently I don't see if it is possible to rebuilt sets $F$ in order to construct Folner sequence for $G$. -The question is related to Amenability of groups - -REPLY [6 votes]: If I am not mistaken the group should be amenable by the following arguments. -As was explained to me by Jesse Peterson, if $\pi:G\rightarrow B(H)$ is a representation of $G$ and there exists unit vector $\xi$ such that for every $g\in G$: $\|\pi(g)\xi-\xi\|\leq C$ then $\pi$ has an invariant vector. -Let $\xi_F=\frac{1}{\sqrt{F}}1_{F}$, then from the property we have $||\lambda(s)\xi_F-\xi_{F}||\leq C$ for every $s\in S$. Then taking $S_i$ as an increasing sequence such that $\cup S_i=G$ and ultralimit of $\lambda$, denote $\lambda_{\omega}:G\rightarrow B(l_2(G)^{\omega})$, we have the existence of a unit vector $\xi$ such that $||\lambda_{\omega}(g)\xi-\xi||\leq C$ for every $g\in G$. Thus $\lambda_{\omega}$ has an invariant vector, which implies that $\lambda$ has a sequence of almost invariant vectors. The last one is equivalent to amenability.<|endoftext|> -TITLE: Space Curves as Determinantal Varieties -QUESTION [11 upvotes]: I read in a paper of Goryunov -(`Functions on space curves', -Journal of The London Mathematical Society, vol. 61 (2000), 807-822; -available on his home page) -that every space curve can be defined as the vanishing -of the N minors of some N by N+1 matrix with entries functions of x, y, z. -How does one prove this? -(I can do it for the rational normal curve, thanks to Harris's book.) - -REPLY [19 votes]: Let $I\subset R = k[x,y,z]$ be the defining ideal of your curve. Then $R/I$ has dimension one and no embedded components, so has projective dimension $2$ by the Auslander-Buchsbaum formula. Therefore $I$ itself has projective dimension $1$, and so can be fit into a short exact sequence: -$$0 \to F \to G \to I \to 0 $$ -with $F,G$ free (a minor point: one needs that projective modules are free here, it is easy if you assume $I \subset (x,y,z) $, since you may as well look at the local ring at the origin). If $N=\text{rank} F$, then $\text{rank} G=N+1$, and the matrix representing the map from $F$ to $G$ is what you want. This is known as the Hilbert-Burch theorem, and details can be found in Chapter 20 of Eisenbud's book "Commutative Algebra with a view..."<|endoftext|> -TITLE: Formally real Jordan algebras -QUESTION [28 upvotes]: In 1934, Jordan, von Neumann and Wigner gave a nice classification of finite-dimensional simple Jordan algebras that are 'formally real', meaning that a sum of squares is zero only if each term in the sum is zero. In 1983 Zelmanov generalized this to all simple Jordan algebras, but unlike the original result, Zelmanov's 'classification' is not a neat list. What about formally real Jordan algebras, not necessarily finite-dimensional? What does the classification of these look like? - -REPLY [13 votes]: The formally real Jordan algebras include the class of JB-algebras, a class of normed Jordan algebra which is the Jordan algebra equivalent of C*-algebras. From this you might imagine that obtaining a complete classification is a rather non-trivial task. In this class you also find JW-algebras, which are JB-algebras with a predual, thus corresponding to von Neumann algebras. Some of the concepts from C*- and von Neumann algebra theory carry over to the Jordan algebra setting, but this margin is too narrow to summarize what is known. With apologies for tooting my own horn here, in 1984 I coauthored a book “Jordan Operator Algebras” with Erling Størmer which pretty much summarized the state of the art at the time. (Since then I have left that field, so I don't know if a lot has happened since.)<|endoftext|> -TITLE: maximal coordinate on a sphere -QUESTION [7 upvotes]: What is the easiest (preferably without calculations) way to see that the mean value of $\max(x_1,x_2,\dots,x_n)$ on the sphere $\mathbb{S}^{d-1}= \{ (x_1,\dots,x_n):\ x_1^2+\dots+x_n^2=1 \}$ behaves like $\sqrt{\log(n)/n}$, or at least that is is much more then $1/\sqrt{n}$ for large $n$? The same (and less or more a priori equivalent) question concerns the standard Gaussian measure and expectation of $\infty$-norm w.r.t. it. -The proofs I know (for example, the one which V. Milman attributes to Figiel) use too many integrals. -And by the way, how to put {,} in math here? \ { does not work for me - -REPLY [15 votes]: For some positive $c$ bounded away from zero, the probability that a standard gaussian variable is larger than $c\sqrt{\log n}$ is $1/n$. It follows that the probability that at least one variable out of $n$ independent standard gaussians is larger than $c\sqrt{\log n}$ is $1-(1-1/n)^n$ which tends to $1-1/e$. From that one gets that the expectation of the maximum of $n$ standard gaussians is at least (basically) $(1-1/e)c\sqrt{\log n}$. -As you said the question about variables on the sphere is equivalent to that.<|endoftext|> -TITLE: Is this sequence of polynomials well-known? -QUESTION [9 upvotes]: While working on a problem in p-adic Hodge theory, and needing to write down a solution to a certain equation involving p-adic power series, I stumbled across a certain sequence of polynomials. Define $h_j(X)$ for $j \ge 0$ by $h_0(X) = 1$ and -$$ h_{j}(X) = \frac{X + 1}{j}\left(- X \frac{\mathrm{d}}{\mathrm{d}\ X} + j\right)h_{j-1}(X)$$ -for $j \ge 1$. -I was interested in these because $h_j(X)$ is the unique polynomial of degree $j$ such that -$$\left(\frac{t}{e^t - 1}\right)^{j+1} \cdot h_j(e^t - 1) = 1 + O(t^{j+1}),$$ -and in fact it follows from the recurrence that -$$\left(\frac{t}{e^t - 1}\right)^{j+1} \cdot h_j(e^t - 1) = 1 + (-1)^j \sum_{n \ge j+1} \binom{n-1}{j} \frac{B_n t^n}{n!}$$ -where $B_n$ are the usual Bernoulli numbers. -Now, I can't believe that these polynomials $h_j$ aren't some terribly classical well-studied thing, but they don't match any of the standard sequences of polynomials I could find on the web. Does anyone recognise these? - -REPLY [16 votes]: The first several are: -$$0! \cdot h_0(x) = 1$$ -$$1! \cdot h_1(x) = x+1$$ -$$2! \cdot h_2(x) = x^2+3 x+2$$ -$$3! \cdot h_3(x) = x^3+7 x^2+12 x+6$$ -$$4! \cdot h_4(x) = x^4+15 x^3+50 x^2+60 x+24$$ -Feeding the sequence $2,3,1,6,12,7,1,24,60$ into the OEIS gives the following page, which contains generating functions, relations, and citations to occurrences of this sequence of polynomials in the literature.<|endoftext|> -TITLE: Diagonalization of Infinite Hermitian matrices -QUESTION [6 upvotes]: We know that $n\times n$ square Hermitian matrices can be diagonalized and have real eigenvalues. -Suppose I have a countable sized Hermitian matrix $A=(a_{ij})$ where the indices $i$ and $j$ run over the natural numbers and the complex entries satisfy $\bar{a_{ij}}=a_{ji}$. -Can such a matrix $A$ be diagonalized? Otherwise, what conditions can I impose that will make it diagonalizable? I'm interested in the case where only finitely many of the eigenvalues are nonzero. - -REPLY [5 votes]: Such a matrix defines an unbounded operator $\mathcal A$ over the Hilbert space $\ell^2(\mathbb N)$, with domain $D(\mathcal A)$, formed of vectors $x$ such that $\mathcal Ax\in H$. It may happen of course that $\mathcal A$ be a bounded operator. The property $\overline{a_{ij}}=a_{ji}$ tells you that $\langle\mathcal Ax,y\rangle=\langle x,\mathcal Ay\rangle$ whenever $x,y\in D(\mathcal A)$. -For an unbounded operator $\mathcal A$, we define the domain of the adjoint operator $\mathcal A^*$ to be the set of $y\in H$ such that the linear form $x\mapsto\langle\mathcal Ax,y\rangle$ is bounded. - -Definition. The operator $\mathcal A$ is self-adjoint if $\langle\mathcal Ax,y\rangle=\langle x,\mathcal Ay\rangle$ whenever $x,y\in D(\mathcal A)$, and in addition $D(\mathcal A^*)=D(\mathcal A)$. - -It turns out that the symmetry property does not imply the equality of the domains. -The theory of unitary diagonalisation of Hermitian matrices basically extends naturally to self-adjoint operators. See the book by Reed and Simon. - -REPLY [2 votes]: As a comment on Denis's answer... You have to be a little bit careful about what you mean by "diagonalisation" though. Consider $[0,1]$ with Lebesgue measure, and define $T:L^2[0,1]\rightarrow L^2[0,1]$ by $$ T(f)(x) = x f(x). $$ -So $T$ is ``multiplication by $x$''. This is easily seen to be self-adjoint. So if we pick an orthonormal basis for $L^2[0,1]$, we get a self-adjoint matrix $A$ of the type you describe (and it induces a bounded map on $\ell^2$ of course). -But notice that $T$ has no eigenvectors or eigenvalues, and thus the same is true of $A$ (assuming we ask for our eigenvectors to be in $\ell^2$, but otherwise I don't see how we can talk about ``eigenvalue''). -Moral: you have to give up "diagonalisation" in favour of the more general idea of being "unitarily equivalent to a (real-valued) multiplication operator".<|endoftext|> -TITLE: Moving to academia from industry -QUESTION [24 upvotes]: I hope that the following question is suitable for MO. There have been several others in a similar vein, including this one about moving from maths into industry, this one about careers advice for mathematicians and this about returning to graduate school after a couple of years in industry (among others). My situation is a bit different from those discussed in the referenced questions, as I am interested in moving from an already well-established career into academia. -I'll give a bit of background about myself. Naturally, this will be rather personal to my situation but, maybe, the general question could also be of interest to others. -I already have a PhD in maths from a top university and, now, have quite a successful and well paid career in industry. The reason for moving in the first place was complicated and influenced by several factors. Money was certainly one of them, as I was virtually skint at the time, but the main reason was that I was just not happy with my lifestyle at the time and was wanting a complete change. This was over ten years ago. -However, after family, my main interest in life is maths. While it was fun initially to work in a highly technical field in industry, over the past few years I have found myself getting bored. Throughout my time working, I have been keeping up my interest in maths, mainly by reading books and papers, but also by writing a small number of papers independently. I had a couple published, and a couple more accepted but, between family life and work, have not had time to pursue this further. More recently, I have started writing a blog, and have been contributing to this site. I would like to say that I have been attending lectures or seminars at one of the top universities around here, as it is something that I have always been intending to do, but have had very little time to actually organize doing. -My question then, is what are the options for moving back into academia after spending a significant time in industry? Also, as time is moving on, I think that moving back into academia at the point where I left off - having just completed a PhD - would not be ideal. I certainly feel ready to contribute to the mathematical community at the highest level. -One answer, I suppose, is simply to speak to the people I worked with during my PhD. That is not so simple though. The people I worked most closely with have since moved on (either into industry themselves, or retired) and I am not so sure about suddenly contacting people who I never worked very closely with after such a large gap. That is certainly something I am considering, but would first like to get some opinions from the wider mathematical community. I am based in the UK, although any opinions based on experience elsewhere could also be valuable. -Finally. I have been quite active on this website under my real name. However, I do not think it is wise to openly talk on the internet about possibly leaving my current career, so am posting this question anonymously. - -REPLY [4 votes]: Perhaps an option is to first become a (part-time?) tutor at an Oxford or Cambridge University college (provided these are not too far away); some of the colleges may indeed appreciate your industry experience. -There might be similar positions at other English universities, of course.<|endoftext|> -TITLE: The category of abelian group objects -QUESTION [16 upvotes]: Let $C$ be a category, say with finite products. What can be said about the category $\operatorname{Ab}(C)$ of abelian group objects of $C$? Is it always an abelian category? If not, what assumptions on $C$ have to be made? What happens when $C$ is the category of smooth proper geometrically integral schemes over some locally noetherian scheme $S$? -For example if $C=\mathrm{Set}$, we get of course the abelian category of abelian groups. If $C=\mathrm{Ring}/R$ for some ring $R$, then we get the abelian category $\operatorname{Mod}(R)$ (cf. nLab). In general, I have already trouble to show that $\operatorname{Hom}(A,B) \times \operatorname{Hom}(B,C) \to \operatorname{Hom}(A,C)$ is linear in the left coordinate if $A$, $B$, $C$ are abelian group objects. - -REPLY [2 votes]: I'm not sure why this hasn't been mentioned yet, but the category of abelian group objects in smooth proper geometrically integral schemes is not abelian, already if $S = \operatorname{Spec} k$. Indeed, if $A$ is any abelian variety, then the map $[n] \colon A \to A$ for $n > 1$ is a monomorphism and an epimorphism: - -As the OP explained at What are the epimorphisms in the category of schemes?, it is an epimorphism of schemes since it is surjective on points and injective on sections (being a dominant morphism of integral schemes). Thus it is also an epimorphism of group schemes. -If $f \colon G \to A$ is a map such that $G \to A \stackrel{[n]}\to A$ is trivial, then $f(G)$ lands in $A[n]$ scheme-theoretically. But if $G$ is geometrically integral, this implies $f(G)$ lands in the trivial subgroup $\boldsymbol 1$, again scheme-theoretically. - -Thus, $[n] \colon A \to A$ is both a monomorphism and an epimorphism, but it clearly does not have an inverse (even as schemes) when $n > 1$. -Dropping "geometrically integral" doesn't help: if $\operatorname{char} k = p > 0$ and $A$ is supersingular, the same example with $n = p$ shows that looking at smooth abelian group schemes is not enough. Dropping both does work if you further impose finite type hypotheses: the category of abelian group schemes of finite type over a field $k$ is abelian; see Is the category of commutative group schemes abelian? and the references therein. Then the proper ones form a Serre subcategory, hence an abelian category. -It would be interesting to see what versions are true over a general base. It seems hard to me to make quotients if there are no flatness assumptions, but maybe kernels or cokernels of flat group schemes don't want to be flat. I should probably try to read some SGA3….<|endoftext|> -TITLE: Gelfand-Tsetlin bases for Lie groups over finite fields -QUESTION [7 upvotes]: There is a method of constructing representations of classical Lie algebras via Gelfand-Tsetlin bases. It has also been applied to Symmetric groups by Vershik and Okounkov. Does anybody know of any application of the method to complex representations of $GL_n(\mathbb F_q)$? Or, at least, any results in this directions, like what is the centralizer of $GL_{n-1}$ in $\mathbb C[GL_n]$? - -REPLY [3 votes]: My earlier comment was not at all well-focused. After more thought, I'm inclined to be pessimistic about using a Gelfand-Tsetlin approach here (even if it has some success for symmetric groups). Though of course it would be interesting to be proven wrong. -As Matt Davis reminds me, my offhand reference to Schur-Weyl duality is not helpful here since the work of Benson, Doty, and others deals mainly with the representations of various groups over fields of prime characteristic. (See especially Doty's papers on arXiv.) Irreducible representations of finite general linear groups over $\mathbb{C}$ are very difficult to construct directly and have very little in common with the finite dimensional representations of general linear groups or their Lie algebras in characteristic 0. Instead, the theory imitates more closely the infinite dimensional Harish-Chandra approach to Lie group representations in which parabolic induction is exploited together with a study of "discrete series". -J.A. Green's 1955 TAMS paper followed somewhat this pattern in developing combinatorially the character theory of finite general linear groups. But there is little insight here into constructing the elusive discrete series characters; instead orthogonality relations and the like are exploited. The best approach to an actual construction of discrete series representations was given in Lusztig's 1974 Annals of Mathematics Studies No. 81. Soon after that, Deligne and Lusztig pioneered a more sophisticated method for constructing generalized characters of arbitrary finite groups of Lie type. This has become the dominant influence in the subject, since Lusztig's earlier techniques don't go far enough beyond the finite general linear case.<|endoftext|> -TITLE: Almost complex structures in Floer theory -QUESTION [6 upvotes]: When defining the Floer cohomology $HF(L_0,L_1)$ of 2 Lagrangians in a symplectic manifold $(M,\omega)$, one first has to choose some extra data such a 1-parameter family of almost complex structures $(J_t)$. Usually one requires that $J_t$ be compatible with $\omega$, ie that $g(u,v)=\omega(u,J_tv)$ defines a Riemannian metric. -However there is also the related notion of a tame $J$, one such that $\omega(u,Ju)>0$ for all nonzero $u$. My question is: -What goes wrong if we try to use tame but not necessarily compatible $J_t$ to define $HF(L_0,L_1)$? - -REPLY [9 votes]: As far as I know, nothing goes wrong if one uses tame instead of compatible almost complex structures (if anyone knows better please correct me). Either one of these conditions implies that the area of a holomorphic curve is bounded in terms of the integral of the symplectic form on it. This is what one needs to get Gromov compactness and to set up the Novikov ring. I don't think tameness or compatibility is needed elsewhere in the theory.<|endoftext|> -TITLE: Aleph 0 as a large cardinal -QUESTION [19 upvotes]: The first infinite cardinal, $\aleph_0$, has many large cardinal properties (or would have many large cardinal properties if not deliberately excluded). For example, if you do not impose uncountability as part of the definition, then $\aleph_0$ would be the first inaccessible cardinal, the first weakly compact cardinal, the first measureable cardinal, and the first strongly compact cardinal. This is not universally true ($\aleph_0$ is not a Mahlo cardinal), so I am wondering how widespread of a phenomenon is this. Which large cardinal properties are satisfied by $\aleph_0$, and which are not? -There is a philosophical position I have seen argued, that the set-theoretic universe should be uniform, in that if something happens at $\aleph_0$, then it should happen again. I have seen it specifically used to argue for the existence of an inaccessible cardinal, for example. The same argument can be made to work for weakly compact, measurable, and strongly compact cardinals. Are these the only large cardinal notions where it can be made to work? (Trivially, the same argument shows that there's a second inaccessible, a second measurable, etc., but when does the argument lead to more substantial jump?) -EDIT: Amit Kumar Gupta has given a terrific summary of what holds for individual large cardinals. Taking the philosophical argument seriously, this means that there's a kind of break in the large cardinal hierarchy. If you believe this argument for large cardinals, then it will lead you to believe in stuff like Ramsey cardinals, ineffable cardinals, etc. (since measurable cardinals have all those properties), but this argument seems to peter out after a countable number of strongly compact cardinals. This doesn't seem to be of interest in current set-theoretical research, but I still find it pretty interesting. - -REPLY [3 votes]: There is a sense in which rank-into-rank axioms could be considered generilzations of $\omega$. Woodin discussed this in his paper Suitable Extender Models $I$, given the following conjecture, which I denote $(^*)$. - -$(^*)$ Assume $V=\text{Ultimate}-L$. Then there is some $j: V_{\lambda+1}\prec V_{\lambda+1}$ if and only if $L(P(\lambda))\nvDash AC$. - -In the same vein is the following conjecture, described by Vincenzo Dimonte in $I0$ and rank-into-rank axioms. - -Question 11.2: Is it true that $\text{Ultimate}-L\vDash I0(\lambda)$ if and only if $\text{Ultimate}-L\vDash L(V_{\lambda+1})\nvDash AC$? - -If either one of these were true, then it would be a shocking affirmation of the existence of $I1$ and $I0$ cardinals respectively (Assuming the proof does not show that there are none). The reason for this is that if $V=\text{Ultimate}-L$, then there is a proper class of Woodin cardinals and so in particular $L(\mathbb R)=L(P(\omega))=L(V_{\omega+1})\vDash AD$ and so $L(P(\omega))=L(V_{\omega+1})\nvDash AC$. -This corresponds with the sense in which it is not the critical point $\kappa$ of $j: V_{\lambda+1}\prec V_{\lambda+1}$ which is important, but the $\lambda$ itself. Note that $\lambda$ is a strong limit of cofinality $\omega$, and also $\omega$ is a strong limit of cofinality $\omega$, and finally a theorem of Shelah is that if $\lambda$ is a strong limit of uncountable cofinality, then $L(P(\lambda))\vDash AC$.<|endoftext|> -TITLE: Algebraic analogue of the Moebius bundle over the circle -QUESTION [15 upvotes]: Let $R$ be the ring $R[X,Y]/(X^2+Y^2−1)$. The space of $\mathbb{R}$-rational points of the affine scheme associated to $R$ is the topological circle $S^1$. -An algebraic vector bundle over $R$ is an $R$-algebra $A$ with certain properties or equivalently a finitely generated projective $R$-module $A$. -My first question is: What is the explicit projective $R$-module $A$ corresponding to the topological Moebius bundle over $S^1$? (By 'corresponding' I mean in particular that the $\mathbb{R}$-rational points should induce the topological Moebius bundle and that it has rank one.) -My second question is motivated by the fact that topologically there are only two non-isomorphic rank one bundles over $S^1$: the trivial bundle and the Moebius bundle. What is the analogous algebraic situation? Are there more than two non-isomorphic and rank one projective modules over $R$? -Thank you! -(I apologize that I first asked this question on math.stackexchange, I deleted it there.) - -REPLY [16 votes]: The internal note "Le ruban de Moebius comme représentation d'un idéal non principal" (Moebius band, as a non-principal ideal), by Daniel Ferrand, contains more or less the material you want. Daniel is retired and doesn't have a website, but, thank you Google!, -the note is downloadable from http://www.math.unibas.ch/~giordano/Moebius.pdf -The non-principal ideal you want is the ideal $(1+x,y)$ in the ring $R[x,y]/(x^2+y^2-1)$. - -REPLY [7 votes]: If you want to describe the projective module as a direct summand of a free module, you can use the 2 by 2 projection matrix -$\frac{1+X}{2}$ $\frac{Y}{2}$ -$\frac{Y}{2}$ $\frac{1-X}{2}$ -Since this ring is a Dedekind domain, the projective modules of rank one correspond to the ideal classes. (And see Pete Clark's answer for what that works out to be. I think it's a fun exercise.)<|endoftext|> -TITLE: Examples of nice families of irreducible polynomials over Z -QUESTION [6 upvotes]: Hi, -i search for irreducible polynomials over Z which have variable coefficients you can "choose". -Since I found nearly nothing in books or the internet i hope you can help me. -Here 3 examples: -Let g a polynomial over Z with degree smaller then n/2 ,then: -$g* (\prod_{i=1}^n (x-a_i)) -1 $ is irreducible if the a_i are all distinct. -Here you can choose n the coefficients of g and the a_n so its a nice example. -Another one,I found, is from Furtwängler : $x^4 (\prod_{i=1}^{n-4} (x-b_i)) -(-1)^n *(2x+4) $ where the b_i are strictly increasing.Can you generalize this example ?I think it should work also for some integers other than 2 and 4. -Here is another nice example : -Polynomial with the primes as coefficients irreducible? -I try to find examples where its easy to control zeros modulo p of the some irreducible polynomials and its derivation for another problem. - -REPLY [2 votes]: If $m$ and $n$ are odd then $x^5+mx^2+n$ is irreducible over the rationals (as one finds out by reducing it mod 2). -Michael Filaseta (and various co-authors) has a string of results on families of irreducible polynomials. E.g., Filaseta, Finch and Leidy, T N Shorey's influence in the theory of irreducible polynomials, has results on irreducibility of Laguerre polynomials. Filaseta, Kumchev and Pasechnik, On the irreducibility of a truncated binomial expansion, does what the title says. Filaseta, Luca, Stanica and Underwood, Two diophantine approaches to the irreducibility of certain trinomials, proves that $x^{2p}+bx^p+c$ is irreducible whenever $p$ is an odd prime and $\gcd(p,q^2-1)=1$ for all prime $q$ dividing $181b$. There's more where that came from. If you have access to MathSciNet you might try typing in Filaseta and irreducible.<|endoftext|> -TITLE: Does the category of topological symmetric spectra satisfy the monoid axiom ? -QUESTION [8 upvotes]: In the paper "Symmetric spectra"by Hovey, Smith and Shipley, they say that they don't know if the monoid axiom holds for topological symmetric spectra. This paper was written in 1998 so I am wondering what is the situation on this question today. - -REPLY [20 votes]: The monoid axiom for symmetric and orthogonal spectra of spaces is -Proposition 12.5 of Mandell, May, Schwede, and Shipley's paper -``Model categories of diagram spectra''.<|endoftext|> -TITLE: "Closed-form" functions with half-exponential growth -QUESTION [44 upvotes]: Let's call a function f:N→N half-exponential if there exist constants 1 1/(e \ln \beta)$, then the demi-exponential function has no real-valued fixed points, but rather develops jump-type singularities. In particular, the seemingly reasonable parameters $\beta=e$ and $\gamma=1$ have no smooth demi-exponential function associated to them (at least, that's the numerical evidence). -Perhaps this is one reason that demi-exponential functions have a reputation for being difficult to construct ... it is indeed very difficult to construct smooth functions for ranges of parameters such that no function has the desired smoothness! -It might be feasible (AFAICT) to write an article On demi-exponential functions associated to the Lambert W Function, and to include these functions in standard numerical packages (SciPy, MATLAB, Mathematica, etc.). -Some tough challenges would have to be met, however. Especially, there is at present no known integral representation of the demi-exponential functions (known to me, anyway), and yet such a representation would be very useful (perhaps even essential) in rigorously proving the analytical structures that the numerical Pade approximants show us so clearly. -Mathematica script here (PDF). - -Here's what these functions look like: -halfexpPicture http://faculty.washington.edu/sidles/Litotica_reading/halfexp.png - -Final note: Inspired by the recent burst of interest in these demi-exponential functions, and mainly for my own recreational enjoyment, I have verified (numerically) that demi-exponential functions $d$ having (1) fixed point $z_f = d(z_f) = 1$, and (2) any desired asymptotic order, gain, and base can readily be constructed. -I'd be happy to post details of this construction ... but it's not clear that anyone has any practical interest in computing numerical values of demi-exponential functions. -What folks mainly wanted to know was: (1) Do smooth demi-exponential functions exist? (answer: yes), (2) Can demi-exponential functions be computed to any desired accuracy? (answer: yes), and (3) Do demi-exponential functions have a tractable closed form, either exact or asymptotic? (answer: no such closed-form expressions are known).<|endoftext|> -TITLE: Word maps on compact Lie groups -QUESTION [21 upvotes]: Let $w=w(a,b)$ be a non-trivial word in the free group $F_2 = \langle a,b \rangle$ and $w_G \colon G \times G \to G$ be the induced word map for some compact Lie group $G$. -Murray Gerstenhaber and Oskar Rothaus showed in 1962 (see M. Gerstenhaber and O.S. Rothaus, The solution of sets of equations in groups, Proc. Nat. Acad. Sci. U.S.A. 48 (1962), 1531–1533.) that $w_G$ is surjective in a very strong sense for $G=U(n)$ whenever the exponent-sum of $a$ or $b$ in $w$ is non-zero. Using degree-theory and the calculations of the homology of compact Lie groups due to Heinz Hopf, they showed that if the exponent sum of $a$ is non-zero, then $u \mapsto w_{U(n)}(u,v_0)$ is surjective for any fixed $v_0 \in U(n)$. -Earlier, in 1949, Morikuni Gotô (J. Math. Soc. Japan, 1949 vol. 1 pp. 270-272) already showed that the commutator $w=[a,b]$ induces a surjective map $w \colon G \times G \to G$ for any simple compact Lie group $G$. His proof proceeded as follows: Since the commutators are a conjugation invariant set, one can assume that $u$ is in a fixed maximal torus. We can now take a suitable $a$ in that torus and $b$ an element the Weyl group $W_G$ of this torus, such that $b$ does not have $+1$ as an eigenvalue (when acting on the universal cover of the maximal torus). This implies that the action of $1-b$ on the universal covering is an isomorphism. Hence, every element in the maximal torus can be obtained as $a \cdot ba^{-1}b^{-1}$ for some $a$. This strategy can be used for a few other words but does not lead any further. - -Question: Are there any other techniques to show that $w_{G}$ is surjective for some $w$ and $G$? - -The easiest word for which I cannot answer this is $w= [[a,b],[a^2,b^2]]$. - -Question: Is the word-map $w_{PSU(n)}\colon PSU(n) \times PSU(n) \to PSU(n)$ surjective for the word $$w = [[a,b],[a^2,b^2]]?$$ - -REPLY [11 votes]: In this paper of Borel it is shown that any non-trivial word is a dominant map from $G \times G$ to $G$ whenever $G$ is a compact connected semisimple Lie group. So the answer to your second question is "yes". I don't recall offhand exactly what techniques are used, but I vaguely recall that one starts with the $SL_2$ case (which contains free subgroups that one can play with) and builds up from there. -EDIT: Dominance would imply surjectivity (or at least that the image is Zariski dense) in an algebraically closed field, but I didn't realise that the question is over the reals, and so Borel's result does not fully resolve the question.<|endoftext|> -TITLE: Non-vanishing cohomology of line bundles on projective varieties in prime characteristic? -QUESTION [13 upvotes]: This is a somewhat naive question about the expected non-vanishing behavior of sheaf cohomology groups $H^i(X, \mathcal{L})$, where $X$ is a smooth projective variety of dimension $d$ over an algebraically closed field of characteristic $p>0$ and $\mathcal{L}$ is a line bundle on $X$. Here one starts with a few general principles: cohomology is finite-dimensional and vanishes in degrees outside the interval $0 \leq i \leq d$, while Serre duality holds. -Though my knowledge of algebraic geometry is sketchy at best, I am motivated by the case when $X$ is a flag variety for a connected reductive algebraic group $G$ and the line bundle comes from a character of some maximal torus of $G$ (lifted to a Borel subgroup). Unlike the nice classical theory in characteristic 0 where cohomology can be nonzero in at most one degree and then affords an irreducible representation of $G$, the situation in prime characteristic becomes intricate. Results here, due mainly to Kempf and Andersen, provide a few important general theorems as well as some low rank examples. But the associated representation theory is poorly understood and seems to be determined somehow in terms of Kazhdan-Lusztig theory for the affine Weyl group of Langlands dual type relative to $p$ in place of the usual Weyl group. -Two features occur frequently in low ranks and may hold in general, so I wonder if they are seen elsewhere in the world of smooth projective varieties. Here it's just a question of nonvanishing cohomology rather than more subtle module structure: -(A) Connectedness of the interval of degrees between 0 and $d$ in which non-vanishing occurs. -(B) Occurrence of non-vanishing in no more than $\frac{d}{2} +1$ degrees. - -Is either (A) or (B) usual/unusual in other cases of non-vanishing cohomology in prime characteristic? - (And are there relevant examples, references?) - -ADDED: The responses show that in some respects my question is too broad and therefore naive, since projective varieties come in so many shapes and sizes. Flag varieties are homogeneous spaces for a nice algebraic group (in any characteristic), though the cohomology of line bundles mysteriously gets far more complicated when the characteristic is prime. It's hard to trace the breakdown of most of the classical Borel-Weil-Bott theory, which motivates my question. At the same time, it's hard to see the role of geometry here apart from the related group theory. -By the way, (A) above is more speculative than (B) due to possible over-reliance on low rank examples. But neither property of non-vanishing has an obvious source in the geometry of the situation. I'm still interested in seeing similar examples elsewhere involving constraints on degrees where non-vanishing can occur. - -REPLY [2 votes]: I am not sure to which extent your questions are really related to positive -characteristic. -The obvious difference between positive characteristic and characteristic zero -related to the question is of course failure of Kodaira vanishing in positive -characteristic. However, that does not come up for flag varieties as Kodaira -vanishing is true for them, either by the Deligne-Illusie result mentioned by -Francesco or by Frobenius splitting (though interestingly enough one get -examples of Kodaira non-vanishing by looking at "inseparable flag spaces"). The -examples that the OP mentions are non-ample line bundles. -If one drops (or never introduces it in the first place) the ampleness condition -that almost anything can happen but that is true in all characteristics. -To answer questions A and B one can take the product $X:=X_1\times -X_2\times\cdots\times X_k$, with $n_i:=\dim X_i$, of smooth hypersurfaces in -projective spaces of high enough degree so that $H^{n_i}(X_i,\mathcal -O_{X_i})\ne0$, we always have $H^j(X_i,\mathcal O_{X_i})=0$ for $0 -TITLE: Are the Platonic solids shadows of 4-polytopes? -QUESTION [15 upvotes]: Say that a 3D shadow of a 4-polytope is a parallel projection to 3-space, not necessarily orthogonal to that 3-space (that would make it an orthogonal projection). -I am wondering if each of the five regular polyhedra in 3D are shadows of regular 4-polytopes. -The 4-simplex can project to a regular tetrahedron, the tesseract to a cube, the 16-cell can parallel project to a cube or a regular octahedron. But I do not know if the 120-cell can project to a regular dodecahedron, or if the 600-cell can project to a regular icosahedron. (I think not?) Perhaps under perspective projection they can? Failing that, perhaps irregular convex 4-polytopes have the regular dodecahedron and icosahedron as shadows? -I am not as familiar with the regular 4-polytopes as I should be; otherwise the answers would be evident. I am sure those more knowledgeable can answer my question easily. Thanks! -Addendum. The confusion in my question (sorry!) caused a confusing welter of comments (some now deleted), but I think matters are clearer now. Three of the Platonic solids can be achieved as parallel-projection shadows -(light at $\infty$): tetrahedron, cube, octahedron. -As Theo explains in his answer, the dodecahedron and icosahedron cannot be so achieved. -However, the dodecahedron can be achieved as a perspective-projection shadow (light at a finite point). -The analogy with 3D is as follows. If a light is placed close to a face of the dodecahedron, a pentagon shadow results. Similarly, if a light is placed close to dodecahedral facet of the 120-cell, a dodecahedral shadow results (as Theo says). -What remains unclear to me is if the icosahedron can be achieved as a shadow. If one could place the light directly on a vertex of the 600-cell, then the shadow should be its vertex figure, which is an icosahedron. But if the light is exterior to the polytope, the shadow is more complicated. - -REPLY [3 votes]: Just for fun, here's a non-answer all about the case of 3 projected to 2. i never thought all of this through before. For each of the five Platonic solids you can ask what do you see if you look at it from a face direction or from a corner direction. The answer usually depends on how far away it is. (I'm thinking of an eyeball instead of light source, so "what you see from there" rather than "what the shadow looks like".) Taking the obvious ones first, we have: -4-hedron face or corner view: Always a regular 3-gon -6-hedron face view or 8-hedron corner view: Always a regular 4-gon -8-hedron face view: From infinitely far, a regular 6-gon. From large finite distance, a 6-gon with all sides equal but only regular 3-gon symmetry. At a certain distance (namely where you suddenly can see only 1 face rather than 1+3) a regular 3-gon, and so it remains until it hits you like a cream pie. -6-hedron corner view: Ditto, except that the 6-gon doesn't turn into a 3-gon until the final moment when it pokes you in the eye. -12-hedron face view: Exactly like the 8-hedron face view, except change 6 to 10 and 3 to 5. -20-hedron corner view: Just like 6-hedron corner view ($\times \frac{5}{3}$) except that the 10-gon does not wait until the final moment to become a 5-gon. -20-hedron face view: Always has at least the symmetry of a regular 3-gon. Initially a regular 6-gon, then an irregular 6-gon with equal sides, then suddenly the angles reverse their direction of change. Passes through a regular 6-gon again, then back to an irregular 6-gon and eventually turns into a regular 3-sided cream pie. -12-hedron corner view: Always has at least the symmetry of a regular 3-gon. Initially a 12-gon with the symmetry of a regular 6-gon, equal angles but not equal sides. Then an irregular 12-gon with merely 3-gon symmetry. For one moment a 6-gon (not regular, I'm sure) then an irregular 9-gon until it pokes you in the eye. -I don't know what the corresponding story is for the thing whose faces are 600 4-hedra, or the thing whose faces are 120 12-hedra. I started thinking about the thing whose faces are 24 8-hedra.<|endoftext|> -TITLE: What are the odds two permutations in S_n do NOT generate the whole group? -QUESTION [5 upvotes]: What are the odds two uniformly chosen elements of S_n span the whole group (or just the alternating group)? Mathematica experements suggest those odds approach 1 - this might have been proven a long time ago. How likely is it to get the alternating group or something much smaller? -Also, how can you efficiently find the size of the subgroup $\langle a,b\rangle$ in S_n ? My crude tests consists of randomly multiplying the two permutations and seeing how many different elements you get. Maybe there's a more efficient way to generate all the elements spanned by two permutation. - -You can generate the whole permutation group using a swap (12) and a shift (12...n). I wonder if all two element generating sets are conjugate to this. - -REPLY [7 votes]: Concerning the second part of the OP's question: - -Also, how can you efficiently find the size of the subgroup $\langle a,b \rangle$ in - $S_n$ ? My crude tests consists of randomly multiplying the two permutations and seeing - how many different elements you get. Maybe there's a more efficient way to generate all - the elements spanned by two permutation. - -There is a "classical" polynomial-time method known as the "Schreier-Sims" algorithm for finding the order of the subgroup of $S_n$ generated by a given set of permutations - just google it for further details. It has a number of improvements for dealing with groups of very large degree. Refinements of this were used by Sims to prove the existence of some of the large sporadic finite simple groups, including the Lyons group and the Baby Monster. -There are also very fast "one-sided Monte-Carlo" probabilistic algorithms for verifying that the group generated by a given set of permutations is $A_n$ or $S_n$. If they do, then "yes" will be returned rapidly with high probability. If they do not, then the algorithm does not terminate, so normally you would give up and use Schreier-Sims instead. This method is based on old results due to Jordan and others, which say that if $G \le S_n$ is transitive and contains an element of prime order $p$ with $n/2 < p < n-2$, then $G = A_n$ or $S_n$. -For further details, see Akos Seress' book "Permutation Group Algorithms" Cambridge University Press, 2003, or my own book, with B. Eick and E.A. O'Brien, "Handbook of Computational Group Theory", CRC Press, 2005.<|endoftext|> -TITLE: Big O notation and the maximal set of comparable functions -QUESTION [13 upvotes]: One can easily find a set of functions that are comparable with respect to the big O notation that is, -$$f \leq g \Leftrightarrow \exists c \exists x_0 \forall x\geq x_0: |f(x)| \leq c|g(x)|,$$ -for $f,g \in \mathbb{R}\rightarrow \mathbb{R}$ and continuous. -For example, a set containing all products of polynomials, logarithms, exponential functions and factorial is good. However, it is not a maximal one (e.g. $\ln \ln x$ is lesser than all of its elements). -According to Kuratowski-Zorn lemma, there is a maximal set of comparable functions. My questions are: - -Is there one canonical maximal set of comparable functions with respect to big O notation? -Is there an explicit construction of any such set? - -Edit: - -There was $\lim_{x\rightarrow\infty} |f(x)| \leq C|g(x)|$ which was in my intention an informal notation for the above ($\lim$ about the whole inequality, not only the left hand site). -However, I do not cling to the definition (especially if some tinkering is going to give more interesting results). - -REPLY [7 votes]: Let me assume that you mean the order on functions $f$ and -$g$ by which $f\leq g$ if and only if $\exists C\exists -x_0\forall x\geq x_0$ $f(x)\leq C\cdot g(x)$. In other -words, $f(x)$ is eventually less than $C\cdot g(x)$. This -order is a linear smoothing out of the usual -eventually-less-than order on functions, which has been -considered in many other questions here on MO. This -relation is more properly called a pre-order than an order, -since we can have $f\leq g\leq f$ for distinct $f$ and $g$, -but there is an underlying equivalence relation. We may say -that $f\lt g$ if $f\leq g$ but $g\not\leq f$. Finally, let -me say that much of the interesting phenomenon in this -order arises already in the case of functions -$f:\mathbb{N}\to\mathbb{N}$ rather than $f:\mathbb{R}\to -\mathbb{R}$. -(Note that this way of defining the order does not presume -that the limit $\lim_{x\to\infty} \frac{f(x)}{g(x)}$ -exists, and this makes a huge difference in the nature of -the order. For example, if you insist that the limit exist, -then even a function $f$ that is everywhere less than $g$ -will not necessarily be less in the order, if $f$ -periodically jumps up nearly to $g$ and then down to $0$ in -such a way that prevents the limit from converging.) -This is a partial order on the function space, and you are -seeking a natural linearly ordered family of functions that -is maximal, in the sense that no additional functions can -be added to it while preserving pairwise -order-comparability of the elements. I claim that there -will be no nice such family along the lines that you seek, -even in the case just of functions -$f:\mathbb{N}\to\mathbb{N}$. -First, as observed by Yuval Filmus, there is no countable -maximal linearly ordered subset. He explains that one can -always exceed any given countable family with a higher rate -growth. This observation can be refined to show a bit more: -if $f_n\lt g$ for all $n$, then there is $f\lt g$ with -$f_n\lt f$ for all $f$. That is, we can exceed all the -$f_n$ even while staying below $g$. To see this, observe -that $f_n$ is eventually less than $c_n g_n$ for some -constant $c_n$. We may assume that $c_n=1$ by absorbing the -constant $\frac 1{c_n}$ into the function $f_n$. Let $d_n$ -be the point beyond which $f_n$ is less than $c_n g$. Now -build a function $f$ which at value $m$ is the maximum of -the $f_n(m)$ for which $d_n\leq m$. Thus, $f$ is eventually -bounding every $f_n$ and if $g$ is eventually below $c\cdot -f$, then it is also eventually below many $c\cdot f_n$ for -large enough $n$. So $f$ is as desired. This argument is -essentially the same as Hausdorff used to show that -countable Hausdorff gaps can always be filled, as I explain -in this MO -answer. -The previous observation shows that the order has no cuts -of order type $(\omega,\omega)$. That is, any partition of -the order into a lower family and an upper family, each -countable, can be extended by placing additional functions -in the middle. For example, you can continually add -functions in this way to the lower family. -My main observation now is that, because of this, there can -be no maximal linearly ordered family that is parameterized -by reals $f_c$ or by finite sequences of reals $f_{\vec -c}$, in such a way that increasing the parameters makes a -higher function. (This is true even for the functions -$\mathbb{N}\to\mathbb{N}$.) The reason is that the real -parameters all have countable cofinality, and so as we -increase parameters from below and decrease them from -above, we can find a countable cofinal subfamily. Our -parameterized family will have just one function in the -gap, but the argument above shows that we can fill this gap -with uncountably many. The problem is that every point in -$\mathbb{R}$ is approachable by a countable sequence, but -the order $\leq$ on functions is not at all like that. -There is indeed a rich set-theoretic interaction with the -possible cofinalities that arise in the order (and this is -the reason I suggested the set-theory tag). In particular, -although the observation above shows that uncountable -cofinalities must arise, the particular cardinals that -arise as the cofinality of the entire order are independent -of ZFC. This phenomenon is studied in the theory of -cardinal characteristics of the continuum via such concepts -as the bounding number and the dominating number. -Finally, despite all this, let me say that it is consistent -with ZFC that there is a definable, constructible maximal -linearly ordered subset of your order, because in Goedel's -constructible universe $L$ there is a -$\Delta^1_2$-definable well-ordering of the reals, and one -can use this order to produce a canonical family by -transfinite recursion, whose definition is fairly low by -descriptive set-theoretic standards.<|endoftext|> -TITLE: Density of values of polynomials in two variables -QUESTION [40 upvotes]: This question is a reposting of a comment I made on Polynomial representing all nonnegative integers. Let $f(x,y)\in \mathbb Q[x,y]$ such that $f(\mathbb Z\times \mathbb Z)$ is a subset of $\mathbb N$ (the nonnegative integers). Let $g(n)$ be the number of elements of $f(\mathbb Z\times \mathbb Z)\cap \lbrace 0,1,\dots, n\rbrace$. How fast can $g(n)$ grow? Is it always true that $g(n)=O(n/\sqrt{\log(n)})$? If true this is best possible since if $f(x,y)=x^2+y^2$ then $g(n)\sim cn/\sqrt{\log(n)}$. - -REPLY [8 votes]: I would certainly expect $g(N)$ to be rather small when $f$ has larger degree. Let us consider the special case where $f(x,y)=x^d+y^d$, for $d\geq 3$. Consider -the arithmetic function $r(n)$, which counts the number of $(x,y)\in \mathbb{N}^2$ such that $n=f(x,y)$. The first moment of $r(n)$ is easily understood via the geometry of numbers. The second moment was looked at by Hooley (On another sieve method and the numbers that are a sum of two $h$th powers. Proc. London Math. Soc. 43 (1981), 73-109). -As a consequence of this there exists an explicit constant $c>0$ such that there are asymptotically $c N^{2/d}$ integers $n\leq N$ which can be written as $x^d+y^d$, and furthermore, almost all of these have essentially just one representation.<|endoftext|> -TITLE: Number of Longest Decreasing subsequences and RSK -QUESTION [6 upvotes]: It is well known via the RSK-correspondence that the length of the longest decreasing subsequence in a permutation $\pi \in S_n$ is the length of the longest column of the insertion tableau of $\pi$. ( The insertion tableau and the recording tableau produced by this algorithm have the same shape) -link text -My question is what else can be gleaned from the RSK correspondence in terms of ,say, -a) the length of the next longest decreasing subsequence in $\pi$ ? -b) the number of longest decreasing subsequences in $\pi$, given the fact that there is exactly one column of maximum length ? -c) can we say more about the above two questions if we knew that $\pi$ was an involution? (If $\pi$ happens to be an involution, then insertion tableau and recording tableau produced are equal) - -REPLY [6 votes]: For a version of a), see C. Greene, An extension of Schensted's theorem, Advances in Mathematics, 1974. Note: the result is beautiful, but the statement is a bit delicate.<|endoftext|> -TITLE: Formally smooth morphisms, the cotangent complex, André-Quillen cohomology, and representability of nilpotent extensions as trivial extensions over a cofibrant replacement -QUESTION [9 upvotes]: Recall that an $R$-algebra $R\to S$ is called formally smooth (resp. formally unramified resp. formally étale) if given any lifting problem of the form -$$\begin{matrix} -R&\to &T\\ -\downarrow&{}^?\nearrow&\downarrow\\ -S&\to&T/J\end{matrix}$$ -where $J$ is a square-zero nilpotent ideal of $T$, there exists at least one (resp. at most one, resp. exactly one) lift $S\to T$ making the diagram commute. -In Quillen's paper "Homology of Commutative Rings", he cites the following construction (Not sure who came up with it, but my copy of the paper has "Hopf?" written next to the relevant proposition): -Given an $R$-algebra $S$ and an $S$-module $M$, we let $S\oplus M$ be the commutative ring on the underlying $S$-module $S\oplus M$ with the multiplication given by the composite -$$(S\oplus M)\otimes_R (S\oplus M)\cong (S\otimes_R S) \oplus (S\otimes_R M) \oplus (M\otimes_R S) \oplus (M\otimes_R M)\to S\times M\cong S\oplus M$$ -Where: -$S\otimes_R S\to S$ is the multiplication of $S$ -$S\otimes_R M\to M$ is the left action of $S$ on $M$ -$M\otimes_R S\to M$ is the right action of $S$ on $M$ -$M\otimes_R M\to M$ is the zero map $M\otimes_R M\to 0\to M$ -This ring is commutative, unital, and is equipped with a canonical $R$-linear projection $\phi:S\oplus M\to S$ with kernel $S$-linearly isomorphic to $M$. Also notice that $ker(\phi)$ is square-zero. -Interestingly, given any $R$-algebra $A$ extending $S$, i.e. $A\in (R\operatorname{-Alg}\downarrow S)$, it is easily verified that $\operatorname{Hom}_{(R\operatorname{-Alg}\downarrow S)}(A,S\oplus M)\cong \operatorname{Der}_R(A,Res_{A\to S}(M))$, that is, lifts $A\to S\oplus M$ are in bijective correspondence with derivations $A\to Res_{A\to S}(M)$ where $Res_{A\to S}$ is the restriction of scalars by the map $A\to S$. -In particular, it seems, at least to my untrained eye, that there's a connection between this construction and the definition of formally smooth (resp. formally unramified resp. formally étale) algebras. The lifts we see, at least when $T$ is of the form $U\oplus M$ for some $U$-module $M$ correspond exactly to $R$-derivations. -Quillen then proves that the exact objects corresponding to the modules of the form $S\oplus M$ are the abelian group objects of the category $(R\operatorname{-Alg}\downarrow S)$. In particular (back to our original notation), this does not necessarily hit all square-zero extensions of $T/J$, since it misses those square-zero extensions that don't admit a section $T/J\to T$. Since the module of relative Kähler differentials carries all of the data of these derivations, this appears to explain exactly why Kähler differentials are only sufficient to characterize formally unramified algebras. -So far, I follow. -So here's the question: Why, morally, do we need to look at the $S$-modules over the cofibrant replacement of $S$ (we can give a functorial cofibrant replacement by looking at a simplicial resolution of $S$ by defining a simplicial ring consisting of free algebras over $A$ in every degree (c.f. André 1974)) (with all of the homotopical baggage we need to characterize the model structure) to capture the lifting data from the rest of the square-zero extensions that we would need to characterize formal smoothness (resp. formal étaleness)? -That is, similar to how every trivial extension is of the form $S\oplus M$ for an $S$-module M, can we give a homotopical version of the trivial extension such that the nilpotent extensions of $S$ are exactly those that that are trivial over some cofibrant replacement $QS$? - -REPLY [7 votes]: Your question is: - -Why, morally, do we need to look at the S-modules over the cofibrant replacement - of S to capture the lifting data from the rest of the square-zero extensions - that we would need to characterize formal smoothness (resp. formal étaleness)? - -Here's how I think about it, though I don't know if it addresses what's bothering you about this. -You'd like to understand surjective maps of rings $f:T\to T'$ whose kernel $M$ (an $S$-module) is square-zero, but where $f$ might not admit a section. -There's a universal example of such an extension (in $R$-algebras over $S$); it lives in simplicial rings, i.e., in $(s(R\operatorname{-Alg})\downarrow S)$. Namely, form the simplicial $S$-module $BM$ by shifting (so $\pi_1BM=M$, $\pi_iBM=0$ for $i\neq1$), then consider the simplicial ring $S\oplus BM$, formed with trivial multiplication on the $BM$ part. -Now let $g: S\to S\oplus BM$ be the evident inclusion map into the first factor. The "kernel" of this map of simplicial rings is the homotopy fiber of $g$ (calculated in simplicial $S$-modules), and this homotopy fiber is equivalent to $M$ (concentrated in dimension $0$). -So $g$ has "kernel" $M$. -The claim is that isomorphism classes of square-zero extensions (possibly not split) over $S$ is the same as homotopy classes of maps $S\to S\oplus BM$ in $(s(R\operatorname{-Alg})\downarrow S)$. Calculating homotopy classes requires using a cofibrant model for the domain $S$. (In fact, there are no nontrivial maps $S\to S\oplus BM$ at all.) -Given such a map $f:QS\to S\oplus BM$ from a cofibrant replacement, pull back $g$ along it to get an extension $g'\colon E\to QS$. The simplcial ring $E$ has homotopy concentrated in degree $0$, and in fact there's an exact sequence $0\to M\to \pi_0 E\to \pi_0 QS=S$. -(Note that $(s(R\operatorname{-Alg})\downarrow QS)$ and $(s(R\operatorname{-Alg})\downarrow S)$ are Quillen equivalent model categories, so in fact giving an "extension" over $QS$ is the same as giving one for $S$.) -This story is morally just like the one you can tell for extension of $R$-modules: to compute extensions of $M$ by $N$, note that there's a universal extension by $N$ in the derived category of $R$-modules, of the form $N\to C\to N[1]$ (where $H_*C=0$). So extensions over $M$ are $\mathrm{Hom}_{D(R)}(M, N[1])$, and you can compute this group by replacing $M$ with a projective resolution (i.e., cofibrant replacement).<|endoftext|> -TITLE: Can the fundamental group of any manifold be realized as the fund grp of a finite space? -QUESTION [30 upvotes]: Recently, I was asked to calculate the fundamental group of the space $X= \{a,b,c,d\}$ with open sets generated by $\{ a, c, abc, acd \}$. -Turns out, $\pi_1(X)\cong \mathbb Z$ and in fact, $X$ is the quotient of $S^1$ (with the northern and southern hemispheres identified). But the result was not so easy to prove and this motivates the questions: - -Is the fundamental group of every connected manifold (other restrictions / generalizations on the manifold are welcome) the fundamental group of a finite space? (Of course, it would not be Hausdorff). (I observe that there are many redundant points on a manifold where homotopy-equivalent loops need not traverse) -Is calculating $\pi_1$ of such finite spaces easier than for the given space? (If yes, this gives a method to calculate fundamental groups of many familiar spaces) - -Perhaps the fact that -- $\pi_1$ of any CW complex just depends on its 1-skeleton [edit:2 skeleton]-- may be helpful. - -REPLY [67 votes]: In fact, there is the following theorem: Every finite CW complex is weakly homotopy equivalent to a finite topological space, and vice versa. -For simplicial complexes, this theorem is realized by mapping a complex to its face poset, and using the correspondence between finite posets and finite topological spaces. In the other direction, one maps a poset to its order complex. -In general it is not easy to compute homotopy groups of a finite topological space. I know that there are some techniques in Jonathan Barmak's Ph.D. thesis.<|endoftext|> -TITLE: The logic of convex sets -QUESTION [31 upvotes]: Let me start with Helly's theorem: Let $A_1$, $A_2$, ..., $A_{n+2}$ be $n+2$ convex subsets of $\mathbb R^n$. If any $n+1$ of these subsets intersect (this means: have nonempty intersection), the so do all $n+2$. -This assertion is, logically speaking, a definite clause: All conditions are of the form "some subsets intersect", and so is the assertion. Generally, a definite clause about intersection of convex sets looks like this: Given some sets $S_1$, $S_2$, ..., $S_k$ and $T$ and a family $\left(F_a\right)_{a\in S_1\cup S_2\cup ...\cup S_k\cup T}$ of convex subsets of $\mathbb R^n$ indexed by the elements of $S_1\cup S_2\cup ...\cup S_k\cup T$, we claim that if every $i\in\left\lbrace 1,2,...,k\right\rbrace$ satisfies $\bigcap\limits_{a\in S_i}F_a\neq\emptyset$, then $\bigcap\limits_{a\in T}F_a\neq\emptyset$. -Now it is an easy exercise to see that every tautological (= true for every choice of convex subsets) definite clause about intersection of convex sets is derivable using trivial properties of intersection (such as $A\cap A=A$, $A\cap B=B\cap A$ and $A\cap \left(B\cap C\right)=\left(A\cap B\right)\cap C$) and Helly's theorem only. (Note that we consider $n$ as given a priori and fixed during our proof, so we can't project on a subspace and use Helly for a smaller $n$, for example. We really only can manipulate intersections and use Helly for various intersection of convex subsets. I could write up what we can do as a natural deduction system, but it is pretty obvious.) -Now, what if we allow indefinite clauses? These are statements of the form: Given some sets $S_1$, $S_2$, ..., $S_k$ and $T_1$, $T_2$, ..., $T_l$ and a family $\left(F_a\right)_{a\in S_1\cup S_2\cup ...\cup S_k\cup T_1\cup T_2\cup ...\cup T_l}$ of convex subsets of $\mathbb R^n$ indexed by the elements of $S_1\cup S_2\cup ...\cup S_k\cup T_1\cup T_2\cup ...\cup T_l$, we claim that if every $i\in\left\lbrace 1,2,...,k\right\rbrace$ satisfies $\bigcap\limits_{a\in S_i}F_a\neq\emptyset$, then at least SOME $j\in\left\lbrace 1,2,...,l\right\rbrace$ satisfies $\bigcap\limits_{a\in T_j}F_a\neq\emptyset$. -What set of "axioms" (such as Helly's theorem) do we need in order to prove such indefinite clauses, if they are tautological? By "prove" I mean prove without using the convexity of the sets or that the sets are set at all (a pointfree approach, so to speak) - only using formal properties of $\cap$, logic (let's say constructive) and these axioms. Obviously Helly alone is not enough anymore; for example, for $n=1$ we have this here: If $A$, $B$, $C$, $D$ are four convex subsets of $\mathbb R^n$ such that $A\cap B$, $B\cap C$, $C\cap D$ and $D\cap A$ are nonempty, then at least one of the sets $A\cap C$ and $B\cap D$ are nonempty. -A connection to temporal logic is possible, but to be honest I have no idea about temporal logic; if somebody could point me to a reference that is of help here this might change... - -REPLY [10 votes]: This is an interesting question. (Even if there is no finite list of "axioms".) -For example the following is true: Suppose you have a (d+1)-dimensional polytope P. Associate a convex set in $R^d$ to every facet of P, and suppose that every non empty intersection among the facets implies a non empty intersection for the corresponding sets. Then there are some set of facets of the polytope with empty intersection so that the corresponding convex sets have non empty intersection. This is a large collection of statements of the kind you asked that already includes Helly's theorem. -Note that the nerve theorem gives you a lot of information on intersection patterns that cannot be realized. (Including the statement made above.) -A stronger statement for the case described above goes as follows: Suppose that 0 is in the interior of P. Then you can find two faces of F and G of P so that 0 is in their convex hull and the intersection of all sets corresponding to facets containing F and all sets corresponding to facets containing G is nonempty. This can be derived from the Borsuk-Ulam theorem. -Another well known result of the kind you ask about is the colorful Helly theorem (proved by Lovasz and a dual Caratheodory version was proved independently by Barany). It sais that if you have d+1 collections of convex sets in R^d, and if for every choice of one set from each collection there is a point common to all chosen sets then there is a collection all whose members have a point in common. -For n=1 there is a theorem by Lekkerkerker and Boland characterizing interval graphs. The characterization is based on two requirements: (1) there are no induced cycles of length >3. (This leaves us with chordal graphs.) (2) for every 3 vertices, one of the vertices is a neighbor of a vertex in every path between the other two. This gives an infinite set of "axioms" in the sense asked in the question. I suppose there is no finite list. -So overall there are many interesting theorems of the kind you described, and this is a rich area. But a complete description for dimension > 1 does not look realistic. -A more general question is to study the situation where you allow both operation: (1) taking the intersection (2) taking convex hull. This makes the problem even much more complicated and not much is known. An example of a recent result in this more general setting is the following theorem by Novick: Given 7.2k pairwise disjoint convex sets in the plane there is a set in the family that is disjoint to the convex hull of k other sets in the family.<|endoftext|> -TITLE: Martin's cone theorem and recursion theory -QUESTION [9 upvotes]: Martin's remarkable cone theorem in the theory of determinacy says the following: - -Suppose $A\subseteq \omega^\omega$ is Turing invariant and determined. If $\forall x\exists y(x\le_T y\& y\in A)$ then $A$ contains a cone. - -Let me explain what this means: $A$ is Turing invariant iff $\forall x\in A\forall y(x\equiv_T y\Rightarrow y\in A)$. Here, $\le_T$ is the relation of Turing reducibility and $\equiv_T$ is the corresponding equivalence relation. -"Determined" is in the usual sense of infinite games on integers. -A cone is a set of the form $$C_y=\{x\mid y\le_T x\}.$$ Clearly, cones are Turing invariant. We say that $y$ is the base of the cone $C_y$. -If $\forall x\exists y(x\le_T y\& y\in A)$ we say that $A$ is cofinal. -When Martin proved his theorem, he thought that it would be a quick way of showing determinacy fails (in ZF), by considering explicit sets coming from recursion theory. Instead, he found several results in recursion theory as a consequence. -Here are some examples: - -For every $x$ we have $x<_T x'$, where $x'$ is the Turing jump of $x$. This means that the set of jumps is cofinal. By Borel determinacy, it follows that there is a $y$ such that if $y\le_T x$, then $x\equiv_T z'$ for some $z$. Well known recursion theoretic results show that in fact we can take $y=0'$. -Again by Borel determinacy, there is a real $x$ such that any $y$ with $x\le_T y$ is a minimal cover above some $z$. Again, recursion theoretic arguments show that we can take $x=0^{(\omega)}$. - -I do not know many examples coming from recursion theory, but maybe somebody here does. - -Are there (natural) examples of sets $A$ defined recursive theoretically that we know need to contain a cone, but for which we do not know of a (natural) base? - -"Need to contain a cone" could be taken to mean, say, that they are Turing invariant and cofinal and Borel. -Naively, a negative answer would mean we have very strong abstract basis results. But I would be interested in natural candidates for a positive answer as well. - -REPLY [8 votes]: I would like to add another example. -Given a sentence $\phi$ from partial order language, then for any Turing degree $x$, either $D(\leq x)\models \phi$ or $D(\leq x)\models \neg\phi$. By the BD, there is a Turing degree $x_{\phi}$ so that either for all $y\geq_T x_{\phi}$, $D(\leq y)\models \phi$ or for all $y\geq_T x$, $D(\leq y)\models \neg \phi$. -Let $z$ be a Turing degree above all the $x_{\phi}$'s, then for every $y\geq_T z$, $D(\leq y)$ is elementary equivalent to $D(\leq z)$. -I don't know a natural base for this.<|endoftext|> -TITLE: Is there an extension of the Arzela-Ascoli theorem to spaces of discontinuous functions? -QUESTION [14 upvotes]: The Arzela-Ascoli function basically says that a set of real-valued continuous functions on a compact domain is precompact under the uniform norm if and only if the family is pointwise bounded and equicontinuous. -Are there any analogs of this kind of result for spaces of noncontinuous functions? The specific set I have in mind is the càdlàg functions, which are right continuous and have left limits. Essentially, I want to know if there is any relationship between compactness and equicontinuity. Here, equicontinuity would obviously have to be relaxed to account for jumps, and compactness in the uniform topology could be relaxed to compactness in, say, the Skorokhod topology, or something weaker than uniform. - -REPLY [7 votes]: In $L^p(\Omega)$ spaces ($1 \leq p < \infty$) , this theorem by A.Kolmogorov asserts that a family of functions $\mathcal{F}$ is compact in $L^p(\Omega)$ if and only if -1) $\mathcal{F}$ is bounded. -2) For every $\varepsilon > 0$, there exist $\delta > 0$ such that for each $f \in \mathcal{F}$ we have: -if $\| h \| < \delta $ then $ \| f(\cdot + h) - f(\cdot) \|_p < \varepsilon$ -(this is equivalent to the equicontinuity condition but in p-mean). -3) For any $\varepsilon > 0$ , there exists a Borel set $B \subseteq \Omega$ such that for every $f \in A$ we have -$$ \left( \int_{\Omega \setminus B} | f(x) |^p dx \right)^{\frac{1}{p}} < \varepsilon$$ -I don't have a reference since I've got this from my measure theory notes. The proof is somewhat technical and in the kernel of it the Arzelá Ascoli theorem is used.<|endoftext|> -TITLE: The diameter of the Erdös component of the collaboration graph -QUESTION [10 upvotes]: This site claims that the diameter of the Erdös component of the collaboration graph in 2004 was 23. What is it now? Is it increasing or decreasing with time? Recall that the vertices of the collaboration graph are mathematicians and two vertices are connected if the mathematicians co-author a paper. MathSci allows one to find the collaborative distance between any two mathematicians. So in principle one can find the diameter just by using MathSci. In general (and more seriously) is there a mathematical theory which describes the growth of "real life" networks like the collaborative graph? - Update 1 A process that I had in mind is something like this. At every step one of the following things can happen. - -A new vertex $a$ is born with some probability $p$. Usually that vertex is a "student" of some other vertex $x$. We can connect $a$ and $x$ by an edge. -A randomly chosen vertex $y$ gets connected with a randomly chosen vertex $z$. But the probability for choosing $z$ is not uniform. Those vertices that are closer to $y$ have more chances getting connected to $y$. -A randomly chosen vertex "dies" (with some probability $q$) meaning it does not participate in new edges any longer. - - Update 2. Many thanks to Balazs and Joseph for their answers. But the first question still remains: what is the diameter of the Erdös component now? - -REPLY [3 votes]: In the long run, the diameter will grow. Let k be an upper bound to the number of years between a mathematician's first and last publication. Then nk years after Erdos' last publication, all new nodes in the Erdos component will have Erdos number at least n, so the diameter will be at least n. -(I am assuming here that there will be new nodes, which is not true for very large values of n. Also, I am assuming that k is constant, but only to keep the estimate "nk" simple.)<|endoftext|> -TITLE: How small can the set of $p$ such that the $L^p$ norms are different for two fixed functions? -QUESTION [6 upvotes]: What does it tell you about two functions if their $L^p$ norms are the same for all $p\in[1,\infty]$? Certainly they could be related by composition with a diffeomorphism with Jacobian of norm 1, or even one could be a "pulled apart version of the other one" in the sense of -$x^2\chi_{[0,2]}$ vs $x^2 \chi_{[0,1]} + (x-1)^2 \chi_{[2,3]}$. To try to ignore the second type of issue, I'll restrict to smooth functions, and ask the following precise question: -Given $f,g\in C^\infty(\mathbb{R})$ such that $\Vert f \Vert_{L^p(\mathbb{R})} = \Vert g \Vert_{L^p(\mathbb{R})} <\infty$ for all $p\in [1,\infty]$ is it necessarily true that $f(x) = g(\pm x + C)$ for some constant $C$ and for a choice of either $+$ or $-$ for all $x \in \mathbb{R}$? -As Qiaochu Yuan shows in his answer, smoothness doesn't solve the issue of "pulling apart" at all. Thus, I am interested in the following: -What is the "smallest" (in whatever sense) but still nonempty subset of $S \subset [1,\infty]$ such that there is $f,g\in C^\infty(\mathbb{R})$ such that $S$ is the set of $p$ such that -$\Vert f \Vert_{L^p} \neq \Vert g \Vert_{L^p}$? - -REPLY [12 votes]: The complement of $S$ in $(1,\infty)$ can't contain an interval or even a sequence converging to a point in $(1,\infty)$. Let $f$ and $g$ be two real functions on ${\mathbb{R}}$ all whose moments exist. Assume $\int|f|^p=\int|g|^p$ for all $p\in S^c$. put $h(z)=\int|f|^z-\int|g|^z$. where $z\in \mathbb{C}$. $h$ is an analytic functions in $\{Real z>1\}$ which we assume is not constantly zero (since $S$ is not empty) so can't vanish on a converging sequence.<|endoftext|> -TITLE: What topics should be included in a calculus-for-the-liberal arts course? -QUESTION [9 upvotes]: I have in mind a course taken by liberal-arts students who will probably never take another math course. I would like such a course to convey some of the way mathematical thinking is done (i.e. not a cookbook course) without getting "rigorous" (since such students cannot be assumed to understand that and can learn rather little of it during a one-year calculus course). Apparently I am not the first to think of excluding the mean value theorem, since one of James Stewart's books does that. I would also like to include some of the ways in which differential and integral calculus have played a role in the history of science. -I'm teaching a course in which I began with this and I use that idea repeatedly in exercises. It will of course be used in explaining the fundamental theorem. One of various places where I've used that proposition so far is #3 in this assignment, where I was told by multiple students that no one else who teaches math ever asks students to think through steps like this. They "know" very well that that's not at all how math is done. Hence, they say, it is quite confusing. I do some topics that might normally be done only in "rigorous" course, such as things like #1 in this, but as you see, I don't do it in the way in which rigorous arguments are written. -I'd like to see skills taught in such a course only to the extent to which they aid thinking, and I like to have students write carefully about that thinking. This contrasts with a practice that perhaps few if any mathematicians intend to do, but which is widespread, and that is that students in such courses are taught that mathematics consists entirely of skills. This leaves no place for things like one that I like to include: What is "natural" about the number $e$? (Here is how I begin the treatment of that question.) -It seems as if mathematical thinking is often reserved for advanced courses rather than freshman calculus or the like, despite what is probably overwhelming empirical evidence that it can be done even at the most elementary levels, e.g. teaching graph theory to 4th-graders. -The question here is: Which specific topics should be included in a course consistent with the ideas sketch above and why? In particular, which that are now customarily not included should be there, and vice-versa? - -REPLY [9 votes]: Several years ago the liberal-arts-ish university where I was at the time was pushing to have more interaction between the sciences and the humanities. In that spirit I volunteered to give an hour-long seminar entitled, "So You Think You're Educated, But You Don't Know Calculus: A Brief Introduction to One of Humanity's Greatest Inventions." It was aimed at the humanities faculty. My goal was to explain the big ideas behind calculus and place them in their historical and philosophical context for an audience of very smart people with weak math backgrounds. You might be able to use the historical and philosophical context part of the talk for the "ways in which differential and integral calculus have played a role in the history of science" aspect of your question. You are welcome to borrow freely from my presentation. -In retrospect the title may have been a bit too audacious, but the talk went much better than I had expected. A few of the scientists showed up for fun, but most of the audience were folks from the humanities and social sciences. They were engaged, and they peppered me with questions for half an hour after the talk was over. After I left there were still people who stayed behind to discuss the seminar. Later I even got an email from the provost (a religion scholar) who wanted me to meet with him to discuss the ideas in the talk! It was, frankly, the most successful academic talk I've ever given - and much more so than the one I gave three days ago at a math conference that was attended by eight people in a room that could hold hundreds and yielded no questions. :( -One caveat: When I discuss the philosophical implications of calculus, I'm doing so as I think they appeared to people at the time, not today. Clearly, humanity's consensus on these big questions has changed in the last 300 years. -The other thing I would say is to second Deane Yang's recommendation to look at the Hughes-Hallet, et al, calculus texts. I know there are strong opinions on the calculus reform movement, and I don't want to wade into that. But what the Hughes-Hallet texts do well (in my opinion) is to emphasize ideas and mathematical thinking over rote computation. Since you're after the former, looking at what they've done may be helpful.<|endoftext|> -TITLE: Covering a random graph with spanning trees. -QUESTION [10 upvotes]: Let $G=(V,E)$ be a connected graph, say $V=\{1,\ldots,n\}$. Let $F=(V,E')$ be a uniformly random forest in $G$. (In other words, $E'$ is a subset of edges $E$ not containing a cycle, and it is uniformly chosen over all such sets.) -Associated to the random forest $F$ are marginals $\{p_e:e \in E\}$, where $p_e = \mathbb{P}(e \in E')$. Now let $E^*$ be a random subset of $E$, chosen by independently including each edge $e \in E$ with probability $p_e$. -Finally, let $N$ be the (random) smallest number of spanning trees of $G$ whose union contains $E^*$. - -What is known about the distribution of $N$? How does $\sup_{G} \mathbb{E}(N)$, the largest expected value of $N$ over all $n$-vertex graphs, grow? Is it $O(\log n)$? Is it $O(1)$? - -Edit: is it $O(\sqrt{\log n})$? Fedor has a nice example showing that it is not $O(1)$. I believe optimizing Fedor's example yields a lower bound of order $(\log n/\log\log n)^{1/2}$. -Note: the question also makes sense if $E'$ is the edge set of a uniformly random spanning tree, and Fedor's example applies in either case. - -REPLY [2 votes]: Consider the path P of length n. Now replace every edge by n parallel edges. A random tree has $p_e=1/n$ and $p_e=1/(n+1)$ in a random forest for each edge $e$. Now, consider the set E^*. There is a constant probability that $\Omega(\frac{\log n}{\log \log n})$ copies of some parallel edge will show up. Therefore, the expectation of $N$ will be $\Omega(\frac{\log n}{\log \log n})$. Am I missing anything here? -I believe the upperbound will be the same as well.<|endoftext|> -TITLE: Can Vopenka's principle be violated definably? -QUESTION [18 upvotes]: One form of Vopenka's principle (a large cardinal axiom) states that no locally presentable category contains a full subcategory which is large (= a proper class) and discrete (= contains no nonidentity morphisms). In terms of this definition, my question is: - -Can one define a particular locally presentable category C and write down an explicit formula $\phi(x)$ in the first-order language of set theory such that if Vopenka's principle fails, then $\{ x | \phi(x) \}$ is a large discrete full subcategory of C? - -But feel free to use any equivalent statement of Vopenka's principle and answer a suitably equivalent version of the question. - -REPLY [21 votes]: Update. My new article grows out of and extends my 2010 answer to this question. The new part is the conservativity result, showing that the Vopěnka principle has the same first-order consequences as the strictly weaker Vopěnka scheme. - -Joel David Hamkins, The Vopěnka principle is inequivalent to but conservative over the Vopěnka scheme, manuscript under review. (arxiv) - -Abstract. The Vopěnka principle, which asserts that every proper class of first-order structures in a common language admits an elementary embedding between two of its members, is not equivalent over GBC to the first-order Vopěnka scheme, which makes the Vopěnka assertion only for the first-order definable classes of structures. Nevertheless, the two Vopěnka axioms are equiconsistent and they have exactly the same first-order consequences in the language of set theory. Specifically, GBC plus the Vopěnka principle is conservative over ZFC plus the Vopěnka scheme for first-order assertions in the language of set theory. -The Vopěnka principle is the assertion that for every proper class $\mathcal{M}$ of first-order $\mathcal{L}$-structures, for a set-sized language $\mathcal{L}$, there are distinct members of the class $M,N\in\mathcal{M}$ with an elementary embedding $j:M\to N$ between them. In quantifying over classes, this principle is a single assertion in the language of second-order set theory, and it makes sense to consider the Vopěnka principle in the context of a second-order set theory, such as Godel-Bernays set theory GBC, whose language allows one to quantify over classes. In this article, GBC includes the global axiom of choice. -In contrast, the first-order Vopěnka scheme makes the Vopěnka assertion only for the first-order definable classes $\mathcal{M}$ (allowing parameters). This theory can be expressed as a scheme of first-order statements, one for each possible definition of a class, and it makes sense to consider the Vopěnka scheme in Zermelo-Frankael ZFC set theory with the axiom of choice. -Because the Vopěnka principle is a second-order assertion, it does not make sense to refer to it in the context of ZFC set theory, whose first-order language does not allow quantification over classes; one typically retreats to the Vopěnka scheme in that context. The theme of my article is to investigate the precise meta-mathematical interactions between these two treatments of Vopěnka's idea. -Main Theorems. - -If ZFC and the Vopěnka scheme holds, then there is a class forcing extension, adding classes but no sets, in which GBC and the Vopěnka scheme holds, but the Vopěnka principle fails. -If ZFC and the Vopěnka scheme holds, then there is a class forcing extension, adding classes but no sets, in which GBC and the Vopěnka principle holds. - -It follows that the Vopěnka principle VP and the Vopěnka scheme VS are not equivalent, but they are equiconsistent and indeed, they have the same first-order consequences. -Statement 1 is proved by class forcing to add a club class $C$ avoiding the regular cardinals. This destroys the assertion "Ord is Mahlo" and therefore destroys the Vopěnka principle, while preserving the Vopěnka scheme because it does not add sets. -Statement 2 is proved by class forcing of the global axiom of choice. The difficult part is to show that the Vopěnka principle holds true with respect the new classes definable from the generic filter. The proof involves the concept of a stretchable set $g\subset\kappa$ for an $A$-extendible cardinal, one which has the property that for every cardinal $\lambda<\kappa$ and every extension $h\subset\lambda$ with $h\cap\kappa=g$, there is an elementary embedding $j:\langle V_\lambda,\in,A\cap V_\lambda\rangle\to\langle V_\theta,\in,A\cap V_\theta\rangle$ such that $j(g)\cap\lambda=h$. Thus, the set $g$ can be stretched by an $A$-extendibility embedding so as to agree with any given $h$. This stretchability property is the $A$-extendibility analogue of the master condition technique in other large cardinal contexts. -Corollaries. - -Over GBC, the Vopěnka principle and the Vopěnka scheme, if consistent, are not equivalent. -Nevertheless, the two Vopěnka axioms are equiconsistent over GBC. -Indeed, the two Vopěnka axioms have exactly the same first-order consequences in the language of set theory. Specifically, GBC plus the Vopěnka principle is conservative over ZFC plus the Vopěnka scheme for assertions in the first-order language of set theory. $$\text{GBC}+\text{VP}\vdash\phi\qquad\text{if and only if}\qquad\text{ZFC}+\text{VS}\vdash\phi$$ - -See the edit history for my 2010 answer.<|endoftext|> -TITLE: Does the formal power series solution to $f(f(x))= \sin( x) $ converge? -QUESTION [46 upvotes]: I have spent some time using gp-pari. There is, of course, a formal power series solution to -$ f(f(x)) = \sin x.$ It is displayed below, identified by the symbol $g$ because I am not entirely sure whether it is a function of anything. -On the other hand, should the coefficients continue to (by and large) decrease, this suggests a nonzero radius of convergence. If the radius of convergence is nonzero, then inside that, not only is a function defined and, you know, analytic, but the functional equation is satisfied. Indeed, all that is necessary is radius of convergence strictly larger than $\frac{\pi}{2}$ owing to certain symmetries. For instance, given my polynomial $g,$ it seems we have $g=1$ at about $x \approx 1.14.$ Then we seem to have a local maximum at $x =\frac{\pi}{2},$ and apparently there $g \approx 1.14,$ strictly larger than 1 which is an important point. So everything would fall into place with large enough nonzero radius of convergence. -$$ -\begin{array}{lll} -g & = & x - \frac{x^3 }{ 12} - \frac{x^5 }{ 160} - \frac{53 x^7 }{ 40320} - \frac{23 x^9 }{71680} - \frac{92713 x^{11}}{1277337600} - \\\ - & & \\\ -& & \frac{742031 x^{13} }{79705866240} + \frac{594673187 x^{15} }{167382319104000} + \frac{329366540401 x^{17} }{91055981592576000} + \\\ -& & \\\ -& & \frac{104491760828591 x^{19} }{62282291409321984000} + \frac{1508486324285153 x^{21} }{4024394214140805120000} + \cdots -\end{array} -$$ -Note that the polynomial $g$ is smaller than $x$ but larger that $\sin x,$ for, say, - $0 < x \leq \frac{\pi}{2}.$ -So, that is the question, does the formal power series beginning with $g$ converge anywhere -other than $x = 0$? -EDIT: note that the terms after the initial $x$ itself have all turned out to be -$$ \frac{a_{2 k + 3} x^{2 k + 3} }{2^k ( 2 k + 4)!} $$ -where each $a_{2 k + 3}$ is an integer. This much seems provable, although I have not tried yet. -EDIT, Friday 12 November 2010. It now seems really unlikely that this particular problem gives an analytic answer. I suspect that the answer is $C^\infty$ and piecewise analytic, with failure of analyticity at only the points "parabolic" where the derivative has absolute value as large as 1, those points being $0,\pi, 2 \pi, \ldots.$ However, we need the anchor point at the fixpoint 0, otherwise how to begin? And I do think the power series will serve as an asymptotic expansion around 0. -Given the problem with the size of the derivative, now I am hoping for great things, and an obviously periodic and analytic solution, to the easier variant $f(f(x)) = g(x) = (1/2) \sin x.$ I would like both a nice power series and a nice answer by methods summing iterates $ g^{[k]}(x),$ which for the moment is an entirely mysterious method to me, but attractive for periodic target functions as periodicity would be automatic. - -REPLY [4 votes]: This is not a new answer but additional info for Will Jagy's answer about the computation of the Abel-function with the method of J. Écalle. - -[update] With some examples it comes out, that the "standard" computation of the fractional iteration via the formal power series -for the iterative logarithm and its truncation to the leading terms -provides the same values as Écalle's method using the Abel function -as described below. With the same truncation to 64 terms of the power -series and the same shift of z_0 to z_h towards zero the -difference between the two methods is smaller than 1e-40 , which -is the accuracy which is also achieved by each method alone. [end -update] - -I computed the formal Laurent-series for the Écalle-type Abel-function using Pari/GP to 509 coefficients in exact rational numbers which means the coefficients from $z^{-2}$ to $z^{506}$. -The last coefficient has numerator with 1423 digits and denominator with 1247 digits amounting to an absolute value of about 175 digits, roughly -2.66945040282 E175 , so the series has in the same way as the comparable series for the fractional iteration of $\exp(x)-1$ zero-radius of convergence and when we plot the curve showing the number of digits of the nonzero coefficients by $\log_{10}(|a_k|)$ we get the typical shape of the upright hockeystick. -Here are the leading 11 nonzero terms of the Laurent series for the Abel-function (which I call here "incomplete Abel function" so far because the "complete" Abel-function needs also the term for the logarithm and the term for the iteration-height h (this is index n in Will's answer): - Laurent-series in z: 3 *z^-2 - + 79/1050 *z^2 - + 29/2625 *z^4 - + 91543/36382500 *z^6 - + 18222899/28378350000 *z^8 - + 88627739/573024375000 *z^10 - + 3899439883/142468185234375 *z^12 - - 32544553328689/116721334798818750000 *z^14 - - 4104258052789/1554729734250000000 *z^16 - - 119345896838809094501/141343700374629565312500000 *z^18 - + 745223773943295679/3505548124370772949218750 *z^20 - + O(z^22) - -$ \qquad \qquad$ (remark: see an overview characterizing the growthrate at end (§2)) -This gives the "incomplete Abel function" in terms of its coeffs up to some truncation n (all formulae in Pari/GP notation): - abel_inc(z,n=64) = sum(k=1,n, coeff[k]*z^(k-3) ) - -The complete Abel-function is then : - {abel(z,h=32,n=64) = local(z_h,a); \\ give some sufficient default values - \\ in h and n for the required numerical - \\ precision of the approximate results - z_h = sin_iter(z,h); \\ sin_iter prev. defined as iterable sin() - a = abel_inc(z_h,n) + 6/5*log(z_h) - h ; - return(a); } - -The inverse abel-function must be implemented by some root-solver. In Pari/GP I used the following, where the inverse Abel-function is included in the body of the full fractionally-iterable sin_h() function: - {sin_h (h = 0,z_0=1) = local(a_0,z_h,a_h); \\ restriction abs(h)<1 - a_0 = abel(z_0, 32, 64); \\ get the Abel-value for z_0 - \\ with meaningful precision - a_h = a_0 + h ; \\ comp Abel-value for z_h - \\ the following is the implementation of - \\ the inverse Abel-function: - z_h = solve(z = sin(z_0),z_0, abel(z,32,64) - a_h); - return(z_h); } - -The following is done to apply the above to some example, reproducing additivity of the iteration heights 0.5 and 0.5 to integral height 1 with more than 40 digits precision: - \\ Pari-output - z_0 = 1 \\ %529 = 1 - z_05 = sin_h(0.5,z_0 ) \\ %530 = 0.908708429743 - z_1 = sin_h(0.5,z_05) \\ %531 = 0.841470984808 - z_1 - sin(z_0) \\ %532 = -6.38920219348 E-42 - -Below I show the recomputed list of computations in Will's answer with 40 digits correct: - step z0=Pi/2 - step abel(z0) z05=sin_h(0.5,z0) z1=sin_h(0.5,z05) z1 - sin(z0) - 0.00 1.57079632679 2.08962271973 1.14017947617 1.00000000000 -2.89445031739E-41 - 0.05 1.52079632679 2.09536408453 1.13806963935 0.998750260395 -2.86591796888E-41 - 0.10 1.47079632679 2.11273622895 1.13178674818 0.995004165278 -2.78164697945E-41 - 0.15 1.42079632679 2.14218948912 1.12146458427 0.988771077936 -2.64553725829E-41 - 0.20 1.37079632679 2.18449553252 1.10730765183 0.980066577841 -2.46383393292E-41 - 0.25 1.32079632679 2.24078077607 1.08956885996 0.968912421711 -2.24476553049E-41 - 0.30 1.27079632679 2.31257688904 1.06852649593 0.955336489126 -1.99807394218E-41 - 0.35 1.22079632679 2.40189260763 1.04446448663 0.939372712847 -1.73446474837E-41 - 0.40 1.17079632679 2.51131312355 1.01765794736 0.921060994003 -1.46500647333E-41 - 0.45 1.12079632679 2.64413616528 0.988364216777 0.900447102353 -1.20050550750E-41 - 0.50 1.07079632679 2.80455803137 0.956818478819 0.877582561890 -9.50882282773E-42 - 0.55 1.02079632679 2.99792899241 0.923232674366 0.852524522060 -7.24576289372E-42 - 0.60 0.970796326795 3.23110684637 0.887796468526 0.825335614910 -5.28014362671E-42 - 0.65 0.920796326795 3.51295197372 0.850679308887 0.796083798549 -3.65188391373E-42 - 0.70 0.870796326795 3.85503037983 0.812032915560 0.764842187284 -2.37402622132E-42 - 0.75 0.820796326795 4.27262886030 0.771993802047 0.731688868874 -1.43260703471E-42 - 0.80 0.770796326795 4.78624925852 0.730685613103 0.696706709347 -7.89576195851E-43 - 0.85 0.720796326795 5.42385666222 0.688221187210 0.659983145885 -3.89074331205E-43 - 0.90 0.670796326795 6.22434753781 0.644704322722 0.621609968271 -1.66626510284E-43 - 0.95 0.620796326795 7.24305478745 0.600231264287 0.581683089464 -5.96979699941E-44 - 1.00 0.570796326795 8.56077779381 0.554891942675 0.540302305868 -1.69831000319E-44 - -A picture of $y=\sin(x)$, the half-iterate $y=\sin^{\circ 0.5}(x)$ , $y=\sin^{\circ 1/3}(x)$ and $y=x$: - -Remark: at x, where sin(x)=0 the computation of the Abel-function runs in singularities and the value for the function is (interpolated from its neighbourhood) assumed to be zero. - -In the Abel-function abel(z,h=32,n=64)=... there is the parameter h which allows to control the quality of approximation. The formal exact solution is given as limit when h goes to infinity - but we use only finite approximations here. They key is, that h controls the implicite iteration of the argument z towards the fixpoint zero, so the numerical evaluation of the (truncated to n coefficients) Laurent-series gives a better approximation to the true value - although actually the convergence-radius is still zero! The purpose of those iterations shifting z_h towards zero is to shift the position, from where the Laurent series with the argument z_h begins to diverge, to higher indexes and thus to get more accuracy. A combination of h=32 and n=64 for arguments $|z| \le 1$ is apparently enough for 40 correct digits. (see remark (§1)) -Finally, to show the effect of the h=32 iteration at work I provide below the partial sums of the Laurent-series for z=1 in comparision to h=4. -In the first example I use h=4 and in the second example I use h=32 . -In the table k is the index of the coefficient up to where the partial sums are computed. ps_k indicates the partial sum using z_h which is the h 'th iterate from z_0=1 . But for convenience the term for the logarithm and the h-term are always included so we can compare the sum up to this term with the accurate value a_1 for the Abel-function at z_1 : - k ps_k error: a_1 - ps_k iteration height h=4 - 0 3.05810608515 -0.0315166345810 - 2 3.05810608515 -0.0315166345810 - 4 3.08773833843 -0.00188438129901 - 6 3.08945198975 -0.000170729978211 - 8 3.08960570369 -0.0000170160371392 - 10 3.08962115403 -0.00000156570332243 - 12 3.08962261968 -0.000000100050871450 - 14 3.08962272183 0.00000000210083986271 - 16 3.08962272142 0.00000000169099804938 - 18 3.08962271989 1.62746538183E-10 - 20 3.08962271970 -2.97721970306E-11 - ... - 50 3.08962271973 -3.98604755990E-18 - 52 3.08962271973 7.74229820435E-19 - 54 3.08962271973 1.21098784690E-18 - 56 3.08962271973 -6.22150631919E-20 - 58 3.08962271973 -3.98357488277E-19 - 60 3.08962271973 -3.38541477910E-20 - 62 3.08962271973 1.42850133024E-19 - -We see, that with iteration-height h=4 we arrive at an absolute error smaller than 1e-18 at the 64. term. -And below, iteration-height h=32 provides accuracy to an absolute error of smaller than 1e-40 with that 64 terms used: - k ps_k error: a_1 - ps_k iteration height h=32 - 0 3.08337725463 -0.00624546510435 - 2 3.08337725463 -0.00624546510435 - 4 3.08954701281 -0.0000757069234782 - 6 3.08962130264 -0.00000141708899538 - 8 3.08962269011 -0.0000000296188288642 - 10 3.08962271915 -0.000000000581829933894 - 12 3.08962271972 -8.31025344698E-12 - ... ... ... - 52 3.08962271973 -3.06907747463E-37 - 54 3.08962271973 5.27409063179E-37 - 56 3.08962271973 2.10119895640E-38 - 58 3.08962271973 -6.82487772781E-39 - 60 3.08962271973 -5.39925105785E-40 - 62 3.08962271973 9.44571568505E-41 - - -(§1): a Noerlund-summation, as I have it proposed in some treatizes for the evaluation of the fractional iterates of the $\exp(x)-1$ might give arbitrary approximations as well, but it seems now to me that such a summation-procedure were at most needed here for theoretical reasons to prove the summability of the Laurent-series for the Abel-function. -(§2): A short overview over the first 512 coefficients of the abel_inc()-series: - index value index value index value index value - 0 3.000000000 47 0 92 -0.005185699555 496 4.633504372E168 - 1 0 48 -0.00000003870320993 93 0 497 0 - 2 0 49 0 94 0.01347223160 498 -4.983759375E169 - 3 0 50 0.000000006386371562 95 0 499 0 - 4 0.07523809524 51 0 96 0.03559427183 500 -8.187596780E170 - 5 0 52 0.00000006229599636 97 0 501 0 - 6 0.01104761905 53 0 98 -0.06747379661 502 8.333103850E171 - 7 0 54 0.00000001451248843 99 0 503 0 - 8 0.002516127259 55 0 100 -0.2528544049 504 1.467790435E173 - 9 0 56 -0.0000001074166810 101 0 505 0 - 10 0.0006421408926 57 0 102 0.3439480705 506 -1.412786474E174 - 11 0 58 -0.00000007200630916 103 0 507 0 - 12 0.0001546666126 59 0 104 1.879638019 508 -2.669450403E175 - 13 0 60 0.0000001982539503 105 0 ... ... - 14 0.00002737060121 61 0 106 -1.706858981 - 15 0 62 0.0000002440284845 107 0 - 16 -0.0000002788226624 63 0 108 -14.69827943 - 17 0 64 -0.0000003845753696 109 0 - 18 -0.000002639853064 65 0 110 7.295584305 - 19 0 66 -0.0000007917263057 ... ... - 20 -0.0000008443665796 67 0 - ... ... ... ...<|endoftext|> -TITLE: Categorical Schur's Lemma -QUESTION [10 upvotes]: In attempt to prove (and compute) a formula for the dimensions of the HOMFLY homology -of the (p,q)-torus knot one could try to follow original proof by Jones of a formula for -HOMFLY polynomial of the torus knot. One of the key moments of the Jones proof -is Schur-Weyl duality and Schur's Lemma. Then the question: -What is a categorical analogues of the Schur's Lemma and Schur-Weyl duality. -In general what are the obstacles for finding formula for the HOMLFLY homology -of the (p,q)-torus knot. -Any refs and suggestions would be greatly appreciated. - -REPLY [2 votes]: One categorical analogue of Schur's lemma is the definition of a simple object in, say, an abelian category. This is essentially an object $V$ for which a generalization of Schur's lemma holds. In particular, all morphisms $f \colon V \to V$ are either isomorphisms or equal zero. -If the category is enriched over vector spaces over some field $k$, it follows that the vector space $\mathrm{hom}(V,V)$ becomes a division algebra over $k$. If $k$ is algebraically complete, it then follows that $\mathrm{hom}(V,V) \cong k$.<|endoftext|> -TITLE: Is there a "classical" proof of this $j$-value congruence? -QUESTION [64 upvotes]: Let $j: \mathbf{C} - \mathbf{R} \rightarrow \mathbf{C}$ denote the classical $j$-function from the theory of elliptic functions. That is, $j(\tau)$ is the $j$-invariant of the elliptic curve $\mathbf{C}/(\mathbf{Z} + \mathbf{Z}\tau)$, so this has the familiar expansion $1/q_{\tau} + 744 + \cdots$ where $q_{\tau} = \exp(2\pi i_{\tau} \cdot \tau)$ with $i_{\tau}$ denoting the square root of $-1$ lying in the same connected component of $\mathbf{C} - \mathbf{R}$ as $\tau$ does. (In particular, $\overline{j(\tau)} = j(\overline{\tau})$ and -$j(-\tau) = j(\tau)$.) -Let $\zeta$ denote a primitive cube root of unity in $\mathbf{C}$, and let $n \ge 1$ be an integer. The value $j(n \zeta)$ is an algebraic integer (lying in the ring class field of conductor $n$ over $\mathbf{Q}(\zeta)$, by CM theory). In more conceptual terms, this is the $j$-invariant of the elliptic curve $\mathbf{C}/\mathcal{O}_n$ where $\mathcal{O}_n$ is the unique order of conductor $n$ in the subfield of $\mathbf{C}$ generated by the cube roots of unity. -I was recently asked by someone (based on numerical evidence for $n$ up to 20 or so) to prove that $j(n \zeta) \equiv 0 \bmod 81$ for $n \equiv 1 \bmod 3$ and $j(n \zeta) \equiv -27 \bmod 81$ for $n \equiv -1 \bmod 3$ (congruence modulo 81 in the ring of algebraic integers in $\mathbf{C}$); beware that this is sensitive to the fact that $n > 0$ but is -independent of the choice of $\zeta$ in $\mathbf{C}$ (contemplate the behavior of $j$ with negation, as noted above). For $n = 1, 2$ these congruences are easily verified by hand. I eventually came up with an affirmative proof in general using the Grothendieck-Messing crystalline Dieudonne theory for $3$-divisible groups and some concrete calculations. -This leads to the following question. It is natural to use deformation theory (which is what the Grothendieck-Messing theory is part of) to prove congruential properties of $j$-values, but hauling out such high-powered theoretical machinery to prove something as down to earth as a mod 81 congruence on $j(n \zeta)$ for $n > 0$ (not divisible by 3) may seem like killing a fly with a sledgehammer. So...does anyone see a way to determine $j(n \zeta) \bmod 81$ (for all integers $n > 0$ not divisible by 3) by using pre-Grothendieck technology? -[Note that it is equivalent to prove that for $n > 0$ not divisible by 3, $j(n \zeta) \bmod 81$ only depends on $n \bmod 3$, as we can then compute for $n = 1, 2$ to conclude. But it seems not obvious at the outset that this congruence class in the ring of algebraic integers in $\mathbf{C}$ is represented by a rational integer, let alone one that only depends on $n \bmod 3$. It is also relatively easy to prove that $j(n\zeta) \equiv 0 \bmod 27$, so the -real difficulty lies in improving things to work modulo 81.] - -REPLY [48 votes]: Since I wasn't yet reading Mathoverflow at the time, I didn't see this question until Brian e-mailed it to me in January. I was eventually able to give a more elementary proof by applying formulas of Vélu to the $n$-isogeny from a curve with $j=0$ to one with $j=j(n\zeta)$, which in fact determined $j(n\zeta) \bmod 3^{9/2}$. With some more work I then proved a congruence $\bmod 3^5$ in the case $n \equiv -1 \bmod 3$, and even obtained some information $\bmod 3^6$ for $n \equiv +1 \bmod 3$. Namely: -@ if $n \equiv -1 \bmod 3$ then $j(n\zeta) \equiv -54 \bmod 3^5$; and -@ if $n \equiv 1 \bmod 3$ then $j(n\zeta)$ has valuation at least $9/2$ at $3$, and every conjugate is congruent to either $0$ or $\pm 324 \sqrt{3} \bmod 3^6$. -Brian soon replied that he can understand these refined congruences using the same "Grothendieck-Messing crystalline Dieudonné theory" that he applied to get the congruences $\bmod 3^4$. -Here's a link to a talk I gave here in mid-February that recounts this story and outlines the proofs and some additional information: http://www.math.harvard.edu/~elkies/j3.pdf<|endoftext|> -TITLE: Pushforward and pullback. -QUESTION [20 upvotes]: Often I say "pushforward" or "pullback", but I do not know exact -meaning of these words. -Each time I see a map $f\colon X \to Y$, plus: - -I have some object $O_X$ associated with $X$ (say measure or subset); -the map $f$ gives me a natural way to find -corresponding object $O_Y$ associated with $Y$. - -Then I say "$O_Y$ is pushforward of $O_X$" and I write $O_Y=f_*O_X$. -If I switch $X$ and $Y$ in (1) and (2), I say "$O_X$ is pullback of -$O_Y$" and I write $O_X=f^*O_Y$. -I do this all the time, and no one complains, but I do not feel that -it is right... - -Could someone explain the right way to think about "pushforward" and "pullback"? - -REPLY [4 votes]: I think that in general, the word "pullback" is associated to (some) contravariant functors, and "pushforward" is associated to (some) covariant functors. -To be more specific, consider the measure example. There is an endfunctor $M$ of the category $Set$, which maps a set $X$ to the set $M(X)$ of all measures on $X$, and a function $f:X\to Y$ to a function $M(f)=f_*:M(X)\to M(Y)$, defined so that $f_*(\mu)$ is a measure on $Y$, given by $f_*(\mu)(U)=\mu(f^{-1}(U))$ for all $U\subseteq Y$ for which the right-hand side is defined. Since $M$ is a covariant functor, we might call $M(f)(\mu)=f_*(\mu)$ a pushforward of a measure $\mu$ on $X$. -As an example of pullback, consider the pullback from cohomology. -There is the functor $H^*$ from the category $Top$ of topological spaces to $Set$, which maps a topological space $X$ to the set (in fact, a graded ring) $H^*(X)$ of cohomology classes of $X$, and a continuous function $f:X\to Y$ to the function (in fact, a homomorphism of graded rings) $H^*(f)=f^*:H^*(Y)\to H^*(X)$ on cohomology classes. Since $H^*$ is a contravariant functor, we may call $H^*(f)(\sigma)=f^*(\sigma)$ a pullback of a cohomology class $\sigma$ on $Y$.<|endoftext|> -TITLE: Is there a q-analog to the braid group? -QUESTION [24 upvotes]: The braid group $B_n$ on $n$ strands fits into a short exact sequence of groups: -$$ 1 \longrightarrow P_n \longrightarrow B_n \longrightarrow S_n \longrightarrow 1,$$ -where $S_n$ is the symmetric group on the strands, and $P_n$ is the normal subgroup of braids that do not permute the strands. -Since symmetric groups are, in some sense, ``general linear groups over the field with one element,'' perhaps there is some corresponding short exact sequence ending with $GL_n(F_q)$ that specializes to the exact sequence above as $q \rightarrow 1$. -In the spirit of the exact sequence above, is there a $q$-analog to the braid group? - -REPLY [4 votes]: There is a nice paper by Etingof and Rains about a non-standard q-deformations of S_n: -http://xxx.lanl.gov/abs/math/0409261 -They deform not quadratic relations but the braid relations. -It seems to be relevant to the discussion.<|endoftext|> -TITLE: Infinite-dimensional normed division algebras -QUESTION [30 upvotes]: Let's say a normed division algebra is a real vector space $A$ equipped with a bilinear product, an element $1$ such that $1a = a = a1$, and a norm obeying $|ab| = |a| |b|$. -There are only four finite-dimensional normed division algebras: the real numbers, the complex numbers, the quaternions and the octonions. This was proved by Hurwitz in 1898: - -Adolf Hurwitz, Über die Composition der quadratischen Formen von beliebig vielen Variabeln, Nachr. Ges. Wiss. Göttingen (1898), 309-316. - -Are there any infinite-dimensional normed division algebras? If so, how many are there? - -REPLY [27 votes]: A MathSciNet search reveals a paper by Urbanik and Wright (Absolute-valued algebras. Proc. Amer. Math. Soc. 11 (1960), 861–866) where it is proved that an arbitrary real normed algebra (with unit) is in fact a finite-dimensional division algebra, hence is one of the four mentioned in the OP. A key piece of the argument (Theorem 1) is to show that such an algebra $A$ is algebraic, in the sense that if $x \in A$, then the subalgebra of $A$ generated by $x$ is finite-dimensional. The authors then invoke a theorem of A. A. Albert stating that a unital algebraic algebra is a finite-dimensional division algebra.<|endoftext|> -TITLE: local behavior of a finite Borel measure -QUESTION [9 upvotes]: Let $\mu$ be a finite Borel measure on $\mathbb{R}^n$. I am interested in how does $\mu(B(x,r))$ behave, where $B(x,r)$ is the open ball of radius $r$ centered at $x$. For instance, as far as I recall, for each $\alpha \in [0,n]$, there exists finite $\mu$ so that $\mu(B(x,r)) \sim r^{\alpha}$ for $\mu$-a.e. $x$, which are called dimensionally-regular measures (with constant local dimension $\alpha$). -Here is my question: does there exist finite Borel measure$\mu$ such that that $\mu(B(x,r))$ vanishes superpolynomially fast, say, $\sim e^{-\frac{1}{r}}$, i.e., -$\mu\left(\left\{x: \liminf_{r \to 0} r |\log \mu (B(x,r)) | > 0 \right\}\right) > 0$? - -REPLY [3 votes]: If you are willing to replace $\mathbb{R}^n$ by a Banach space $B$, then it is perfectly possible. For example, the Wiener measure $\mu$ on the space of continuous function from $[0,1]$ to $\mathbb{R}$ satisfies that for $\mu$-almost every $x$, $\mu(B(x,r)) \sim \exp(-1/r^2)$ for small $r$. (The keyword for this is "small ball estimates".) If you are willing to forgo separability of $B$, then you can even construct probability measures with the property that, for every $x \in B$, $\mu(B(x,1)) = 0$! -The canonical example for this is to take for $\mu$ the law of a sequence of i.i.d. normal random variables and for $B$ the space of sequences $x = (x_n)$ such that -$$ -\|x\| := \sup_{n\ge 1} \Big(2^{-n}\sum_{k=2^{n-1}}^{2^n-1} x_k^2\Big)^{1/2} < \infty\;. -$$ -The claim then follows from the strong law of large numbers.<|endoftext|> -TITLE: Transfer homomorphisms with coefficients -QUESTION [6 upvotes]: In group cohomology, for $H$ a finite-index subgroup of $G$ and $M$ a $G$-module, there is a transfer (or corestriction) map $Cor : H^* (H;M) \to H^*(G;M)$. -In homotopy theory, there is a transfer map for finite covering spaces $\bar{X} \to X$, and it exists for all coefficient systems on $X$. It is given on homology (say) by sending a small simplex in $X$ to its finitely-many lifts to $\bar{X}$. The group transfer is obtained from the topological one by the finite covering space $BH \to BG$. -There is a fancier transfer, due to Becker and Gottlieb, for any map $E \to B$ whose homotopy fibre is stably equivalent to a finite complex. Can this be extended to give a transfer map for coefficient systems on $B$? - -REPLY [3 votes]: I think what Charles has written up above is closely related to what Becker and Gottlieb did in -Becker, J. C.; Gottlieb, D. H. -Transfer maps for fibrations and duality. -Compositio Math. 33 (1976), no. 2, 107–133. -They work fiberwise, but that is equivalent to working equivariantly in the sense that a $G$-space $X$ gives rise to a fibration $X \times_G EG \to BG$ (Borel construction; this induces a Quillen equivalence).<|endoftext|> -TITLE: Borromean braids -QUESTION [12 upvotes]: Consider the Kernel $K_n$ of the natural group homomorphism from the $n$-th braid group to the symmetric group. Then one can delete the $m$-th braid. This is a well defined homomorphism $d_m:K_n\rightarrow K_{n-1}$. So is there for every $n\in \mathbb{N}$ a braid $1\neq b\in K_n$ with $d_m(b)=0$ for all $m$. -This is clearly true for $n=2$, as $K_1$ is trivial and it is also true for $n=2$ (The "standard" braid does the job). What about higher $n$. Is there a nice construction, that works for every $n$ ? - -REPLY [7 votes]: Ted Stanford has a paper on this topic, Brunnian braids and some of their generalizations (arXiv:math/9907072), giving a set of generators for the kernel.<|endoftext|> -TITLE: Playing an (invertible) matrix game with two players -QUESTION [16 upvotes]: Players $A$ and $B$ take an empty $n \times n$ matrix and place, one by one, an element (say, a rational number) in an unoccupied place of this matrix. Player $A$ starts. The game ends if there is no move left. Player $A$ wins if the matrix is invertible; player $B$ wins if it is not. -For a given $n > 0$, is there a winning strategy for one of the two players? -It is not hard to show that for $n = 3$, player $A$ can win. Also if $n$ is even player $B$ has a winning strategy. But what if $n > 3$ is odd? - -REPLY [3 votes]: Player A wins the trivial n=1 case by playing any non-zero number in (1,1). -For all even n, player B wins by using symmetry a la a horizontal mirror. -As Ben points out in the comments, if n = 3, player B can force a win. I had a long demonstration written out, but I decided against it (if you want, I can put it in later). -Anyway, as for the general case, after a little searching, I found a paper called "A determinantal version of the Frobenius - König Theorem" by D. J. Hartfiel and Raphael Loewy, which can be purchased here. -The abstract, at least, says that given an n by n matrix A of, say, rational numbers, if the determinant is zero, then A must contain an r by s submatrix B such that r + s = n + p, and rank(B) ≤ p - 1 (no more than p - 1 linearly independent rows), for some positive integer p. This means that if we have, say, a 5x5 matrix whose determinant is zero, then there exists a submatrix B in A such that B is: - -a 1x5, 2x4, 3x3, 4x2, or 5x1 matrix of 0s -a 2x5, 3x4, 4x3, or 5x2 matrix whose rows are all scalar multiples of each other -a 3x5, 4x4, or 5x3 matrix with no more than two linearly independent rows -a 4x5 or 5x4 matrix with no more than three linearly independent rows -a 5x5 matrix with no more than four linearly independent rows (duh) - -While it doesn't say so explicitly, I think that it's a biconditional, so if player B manages to get one of these in the matrix, then she will win. However, even if it isn't biconditional, if player A can prevent any of those forming, he will win. -Of these two, I believe it would be easier for player A to prevent any of these forming than it would be for player B to force one of these, but I haven't given that in particular a great deal of thought. I hope this is helpful.<|endoftext|> -TITLE: Abelian groups as fundamental groups of topological groups -QUESTION [8 upvotes]: Hi, -It is well known that the fundamental group of a topological group is abelian, and that every group is the fundamental group of some topological space. -My question is: Does every abelian group arise as the fundamental group of some topological group or are there other restrictions? - -REPLY [9 votes]: I'd like to add a quick little explanation to Georges's already sufficient answer (I'm sure this explanation is in Baez's post, but it can be said in a few lines here). -One of the morals of the famous Eckmann-Hilton lemma is that an abelian group is an abelian group object in the category of groups. Now, the classifying space functor $B: Grp \to Top$ preserves products (here $Top$ should be a convenient category of spaces like compactly generated weak Hausdorff spaces; see nLab). But product-preserving functors take algebraic gizmos (like abelian group objects) to algebraic gizmos. So $B$ takes abelian groups $G$ = abelian group objects in $Grp$ to abelian group objects in $Top$, i.e., topological abelian groups. And since we have a universal covering fibration $G \to EG \to BG$, we get $\pi_1(BG) \cong G$. -A refinement of this observation shows that we have a classifying space functor -$$B: TopAb \to TopAb$$ -which can be used to realize topological abelian groups as infinite loop spaces.<|endoftext|> -TITLE: Four polynomials representing all integers modulo m -QUESTION [5 upvotes]: I would like to classify the integers $m \geq 2$ for which the four quadratic polynomials -$3k^2$, $3k^2+2k$, $3k^2+3k+1$, and $3k^2+5k+2$ together represent all integers modulo $m$. That is, every integer modulo $m$ should be in the range of at least one of these polynomials (where all operations are carried out modulo $m$). Computer evidence suggests that this holds if and only if $m$ is one of the following: -$7, 10, 19, 2^j, 3^j, 5^j, 11^j, 13^j, 41^j, 2\cdot3^j, 5\cdot3^j$, where $j \geq 1$. -Does someone see how to prove this? Thank you. - -REPLY [7 votes]: For a prime $p>2$, fix a nonsquare $c$. If you find $y$ such that $y/3$ is a non-square (i.e. $y/3=cx^2, x\ne0$) and $y/3 - 1/9 = cz^2, z\ne 0$, then $y$ is not represented by the first two polynomials and I can't be bothered completing the square to write the conditions for the other two. Bottom line is, you find such a $y$ if you can find a point on a curve over the finite field -$\mathbb{F}_p$. By Weil, this will happen as soon as $p$ is large enough. So your $m$ can only have prime factors from a finite set. Should be downhill from here.<|endoftext|> -TITLE: Cohomology of the unitary group -QUESTION [8 upvotes]: The de-Rham cohomology ring of U(n) is the exterior algebra generated by the odd-dimensional classes x_1, x_3, ..., x_(2n-1). Moreover, on a Lie group every cohomology class is represented by a unique invariant form (both left and right). I ask two questions: -1) if we represent U(n) with matrices U = [z_ij], what is an explicit expression of a generator as an invariant form of U(n), in terms of the differentials dz_ij, for each odd degree between 1 and 2n-1? -2) in the 1-dimensional space of the invariant forms of a fixed degree (multiples of x_i), which are the two (opposite to each other) which represent the real image of a generator of the integral cohomology? - -REPLY [10 votes]: Here's an answer to the first question. -Let $\theta = U^{-1} dU$ be the left-invariant Maurer-Cartan one-form: it is a matrix of one-forms. Then -$$\omega_{2n+1} = \mathrm{Tr} \theta^{2n+1}$$ -is the desired bi-invariant form. Here $\theta^{2n+1}$ stands both for the wedge and matrix product of $\theta$ with itself $2n+1$ times. -I'm not sure I understand the second question: what "two" representatives are you talking about? I think the canonical representative will be certain multiple of $\omega_{2n+1}$. I'll try to fish out a reference later on.<|endoftext|> -TITLE: A Question concerning the Fourier Transform of $\mathbb{R}$ -QUESTION [7 upvotes]: Consider the classical Schwartz space $\mathcal{S}(\mathbb{R})$ together with the Fourier transform $\mathcal{F} : \mathcal{S}(\mathbb{R}) \rightarrow \mathcal{S}( \mathbb{R})$. -Consider the subspace $V$ of the even, smooth functions on the interval $[-1,1]$. -Can you construct a (bounded) operator $D:\mathcal{S}(\mathbb{R}) \rightarrow \mathcal{S}(\mathbb{R}) $ such that -$$ D \mathcal{F} v = 0, \quad Dv=v \qquad\forall v \in V ?$$ -Observe that by Paley-Wiener, the intersection $\mathcal{F}V \cap V =0$ is trivial. -What is the associated Schwartz kernel? - -REPLY [5 votes]: You don't need to do things the rough way; there is enough freedom for the smooth approach. -Take any even $C_0^\infty$ descent $\Phi$ from $[-1,1]$ and define $Pf=\Phi f$ and $Qf=\mathcal F^{-1}(\Phi\mathcal F f)$. Now take the standard $D=I-(I-PQ)^{-1}(1-P)$. This works in $L^2$ for the same reason as it does with orthogonal projections: $PQ$ is a contraction. The good news is that $PQ$ maps $L^2$ to $S$ continuously and $(I-PQ)^{-1}=I+PQ(I-PQ)^{-1}$, so the resulting operator is bounded from $S$ to $S$ as well. -The kernel can be "found" by expanding $D$ into power series that converges geometrically but, since this construction involves an arbitrary smooth cutoff, to write an explicit formula seems quite hopeless.<|endoftext|> -TITLE: Maximal dimension of linear system of curves of fixed genus on a surface -QUESTION [6 upvotes]: To a (projective smooth) algebraic surface $S$ over an algebraically closed field and a divisor $D$ of $S$, we can associate $n= \dim |D|$ and $g$, the genus of a generic member of $|D|$. I would like to fix $g$ and vary $S,D$ so as to make $n$ as large as possible. Is $n$ unbounded and, if not, what is the optimal bound? I am mainly interested in the case of positive characteristic but an answer over the complex numbers would be welcome too. - -REPLY [11 votes]: $n$ is unbounded, as it is shown by the following example. -EDIT: my example did not work, as pointed out by quim in the comments. -The example he suggests however works: take $S$ the blowup of $P^2$ at a point $x$ and $D$ the strict transform of a curve of degree $d$ with a singular point of multiplicity $d-1$ at $x$. The general $D$ is smooth and $|D|$ has dimension $2d$.\ -Of course, a similar construction can be used to construct examples with $g>0$. -On the other hand, if the Kodaira dimension of $S$ is $\ge 0$ and $D$ is irreducible, then $n=\dim|D|$ is bounded by $g$ by the following argument. -Up to blowing up $S$ we may assume that the general $D$ is smooth. -Let $m>0$ be such that $mK_S\ge 0$. If $n>0$, then $D$ is not in the fixed part of $|mK_S|$, hence $K_SD\ge 0$. Hence by the adjunction formula $D|_D$ is a divisor of $D$ of degree $\le 2g-2$ and therefore it satisfies $2\dim|D|_D|\le \deg D$ (this is Clifford's theorem if $D$ is special and it is trivially true otherwise). -So we have $\dim |D|\le \dim|D|_D|+1\le D^2/2+1\le (D^2+K_SD)/2+1=g$. -I, hence $K_S|_D$ is effective and $D|_D$ is special. Then Clifford's theorem and the adjunction formula give $2(n-1)\le D^2\le D^2+K_SD=2g-2$.<|endoftext|> -TITLE: Expressions involving Eulerian numbers of the second kind: trying to show $\sum_{m=0}^{n} (-1)^m(m)m!(2n-m-2)!\left\langle\left\langle n\atop m\right\rangle\right\rangle\neq0$ for even $n$. -QUESTION [10 upvotes]: Considering the success of a previous question involving Eulerian numbers, I thought I might throw this question into the mix. It comes from some localization computations in GW theory, but in this form is purely combinatorial. -Eulerian numbers of the second kind are defined by the recursion relation -$$\left\langle\left\langle n\atop m\right\rangle\right\rangle = (m+1)\left\langle\left\langle n-1\atop m\right\rangle\right\rangle+(2n-m-1)\left\langle\left\langle n-1\atop m-1\right\rangle\right\rangle$$ -with the initial conditions $\left\langle\left\langle n\atop 0\right\rangle\right\rangle=1$ and $\left\langle\left\langle n\atop m\right\rangle\right\rangle$ = 0 for $m\geq n$. -For references, see: -http://en.wikipedia.org/wiki/Eulerian_number and http://oeis.org/classic/A008517. -The following three statements are known: - -$\sum_{m=0}^{n} - (-1)^mm!(2n-m-2)!\left\langle\left\langle - n\atop - m\right\rangle\right\rangle=0$ for -all $n$; -$\sum_{m=0}^{n} -(-1)^m(m)m!(2n-m-2)!\left\langle\left\langle -n\atop -m\right\rangle\right\rangle=0$ for -odd $n$; -$\sum_{m=0}^{n} -(-1)^m(m+1)!(2n-m)!\left\langle\left\langle -n\atop -m\right\rangle\right\rangle=0$ for -even $n$; - -(2 and 3 are equivalent). -Question: Show that the expression in the second statement is non-zero for even $n$, i.e. show -$\sum_{m=0}^{n} -(-1)^m(m)m!(2n-m-2)!\left\langle\left\langle -n\atop -m\right\rangle\right\rangle\neq0$ for - even $n$. -Certainly we've been checking this on a computer for modest values of $n$. - -REPLY [14 votes]: Surprise: your expression is a multiple of a Bernoulli number: -$$A(n)=-\frac{(2n)!}{n}B_n $$ -for all $n>1$, and the Bernoulli numbers are indeed vanishing exactly for all odd integers greater than 1. This follows computing the generating function for the $A(n)$, on the lines of the preceding answer and comments. Note that your identity can be written in the form -$$\sum_{m=0}^n (-1)^n\frac{ \left\langle \!\!\left\langle n\atop m\right\rangle \!\!\right\rangle }{\left( 2n+1\atop m+1\right)}=2B_{n+1}\, .$$ -After that, I had a vague memory of a similar relation about the Eulerian numbers of first kind, that it is the fourth identity reported here: -$$\sum_{m=0}^n (-1)^n\frac{ \left\langle n\atop m\right\rangle }{\left( n\atop m\right)}=(n+1)B_n\, .$$ -Checking the source of the latter may be useful for you, and will possibly give a reference for your identity (I'm quite confident that it should be written somewhere; hopefully some expert may recognize and provide a reference). In case, I will add the details of the (quite standard) computation of the generating function of yours $A(n)$ . -Details. Here is a generating function for yor sequence as a real function. This should allow a nice proof of the Bernoulli number formula, provided one is able to compute the last integral. -For $t\in\mathbb{R}$ and $x\in [0,\infty)$ let $$\psi(t,x):=W(x e^{x+t})=L^{-1}\big(t+L(x)\big)$$ -where $W$ is the Lambert function and $L:\mathbb{R_+}\to\mathbb{R}$ is the invertible function $L(x):=x+\log(x)$. So $\psi$ is the general solution of the Cauchy problem for the autonomous ODE (the flow) -$$\psi_t=\frac \psi {1+\psi} $$ -$$\psi(0,x)=x$$ -Thus it also solves the linear first-order PDE -$$\psi_t- \frac x {1+x} \psi_x =0$$ -and all derivatives w.r.to $t$ have the form -$$\partial_t^n\psi= v_n(\psi)$$ -for a recursively defined sequence $v_n(x)$ -$$v_0(x)=x$$ -$$v_{n+1}(x)=\frac x {1+x}v'_n(x)$$ -So the $v_n$ are rational functions with poles at $-1$; $v_n(x)=O(x^{-n})$ as $x\to+\infty$ and in fact -$$v_n(x)=-x(1+x)^{-2n+1}E_n(-x)$$ -where $E_n$ are the Eulerian polynomial of second kind (these $v_n$ are just a simple modification of to the previously defined sequence $U_n$ ). -For all $t\in\mathbb{R}$, all $r>0$ and $n\ge2$, it's easy to see that -$$\sup_{|t|\le r}\, \big| v_n\big(\psi(t,x)\big) \big|:=g_{r,n}(x)\in L^1(\mathbb{R}_+),$$ -which allows (by the dominated convergence theorem) to differentiate under the sign of integral the function -$$h(t):=\int_0^\infty \psi_{tt}(t,x)dx=\int_0^\infty v_2\big(\psi(t,x)\big)dx\, ,$$ -so that -$$h^{(n)}(t)=\int_0^\infty v_{n+2}\big(\psi(t,x)\big)dx,$$ -and in particular we have for $n\ge 1$ -$$h^{n-1}(0)=\int_0^\infty v_{n+1}(x)dx=-\frac {A(n)}{(2n)!}$$ -The relation $$-\frac {A(n)}{(2n)!} = \frac {B_n} n$$ for all $n\ge 1$ now writes: -$$\int_0^\infty \frac{W(xe^{x+t})}{\big(1+W(xe^{x+t})\big)^3}dx=\frac1 t - \frac 1 {e^t - 1}.$$<|endoftext|> -TITLE: Heat kernel estimates and Gaussian estimates for semigroups, good reference? -QUESTION [6 upvotes]: Hi, it seems like a big field and I'm having trouble getting some solid/classic references to get me started. -If $U \subset \mathbb{R}^d$ is a bounded domain with, say, $C^2$-boundary $\partial U$ and $(S(t),t \ge 0)$ is the Dirichlet heat semigroup on $L^p(U)$ then $(S(t) f)(x) = \int_U G_U(t,x,y) f(y)\,dy$ for $f \in L^p(U)$ where $G_U$ is the Dirichlet heat kernel. -I would like to find a good reference for bounds of the type: -$$ -\left|\frac{\partial G_U}{\partial \nu_y} (t,x,y)\right| \le C_1 t^{-(d + k)/2} \exp\left(- \frac{|x-y|^2}{C_2 t}\right) -$$ -It seems like it should be classic result? I found Aronson's 1968 paper but it only contains estimates for the kernel and not the 'derivatives'. I can find lots of recent papers on manifolds and such but I am just looking for a solid and accessible reference for my simple case. -Further, if one has a semigroup $(T(t), t \ge 0)$ with a kernel $k(t,x,y)$ that has a pointwise Gaussian estimate $|k(t,x,y)| \le c_1 e^{\omega t} t^{-d/2} e^{-|x-y|^2/(c_2 t)}$, do the estimates for the 'derivatives' as above follow readily? or are more assumptions needed? Again, a reference to point me in the right direction would be greatly appreciated. -Thanks. - -REPLY [4 votes]: I would like to know a complete answer to this question myself, since this is useful in a number of situations. From what I know, contrary to estimates for the kernel which are a quite general phenomenon (see the classical book by Davies for your case), estimates for the derivatives are much more subtle and are connected with several properties of the generator such as $L^p$ boundedness for the associated Riesz operator when $p>2$ (while when $p<2$ the estimates for the derivatives are not needed). -This paper by ter Elst et al. gives some info in the case of elliptic operators, and on manifold this is studied in a 2004 paper by Auscher, Coulhon, Duong and Hofmann (Riesz transform on manifolds and heat kernel regularity) which should be on arXiv.<|endoftext|> -TITLE: Harmonic level sets and boundary data -QUESTION [8 upvotes]: This is probably a classic problem, so a good reference book or paper to get me started on this type of question would be great: -Let $\mathbb{D} \subset \mathbb{C}$ be the unit disk with boundary $\mathbb{T}$. Take a function $f : \mathbb{T} \to \mathbb{R}$ and let $u_f$ be the harmonic extension to $\mathbb{D}$ using the Poisson kernel, i.e. the Poisson integral of $f$. - -What can be said about the levels sets of $u_f$ in relation to the data $f$? - -For example, If the data $f$ has certain behaviour such as, say, oscillatory taking positive and negative values. - -REPLY [2 votes]: I know that's been a while now that the question has been asked, but as I'm looking more or less into this topic, I think I should share some of my discoveries in the literature. I'm somewhat amazed at the really small number of occurrences of this theme. - -I. De Carli & S. Hudson, Geometric remarks on the level curves of harmonic functions (2010), MR2586969 -L. Flatto, D. Newman & H. Shapiro, The level curves of harmonic functions (1966), MR197755 -W. Boothby, The topology of the level curves of harmonic functions with critical points (1951), MR43456<|endoftext|> -TITLE: Is "subamenable" the same as amenable? -QUESTION [12 upvotes]: Let $G$ be a finitely generated group. Does the following condition imply the amenability of $G$: -there is a function $\mu:\mathcal{P}(G) \to [0,1]$ such that: - (subadditive) $\mu(G) = 1$, $\mu(A \cup B) \leq \mu(A) + \mu(B)$, and $A \subset B$ implies $\mu(A) \leq \mu(B)$, - (invariant) $\mu(A \cdot g) = \mu(A)$ for all $g$ in $G$, - (exhaustive) if $\{A_n : n <\infty\}$ is pairwise disjoint, then $\inf_n \mu(A_n) = 0$. -A group distinguishing this condition from amenability cannot contain $F_2$. -I am aware of the relationship to the (now solved) Maharam problem. -I am not expecting an answer so much as asking whether (and where) this question has been studied. - -REPLY [7 votes]: Modulo a (possibly severe) measurability issue, the only groups which admit an invariant submeasure are the amenable groups. -Let $G$ be a non-amenable group and $\mu$ be a $G$-invariant submeasure on $G$. Damien Gaboriau and Russell Lyons showed that every non-amenable group contains a random free subgroup. One way of putting this is to say that there exists a free and measure preserving Borel action of $G$ and $F_2$ on some standard probability space $(X,\alpha)$ and a cocycle -$$c \colon F_2 \times X \to G$$ -such that $c_x$ is injective for each $x \in X$ and -$$c(g,x) \cdot x = g\cdot x, \quad \forall g \in F_2,x \in X.$$ -Considering now the measure space $Z=G \times X$, we see that the action of $F_2$ on $Z$ given by $g(t,x) = (c(g,x)t,g \cdot x)$ admits a fundamental domain $Y \subset Z$. (The measure of $Y$ is typically not finite. To get a picture, $Y$ plays the role of a set of representatives of cosets in case $F_2$ is an honest subgroup of $G$.) -Let $W$ be a measurable subset of $Z$ and set $$\mu'(W) = \int_X \mu(W_x) d \alpha(x).$$ -Moreover, for every element $z$ in the full group of the equivalence relation generated by $G$, we see that $\mu'(zA) = \mu'(A)$. Indeed, the action of each such element on $Z$ is given by a measure preserving shuffle of the $x$-coordinate and an $x$-dependent shift in the $G$-coordinate. Both operations preserve the integral. -Let now $A \subset F_2$ and define -$\mu''(A) := \mu'(A \cdot Y).$ Since $F_2$ lies in the full group of the equivalence relation generated by the action of $G$, we can conclude that $\mu''$ is $F_2$-invariant. -It remains to analyze the implications of the assumption that $\mu$ is exhaustive. Let us assume that there exists an infinite disjoint family $(A_n)$ of subsets of $F_2$ such that $\mu''(A_n) \geq \delta>0$. Then, the functions $x \mapsto \mu(A_{n,x})$ are greater $\delta/2$ on a set of measure at least $\delta/2$. This, with an additional argument, (I did not check every detail) seems to contradict that $\mu$ is exhaustive. Hence, $\mu''$ is exhaustive as well. -This is a contradiction. Hence, every group which admits a $G$-invariant submeasure is amenable. -(The problem with this argument is that $x \mapsto \mu(W_x)$ need not be Borel, so that an additional argument is needed to make sense of the integral in the definition of $\mu'$. Maybe this can be cured, but I do not have time to think about it right now.)<|endoftext|> -TITLE: Philosophical consistency proof for set theory -QUESTION [20 upvotes]: In his ASL Gödel lecture (Las Vegas, Nevada, 2002), Harvey Friedman asked the following question: -Are there fundamental principles of a general philosophical nature which can be used to give consistency proofs of set theory, including the so-called large cardinal axioms? -Recently, he is able to isolate the following two fundamental philosophical principles and use them to prove the interpretability of various common sense thinking and set theory (with large cardinals). -(i) Plenitude Principle (PP): Anything that can happen will. -(ii) Indiscernibility Principle (IP): Any two horizons are indiscernible to observers on the basis of their extent. -(See his Concept Calculus article, e.g. link text) - -My main questions are: How confident are we that PP and IP are “true” ? More specifically, is it possible to “prove” or justify PP and IP rigorously? If yes, how? If not, why not? - -In my view, the ultimate justification of PP and IP would be to construct a (meta) system S based on PP and IP, and then prove its consistency and completeness. In the light of Gödel’s incompleteness theorem, I’m not sure that this can be done. But perhaps S is not recursively axiomatizable, and so Gödel’s incompleteness theorem would not apply to S. -My secondary question is: are there any logician (beside Friedman) who are working on this kind of research? -Update: July 2011 -Here is a rephrasing of the question by Timothy Chow that makes it closer to mathematical logic: - -Is there some precise mathematical statement, that has the flavor of IP or PP, which proves the consistency of all (or most) set-theoretic axioms that are generally accepted today (e.g., large cardinal axioms)? - -Update: The question has now been reopened. It is now time for people who can relate to the problem to answer it. - -REPLY [6 votes]: In a more recent paper of Friedman -Friedman, Harvey M., Concept calculus: much better than, Heller, Michael (ed.) et al., Infinity. New research frontiers. Based on the conference on new frontiers in research on infinity, San Marino, August 18–20, 2006. Cambridge: Cambridge University Press (ISBN 978-1-107-00387-3/hbk). 130-164 (2011). ZBL1269.03008, -the author defines a mathematically precise system MBT (much better than) and proves it and ZF have mutual interpretability. This establishes that if either is consistent they both are. These axioms have some of the flavor of IP and PP, but of course these axioms are not implied by IP and PP. -At the end of the paper Friedman claims a to be published result. STAR is defined as: - -There exists a star. I.e., something which is better -than something, and much better than everything it is better than. - - -We have shown that MBT + STAR can be interpreted in some large cardinals compatible with V = L, and some large cardinals compatible with V = L are interpretable in MBT + STAR." - -For PP and IP to be true, in a sense that can prove mathematics, they need to be stated precisely like the axioms in MBT. That formulation is much more complex than PP and IP as it must be to interpret ZF. -It is important to keep in mind that consistency does not imply truth. The statement that a formal system is consistent is equivalent to a statement of the form $\forall_{n\in\omega} r(n)$ where $r$ is a recursive relationship. This is equivalent to the halting problem for a particular Turing machine. -The following quote from Friedman is, I suspect, a big part of his and others interest in this work: - -STARTLING OBSERVATION. Any two natural theories S,T, known to interpret PA, are known (with small numbers of exceptions) to have: S is interpretable in T or T is interpretable in S. The exceptions are believed to also have comparability. - -It is an interesting and even startling observation, but it is worth keeping in mind that that rigorous theories are, among other things, recursive processes for enumerating theorems. To say that one theory is interpretable in another is to say a subset of one processes outputs are, in a specific well defined sense, isomorphic to the outputs of the process defined from the theory being interpreted. Whatever other significance it may have, this is a statement about unbounded recursive processes. -My personal view (see what is Mathematics About?) is that the only mathematics that can be interpreted as a properties of recursive processes is objectively true or false. This is based on the old idea that infinite is a potential that can never be realized. In this view Cantor's proof that the reals are not countable is an incompleteness theorem. The cardinal hierarchy is a hierarchy of the ways the real numbers provably definable in a formal system can always be expanded. Because of the Lowheheim Skolem theorem, we know such an interpretation exists. Interpretations that assume the absolutely uncountable are inevitably ambiguous at least as far as they can be expressed formally in the always finite universe that we seem to inhabit.<|endoftext|> -TITLE: What's the name of this flavor of n-category? -QUESTION [6 upvotes]: I'm looking for the name of a certain n-category definition. (Someone explained it to me a couple of years ago. I remember the definition, but not the name. Without the name it's difficult to search for a citation. I want the citation in order to explain something we're not doing in a paper.) -For background, consider the Moore loop space $\Omega_r$ of loops of length $r$ (that is, parameterized by the interval $[0,r]$). We have a strictly associative composition $\Omega_r\times \Omega_s\to \Omega_{r+s}$. The main idea of an "xxxx" n-category is to imitate this idea in higher dimensions. The $k$-morphisms are parameterised by $k$-dimensional rectangles with sides of lengths $r_1,\ldots,r_k$. Gluing rectangles together gives $k$ different strictly associative ways to compose $k$-morphisms. -Question: What is "xxxx" above? -Bonus question: What's the best (or any) citation for this idea? - -EDIT: It turns out the definition I was trying to remember is unpublished work of Ulrike Tillmann. But the version from Ronnie Brown linked to in David Roberts' answer is pretty similar (for my purposes, at least). - -REPLY [2 votes]: Simpson-semistrict $n$-categories could be what you're after: $n$-categories where everything except the unit laws holds strictly, generalising one of the crucial properties of Moore path spaces? It's not a specific definition of $n$-category, but a strictness property which can be applied within various definitions. -Carlos Simpson has conjectured that these are enough to model homotopy types; Moore path space show this in dimension 1. I know very little about the details of this myself, I'm afraid, but what I have read about it is mostly from these sources plus their links and discussions: - -Simpson, Homotopy types of strict 3-groupoids. -nlab: semi-strict $\infty$-category -nlab: Simpson’s conjecture (I can't figure out how to link this directly; the single-quote in the url seems to confuse markdown) -n-Category Café: Urs Schreiber, Semistrict Infinity-Categories and ω-Semi-Categories - -I believe several people have been making some progress on it recently; eg Makkai mentioned some results along these lines at the latest Octoberfest.<|endoftext|> -TITLE: Does pointwise convergence imply uniform convergence on a large subset? -QUESTION [77 upvotes]: Suppose $f_n$ is a sequence of real valued functions on $[0,1]$ which converges pointwise to zero. - -Is there an uncountable subset $A$ of $[0,1]$ so that $f_n$ converges uniformly on $A$? -Is there a subset $A$ of $[0,1]$ of cardinality the continuum so that $f_n$ converges uniformly on $A$? - -Background: Egoroff's theorem implies that the answer to (2) is yes if all $f_n$ are Lebesgue measurable. It is not hard to show that the answer to (1) is yes if you change "uncountable" to "infinite". -Motivation: I thought about this question while teaching real analysis this term but could not solve it even after looking at some books, googling, and asking some colleagues who are much smarter than I, so I assigned it as a problem (well, an extra credit problem) to my class. Unfortunately, no one gave me a solution. -ADDED 11-12-10: Thanks for all the great answers. I accepted Jonas' answer since it was the first one. - -REPLY [5 votes]: Also, Shinoda (1973) proved that if Martin's Axiom and $\neg CH$ hold, then the answer to 1) si affirmative.<|endoftext|> -TITLE: Minimum of Milnor number for the curve singularities of fixed multiplicity -QUESTION [10 upvotes]: An element $F\in \mathbb{C}[[x,y]]$ defines a germ of plane curve. -We assume $F(0,0)=0$. -The multiplicity $mult$ of the germ is defined to be a minimal number $i$ -such that $F\in m^i$ where $m=(x,y)$ is the maximal ideal in $\mathbb{C}[[x,y]]$. -Other standard invariants of the germ are Milnor number: -$$ -\mu=\dim \mathbb{C}[[x,y]]/(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}) -$$ -and delta invariant: -$$ \delta=\frac{\mu+r-1}{2}$$ -where $r$ is number of branches of curve $F=0$ at $(0,0)$. -Question: What is the minimum of $\delta$ and $\mu$ among the germs -of given multiplicity? - -REPLY [7 votes]: As Roy remarked in his answer, the delta invariant of the germ of plane curve singularity $f(x,y)=0$ at $p$ is equal to -$\delta(f) = \sum \frac{m_q(m_q-1)}{2}$, -where the sum is extended over all the points $q$ which are "infinitely near" to $p$ and $m_q$ denotes the multiplicity at $q$. -Then $\delta(f)$ is minimal among germs of a given multiplicity when there are no infinitely near points, in other words when the first blow-up of the germ is smooth. Of course there can be many analitically distinct germs satisfying this property: for instance, both the node and the ordinary cusp do the job among double points. -Now, since the Milnor number is equal to -$\mu(f)=2 \delta(f)-r(f)+1$, -where $r(f)$ is the number of branches, it follows that $\mu(f)$ is minimal among plane singularities of given multiplicity $n$ when $\delta(f)$ is minimal and the number of branches is maximal, in other words when $f=0$ is the "ordinary" $n$-ple point. -As an example, for the ordinary double point (node) $y^2=x^3+x^2$ we have -$(\delta, \mu)=(1,1)$, -whereas the ordinary cusp $y^2=x^3$ satisfies -$(\delta, \mu)=(1,2)$. -Summing up, the ordinary $n$-ple point is the only germ of plane curve singularity which minimizes both $\delta$ and $\mu$, and the corresponding values are -$(\delta, \mu)= (\frac{n(n-1)}{2}, (n-1)^2)$.<|endoftext|> -TITLE: Undergraduate math research -QUESTION [52 upvotes]: I believe this is the right place to ask this, so I was wondering if anyone could give me advice on research at the undergraduate level. -I was recently accepted into the McNair Scholars program. It is a preparatory program for students who want to go on to graduate school. I am expected to submit a research topic proposal in the middle of the spring semester and study it during the summer with a mentor. -Since I am currently in the B.S. Mathematics program and I want to get my Masters later. I figured that while my topic can be in any area, it should be in math since it is my main interest as well. -I am a junior at the moment and taking: One-Dimensional Real Analysis, Intro to Numerical Methods, and Abstract Algebra. I frequently search MathWorld and Wikipedia for topics that interest me, although I don't consider myself a brilliant student or particularly strong. I have begun speaking with professors about their research also. -I have not met any other students doing undergraduate math research and my current feeling is that many or all the problems in math are far beyond my ability to research them. This may seem a little defeatist but it seems mathematics is progressively becoming more specialized. I know that there are many areas emerging in Applied mathematics but they seem to be using much higher mathematics as well. -My current interest is Abstract Algebra and Game Theory and I have been considering if there are possibilities to apply the former to the latter. -So my questions are: -1) Are my beliefs about the possibilities of undergraduate research unfounded? -2) Where can I find online math journals? -3) How can I go about finding what has been explored in areas of interest. Should I search through Wikipedia and MathWorld bibliographies and or look in the library for research? -Thanks I hope someone can help to clarify and guide me. - -REPLY [3 votes]: Since you are interested in game theory, one area you could consider is "Algorithmic Game Theory" (basically Algorithm Design + Game Theory) It is a now a fairly hot area in theoretical computer science but still seems relatively approachable to an undergraduate with knowledge of game theory. If you can find someone willing/able to mentor you in this area I think there is good potential for a productive experience. -There is a free textbook online and the blog of Noam Nisan (a leader in the field) is a good place to follow the latest developments.<|endoftext|> -TITLE: Use of everywhere divergent generating functions -QUESTION [12 upvotes]: Generating functions are well-known to be much useful in combinatorics. But, maybe just since I am illiteral, all the applications coming in mind deal with power series, which are not just formal, but have non-zero radius of convergence. So, for sequences of super-exponential growth exponential generating functions $\sum a_nx^n/n!$ are considered, which already converge somewhere, and so on. But for considering formal power series, when we do not use analysis features (sometimes we use, but often do not), convergence is not necessary, and looks like a "coincidence". -Here comes the question: are there any less or more nice and natural application of formal power series like $\sum n! x^n$, which converge only for $x=0$? - -REPLY [14 votes]: Formal power series with radius of convergence 0 often arise in counting labeled graphs. For example, the exponential generating function for labeled connected graphs is $\log G(x)$, where $$G(x) = \sum_{n=0}^\infty 2^{\binom{n}{2}} \frac{x^n}{n!},$$ which has radius of convergence 0. -As Aaron noted, series like $\sum_{n=0}^\infty n! x^n$ arise in the theory of continued fractions; this series has the continued fraction expansions -$$ -\frac{1}{1-\displaystyle\frac{\mathstrut x}{1- -\displaystyle\frac{\mathstrut x}{1- -\displaystyle\frac{\mathstrut 2x}{1- -\displaystyle\frac{\mathstrut 2x}{1- -\displaystyle\frac{\mathstrut 3x}{1- -\displaystyle\frac{\mathstrut 3x}{1-\cdots -}}}}}}} -$$ -and -$$ -\frac{1}{1-x-\displaystyle\frac{\mathstrut x^2}{1- 3x - -\displaystyle\frac{\mathstrut 2^2x^2}{1-5x - -\displaystyle\frac{\mathstrut 3^2x^2}{1-7x -\cdots -}}}} -$$ -Similar continued fractions exist for ordinary generating functions (with radius of convergence 0) for Bell numbers, Eulerian polynomials, matchings, and more generally, moments of orthogonal polynomials. A very nice combinatorial approach to these continued fractions has been given by Philippe Flajolet, Combinatorial aspects of continued fractions. -It is true that most, if not all, of these examples of nonconverging power series can be refined to power series in more than one variable that do converge for some values of the parameters. For example, the exponential generating function for labeled connected graphs by edges is $\log G(x,t)$, where -$$G(x,t) = \sum_{n=0}^\infty (1+t)^{\binom{n}{2}} \frac{x^n}{n!};$$ -this converges for $|1+t|<1$. On the other hand, the exponential generating function for strongly connected tournaments is $1-1/G(x)$, and this doesn't seem to generalize since $1-1/G(x,t)$ has some negative coefficients.<|endoftext|> -TITLE: Computational algebra: where? -QUESTION [5 upvotes]: I'm on my last semester of a math B.Sc. and about to start studying for a math M.Sc in the same institute. -It now seems like a good time to start thinking of a PhD. -I'm interested in both algebra and algorithms. So I read a little in some computational algebra books (comp. group theory to be precise) and it looks like a great choice for me. -My questions: -1. Is this a "hot topic" in mathematics? Are there many serious people researching it? -2. Where are the major places in the world to research comp. algebra and to get a PhD? -Thank you! - -REPLY [3 votes]: Some other suggestions: -The University of Auckland (Eamonn O'Brien and Marston Conder) -The University of Warwick (Derek Holt) -I did my MSc with Eamonn O'Brien and my PhD with Derek Holt, and can recommend them both as supervisors. You are welcome to contact me by email (which you can find on my website) if you have any specific questions about either university.<|endoftext|> -TITLE: Resolution of singularities -QUESTION [6 upvotes]: What is the relation between crepant resolutions and minimal resolutions? Are they the same thing? - -REPLY [7 votes]: They are definitely not the same thing, far from it: -Minimal resolution usually refers to surfaces, and every surface has one: you can get it from an arbitrary resolution by consecutively blowing down $(-1)$-curves. In general, one usually speaks about minimal models. Actually, one could say minimal resolution in arbitrary dimension as well, but have to keep in mind that that is usually a) not actually a resolution (minimal models are not necesssarily smooth) b) not unique. -Crepant resolution exists in every dimension, but not for everything. In fact, admitting a crepant resolution is a very special property of the variety. In particular, it means that it has (strictly) canonical singularities, that is, as soon as it has either worse or better (say $\mathbb Q$-factorial terminal, but not smooth) singularities, then there is no crepant resolution. -As Karl shows, not every minimal resolution of surfaces is crepant, but it is easy to see that every crepant resolution of a surface is necessarily minimal. -EDIT: Added "$\mathbb Q$-factorial" to rule out small resolutions following Karl's comment. -EDIT2: Added comment about minimal resolutions in arbitrary dimension.<|endoftext|> -TITLE: Anti-concentration bound for permanents of Gaussian matrices? -QUESTION [40 upvotes]: In a recent paper with Alex Arkhipov on "The Computational Complexity of Linear Optics," we needed to assume a reasonable-sounding probabilistic conjecture: namely, that the permanent of a matrix of i.i.d. Gaussian entries is "not too concentrated around 0." Here's a formal statement of our conjecture: - -There exists a polynomial $p$ such that for all $n$ and $\delta>0$, - -      $\Pr_{X\sim\mathcal{N}\left( 0,1\right) _{\mathbb{C}}^{n\times n}}\left[\left\vert \operatorname*{Per}\left( X\right) \right\vert \leq \frac{\sqrt{n!}}{p\left( n/\delta\right) }\right] \leq \delta. $ - - -This conjecture seems interesting even apart from our application, so I wanted to bring it to people's attention -- maybe there's a simple/known proof that we're missing! -Here's what we do know: - -The expectation of Per(X) is of course 0 (by symmetry), while the standard deviation is $\sqrt{n!}$. Thus, our conjecture basically says that "Per(X) is polynomially smaller than its standard deviation only a 1/poly(n) fraction of the time." - -Recently, Terry Tao and Van Vu proved a wonderful anti-concentration bound for the permanents of Bernoulli matrices, which can be stated as follows: for all $\varepsilon > 0$ and sufficiently large n, - -$\Pr_{X\in\left\{ -1,1\right\} ^{n\times n}}\left[ \left\vert -\operatorname*{Per}\left( X\right) \right\vert \leq \frac{\sqrt{n!}% -}{n^{\varepsilon n}}\right] \leq \frac{1}{n^{0.1}}.$ - -Unfortunately, their result falls short of what we need in three respects. First, it's for Bernoulli matrices rather than Gaussian matrices. (Though of course, the Gaussian case might well be easier than the Bernoulli case, which is our main reason for optimism!) Second, and most important, Tao and Vu only prove that Per(X) is at least a 1/nεn fraction of its standard deviation with high probability, whereas we need that it's at least a 1/poly(n) fraction. Third, they upper-bound the probability of a "bad event" by 1/n0.1, whereas we'd like to upper-bound it by 1/p(n) for any polynomial p. - -The numerical evidence that we've obtained is strongly consistent with our conjecture being true (see figure below). - -We can prove that our conjecture holds with the determinant in place of the permanent. To do so, we use the fact that if X is Gaussian, then because of the rotational invariance of the Gaussian measure, there's an explicit formula for all the moments of Det(X) -- even the fractional and inverse moments. - -One might wonder if we can also calculate the higher moments of Per(X), and use that to prove our conjecture. Indeed, we can show that $\operatorname*{E}_{X\sim\mathcal{N}\left( 0,1\right) _{\mathbb{C}}^{n\times n}}\left[ \left\vert \operatorname*{Per}\left( X\right)\right\vert ^{4}\right] =\left( n!\right) ^{2}\left( n+1\right)$, which then implies the following weak anti-concentration bound: for all β<1, $\Pr_{X\sim\mathcal{N}\left( 0,1\right) _{\mathbb{C}}^{n\times n}}\left[ \left\vert \operatorname*{Per}\left( X\right) \right\vert \geq -\beta\sqrt{n!}\right] \geq\frac{\left( 1-\beta^{2}\right) ^{2}}{n+1}$. Unfortunately, computing the 6th, 8th, and higher moments seems difficult. - -See section 8 of our paper for the proofs of 4 and 5. -Short of proving our anti-concentration conjecture, here are two easier questions whose answers would also greatly interest us: - -Can we at least reprove Tao and Vu's bound for Gaussian matrices rather than Bernoulli matrices? In their paper, Tao and Vu say their result holds for "virtually any (not too degenerate) discrete distribution." I don't think the Gaussian distribution would present serious new difficulties, but I'm not sure. -Does the pdf of Per(X) diverge at the origin? (We don't even know the answer to that question in the case of Det(X).) I don't know of any formal implications between this question and the anti-concentration question, but it would be great to answer anyway. - - (source: Wayback Machine) - -REPLY [39 votes]: I did a preliminary feasibility analysis of our methods and it appears possible that one may be able to tighten our $n^\epsilon$ loss to something more like $\exp( \sqrt{n} )$ in the Gaussian case, but this is still well short of what you want. The main obstacle is potential coupling between permanents of minors, which we were not able to fully avoid. -Here's the heuristic calculation. Suppose that a Gaussian $k \times k$ permanent has some distribution $P_k$. Then a Gaussian $k+1 \times k+1$ permanent $P_{k+1}$, by cofactor expansion, looks like -$$ P_{k+1} = \sum_{i=1}^{k+1} (-1)^i x_i P_k^{(i)}$$ -where $x_i$ are iid Gaussians and the $P_k^{(i)}$ are copies of $P_k$ corresponding to various $k \times k$ minors of the $k+1 \times k+1$ matrix. -As the entries of the $k+1 \times k+1$ matrix are iid, the $x_i$ are independent of the $P_k^{(i)}$, and so for fixed values of $P_k^{(i)}$, we see that $P_{k+1}$ is distributed normally with mean zero and variance $\sum_{i=1}^{k+1} |P_k^{(i)}|^2$, which we can rewrite as -$$ P_{k+1} = (\sum_{i=1}^{k+1} |P_k^{(i)}|^2)^{1/2} \cdot N_{k+1}\qquad (1)$$ -where $N_{k+1}$ is a standard normal random variable (independent of the $P_k^{(i)}$). -Now we come up against the key problem: the $P_k^{(i)}$ are identically distributed, but are not jointly independent, because there is a huge amount of overlap between the $k \times k$ minors. So, while heuristically one expects concentration of measure to kick in and make $(\sum_{i=1}^{k+1} |P_k^{(i)}|^2)^{1/2}$ more concentrated than any one of the $P_k^{(i)}$, we don't know how to prevent huge correlations from happening. In the worst case, all the $P_k^{(i)}$ are perfectly correlated to each other, and then (1) could become something more like -$$ P_{k+1} = (k+1)^{1/2} |P_k| \cdot N_{k+1}.$$ -This multiplicative normal process would lead to $P_n$ to concentrate between $\sqrt{n!} \exp(-O(\sqrt{n}))$ and $\sqrt{n!} \exp(O(\sqrt{n}))$, as can be seen by taking logs and applying the central limit theorem. -But this worst case can't actually be the truth - among other things, it contradicts the second moment calculation. So there should be some way to prevent this correlation. Unfortunately, my paper with Van completely evades this issue - we try to get as far as we can just from the obvious fact that disjoint minors are independent from each other. This is why our bounds are significantly worse than $\exp(O(\sqrt{n}))$ from the truth. -As you say, the situation is much better for the determinant of a Gaussian iid matrix. Here, we can use the base-times-height formula to express the determinant as a product $\prod_{i=1}^n \hbox{dist}(X_i,V_i)$, where $X_i$ is the $i^{th}$ row and $V_i$ is the span of the first $i-1$ rows. With everything being Gaussian, $\hbox{dist}(X_i,V_i)^2$ is a chi-squared distribution, which is a martingale in $i$, and as a consequence one can get the determinant within $\exp(O(\sqrt{\log n}))$ of $\sqrt{(n-1)!}$, which would give you what you want. Unfortunately, there is nothing like the base-times-height formula for the permanent... -Finally, I am fairly certain that at the level of $n^{-\epsilon n}$ losses, one can replicate our paper in the Bernoulli case to the Gaussian case. I don't think the $n^{-\epsilon n}$ loss gets much better in that case, but the $\frac{1}{n^{0.1}}$ bound should improve substantially.<|endoftext|> -TITLE: Can assignment solve stable marriage? -QUESTION [29 upvotes]: This is an excellent question asked by one of my students. I imagine the answer is "no", but it doesn't strike me as easy. -Recall the set up of the stable marriage problem. We have $n$ men and $n$ women. Each man has sorted the women into some order, according to how much he likes them, and each women has likewise ranked the men. We want to pair off the men with the women so that there do NOT exist any pairs $(m,w)$ and $(m', w')$ where - -$m$ prefers $w'$ to $w$ and -$w'$ prefers $m$ to $m'$. - -It is a theorem of Gale and Shapley that such an assignment is always possible. -Here is a potential way you could try to find a stable matching. Choose some function $f: \{ 1,2,\ldots, n \}^2 \to \mathbb{R}$. Take the complete bipartite graph $K_{n,n}$, with white vertices labeled by men and black vertices by women, and weight the edge $(m,w)$ by $f(i,j)$ if $w$ is $m$'s $i$-th preference, and $m$ is $w$'s $j$-th preference. Then find a perfect matching of minimal weight, using standard algorithms for the assignment problem. - -Is there any function $f$ such that this method works for all preference lists? - -REPLY [17 votes]: David's question is sufficiently more precise than the question of Donald Knuth referred to in the answer below. I believe it can be answered in the negative for $n \geq 3$, as follows. -Let $\{m_1,m_2,\ldots,m_n\}$ and $\{w_1,w_2,\ldots,w_n\}$ be the vertices of the parts of our graph $K_{n,n}$. Consider a choice of preferences where for $i\geq 3$ the man $m_i$ and the woman $w_i$ are each other's top choice. Then any stable marriage must match them to each other. -We consider now the preferences of $m_1,m_2,w_1$ and $w_2$. I will write $m_1 : w_1 \to 1, w_2 \to 3$ to denote that $w_1$ is $m_1$-th first choice and $w_2$ is his third, etc. - -$m_1 : w_1 \to 1, w_2 \to 2,$ -$m_2 : w_1 \to 2, w_2 \to 3,$ -$w_1 : m_1 \to 2, m_2 \to 3,$ -$w_2 : m_1 \to 1, m_2 \to 2.$ - -Here $m_1w_1, m_2w_2$ is a stable matching and therefore if the function $f$ is as desired then we must have -$f(1,2)+f(3,2) < f(2,1)+f(2,3),$ -where on the left we have the weight of the stable matching and on the right is the weight of the matching $m_1w_2, m_2w_1$. -But the left and right side of the inequality above are symmetric and so by switching the $m$'s and $w$'s we can design another set of preferences implying -$f(2,1)+f(2,3) < f(1,2)+f(3,2).$ -It follows that the function $f$ as specified in the question can not exist.<|endoftext|> -TITLE: Are there examples of non-orientable manifolds in nature? -QUESTION [138 upvotes]: Whilst browsing through Marcel Berger's book "A Panoramic View of Riemannian Geometry" and thinking about the Klein bottle, I came across the sentence: -"The unorientable surfaces are never discussed in the literature since the primary interest of mathematicians in surfaces is in the study of one complex variable, number theory, algebraic geometry etc. where all of the surfaces are oriented." -(I won't give context, other than a page number: 446.) This got me thinking, perhaps non-orientability is purely an invention of topologists. Surely non-orientable manifolds play an important role in other areas of mathematics? Are there "real world" examples of non-orientability phenomenon in the natural sciences? - -REPLY [5 votes]: Not exactly nature, but not designed like that intentionally. Or so I hope.<|endoftext|> -TITLE: What is this subgroup of $\mathfrak S_{12}$? -QUESTION [26 upvotes]: On some occasion I was gifted a calendar. It displays a math quizz every day of the year. Not really exciting in general, but at least one of them let me raise a group-theoretic question. -The quizz: consider an hexagon where the vertices and the middle points of the edges are marked, as in the figure -$\hskip11em$ -One is asked to place the numbers $1,2,3,4,5,6,8,9,10,11,12,13$ (mind that $7$ is omitted) on points $a,\ldots,\ell$, in such a way that the sum on each edge equals $21$. If you like, you may search a solution, but this is not my question. -Of course the solution is non unique. You may apply any element of the isometry group of the hexagon. A little subtler is the fact that the permutation $(bc)(ef)(hi)(kl)(dj)$ preserves the set of solutions (check this). - -Question. What is the invariance group of the solutions set ? Presumably, it is generated by the elements described above. What is its order ? Because it is not too big, it must be isomorphic to a known group. Which one ? - -REPLY [37 votes]: The invariance group of the solutions set can be given a geometric interpretation as follows. Note that $\mathfrak{S}_4 \times \frac{\mathbf{Z}}{2\mathbf{Z}}$ is none other than the group of isometries of the cube. -It is known that if one cuts a cube by the bisecting plane of a space diagonal, the cross-section is a regular hexagon (see the picture at the middle of this page). The vertices of this hexagon are midpoints of (some) edges of the cube. Let $X$ be the set of corners and middles of this hexagon (it has cardinality $12$). Let us consider the following bijection between $X$ and the set $E$ of edges of the cube : if $[AB]$ is a side of the hexagon, with midpoint $M$, we map $A$ (resp. $B$) to the unique edge $e_A$ (resp. $e_B$) in $E$ containing it, and we map $M$ to the unique edge $e_M \in E$ such that $e_A$, $e_B$ and $e_M$ meet at a common vertex of the cube. -Given any solution of the initial problem, we can label the edges of the cube using the above bijection. This labelling has the following nice property : the sum of three edges meeting at a common vertex is always 21. Proof : by construction, six of these eight summing conditions are satisfied. The remaining two conditions read $b+f+j=d+h+\ell=21$ using Denis' notations, and are implied by the first six conditions. -So we found an equivalent ($3$-dimensional) formulation of the problem, namely labelling the edges of a cube. It is now clear that the symmetry group of the cube acts on the set of solutions. It remains to prove that the solution is unique up to isometry, which can be done by hand, here is how I did it : note that only two possible sums involve $1$ (resp. $13$), namely $1+8+12$ and $1+9+11$ (resp. $2+6+13$ and $3+5+13$). Therefore $1$ and $13$ must sit on opposite edges. Then $4$ and $10$ must sit on the unique edges which are parallel to $1$ and $13$. It is the easy to complete the cube. -The resulting labelling has some amusing properties For example, the sum of edges of a given face is always $28$. The sum of two opposite edges is always $14$. Finally, the sum of edges along a cyclohexane-like circuit is always $42$.<|endoftext|> -TITLE: What is the current status of Agrawal's conjecture? -QUESTION [35 upvotes]: In their famous 'Primes is in P' paper Agrawal, Kayal and Saxena stated the following conjecture: - -If for coprime integers $n$ and $r$ the equality $(X-1)^n = X^n - 1$ holds in $\mathbb{Z}_n[X]/(X^r-1)$ then either $n$ is prime or $n^2 = 1 \pmod{r}$. - -If true this would give a beautiful characterization of primes that could be easily transformed into a fast ($O(\log^{3+\epsilon}{n})$) and deterministic primality test. -Shortly after publishing 'Primes is in P' Hendrik Lenstra noticed that the conjecture may not be valid for $r=5$ and $n$ of the very special form (see Lenstra's and Pomerance's note, p.30). It was unknown whether any such $n$ existed but Carl Pomerance gave a heuristic argument convincing that there should be infinitely many $n$'s sharing these, apparently rare, properties. I'm not aware of any strict proof for this. -It may also happen that the conjecture in a modified form (if we restrict $r$ to be greater than $\log{n}$) can be still true. -Martin Mačaj (see Some remarks and questions about the AKS -algorithm and related conjecture) gave another version of this conjecture together with a proof that relied on yet another unsolved problem. -Does anyone know if there were any advances in this area in the recent years? - -REPLY [6 votes]: About primaboinca -PRIMABOINCA is a research project that uses Internet-connected computers to search for a counterexample to some conjectures. -This project concerns itself with two hypotheses in number theory. Both are conjectures for the identification of prime numbers. The first conjecture (Agrawal’s Conjecture) was the basis for the formulation of the first deterministic prime test algorithm in polynomial time (AKS algorithm). Hendrik Lenstras and Carl Pomerances heuristic for this conjecture suggests that there must be an infinite number of counterexamples. So far, however, no counterexamples are known. This hypothesis was tested for n < 1010 without having found a counterexample. The second conjecture (Popovych’s conjecture) adds a further condition to Agrawals conjecture and therefore logically strengthens the conjecture. If this hypothesis would be correct, the time of a deterministic prime test could be reduced from O(log N)6 (currently most efficient version of the AKS algorithm) to O(log N)3. -You can participate by downloading and running a free program on your computer. -http://www.primaboinca.com/<|endoftext|> -TITLE: Triangulations of polyhedra -QUESTION [13 upvotes]: A topologist came to me with this question, but everything I think should work doesn't. -How many triangulations are there of a polyhedron with n vertices? -By a "triangulation" of a polyhedron P we mean a decomposition of P into 3-simplices whose interiors are disjoint, whose vertices are vertices of P, and whose union is P. Since this obviously depends on the polyhedron, let's say that P is the convex hull of n points on the curve (t, t^2, t^3). (I think this is general, but a proof of that would be nice too.) In particular, this means that all of the faces are triangles, since no four vertices are coplanar. -Since triangulations of a polygon are counted by the Catalan numbers, a reasonable first guess is that these are counted by the generalized Catalan numbers $C_{n,k} = \frac{1}{(k-1)(n+1)} {kn \choose n}$, which count k-ary trees (among other things). But just at n=5 we run into trouble: there are 2 (not 3) such triangulations, and they don't even contain a fixed number of pieces: one of them triangulates P into two tetrahedra, and one breaks it into three. -This seems obvious enough that someone would have asked it before, but I'm not finding anything. Of course, answers to the obvious generalization (triangulations of k-polytopes whose vertices lie on (t, t^2, ..., t^k)) are welcome as well. - -REPLY [4 votes]: Although the answer is provided by Igor's pointers to the Triangulations book, it might be useful -to supplement those pointers with the explicit bounds. -The lower bound is due to Gil Kalai, and the upper bound to Tamal Dey. -For fixed dimension $d$, the cyclic polytope has at least $\Omega( 2^{n^{ \lfloor d/2 \rfloor }})$ -triangulations, and for $d$ odd, at most $2^{ O( n^ {\lceil d/2 \rceil} ) }$ triangulations. -So, for $d=3$, the case posed in the question, the bounds are between $c^n$ and $c^{n^2}$. -See Section 8.4 (pp. 396-398) of Triangulations.<|endoftext|> -TITLE: Subfactor theory and Hilbert von Neumann Algebras -QUESTION [7 upvotes]: There seem to be intimate connections between the different definitions of von Neumann module. The two that I'm aware of are Hilbert von Neumann modules and correspondences (in the sense of Connes). I was wondering if there is any significant interplay between the two - for instance if any of the ideas around Jones' submodule theory (which uses correspondences of von Neumann algebras) have been extended or transfered to Hilbert von Neumann modules? -Some observations: Suppose we have a left Hilbert $N$-module (so inner-product $N$-linear in the first variable) with a faithful state $\varphi$ on $N$. Then define a complex-valued inner product on $E$ via -$$\langle x,y\rangle_\varphi:=\varphi(\langle x,y\rangle).$$ -If we complete then we get a vector space $\widetilde{E}$ and a von Neumann algebra $N$ acting on it - so a left von Neumann module in a sense similar to that of correspondences. -Questions: -(i) Given a faithful representation of $N$ on a Hilbert space $H$ can we always find a state $\varphi$ and left Hilbert $N$-module $E$ so that the representation of $N$ on $\widetilde{E}$ is equivalent to the one on $H$? -(ii) Can this idea be taken over to the case of correspondences and Hilbert bimodules (for instance in the case of $II_1$ factors)? -(ii) What results (if any) can be pulled up to the Hilbert bimodule level? -I'm aware that question (iii) is vague, but any input or references would be appreciated! - -REPLY [7 votes]: Answers: (i) Yes, if we replace states by weights (not every -von Neumann algebra admits a faithful state); -(ii) Yes (for all von Neumann algebras); (iii) All of them. -Suppose M is an arbitrary von Neumann algebra and p≥0 is a real number. -Then we define a right L_p(M)-module as a right M-module equipped -with an inner product with values in L_{2p}(M), satisfying the same algebraic -properties as for Hilbert W*-modules together with the appropriate completeness -condition (we require completeness in the measurable topology, -which coincides with the σ-weak topology for p=0 and with the norm topology for p>0). -Here L_p(M)=L^{1/p} denotes the L_p-space of M, in particular, -L_0(M)=L^∞(M)=M, L_1(M)=L^1(M)=M_* (the predual), L_{1/2}(M)=L^2(M)=the Hilbert -space of half-densities on M. (The subscript notation is much more natural than the superscript notation -because L_p-spaces form a graded algebra, p being the grading.) -A morphism of right L_p(M)-modules is defined as a morphism of algebraic right M-modules -that is continuous in the measurable topology. -It turns out that right L_p(M)-modules form a W*-category. -We observe that the category of representations of M on Hilbert spaces is equivalent -to the category of right L_{1/2}(M)-modules. -If we have a right L_{1/2}(M)-module X with an inner product x,y↦(x,y)∈L_1(M), -then x,y↦tr(x,y)∈C is a complex-valued inner product on X, -which turns X into a Hilbert space together with an action of M. -Vice versa, if X is a Hilbert space equipped with an action of M, -then x,y→(w∈M↦(x,yw)∈C)∈L_1(M) is the corresponding L_1(M)-valued inner product. -Suppose 0≤p≤q are real numbers. We define a functor from the category -of right L_p(M)-modules to the category of right L_q(M)-modules -by sending a right L_p(M)-module X to X⊗L_{q-p}(M). -Here ⊗ denotes the algebraic tensor product, without any kind of completion. -Although it is non-obvious, in the end this tensor product turns out to be complete. -Likewise, we define a functor from the category of right L_q(M)-modules -to the category of right L_p(M)-modules by sending a right L_q(M)-module Y -to Hom_M(L_{q-p}(M),Y). -Here Hom_M denotes the space of algebraic homomorphisms preserving the right action of M, -without any kind of continuity property. -Again it is a non-obvious fact that this space is actually a right L_p(M)-module. -One can prove that the two functors defined above form an adjoint unitary equivalence -of the W*-categories of right L_p(M) and L_q(M) modules. -In particular, the category of Hilbert W*-modules over M and the category -of representations of M on Hilbert spaces are equivalent. -The result above extends to bimodules. -An M-L_p(N)-bimodule is a right L_p(N)-module X together -with a morphism of von Neumann algebras A→End_N(X). (The algebra -of endomorphisms of any object in a W*-category is a von Neumann algebra.) -Since the above equivalence is an equivalence of W*-categories, -we can immediately extend it to an equivalence of categories of M-L_p(N) and M-L_q(N) bimodules. -In particular, the category of Hilbert W*-bimodules from M to N is equivalent to the category -of Connes' correspondences from M to N. -Moreover, one can observe that the bicategory of von Neumann algebras, Connes' correspondences, which compose via Connes' fusion, and their intertwiners -is equivalent to the bicategory of von Neumann algebras, Hilbert W*-bimodules, -which compose via the completed tensor product, and their intertwiners. -This result is also valid for arbitrary p. -References: - -The equivalence in the last paragraph of the answer was apparently first proven by Baillet, Denizeau, and Havet in their 1988 paper Indice d'une espérance conditionnelle. -L_p(M) modules were defined by Junge and Sherman in their 2005 paper Noncommutative L^p modules. -I am not aware of any paper that proves the above equivalences for arbitrary p, but I will include a proof of these statements in my thesis.<|endoftext|> -TITLE: How can you compute the number of topological sorts in a DAG? -QUESTION [10 upvotes]: If you have a DAG, G, a topological sort is just an ordering of the vertices such that if an edge x->y exists in G, then the index of x is less than the index of y. -It's not hard to figure out how a topological sort can be given, but how efficiently can one compute the total number of topological sorts that exist for a given acyclic graph? - -REPLY [7 votes]: This problem is #P-complete. See "Counting linear extensions is #P-complete", G. Brightwell and P. Winkler, Proc. 23rd ACM Symposium on the Theory of Computing, 1991<|endoftext|> -TITLE: Space-discriminating injective curve -QUESTION [7 upvotes]: Let $f\colon \mathbb R^1\to \mathbb R^3$ be a continuous and injective map. Is $\mathbb R^3\setminus f(\mathbb R^1)$ a path-connected space? - -REPLY [14 votes]: Yes. In fact, $\mathbb R^3$ could be any 3-manifold and $f(\mathbb R^1)$ any countable union of embedded segments. -Lemma 1. Let $U$ be an open ball in $\mathbb R^3$ and $f:I\to\mathbb R^3$ an embedding. Then $U\setminus f(I)$ is path-connected. -Proof. See Hatcher, Proposition 2B.1 on page 169. This proposition proves (among other things) that $S^3$ minus any embedded segment is path-connected. Extract the proof of this particular statement and repeat it word-by-word with $U$ in place of $S^3$. (The proof uses only the following facts about $U$: $U$ minus any closed sub-segment of $f(I)$ is open; $U$ minus any point of $\mathbb R^3$ is simply connected.) -Lemma 2. Let $f:I\to\mathbb R^3$ be an embedding and $s:I\to\mathbb R^3$ a path with endpoints outside $f(I)$. Then for any $\varepsilon>0$ there exists a path $s_\varepsilon:I\to\mathbb R^3$ which is $\varepsilon$-close to $s$ (in $C^0$), connects the same endpoints, and avoids $f(I)$. -Proof. Divide the domain of $s$ into intervals so small that the diameters of their images are less than $\varepsilon/10$. Let $p_0,p_1,\dots,p_n\in\mathbb R^3$ be the images of the division points ($p_0$ and $p_n$ are the endpoints of $s$). For each $p_i$, let $q_i$ be a point outside $f(I)$ such that $|p_i-q_i|<\varepsilon/10$ (for $i=0$ and $i=n$, choose $q_i=p_i$). Such a point $q_i$ exists since $f$ is injective and hence cannot cover a set with nonempty interior. Let $U_i$ be the ball of radius $\varepsilon/3$ centered at $p_i$. By Lemma 1, $U_i\setminus f(I)$ is path-connected. Therefore we can connect $q_i$ to $q_{i+1}$ by a path contained in $U_i$ and avoiding $f(I)$. The (suitably parametrized) product of these paths is the desired $s_\varepsilon$, q.e.d. -Now return to the original problem. The set $f(\mathbb R)$ is a countable union of sets $J_k$, $k\in\mathbb N$, where each $J_k$ is an injective image of a segment. We want to prove that any two points $p,q\in\mathbb R^3$ can be connected by a path avoiding $\bigcup J_k$. -Let $s_1$ be a path from $p$ to $q$ avoiding $J_1$ (such a path exists by Lemma 1). Let $\varepsilon_1$ be the minimum distance from $s_1$ to $J_1$, divided by 10. By Lemma 2, there exists a path $s_2$ which is $\varepsilon_1$-close to $s_1$ and avoids $J_1$. Let $\varepsilon_2$ be the minimum of $\varepsilon_1$ and the minimum distance from $s_1$ to $J_1$, divided by 10. Then there is a path $s_3$ which is $\varepsilon_2$-close to $s_2$ and avoids $J_2$, and so on. The resulting sequence $s_1,s_2,\dots$ converges (since $C^0$ is complete) to some path $s$ which avoids every set $J_k$. -Remark. The closure of $f(\mathbb R)$ can separate the space, for example, $f(\mathbb R)$ can be a dense subset of the torus.<|endoftext|> -TITLE: Expressing a number field as a composite of extensions ramified at one place -QUESTION [13 upvotes]: If $K$ is a number field, is it always possible to find a finite extension $L/K$ such that $L$ is the composite of fields $L_1,\ldots, L_n$, with the property that at most one prime ramifies in $L_i/\mathbb{Q}$? Equivalently, is the composite of all extensions ramified at $\leq 1$ place all of $\overline{\mathbb{Q}}$? -The first question has an affirmative answer when $K$ is abelian, but for the general case, the equivalent second question sounds too strong to be true. Any ideas? - -REPLY [15 votes]: I believe the answer is `No', and Franz Lemmermeyer's example $K=Q(2^{1/3})$ and his strategy of the proof do the trick. -Suppose this particular $K$ is contained in the compositum $F$ of $L_i$, with every $L_i$ ramified at only one prime. Assume each $L_i$ is Galois over $Q$ (otherwise replace it by its Galois closure) and that no two $L_i$ ramify at the same prime $p$ (otherwise replace this pair by their compositum). The $L_i$ are then linearly disjoint over $Q$, since their pairwise intersections would have to be unramified at all primes. So $G=Gal(F/Q)$ is the direct product of $Gal(L_i/Q)$'s. -Now the group $G$ has a 2-dimensional irreducible representation $\rho$, the one that factors through the Galois closure of $K/Q$ (an $S_3$-extension of $Q$). As $G$ is the direct product of groups, we can write $\rho=\rho_1\otimes...\otimes\rho_n$ uniquely, with $\rho_i$ irreducible representations of $Gal(L_i/Q)$. Moreover, $\rho$ is self-dual, so all the $\rho_i$ are self-dual as well. Of these $\rho_i$ one must be 2-dimensional and the others are 1-dimensional. Because 1-dimensional self-dual characters have order 2, this shows that at all primes $p$ except at most one (the one corresponding to the 2-dimensional $\rho_i$) inertia $I_p$ acts on $\rho$ through a quotient of order 2. But $I_2$ and $I_3$ act through quotients of order 3 and 6 respectively, contradiction!<|endoftext|> -TITLE: Examples and importance of Embedding (and Non-Embedding) Theorems -QUESTION [5 upvotes]: An embedding is an injective map into a universal, simpler model object. Many embedding theorems are without obstruction, in the sense that every object which you wish to embed can be embedded. Examples of such theorems are Yoneda lemma, algebraic closure of fields, Nash embedding theorem for Riemannian manifolds. unconditional. -I'm interested in embedding theorems with obstruction. Do you have examples of theorems that give an obstruction to embedding? In the case where there is an obstruction, would you consider the obstruction to be local or instrinsic? Most embedding problems are possible locally, but there is often a local-global obstruction. -The example that led me to this question is Kodiara's embedding theorem that gives an obstruction for a complex manifold to be a submanifold of complex projective space. Here the obstruction is that the manifold must carry a positive line bundle. Positivity of curvature is a local criterion. -PS. Sorry, but I really don't know how to tag this question. - -REPLY [4 votes]: I'm not sure whether you look for such an answer, because it comes from analysis. Analysts use various functional spaces, especially the Sobolev spaces. $W^{s,p}(\Omega;\mathbb R)$ is, roughly speaking, the set of functions with $s$ derivatives in $L^p(\Omega)$ (but $s\ge0$ needs not be an integer). - -Sobolev embedding. If $\Omega$ is an open subset with a smooth boundary, and if $\frac1q=\frac1p-\frac{s}{n}$ with $1\le p< q<\infty$, then $W^{s,p}(\Omega;\mathbb R)$ embeds into $L^q(\Omega)$. If instead $sp>n$, then $W^{s,p}(\Omega;\mathbb R)$ embeds into ${\mathcal C}^\alpha(\bar\Omega)$ where $\alpha:=s-\frac{n}{p}$, unless this exponent is an integer. - -When the target $\mathbb R$ is replaced by a manifold, the situation may not be so nice. Embedding theorems are related to norm inequalities, which are usually proved first for ${\mathcal C}^\infty$-fields, then extended by means of density of ${\mathcal C}^\infty$ in $W^{s,p}$. - -Obstruction (Bethuel 1991). Assume that $p< n$, and let $N$ be a compact manifold of dimension $k$. Then ${\mathcal C}^\infty(\Omega,N)$ is dense in $W^{1,p}(\Omega;N)$ if and only if $\pi_{[p]}(N)=0$, where $[p]$ is the largest integer $\le p$. - -The consequence of this is that in some situations, there is a discrepency between $W^{s,p}$ and the closure of ${\mathcal C}^\infty$ under the $W^{s,p}$-norm.<|endoftext|> -TITLE: Shortest grid-graph paths with random diagonal shortcuts -QUESTION [24 upvotes]: Suppose you have a network of edges connecting -each integer lattice point -in the 2D square grid $[0,n]^2$ -to each of its (at most) four neighbors, {N,S,E,W}. -Within each of the $n^2$ unit cells of this grid, -select one of the two diagonals at random to -add to the network. -These diagonals serve as local "short cuts." -One could ask many questions about this model, -but let me start with this one: - -What is the expected length of the shortest path - in such a network from $(0,0)$ to $(n,n)$? - -Here is an example for $n=25$: - -       - -Here a shortest path has length -$10+20 \sqrt{2} \approx 38.3$ -in comparison to the shortest possible length, -$25 \sqrt{2} \approx 35.4$. -For small $n$, the growth rate of the length of the shortest -path appears to be linear in $n$, with a slope of about -1.52315. - -       - -I would appreciate learning if anyone recognizes -this model and/or knows the true growth rate. Thanks! -Edit. One more figure to address a question raised by jc. -This shows 10 shortest paths for $n=50$, for different random diagonal -choices (not shown). But be aware that usually several shortest paths are tied as equally long, -and my code selects one with a systematic bias toward the lower right corner. -But the figure provides a sense of the variation. - -REPLY [7 votes]: I also think, as already said, that this is exactly a problem of First Passage Percolation. I precise Tom's idea: take the whole lattice $\mathbb{Z}^2$ and add S0-NE diagonals (this is in fact a triangular lattice). Then put deterministic edge-weights equal to $1$ on the horizontal and vertical edges, and put independently on each diagonal edge a weight equal to $\sqrt{2}$ with probability $1/2$ and to $2$ (or anything larger than $2$) with probability one half. I think this is exactly equivalent to your model, no ? -The edge-weights are not i.i.d, but they are independent and stationnary (and thus ergodic). Kingman's result applies to show the existence of an almost sure limit. -Concerning fluctuations, Benjamini, Kalai and Schramm's argument should apply as is (it works on any lattice), but this should not be optimal as mentioned by Tom. -For the possibility to compute the value of the limit, maybe take a look at this article: - -Timo Seppäläinen, Exact limiting shape for a simplified model of first-passage percolation on the plane, Ann. Probab. 26 (3) (1998) 1232–1250, https://doi.org/10.1214/aop/1022855751 - -Anyway, this is a nice model ! -@ Omer: You said "LPP on $\mathbb{Z}^2$ with Bernoulli weights (for which the limit shape is known)" ... are you sure the shape is known ?<|endoftext|> -TITLE: Divisibility of a binomial coefficient by $p^2$ -- current status -QUESTION [6 upvotes]: While skimming the book Concrete Mathematics, (edit: first edition) I came across the following problem, which is listed there as a Research Problem: (Chapter 5, Exercise 96) - -Is ${2n \choose n}$ divisible by the square of a prime for all $n > 4$. - -This problem looked to me much simpler than a divisibility problem that I found on MO (look here), but then again, I guess in number theory, the simpler the problems looks, the harder it usually is! -The nice form of this problem has made me very curious to find out more about it. But because I do not have more than a fleeting acquaintance with number theory, I don't know what search keywords would be useful to gain more information about this problem. -Thus, could somebody please tell me more about this problem and its current status? - -REPLY [13 votes]: This is/was known as the Erdős square-free conjecture, and seems to now be solved. See the bottom of this page.<|endoftext|> -TITLE: A conceptual proof that local fibrations over paracompact spaces are global fibrations? -QUESTION [17 upvotes]: I'm teaching some homotopy theory at the moment, and I'm discovering a number of things I've never before learned properly. (I guess everybody who's ever taught knows that feeling.) One of those things is that a local Hurewicz fibration is a global Hurewicz fibration if the base space is paracompact. This result goes back to Hurewicz and Huebsch, is proved in Spanier's book, and there's a paper by Dold where he compares various local and global properties of projection maps, including being a fibration. None of these proofs are what I would call conceptual. There's a lot of gluing together of "extended" lifting functions, dividing up intervals in three parts, etc etc, and they use the axiom of choice. Since paracompactness needs to get used, any possible proof will probably require some gluing, but does anyone have a more conceptual proof of the result? Has anyone taught this to students, and how was that? Is the axiom of choice a necessary input? - -REPLY [4 votes]: You should look at -Dyer, Eldon and Eilenberg, Samuel. Globalizing fibrations by schedules, Fund. Math., 130, (1988), 125--136. -They deduce the theorem on globalising fibrations from a more general theorem about the existence of a continuous function $f$ from the path space of $X$ to a space of "schedules" on $X$, given a numerable cover on $X$, such that any path $a$ "fits" the schedule $f(a)$. The intuitive interpretation is as follows: we know by the Lebesgue covering lemma, that for an open cover of $X$ and a path in $X$ we can find a subivision of the path so that each part of the subdivision lies in a set of the cover. The aim is to do this trick globally over all paths and "continuously". -I once tried to follow Spanier's proof in the first edition of his book,but found that a function given in a key proposition was not well defined. I managed to find and send him a definition which was well defined, see the second edition, but I could not prove it was continuous! (He does not give a proof of continuity.) Probably I'm not clever enough?<|endoftext|> -TITLE: Does Arzelà-Ascoli require choice? -QUESTION [26 upvotes]: Inspired by a recent Math.SE question entitled Where do we need the axiom of choice in Riemannian geometry?, I was thinking of the Arzelà--Ascoli theorem. Let's state a very simple version: - -Theorem. Let $\{f_n : [a,b] \to [0,1]\}$ be an equicontinuous sequence of functions. Then a subsequence $\{f_{n(i)}\}$ converges uniformly on $[a,b]$. - -The proofs I have seen operate as follows: Take a countable dense subset $E$ of $[a,b]$. Use a "diagonalization argument" to find a subsequence converging pointwise on $E$. Use equicontinuity to conclude that this subsequence actually converges uniformly on $[a,b]$. -The "diagonalization" step goes like this: Enumerate $E$ as $x_1, x_2, \dots$. $\{f_n(x_1)\}$ is a sequence in $[0,1]$, hence has a convergent subsequence $\{f_{n_1(i)}(x_1)\}$. $\{f_{n_1(i)}(x_2)\}$ now has a convergent subsequence $\{f_{n_2(i)}(x_2)\}$, and so on. Then $\{f_{n_i(i)}\}$ converges at all points of $E$. -Of course, to do this, at each step $k$ we had to choose one of the (possibly uncountably many) convergent subsequences of $\{f_{n_{k-1}(i)}(x_k)\}$, so some sort of choice is needed here (I guess dependent choice is enough? I am not a set theorist (IANAST)). Indeed, we have proved that $[0,1]^E$ is sequentially compact (it is metrizable so it is also compact). -On the other hand, we have not used (equi)continuity in this step, so perhaps there is a clever way to make use of it to avoid needing a choice axiom. -So the question is this: - -Can the Arzelà--Ascoli theorem be proved in ZF? If not, is it equivalent to DC or some similar choice axiom? - -REPLY [4 votes]: I should mention that the theorem -Theorem. Let $\{f_n:[a,b] \to [0,1]\}$ be an equicontinuous sequence of functions. Then a subsequence $\{f_{n(i)}\}$ converges uniformly on $[a,b]$. -holds without any choice (for $a, b \in \mathbb R$). This is proven by using Heine-Cantor for equicontinuous sets and choosing the subsequence in a clever way (see http://arxiv.org/abs/1509.01078 for a proof with a compact subset of $\mathbb R^d$ instead of $[a, b]$; it's not quite short (7 pages), thus I refer to the paper). -The problematic direction is: -Theorem. Let $F \subset \mathcal C(C)$ be sequentially compact ($C \subset \mathbb R^d$ being a compact set). Then $F$ is bounded and equicontinuous. -That's what is equivalent to countable choice for subsets of the real numbers.<|endoftext|> -TITLE: Is Illusie's generalization of the cotangent complex to arbitrary ringed toposes necessary in algebraic geometry? -QUESTION [10 upvotes]: André and Quillen both gave constructions of the relative cotangent complex for commutative rings, so pretty immediately that gives us that we understand the cotangent complex for affine schemes. Illusie generalized the cotangent complex construction from "rings over A" for a ring A to "rings over $\mathcal{O}_X$" for a base ring object of an arbitrary Grothendieck topos. At least for ordinary schemes, it doesn't seem too hard to believe that we could glue together relative cotangent complexes along affine opens, but for things like algebraic spaces and/or formal schemes it seems conceivable to me that it might be substantially harder to glue the local modules together while preserving their simplicial structure. -What difficulties with globalizing the local definition of the cotangent complex lead to the topos-theoretic approach used by Illusie? (This is not a history question. I'm just wondering what the motivation is for the greater generality, since I'm currently reading André's book, which only covers the "classical" case of a commutative $A$-algebra for set-theoretic commutative ring $A$.) - -REPLY [11 votes]: As BCnrd points out, gluing cotangent complexes is a nontrivial thing. You might still ask whether it is really necessary for Illusie to work in the generality of a ringed topos. Would using a ringed space suffice? For standard deformation problems (deformation of a morphism or deformation of a scheme) working on the underlying ringed space would be enough. For more "interesting" deformation problems, like deformation of a morphism $X \rightarrow Y$, where $X$, $Y$, and the morphism are all allowed to vary, one needs something more sophisticated. Illusie constructs a ringed topos that encodes all of these data and then applies the machinery for ringed topoi that he has already developed.<|endoftext|> -TITLE: Proofs of Rohlin's theorem (an oriented 4-manifold with zero signature bounds a 5-manifold) -QUESTION [17 upvotes]: A celebrated theorem of Rohlin states the following - -An oriented closed 4-manifold $M^4$ bounds an oriented 5-manifold if and only if the signature of $M^4$ is zero. - -Simple homological arguments based on Lefschetz duality show that the vanishing of the signature is a necessary condition. Showing that it is also sufficient is however harder. -I know two proofs of this fact. Each is a variation of Rohlin's proof of the simpler 3-dimensional case, which says - -An oriented closed 3-manifold $M^3$ bounds an oriented 4-manifold. - -I was wondering if someone knows a more elementary proof, for instance based on Kirby calculus. The two proofs I know start as follows. - -By Whitney's theorem we can embed any closed oriented n-manifold $M^n$ in $\mathbb R^{2n}$ and we can immerse it in $\mathbb R^{2n-1}$. The immersion self-intersects into circles, and by accurately surgerying $M^n$ we can eliminate these self-intersections. Surgerying changes $M^n$ via a $(n+1)$-dimensional cobordism, hence we can suppose that $M^n$ itself embeds in $\mathbb R^{2n-1}$. -As for knots in 3-space, any codimension-2 closed oriented manifold $M^n \subset \mathbb R^{n+2}$ bounds an oriented "Seifert" $(n+1)$-manifold $W^{n+1}$. - -When $n=3$ these two facts imply that every closed oriented 3-manifold bounds an oriented 4-manifold. When $n=4$ we only obain that every closed oriented 4-manifold is cobordant to a codimension-3 embedded $M^4 \subset \mathbb R^7$ and more work has to be done. - -In his original proof, Rohlin shows that up to blowing up $M^4$ in some points (i.e. making connected sums with $\pm\mathbb {CP}^2$) we can suppose that $M^4$ bounds a 5-cycle in $\mathbb R^7$, which can be subsequently smoothed to an oriented 5-manifold (blow-ups are needed in both steps!). This proof is explained in A la recherche de la topologie perdue. -In Kirby's book The topology of 4-manifolds, he proves that up to cobordism the 4-manifold $M^4$ can be immersed in $\mathbb R^6$. Such an immersion has double and triple points, like a surface in $\mathbb R^3$. Triple points have signs. He proves a nice theorem which says that the number of triple points counted with sign equals (up to a factor) the first Pontryagin number, which in turns equals (up to a factor 3) the signature thanks to the Hirzebruch formula! Therefore if $M$ has signature zero we can pair double points with opposite signs and destroy them by surgery. Finally we obtain an embedded cobordant 4-manifold $M^4 \subset \mathbb R^6$. Now codimension is two and there is a "Seifert" 5-manifold bounding $M^4$. - -Finally, here is my question: - -Do you know any other proof different from these ones? For instance, a proof which does not use embeddings in Euclidean space? - -References are of course welcome. - -REPLY [11 votes]: It depends on what you consider elementary. Gompf-Stipcisz has something like this: Morse theory gives a handle decomposition. Surger circles (the trace gives a bordism) to kill the 1-handles (this introduces 2-handles). Turn upside down and kill the 3-handles. 2 handles are attached along a framed link whose surgery gives an $S^3$ since the 4-manifold is closed. Kirby calculus says your diagram is Kirby move equivalent to the empty framed link. So do the Kirby moves, but every time you blow up a +1 also blow up a -1 (off in a corner) and note that $CP^2 \# -CP^2$ bounds (it's the boundary of a $D^3$ bundle over $S^2$.). Handle slides are diffeos of the 4-manifold. Don't blow down, just move extra $\pm 1$ unknots aside. When you are done, your picture is a unlink with framings $\pm 1$, the same number of each since the signature is zero. This shows your manifold is bordant to a connected sum of $CP^2\# -CP^2$.<|endoftext|> -TITLE: Lattice-ordered commutative monoids -QUESTION [8 upvotes]: By a lattice-ordered monoid, I mean a structure $(A,0,{+},{\vee},{\wedge})$ such that $(A,0,{+})$ is a (not necessarily commutative) monoid, $(A,{\vee},{\wedge})$ is a lattice, and the two distributive laws -$$(a+x+b)\wedge(a+y+b) = a + (x\wedge y) + b$$ -and -$$(a+x+b)\vee(a+y+b) = a + (x\vee y) + b$$ -both hold. A lattice-ordered group is defined similarly. -It is well known that the lattice reduct $(A,{\vee},{\wedge})$ of a lattice-ordered group is always distributive, but this result does not extend to lattice-ordered monoids. - -Are there quasi-equational conditions on the monoid $(A,0,{+})$ that guarantee that the lattice reduct $(A,{\vee},{\wedge})$ is distributive? I am particularly interested in the case where the monoid is commutative. - -I'm not looking for an exact characterization (which is probably impossible in terms of quasi-identities) but for sufficient conditions. - -REPLY [4 votes]: It is highly tempting to expand the signature just a bit so that we can talk about, e.g., residuated lattices. Thus, the discussion in this paper would seem generally relevant; I direct your attention particularly to theorem 15 (page 15) and to the remark near the top of page 20 that the algebras of the quasi-variety described in theorem 15 have distributive lattices as their underlying lattices. So even though such structures go outside the strict confines of your question, perhaps they give some food for thought. -(Or, perhaps you already knew of this paper?) -Just a remark that residuated lattices are important in categorical logic (cf. the concept of quantale) and have nice varietal properties, e.g., they form a Mal'cev variety. -I'll add that I have no particular expertise here, but the question looked interesting.<|endoftext|> -TITLE: Is there a "quantum" Riemann zeta function? -QUESTION [37 upvotes]: Occasionally I find myself in a situation where a naive, non-rigorous computation leads me to a divergent sum, like $\sum_{n=1}^\infty n$. In times like these, a standard approach is to guess the right answer by assuming that secretly my non-rigorous manipulations were really manipulating the Riemann zeta function $\zeta(s) = \sum_{n=1}^\infty n^{-s}$ and its cousins. Then it's reasonable to guess that the "correct" answer is, for example, $\sum_{n=1}^\infty n = \zeta(-1) = -\frac1{12}$. Thus the zeta function and its cousins are a valuable tool for other non-number-theoretic problem solving: it's always easier to rigorously prove that your guess is correct (or discover, in trying to prove it, that it's wrong) than it is to rigorously derive an answer from scratch. -I recently found myself wishing I could do something similar for the sum of the quantum integers. Recall that at quantum parameter $q = e^{i\hbar}$, quantum $n$ is the complex number $$[n]_q = \frac{q^n - q^{-n}}{q - q^{-1}} = q^{n-1} + q^{n-3} + \dots + q^{3-n} + q^{1-n}.$$ The point is that $[n]_1 = n$. - -Question: Are there established methods to sum the divergent series $\sum_{n=1}^\infty [n]_q $ and its cousins? For example, is there some well-behaved function $\zeta_q(s)$ for which the series is naturally the $s=-1$ value? - -Note that when $n$ is a root of unity, the series truncates, and it would be nice (but maybe too much too hope for) if the regularized series agreed with the truncated series at these values. -I should mention also that I consider the following answer tempting but inaccurate, as it definitely doesn't work at roots of unity, which I do care about: -$$ \sum_{n=1}^\infty [n]_q = \frac1{q-q^{-1}} \sum_{n=1}^\infty (q^n - q^{-n}) = \frac1{q-q^{-1}} \left( \sum_{n=1}^\infty q^n - \sum_{n=1}^\infty q^{-n}\right) = $$ -$$ = \frac1{q-q^{-1}} \left( \frac{q}{1-q} - \frac{q^{-1}}{1-q^{-1}}\right) = \frac{q+1}{(q-q^{-1})(q-1)}$$ - -REPLY [11 votes]: Here is another article dealing with similar functions: -q-analogue of Riemann’s ζ-function and q-Euler numbers. by Junya Satoh. -There are also many articles by Taekyun Kim on related functions. -One key point is that the value of the function $\zeta_q$ at negative integers is a fraction which has no limit when $q$ goes to $1$. One can obtain a relation to the $q$-Bernoulli numbers introduced by Carlitz in 1948, by taking a difference with the value of a modified $\zeta_q$ function.<|endoftext|> -TITLE: Interesting examples of vacuous / void entities -QUESTION [103 upvotes]: I included this footnote in a paper in which I mentioned that the number of partitions of the empty set is 1 (every member of any partition is a non-empty set, and of course every member of the empty set is a non-empty set): - -"Perhaps as a result of studying set theory, I was surprised when I learned that some respectable combinatorialists consider such things as this to be mere convention. One of them even said a case could be made for setting the number of partitions to 0 when $n=0$. By stark contrast, Gian-Carlo Rota wrote in \cite{Rota2}, p.~15, that 'the kind of mathematical reasoning that physicists find unbearably pedantic' leads not only to the conclusion that the elementary symmetric function in no variables is 1, but straight from there to the theory of the Euler characteristic, so that 'such reasoning does pay off.' The only other really sexy example I know is from applied statistics: the non-central chi-square distribution with zero degrees of freedom, unlike its 'central' counterpart, is non-trivial." - -The cited paper was: G-C.~Rota, Geometric Probability, Mathematical Intelligencer, 20 (4), 1998, pp. 11--16. The paper in which my footnote appears is the first one you see here, doi: 10.37236/1027. -Question: What other really gaudy examples are there? -Some remarks: - -From one point of view, the whole concept of vacuous truth is silly. It is a counterintuitive but true proposition that Minneapolis is at a higher latitude than Toronto. "Ex falso quodlibet" (or whatever the Latin phrase is) and so if you believe Toronto is a more northerly locale than Minneapolis, it will lead you into all sorts of mistakes like $2 + 2 = 5,$ etc. But that is nonsense. -From another point of view, in its proper mathematical context, it makes perfect sense. -People use examples like propositions about all volcanoes made of pure gold, etc. That's bad pedagogy and bad in other ways. What if I ask whether all cell phones in the classroom have been shut off? If there are no cell phones in the room (that is more realistic than volcanoes made of gold, isn't it??) then the correct answer is "yes". That's a good example, showing, if only in a small way, the utility of the concept when used properly. -I don't think it's mere convention that the number of partitions of the empty set is 1; it follows logically from some basic things in logic. Those don't make sense in some contexts (see "Minneapolis", "Toronto", etc., above) but in fact the only truth value that can be assigned to $\text{“}F\Longrightarrow F\text{''}$ or $\text{“}F\Longrightarrow T\text{''}$ that makes it possible to fill in the truth table without knowing the content of the false proposition (and satisfies the other desiderata?) is $T.$ That's a fact whose truth doesn't depend on conventions. - -REPLY [3 votes]: Pfister's local-global theorem has an interesting "vacuous" instance: -Remember that quadratic forms over a field $F$ of characteristic $\neq 2$ are diagonalizable, and hence a nondegenerate quadratic form can be represented by a sequence $\langle a_1, \ldots, a_m \rangle$ of elements $a_i \in F^*$. -One can "add" diagonal quadratic forms by declaring -$\langle a_1, \ldots, a_m \rangle \oplus \langle b_1, \ldots, b_k \rangle :=\langle a_1, \ldots, a_m, b_1, \ldots, b_k \rangle$ and one can multiply quadratic forms with natural numbers by declaring $n\cdot\langle a_1, \ldots, a_m \rangle:=\underbrace{\langle a_1, \ldots, a_m \rangle\oplus \ldots \oplus \langle a_1, \ldots, a_m \rangle}_{n \text{ times}}$ -If "$<$" is an ordering on $F$ (compatible with $+$ and $\cdot$), then to a quadratic form $f=\langle a_1, \ldots, a_m\rangle$ one can assign its signature $\sigma(f)(<):=\#\{ a_i \mid 0 < a_i\} - \#\{a_i \mid a_i < 0\}$. That is, the signature is the number of positive minus the number of negative diagonal entries, according to the ordering "$<$". -One can show that this signature is independent of the choice of diagonalization. -Thus, if we denote by $X(F)$ the set of all orderings of $F$, from a fixed quadratic form $f$ we obtain a well-defined map $X(F)\to \mathbb{Z}, \ < \,\mapsto \sigma(f)(<)$. -Pfister's local-global theorem says something about how much the map $\sigma(f)$ determines $f$: -Let $f,g$ be quadratic forms of the same rank over a field $F$. -If $\sigma(f)=\sigma(g)$, then there exists $\ell \in \mathbb{N}$ such that $2^\ell\cdot f \cong 2^\ell \cdot g$ (i.e. the multiplied forms are isometric). -[The theorem says more, but this is the relevant part.] -When $F$ is not orderable at all (e.g. if $F$ is algebraically closed or of characteristic $>0$, or $F=\mathbb{Q}_p$), then $X(F)=\emptyset$. Therefore there exists only one map $S\colon X(F) \to \mathbb{Z}$, i.e. any two quadratic forms $f,g$ satisfy $\sigma(f)=\sigma(g)$. Thus, for a non-orderable field, any two quadratic forms have isometric multiples: $2^\ell\cdot f \cong 2^\ell \cdot g$ for some $\ell$. -I learned this from a wonderful expository note by Pete L. Clark.<|endoftext|> -TITLE: Is every real n-manifold isomorphic to a quotient of $\mathbb{R}^n$? -QUESTION [18 upvotes]: I'm curious about the following: -Is every real $n$-manifold isomorphic to a quotient of $\mathbb{R}^n$? -Thanks. -EDIT: As Tilman points out, the manifold should be connected. Also, yes, I'm thinking about topological quotients. Specifically, is there a surjective map $\mathbb{R}^n\to M$ such that $M$ has the quotient topology? -EDIT': I guess an interesting addendum to the question is "when is it true?" - -REPLY [33 votes]: Hahn–Mazurkiewicz Theorem: Suppose $X$ is a nonempty Hausdorff topological space. -Then the following are equivalent: - -there is a surjection $[0,1]\to X$, -$X$ is compact, connected, locally connected and second-countable. - -It follows that a Hausdorff space satisfying the conditions of (2) is a quotient -of $I = [0,1]$. -Cor: Every connected compact manifold is a quotient of $I$. -Since $I$ is a quotient of $\mathbb{R}^n$, we have your answer. -Cor: Every compact connected $m$-manifold is a quotient of $\mathbb{R}^n$ for any $n\geq 1$.<|endoftext|> -TITLE: Smallest eigenvalue of a tricky random matrix -QUESTION [19 upvotes]: While experimenting with positive-definite functions, I was led to the following: -Let $n$ be a positive integer, and let $x_1,\ldots,x_n$ be sampled from a zero-mean, unit variance gaussian. Consider the (positive-definite) matrix $$M_{ij}=\frac{1}{1+|x_i-x_j|}.$$ -Now I wish to know: - -How do I obtain an estimate for the smallest eigenvalue $\lambda_n$ of $M$? - -Preliminary experiments (see plot; x-axis: $n$, y-axis: $\lambda_n$) suggest that $\lambda_n \approx 1/n^2$, but how do I prove that or perhaps a more accurate result? - -REPLY [16 votes]: I think I can get an upper bound of $O(1/n^2)$ by exhibiting a vector $v$ of magnitude comparable to $1$ which gets mapped to a vector of magnitude $O(1/n^2)$. The basic idea is to exploit the birthday paradox to find (with high probability) two indices $i \neq i'$ such that $x_i-x_{i'} = O(1/n^2)$. It should also be possible to then find another additional index $i''$ such that $x_{i''} = x_i + O(1/n)$. -Now look at the $i^{th}$ and $(i')^{th}$ rows, which have components $1/(1+|x_i-x_j|)$ and $1/(1+|x_{i'}-x_j|)$. These rows differ by $O(n^{-2})$ in each coefficient. This already gives an upper bound of $O(n^{-3/2})$ for the smallest eigenvalue, but one can do better by using Taylor expansion to note that the difference between the two components is $(x_i-x_{i'}) \hbox{sgn}( x_i-x_j ) / (1 + |x_i-x_j|)^2 + O(n^{-4})$ except when $x_j$ is very close to $x_i$, at which point we only have the crude bound of $O(n^{-2})$. Similarly, the difference between the $i''$ and $i$ rows is something like $(x_i-x_{i''}) \hbox{sgn}( x_i-x_j ) / (1 + |x_i-x_j|)^2 + O(n^{-4})$ except when $x_j$ is too close to $x_i$. So we can use a multiple of the second difference to mostly cancel off the first difference, and end up with a linear combination of three rows in which most entries have size $O(n^{-4})$ and only about $O(1)$ entries have size $O(n^{-2})$. This seems to give an upper bound of $O(n^{-2})$ for the least eigenvalue (or least singular value), though I have not fully checked the details. -To get a matching lower bound is trickier. One may have to move to a Fourier representation of the matrix as this would more readily capture the positive definiteness of the matrix (as suggested by Bochner's theorem).<|endoftext|> -TITLE: Reference request: moduli spaces of vector bundles -QUESTION [9 upvotes]: I am trying to study the moduli spaces of holomorphic vector bundles quickly, and I'm primarily interested in understanding: - -Why and where the stability condition is used. -How are the moduli spaces constructed. -What are the examples, especially in the case of vector bundles on curves. - -I was looking for some references but I could not find any reasonable source online. -I have the base for that and I don't want some thing very long and full of details. -I just want to see all the ideas very clearly without to much details. -Please tell me if you know any lecture notes or a book which contains this stuff. -Thanks in advance - -REPLY [2 votes]: Twelve years after Atiyah's article classifying vector bundles on curves of genus one, Narasimhan and Ramanan published a lovely paper in the Annals, (89) no.2, 1969, p.14, where they solved the case of semi stable rank 2 vector bundles on genus two curves. This case is perhaps more typical of the higher genus situation. Basically, a rank two vector bundle is analyzed by producing a sub line bundle, whose quotient is also a line bundle, and then studying how the vector bundle is reconstructed as a twisted sum of those two line bundles. I cannot improve on the wonderful references given by Georges Elencwajg above, but I have a short 4 or 5 page set of notes from a lecture given by Daniele Arcara, in a graduate class of mine, summarizing the status of moduli of rank 2 bundles on curves in 2001, if there is some way to send it to you or attach it here as a pdf file.... -I sent them to your email address at Princeton.<|endoftext|> -TITLE: Wick rotation and the Riemann zeta function -QUESTION [12 upvotes]: The goal of this question is to conceptualize in some way the fact that the Riemann zeta function $\zeta(s)$, and other zeta functions like it, have analytic continuations. -Background -I have by now convinced myself that the following is a reasonable conceptualization of why the Riemann zeta function for real $s > 1$ is a natural object of study. First, the probability distributions $\mathbb{P}(X = n) = \frac{1}{\zeta(s)} \left( \frac{1}{n^s} \right)$ on $\mathbb{N}$ are the unique ones satisfying the following two conditions: - -Given that $n | X$, the probability distribution on $\frac{X}{n}$ is the same as the original distribution, and -$\mathbb{P}(X = n)$ is monotonically decreasing. - -Second, if one looks at the sequence of measures $\mu_s(A) = \sum_{a \in A} \mathbb{P}(X = a)$ of a subset $A \subset \mathbb{N}$ with respect to the above distribution, then the $s \to 1^{+}$ limit is the logarithmic density of $A$, which agrees with the natural density of $A$ if it exists. -One can also use statistical-mechanical language to describe the above distribution. There is a statistical-mechanical system, the Riemann gas, whose states are the positive integers $n$ and whose energies are the numbers $\log n$, and $\zeta(s)$ is its partition function (which then determines a distribution on $\mathbb{N}$ in the usual way). This explanation conceptualizes, among other things, the von Mangoldt function, whose Dirichlet series is just the average energy of the above system. -However, the language of probability distributions is insufficient for talking about $\zeta(s)$ for $s \le 1$ or for complex $s$. -Question -Is there a way to conceptualize the values of the zeta function at complex values of $s$ as a "Wick rotation" of its values at real $s$? That is, is there some reasonable quantum-mechanical interpretation of numbers like the "formal" measure -$$\mu_{s+it}(A) = \frac{1}{\zeta(s+it)} \sum_{a \in A} e^{-(s + it)\log a}$$ -(for $s, t$ real) as a probability amplitude, or something along those lines? Does this reasonable quantum-mechanical interpretation single out the critical line $s = \frac{1}{2}$? - -REPLY [8 votes]: The Riemann zeta function $\zeta(s)$ at complex $s$ has the statistical physics interpretation of a partition function at complex temperature. This has no direct physical meaning in general, but for certain models it does. A notable example is the Ising model, where the real and imaginary temperature axes are related by a transformation from an hexagonal to a triangular lattice. -Quite generally, the zeroes of the partition function in the complex plane fall on lines rather than in areas. For ferromagnetic models this is the content of the Yang-Lee theorem. It is therefore natural to expect the Riemann hypothesis to hold, although the Yang-Lee theorem does not cover this case. -An overview of the older literature on complex temperature partition functions is: -"Location of zeros in the complex temperature plane: Absence of Lee-Yang theorem", W. van Saarloos and D. A Kurtze, J. Phys. A: Math. Gen. 17 (1984) 1301-1311. -A more recent paper is -"Complex-temperature partition function zeros of the Potts model on the honeycomb and kagome ́ lattices", H. Feldmann, R. Shrock, and S.-H. Tsai, Phys. Rev. E 57, 1335 (1998). -There are many more papers, it is a quite active field of study. -A very recent paper is http://arxiv.org/pdf/1110.0942<|endoftext|> -TITLE: Is there an additive functor between abelian categories which isn't exact in the middle? -QUESTION [20 upvotes]: Suppose $F: C\to D$ is an additive functor between abelian categories and that -$$0\to X\xrightarrow f Y\xrightarrow g Z\to 0$$ -is and exact sequence in $C$. Does it follow that $F(X)\xrightarrow{F(f)} F(Y)\xrightarrow{F(g)} F(Z)$ is exact in $D$? In other words, is $\ker(F(g))=\mathrm{im}(F(f))$? -Remark 1: If the answer is no, a counterexample must use a non-split short exact sequence. This is because additive functors send split exact sequences to split exact sequences. A splitting is a pair $s:Y\to X$ and $r:Z\to Y$ so that $id_Y=f\circ s+r\circ g$, $id_X=s\circ f$, and $id_Z=g\circ r$. An additive functor preserves these properties, so $F(s)$ and $F(r)$ will split the sequence in $D$. -Remark 2: You probably know you know lots of left exact and right exact additive functors, but you also know lots of exact in the middle additive functors. $H^i$ and $H_i$ for any (co)homology theory are neither left or right exact, but they are exact in the middle by the long exact sequence in (co)homology. - -REPLY [13 votes]: As far as I remember, there is an important example of a functor that transforms mono- and epimorphisms into mono- and epimorphisms, respectively, but is not half-exact; this is the functor of intermediate extension of perverse sheaves.<|endoftext|> -TITLE: Serre's FAC versus Hartshorne as an introduction to sheaves in algebraic geometry -QUESTION [11 upvotes]: I just found an English translation of Serre's FAC at Richard Borcherds' Algebraic Geometry course web page. I really want to read it sometime. I am beginner in Algebraic Geometry, just started learning Scheme theory (Hartshorne Ch II). My question is : -Shall I read coherent sheaves and the cohomology from the translation of Serre's FAC or from the Hartshorne, this is going to be my first encounter with Coherent sheaves. Do you recommend FAC as an introduction to Coherent Sheaves. -My prime interest at present is the derived categories of coherent sheaves on projective varieties, specially the work of Tom Bridgeland and the Mirror Symmetry. - -REPLY [18 votes]: As always, the source you use may be related to what your goals are. To give some perspective, recall there are several ways to define sheaf cohomology, and Serre and Hartshorne feature different methods. Serre used Cech cohomology, and there the important long exact sequence property does not always hold. He was able to prove however that it does hold for "coherent" sheaves. One big advantage of Cech theory is its easier computability in specific cases, such as on projective space. Hartshorne presents first Grothendieck's theory of derived functor cohomology, but then proves it agrees with Cech cohomology before using that theory to compute the cohomology of coherent sheaves on projective space. But if you want to learn the derived functor theory you must choose Hartshorne over Serre. -The distinction made above between schemes and varieties is also relevant. Serre teaches Cech cohomology on varieties,and Hartshorne presents derived functor cohomology on schemes. If you are only interested in varieties, or prefer learning cohomology in the easier setting of varieties, then you may prefer Serre's FAC. Another good source is the book Algebraic Varieties by George Kempf, where the derived functor theory is presented on varieties and used for basic computations, including coherent cohomology of projective space and even the full Riemann Roch theorem, before being linked with Cech theory. So if you want to learn to make computations with the abstract derived functor theory you might prefer George's treatment, although some details are missing there, and some misprints exist. -Finally there are slight differences in Serre's and Hartshorne's results which can be relevant in some settings. E.g. in Beauville's book on surfaces, he uses Cech theory to relate rank two vector bundles on curves with ruled surfaces. To prove that all ruled surfaces arise from vector bundles he then uses Serre's result that Cech H^2 vanishes on a curve with coefficients in any sheaf coherent or not. (He also gives a second argument.) But this vanishing theorem for Cech cohomology does not follow from Hartshorne's treatment, since he proves vanishing for derived functor cohomology but does not relate derived functor and Cech cohomology on non coherent sheaves above degree one. -There is a sentence in Hartshorne, at the end of chapter III, section 2, page 212 in the 1977 edition, which says that Serre proved vanishing for "coherent sheaves on algebraic curves and projective algebraic varieties", whereas the correct statement would be that he proved it "for curves, and for coherent sheaves on projective algebraic varieties". Since Robin is very careful, one wonders whether some well intentioned copy editor did not change this sentence's meaning unwittingly to make it flow better.<|endoftext|> -TITLE: Regularity of asymptotic cones -QUESTION [6 upvotes]: Are there any general conditions guaranteeing that the asymptotic cone of a group/graph is "regular" in some sense? E.g. for $\mathbb{Z}^d$ we get $\mathbb{R}^d$ as the asymptotic cone, which is even a manifold, but for general groups we only get a metric space without additional structure. Does knowing that asymptotic cone is regular (e.g. a manifold) imply any properties of the original group? - -REPLY [5 votes]: I would just like to add to the answer by Simon Thomas that --if a group is virtually nilpotent, its asymptotic cones are very regular: they have a Lie group structure and their metric is of Carnot-Caratheodory type (these metrics are described in the wikipedia article "Sub-Riemannian manifold"). Also, the asymptotic cones do not depend on the scaling factor. --if a group is not virtually nilpotent, its asymptotic cones tend to be VERY large objects. For example, the asymptotic cones of each non-virtually cyclic hyperbolic group are real trees with valency $2^{\aleph_0}$ at each point (those groups have exponential growth, I have to admit that I know very little about asymptotic cones of groups of intermediate growth).<|endoftext|> -TITLE: What is the German translation for intertwiner? -QUESTION [5 upvotes]: I'm searching for a translation for the term "intertwiner" in German. - -REPLY [11 votes]: I (native speaker) learned the term "Vertauschungsoperator" in my undergraduate courses. Unfortunately I cannot cite any reference right now, except the fact that the lecturer of the courses, Prof H.S. Holdgruen, is very sensible in his use of the german language. Therefore I estimate the probability very high that this is a common term in german texts on representation theory.<|endoftext|> -TITLE: Explicit Hamel basis of real numbers -QUESTION [20 upvotes]: Is there an explicit construction of a Hamel basis of the vector space of real numbers $\mathbb R $ over the field of rational numbers $\mathbb Q $? - -REPLY [3 votes]: Re Andreas' comment, I have a writeup of a generalization of Miller's result: https://ivv5hpp.uni-muenster.de/u/rds/talks_kiel.pdf<|endoftext|> -TITLE: How small can a sum of a few roots of unity be? -QUESTION [104 upvotes]: Let $n$ be a large natural number, and let $z_1, \ldots, z_{10}$ be (say) ten $n^{th}$ roots of unity: $z_1^n = \ldots = z_{10}^n = 1$. Suppose that the sum $S = z_1+\ldots+z_{10}$ is non-zero. How small can $|S|$ be? -$S$ is an algebraic integer in the cyclotomic field of order $n$, so the product of all its Galois conjugates has to be a non-zero rational integer. Using the utterly crude estimate that the magnitude of a non-zero rational integer is at least one, this gives an exponential lower bound on $S$. On the other hand, standard probabilistic heuristics suggest that there should be a polynomial lower bound, such as $n^{-100}$, for $|S|$. (Certainly a volume packing argument shows that one can make $S$ as small as, say, $O(n^{-5/2})$, though it is unclear to me whether this should be close to the true bound.) Is such a bound known? Presumably one needs some algebraic number theoretic methods to attack this problem, but the only techniques I know of go through Galois theory and thus give exponentially poor bounds. -Of course, there is nothing special about the number $10$ here; one can phrase the question for any other fixed sum of roots, though the question degenerates when there are four or fewer roots to sum. - -REPLY [19 votes]: This question grabbed my attention a couple of years ago and I've just put a paper on the arXiv with new upper bounds for $k=5$. I began by computing lots of data, then teased out the structure of particularly well-performing configurations. The headline is that $f(5,n) = O(n^{-4/3})$, improving to $O(n^{-7/3})$ infinitely often. -The first picture that really caught my eye was this one, which shows a large dip in the minimum length for $n \approx 10 000$ (log-log scales) when we only look at $n \equiv 11 \mod 12$. - -Looking at the points in the dip led to me scribbling this page. - -These small sums arise from two ideas: that $1 + \omega + \omega^2 + i + (-i) = 0$ (writing $\omega$ for the primitive third root), and that perturbations of this configuration can also be small when they are related to close rational approximations of $\sqrt 3$. -There are also places with dips corresponding to perturbations of the set of fifth roots of unity, relying on close rational approximations to the golden ratio $\phi$. This is easier to work with than $\sqrt 3$, so most of the arguments in the paper are in that setting. (It's easier to work in prescribed congruence classes, for example, because the convergents to $\phi$ have a very simple structure.) I also mention perturbing other configurations which sum to zero, with no concrete improvements but greater possibilities for analysis by more sophisticated means. -My code and data aren't anywhere online right now, but that's only because I haven't decided on a good stable place to post them. I'll happily share them with anyone who's interested. -e: The perfect is the enemy of the good, so my code and data are at least temporarily available at https://babarber.uk/583/small-sums-of-five-roots-of-unity/<|endoftext|> -TITLE: Model structure on Simplicial Sets without using topological spaces -QUESTION [27 upvotes]: The category of simplicial sets has a standard model structure, where the weak equivalences are those maps whose geometric realization is a weak homotopy equivalence, the cofibrations are monomorphisms, and the fibrations are Kan fibrations. -Simplicial sets are combinatorial objects, so morally their model structure should not be dependent on topological spaces. Are there any approaches to this model structure which do not use the geometric realization functor, and do not use topological spaces? - -REPLY [29 votes]: Quillen's original proof (in Homotopical Algebra, LNM 43, Springer, 1967) is purely combinatorial (i.e. does not use topological spaces): he uses the theory of minimal Kan fibrations, the fact that the latter are fiber bundles, as well as the fact that the classifying space of a simplicial group is a Kan complex. This proof has been rewritten several times in the literature: at the end of -S.I. Gelfand and Yu. I. Manin, Methods of Homological Algebra, Springer, 1996 -as well as in -A. Joyal and M. Tierney An introduction to simplicial homotopy theory -(I like Joyal and Tierney's reformulation a lot). However, Quillen wrote in his seminal Lecture Notes that he knew another proof of the existence of the model structure on simplicial sets, using Kan's $Ex^\infty$ functor (but does not give any more hints). -A proof (in fact two variants of it) using Kan's $Ex^\infty$ functor is given in my Astérisque 308: the fun part is not that much about the existence of model structure, but to prove that the fibrations are precisely the Kan fibrations (and also to prove all the good properties of $Ex^\infty$ without using topological spaces); for two different proofs of this fact using $Ex^\infty$, see Prop. 2.1.41 as well as Scholium 2.3.21 for an alternative). For the rest, everything was already in the book of Gabriel and Zisman, for instance. -Finally, I would even add that, in Quillen's original paper, the model structure on topological spaces in obtained by transfer from the model structure on simplicial sets. And that is indeed a rather natural way to proceed.<|endoftext|> -TITLE: If $H$ is a separable Hilbert space, is $L^2(H)$ separable? -QUESTION [7 upvotes]: Let $H$ be a separable Hilbert space, and let $\gamma$ be a Radon probability measure on $H$ with mean zero and covariance operator the identity $I$. -Is the Hilbert space $L^2(H,\gamma)$ separable? - -REPLY [14 votes]: By Example 7.14.13 in Volume 2 of Bogachev's Measure Theory, every Radon measure on $H$ is separable, so that $L^2(H,\gamma)$ is also separable. It is not necessary that $H$ is a Hilbert space, just that every compact subset of $H$ be metrizable.<|endoftext|> -TITLE: Conformal Mappings for hyperbolic polygon -QUESTION [8 upvotes]: I am searching for a conformal mapping from the upper halfplane onto a hyperbolic polygon, i.e. the sides of the polygon have to be geodesics. -The classical Schwarz Christoffel theorem does the job for euclidean polygons (see e.g. http://en.wikipedia.org/wiki/Schwarz-Christoffel_mapping). -Does anybody know of a similar constructions in hyperbolic geometry? -Does anybody know of similiar constructions for any other domains? -Any idea will be very wellcomed! I am far from being an expert in conformal mappings and do only know some isolated examples! - -REPLY [2 votes]: There is a theory of conformal map for circular polygons (polygons bounded by arcs of circles). -But in this case, instead of an integral in the Schwarz-Christoffel formula, you obtain a linear -differential equation. In the case of a circular triangle, the equation is hypergeometric and you -have an explicit representation of your mapping. The paper of Harmer and Martin mentioned in the previous answer deals mainly with the case of a triangle. The most comprehensive treatment of triangles -is in the second volume of Caratheodory's textbook on complex variables, and in other books -on hypergeometric functions. The case of quadrilateral is -the simplest case when there is no explicit formula. It was subject of much research. -See, for example, arXiv:1110.2696, arXiv:1111.2296, and references in these papers.<|endoftext|> -TITLE: Picard group of $\mathfrak{M}_g$ -QUESTION [9 upvotes]: Let $\mathfrak{M}_g$ denote the moduli stack of Riemann surfaces of genus $g$, it is a smooth complex analytic stack, and is the analytic stack underlying $\mathsf{M}_g$, the moduli stack of complex algebraic curves of genus $g$. Alternatively, it is the quotient stack $\mathcal{T}_g // \Gamma_g$ of the mapping class group acting on Teichmuller space (by biholomorphisms). -There are two things we might mean by the Picard group of $\mathfrak{M}_g$, using either holomorphic line bundles or algebraic line bundles. The first is $Pic_{hol} := H^1(\mathfrak{M}_g;\mathcal{O}^\times_{\mathfrak{M}_g})$ and the second is $Pic_{alg} := H^1(\mathsf{M}_g;\mathcal{O}^\times_{\mathsf{M}_g})$. As $\mathfrak{M}_g$ has a finite cover by the quasiprojective variety $\mathfrak{M}_g[3]$ of curves with a level 3 structure, I believe the latter group can also be defined to be the subgroup of $Pic_{hol}$ of those holomorphic line bundles which become algebraic on $\mathfrak{M}_g[3]$. If not, this defines yet another notion of Picard group. -It seems to me that it is $Pic_{alg}$ which is usually discussed, for example in Mumford's paper on Picard groups of moduli problems. Is $Pic_{hol}$ discussed explicitly anywhere? On the other hand, presumably these groups agree (certainly $Pic_{alg} = \mathbb{Z}$ is a summand of $Pic_{hol}$). Is this explicitly stated and proved anywhere? -Any other observations about this distinction are welcome too. -EDIT: As a related question, suppose that one only knew about the analytic object $\mathfrak{M}_g$. Can one directly show that $c_1 : Pic_{hol} \to H^2(\mathfrak{M}_g;\mathbb{Z})$ is injective? (Granted $H^1(\mathfrak{M}_g;\mathbb{Q})=0$, say.) - -REPLY [9 votes]: This will be a bit sketchy, but hopefully you can fill in the necessary steps. -If you see a problem, let me know. -To avoid getting distracted with stacks, let me use a level $n\ge 3$ structure, and -write $M=M_{g}[n]$. Let $j:M\hookrightarrow \bar M$ be the Satake (not Deligne-Mumford) compactification. -This is the normalization of the closure of the image of $M$ in the Satake compactification -of $A_{g}[n]$. The point is that $codim (\bar M-M)\ge 2$. So given a holomorphic line -bundle $L$ on $M$, $j_*L$ would be a coherent analytic sheaf. Therefore by GAGA -$j_*L$ is coherent algebraic, so that $L$ is an algebraic line bundle. This -implies surjectivity of map $Pic_{alg}\to Pic_{hol}$ in your notation. -This should finish it, since as you said, -you already know injectivity of the above map.<|endoftext|> -TITLE: Subgroups of a finite abelian group -QUESTION [27 upvotes]: Let $$G=\mathbb{Z}/p_1^{e_1}\times\cdots\times\mathbb{Z}/p_n^{e_n}$$ be any finite abelian group. -What are $G$'s subgroups? I can get many subgroups by grouping the factors and multiplying them by constants, for example: If $$G=\mathbb{Z}/3\times \mathbb{Z}/9\times \mathbb{Z}/4\times \mathbb{Z}/8,$$ then I can take $$H=3(\mathbb{Z}/3\times \mathbb{Z}/9)\times 2(\mathbb{Z}/4) \times \mathbb{Z}/8.$$ Do I get all subgroups that way? (I'm interested in all subgroups, not just up-to-isomorphism). -Which are the subgroups $H$ in $G$ for which $G/H$ is primary cyclic? - -REPLY [6 votes]: It's worth noting this special case: if $G$ is isomorphic to $(\mathbb{Z}/p)^r$ then it can be regarded as a vector space of dimension $r$ over the field $\mathbb{Z}/p$, and the subgroups are just the subspaces. The number of linearly independent lists of length $n$ is -$(p^r-1)(p^r-p)\dotsb(p^r-p^{n-1})$ -(choose a nonzero vector $v_1$, then a vector $v_2$ not in the one-dimensional space spanned by $v_1$, then $v_2$ not in the 2-dimensional space spanned by $v_1$ and $v_2$, and so on). For any subspace $V$ of dimension $n$, the number of bases is -$(p^n-1)(p^n-p)\dotsb(p^n-p^{n-1})$ -by the same argument. It follows that the number $N_n$ of subspaces of dimension $n$ is -$ N_n = \frac{(p^r-1)\dotsb(p^r-p^{n-1})}{(p^n-1)\dotsb(p^n-p^{n-1})} - = \frac{(p^r-1)(p^{r-1}-1)\dotsb(p^{r-n+1}-1)}{(p^n-1)(p^{n-1}-1)\dotsb(p-1)} -$<|endoftext|> -TITLE: $G$-bundles on affine spaces -QUESTION [11 upvotes]: Let $G$ be a connected algebraic group. Is it true that every $G$-bundle on -${\mathbb A}^n$ is trivial? What is the reference? -I am actually only interested in the case $n=2$. - -REPLY [3 votes]: In characteristic $p$ you can make an easy counterexample with $n=1$, right? An exact sequence of commutative algebraic groups $0\to E\to X\to \mathbb A\to 0$ with $E$ an elliptic curve.<|endoftext|> -TITLE: Multiplicative order of zeros of the Artin-Schreier Polynomial -QUESTION [12 upvotes]: This question was asked on NMBRTHRY by Kurt Foster: -If $p$ is a prime number and $\mathbb{F}_p$ the field of $p$ elements, the zeroes of the Artin-Schreier polynomial -$x^p - x - 1 \in \mathbb{F}_p[x]$ -obviously have multiplicative order dividing $1 + p + p^2 + \dots + p^{p-1} = (p^p - 1)/(p-1)$ (express the norm as the product of the compositional powers of the Frobenius map) -Once upon a time, long long ago, I read that it had been conjectured (by Shafarevich IIRC) that this is the exact multiplicative order for every prime $p$. Can anyone supply a reference? - -REPLY [10 votes]: The current computational status is that this is known for all p < 126 and also for p = 137, 149, 157, 163, 167 and 173. See Peter L. Montgomery, Sangil Nahm and Samuel S. Wagstaff Jr., "The period of the Bell numbers modulo a prime", Math. Comp., 79, 271, July 2010, 1793-1800. The method used requires a factorization of (p^p-1)/(p-1) which is hard for larger p :-)<|endoftext|> -TITLE: Does "taking the dual space" stabilize? -QUESTION [14 upvotes]: Every book which treats dual spaces of normend spaces states that $(c_0)' = \ell^1$ and $(\ell^1)' = \ell^\infty$ and some also describe $(\ell^\infty)'$. -However, is anything known about higher order duals in general? Does taking the normed dual of a given normed space $X$ stabilize? In other words: Defining $X^{(1)} = X'$ and -$X^{(n)} = (X^{(n-1)})'$. Is there always an integer $n$ such that $X^{(n+1)} = X^{(n)}$? - -REPLY [16 votes]: As noted previously, we can restrict our attention to Banach spaces. -A Banach space $X$ is reflexive if the canonical embedding into its double-dual is onto. -Now let $X$ be any Banach space. We define $X^{(\alpha)}$ for all ordinals $\alpha$ as follows: -$X^{(0)}=X$, $X^{(\alpha+1)}=(X^{(\alpha)})^\prime$, and for limit ordinals $\delta$ let $X^{(\delta)}$ be the direct limit of the $X^{(\alpha)}$, $\alpha<\delta$, $\alpha$ is even. -For a given space $\alpha$, we can ask whether $X^{(\alpha)}$ will ever become reflexive. -The key fact is that a closed subspace of a reflexive space is again reflexive. -(It was pointed out before that $X$ is reflexive iff its dual is. But for the following discussion we have to think about subspaces of reflexive spaces.) -This implies that if $X$ is not reflexive, then the sequence $X^{(\alpha)}$, $\alpha$ ordinal, will never stabilize, i.e., no $X^{(\alpha)}$ will be reflexive, like in the $\ell^\infty$ example above (which can be iterated through all the ordinals). -I find this fact rather striking. It was pointed out to me by either Dirk Werner or Ehrhard Behrends. I don't remember exactly who. -Now remarkably, this needs the axiom of choice (which comes from the use of the Hahn-Banach Theorem in the proof). I have convinced myself a while ago that if you don't have a nontrivial atomless finitely additive probability measure on $\mathbb N$ (which can happen if AC fails), -then the sequence for $X=c_0$ is $c_0$, $\ell^1$, $\ell^\infty$, $\ell^1$ and so on. -I.e., $\ell^1$ and $\ell^\infty$ are both reflexive in this situation and duals of each other. -This is because without the nontrivial measure, every functional on $\ell^\infty$ that vanishes on $c_0$ vanishes on all of $\ell^\infty$. -If I remember correctly, there is some remark about this in Solovay's paper on the model of set theory in which all sets of reals are measurable. In this model there is no nontrivial -finite, finitely additive, atomless measure on the integers.<|endoftext|> -TITLE: Finding constant curvature metrics on surfaces for the case of positive Euler characteristic -QUESTION [10 upvotes]: We approach the problem of finding a metric of constant curvature on a surface (i.e. a $C^\infty$ 2-manifold). Specifically, what we want to do is, given a surface $M$ and a metric $g_0$, show that there exists a new metric $g$ of the form $g=e^{2u}g_0$ for some $u\in C^\infty (M)$ such that $g$ has constant curvature. It is well known that if we let $K_0$ and $K$ be the curvatures associated to $g_0$ and $g$ respectively, then they must satisfy the curvature equation -\begin{equation} -K=(K_0-\triangle u)e^{-2u} -\end{equation} -or equivalently, -\begin{equation} -\triangle u-K_0+Ke^{2u}=0 -\end{equation} -where $\triangle$ stands for the Laplace-Beltrami operator associated to the original metric $g_0$. Thus, solving our problem amounts to solving the curvature equation (i.e. finding $u$) for different values of $K$. -We have essentialy three cases (thanks to the Gauss-Bonnet theorem), which are $K=0$, $K=-1$, and $K=1$. The first two cases are fairly straightforward (simple solutions can be found in an article by Melvin S. Berger 1971, or the book on PDE's by Taylor), but the case when $K=1$ is more challenging. I have read solutions which use the Riemann-Roch theorem on one hand or Ricci flow on the other. -The question is: is there any other way to find a solution to the curvature equation (in the case $K=1$)? The reason I want to know this is because I am writing my thesis and I would like to give the simplest proof of this result to make it more accessible. Thanks! - -REPLY [2 votes]: If you relax the fact the metrics $g$ and $g_0$ have to be point-wise conformal to be globally conformal, i.e there exists $\phi$ a diffeomorphism of $S^2$ such that $g=e^u \phi^*(g_0)$. -Then the existence of a metric conform to $g_0$ with constant curvature is equivalente to the fact that there is only one conformal class on $S^2$. Which can be proved using the fact every surface is locally conformaly flat and the fact that $S^2$ is simply connected.<|endoftext|> -TITLE: Lower bound on number of tetrahedra needed to triangulate a knot complement -QUESTION [10 upvotes]: Following along a similar line to the question asked here: Is there an explicit bound on the number of tetrahedra needed to triangulate a hyperbolic 3-manifold of volume V? -Let $K$ be a (hyperbolic) knot in $S^3$. Let $n$ be the minimal number of crossings of any diagram of $K$ and let $M = S^3 \backslash K$ be its complement. By Moise’s theorem the 3-manifold $M$ can be triangulated by tetrahedra. But is there any known bound on the number of tetrahedra needed to triangulate $M$ as a function of $n$? I am particularly interested in any known lower bounds, but upper bounds would also be interesting. - -REPLY [12 votes]: Revision -Let $t(K)$ be the minimal number of tetrahedra needed to triangulate a knot complement. Let $c(K)$ be the minimal crossing number of a knot. -As Ryan Budney points out in the comments, $t(K)\leq C c(K)$ for some constant $C$. One may show that this lower bound is optimal, indirectly using a result of Lackenby. -For alternating knots, one obtains a linear lower bound on the hyperbolic volume of the knot complement, and therefore on the number of tetrahedra needed to triangulate the complement (as noted in Thurston's answer to the other question), in terms of the twist number of the alternating knot, by work of Lackenby. One may find hyperbolic alternating knots $K_i$ where the twist number is equal to the crossing number $c(K_i)$, and thus $ c(K_i)\leq c_1 Vol(S^3-K_i) \leq c_2 t(K_i)$ for some constants $c_1, c_2$. This shows that the linear lower bound is optimal. -For the other direction, -of course, there is some sort of upper bound, since there are only finitely many knot complements with a triangulation by $\leq n$ tetrahedra, so just take the one with the maximal crossing number! We may attempt to make this relation more explicit, using a result of Simon King. I think I can show that $c(K)\leq e^{p(t(K))}$, for some polynomial $p(n)$. -Inverting this inequality gives a lower bound on $t(K)$ in terms of $c(K)$. -King shows that if one has a triangulation $\tau$ of $S^3$ with $m$ tetrahedra, and a knot $K$ in the 1-skeleton of $\tau$, then the crossing number $c(K)$ is bounded by $C^{m^2}$ for some $C$. -Given a triangulation of the knot complement, one wants to estimate the crossing number. To apply King's result, one must extend the triangulation of $S^3-K$ to a triangulation of $S^3$. The difficulty is that the meridian of the knot may be a very complicated curve in the triangulation of the torus boundary of the knot complement. Work of Jaco and Sedgwick allow one to (in principle) estimate the combinatorial length of the meridian in the triangulation of the boundary torus. Their work also allows one to extend the triangulated knot complement along a solid torus to get a triangulation of $S^3$ with the knot in the 1-skeleton. I think one could obtain from their work an estimate on the number of tetrahedra of $\tau$ which is polynomial in the number of tetrahedra $t(K)$. Together with Simon King's result, this should give a bound $c(K)\leq e^{p(t(K))}$ ($p(n)$ a polynomial) on the crossing number in terms of the number of tetrahedra. I expect the answer to be exponential though. -Exponential lower bounds are realized, for example, by the torus knots. The crossing number is bounded linearly below by the genus. One may find sequences of torus knot complements triangulated by $n$ tetrahedra, but with genus growing exponential in $n$, and therefore crossing number growing exponentially. Estimates of the number of tetrahedra needed to triangulate Seifert fibered spaces were given by Martelli and Petronio, in Complexity of geometric three-manifolds. The point is that you can get a triangulation of a $(p,q)$ torus knot with the number of tetrahedra growing like the continued fraction expansion of $p/q$, which can be like $log(|p|+|q|)$, but the genus is $(p-1)(q-1)/2$.<|endoftext|> -TITLE: Chern classes generating cohomology -QUESTION [5 upvotes]: The fact that Chern classes are Hodge classes (and are rational combinations of algebraic cycles) is a part of the proof of the "Gauss Bonnet theorem" (as given in Griffiths and Harris). So my question is why is the fundamental class of every algebraic variety a rational combination of them? - -REPLY [6 votes]: I am a novice, but here is what I gather from perusing the literature. The Hodge conjecture asserts that all Hodge classes are spanned by algebraic classes. The fact that all algebraic classes are spanned by the chern classes of (holomorphic or algebraic) vector bundles, is proved by resolving the structure sheaf of a subvariety by a finite complex of such vector bundles. I.e. one defines formal "K groups" generated either by isomorphism classes of locally free algebraic sheaves (vector bundles), or more generally by coherent algebraic sheaves, with an equivalence relation defined by formal alternating sums of sheaves occurring in exact sequences. The fundamental result that all coherent algebraic sheaves have resolutions by locally free algebraic sheaves shows these two K groups are isomorphic. Hence the subgroup of algebraic cohomology classes, which is the image of the chern character map on the K group of algebraic coherent sheaves, equals the image of this map on locally free sheaves as well. Since the chern character is generated by chern classes, the result follows. The 1974 book Topics in algebraic and analytic geometry, Princeton mathematical notes #13, by Phillip Griffiths and John Adams discusses this in detail.<|endoftext|> -TITLE: Subdividing a polyhedral space into convex simplices -QUESTION [16 upvotes]: A (Euclidean) polyhedral space is a metric space obtained by "gluing together" several (let's assume finitely many) Euclidean simplices (of varying dimensions) by identifying some faces via isometries. In other words, a polyhedral space is a simplicial complex where every simplex is equipped with a Euclidean metric and these Euclidean metrics agree on common faces. Such a space is naturally equipped with the quotient metric (i.e. the maximal metric that does not exceed the original Euclidean metric on every simplex); this makes it a geodesic metric space. -Including a simplex in a polyhedral space can make some distances smaller. For example, consider a circle made from two segments of different lengths: the distance between the endpoints of the longer one is smaller in the circle that in the original segment. Let me call a simplex in a polyhedral space convex if this does not happen to it (i.e. the distances induced from the polyhedral space on this simplex are the original Euclidean distances). It is obvious that simplices of the original triangulation can be subdivided into convex ones, isn't it? (For example, one can do sufficiently many barycentric subdivisions.) -Unfortunately, this obvious fact is not that easy to prove, and I was not able to locate any source where a proof can be found. (Although it is asserted without proof in several texts that I checked.) Well, I think I know a working plan of a proof, but it would be so long and full of nasty details... So here are my questions: -1) Does this fact (that every polyhedral space admits a subdivision into convex simplices) have a written proof somewhere (or maybe a slick proof that fits in MO)? -2) Is it true that this can be done by iterated barycentric subdivision? -ADDED. The existence of a desired subdivision follows from the fact that every polyhedral space admits a piecewise-linear path-isometric map to a Euclidean space. Unfortunately this fact does not seem to be written in citeable sources either. And it does not answer Q2 anyway. - -REPLY [6 votes]: Your definition (if made precise) is quite long, and it does not worth to define polyhedral space this way. -You may do the following two things: - -Define it as a finite (or locally finite) simplicial complex with a length metric such that each simplex is isometric to Euclidean simplex. -Define it as a compact (or locally compact) metric space such that each vertex adimits a cone neighborhood. - -The second definition is weaker than yours and the first is stronger. -The equivalence of these two definitions is written, -see Local characterization of polyhedral spaces by Lebedeva and me.<|endoftext|> -TITLE: Adding a formal inverse of an element to a free monoid -QUESTION [10 upvotes]: Let $FM_2=\langle a,b\rangle$ be the free monoid of rank 2. If we add a formal inverse to the word $aba$, we get the free group $F_2$ (because both $a$ and $b$ will have inverses). -Question: For which other words $w=w(a,b)$, adding a formal inverse to $w$ turns the free monoid into the free group? -I need a complete description, not just examples. - Update question: The same question for $FM_k$, the free monoid of rank $k\ge 3$. - Edit I moved my answer from here to the answer box below. - -REPLY [2 votes]: OK, I will move my partial answer here as an answer to my question. If anybody can improve that answer, it would be good. - A possible solution. I think I found a solution but it is not very explicit, so a more explicit description is welcome. Let $w$ be a word. Let $S_0=\{w\}$ . We shall construct sets of words $S_n$, $n=0,1,2,...$ by induction. Suppose $S$ is already constructed. If $S_n$ contains two words $u,v$ of the form $pu', v'p$, then we replace these pair of words ($u$ and $v$ may be the same) by the three words $p, u', v'$. Also if we have a pair of words $p, pu$ or $p,up$, then we replace it by $p,u$. This way we get a new set $S_{n+1}$. If we cannot do any changes, the process terminates. Clearly the process eventually terminates because the lengths of the words can only get smaller. - Claim. Inverting $w$ gives us a group iff the last set $S_n$ contains all the generators. - Proof. It is clear that if $S_n$ contains all generators, then the result is a group. Assume that a generator $x_1$ is not in the set. Then $x_1$ is either not the first letter of any word in $S_n$ or not the last letter of any of these words (otherwise $S_n$ is not the terminal set of words). Suppose the former holds (WLOG). Let $S_n=\{u_1,...,u_k\}$. Then adding inverse to $w$ implies adding inverses to $u_1,...,u_k$. Hence the resulting monoid is a quotient of the following monoid: -$$G_t=\langle x_1,...,x_n, t_1,...,t_k\mid u_it_i=1, t_iu_i=1, i=1,...,k\rangle.$$ Moreover it is clear that $G_t$ is in fact islomorphic to the monoid obtained by adding the inverse to $w$. Now the fact that $S_n$ is terminal set means that the presentation of $G_t$ is ``complete" (i.e. confluent and terminating because there are no overlaps, see any book on string rewriting). Now suppose that $x_1$ has an inverse. Then $x_1v=1$ in $G_t$ for some $v$. But this relation cannot be deduced by applying the defining relations of $G_t$ from left to right since $x_1$ is not the first letter of any word in $S_n$, a contradiction.<|endoftext|> -TITLE: What are unramified morphisms like? -QUESTION [10 upvotes]: I'm wondering if finite unramified morphism between reduced schemes decomposes as closed immersions and etale morphisms. Suppose I have a morphism between reduced schemes which is finite, surjective and unramified, is it necessarily etale? I think this is certainly true if both source and target are curves, but I'm not sure about higher dimensional examples. Thanks -EDIT: to avoid trivial example let's assume the source and target are connected. What I'm wondering is precisely when one can deduce flatness from these conditions. - -REPLY [13 votes]: Finite, surjective, and unramified does not imply etale. E.g. suppose that $Y$ is a proper closed subscheme of $X$, and we consider the map $X \coprod Y \to X$ defined as the disjoint union of the identity on $X$, and the given closed immersion $Y \to X$ -on $Y$. -Then this map is finite, unramified, and surjective, but not etale. (See Sandor's answer for the missing condition, -which is flatness!) -Added: A more interesting example is given by letting $X$ be a nodal cubic, letting -$\tilde{X}$ be the normalization, and considering the natural map $\tilde{X} \to X.$ -This map is not flat and certainly not etale, but it is unramified. (Formally, each branch -through the node maps by a closed immersion into the nodal curve.) - -REPLY [8 votes]: You also need flat. See Hartshorne, Ex. III.10.3.<|endoftext|> -TITLE: How to sample pairwise independent gaussians -QUESTION [10 upvotes]: If $X_1, \ldots , X_k$ are i.i.d normal random variables with mean $0$ and variance $1$, then is there a way to sample $Y_1, \ldots , Y_m$ for $m=\omega(k)$ such that each of the $Y_i$'s is a normal random variable with mean $0$ and variance $1$ and they are pairwise independent? - -REPLY [9 votes]: Here is the answer I promised in my last comment. -Instead of considering ${\rm N}(0,1)$ variables, we may consider uniform$[0,1)$ variables. -Indeed, if $Z_i$ are i.i.d. ${\rm N}(0,1)$ variables, then, with $\Phi(\cdot)$ denoting the ${\rm N}(0,1)$ distribution function, $U_i := \Phi (Z_i)$ are i.i.d. uniform$[0,1)$ variables. In turn, if $\tilde U_i$ are pairwise independent uniform$[0,1)$ variables, then $\tilde Z_i := \Phi^{-1} (\tilde U_i)$ are pairwise independent ${\rm N}(0,1)$ variables. -The rest of this answer is based on the recent paper "Recycling physical random numbers", -available at 1 or 2. Henceforth, we use the same letters as in that paper. Suppose that $U_1,\ldots,U_n$ are independent uniform$[0,1)$ variables. Fix $2 \leq m \leq n$, and define $N_m = {n \choose m}$. Now let $X_i$, for $i = 1,\ldots,N_m$, comprise all $N_m$ distinct sums of the form $U_{r_1 } \oplus U_{r_2 } \oplus \cdots \oplus U_{r_m }$, for $1 \le r_1 < r_2 < \cdots < r_m \le n$. Here $U_{r_1 } \oplus U_{r_2 } \oplus \cdots \oplus U_{r_m }$ is the sum modulo $1$ of the $U_{r_i}$, given explicitly by -$$ -U_{r_1 } \oplus U_{r_2 } \oplus \cdots \oplus U_{r_m } = U_{r_1 } + U_{r_2 } + \cdots + U_{r_m } - \left\lfloor {U_{r_1 } + U_{r_2 } + \cdots + U_{r_m } } \right\rfloor , -$$ -where $\left\lfloor \cdot \right\rfloor$ is the floor function. Then, the $X_i$ are pairwise independent uniform$[0,1)$ variables. In particular, by letting $m=2$, we can efficiently construct $n(n-1)/2$ pairwise independent uniform variables from $n$ independent ones. -Finally, for general purposes it might be worth stating the following simple fact (Proposition 2 in the aforementioned paper). For $N \geq 2$, let $Y_1,\ldots,Y_N$ be pairwise independent random variables with common mean $\mu$ and common variance $\sigma^2 < \infty$. Define $\bar Y = \frac{1}{N}\sum\nolimits_{i = 1}^N {Y_i }$ and $s^2 = \frac{1}{{N - 1}}\sum\nolimits_{i = 1}^N {(Y_i - \bar Y)^2 }$. Then, ${\rm E}(\bar Y) = \mu$, ${\rm Var}(\bar Y) = \sigma^2/N$, and ${\rm E}(s^2) = \sigma^2$. Combined with the previous paragraph, a straightforward implication is that for a square-integrable function $f$ defined on $[0,1)$, we can approximate the integral $\mu = \int_{[0,1)} {f(x)\,{\rm d}x}$ using a modest number $n$ of independent random inputs. Indeed, note that $n$ independent random inputs can be used to get unbiased Monte Carlo estimates for $\mu$ with the same variance as with $N_m = {n \choose m}$ independent random inputs.<|endoftext|> -TITLE: How does one motivates the method of separation of variables when teaching PDE's? -QUESTION [15 upvotes]: I'm not sure if this question is appropriate for MO. Add comments if it is not. Thanks. -How to explain/motivate the method of separation of variables for PDEs to undergraduates? What's the real math behind it? It's not just because the guy who fancied it is very smart, right? (Although I feel like it does give students this impression...) -(Background: At Berkeley there is a course called Math 54, in which students learn linear algebra, linear ODEs and then 1 or 2 weeks of PDEs. Teaching Separation of variables is always my nightmare...) - -REPLY [2 votes]: At the level of your Math 54 I suggest the following: By the time you -get to a heat or wave equation, the students are used to seeing functions -such as $ce^{at}$, $c_1\cos(at)+c_2\sin(at)$, as solutions to simple ODE. It is -somewhat natural then to let the constants $c$ become functions of $x$. -It becomes a reasonable thing to try, an extension of ``undetermined coefficients''. - (The Notes on Differential Equations -on my web page uses this description.)<|endoftext|> -TITLE: Basepoints in the Canonical System of Algebraic Surfaces -QUESTION [6 upvotes]: Let X be an smooth projective variety defined over $\mathbb{C}$. In the context of the minimal model program it is often important to understand the geometry of the maps defined by the complete linear systems associated to powers of the canonical class $nK_{X}$. I would like to understand some examples of how intricate these maps can be for small $n$. -For simplicity let's restrict to the case of algebraic surfaces. We assume that $X$ is of general type and minimal. A concrete question that I have in mind is the following. One expects that for large $n$ the complete linear systems become basepoint free, but how large is large? What explicit examples are known where the systems $nK_{X}$ all have a basepoint for $n \leq N$ but become free when $n>N$ ? Is there some absolute bound on $N$, and if so for what surface is the bound saturated? - -REPLY [14 votes]: For a minimal complex surface of general type, to the best of my knowledge, the situation is as follows: - -$|nK_X|$ is free for $n\ge 4$; -$|3K_X|$ is free, except possibly for $K_X^2=1$; -If $K^2_X=1$, then $|2K_X|$ has base points. In the remaining cases $|2K_X|$ is free, except possibly for $K^2_X=2,3,4$ and $p_g(X)=0$. (Examples with $K^2=2$, $p_g=0$ and $|2K_X|$ not free are known). - -Statements 1),2) and statement 3) for $K^2_X>4$ are essentially due to Bombieri (MR0318163) and can be easily reobtained using Reider's Theorem. The proof that $|2K_X|$ is free if $p_g(X)>0$ is due to P. Francia (MR1273376) except for the case $p_g=q=1$ which was settled by Catanese-Ciliberto (MR1273372).<|endoftext|> -TITLE: Relativistic Cellular Automata -QUESTION [14 upvotes]: Cellular automata provide interesting models of physics: Google Scholar gives more than 25,000 results when searching for "cellular automata" physics. -Google Scholar still gives more than 2,000 results when searching for "quantum cellular automata". -But it gives only 1 (one!) result when searching for "relativistic cellular automata", i.e. cellular automata with a (discrete) Minkoswki space-time instead of an Euclidean one. - -How can this be understood? -Why does the concept of QCA seem more -promising than that of RCA? -Are there conceptual or technical barriers for a thorough treatment of RCA? - -REPLY [11 votes]: What Willie's answer shows is that, for some non-trivial Lorentz-translatable cellular automaton, every cell would need an infinite number of neighbors. This cellular automaton couldn't work if any cell had an infinite number of live neighbors. As was pointed out in the comments (2018), it could work if there were only a finite number of live neighbors of any cell. One would need to impose constraints on which kinds of configurations are allowed to ensure this, though. -There's a way of getting around this, however. You could make each cell correspond to a point in space-time and also a boost (a boost is essentially a velocity in the Lorentz group). Then, cells would interact with cells both close to them in space-time and also close in boost. I don't know whether anybody has considered cellular automata like this. -In order for this to have a correspondence to realistic quantum field theories, it would have to be the case that when two particles interact at a high boost, the interaction strength goes to 0 as the boost goes to infinity. I don't know whether this is true, although the thought experiment of considering particles falling into a black hole through a sea of Hawking radiation makes it seem like it might be.<|endoftext|> -TITLE: Structure theorem of f.g. modules over a (non) PID -QUESTION [5 upvotes]: I am looking for an example of a commutative ring with $1$, in which every ideal is generated by a single element, for which the conclusion of the structure theorem for finitely generated modules is wrong. (The ring is allowed to have zero divisors, so it is not a PID). -Are there any examples? What happens if one drops the other conditions instead (commutativity, $1\in R$)? Does then the structure theorem still fail ? - -REPLY [3 votes]: The note linked to in Timothy Wagner's answer has been replaced by another one, which only shows the structure theorem for PIDs, so it may be worth to point out that the structure theorem holds for any principal ideal ring (PIR), possibly with zero divisors. Namely, a theorem of Zariski-Samuel tells us that a PIR is a direct product of PIDs and local artinian PIRs. For these, the structure theorem holds and one has uniqueness (for the latter, see Keenan Kidwell's question he mentioned in the comment). Since a module $V$ over a ring $R= R_1 \times \dotsb \times R_n$ decomposes canonically as $V= Ve_1\oplus \dots Ve_n$, where the $e_i$ are the obvious idempotents, and $Ve_i$ is an $R_i$-module, we are done.<|endoftext|> -TITLE: Is there a good definition of the universal cover for non-connected Lie groups? -QUESTION [7 upvotes]: It is well-known that the universal cover $\tilde G$ of a connected Lie group $G$ has a Lie group structure such that the covering projection $\tilde G\to G$ is a Lie group morphism. Of course $\tilde G$ might not be linear even though $G$ is, but this is not the point here. -My question is: assume that $G$ is a not necessarily connected Lie group. Does there exist a Lie group $\tilde G$ and an onto Lie group morphism $\tilde G\to G$ whose restriction to the identity component of $\tilde G$ is the universal cover of the identity component of $G$? -I assume that the answer is "no" in general, but I could not find any counter-example. - -@Jim: Of course, the terminology "universal cover" would have been inappropriate even though such a cover existed (which as you and André pointed out, is not the case). -I came to this question from some other direction. Namely, the $PSL_2(R)$ action on $RP^1$ lifts to a $\widetilde{PSL_2(R)}$ action on the universal cover of $RP^1$, and this action extends to an action of a two-sheeted cover of $PGL_2(R)$. It is tempting to denote this cover by $\widetilde{PGL_2(R)}$, and I wondered whether such a construction was standard. - -REPLY [8 votes]: Expanding on Andre's answer, there is an obstruction class in $H^3(\pi_0(G),\pi_1(G,e))$ (due to R.L. Taylor, Covering groups of non connected topological groups, Proc. Amer. Math. Soc. 5, -pp753-768, 1954) to the existence of a universal covering space. There is a University of Wales thesis by Mucuk with the main results contained in this paper by Brown and Mucuk which detail when you can get a universal covering space that acts as you want.<|endoftext|> -TITLE: Density of congruence classes covered by a set -QUESTION [9 upvotes]: Thinking about the question Four polynomials representing all integers modulo m lead me to the following complementary question: -If $S$ is a set of positive integers, say that a positive integer $m$ is covered by $S$ if every congruence class $\bmod m$ has a representative in $S$. Denote by $C(S)$ the set of positive integers covered by $S$. If $x>0$ let $S(x) = \{ k \in S : k \le x \}$ and the lower density of $S$, $\ell(S) := \lim \inf_{x \rightarrow \infty} |S(x)|/x$. My question: is there a non-trivial lower bound on $\ell(C(S))$ in terms of $\ell(S)$? That is, is there a continuous function $f : [0,1] \rightarrow [0,1]$, not identically 0, such that $\ell(C(S)) \ge f(\ell(S))$. -The set in the question I referred to has density 0, so my question wouldn't apply to it. However, I wondered if there were a simple argument in the case of positive lower density. This has the smell of the kind of question that Erdos would ask, so I wouldn't be surprised to see it there. - -REPLY [3 votes]: Denote by $P$ the set of prime powers not covered by $S$ (for each prime $p$, take only the smallest non-covered its power). If $\sum_{x\in P} 1/x=+\infty$, then $\prod_{x\in P} (1-1/x)$ is 0, so the product over some finite subset is arbitrarily small. But this product is a density of numbers without forbidden remainders modulo respective prime powers. So, $S$ has density 0. A contradiction. Hence $a=\prod_{x\in P} (1-1/x)$ is positive and $\ell(S)\leq a$. But then complement of $C(S)$ is the set of numbers divisible by at least one element of $P$. Density of such numbers equals $1-a$ (this is rather technical, but standard). So, we get that $\ell(C(S))\geq \ell(S)$.<|endoftext|> -TITLE: Submission of a paper with a serious error to a good journal -QUESTION [45 upvotes]: I recently obtained (or so I thought) a good result, and after a month of reading and rereading what I'd written, submitted my paper to a very good journal. I'm early in my career (got my Ph.D. a few years ago) and have published 2 papers in good journals; so I figured I'd aim for a top journal this time. -Unfortunately, I just got an email from the journal indicating that the anonymous referee had found a serious, probably unfixable error in my paper, and that (obviously) they don't recommend it for publication. After looking at it, I realize that I really should have caught this error: although it is a relatively subtle error buried inside of a technical lemma, it still is an obvious error once you notice it. I'm frustrated that I missed this mistake and I'm embarrassed to have wasted the time of the referee, who probably spent a lot of time combing through my paper before they found the error. -So here are my questions. -1) Has this happened to anyone else? Is this a relatively common occurrence, or am I just sloppy? -2) The anonymous referee is probably someone distinguished in my field. Do they now have a bad impression of me? (This probably is not a question that can easily be answered . . . .) -3) If I manage to patch up this paper, is it reasonable to resubmit it to this journal, or have I burned my bridges there? -I'm going to make this community wiki, since I don't know if there's a "right" answer. - -REPLY [31 votes]: 1) Has this happened to anyone else? Is this a relatively common occurrence, or am I just sloppy? -It is not very common (the usual preventive techniques include showing the draft to a few experts/friends, putting it on ArXiV, letting it lie for a month or two and then rereading it, etc. before sending it to a "top" journal) but it happens now and then. What is common is severe difficulty with finding an error in one's own work. I would say that affects more than a half of mathematicians I know. The reason is that you read not what is written but rather what you believe should be there when you just finish typing the manuscript and start proofreading. The main trick of good proofreading is to turn yourself into a complete idiot, who doesn't see a single step ahead, has no idea of the overall structure of the argument, takes everything literally, and is not convinced of anything that is not clearly put in a modus ponens form. Needless to say, it is about as hard for a shrewd person to read like that as for a genuinely stupid one to read between (or over) the lines. And even if you know all that, you are still destined to submit or even publish papers with errors. Just a few months ago, I was informed about an error in one of mine published papers. It was just a remark and the statement was actually correct, but the proof wasn't. So, to have this kind of public shame now and then is almost inevitable whether you are an unknown postdoc, or Andrew Wiles, or something in between. -I'm not sure if Poincare published a single formally correct proof in his lifetime and people still are completely puzzled by some passages in Linnik's works, so you are in a good company. -2) The anonymous referee is probably someone distinguished in my field. Do they now have a bad impression of me? (This probably is not a question that can easily be answered . . . .) -It is actually easy to answer: most likely, for him you are nobody, your name is just a random combination of letters, and your "initial value" is zero. An erratic paper leaves it this way, so nothing is lost. We are all getting worthless papers to referee every month and I challenge everyone to recall the name of the author of some bad paper he rejected 6 months ago. The only scenario in which "someone distinguished" would bother to take a mental record of your name after looking at a single opus of yours is when he finds something interesting and unusual in your work. Then your value for him is currently positive, though, of course, not as high as it would be if you solved the problem. So, again, there is absolutely nothing to worry about. -3) If I manage to patch up this paper, is it reasonable to resubmit it to this journal, or have I burned my bridges there? -Of course, it is. What matters is not how many mistakes you made on the road and who saw them but whether you finally reached your destination and whether other people consider that destination worth reaching. The theorem and its correct proof lose nothing in value if somebody published or tried to publish 20 false proofs before that. That some of those false proofs might be proposed by a person with the same name and biometrical characteristics as those of the one who finally found a correct proof changes nothing in the grand scheme of events. So, if you manage to fix the error and make sure that the argument is, indeed, correct, I see absolutely nothing wrong with submitting it again because from a purely logical standpoint, it is a different paper. If you get it returned solely on the grounds that the previous version was incorrect and not based on the merit considerations (even correct and good papers get rejected sometimes for various reasons), it'll merely tell you that the jornal is not really "top" but just "snobbish", in which case I would avoid it altogether in the future (at least, until they change the editorial board).<|endoftext|> -TITLE: Is the boundary $\partial S$ analogous to a derivative? -QUESTION [130 upvotes]: Without prethought, I mentioned in class once that the reason the symbol $\partial$ -is used to represent the boundary operator in topology is -that its behavior is akin to a derivative. -But after reflection and some research, -I find little support for my unpremeditated claim. -Just sticking to the topological boundary (as opposed to the boundary -of a manifold or of a simplicial chain), -$\partial^3 S = \partial^2 S$ -for any set $S$. -So there seems to be no possible analogy to -Taylor series. Nor can I see an analogy with the -fundamental theorem of calculus. -The only tenuous sense in which I can see the boundary -as a derivative is that $\partial S$ is a transition between $S$ -and the "background" complement $\overline{S}$. -I've looked for the origin of the use of the symbol $\partial$ in topology without luck. -I have only found references -for its use in calculus. -I've searched through History of Topology (Ioan Mackenzie James) -online without success (but this may be my poor searching). -Just visually scanning the 1935 Topologie von Alexandroff und Hopf, I do not -see $\partial$ employed. -I have two questions: - -Q1. Is there a sense in which the boundary operator $\partial$ is - analogous to a derivative? -Q2. What is the historical origin for the use of the symbol - $\partial$ in topology? - -Thanks! -Addendum (2010). Although Q2 has not been addressed -[was subsequently addressed by @FrancoisZiegler], it seems appropriate to accept one -among the wealth of insightful responses to Q1. Thanks to all! - -REPLY [54 votes]: Q1. Is there a sense in which the boundary operator $\partial$ is analogous to a derivative? - -It may be worth noting that de Rham in (1936) and especially (1938, pp. 317–323) already spelled out most of the other replies’ analogies. Indeed the latter answers the title question with (my bold): - -We already observed (...) an analogy between the boundary operation and differentiation: Both are linear, and performed twice, both give zero: - $$ -\begin{align} -f(c_1\pm c_2)&=fc_1\pm fc_2,& f(fc)&=0,\tag1\\ -d(\omega_1\pm\omega_2)&=d\omega_1\pm d\omega_2,& d(d\omega)&=0.\tag2 -\end{align} -$$ - This analogy actually goes much further (...). The differential of a product is given by the rule - $$ -d(\omega_1\omega_2)= d\omega_1\omega_2+(-1)^p\omega_1d\omega_2, -%\quad -%\style{font-family:sans-serif}{\text{where }} p= -%\style{font-family:sans-serif}{\text{degree of }}\omega_1. -%\style{font-family:sans-serif}{\text{ ...}} -\tag3 -$$ - where $p=$ degree of $\omega_1$. (...) Our analogy’s starting point is the fact that chains, like forms, make an alternating ring (...) and to the differentiation rule for products corresponds the formula which gives the boundary of an intersection: - $$ -f(c_2c_1)=c_2fc_1+(-1)^{n-p}fc_2\cdot c_1. -\tag4 -$$ - The formula is an exact match if we let the intersection $c_2\cdot c_1$ of two chains of dimensions $p_2$ and $p_1$ correspond to the product $\omega_1\omega_2$ of two forms of degrees $n-p_1$ and $n-p_2$. I followed the usual definitions, which is why one must reverse the order of factors. (...) To the index [sum of coefficients] of a $0$-chain there corresponds the number $I(\omega)=\smash{\int_V\omega}$ attached to any $n$-form, and one can show that all matching properties hold. -But I would now like to show that, much more than a mere formal duality, there is a deeper identity between $p$-chains and $(n-p)$-forms; or in other words, that on an $n$-dimensional manifold, a $p$-chain and an $(n-p)$-form are two instances of one and the same concept. This idea (...) is suggested by physical considerations, actually cases where physical quantities occur which can be represented, sometimes by a chain, sometimes by a form. - -As @BS. said, de Rham goes on to unify both in the idea of a current $(c,\omega)$ (pre-distribution theory). Also, the formulas of @DenisSerre and @VaughnClimenhaga are ungraded versions of (4) and -$$ -f(c_1\times c_2)=c_1\times fc_2+(-1)^{\dim c_1}fc_1\times c_2. -\tag5 -$$ -Both (4) and (5) are attributed to Lefschetz (1925, §1; 1926, 16.1, 49.2; cf. 1942, IV.2.3, IV.5.6, V.8.4) by e.g. van der Waerden (1929, p. 343; 1930, p. 129), de Rham (1932), Steenrod (1957, pp. 29, 40). But of these authors, only Steenrod uses $\partial$ and not some variant of $B$, $F$, or $R$. So... - - -Q2. What is the historical origin for the use of the symbol $\partial$ in topology? - -Michèle Audin writes in Henri Cartan & André Weil: du vingtième siècle et de la topologie (2012, p. 46): - -$d$ is for differential, no doubt, but what is $\partial$ for? Neither boundary nor frontier, in neither French nor German (nor even English), so? Well, Rand, in German, written in Gothic and abbreviated $\mathfrak{Rd}$, then $\mathfrak d$ (two letters, that was too much), then $\partial$ ! - -She doesn’t quote sources, but indeed we find in §16 of Seifert & Threlfall’s Lehrbuch der Topologie (1934), the following: - -der Rand von $E^k$ [ist] hiernach die $(k-1)$-Kette - $$ -\mathscr R\partial E^k = \varepsilon\sum_{i=0}^k(-1)^i(P_0P_1\cdots P_{i-1}P_{i+1}\cdots P_k)\tag6 -$$ - (...) Der Rand wird mit dem Symbole $\mathscr R\partial$ bezeichnet. - -They still use the same notation in Variationsrechnung im Grossen (1938, p. 17 sq). The earliest I find using $\partial$ alone in combinatorial topology are Eilenberg (1938, p. 180) and Whitney (1937, pp. 38, 40, 46, 51). In point-set topology — your initial concern — it may also be Whitney: (1944, pp. 221, 247). -Question: has anyone anywhere seen the Gothic (Fraktur) spellings Audin alludes to?<|endoftext|> -TITLE: Generalizations of Belyi's theorem -QUESTION [13 upvotes]: Belyi's theorem states that the following properties of a nonsingular projective algebraic curve $X$ are equivalent: -1) $X$ is defined over $\overline{\mathbb{Q}};$ -2) There exists a meromorphic function $\phi: X\to\mathbb{P}^1\mathbb{C} $ ramified at most at $0,1,$ and $\infty$; -3) $X$ is isomorphic to $\Gamma \backslash \mathbb{H}$ (compactified at cusps) for a finite index subgroup $[\mathrm{PSL}_2(\mathbb{Z}), \Gamma]<\infty.$ -The remaining question is: -$$\boxed{\text{ -Is there a way to treat singularities in this or a similar framework? }}$$ -The following of my original questions have been answered: -Can this be generalized to arbitrary projective nonsingular varities of higher dimensions? (I discussed this with one professor here in Goettingen. That seems to be ongoing research. Please see also comment of David Roberts.) -What compactification do they mean here? $\Gamma \backslash \mathbb{H} \cup \mathbb{Q}$! (see the answer of Robin Chapman) -What is a nice reference for the proof of Belyi's theorem? (see answer of YBL and Koeck + the comments of Emerton) -How does the Galois group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ enter the picture? (see comment of Ariyan and answer of YBL) -Where can I find nice examples where these computations have been done explicitly? -(see answer of Andy Putnam and JSE) - -REPLY [5 votes]: Recently I was wondering about generalizations of Beyli's theorem to higher dimensions and did some googling. As this issue is only discussed briefly in David Roberts' comment, I thought I contribute what references I found hoping someone might find it useful: -There is one direction of research which looks for actions of $\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ on algebraic objects other than curves. A general criterion for an algebraic variety to be defined over a number field has been provided in -this paper of Gabino González-Diez. In the special case of surfaces, there was a question of Catanese on the Galois action on moduli surfaces. This question has been answered in papers of Easton and Vakil (published in International Math. Res. Notices 20, 2007) and of Bauer, Catanese and Grunewald. -There is an interesting -survey of Goldring (published in the Serge Lang memorial proceedings "Number theory, analysis and geometry"), which also discusses higher-dimensional generalizations of Belyi's theorem. Apparently one way of generalizing Belyi's theorem is - -Braungardt's question: Is every connected quasi-projective variety $X$ - that is defined over $\overline{\mathbb{Q}}$ birational to a finite - étale cover of some moduli space of curves $\mathcal{M}_{g,n}$? - -It was formulated in the paper: V. Braungardt: Covers of moduli surfaces. Compositio Math. 140 (2004) 1033-1036. In this paper, there are also some partial results on this question. There is also a paper of Paranjape on realization of surfaces defined over $\overline{\mathbb{Q}}$ as branched covers of $\mathbb{P}^2$.<|endoftext|> -TITLE: Unexpected applications of Dvoretzky's theorem -QUESTION [21 upvotes]: Dvoretzky's theorem is a classic of convex geometry. Recently at a conference in quantum information I learned (from Patrick Hayden's talk) about a nontrivial application of the theorem to a problem in quantum cryptography (which was solved previously, but using more complicated tools). What are the unexpected applications of Dvoretzky's theorem that you have heard of, if any? -(by "unexpected" I mean applying it to a problem which is not directly connected to convex geometry, functional analysis etc. or perhaps is connected, but requires phrasing the problem in a different language in a non-obvious way) - -REPLY [7 votes]: An exotic application to cosmology was suggested by K. Villaverde, O. Kosheleva, and M. Ceberio, Why Ellipsoid Constraints, Ellipsoid Clusters, and Riemannian Space-Time: Dvoretzky's Theorem Revisited. - -Modern cosmology asserts that space-time is high-dimensional with - some complex metric. Still, we observe a locally Euclidean metric in - our four-dimensional world. - Dvoretzky’s theorem explains this phenomenon: we observe a section of a - high-dimensional convex unit ball, and such a section is close to an - ellipsoid, implying a locally Euclidean metric.<|endoftext|> -TITLE: If Q is a subset of the plane of size less than continuum, then does every closed F in Q extend to a closed connected G in the plane with the same trace on Q? (Or is this independent of ZFC?) -QUESTION [11 upvotes]: This question arises in connection with this MO -question -and especially with Sergei Ivanov's wonderful -answer, -which showed that for any countable set -$Q\subset\mathbb{R}^2$ and every closed set $F\subset Q$, -there is a closed connected $G\subset\mathbb{R}^2$ with -$G\cap Q=F$. (In fact, he makes $G$ path-connected.) -My question is about the extent to which this phenomenon -might generalize to higher cardinals, when the Continuum -Hypothesis fails. For example, if the continuum $2^\omega$ -is very large, then can we hope to handle uncountable sets -$Q$ in the way Sergei handled the countable sets, provided -that they have size less than the continuum? Or perhaps the -best possible is always just the countable sets? Or is this -independent of ZFC? -It seems sensible to introduce what seems to be a new -cardinal characteristic here. Specifically, let $\kappa$ be -the size of the smallest counterexample, that is, the -smallest cardinal size of a set $Q\subset\mathbb{R}^2$ -having a closed subset $F\subset Q$ for which there is no -closed connected $G\subset\mathbb{R}^2$ with $G\cap Q=F$. -Sergei proved that this cardinal $\kappa$ is uncountable, -and obviously $\kappa$ is at most the continuum (it is easy -to make counterexamples of size continuum), and so -$$\omega_1\leq\kappa\leq 2^\omega.$$ -So the question is, what can we say about -$\kappa$ in ZFC? -If the Continuum Hypothesis holds, of course, then the two -endpoints above are identical and so $\kappa=2^\omega$. But -is it consistent with ZFC that -$$\omega_1\leq \kappa\lt 2^\omega?$$ -Perhaps one can achieve particular values of $\kappa$ by -forcing? Is the cardinal $\kappa$ related to other -well-known cardinal characteristics? Perhaps the value of -$\kappa$ is pushed up to the continuum $2^\omega$ by some -of the standard forcing axioms? - -REPLY [3 votes]: In fact, the MO user @trutheality was right saying that the cardinal $\kappa$ is equal to the continuum. This follows from -Theorem. For any set $X\subset\mathbb R^d$ of cardinality $<\mathfrak c$ in the Euclidean space of dimension $d\ge 2$ and any closed subset $F\subset X$ there exists a closed path-connected set $P\subset \mathbb R^n$ such that $P\cap X=F$. -Proof. First we prove a partial case of this theorem for $X$ contained in the power $\mathbb I^d$ of the set $\mathbb I:=\mathbb R\setminus\mathbb Q$ of irrational numbers. -Lemma. For any subset $X\subset\mathbb I^d$ there exists a closed path-connected set $P\subset \mathbb R^d$ such that $P\cap X=F$. -Proof. Consider the open cover $$\mathcal C:=\{x+(0,1)^d:x\in\mathbb Z^d\}$$ of $\mathbb I^d$ by integer translations of the open unit cube $(0,1)^d$. -Next, for every $n\in\omega$ consider the homothetic copy $\mathcal C_n:=\{\tfrac1{2^n}C:C\in\mathcal C\}$ of the cover $\mathcal C_n$ and observe that each point $x\in \mathbb I^n$ is contained in a unique cube $C_n(x)\in\mathcal C_n$. Observe also that $(C_n(x))_{n\in\omega}$ is a decreasing neighborhood base at $x$. -Now fix any closed set $F\subset X$ and for every $x\in X\setminus F$ find the smallest number $n_x\in\omega$ such that $C_{n_x}(x)\cap F=\emptyset$. Observe that for any points $x,y\in\mathbb I^d\setminus X$ the cubes $C_{n_x}(x)$ and $C_{n_y}(y)$ are either disjoint or coincide. Indeed, if $C_{n_x}(x)$ intersects $C_{n_y}(y)$ and $n_x\le n_y$, then $C_{n_y}(y)\subset C_{n_x}(x)$ and then $C_{n_x}(y)=C_{n_y}(y)$ does not intersect $F$, which implies that $n_x=n_y$ by the minimality of $n_y$. -Then $E:=\mathbb R_d\setminus \bigcup_{x\in X}C_{n_x}(x)$ is a closed path-connected set in $\mathbb R^d$ such that $E\cap X=F$. The path-connectedness of $E$ can be proved exactly as in the answer of Sergei Ivanov (using the fact that the cover $\{C_{n_x}(x):x\in X\setminus F\}$ is disjoint and consists of open cubes with path-connected boundary in $\mathbb R^d$).$\square$ -Now we consider the general case. Given any subset $X\subset \mathbb R^d$ of cardinality $<\mathfrak c$, we can find a subset $Y\subset\mathbb R$ of cardinality $<\mathfrak c$ such that $X\subset Y^n$. -Choose any countable dense set $Q\subset \mathbb R$ such that $Q\cap Y=\emptyset$. By the countable dense homogeneity of $\mathbb R$, there exists a homeomorphism $h:\mathbb R\to\mathbb R$ such that $h(Q)=\mathbb Q$ and hence $h(Y)\subset h(\mathbb R\setminus Q)=\mathbb R\setminus\mathbb Q=\mathbb I$. -The homeomorphism $h$ induces a homeomorphism $g:\mathbb R^n\to\mathbb R^n$, $g:(x_i)_{i=1}^n\mapsto (h(x_i)_{i=1}^n)$, such that $g(X)\subset g(Y^n)=h(Y)^n\subset\mathbb I^n$. Given any closed set $F\subset X$, consider the its image $g(F)\subset g(X)\subset\mathbb I^d$ and using Lemma, find a closed path-connected set $E\subset\mathbb R^d$ such that $g(F)=g(X)\cap E$. -Then $D:=g^{-1}(E)$ is a closed path-connected subset of $\mathbb R^d$ such that $F=X\cap D$.<|endoftext|> -TITLE: union of regular polygons -QUESTION [9 upvotes]: it is motivated by Density of congruence classes covered by a set -Let say just "$n$-gon" for the set of vertices of a regular $n$-gon inscribed in a unit circle, "2-gon" for the set of two opposite points. -is it true that for given positive integers $1 < b_1 < b_2 < \dots < b_k$ the union of $b_i$-gons has the minimal cardinality when they all have a common vertex? -In other words, if $G_i$ are subgroups of the finite cyclic group $G$, is it true that -$$|\cup_{i=1}^k x_iG_i|\ge |\cup_{i=1}^k G_i|$$ -for arbitrary cosets $x_iG_i$? - -REPLY [4 votes]: Let me attempt a proof using the group-theoretic formulation. I will use the additive notation for the group operation. -The proof is by induction on $n=|G|$, with the base being trivial. Let $n=p^rm$ for some prime $p$ with $\gcd(p,m)=1$. -Consider $G' = pG$. Our goal is to reduce the problem for $G$ to its instance for $G'$. Let $R_j:=G' + jm$. Then $\{R_0,R_1,\ldots, R_{p-1}\}$ is a partition of $G$ into $G'$-cosets. -Let $G_i=d_iG$ be a subgroup of $G$ with $d_i | n$. If $p|d_i$ then $G_i \subseteq G'$ and every $G_i$ coset belongs to some $R_j$. In this case -we say that $G_i$ is of the first kind. Otherwise, $d_i | m$, and translating $G_i$ by $m$ does not change $G_i$. We say that such $G_i$ is of the second kind. -Let $S = \cup_{i=1}^k (G_i+x_i)$ be the union under consideration, let $S_1$ be the union of cosets of the first kind among cosets comprising $S$, and $S_2$ -- of the second. Let $T_j=S_1 \cap R_j$ for $0\leq j \leq p-1$. Then $T_j$ is actually union of some of our cosets, and the sets $T_0,T_1,\ldots,T_{p-1}$ are disjoint. Let $T_j'=T_j-jm$: we shift all the cosets in $T_j$ from $R_j$ to $R_0=G'$. Note that $S_2-jm=S_2$ and therefore the intersection of $T_j$ and $S_2$ shifts with $T_j$. In particular, $|T_j'-S_2|=|T_j-S_2|$. -The set $S'=S_2 \cup (\cup_{j=0}^{p-1}T'_j) $ is still a union of cosets of $G_i$'s. We have -$|S'| \leq |S_2|+ \sum_{j=1}^{p-1}|T'_{j}-S_2| = |S_2|+ \sum_{j=1}^{p-1}|T_{j}-S_2|=|S|$. -Thus it suffices to consider $S'$, which means that we may assume that $S_1 \subseteq G'$. Let $G_i'=G_i \cap G'$ and let -$x_i'$ be chosen so that $G_i'+x_i'= (G_i+x_i) \cap G'$. By the induction hypothesis applied to $G'$, we get -$a_1:=| \cup_{i=1}^k G_i' | \leq | \cup_{i=1}^k (G_i' + x_i')|=: b_1$ -and also -$a_2:=|\cup_{i: G_i \not \subseteq G'} G_i'| \leq |\cup_{i: G_i \not \subseteq G'} (G_i' + x_i')|=:b_2$, -where in this second inequality we restrict our attention to $G_i$'s of the second kind. The set $S_2$ is the disjoint union of $p$ translates of its intersection with $G'$, which intersection is present on the right side of the inequality directly above. It follows that -$|S|=|S \cap G'|+|S_2 - G'|=b_1 + (p-1)b_2,$ -while similarly we have -$|\cup_{i=1}^k G_i|=a_1 + (p-1)a_2$. -It follows that $|S| \geq |\cup_{i=1}^k G_i|$, as desired. - -Finally, let me note that the inequality does not hold for non-cyclic groups. Already for $G = \mathbf{Z}_2 \times \mathbf{Z}_2$ the union of three distinct subgroups of $G$ of size $2$ is $G$, while it is possible to choose their cosets with the union of size $3$.<|endoftext|> -TITLE: Definition of the Kervaire invariant for normal maps (as in Browder's book) -QUESTION [5 upvotes]: Browder's book "Surgery on simply-connected manifolds" defines the Kervaire invariant in a very general setting. My question is: how does one get the more usual definition of the invariant for a framed (or Wu-orinented) manifold from Browder's general definition? -Browder's setting: For Poincare pairs $(X,Y)$ and $(A,B)$ with specified Spivak normal bundles, a normal map is a map $f: (X,Y) \to (A,B)$ that has degree 1 and is covered by a bundle map $b$ from one Spivak normal bundle to the other. Browder manages to define the $\mathbb Z /2$-valued Kervaire invariant $\sigma$ for any such map (no additional information needed), at least as long as $f^*$ takes the Wu class of $A$ to the Wu class of $X$. I find it hard to understand his definition. -Some simplification: For the sake of understanding, it should be safe to replace all the Poincare pairs with closed manifolds, and Spivak normal bundles with normal bundles. Then, a normal map becomes a degree-1 map of manifolds embedded in a high-dimensional Euclidean space that is covered by a bundle isomorphism of normal bundles. If the manifolds can be framed (or Wu-oriented), the Wu classes will be zero, so there's no reason to worry about them. -Question: How does this situation relate to the more usual situation of having a manifold with a framing or Wu orientation (which are, of course, necessary to define the Kervaire invariant in the usual way)? I'd imagine that there is some standard thing I could fix $(A,B)$ to be so that Browder's version of Kervaire invariant computes the usual Kervaire invariant of $(X,Y)$. I couldn't figure out what it should be. This choice of $(A,B)$ should somehow encode the Wu orientation on $(X,Y)$. -Please tell me if there are any mistakes in the above. Thank you! - -REPLY [4 votes]: My algebraic theory of surgery gives the following approach to the definition of the Kervaire invariant. Let $Q_n(C,\gamma)$ be the Weiss twisted quadratic $Q$-group defined for any chain bundle $(C,\gamma)$. A spherical fibration $\nu:X \to BG(k)$ determines a chain bundle $(C(X),\gamma(\nu))$ with a Hurewicz-style group morphism -$$h~:~\pi^S_{n+k}(T(\nu)) \to Q_n(C(X),\gamma(\nu))$$ -from the stable homotopy groups of the Thom space $T(\nu)$. The image $h(\rho)$ of a stable homotopy class $\rho$ relates the evaluations on the Hurewicz-Thom image fundamental homology class -$$[X]~=~[\rho] \in H_{n+k}(T(\nu))~=~H_n(X)$$ -of the Steenrod squares of $X$ and the cup products with the Wu classes $v_r(\nu) \in H^r(X)$, verifying on the chain level the formula of Wu and Thom -$$\langle v_r(\nu) \cup y,[X] \rangle = \langle Sq^r(y),[X] \rangle~(y\in H^{n-r}(X))~.$$ An $n$-dimensional geometric Poincare complex $X$ (e.g. an $n$-dimensional manifold) has a canonical class of pairs -$(\nu_X:X \to BG(k),\rho_X:S^{n+k} \to T(\nu_X))$ with $\nu_X$ the Spivak normal fibration (= sphere bundle of the normal bundle of an embedding $X \subset S^{n+k}$ for a manifold $X$). A fibre homotopy trivialization $b:\nu_X \simeq *:X \to BG(k)$ (e.g. one determined by a framing of a manifold) determines a morphism $Q_n(C(X),\gamma(\nu_X)) \to {\mathbb Z}_2$ such that the image of $h(\rho_X)$ is the Kervaire invariant $K(X,b)\in {\mathbb Z}_2$. More generally, a Wu-orientation $b$ of $X$ (for which Browder's 1969 Annals paper The Kervaire invariant of framed manifolds and its generalization is a good reference) determines a morphism $Q_n(C(X),\gamma(\nu_X)) \to {\mathbb Z}_8$ such that the image of $h(\rho_X)$ is the Brown generalized Kervaire invariant $K(X,b)\in {\mathbb Z}_8$. Most of this is already explained in my paper Algebraic Poincare cobordism.<|endoftext|> -TITLE: Non-Kahler Complex manifolds -QUESTION [11 upvotes]: For a non-Kahler complex manifold $M$, we still have the decomposition of differential forms into differential forms of type $(p,q)$ and we can write $d=\partial+\bar\partial$ and we can define cohomology classes $H^{p,q}_{\bar\partial}(M)$. -In general is there any relation between $h^r(M)$ and $h^{p,q}(M)$? - -REPLY [15 votes]: Let $X$ be a compact complex manifold of complex dimension $n$. The Hodge-Frölicher spectral sequence starts with -$$ -E_1^{p,q}=H^{p,q}(X,\mathbb C) -$$ -and the limit term $E^{p,q}_\infty$ is the graded module associated to a filtration of the de Rham cohomology group $H^{p+q}_{\text{dR}}(X,\mathbb C)$. In particular -$$ -b_k=\sum_{p+q=k}\dim E_\infty^{p,q}\le\sum_{p+q=k}\dim E_1^{p,q}=\sum_{p+q=k}h^{p,q}. -$$ -The equality is equivalent to the degeneration of the spectral sequence at the $E_1^\bullet$ level. -On the other hand, an elementary lemma on bounded complexes of finite dimensional vector spaces applied to $E^\bullet_r$, tells you that you always have equality for the (topological) Euler characteristic: -$$ -\chi_{\text{top}}(X)=\sum_{k=0}^{2n}(-1)^kb_k=\sum_{p,q=0}^n(-1)^{p+q}h^{p,q}. -$$<|endoftext|> -TITLE: Hyperbolicity for Algebraic Varities and relation to curves on them -QUESTION [6 upvotes]: My question is related to several notions of hyperbolicity, applied to Kahler manifolds (projective, in general). Kahler hyperbolicity was introduced in this paper of Gromov's. He calls a Kahler manifold Kahler hyperbolic if the lift to the universal cover of the symplectic form(=imaginary part of hermitian metric) is the differential of a bounded $1$-form. A $1$-form is bounded if its norm is pointwise absolutely bounded (the norm is induced by the pull-back of the metric to the tangent space). -As he notes in the introduction, this notion implies Kobayashi hyperbolicity. My question is regarding the converse, namely to find an example of a Kobayashi hyperbolic manifold that isn't Kahler hyperbolic. - -My guess is that an example of an algebraic variety with the properties described below exists (so it would be nice if an algebraic geometer could say a few things here). -For a Kahler hyperbolic manifold, Gromov proves that its Euler characteristic is $(-1)^n$, where $n$ is its complex dimension. He also shows a Kahler hyperbolic manifold has quasi-ample canonical bundle (namely, it has Kodaira dimension $n$). Another observation is that a Kahler hyperbolic manifold cannot have amenable fundamental group. So any algebraic variety which fails one of the above tests but still is Kobayashi hyperbolic would fit the bill. -For somewhat different reasons, I would be happier if this would be a projective variety, which also doesn't have any $(2,0)$ cohomology. As a side question, just to check my understanding, is it true that for a variety with no $(2,0)$ classes in cohomology, any homology class that comes from a map of a surface (pushing forward the fundamental class) can in fact be realized by some algebraic curve in the variety? - -REPLY [7 votes]: Take a very general hypersurface $j \colon X \hookrightarrow \mathbb{P}^n$ of sufficiently high degree, $3 \leq n \leq 4$. Then $X$ is Kobayashi hyperbolic (Kobayashi actually conjectured that this is true for all $n$ and Siu recently outlined a strategy for proving this, see Diverietti's comment below). -On the other hand, Lefschetz hyperplane theorem says that the natural map -$\pi_1(X) \stackrel{j_*}{\longrightarrow} \pi_1(\mathbb{P}^n)$ -is an isomorphism. Then $X$ is simply connected, in particular its fundamental group is amenable and so $X$ cannot be Kahler hyperbolic. -If $n=4$, the projective variety $X$ also satisfies your second request. If fact, again by Lefschetz theorem, see for instance [Dimca, Singularities and topology of hypersurfaces, Theorem 2.6 p. 151], for all $n \geq 4$ we also have an isomorphism -$H^2(\mathbb{P}^n) \stackrel{j^*}{\longrightarrow} H^2(X)$. -Since $H^2(\mathbb{P}^n)=\mathbb{C}$, Hodge decomposition yields -$H^{2,0}(X)=H^{0,2}(X)=0, \quad H^{1,1}(X)=\mathbb{C}$. -If Kobayashi's conjecture were true, then the hypersurfaces $X$ would provide examples in all dimensions.<|endoftext|> -TITLE: Explicit uses of alephs above 'small ones' -QUESTION [7 upvotes]: In a paper placed on the arXiv today Shelah references theorem 0.9 from this paper (also Shelah) that uses $\aleph_{736}$ as an upper bound. This strikes me as analogous to Skewes' number. Are there any other examples where explicit mentions of 'large alephs' are used in proofs or theorems? Large in this instance is $\aleph_n$ for $n > 3$, say. -Edit: The first version of this question asks for $n$ a natural number, because clearly there are uses of $\aleph_\omega$ that are not all that uncommon, I imagine. But if there are uses of alephs - in isolation, not as part of a general transfinite induction scheme - that have $n$ an infinite ordinal like $\omega^2\cdot 5 + 45$ or some specific polynomial in $\omega$ which is 'not boring' (e.g. $\omega, \omega+1$), then I'd like to hear those as well. -Bonus points, even though this is a separate question: where does the 736 come from? Is it due to some sort of Goedelian numbering scheme that encodes the statement “G is a free abelian group”, which is part of the theorem? - -REPLY [17 votes]: David: - -As for the "736" in Shelah's paper, this is just his way of saying "for every $n>1$, there is a group as in the theorem which is free iff the continuum is below $\aleph_n$", there is nothing privileged or special about $\aleph_{736}$ in this result, other than it is ``random''. -Of course, what should be by far the most famous occurrence of a "large" $\aleph_n$ is $\aleph_4$, in a result due to Shelah, which is a cornerstone of modern cardinal arithmetic. The result is usually stated about exponentiation, although it is really about the pcf structure. For example: - - -If $2^{\aleph_0}<\aleph_\omega$, then $\aleph_\omega^{\aleph_0}<\aleph_{\aleph_4}$. - -This is quite strange, really. We do not know at the moment that $\aleph_4$ is needed. The best results (due, once again, to Shelah) show that it cannot be anything below $\aleph_1$. Some people suspect $\aleph_2$ should be the right number, but at the moment $\aleph_4$ is what we get. And we do not understand it well. When Shelah found the result, he told Leo Harrington about it, and Leo asked, quite naturally "Why the HELL is 4?" So, of course, when Shelah wrote his "Cardinal Arithmetic" book, that was the title of the relevant section. -3. Though they are not strictly $\aleph_n$s, you get an interesting collection of possible cardinalities as the uncountable spectrum of a countable theory. Given a consistent complete first-order theory $T$ in a countable language which does not admit finite models, let $I(T,\kappa)$ denote the number of isomorphism classes of models of $T$ of size $\kappa$. -The possible values of $I(T,\aleph_0)$ are not known (Vaught's conjecture). But for $\kappa$ uncountable, the possibilities for $I(T,\kappa)$ are known. This is essentially due to Shelah, but the last pieces of the result were obtained in Bradd Hart, Ehud Hrushovski, Michael C. Laskowski, "The uncountable spectra of countable theories", Ann. of Math. (2) 152 (2000), no. 1, 207--257:<|endoftext|> -TITLE: Beginners text on calculus of variations -QUESTION [10 upvotes]: I want to begin learning Calculus of Variations. What texts would MathOverflow recommend? Amazon shows up quite a few options: -http://tinyurl.com/36koaq4 -I work on Machine Learning, and that where I intend to apply this. -Thanks! - -REPLY [4 votes]: One of my favourite books for calculus of variations is the following: -"Introduction to the Calculus of Variations" by Bernard Dacorogna -http://www.amazon.co.uk/Introduction-Calculus-Variations-Bernard-Dacorogna/dp/1848163347/ref=sr_1_2?ie=UTF8&qid=1430733426&sr=8-2&keywords=introduction+to+the+calculus+of+variations -It is a great introductory book. It has plenty of solved problems to get familiar with the material. -It starts with the basics on function spaces, and then introducing classical and direct methods. It then moves onto minimal surfaces and isoperimetric inequalities. -It is a little expensive but it is worth it.<|endoftext|> -TITLE: Numeric equality testing? -QUESTION [6 upvotes]: Suppose we have two closed-form expressions with $k$ unknowns which are hard to test for equality but easy to evaluate numerically over $\mathbb{R}^k$. One could then approach the problem of equality testing by checking equality numerically at several points. The interesting questions are then -- for which kinds of expressions can you do it, how to pick sampling points and how many points are needed. -Google Scholar gives 0 hits for "numeric equality testing" -Has this kind of problem been studied before? What are the right keywords to search for? - -REPLY [7 votes]: Maple has a procedure testeq which is a "random polynomial-time equivalence tester". It works in this way. The 1986 paper New results for random determination of equivalence of expressions by Gaston H. Gonnet might be a starting point for checking that out.<|endoftext|> -TITLE: English or French translation of Gauss' "Summatio Quarumdam Serierum Singularium" -QUESTION [8 upvotes]: I'm interested in looking at the details of Gauss' method of determining the sign of the Gauss sum in his "Summatio Quarumdam Serierum Singularium", and I was wondering if anyone knew if there was an English or French translation available at all? I've tried searching on the internet and haven't found anything. Also, are there any modern presentations of the Gauss' method that you would recommend? -Thanks for your help. - -REPLY [2 votes]: An English translation of Baumgart's -exposition of Gauss's fourth proof can be found -here.<|endoftext|> -TITLE: Between mu- and primitive recursion -QUESTION [27 upvotes]: It is well known that primitive recursion is not powerful enough -to express all functions, Ackermann function being probably the best -known example. -Now, in the logic courses (that I have had look at) one always proceeded from primitive recursion to mu-recursion. In computer science terms this basicly means we are jumping from a formalism where programs are quaranteed to halt to a Turing-complete formalism where halting is a non-computable property i.e. we can't say for every program if it will eventually halt. -I got curious if there is any hierarchy between primitive recursion and -mu-recursion. After a while I found a programming language called Charity. In Charity (according to Wikipedia) -all programs are quaranteed to stop, thus its not Turing-complete, but, -on the other hand, it is expressive enough to implement Ackermann function. -This suggests there is at least one level between mu-recursion and primitive recursion. -My question is: does there exists any other halt-for-sure formalisms that are more expressive than primitive recursion? Or, even better, does there exist some known hierarchies between mu-recursive and primitive recursive functions? I'm curious about how "much" we can compute with a formalism that guarantees halting. - -REPLY [5 votes]: I'm aware that I'm replying to a very old question and that the original poster is not even on MO any more, but for the sake of completeness, since the term did not occur in the other answers, a keyword here is "subrecursive hierarchies", and Joel Robbin's doctoral dissertation is entirely about working out the connection between classes of nested recursion, the extended Grzegorczyk hierarchy of fast-growing functions, and transfinitely adding universal functions, all being indexed by ordinal notations. -One standard class between primitive recursive and general recursive/computable is the class of functions provably total in Peano arithmetic, or $\varepsilon_0$-recursive functions.<|endoftext|> -TITLE: Menger's theorem via matroids -QUESTION [13 upvotes]: Let $G=(V,E)$ be an oriented graph, $Y\subset V$ be some fixed set of its vertices. Call $A\subset V$ independent if there exist $|A|$ vertex-disjoint paths starting in $A$ and ending in $Y$. It is not hard to prove this independence system is actually a matroid. Indeed, matroids arising in this way are called gammoids. -Menger's theorem (in Goering's form, I think) states that the rank function of this matroid is given by -$r(A)=$the minimum number vertices which may be deleted so that no path from $A$ to $Y$ remains. -Is there any matroid interpretation, or matroids-assisted proof of this? -I saw some papers in which both Menger's theorem and matroids appear in the title, but on the first glance they deal with usual cycles/cuts graph matroids. - -REPLY [20 votes]: There is indeed a Menger's theorem for matroids first proven by Tutte. The reference is - -Tutte, W. T., Menger’s theorem for matroids, Journal of Research of the National - Bureau of Standards—B. Mathematics and Mathematical Physics, 69B (1965), 49–53. - -A copy of the paper can be found here. -This theorem is nowadays called Tutte's linking theorem, and it is sad that it is not more widely known. I'll take this chance to try and popularize it. First some notation. -Let $M=(E,r_M)$ be a matroid and let $A$ and $B$ be disjoint subsets of $E$. -We define the local connectivity between $A$ and $B$ to be -$\sqcap_M(A,B):=r_M(A)+r_M(B)-r_M(A \cup B)$. -We next define $\lambda_M(A):=\sqcap_M(A,E-A)$, and call $\lambda_M$ the connectivity function of $M$. It is fairly straightforward to check that $\lambda_M$ is symmetric, submodular, invariant under duality, and monotone under taking minors. Finally, we define -$\kappa_M(A,B) = \min(\lambda_M(X) : A \subseteq X \subseteq E-B)$. It is easy to show that for any $C \subseteq E - (A \cup B)$, we have $\sqcap_{M / C} (A,B) \leq \kappa_M(A,B)$. Tutte's linking theorem says that we can always find a $C$ that gives us equality. -Tutte's Linking Theorem. There exists $C \subseteq E - (A \cup B)$, such that -$\sqcap_{M / C} (A,B) = \kappa_M(A,B)$. -The proof is not very difficult, so instead I'll just briefly say why this generalizes Menger's theorem for graphs. The form of Menger's theorem that it generalizes is -Menger's Theorem. Let $a$ and $b$ be non-adjacent vertices in a graph $G$. Let $k$ be the size of a smallest vertex cut separating $a$ and $b$. Then there exist $k$ internally vertex disjoint paths between $a$ and $b$. -Proof (via Tutte's Linking Theorem). Let $A$ and $B$ be the sets of edges incident to $a$ and $b$ respectively. Note that $A$ and $B$ are disjoint since $a$ and $b$ are non-adjacent. Let $k$ be the size of the smallest vertex cut separating $a$ and $b$. Now just apply Tutte's Linking Theorem to $A$ and $B$ together with the following two observations: - -$k = \kappa_{M(G)}(A,B)+1,$ and -there exists $n$ internally vertex disjoint paths between $a$ and $b$ if and only if there exists -$C \subseteq E(G) - (A \cup B)$ such that $\sqcap_{M(G /C)}(A,B) \geq n-1$.<|endoftext|> -TITLE: length of decompositions into elementary matrices -QUESTION [6 upvotes]: The Gaussian algorithm tells us, that for any field $k$ a $n\times n$-matrix over $k$ can written as a product of at most $C$ elementary matrices ($C\sim n^2$). -I am wondering, whether such a constants also exists for other rings - like $\mathbb{Z}$. -Given a matrix $A\in SL_2(\mathbb{Z})$, one can basically use the Euclidean algorithm to find such a decomposition. However if we take a the following matrix involving the Fibonacci numbers, the algorithm takes about $n$-steps and hence we get a decomposition in $\sim n$ factors. But there might still be a better decomposition. -So is there for every $n$ a matrix $A \in SL_2(\mathbb{Z})$, that cannot be written as a product of elementary matrices ? -I guess the construction with the Fibonacci numbers might be a candidate, I don't know how to prove, that it is impossible to decompose it in a better way. - -REPLY [4 votes]: Yes. The group $SL_2(\mathbb Z)$ is virtually free and is not boundedly generated by any finite generating set because of that. You can look at the (very nice) slides of Dave Witte-Morris' talks on bounded generation here.<|endoftext|> -TITLE: Zorn's Lemma and plane geometry -QUESTION [5 upvotes]: Given a graph $G$ and a number $n$, Zorn's Lemma immediately implies the existence of a -maximal partial coloring of $G$. Equivalently, one may assign $n+1$ colors to the nodes of $G$ such that nodes with the first $n$ colors never lie adjacent, but every node of the last color lies adjacent to at least one node receiving each of the other colors. -In particular, the observation above applies to geometrically determined graphs and even more specifically to graphs on, say, ${\Bbb R}^2$, where one declares two points $p_1$ and $p_2$ adjacent provided that their distance $d(p_1,p_2)$ lies in $D$, some prescribed set of distances. -I would like to know what values of $n$ and $D$ make this existence result not a theorem of ZF. If that asks too much, any example of such an $n$ and $D$ would satisfy me, even better, the simplest possible example (say with $|D|$ as small as possible, and $n$ as small as possible given that). - -REPLY [3 votes]: There is some recent material by Michael Payne that is also relevant here: -He considers the following variant of the Shelah-Soifer graph: Two points in the plane form -an edge if the distance is 1, the $x$-coordinates have rational difference, and the $y$-coordinates have rational difference. -The chromatic number of this graph is 2 (in ZFC). The measurable chromatic number (preimage of colors Lebesgue measurable) is at least 5. If all subsets of the plane are measurable (e.g., in Solovay's model), the chromatic number of this graph is at least 5. -Unfortunately I cannot give you a link to the paper at the very moment. -I believe the general problem that you are asking about is open.<|endoftext|> -TITLE: Online Library of Unlabeled Connected Graphs on n Vertices -QUESTION [5 upvotes]: Does anyone know of the link to an online library of of unlabeled, connected graphs on n vertices? I remember looking at such an archive a few years ago while at a Macaulay 2 workshop, but I've been unable to find it (or any other one) since then. -The page I remember seeing only had enumerated unlabeled graphs up to n=11,12, or 13 vertices, and the graph I'm looking for data on is much larger, so links to repositories of larger (special) graphs. -The most specific part of this request: The graph I'm looking to find a list of edges for is the 1-skeleton of the 600-cell, if anyone happens to just have that information on-hand (or readily available.) - -REPLY [2 votes]: Is it possible that this is what you remember seeing? -http://wwwteo.informatik.uni-rostock.de/isgci/smallgraphs.html<|endoftext|> -TITLE: How often does suspension define an action of Z/2 on a category of module spectra? -QUESTION [13 upvotes]: Let R be the 2-periodic complex K-theory spectrum, or any other naturally occuring 2-periodic E-infty ring spectrum. The suspend-once functor gives an autoequivalence of the category of R-module spectra, and since R is 2-periodic applying it twice is isomorphic to the identity functor. In some coarse sense this defines an action of Z/2 on the category of R-modules. -There is a better, more complicated and more correct notion of Z/2-action on a collection of module spectra. Module spectra form an infty-category, and we can ask if Z/2 acts on this infty-category. One way to put it: the suspend-once functor and the homotopy between suspend-twice and the identity give us a map from RP^2 into the classifying space of automorphisms of this infty-category. Does this extend to a map from BZ/2 = RP^infty? - -REPLY [12 votes]: As Jacob says, to produce a map $RP^\infty\to B\mathrm{Pic}(KU)$, we should use the fact that there is an infinite loop splitting $\mathrm{Pic}(KU)=\mathrm{Cliff}_{\mathbb{C}} \times Y$, where $Y$ is the 3-connected cover. -So, when I got to do the exercise of computing homotopy classes of maps $RP^\infty\to B\mathrm{Cliff}_{\mathbb{C}}$ (using knowledge of the $\infty$-loop k-invariants, as decribed in the other answers), I get $[RP^\infty, B\mathrm{Cliff}_{\mathbb{C}}] = \mathbb{Z}/8$: each stage in the postnikov tower of $B\mathrm{Cliff}$ adds a factor of $2$, and all the extensions are non-trivial. In particular, the elements which are congruent to $1$ mod 2 in $\mathbb{Z}/8$ are non-trivial on the bottom cell. So, if I've done the calculation right (and that's a big if), there are four distinct ways to make $\mathbb{Z}/2$ act so that the generator acts by suspension. -I find it interesting that I get $\mathbb{Z}/8$, which reminds me of real Bott periodicity. In fact, the calculations make it look like -$$\mathrm{map}_*(RP^\infty, B\mathrm{Cliff}_{\mathbb{C}}) \approx \mathrm{Cliff}_{\mathbb{R}}$$ -as infinite loop spaces. -Is there a way to take a real Clifford algebra $A$, and functorially produce from it a monoidal $\mathbb{Z}/2$-action on (the symmetric monodial Morita $2$-groupoid of) complex Clifford algebras?<|endoftext|> -TITLE: How quickly will billiard trajectories cluster? -QUESTION [7 upvotes]: Suppose you launch $n$ point-particles on -distinct reflecting nonperiodic billiard trajectories -inside a convex polygon. Assume they all have the same speed. -Define an $\epsilon$-cluster as a configuration of the particles -in which they all simultaneously lie within a disk of radius $\epsilon$. -It is my understanding that Poincaré's Recurrence Theorem -implies that at some time after launch, the particles -will form an $\epsilon$-cluster somewhere. -(Please correct me if I am wrong here, in which case the remainder -is moot.) -Picturesquely, if I sit in my office long enough, all the air molecules -will cluster into a corner of the room. :-) -The reason I specify that the trajectories be distinct is to -exclude -the particles being shot in a stream all on the same trajectory. -The reason I specify nonperiodic is to exclude sending -the particles on parallel periodic trajectories whose length -ratios are rational, in which case no clustering need occur. -My question is: - -How long must one wait for an $\epsilon$-cluster to occur? - -Essentially I am seeking a quantitative version of -Poincaré's Recurrence Theorem, quantitative enough -to actually make a calculation. -I would like to put a number of years to the air-molecule example -(air molecules move perhaps 700 mph or 300 m/s). -It could serve as a useful pedagogical anecdote. -I found a beautiful paper that should help -me answer this question: -Benoit Saussol, -"An Introduction to Quantitative Poincaré Recurrence in Dynamical Systems," -Reviews in Mathematical Physics, Volume 21, Issue 08, pp. 949-979 (2009). -But I am having difficulty making the leap from the abstract -theorems to an explicit calculation. -Any help or additional pointers would be appreciated! -Addendum. Vaughn's analysis, although leaving a few loose ends (as he notes), largely -answers my question. Thanks to all for the astute comments and responses! - -REPLY [5 votes]: I'll offer a few partial answers, which may eventually lead to a complete answer. -As observed in Saussol's paper (Theorem 3, Kac's lemma), if you have an ergodic invariant measure $\mu$ then the mean return time to a set $A$ is equal to $1/\mu(A)$. (Note, however, that this is the mean return time for trajectories that begin in $A$, rather than for an arbitrary initial condition, for which you'd need the hitting time). -So we need to know what the ergodic invariant measure $\mu$ is. Many topological dynamical systems have lots of invariant measures lying around, in which case it's a non-trivial task to pick the one you're really interested in. However, if we assume that your convex polygon satisfies the Veech dichotomy (see, for example, this article by John Smillie and Barak Weiss), then it's true that the flow in every direction is either periodic or uniquely ergodic. Regular polygons satisfy the Veech dichotomy, so in that setting your non-periodicity condition would be enough to guarantee that Lebesgue measure is the only invariant measure. -Of course, Lebesgue measure is always invariant for the billiard flow, so the main thing we need this for is the fact that Lebesgue measure is ergodic under these assumptions. (There may be an easier way to get ergodicity, but this is the first thing that came to my mind.) -If we write $X_\theta$ for the phase space of the billiard flow with angle $\theta$, and let $\theta_1,\dots,\theta_n$ be the angles that your $n$ particles are launched at, then the above discussion implies that the billiard flow on each $(X_{\theta_n},\text{Leb})$ is ergodic. The phase space for your entire system is $\prod_n X_{\theta_n}$, and the direct product of the $n$ different Lebesgue measures is an invariant measure for your overall system. -What we'd like to do next is say that this measure is actually ergodic. Unfortunately, it's not quite that simple, since the direct product of two ergodic systems need not be ergodic (just consider $R_\alpha\times R_\alpha$, the direct product of two circle rotations by irrational multiples of $\pi$). So to say that Lebesgue measure is ergodic for your whole system requires something more, which is where your condition that the trajectories be distinct ought to come in. I'm not sure exactly how this step should go, but you should be able to get something along the lines of "for almost every set of angles $(\theta_1,\theta_2,\dots,\theta_n)$, Lebesgue measure is ergodic for the whole system". Billiard flows have something to do with interval exchange transformations (IETs), and so this paper by Jon Chaika may well have the result that's needed at this point. -Once you know that (product) Lebesgue measure is invariant for the whole system, you're in the clear: given a ball of radius $\epsilon$, the (normalised) Lebesgue measure of that ball in each $X_\theta$ is $\pi \epsilon^2 / C$, where $C$ is the area of the polygon, and so the set in $\prod_n X_{\theta_n}$ corresponding to those configurations for which all particles lie in this ball is just $(\pi \epsilon^2 / C)^n$. The inverse of this is your expected return time, provided we have appropriate hypotheses to justify all the above steps. (Expected hitting time might be a different story, I'd need to think a bit.)<|endoftext|> -TITLE: Strategies for digging through literature -QUESTION [19 upvotes]: I hope this question is appropriate for MO - I cannot decidedly tell with soft questions. I was wondering what are the strategies people use when searching for literature on a subject. I shall clarify my question with an example. Suppose, for instance, I want to learn more about $\star$-independence or $\star$-spread and assuming I already do not have a reference for it, how do I go about searching for literature on it. If the latter assumption is relaxed (that is I already have a couple of references on the concept), presumably I can keep looking for cross references and that might, in a small number of steps, exhaust all the literature on the subject. However, I find frequently this is not the case. The problem here is two fold: -1) If I am looking at exploring about a mathematical object/theorem which does not have a name and does not involve any objects which have exotic names, for instance, "A finite union of subspaces of a vector space is a proper subset of the space if the ambient field is infinite" (I hope this is not a bad example and even if it is, that it conveys the underlying issue). Google searching any keywords for an example like this only yields tons of irrelevant entries. This is especially true in cases where the mathematical objects involved have other meanings in english (which can be said about almost every other thing in math e.g. ring, field, ideal, module, etc) and even when this is not the case, it could be a ubiquitous word in mathematics (e.g. vector, space, manifold, etc). So if the concept or theorem does not contain a distinguished word, searching about it is difficult. -2) If the object/theorem is exotic or contains an exotic object like $\star$-independence, google and other search engines suppress special characters and the situation defaults to that in (1). Finally, I find it especially hopeless if you are looking to find say, class number computations of $\mathbb{Q}[\sqrt{2},\sqrt{7}]$. -I would like to know if anyone has any thoughts on getting around these problems. If it's a misplaced question, I would be happy to delete it. - -REPLY [9 votes]: There is a site, which is absolutely perfect for asking questions like "what is known about this problem" or "can you point me to a good reference on this and that".<|endoftext|> -TITLE: relation between toric geometry and log geometry -QUESTION [7 upvotes]: Hello, -I'm trying to understand the relation between the points of view of -log geometry (monoids) and toric geometry (fans). -Suppose that $k$ is a field and $P$ is a finitely generated monoid. -Then $k[P]$ has a natural log structure and furthermore, any choice -of generators $\mathbf N^r\to P$ induces a closed embedding -$Spec(k[P])\subset\mathbf A^r$. -On the other hand, starting from a cone $\sigma$ satisfying some properties in a lattice -$N\otimes\mathbf R$, where $N = \mathbf Z^r$, we obtain a monoid -$P' = \sigma^\vee\cap M$, where $M = Hom(N,\mathbf Z)$ and $\sigma^\vee$ is the set of -all $x\in M\otimes\mathbf R$ such that $x(\sigma) \geq 0$. -Question: if we start with $P$ (and a choice of generators as above), can one write a -corresponding cone so as to recover $P$ by the construction in the previous paragraph? -Thanks! - -REPLY [4 votes]: Not all finitely generated monoids $P$ will come from the cone construction. You need to assume that: - -$P^{gp}$ is torsion-free: If $x \in P^{gp}$ and $n\cdot x = 0$ then $x = 0$. -$P$ is cancellative: If $x + y = x + y'$, then $y = y'$. This is equivalent to saying that the map $P \rightarrow P^{gp}$ is injective. -$P$ is saturated: If $x \in P^{gp}$ and $x^n \in P$, then $x \in P$. Assuming the previous two proporties, this is equivalent to $k[P]$ being normal. - -Here, $P^{gp}$ refers to the group formed by inverting all the elements of $P$. -If $P$ is finitely generated and satisfies 1, then $P^{gp}$ is a lattice, i.e. isomorphic to $\mathbb Z^r$ for some $r$, and this is the lattice $M$ from the cone construction. The dual lattice $N$ is $\textrm{Hom}(M, \mathbb Z)$, and $\sigma$ can be taken to be those $\lambda$ in $N \otimes_{\mathbb Z} \mathbb R = \textrm{Hom}(M, \mathbb R)$ such that $\lambda(x) \geq 0$ for all $x \in P$. - -REPLY [2 votes]: Picking up some of Eric Zaslow's reformulation: Assume $P$ is commutative, saturated, and cancellative, as well as finitely generated. The answer to your question is affirmative if and only if the "groupification" $P^{gp}$ of $P$ is torsion-free. (As mentioned in Dustin's answer, saturated means that for all $p$ in $P^{gp}$, $np \in P$ implies $p\in P$. Cancellative means $p_1+q=p_2+q$ implies $p_1=p_2$.) -All this amounts to $P$ being embeddable as a sub-monoid of ${\Bbb Z}^n$ for some $n$. Then take the subgroup of ${\Bbb Z}^n$ spanned by $P$. This is isomorphic to some ${\Bbb Z}^m$; take the dual of the convex hull of $P$ in ${\Bbb R}^m$ and you've got your cone $\sigma$, just as Eric says. When $P$ is saturated, it is equal to $\sigma^\vee \cap M$; otherwise, this gives the saturation of $P$, corresponding to the integral closure of $k[P]$. -Depending on what references you use, when $P^{gp}$ is torsion-free, $P$ is called either integral or toric. (See, e.g., the toric variety notes on M. Mustata's webpage versus the log geometry notes on Danny Gillam's webpage; both sources are worth looking at.) It seems the latter terminology is more standard in the log geometry world, where "integral" sometimes just means "cancellative".<|endoftext|> -TITLE: The Jacobian ideal generates the socle of a complete intersection -QUESTION [6 upvotes]: This is with reference to theorem 5.20 in Vasconcelos book linked (google books) here: -http://tinyurl.com/2967eov -I shall restate the theorem here for easy reference: "If $A=k[[x_1,x_2,...,x_n]]/I$ is a complete intersection and $dim_k A$ is not divisible by $char(k)$ then the Jacobian ideal generates the socle of $A$". -I am looking for a proof of this theorem. Vasconcelos references three places to look for one. One is a result of Tate - I have looked at this. One is supposed to be in Kunz's - Introduction to commutative algebra and algebraic geometry" - I could not find a result similar to this in there (it's not a pointed reference). Finally there is a Scheja-Storch paper linked below. -http://www.reference-global.com/doi/abs/10.1515/crll.1975.278-279.174 -I am specifically looking for a proof similar to Scheja-Storch (Tate seems to use a different approach), but the above paper is in German and I am not fluent at it. It's probably unlikely, but if anyone has an english reference on this proof, I would really appreciate it. - -REPLY [2 votes]: I'm promoting this comment to an answer, since it appears no one else is jumping in with a proof. Prop. 2 of this paper by Eisenbud attributes it 'essentially' to Berger, and has a sketch of a proof.<|endoftext|> -TITLE: Is there a (hope for) formula for a colored HOMFLYPT invariant in terms of HOMFLYPT invariants colored by the fundamental represenations? -QUESTION [8 upvotes]: Any finite dimensional representation of $SL(n,\mathbb{C})$ can be found as a direct -summand inside of some tensor product of the fundamental representations. -Since we use finite dimensional representations of the quantum $SL(n)$ to construct -the colored HOMFLYPT invariant one can naively hope that the invariants colored by -the fundamental representations plus some independent of knot combinatorics would -produce you colored HOMFLYPT for any color. -How far from true is this naive expectation? -Any suggestions or references are very welcomed. - -REPLY [7 votes]: It depends on what you call "independent of knot combinatorics". Obviously one can get the invariant associated with a representation $V_\lambda\subset V^{\otimes k}$, where $V$ is the fundamental representation, by replacing every strand of the link by $k$ strands (cabling) colored by $V$ and then inserting the projector onto $V_\lambda$; since this projector can be written in terms of Hecke algebra operators (this is the $q$-version of Young symmetrizer), it can be presented graphically as a certain linear combination of $k-k$ tangles. This is what is done in the paper cited by Scott. Not sure if this is the answer you are looking for.<|endoftext|> -TITLE: Infinite direct product of the integers not a free module over the integers -QUESTION [11 upvotes]: Possible Duplicate: -Is it true that, as Z-modules, the polynomial ring and the power series ring over integers are dual to each other? - -Is there an easy proof? I only found citations but have no access. By the way: If we cross over to the rationals every vector space is free (using Zorn's lemma). But can one construct a basis of "Countable infinite product of the rationals"? - -REPLY [2 votes]: See Example 3.5 at http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/dualmod.pdf for an argument.<|endoftext|> -TITLE: Multiple factors of the character of a representation -QUESTION [5 upvotes]: In algebra, various objects admit a unique decomposition into irreducible elements. For instance integers $n\ge1$, univariate polynomials $p\in k[X]$ (even multivariate ones), or characters in representation theory of finite groups. In each situation, an irreducible occurs with a multiplicity. It is interesting, from a theoretical point of view, to have a reduction to the situation where every multiplicity is $1$ (or $0$ if you insist to write the product/sum with all irreducibles of the structure). This can be done explicitly in the case of polynomials, by dividing $p$ by the g.c.d. of $p$ and $p'$, the latter being calculated with the help of the Euclid algorithm. - -Is there something similar for characters in representation theory of finite groups ? Suppose we know only the cardinals of conjugacy classes of $G$, together with the table of multiplication of these classes. But we don't know the table of characters. Given a character $\chi$, is it possible for instance to split it as a sum $\chi_1+\cdots+\chi_r$, where $\chi_\ell$ gathers the irreducible characters entering in $\chi$ with multiplicity $\ell$ ? - -Perhaps the question should be restricted to complex characters; who knows ? Even a weaker property could be interesting, provided it is associated with a finite-time algorithm. -Of course, I have in mind to apply such a property to the regular representation. Then $\chi_\ell$ would be $\ell$ times the sum of irreducible characters of degree $\ell$. - -REPLY [3 votes]: It's important here to work over a field of characteristic 0, since the traditional notion of character gets too complicated otherwise (and is better replaced by the notion of Brauer character). Moreover, the field should at least initially be a splitting field for the group, say algebraically closed. -In the highlighted part of the question, you say: Suppose we know only the cardinals of conjugacy classes of G, together with the table of multiplication of these classes. But we don't know the table of characters. But knowing how the sums over classes multiply would allow you to compute the character table, following ideas which go back to Burnside. See for example Section 33 in the old Curtis-Reiner book Representation Theory of Finite Groups and Associative Algebras (1962). It's rare to know so much in advance anyway when the group is at all complicated, which makes the question look a bit artificial to me. On the other hand, one can certainly study the isotypic components of the regular representation in the spirit of Wedderburn structure theory. For an arbitrary given representation, there is a similar isotypic decomposition. But for character theory the most natural question is how to obtain the full decomposition into irreducibles. -By now there are sophisticated computational methods available for working with characters of fairly large finite groups, but this viewpoint gets farther from the theoretical question here. -ADDED: Though I'm still uncertain what the actual question here is, it's worth pointing to the discussion in Sections 2.6-2.7 of Linear Representations of Finite Groups by the other Serre. There one finds a "canonical" decomposition of a given representation space corresponding to the way its character would decompose into multiples of irreducible characters. How effectively one can find such a decomposition depends of course on precisely what is already known about the finite group in question.<|endoftext|> -TITLE: polynomials with the same discriminant -QUESTION [5 upvotes]: Let $p(x)$ be the chromatic polynomial of a special graph. Performing a certain type of operation on the graph changes $p$ by shifting it and adding a constant, say to: $q(x)=p(x+a) + b$. -I have noticed that this operation always results in $q$ having the same discriminant and splitting field as $p$. This would of course be obvious if we were just shifting $p$ by $a$, but it seems unusual given that we are also adding $b$. -I realise it is difficult to comment without knowing more details, but all I really want to know is whether or not this might be significant. The polynomials in question are irreducible cubics. How common is it that two cubics - which are not just shifted versions of each other - have the same discriminant? Does anybody have any idea what this might signify? -(edit: since posting this I have noticed that the chromatic polynomial in question is of the form $p(u,v,x)$, where $u$ and $v$ are given constants, and the discriminant of $p$ is a symmetric polynomial in $u$ and $v$. The graph operation switches $u$ and $v$...this is why the discriminant remains the same. So my question could be formulated as: what is the significance - if any - of a discriminant which is a symmetric polynomial in 2 variables?) - -REPLY [5 votes]: [Substantial edit: As I mentioned previously, cubic fields can have the same discriminant but not be isomorphic; but I've revised my answer to better address the author's question.] -There are a lot of polynomials that will generate the same cubic field. Here is the basic idea, due to Delone and Faddeev. Write down the cubic form $f(x, y) = ax^3 + bx^2 y + cxy^2 + dy^3$, where setting $y = 1$ gives a generating polynomial for the cubic field. Then there is an action of $GL_2$ on such cubic fields; if $\gamma$ is a 2x2 matrix with entries $\alpha, \beta, \gamma, \delta$ then -$\gamma f(x, y) = f(\alpha x + \gamma y, \beta x + \delta y)$, optionally with a factor of $(det \ \gamma)^{-1}$. -Anyway, if two cubic forms are $GL_2(\mathbb{Q})$-equivalent then the corresponding polynomials (if irreducible and nondegenerate) generate the same field. Moreover, if two cubic forms are $GL_2(\mathbb{Z})$-equivalent then the corresponding polynomials generate the same order over $\mathbb{Z}$. -See here, page 10 of Section 4, for a good reference. -[Previous answer:] -I think the underlying question is how often nonisomorphic cubic fields have the same discriminant. This happens, but it doesn't seem to be very common. -You might find it interesting to browse through this database which lists cubic (and other) fields, their discriminants, and their minimal polynomials. -Theoretically, I believe it is known that this happens infinitely often, although I don't recall the reference. For upper bounds, it is known that the number of cubic fields of discriminant n is $O(n^{1/3 + \epsilon})$, due to Ellenberg and Venkatesh; the "correct" bounds are believed to be much sharper.<|endoftext|> -TITLE: Generators for the Jordan algebra of symmetric 3-by-3 matrices -QUESTION [6 upvotes]: In the vector space $V$ of $3\times 3$ symmetric real matrices, we can define a nonassociative algebra structure by the multiplication -$$A \bullet B = \frac12 (AB +BA).$$ -This turns $V$ into a Jordan algebra. -Question - -What is the minimum number of generators of this Jordan algebra? And could you give me one set of such generators? - -Thanks a lot! - -REPLY [2 votes]: This is already well answered, many choices of two symmetric matrices will do. I'll add this observation: $\{I,A,B,A \bullet A,B \bullet B,A\bullet B\}$ is an additive basis where -$$A=\left[\begin{array}{ccc} 1 & 1 &1 \\ 1 & 0 & 0 \\ 1 & 0 & 0\end{array}\right] \qquad B=\left[\begin{array}{ccc} 0 & 0 &1 \\ 0 & 0 & 1 \\ 1 & 1 & 1\end{array}\right]$$ Note that $A\bullet B$ has rank $3$, not that it matters. I'm not sure if two rank 1 matrices could generate the algebra, but I lean against (I couldn't get past dimension 4), I didn't look hard for a generating set with a rank 1 and a rank 2 matrix.<|endoftext|> -TITLE: How to solve a quadratic equation in characteristic 2 ? -QUESTION [27 upvotes]: What do I do if I have to solve the usual quadratic equation $X^2+bX+c=0$ where $b,c$ are in a field of characteristic 2? As pointed in the comments, it can be reduced to $X^2+X+c=0$ with $c\neq 0$. -Usual completion of square breaks. For a finite field there is Chen Formula that roughly looks like $X=\sum_{m} c^{4^m}$. I am more interested in the local field $F((z))$ or actually an arbitrary field of characteristic 2. - -REPLY [16 votes]: I think this solves $X^2+X+c=0$ over $F((t))$: -I want to assume that $c\in F[[t]]$. If not, say $c=at^{-m}+...$, then the quadratic has no solutions when $m$ is odd or $a$ is not a square, and otherwise the substitution $X\mapsto X+\sqrt{a}t^{-m/2}$ gives a new equation with smaller $m$. So, after finitely many steps $c=c_0+c_1t+...$ is integral. -Because $X^2+X+c$ has derivative $1$, by Hensel's lemma the equation has a solution if and only the constant term $c_0$ is of the form $d^2+d$ for some $d$ in $F$. And if it is, Hensel's approximations are obtained by starting with an approximate solution $x_0=d$ and recursively computing $x_{m+1}=x_m-f(x_m)/f'(x_m)=x_m^2+c$. This gives -$$ - x = d + \sum_{n=0}^\infty (c-c_0)^{2^n} -$$ -as the solution (the partial sums are the $x_m$). Actually, the approach seems to work over any complete field, reducing the problem to the residue field. Hope this helps.<|endoftext|> -TITLE: Every connected planar graph contains adjacent vertices with at most 2 common neighbors -QUESTION [5 upvotes]: I am looking for a reference for the following fact. -Let $G$ be simple undirected connected planar graph with $\geq 2$ vertices. Then $G$ contains an edge $\{u,v\}$ such that $|N(u) \cap N(v)| \leq 2$, in other words the number of common neighbors of $\{u,v\}$ is at most $2$. -Here's a sketch of the proof. Any edge incident on a vertex of degree $\leq 3$ satisfies this requirement. If there is a vertex $v$ of degree $4$ for which none of its edges satisfy the requirement, then $N[v]$ must be a $K_5$ which is not possible; if there is a vertex of degree 5 for which none of its edges satisfy then $N[v]$ must contain a $K_{3,3}$. Since any planar graph contains a vertex of degree $\leq 5$ this is sufficient. -I would like to be able to skip the proof in my paper, using a reference to known work instead. Can anyone give me a reference for the above fact? - -REPLY [3 votes]: it is not what you are asking for (not a reference), but just maybe slightly easier proof, without searching for $K_{3,3}$. We may take any vertex $x$ of degree $d$, if any of $d$ vertices adjacent to $x$ has at least three neighbours between those $d$ vertices, we get at least $d+3d/2$ edges between $d+1$ vertices, hence $5d/2\leq 3(d+1)-6=3d-3$, $d\geq 6$. So degree of any vertex is not less then $6$, what is impossible. -By the way, we may even fix vertex $u$, and then find $v$. Indeed. if $v_1$, $v_2$, $\dots$, $v_d$ are labeled counterclockwise neighbours of $u$, then without loss of generality $v_i$ and $v_{i+1}$ are joined ($v_{d+1}=v_1$), and $u$ is outside the cycle $v_1-v_2-\dots-v_d$. But any triangulation of the (topological) polygon contains a vertex of degree 2.<|endoftext|> -TITLE: Formality of classifying spaces -QUESTION [21 upvotes]: Let $G$ be a compact Lie group (or reductive algebraic group over $\mathbb{C}$), and let $BG$ be its classifying space. Fix a prime $p$. Let $\mathcal{A}$ denote the dg algebra of singular cochains on $BG$ with coefficients in a field or characteristic $p$ (or if you prefer the dg algebra of endomorphisms of the constant sheaf). My question is: - -Is it known for which primes $p$ the dg algebra $\mathcal{A}$ is formal, that is, quasi-isomorphic to a dg algebra with trivial differential? - -I assume / hope that the answer is that this is true if $p$ is not a torsion prime for $G$ (i.e. $p$ arbitrary in types $A$ and $C$, $p \ne 2$ in types $B$, $D$ and $G_2$, $p \ne 2, 3$ in types $F_4$, $E_6$ and $E_7$, and $p \ne 2,3,5$ in type $E_8$.) -Note that we know* that $\mathcal{A}$ is formal in characteristic 0. - -Can one then conclude that it is formal in any characteristic in which the cohomology of $\mathcal{A}$ is torsion free? - -If so I think this would give the above list of primes. -*) for example because $H(BG, \mathbb{Q})$ is a poynomial algebra, and $\mathcal{A}$ admits a graded commutative model using the de Rham complex -- see Bernstein-Lunts "Equivariant sheaves and functors". - -REPLY [10 votes]: This is an old question. But sometimes old questions get answered! -Benson, Greenlees, Formality of cochains on BG -Here is the abstract: -Let $G$ be a compact Lie group with maximal torus $T$. If $|N_G(T)/T|$ is invertible in the field $k$ then the algebra of cochains $C_*(BG;k)$ is formal as an $A_\infty$ algebra, or equivalently as a DG algebra. -(Note that this is not the same condition as I postulate in the question. I still wonder if it is true under the "$p$ not a torsion prime" hypothesis. However in the meantime, I should understand the above paper...) -Update: latest version of paper completely answers question! From Section 5: -"...if $G$ is a connected, simply connected compact Lie group, $k$ is a field of characteristic $p$, and $p$ is not a torsion prime then $H^∗(BG; k)$ is a polynomial ring on even classes, and $C^∗(BG; k)$ is formal."<|endoftext|> -TITLE: Zeta function of monodromy and counting points over C((t)) -QUESTION [7 upvotes]: If $X$ is a smooth, projective variety over $\mathbb{F}_q$, the Weil conjectures tell us: -$$\prod \mathrm{det} (I - TF|_{H^i_c(X)})^{(-1)^{i+1}} = \mathrm{exp}\left(\sum_{m=1}^{\infty} \frac{N_m}{m} T^m \right)$$ -here, $T$ is a formal variable, $H^i_c(X)$ is an appropriate cohomology theory, $F$ is the Frobenius automorphism, and $N_m$ is the number of $\mathbb{F}_{q^m}$ points of $X$. -I would like to replace $\mathbb{F}_q$ with $\mathbb{C}((z))$, on the pretext that both have absolute Galois group $\hat{\mathbb{Z}}$. I am thinking of $\mathbb{C}((z))$ as the ring of functions on a very small punctured disc, and of a variety over $\mathbb{C}((z))$ as a family over this punctured disc. I will also conflate the Frobenius automorphism with the monodromy action. - - -Let $X$ be a variety over $\mathbb{C}((t))$; interpret the LHS of the equation above by understanding $F$ as the monodromy action. In what, if any, sense does the number $N_m$ count points over the field $\mathbb{C}((t^{1/m}))$ ? - - -Note that if one naively takes the cardinality of the set of these points, one would often find $N_m = \mathrm{\infty}$. - - -Is there any structure on the set of $\mathbb{C}((t))$ points which would allow me to take an Euler number? - - -Finally, let me view $\overline{\mathbb{C}((t))}$ as the field over which tropical geometry happens. Making the modifications appropriate to discuss the non-projective case, let me take $X$ to be an affine variety. - - -Can I "count" $\mathbb{C}((t^{1/m}))$ points of $X$ in terms of "counting" $\frac{1}{m} \mathbb{Z}$ points of $\mathrm{Trop}(X)$? - -REPLY [9 votes]: Hi Vivek. You should have a look to the following papers : - -J. Nicaise and J. Sebag (2007). - Motivic Serre invariants, ramification, and the analytic Milnor - fiber. Inventiones mathematicae, - 168 (1), p. 133-173. -J. Nicaise (2009). A trace formula - for rigid varieties, and motivic Weil - generating series for formal schemes. - Math. Ann. 343, p. 285–349. - -They prove a trace formula along the lines you suspect: Let $X$ be a proper $\mathbf C((t))$-variety, let $S(X)$ be its motivic Serre invariant: this is an element of the Grothendieck group of complex varieties modulo the ideal generated by $\mathbf L-1$, where $\mathbf L$ is the class of the affine line; it is computed as the special fiber of any weak Néron model of~$X$ over $\mathbf C[[t]]$. The Euler characteristic of $S(X)$ is well-defined -and $\chi(S(X))$ equals the (alternate sum) of the traces of the action of the monodromy on the étale cohomology of $\bar X$.<|endoftext|> -TITLE: Non-Kahler "Calabi-Yau"? -QUESTION [18 upvotes]: Are there examples of non-Kahler complex manifolds with holomorphically trivial canonical bundle? - -REPLY [4 votes]: More examples are given by nilmanifolds. -Barberis, Dotti and Verbitsky proved in Theorem 2.7 in http://www.ams.org/mathscinet-getitem?mr=2496748 that nilmanifolds endowed with invariant complex structures have trivial canonical bundle. See also Cavalcanti and Gualtieri's Theorem 3.1 in http://www.ams.org/mathscinet-getitem?mr=2131642 -On the other hand, non-tori nilmanifolds never admit a Kaehler structure, because of Benson and Gordon, http://www.ams.org/mathscinet-getitem?mr=976592 , or Hasegawa, http://www.ams.org/mathscinet-getitem?mr=946638 , or ...<|endoftext|> -TITLE: When the Lovász theta-function saturates its upper bound -QUESTION [8 upvotes]: The Lovász $\vartheta$-function of a graph $G$, $\vartheta(G)$, is well-known to be "sandwiched" between the independence number of the graph, $\alpha(G)$, and the chromatic number of its complement, $\chi (\overline{G})$. -When $G$ is perfect then $$\alpha(G)=\vartheta(G)=\chi (\overline{G}).$$ -I would like to know if there are graphs $G$, such that $$\alpha(G)< \vartheta(G)= \chi (\overline{G}).$$ - -REPLY [14 votes]: Suppose $G$ is a $k$-regular graph on $n$ vertices, with least eigenvalue $\tau$. -Lovasz proved that -$$ - \theta(G) \le \frac{n}{1-\frac{k}{\tau}}. -$$ -Further if the automorphism group of $G$ acts arc-transitively, then equality holds. -In fact equality holds if G is a single class in a homogeneous coherent configuration, -for example, if $G$ is a strongly regular graph. -Class 1: Latin square graphs. Take the graph whose vertices are the $n^2$ cells of -and $n\times n$ Latin square, where two cells are adjacent if they are in the same row, -same column, or have the same contents. This graph is regular with valency $3(n-1)$ -and least eigenvalue $-3$ and so $\theta(G)=n$. But if the Latin square is the multiplication -table of a cyclic group of even order then $\alpha(G) < n$ (because cyclic Latin squares -do not have transversals, but you can find a short proof on page 225 of one of my favorite -books on algebraic graph theory). The vertices in a given column form a clique of size -$n$ and so we see that $\chi(\bar{G})=n$ for any Latin square of order $n$. -Class 2: generalized quadrangles: A GQ with parameters $(s,t)$ is a collection of points -and lines satisfying some axioms, in particular each line contains exactly $s+1$ points. -Deem two points adjacent if they are collinear. -If $s,t>1$ this gives a strongly regular graph on $(s+1)(st+1)$ vertices with valency -$s(t+1)$ and least eigenvalue $-t-1$. Hence $\theta(G) = st+1$ and a coclique of size $st+1$ -is known as an ovoid. The points on line forms a clique of size $s+1$, and a set -of lines that partition the point set is called a spread. Hence you want -GQ's with spreads but no ovoids. -For actual examples, I refer you to Payne and Thas's book on GQ's---the GQ's $Q(5,q)$ work.<|endoftext|> -TITLE: Fast trace of inverse of a square matrix -QUESTION [9 upvotes]: Which would be the most efficient way (in computational time) to compute tr(inv(H)), where H is a (dense) square matrix? -In my particular problem I also have a LU decomposition of H already available, which was used in a previous context to solve a system of linear equations. My current approach is to simply use the LU to compute the inverse and then calculate the trace. Is there any other more efficient way to achieve this, considering I already have this matrix factorized? -I could have first computed the inverse and then made a multiplication by the inverse to solve my previous system of linear equations, but I was trying to avoid multiplication by the inverse to avoid numerical inaccuracies. -Edit: Forgot to mention that under normal conditions H should be symmetric. - -REPLY [7 votes]: Given that the poster has specified that his matrix is symmetric, I offer a general solution and a special case: - -Eigendecomposition actually becomes more attractive here: the bulk of the work is in reducing the symmetric matrix to tridiagonal form, and finding the eigenvalues of a tridiagonal matrix is an O(n) process. Assuming that the symmetric matrix is nonsingular, summing the reciprocals of the eigenvalues nets you the trace of the inverse. -If the matrix is positive definite as well, first perform a Cholesky decomposition. Then there are methods for generating the diagonal elements of the inverse.<|endoftext|> -TITLE: Upper bound to the number of generators -QUESTION [7 upvotes]: When defining noetherian ring/module there's no condition on the number of generators of ideals/submodules (apart from being finite). -However, in some cases we can do better: --A noetherian module over a field is a finite vector space, so every submodule can be generated with at most n elements. --A maximal ideal of $\mathbb{K}[X_1,...,X_n]$ where $\mathbb{K}$ is algebraically closed can be generated, via Nullstellensatz, by exactly n generators. -What other examples are there where we can find a system of generators of bounded cardinality? -What happens if we replace maximal ideal by prime ideal in the second example? - -REPLY [12 votes]: Understanding the number of generators is a very subtle problem. I will focus on your second question on ideals, since the first one is a bit broad. By a theorem of Foster-Swan, the problem is local. -There is no absolute upper bound even for a prime ideal of height $2$ in $k[[x,y,z]]$. In this paper, Moh gives a sequence of primes $(P_n)$ such that $\mu(P_n) =n+1$. -OK, so what to do next? One can ask if there are good bounds on $\mu(I)$ if $R/I$ is "nice". If $R/I$ is a complete intersection, then $\mu(I)$ is the height of $I$. In commutative algebra, the next level of "niceness" is being Gorenstein. In this paper Schoutens shows that if $I$ has height $2$ and $R/I$ Gorenstein, then there is a bound only depending on $R$. -If one goes down another notch, and only assume $R/I$ is Cohen-Macaulay, then Moh's examples show there are no hope for bounds independent of $I$. However, there are bounds that depends on invariants of $R/I$, such as the type or multiplicity.<|endoftext|> -TITLE: Is the statement that every field has an algebraic closure known to be equivalent to the ultrafilter lemma? -QUESTION [31 upvotes]: The existence and uniqueness of algebraic closures is generally proven using Zorn's lemma. A quick Google search leads to a 1992 paper of Banaschewski, which I don't have access to, asserting that the proof only requires the ultrafilter lemma. Questions: - -Is it known whether the two are equivalent in ZF? -Would anyone like to give a quick sketch of the construction assuming the ultrafilter lemma? I dislike the usual construction and am looking for others. - -REPLY [32 votes]: As I mentioned in a comment to Eivind Dahl's answer, it -seems that there is also an easy argument directly from the -Compactness theorem of first order logic. Since you said you are looking for alternative constructions, let me expand on the idea here. -The Compactness theorem asserts that if every finite subset -of a first order theory has a model, then the whole theory -has a model. Andres mentioned that this is equivalent to -the Ultrafilter lemma. -Existence. Let $F$ be a field. Let $T$ be the theory -consisting of the field axioms, the atomic diagram of $F$ -(which asserts all the equations and negated equations true -in $F$, using constants for elements of $F$), plus the -assertions that every polynomial over $F$ has a root. This -last assertion is made separately as an assertion about -each particular polynomial over $F$, using the constants in -the language added for the elements of $F$. This theory is -easily seen to be finitely satisfiable, since any finite -subtheory mentions only finitely many polynomials of $F$, -and we can satisfy it in a finite extension of $F$. Thus, -by the Compactness theorem, the whole theory has a model -$K$. If we take the collection of elements of $K$ that are -algebraic over $F$, this will be algebraically closed. -Uniqueness. Let $F$ be a field and let $E$ and $K$ -be algebraic closures of it. Let $T$ be the theory -consisting of the union of the atomic diagrams of $E$ and -$K$, plus the field axioms. (Note that we haven't added any -axioms saying that elements of $E$ and $K$ are distinct, and in the end these constants will be in effect melded together, providing the isomorphism.) -If $T_0$ is a finite subtheory, then only finitely many -elements of $E$ and $K$ appear. Those elements of $E$ and -$K$ appear in some $F[\vec a]\subset E$ and $F[\vec -b]\subset K$. We can embed both of these extensions of $F$ -into a single finite extension $F[\vec u]$, which will -satisfy all assertions in $T_0$. (Note, this embedding -effectively decides a little piece of the isomorphism, by -mapping some of the $\vec b$'s to some of the $\vec a$'s.) -Thus, by Compactness, the whole theory is satisfiable. If -$G$ is a model of $T$, then since $G$ interprets the atomic -diagrams of $E$ and $K$, we get isomorphisms of $E$ and $K$ -into subfields of $G$. These maps agree on $F$ and have a -common range, which is the set of elements of $G$ that are -algebraic over $F$. Thus, the composition is an isomorphism -of $E$ and $K$.<|endoftext|> -TITLE: Functions whose antiderivative behaves like xf(x) -QUESTION [11 upvotes]: I'm wondering if a classification of analytic functions, $f\,$ (it may be that $C^1$ is enough, but I'm not taking any chances, if you have a reason why I only need to consider a larger class of functions, I would enjoy that as well) with the property $F(x):=\int f(x) dx = \mathcal{O}(xf(x))$ as $x\to\infty$ AND $xf(x)=\mathcal{O}(F(x))$ where $\mathcal{O}$ is big-O notation. I'm not sure if it's standard or not, but I'll denote this condition by -$$\mathcal{O}(F(x))=\mathcal{O}(xf(x))\qquad (*)$$ -I'm motivated by the naïve notion of integration from the mistake many first semester students in calculus make when trying to take anti-derivatives and keep on thinking the same thing over and over: the power rule. -It is straightforward to check that $f(x)=x^n\quad n\ne -1$ and $f(x)=\log^n x\quad n\ge 0$ satisfy the condition $(*)$, the first just by checking and the second by induction on $n$ and integration by parts. It's also easy to see that this is not the case for $x^ne^x,\; n\in\mathbb{Z}$ again by integration by parts. -A preliminary investigation yields some interesting first starts: -1) A valid refomulation of the problem is ${F\over f}$ has a slant asymptote, i.e. ${F\over f}=kx+\mathcal{o}(1)$, and if you know that the derivative of this little o function is also little o of 1, and say $k=1$ then you can rephrase this as $1-{d\over dx}(\log f(x))\cdot {F\over f}(x)=1+\mathcal{o}(1)$ -2) If the function is increasing we can get half of this inequality since $F(x)\le xf(x)$ since $F(x)=\int_a^xf(x)$, but the other half fails. -3) A convex $\mathcal{o}(1)$ in (1) might imply that the derivative is also $\mathcal{o}(1)$ - -REPLY [13 votes]: Note: This is a major rewrite of my earlier answer, to include necessary and sufficient conditions applicable to an even wider class of functions. -Instead of expanding to the class of all analytic functions (where the asymptotics can be hard to get control over, due to oscillatory behavior), my inclination would be to focus on large classes of functions with well-behaved asymptotics, including all the functions that arise in ordinary asymptotic analysis. The usual buzz phrase for this is "Hardy field" (mentioned for example in my answer here), which by definition is an ordered field of germs at infinity of $C^\infty$ functions. -I will describe several classes of such functions. The first is the class of all functions which are first-order definable in the structure $(\mathbb{R}, +, \cdot, <, \exp)$ together with all real numbers adjoined as constants. This class contains all functions that are constructible from polynomials, $\exp$, $\log$ and closed under the usual arithmetic operations and composition. It thus contains all the functions that usually arise in asymptotic analysis, and many more besides. This class enjoys the following strong model-theoretic property (as developed more fully in the theory of o-minimal structures): - -(O) The zero set of any function $F: [a, \infty) \to \mathbb{R}$ in this class is a finite union of points and intervals (finite or infinite in extent). - -Condition O ensures that every such function $F$ is either eventually positive ($F(x) > 0$ for all sufficiently large $x$), eventually zero, or eventually negative. As a result, the ring of germs at infinity of the definable functions in this class forms an ordered field, i.e., is a Hardy field. -Also, if $F$ is definable, then $F'$ is also first-order definable (and its domain can be shown to be the domain of $F$ save for finitely many points). Applying condition O to $F'$, every definable $F$ in this class is either eventually increasing, eventually decreasing, or eventually constant. -Proposition: A function $F$ in this class satisfies $F(x) = O(xF'(x))$ and $xF'(x) = O(F(x))$ if and only if there exist $n, N$, both positive or both negative, for which $x^n < |F(x)| < x^N$ for all sufficiently large $x$. -Proof: WLOG we may assume $F$ is eventually positive, and is not eventually constant. Thus $F$ is either eventually increasing or eventually decreasing, say eventually increasing. If $F$ is eventually bounded above by some $x^N$, $N > 0$, then by increasing $N$ if necessary we may assume $F(x)/x^N$ tends to zero, whence it is eventually decreasing. Taking the derivative, we conclude that eventually -$$x^N F'(x) - Nx^{N-1}F(x) < 0$$ -whence $xF'(x) < NF(x)$, i.e., $|xF'(x)| < N|F(x)|$ or $xF'(x) = O(F(x))$. However, if $F(x)$ is eventually bounded above by every positive-power function $x^N$ (think $N$ small!), this also shows $xF'(x) = o(F(x))$, so that $F(x)$ is not $O(xF'(x))$. -By a similar argument, if $F(x)$ is eventually bounded below by some positive-power function $x^n$, we get $nF(x) < xF'(x)$ eventually, so that $F(x) = O(xF'(x))$. This also shows that if $F(x)$ is bounded below by every positive-power function (think $n$ large!), then $F(x) = o(xF'(x))$, so $xF'(x)$ is not $O(F(x))$. -Thus, if $F(x)$ is positive and eventually increasing, a necessary and sufficient condition that $F'$ satisfy condition ($\ast$) in the question is that there exist two positive-power functions that $F$ is eventually squeezed between. An entirely similar analysis shows that if $F(x)$ is positive and eventually decreasing, a necessary and sufficient condition that $F'$ satisfy condition ($\ast$) is that there exist two negative-power functions that $F$ is eventually squeezed between. Thus the proposition is proved. - -The almost freshman-level triviality of this proof testifies to the great power of condition O (which is a special case of the o-minimality axiom), from which all flows. -Thus it is of interest to know of classes of functions which satisfy it. I will mention an extraordinary result in this regard, due largely to Patrick Speissegger (The Pfaffian closure of an o-minimal structure, J. Reine Angew. Math. 508 (1999), 189--211): -There is an o-minimal expansion of the ordered exponential field $\mathbb{R}$ (thus, including the class of functions described above) so large that - -The structure includes the restriction of any analytic function to a compact box, -If $f: [a, \infty) \to \mathbb{R}$ is first-order definable within this structure, then so is any antiderivative $F$ (even though general antiderivatives are not definable by a first-order construction), - -This may fit better with Adam Hughes's formulation in terms of antiderivatives. Since condition O is satisfied (according to the more general o-minimality condition), the same analysis as above applies.<|endoftext|> -TITLE: A question about Transfinite Induction -QUESTION [6 upvotes]: The Transfinite Induction says: Let $\mathbf{P}(x)$ is a property, assume that, for all ordinal numbers $\alpha $ : If $\mathbf{P}(\beta)$ holds for all $\beta < \alpha$, then $\mathbf{P}(\alpha)$ holds. Then $\mathbf{P}(\alpha)$ holds for all ordinals $\alpha$. -My question is: What is the problem if I replace ordinals with cardinals？ I mean could I say that If $\mathbf{P}(\kappa)$ holds for all cardinals $ \kappa< \lambda$, then $\mathbf{P}(\lambda)$ holds. Then $\mathbf{P}(\lambda)$ holds for all cadinals $\lambda$. - -REPLY [8 votes]: @Dong: - -As I pointed out in comments, and Pete mentioned in his answer, well-orderability is all you need, so yes, if $P$ is a property of cardinals as you describe, then it holds of all cardinals. You can see this by taking $P'(\alpha)$ to be: "$\alpha$ is not a cardinal, or else, it is a cardinal and $P(\alpha)$", and applying the usual transfinite induction theorem to $P'$. -In 1., I took "cardinal" to mean "initial-ordinal". Several questions come to mind immediately if we remove choice from the picture: - -a. Suppose $P$ is a property of (not necessarily well-ordered) cardinalities, and it has the property you mention: If for all smaller sizes it holds, it holds for the size under consideration. -This does not suffice for $P$ to hold at all sizes. For example, suppose there is an infinite Dedekind-finite set, and let $P(X)$ be the statement: - -"Either $X$ is not infinite Dedekind-finite, or else, the cardinalities are well-founded below the size of $X$". - -Being Dedekind-finite means that any proper subset has smaller size. It follows that if any smaller set satisfies $P$, so does $X$. But $P(X)$ is plain false if $X$ is infinite Dedekind-finite (and it is consistent with ZF that there are such sets). -b. Suppose that all sets of smaller cardinality than that of $X$ are well-orderable. It does not follow that $X$ itself is well-orderable. For example, consider Solovay's model, and take $X={\mathbb R}$. Since the perfect set property holds, any subset of the reals either has the same size as ${\mathbb R}$, or else it is countable, in which case it is well-orderable. But ${\mathbb R}$ itself is not well-orderable. -c. A significantly harder question is whether, if cardinalities are well-founded, then choice holds. This is an open problem. It was asked independently by T. Forster and D. Savaliev, and I have thought about it on and off for a while.<|endoftext|> -TITLE: Order of the identity map of a Moore space. -QUESTION [8 upvotes]: Write $M_n = S^n \cup_2 D^{n+1}$. I know, as a matter of folklore, that the identity map $\mathrm{id}_M$, considered as an element of the group $[M_n, M_n]$, has order $4$ (for $n > 3$, let's say). -What is a reference for this? And is there a simple and pretty argument? - -REPLY [6 votes]: I don't know who first proved this fact, but a standard reference would be this paper of Toda's. It is stated as Theorem 4.1. Toda's proof is essentially the same as Tom's answer.<|endoftext|> -TITLE: Adjoint of Atkin-Lehner's U_p -QUESTION [13 upvotes]: If $\Gamma=\Gamma_1(N)$, or $\Gamma=\Gamma_0(N)$, the Hecke operator $[\Gamma diag(1,l) \Gamma]$ for $l$ a prime (acting on the space of cusp forms of level $\Gamma$ and some weight $k$) is in general denoted $T_l$ when $l$ does not divide $N$, but $U_l$ otherwise. It is well-known that the Hecke operators $T_l$ are normal (that is commute with their adjoint) for the Petersson's inner product, but not the $U_l$. Or are they ? -Is there any value of the level $N$, the weight $k$, and a prime $l$ dividing $N$ such that $U_l$ is normal ? -Of course, this would imply that $U_l$ is diagonalizable, but this is conjectured (known if $k=2$) to happen if $l^3$ does not divide $N$ (cf. Coleman-Edixhoven, Math. Annalen, 1998). -This question is surely easy, but my problem is that I can't really compute the adjoint of $U_l$. Of course, it is the Hecke operator $\Gamma diag(l,1) \Gamma$ and it's not hard to write this double class as a union of simple class, but the formulas I get look awful. -For example consider this simplest case. Let $f$ be form of weight $k$ and level $1$, normalized eigenform for all $T_l$. Choose a prime $p$. As it is well known, -the space of form of level $\Gamma_0(p)$ with same eigenvalues as $f$ for all the $T_l$ with $l \neq p$ is two dimensional, generated by $f(z)$ and $f(pz)$, or preferably -generated by $f_\alpha(z) = f(z) - \beta f(pz)$ and $f_\beta(z) = f(z)-\alpha f(pz)$ where $\alpha$ and $\beta$ are the two roots of $X^2-a_pX + p^{k-1}$, with $T_p f = a_p f$ (I suppose $\alpha \neq \beta$, which is always conjectured and often known). -The interest of this basis is that $U_p$ is diagonal in it: $U_p f_\alpha = \alpha f_\alpha$ and $U_p f_\beta = \beta f_\beta$ -Can you compute the matrix of the adjoint of $U_p$ in the basis $f_\alpha$, $f_\beta$? - -REPLY [11 votes]: A part of this question was unanswered: the last one, computing the adjoint of $U_p$ on -the two-dimensional space $V_f$ of old forms of level $\Gamma_0(p)$, attached to a normalized eigenform $f$ of weight $2k$ and level $1$. I observe that the reference to Gross given in the comments deals only with the case of new forms, and the same can be said of the reference given in the update of BR's answer. -In the unlikely event that the answer to this technical question might be of interest for someone else than me, here it is: -In the basis $f(z),f(pz)$ of $V_f$, the matrix of the adjoint of $U_p$ for the Peterson's scalar product is $$U'_p = \left( \matrix{ 0 & - p^{-1} \cr p^{2k} & a_p} \right).$$ -Here $a_p$ is the $p$-coefficient of $f$. Note that $U_p$ and its adjoint never commute, which is consistent with BR's answer. -To get the matrix in the basis $f_\alpha=f(z)-\beta f(pz)$, $f_\beta=f(z)-\alpha f(pz)$, -make a change of basis. The result is unpleasant (to my surprise), so I don't give it. -To prove the result, let us consider the action $g \mapsto w(g)=p^{-k}z^{-2k} g(-1/pz)$ -of the Arkin-Lehner involution on $V_f$, the normalization being that of the 1970 paper of Atkin and Lehner at Math. annalen). -Lemma 7 (combined with Lemma 9 and Lemma 14) of that paper tells that for all $g$ modular form for $\Gamma_0(p)$ of weight $2k$, the form $p^{1-k} U_p(g)+w(g)$ is of level $\Gamma_0(1)$, hence a multiple of $f$. One easily deduces that $w(f)= x f(z) + p^{k} f(pz)$ for some scalar $x$, $w(f(pz)) = y f(z)$ for some scalar $y$. Using that the action $w$ is an involution, we easily get the values of $x$ and $y$, i.e. the matrix of the action of $w$ in the basis $f(z),f(pz)$: -$$w=\left(\matrix{0 & p^{-k} \cr p^k & 0}\right).$$ The matrix of $U_p$ in the same basis is as is well-known $$U_p=\left(\matrix{a_p & 1 \cr -p^{2k-1} & 0}\right).$$ Since the adjoint of $U_p$ is $w U_p w^{-1}$, one gets the result given above.<|endoftext|> -TITLE: Aspherical homotopy orbit space of configurations on the 2-sphere -QUESTION [7 upvotes]: The group SO(3) acts naturally on $S^2$ and thus on $Conf(S^2, q)$, the configuration space of $q$ distinct points on the 2-sphere, via the diagonal action on $S^2 \times...\times S^2$. This is a question about $Conf(S^2, q)_{hSO(3)} = ESO(3) \times _{SO(3)} Conf(S^2, q)$, the $SO(3)$-equivariant homotopy type of this space. - -Theorem: $Conf(S^2, q)_{hSO(3)}$ is aspherical for $q\geq 3$. - -Proof (sketch): I will just write $C_q$ for $(Conf(S^2, q))_{hSO(3)}$. Projection to the first three coordinates $p:C_q \to C_3$ is a fibration (an equivariant version of the Fadell-Neuwirth fibration) with fiber $Conf(S^2-$ {3 points}, $q-3)$. But $C_3 \simeq *$, so the inclusion of the fiber $Conf(S^2-$ {3 points},$q-3)$ into $C_q$ is a homotopy equivalence. The fiber is a $K(\pi, 1)$ by an easy induction argument (using Fadell-Neuwirth again). QED. - -Question: So $Conf(S^2, q)_{hSO(3)}$ is a $K(\pi, 1)$. What's $\pi?$ - -REPLY [4 votes]: $\pi$ is an iterated extension of free groups by free groups again -using Fadell-Neuwirth. This is also called the pure braid group of the -sphere minus three points on q-3 braids.<|endoftext|> -TITLE: What is a satisfactory way to format definitions in Latex? -QUESTION [8 upvotes]: There are several ways one may format a definition in latex, but each has their problems. - -Use the amsthm package, and the usual style for theorems. This will result in everything italicized. It is difficult to catch the term you are defining, even if you non-italicize it. -Use the amsthm package, and the style for definitions. This time the term you are defining is the only word/phrase italicized, but the problem is that one does not know where the definition ends. Unlike the proof environment, there is no QED marker, so it is unclear where the definition ends and when the next paragraph starts. -End a definition with a QED-type marker (like a flower or whatever). The problem with this is that there is an over-abundance of markers. Flowers and square boxers. - -How do you format your definitions in latex? - -REPLY [3 votes]: This is how I format my definitions in LaTeX: -\documentclass{article} -\usepackage{color,amsmath,amssymb} -\usepackage{framed} -\makeatletter -\newdimen\errorsize \errorsize=0.2pt -% Frame with a label at top -\newcommand\LabFrame[2]{% - \fboxrule=\FrameRule - \fboxsep=-\errorsize - \textcolor{FrameColor}{% - \fbox{% - \vbox{\nobreak - \advance\FrameSep\errorsize - \begingroup - \advance\baselineskip\FrameSep - \hrule height \baselineskip - \nobreak - \vskip-\baselineskip - \endgroup - \vskip 0.5\FrameSep - \hbox{\hskip\FrameSep \strut - \textcolor{TitleColor}{\textbf{#1}}}% - \nobreak \nointerlineskip - \vskip 1.3\FrameSep - \hbox{\hskip\FrameSep - {\normalcolor#2}% - \hskip\FrameSep}% - \vskip\FrameSep - }}% -}} -\definecolor{FrameColor}{rgb}{0.25,0.25,1.0} -\definecolor{TitleColor}{rgb}{1.0,1.0,1.0} - -\newenvironment{contlabelframe}[2][\Frame@Lab\ (cont.)]{% - % Optional continuation label defaults to the first label plus - \def\Frame@Lab{#2}% - \def\FrameCommand{\LabFrame{#2}}% - \def\FirstFrameCommand{\LabFrame{#2}}% - \def\MidFrameCommand{\LabFrame{#1}}% - \def\LastFrameCommand{\LabFrame{#1}}% - \MakeFramed{\advance\hsize-\width \FrameRestore} -}{\endMakeFramed} -\newcounter{definition} -\newenvironment{definition}[1]{% - \par - \refstepcounter{definition}% - \begin{contlabelframe}{Definition \thedefinition:\quad #1} - \noindent\ignorespaces} -{\end{contlabelframe}} -\makeatother -\begin{document} -\begin{definition}{Quadratic Equation} -A Quadratic Equation is an equation in the form: -\begin{equation} -ax^2+bx+c=0 -\end{equation} -where \(a,b,c\in\mathbb{R}\). -\end{definition} -\end{document} - -and this is what you get (mirror1, mirror2):<|endoftext|> -TITLE: English translation of Voronoi's dissertation -QUESTION [7 upvotes]: I am looking for an English translation of Voronoi's doctoral dissertation, "On a generalization of the Algorithm of Continued Fractions." I can only find it in the original Russian. - -REPLY [12 votes]: Hi Michael and Joel: -there is in fact an English translation of Voronoi's thesis by Emma Lehmer. I have it in printed form. I can have it scanned as PDF and e-mail it to you. E-mail me, contact details at www.math.ucalgary.ca/~rscheidl.<|endoftext|> -TITLE: Why does the Grothendieck group $K_0(R)$ of a ring not depend on our choice of using left modules instead of right modules? -QUESTION [15 upvotes]: I am under the impression that in the definition of the Grothendieck group $K_0(R)$ of a (non-commutative) ring it doesn't matter whether we apply the usual $K_0$ construction to the exact category of all finitely generated projective left $R$-modules, or if we apply the construction to the category of all finitely generated projective right $R$-modules. -Is this obvious? To be honest my first reaction is disbelief but given that certain other notions for rings (such as semisimplicity of a ring or Morita equivalence for rings) are left/right-independent, I guess I can believe it. But it is not clear to me why this is true in the case of $K_0$. -I've taken a look in the standard references for classical algebraic $K$-theory but none of them seem to mention this point and either deal just with left $R$-modules throughout, or make no mention at all of the handedness of their modules. -This circle of ideas leads to a more general and vague follow up question: Does anyone have any intuition for why certain notions for non-commutative rings (such as the examples mentioned above: semisimplicity, Morita equivalence, $K_0$) do not depend on whether we look at properties of the collection of left modules over the ring or whether we look at properties of the collection of right modules over the ring. - -REPLY [22 votes]: Here is an alternative to Andreas proof (which if you unfold it is not so different): We have a functor $M\mapsto \mathrm{Hom}_R(M,R)=:M^\ast$ which gives both a contravariant functor from left $R$-modules to right $R$-modules and vice versa. We also have a naturanl transformation $M\to M^{\ast\ast}$ given by $m\mapsto(f\mapsto f(m))$. Now, as $R^\ast=R$ and $(-)^\ast$ is additive it takes f.g. projective modules to f.g. projective modules and the map $P\mapsto P^{\ast\ast}$ is an isomorphism. This shows that the category of left f.g. right projective modules is anti-equivalent to the category of f.g. left projective modules. However, $K_0(-)$ induces an isomorphism not only for equivalences but also for anti-equivalences. -Unless I am mistaken the same argument works also for Quillen's higher K-groups.<|endoftext|> -TITLE: How commutative is Quillen's Plus-Construction? -QUESTION [7 upvotes]: This question is inspired by this question about the dependence of K-theory on the order of multiplication in the ring. I did not think long about it, so maybe the answer really lies on the surface; but I do not know. -Let $G$ be a discrete group, $G^{op}$ its opposite group (i.e. the one with reversed multiplication). Let $H \subset G$ be a subgroup, such that $H=[H,H]=[G,G]$. Note that $H^{op} \subset G^{op}$ has the same properties. I denote the Quillen Plus-Construction of a space with fundamental group $G$ with respect to $H$ by $X \mapsto X^+_H$ - -Question: Is there a homotopy equivalence between $BG_H^+$ and $B(G^{op})_{H^{op}}^+$, such that the induced map on $\pi_1$ is induced by identity $id: G \to G^{op}$. - -Note that there is clearly a homotopy equivalence between $BG_H^+$ and $B(G^{op})_{H^{op}}^+$ such that the induced map on $\pi_1$ is induced by the inverse $inv: G \to G^{op}$; but that is not the one I am looking for. -However, this shows that one could also ask: - -Question: Is there a homotopy equivalence between $BG_H^+$ and itself, such that the induced map on the abelian group $\pi_1(BG_H^+) = G/H$ is the inversion. - -Given the motivation, any good answer in the case $G=GL_{\infty}(R)$ (for some ring) would be interesting too. -EDIT: Johannes Ebert suggested in a comment a strategy to give a negative answer to Question 2 in general. The Kan-Thurston construction gives a way of obtaining every (finite?) cell complex $X$ as the plus-construction on $BG$ with respect to some perfect subgroup $H$. It seems that the only missing piece is a finite cell-complex $X$ with $\pi_1$ abelian and no $\pi_1$-inversion-inducing self-homotopy equivalence. Then, $X = BG^+_H$ for some group $G$ and some perfect subgroup $H \subset G$. However, since $\pi_1 = G/H$ is abelian, we see that $[G,G] \subset H$ and hence, $[H,H] = [G,G]=H$ as required. - -Question: Can anybody give an example of such a space? - -REPLY [5 votes]: To answer the final question, here is how you can construct a space with no self-equivalence inducing negation on (abelian) $\pi_1$. -If $X \to Y$ is a homotopy equivalence, then for any basepoint it induces isomorphisms $\pi_n(X) \to \pi_n(Y)$ commuting with the action of $\pi_1$. In particular, if we can construct abelian groups $A$ and $B$ with an action of $A$ on $B$ such that there is no isomorphism $\phi: B \to B$ such that ${}^a \phi(b) = \phi({}^{(-a)}b)$, then we can use these to construct the desired space; let $X$ be the homotopy orbit space $K(B,2) \times_A EA$, which has $\pi_1 = A$ and $\pi_2 = B$ with the given action. -If $B$ is cyclic, all endomorphisms commute and so it suffices to construct an action so that ${}^a b = {}^{(-a)}b$ does not hold for all $a, b$. -"Minimal" examples include the action of $\mathbb{Z}/4$ on $\mathbb{Z}/5$ with generator $x \mapsto 2x$ (not isomorphic to $x \mapsto 3x$) and the action of $\mathbb{Z}/3$ on $\mathbb{Z}/7$ with generator $x \mapsto 2x$ (not isomorphic to $x \mapsto 4x$).<|endoftext|> -TITLE: When is a blow-up along a union of subvarieties smooth? -QUESTION [8 upvotes]: Let $V_1$ and $V_2$ be two distinct smooth subvarieties of the smooth variety $X$ which are regularly embedded. I would like to find a reasonable criteria which guaranties the smoothness of the Blow-up $\widetilde{X}$ of $X$ along the union of $V_1$ and $V_2$. -For example, is $\widetilde{X}$ smooth if $V_1$ and $V_2$ meet transversally? - -REPLY [5 votes]: Appearently, the blow-up IS smooth if $V_1$ and $V_2$ intersect transversally. In this case we have that -$$ -Bl_{V_1 \cup V_2} X = Bl_{\bar{V_1}}Bl_{V_2}X =Bl_{\bar{V_2}}Bl_{V_1}X -$$where $\bar{V_i}$ denotes the proper transform of $V_i$. This is essentially Proposition 2.9 in Kiem and Moon's article http://arxiv.org/abs/1002.2461.<|endoftext|> -TITLE: index of a closed subgroup of a profinite group -QUESTION [8 upvotes]: In the book "profinite groups, arithmetic, and geometry" of Shatz, the index $(G:H)$ of a closed subgroup $H$ of a profinite group $G$ is defined to be the supernatural number $lcm\big((G/U):(H/(H\cap U))\big)$ where $U$ runs over the open normal subgroups of $G$. There is an exercise following this definition saying that "$(G:H)=lcm(G:U)$ where $U$ runs over those open normal subgroups of $G$ containing $H$. -If $G$ is a finite group with discrete topology, then the index given is nothing but the number of elements in the coset space $G/H$. However, if we take $G$ to be a finite simple group having a non-trivial proper subgroup $H$, e.g. $Alt_n$ for a suitable $n$, the only normal subgroup contating $H$ is $G$ itself and $\big((G/G):(H/(H\cap G))\big)=1$. -I am not sure if the claim in the exercise is true for infinite profinite groups as they are necessarily non-simple, which means they don't admit trivial counter-examples. But at least the exercise seemed me wrong for finite case. Am I missing something, or this is a well-known misprint which I don't know? - -REPLY [8 votes]: (The first time around I had read your question too quickly and not properly appreciated it. Sorry about that.) -You are right: the exercise on p. 12 of Shatz's book is false, because of the example you suggest. You asked if there were also counterexamples among infinite profinite groups. Certainly: let $n \geq 5$, let $p$ be a prime number greater than $n$, and consider $G = \mathbb{Z}_p \times A_n$. Then the problem persists: take $H = \mathbb{Z}_p \times H'$, where $H'$ is a proper nontrivial subgroup of $A_n$. (Use Goursat's Lemma.) -I checked that this exercise does not appear in Serre's Galois Cohomology. Have you found that it is used at any point of Shatz's book? -It seems plausible to me that you could recover a statement like this by working prime-by-prime with the Sylow subgroups of the groups in question -- certainly there are enough normal subgroups of $p$ groups to detect indices -- but I haven't thought carefully about that.<|endoftext|> -TITLE: Obstructions for embedding a graph on a surface of genus g -QUESTION [21 upvotes]: Kuratowski's theorem tells us the complete graph $K_5$ and the bipartite graph $K_{3,3}$ are the only obstructions to a graph being planar, ie embeddable in the plane with no edge-crossings. - -Is the list of obstructions to being able to embed a graph with no edge-crossings on the surface of genus $g$ known to be finite for all $g$? - -REPLY [5 votes]: For the projective plane, i.e. the nonorientable surface of genus 1, this is known. Look at "A Kuratowski Theorem for the Projective Plane" in the homepage of Dan Archdeacon here: -http://www.emba.uvm.edu/~archdeac/ -This was his PhD thesis. In particular, he found that there are exactly 103 graphs such that if any graph $G$ contains one of these graphs as a subgraph, then $G$ cannot be not embedded into projective plane. You can refer to his original thesis for the list of the 103 graphs, or you can refer to the Appendix A of the book "Graphs on Surfaces" by Bojan Mohar and Carsten Thomassen.<|endoftext|> -TITLE: The Major Families of Quantum Groups -QUESTION [23 upvotes]: If we define a quantum group to be a quasi-triangular or coquasi-triangular Hopf algebra, then what are the major families of quantum groups? -Of couse, to start with we have the h-adic completions of the Drinfeld-Jimbo deformations of the enveloping algebras of complex semi-simple Lie algebras $U_q({\mathfrak g})$, and the dually defined quantised coordinate algebra of the compact semi-simple Lie groups. With these two families really being two sides of the same thing. -Apart from this, the only other main family that I know of are Majid's bicrossproduct Hopf algebras, which he says can be thought of as deformations of solvable groups. -Majid has two other constructions called bosination and double-bosination, but I don't know if these are quantum groups in the sense defined above. - -REPLY [16 votes]: I've been having trouble answering this question because I think your notion of "quantum group" is either too restrictive or too expansive. Hopf algebras suffer from annoying analytic issues as soon as they're infinite dimensional, so you should either be looking at finite dimensional quasitriangular Hopf algebras, or you should pick some particular world of analysis you want to work in (C* quantum groups, h-adic quantum groups, etc.). On the other hand, there's no real reason to restrict your attention to Hopf algebras, lots of things that go under the name "quantum group" (most notably the semisimplified categories at a root of unity which occur in Reshetikhin-Turaev's construction of 3-manifold invariants) are not the category of representations of a Hopf algebra, instead they're a braided tensor category. -Anyway some important constructions that you don't mention include: - -Drinfel'd twists of already known quasi-triangular Hopf algebras -Triangular Hopf algebras (classified by Etingof-Gelaki) -Woronowicz-style quantum groups in the C* setting -Various flavors of quantum groups at roots of unity -Bruguieres-Mueger quotients (often called "modularization" or "deequivariantization) of known braided tensor categories<|endoftext|> -TITLE: Adding a multiple of the Identity to a LU factorized matrix -QUESTION [7 upvotes]: Suppose a square, dense, symmetric matrix $A$ has been factorized into $L$ and $U$ components by performing a LU decomposition. Now let $B = A+\lambda I$. Is there any way to efficiently compute the LU decomposition of $B$ by reusing $L$ and $U$, avoiding the cost of a new decomposition? -If the cost of reusing $L$ and $U$ becomes too similar to the cost of recomputing the entire decomposition, would there be any other decomposition more suitable in this case? -Edit: I have found my question to be very similar to solving series of linear systems with diagonal perturbations. I am also trying to solve a series of linear systems where every system is in the form $(A+\lambda I)x=b$ where $\lambda$ is a real number and all $b$s are the same. However, unlike the aforementioned question, my matrices are always dense. - -REPLY [2 votes]: Such an efficient computation is unlikely. Suppose you can do it, with factors $U_\lambda$ and $L_\lambda$. Then by $\det(A+\lambda I)=\det U_\lambda\cdot\det L_\lambda$, you obtain the characteristic polynomial for free. Thus the cost of the calculation you are looking for cannot be smaller than the cost of the calculation of the characteristic polynomial. -For the latter problem, the best algorithm today is in $O(n^{7/2})$, if you employ standard multiplication of matrices (see the new edition of my book "Matrices ; theory and applications, section 3.10). It becomes $O(n^3)$ if you employ fast methods of multiplication (strictly better than Strassen's), but this is only theoritical, since fast matrix multiplaction is recursive, and extremely uncomfortable (if not impossible) to code.<|endoftext|> -TITLE: Does There Exists a General Quantum Casimir Extending the $U_q({\mathfrak sl}_2$ Case? -QUESTION [5 upvotes]: As is well known (see Kassel), when $q$ is not a root of unity, the centre or the quantum enveloping algebra $U_q({\mathfrak sl}_2)$ of ${\mathfrak sl}_2$ is generated by the element -$$ -C_q = EF + \frac{q^{-1}K+qK^{-1}}{(q-q^{-1})^2}. -$$ -The element is called the quantum Casimir. My questions are as follows: -(i) Does this situation extend to the general setting of $U_q({\mathfrak sl}_N)$? -(ii) If it does, is there a general formula for $C_q$? -(iii) How would this formula relate to the usual formula for the classical Casimir? (The uasual formula I refer to is $\sum X^iX_i$, for some basis $X_i$ and its dual $X^i$, see wikipedia for details.) - -REPLY [8 votes]: The centers of the Drinfeld-Jimbo quantum groups $U_q(\mathfrak{g})$ are well-understood and quite analogous to the classical case. See the book by Klimyk and Schmüdgen, Section 6.3, where in particular the quantum Casimirs are constructed.<|endoftext|> -TITLE: Torsors for finite group schemes -QUESTION [29 upvotes]: Let $k$ be a field of characteristic $p > 0$ (assume $k$ is perfect if it helps). Let $G$ be a connected finite group scheme of height one over $k$. Then $G$ is determined by its Lie algebra $\mathfrak{g}$, as a restricted Lie algebra: it can be recovered as the spectrum of the Hopf algebra dual to the restricted universal enveloping algebra $U( \mathfrak{g} )$. -Let $R$ be a local Artinian $k$-algebra with residue field $R / \mathfrak{m} \simeq k$. -I would like to understand the category of $G$-torsors over $R$ (with respect to the flat topology, say) which are trivialized over $R / \mathfrak{m}$. This category feels very ``infinitesimal,'' so it seems reasonable to expect that there is a way to describe its objects concretely in terms of linear-algebraic data related to $\mathfrak{g}$ (and without ever mentioning the flat topology). Can this be done? (And if so, how?) - -REPLY [16 votes]: Recall that if $G\rightarrow S$ is a flat group scheme, then a $G$-torsor is an -$S$-scheme $X\rightarrow S$ with a $G$-action $G\times_SX\rightarrow X$ such that -$G\times_SX\rightarrow X\times_SX$ given by $(g,x)\mapsto (gx,x)$ is an -isomorphism and such that $X\rightarrow S$ is faithfully flat. This -already gets rid of the flat topology but in the current case where -$S=\mathrm{Spec} R$ a local Artinian ring and we also give ourselves an -isomorphism of $G$-schemes $X\times_S\mathrm{k}=G\times_S\mathrm{Spec}k$ then it -is enough that $X\rightarrow S$ is a flat non-empty $G$-scheme as the fact that -$G\times_SX\rightarrow X\times_SX$ is an isomorphism can be checked upon -reduction to $k$. -As then $X$ is affine we are talking about a finite flat (non-zero) $R$-algebra -$T=R[X]$ which is a $R[G]$-comodule which is also a comodule algebra (i.e., the -product $T\bigotimes_RT\rightarrow T$ as well as the unit $R\rightarrow T$ -are comodule maps) together with a comodule algebra isomorphism -$T\bigotimes_Rk\rightarrow k[G]$. If we specialise further to the actual case at -hand, the category of $U({\frak g}\bigotimes_kR)^\ast$-comodules is isomorphic (for -once this is really an isomorphism) as a tensor category to the category of -${\frak g}\bigotimes_kR$-modules (as $p$-Lie algebra). This gives us a description purely in terms of -${\frak g}\bigotimes_kR$-modules. -Note that there is also a very concrete description of $G$-torsors (for -$G\rightarrow S$ finite flat for simplicity over an affine $S=\mathrm{Spec}R$). -For an $R$-algebra $R'$ we have a tautological map of group schemes -$G(R')\hookrightarrow (R'[G]^\ast)^\times$ (an element $f\in G(R')$ is by -definition an $R'$-homomorphism $R'[G]\to R'$ i.e. an element of $R'[G]^\ast$, it is easily seen to land in -$(R'[G]^\ast)^\times$ and is tautologically a group homomorphism). This gives us -an embedding of flat group schemes $G\hookrightarrow (R[G]^\ast)^\times$. This -gives us a (half-)long exact sequence of cohomology sets associated to -$G\hookrightarrow (R[G]^\ast)^\times\rightarrow (R[G]^\ast)^\times/G$. If $R$ is -Artinian all $(R[G]^\ast)^\times$-torsors are trivial as they correspond to -locally free rank $1$-modules over $R[G]^\ast$ (right modules to be specific) -which are all trivial as $R[G]^\ast$ is also Artinian. Hence $G$-torsors -correspond to sections of $(R[G]^\ast)^\times/G$ modulo the action of the -section of $(R[G]^\ast)^\times$. However, it is in general difficult to get a -concrete description of $(R[G]^\ast)^\times$ and even when you have one the -orbits may be difficult to figure out. -Addendum: In the orbit description I forgot to add the condition that everything should map to the identity in $k$. Note also that the orbit description is no doubt the closest you can get in general to usual description of for instance $\alpha_p$- and $\mu_p$-torsors where one extracts a $p$-th root of a function (resp. an invertible function). In some other cases one also gets simpler descriptions. For instance if $H$ is a smooth $R$-group scheme and $G$ is the kernel of the Frobenius map, then every $G$-torsor with a trivialisation over $k$ is obtained from a section of $H^{(p)}$. -Addendum 1: As an answer to Jacob's followup question, tautologically the image of $G$ in $(R[G]^\ast)^\times$ consists of the elements $f$ for which $\Delta(f)=f\otimes f$, where $\Delta$ is the coproduct. This means (I hope) that if we let $V$ be the graph of $\Delta$ as a subspace of $R[G]^\ast\times R[G]^\ast\bigotimes [G]^\ast$, then $G$ is the stabiliser of $V$. From this we can get a $1$-dimensional subspace by taking the $\dim V$-th exterior power. Mind you this gives a $1$-dimensional subspace not a vector. In many cases one does however get the same result when looking at a fixed vector.<|endoftext|> -TITLE: Can the HOMFLY polynomial be defined for framed unoriented links? -QUESTION [5 upvotes]: Given the diagram of an unoriented knot $K \subset S^3$, one can compute its Jones polynomial by picking an orientation for the knot and then simplifying the diagram using the relations drawn in the "properties" section here (the choice of orientation doesn't matter, although presumably it matters for links). Alternatively, one can give the knot a 0 framing and then simplify using the relations drawn here. -The HOMFLY polynomial of $K$ is computed in a way similar to the first method of computing the Jones polynomial - pick an orientation of the knot and simplify the diagram using the relations here. - -Is there a way of computing the HOMFLY polynomial of $K$ analogous to the second way of computing the Jones polynomial described above, i.e. give $K$ the 0 framing (but no orientation) and simplify the diagram using some relations? - -REPLY [5 votes]: I'm not the best qualified MO-er to answer this question, but since no one else has answered yet I'll give it a shot. This can be thought of as elaboration of Charlie Frohman's comment above. -Modulo a change of variables, the parameters of the HOMFLY polynomial are $N$, for the Lie algebra $sl_N$, and $q$, the deformation parameter of $U_q(sl_N)$. The strands of the link should be thought of as labeled by the basic representation of $sl_N$. For $N=2$ this is the Jones polynomial. For $N>2$ the basic representation is not self-dual, so the orientations of the link components matter a great deal. So the answer to your question is No, there is no unoriented version of the HOMFLY (a.k.a. HOMFLYPT) polynomial. -The basic representation of $sl_2$ is self dual, but importantly it is antisymmetrically self-dual (Frobenius-Schur indicator = -1). Because of the self-duality, the Jones polynomial has an unoriented version (the Kauffman bracket polynomial), but because of the antisymmetry of the self-duality, the unoriented version has some flaws (e.g. loss of positivity and categorifiability). -[And now we see the real reason I'm answering this question. I want to be the first person in the world (so far as I or Google knows) to use the word "categorifiability" in print.] -If you want an unoriented HOMFLY-ish polynomial, check out the Kauffman Dubrovnik polynomial, which is related to BMW algebras and B, C and D type Lie algebras.<|endoftext|> -TITLE: Fair but irregular polyhedral dice -QUESTION [59 upvotes]: I am interested in determining a collection of geometric conditions that will guarantee that a convex polyhedron -of $n$ faces is a fair die in the sense that, upon random rolling, it has an equal $1/n$ probability of -landing on each of its faces. -(Assume the polyhedron is composed of a homogeneous material; i.e., it is not "loaded.") -There has been study of what Grünbaum and Shephard call isohedral polyhedra, -which always represent fair dice: "An isohedron is a convex polyhedron with symmetries acting transitively on its faces with respect to the center of gravity. Every isohedron has an even number of faces." -It is clear such a polyhedral die is fair. -Here is an example of the trapezoidal dodecahedron, an isohedron of 12 congruent faces, -from an attractive web site on polyhedral dice: - -           - -But a clever argument in a delightful paper -by Persi Diaconis and Joseph Keller -("Fair Dice." Amer. Math. Monthly 96, 337-339, 1989) -shows (essentially, by continuity) that there must be fair polyhedral dice -that are not symmetric. -For example, there is no reason to expect that equal face areas is a necessary condition -for a polyhedral die to be fair. -Nor is it reasonable to expect that the distance from each face to the center of gravity -of the polyhedron is alone a determining condition. -Rather it should depend on the dihedral angles between faces, the likelihood of one face -rolling to the next—perhaps a Markov chain of transitions? -My question is: - -Is there a collection of geometric conditions—broader than isohedral—that guarantee - that a (perhaps asymetrical, perhaps unequal-face-areas) convex polyhedron represents - a fair die? - -Sufficient conditions welcomed; necessary and sufficient conditions may be too much to hope for! -Speculations and literature leads appreciated! - -REPLY [2 votes]: This is a tentative to solve the questions that arose in Matt Fayer's answer and comments, regarding the impossibility of a fair experiment. -Appealing to Information Theory, it is established that randomness is equivalent to the lack of information. More formally, uniform measures are the one that maximize Shannon's entropy. Conversely, complete knowledge is equivalent to Dirac measures. From this point of view it is clear that a perfect machine, where perfect means indeed complete knowledge, can do whatever you want and, in particular, is able to land a die always on the same face (this is a Dirac measure!). -After these observations, it seems to me clear that a fair throwing can be simulated imposing the lack of knowledge: the person who throw the die must be blinfolded; (s)he has to throw the die uniformly (i.e., as already suggested, chosing the Haar measure on $SO(3)$) - this can be forced using heavy gloves, but also more naturally using die with a uniform structure (I will come back to this point in a moment). On the other hand, a person who bet on the die is not allowed to use any other tool but his own eyes, in the case of uniform structure (in case of heavy gloves, he's blinfolded either). In this case, the Nash equilibrium of the ideal game between the two players (who throws and who bets) is the uniform measure on the number of faces of the die; i.e. the experiment is fair. -A comment about uniform structures: this probably brings to classical interpretation more or less discussed in other answers: I am thinking of a cube; more specifically, keeping this analogy with the real-world, I am thinking to something with the following property: if you close your eyes and you are allowed only to touch the die, then you cannot distinguish two different faces and the edges are trascurable; i.e. since the set of human beings is finite, take $\epsilon$ small enough such that every imperfection of the die is smaller than $\epsilon$. -P.s. maybe I should write this somewhere, but an important observation to keep in mind is that fairness, randomness, knowledge and so on are subjective. Something can be really random iff it is perfectly isolated.<|endoftext|> -TITLE: Distance to an apartment of the affine building of GL(N) -QUESTION [13 upvotes]: Here $F$ is a locally compact non-archimedean non-discrete field. -Let $X$ be the reduced (affine) Bruhat-Tits building of ${\rm GL}(n,F)$. Fix a maximal split torus $T$. Let $B$ be a Borel subgroup containing $T$ and write $U$ for the unipotent radical of $B$. Let $A$ be the unique apartment of $X$ stabilized by the normalizer $N_G (T)$ of $T$ in $G$. Finally fix a vertex $x$ of $X$. -It is not difficult to see that there exists a unique point $x_U$ of $A$ such that $x_U =u.x\in A$ for some $u\in U$ (use Iwasawa decomposition). Making $B$ vary among the Borel subgroups containing $T$, one gets $n!$ points $x_U$ in $A$. -My first question is: - -What is the link between the points $x_U$ and the projection $x_A$ of $x$ onto the apartment $A$ ($A$ is a closed convex subset of the CAT$(0)$ space $X$ and this projection is well defined) ? (For $n=2$, $x_A$ is the isobarycenter.) - -Now fix $B$ and consider $x$ and $u$ as above. Assume that $x\not\in A$. Then $u$ fixes a sector $C$ of $A$. Let $c$ be a point of $C$. Consider the geodesic segments $[c,x]$ and $[c, x_U ]$ (so that $[c,x_U ]=u.[c,x]$). Let $c_0$ be the unique point of $[c,x]$ such that $[c,c_0 ]\subset A$ and $(c_0 ,x]\cap A =\emptyset$. -My second question is: - -Is the point $c_0$ close to $x_A$ ? Can one choose $c$ such that $c_0 =x_A$ ? - -My third question is : - -Do we have $d(x,x_A )=\frac{1}{2}d(x,x_U )$ ? - -REPLY [9 votes]: I think that the answers to your three questions are negative. Here's an example for $n=3$. -Choose first $x_A$ to be a vertex of $A$. In the link of $x_A$, it is possible to choose a chamber $d$ which is at distance $2$ from $2$ chambers in $A$, and at distance $3$ from the $4$ others. Choose $x$ in the alcove $d$ so that it projects on $x_A$. To fix the ideas, you can take the middle of the side opposite to $x_A$. -Then the geodesic ray from $x$ to $x_A$ can be continuated in only four ways, namely via one of the four alcoves opposite $d$ in the link of $x_A$. This answers negatively to your second question. The retraction from the 4 points towards which these geodesics are going are easily calculated: you just have to retract the geodesic segment from $x$ to $x_A$ from somewhere further on your geodesic ray; the four points you get are the four points in the alcove opposite to the four alcoves which were opposite to $d$ (and, of course, on the middle of the side opposite to $x_A$). -From this, it is possible also to answer negatively to the third question: there is some $x_U$ which is on in one of the two alcoves which are at distance $2$ from $d$. Then the distance from $x$ to this $x_U$ is the distance between the middle of two sides of a regular hexagon which are at distance $2$, and the distance from $x$ to $x_A$ is the distance to the center of this hexagon, so we do not have $d(x,x_A)=\frac 1 2 d(x,x_U)$. -The calculation of the retractions centered at the two other points is not so easy. Let $\xi_1$ be one of them. The sector from $x_A$ to $\xi_1$ starts with some alcove $c$ which is at distance $2$ from $d$. Let $d'$ be the alcove adjacent to both $c$ and $d$, and let $H$ be the wall of $A$ which separates $c$ and $d'$. $H$ separates $A$ in two half-planes, say $\alpha$ and $-\alpha$, with $\xi_1$ in $\alpha$. Let $\beta$ be another half-plane bounded by $H$, such that $d'\in \beta$. Then $\beta\cup\alpha$ is another apartment $A_1$. -The retraction on $A_1$ centered at $\xi_1$ sends $d$ to a chamber which is at distance $2$ from $c$ and adjacent to $d'$, so it is the alcove in $\beta$ which is adjacent to $d'$. The retraction of $d'$ on $A$ is then the retraction of this alcove is the alcove which is at distance 2 from $c$ and in $-\alpha$. Since this alcove is already opposite to an alcove which is opposite to $d'$, we get the same point as some other $x_U$. -Of course, the retraction centered at the last point is treated in a similar way. So, in conclusion, there are only 4 different retractions of $x$. The two "double" points form a segment whose middle is $x_A$. The two other are symmetric with respect to $H$, but are not in the same sector. So the barycenter of our 6 points is some point of $H$ which is not $x_A$.<|endoftext|> -TITLE: For which categories does one have a Goursat Lemma? -QUESTION [19 upvotes]: Background -One of my favourite elementary results in group theory is Goursat's Lemma. This lemma characterises the subgroups of a direct product of groups in terms of fibred products. -Indeed, let $L$ and $R$ be groups and let $G < L \times R$ be a subgroup of their direct product. We have natural projections $\pi_L : L \times R \to L$ and $\pi_R: L \times R \to R$. We may assume without loss of generality that $\pi_L$ and $\pi_R$ are surjective when restricted to $G$. Let $L_0 = \pi_L(G \cap \ker\pi_R)$ and $R_0 = \pi_R(G \cap \ker\pi_L)$ denote, respectively, normal subgroups of $L$ and $R$. Goursat's Lemma is the observation that $G$ defines an isomorphism $L/L_0 \cong R/R_0$. -Proof: If $x \in L$, let $y \in R$ be such that $(x,y) \in G$. Such a $y$ exists because of surjectivity of $\pi_L : G \to L$, but it need not be unique. Nevertheless the coset $y R_0$ is well defined. This procedure then defines a homomorphism $L \to R/R_0$ whose kernel is precisely $L_0$ and which is surjective because $\pi_R : G \to R$ is. -If we let $\lambda: L/L_0 \to F$ and $\rho: R/R_0 \to F$ be isomorphisms to the same abstract group $F$, then we may identify $G$ with the fibred product -$$ G = \lbrace (x,y) \in L \times R ~|~ \lambda(x L_0) = \rho(y R_0) \rbrace .$$ -There are similar results for Lie subalgebras of a direct sum of Lie algebras, and probably also in other categories. This suggests the following categorical slogan: - -"subobjects of a product are pullbacks" - -(Well, at least subobjects with the property that the composition with the epis in the product are also epis.) -Of course, this is not going to be true in all categories, which prompts the following question. -Question -Let $\mathcal{C}$ be a category and $L,R$ be objects whose product $L \times R$ exists. Let $G \to L \times R$ be a monomorphism such that the compositions $G \to L \times R \to L$ and $G \to L \times R \to R$ are epimorphisms. -What must we demand of $\mathcal{C}$ so that there exist epimorphisms $L \to F$ and $R \to F$ such that -$$\begin{matrix} G & \rightarrow & L \cr - \downarrow & & \downarrow \cr - R & \rightarrow & F \end{matrix} -$$ -is a pullback? -Epilogue -One would be tempted to call these categories Goursat categories, but alas the name seems to be taken already for what seems like a different concept. -Thanks in advance. - -REPLY [16 votes]: To make life simple, suppose that finite limits and finite colimits exist. If we work with regular epimorphisms rather than epimorphisms then your condition is equivalent to saying that for any two (regular) epimorphisms $G\to L$ and -$G\to R$, if you form the pushout $F$ then the canonical comparison from $G$ to the pullback $L\times_F R$ is a regular epimorphism. -This is true in any exact Mal'cev category: see Theorem 5.7 of - -Carboni, Kelly and Pedicchio, Some remarks on Maltsev and Goursat categories, Applied Categorical Structures 1:385-421, 1993. - -Here exact means that the category -(i) has finite limits -(ii) has regular epi-mono factorizations -(iii) the pullback of a regular epi is a regular epi -(iv) any equivalence relation is the kernel pair of some map (one can choose the map to be the coequalizer) -and Mal'cev can be characterized in many ways. For example, it says that if R and S are equivalence relations on some object A, then RS=SR. -In fact if the category is regular, in the sense that (i)-(iii) hold, then your condition is equivalent to being exact and Mal'cev. -By the way, the Goursat categories you mention are only slightly weaker: they have -RSR=SRS rather than RS=SR. You can still prove your condition for Goursat categories if you suppose that at least -one of $G\to L$ and $G\to R$ is a split epimorphism (i.e. has a section), and in fact this can be used to characterize Goursat categories. See link text<|endoftext|> -TITLE: What is the difference between hard and soft analysis? -QUESTION [16 upvotes]: I have heard references to "hard" vs. "soft" analysis. What is the difference? It seems to do with generality versus nitty-gritty estimates, but I haven't gotten any responses more clear than that. - -REPLY [6 votes]: Consider the problem of shuffling a deck of cards using some shuffling technique. One may wonder, "If I use this or that shuffling technique, will I shuffle the deck?" The problem can be transferred to a question of whether a particular finite Markov process converges to the uniform distribution or not. Omitting some details, a classical theorem says that, yes, the process will converge (to the uniform distribution) as long as your technique is reasonable. Not only that, the convergence will eventually be exponential. This seems like a useful theorem, but it is actually rather deficient. The problem is that your practical side may wonder how many shuffles are required to get the deck reasonably randomized, and this theorem doesn't help. So you might say that the original analysis was soft because the result does not help in solving this quantitative problem. It tells you that shuffling is a good idea, but it doesn't give you any clue whether a given technique could shuffle the deck in your lifetime or not. A hard analysis would tell you, for example, "If one defines reasonably randomized by measure blahblah, then $2\log_2(52)$ riffle shuffles are sufficient to randomize the deck."<|endoftext|> -TITLE: numbers divisible by at least one of many numbers -QUESTION [14 upvotes]: Is the following true? -For any $c\in (0,1)$ there exists $f(c)>0$ such that for any subset $A\subset \{1,2,\dots,n\}$ of cardinality $|A|\geq cn$, the set - $$B=\left\{ k \in \{1,2,\dots,n!\} \colon \text{ there is } a \in A \text{ that divides } k\right\}$$ -of numbers having at least one divisor in $A$ satisfies $$|B|\geq f(c) n!.$$ -Of course, $n!$ may be replaced to the any common multiple of elements of $A$, or we may ask about densities or probabilities. -I do not know the answer even for $A=\{ cn\ $consecutive integers$\}$ - -REPLY [3 votes]: The asymptotic version is certainly false. Let $\varepsilon (x,y)$ be the density of numbers having a divisor in the interval $[x,y]$, then Besicovitch proved in "On the density of certain sequences of integers" that $$\liminf_{x\to \infty} \varepsilon (x,2x)=0.$$ -Later, Erdos improved this to $$\varepsilon(x,2x)\sim(\log x)^{-\delta +o(1)} , \delta \approx 0.086.$$ Denoting $H(x,y,z)$ to be the number of elements in $\{1,2,\dots,x\}$ which have a divisor in the interval $[y,z]$, some sharper estimates are given in this paper of Kevin Ford.<|endoftext|> -TITLE: Finitely generated, infinite, residually finite groups whose finite quotients are $p$-groups. -QUESTION [10 upvotes]: This question grew out of this post. - -Question: Is there a finitely generated, infinite, residually finite group such that every finite index subgroup has $p$-power index for a fixed prime $p$? - -The $p$-adic integers $\mathbb Z_p$ give an example of a non-finitely generated such group. This is not entirely obvious, but follows from a result of Jean-Pierre Serre, which states that every finite index subgroup of $\mathbb Z_p$ is closed. -EDIT: André Henriques has pointed out that Rostislav Grigorchuk constucted a finitely generated, infinite, residually finite $2$-group. All finite quotients of this group have to be $2$-groups. Colin Reid asked in a comment whether there is a torsionfree example. So let me take the freedom to extend my question: - -Question: Is there a torsionfree example? - -REPLY [10 votes]: Yes, there are torsion free examples. I do not know who constructed them first, but some examples can be found in papers by Grigorchuk and his co-authors. For example Bartoldi and Grigorchuk proved that a certain Fabrykowski-Gupta group $\Gamma $ has the following properties (see Propositions 6.4 and 6.5 in arXiv:math/9911206): -(a) It is a subgroup of the automorphism group of a rooted tree. -(b) It is virtually torsion free. -(c) It satisfies the following 'congruence property': every finite index subgroup of $\Gamma $ contains a level stabilizer (i.e., the stabilizer of a level of the tree) and the index of every level stabilizer is a power of 3. -By (a) $\Gamma $ is residually finite. By (b) there is a torsion free subgroup $K$ of finite index in $\Gamma $. Since every finite index normal subgroup of $K$ contains a finite index normal subgroup of $\Gamma $, (c) implies that every finite quotient of $K$ is a 3-group.<|endoftext|> -TITLE: Cohomology of a configuration space -QUESTION [9 upvotes]: The symmetric group $\Sigma_k$ acts on $X=F({\mathbb R}^n,k)$, - the ordered configuration space of $k$ points in - ${\mathbb R}^n$. - If $n$ is odd, the cohomology $H^*(X;{\mathbb Q})$ is - a rank-one free module over ${\mathbb Q}[\Sigma_k]$. -This is checked by calculating the character of this representation. - (Note also that this is obvious for $n=1$.) -Are there known (one-vector) bases of this module? - -REPLY [4 votes]: This isn't exactly an answer to your question, but here's how I like to think about the fact that you quoted. -Let's assume for a minute that $n=2$, so that we can think of $X$ as the complement of the braid arrangement in $\mathbb{C}^k$. Let $G = \mathbb{Z}/2\mathbb{Z}$, which acts on $X$ by complex conjugation. -Replace $\mathbb{Q}$ with a field $F$ of characteristic $2$. The $G$-equivariant cohomology ring $H^*_G(X; F)$ is a free module over $H^*_G(pt; F) \cong F[x]$ with the property that specializing at $x=0$ gives $H^* (X; F)$ and specializing at $x=1$ gives $H^*(X^G;F)$. -Thus we have a family of $\Sigma_k$ representations over the $F$-affine line interpolating between $H^* (X; F)$ and $H^*(X^G; F)$. Since the category of $\Sigma_k$ representations is semisimple, these two representations have to be isomorphic. The fact that $H^*(X^G; F)$ is the regular representation is obvious. -This is a good way to see that $H^*(X; F)$ is isomorphic to the regular representation of $\Sigma_k$. I'm not sure how to modify this argument to get $H^* (X; \mathbb{Q})$. I'm also not sure if this will help you find a cyclic vector, since it does not give you an explicit isomorphism between $H^* (X; F)$ and $H^* (X^G; F)$. -By the way, for $n>2$ you can do something similar, where $G$ acts on $\mathbb{R}^n$ by negating the last $n-1$ coordinates.<|endoftext|> -TITLE: Is every locally connected subset of Euclidean space R^n locally path connected ? -QUESTION [21 upvotes]: This is not actually a question asked by me. But since I do not know the answer, I would love to know if someone here could answer it. - -REPLY [4 votes]: An easy counterexample: -The basic space that is connected but not locally connected is the closure of the graph $y=\sin(1/x)$ ("the topologist sine curve"). -Call it $S$. Pick a countable set $D$ on the graph $y=\sin(1/x)$ whose closure is $D\cup \{0\}\times [-1,1]$. For each point $z \in D$ add a copy $S_z$ of $S$ (in $R^3$) starting from $z$ and ending on an interval in $0 \times [-1,1]$ so that the diameter of $S_z$ is at most twice the distance from $z$ to the $y$-axis and $S_z$ intersects $S_w$ only along the $y$-axis if $z$ is different from $w$. The union $C$ of $S$ and all $S_z$, $z \in D$, is clearly locally connected. -Any arc from $w$ in $D$ to the $y$-axis contained in $C$ would have to be contained in $S$ (it intersects each $S_z$ at most in $z$), a contradiction. -Added after reading comments: -Let's pick a sequence of planes $\Pi_n$ containing the $y$-axis and converging from above (from the point of view of the half-plane $x\ge 0$) to the $xy$-plane where $S$ resides. Enumerate points $z\in D$ as $z_n$, $n\ge 1$. -From each $z_n$ go up until you hit $\Pi_n$ and then create a scaled copy of $S$ lying on $\Pi_n$ and ending on the $y$-axis. -That should guarantee $S_z\cap S_w \subset y-\text{axis}$. -Why is $C$ locally connected? -The issue is only with point on the $y$-axis. Given any $\epsilon > 0$ and $-1\leq y\leq 1$, consider the union of all $S_z$ that intersect $0\times [y-\epsilon,y+\epsilon]\times 0$ and are of diameter less that $\epsilon$. Add $\epsilon$-neighborhoods of end-points of $S_z$s in $S$. That set is connected, of small diameter, and it contains $(0,y,0)$ in its interior rel.C.<|endoftext|> -TITLE: Does there exist a functorial splitting for the weight filtration (of singular cohomology)? -QUESTION [7 upvotes]: There are plenty of examples of varieties whose singular cohomology with rational coefficients considered as a mixed Hodge structure does not decompose as the direct sum of its pure (weight) factors. Yet if we consider the cohomology just as filtered vector spaces over rationals, such a decomposition certainly exist (for any variety). -My question is: could there exist a functorial decomposition like this (say, for the singular cohomology as a functor from the category of all smooth complex varieties), or does there exist some obstruction for such a functorial splitting? - -REPLY [2 votes]: There is the Deligne splitting. I take this from Peters and Steenbrink's book, Section 3.1. -For a complex variety $X$, we define $I^{p,q} \subseteq H^{\ast}(X, \mathbb{C})$ by -$$I^{p,q} := F^p \cap W_{p+q} \cap \left( \overline{F}^q \cap W_{p+q} + \sum_{j \geq 2} \overline{F^{q-j+1}} \cap W_{p+q-j} \right).$$ -Then $H^{\ast}(X, \mathbb{C}) = \bigoplus I^{p,q}$ and $W_k \otimes \mathbb{C} = \bigoplus_{p+q \leq k} I^{p,q}$ and $F^p = \bigoplus_{r \geq p} \bigoplus_s I^{r,s}$. -In particular, defining $U_k = \bigoplus_{p+q=k} I^{p,q}$ gives a splitting of the weight filtration tensored with $\mathbb{C}$. If I am not mistaken, it is functorial. -The Deligne splitting only exists with $\mathbb{C}$ coefficients. I think Donu Arapura's answer here is fairly convincing that there is no splitting with $\mathbb{Q}$ or $\mathbb{R}$ coefficients. -Disclaimer: I just learned about this today, so I might be missing something.<|endoftext|> -TITLE: Is the cotangent bundle to a Kahler manifold hyperkahler? -QUESTION [13 upvotes]: Let me be more specific. Let $M$ be a Kahler manifold with Riemannian metric $g$ and complex structure $I$. Then $T^\ast M$ will also be Kahler with metric and complex structure induced from $M$ (I will give them the same name). It is also holomorphic symplectic, with canonical holomorphic symplectic form $\Omega _\mathbb C$. -If $M$ was an affine space with the standard metric I could define $\omega _J$ and $\omega _K$ on $T^\ast M$ by taking the real and imaginary parts of $\Omega _\mathbb C$ which would define a hyperkahler structure on $T^\ast M$ (everything is covariantly constant with constant coefficients). -Question 1: Does this work for a general kahler manifold $M$? -It seems a bit unreasonable to me, as the construction of $\Omega _\mathbb C$ does not depend on the metric (but does depend on the complex structure, which is compatible with the metric...) -I also know that every hyperkahler manifold is holomorphic symplectic (with $\Omega _\mathbb C = \omega _J + I\omega _K$) and Yau's theorem implies that every compact holomorphic symplectic manifold is hyperkahler. -Question 2: Does $T^\ast M$ admit a hyperkahler metric, with the associated holomorphic symplectic form the canonical one (coming from the cotangent bundleness)? -Question 3: Is $g$ a hyperkahler metric for $T^\ast M$ at all? Or, does $T^\ast M$ admit a hyperkahler metric at all? -I don't know much about this sort of thing, but it seemed like a natural question to me, and I couldn't find an answer anywhere. - -REPLY [15 votes]: Such hyper Kaehler metrics do exist near the zero section, e.g. in a formal or an analytic tubular neighborhood of the zero section. After that one can use some homogeneity to spread them on the whole cotangent bundle but typically the resulting metrics are non-complete. One gets nice global metrics on the cotangent bundles of Hermitian symmetric spaces but this is pretty much it. This question was studied extensively. There are two different proofs of the existence: in this work of Birte Feix and - this work of Dima Kaledin.<|endoftext|> -TITLE: applications of Plancherel formulae -QUESTION [7 upvotes]: I've learned a few things about harmonic analysis on semisimple Lie groups recently and the amount of effort that goes into the proof of the Plancherel formula seems overwhelming. Of course it has led to great discoveries in representation theory, but I was wondering whether there are some direct applications of the Plancherel formula. And now when I'm thinking about it, I don't even recall any applications of the classical Plancherel formula. - -REPLY [7 votes]: The most basic and useful application of the classical Plancherel formula is to define the Fourier transform on $L^2$ by density. It is then customary to illustrate Plancherel formula by computing a few basic integrals. For example, from the easy fact that -$$ -\widehat{{\bf 1}_{[-1, 1]}}(\xi) = \int_{-1}^1 e^{-i\xi x} \, dx -= 2 \, {\sin \xi \over \xi}, -$$ -we compute the not totally obvious yet classical integral -$$ -\int_{\bf R} \Bigl({\sin \xi\over \xi}\Bigr)^2 d\xi -= {\pi \over 2} \int_{\bf R} ({\bf 1}_{[-1, 1]}(x))^2\,dx = \pi. -$$ -From the easy -$$ -\widehat{e^{iax}{\bf 1}_{[0,\infty[}(x)}= {-i \over \xi -a} -$$ -and the standard formula for the derivative, we get the not so easy Wallis integral -$$ -\int_{\bf R} {dx\over (1+x^2)^n} -= \biggl\Vert {1\over (x-i)^{n}}\biggr\Vert_2^2 -= {2\pi\over (n-1)!^2} \ - \Bigl\Vert \xi^{n-1}\,e^{\xi}\,{\bf 1}_{{\bf R}_-}(\xi)\Bigr\Vert_2^2 -% = {4\pi^2\over (n-1)!^2} \int_0^\infty \xi^{2n-2} e^{-\xi} d\xi -= {(2n-2)!\over 2^{2n-2}(n-1)!^2} \, \pi -$$ -and so on. -For the Lie group case, see the thesis of R. Gomez<|endoftext|> -TITLE: Covering the primes by arithmetic progressions -QUESTION [16 upvotes]: Define the length of a set of arithmetic progressions -of natural numbers -$A=\lbrace A_1, A_2, \ldots \rbrace$ -to be $\min_i | A_i |$: the length of the shortest sequence -among all the progressions. -Say that $A$ exactly covers a set $S$ -if $\bigcup_i A_i = S$. -Let $P'$ be the primes excluding 2. - -What is the longest set of arithmetic progressions - that exactly covers the primes $P'$? - -In other words, I want to maximize the length of -a set of such arithmetic progressions. -Call this maximum $L_{\max}$. -$L_{\max} \ge 2$ because -$$ -P' \;=\; -\lbrace 3,5 \rbrace -\cup -\lbrace 7,11 \rbrace -\cup -\cdots -\cup -\lbrace 521,523 \rbrace -\cup -\cdots -$$ -Perhaps it is possible that -$$P' \;=\; -\lbrace 3, 11, 19 \rbrace -\cup -\lbrace 5, 17, 29,41,53 \rbrace -\cup -\lbrace 7,19,31,43 \rbrace -\cup -\cdots -\;,$$ -but I cannot get far with sequences of length $\ge 3$. -(I know Green-Tao establishes that there are arbitrarily -long arithmetic progressions in $P$, but I don't know -if that helps with my question.) -I am number-theoretically naïve, -and apologize if this question is nonsensical or trivial. -In any case, I appreciate the tutoring! -Addendum. Although my question should be revised (as Idoneal suggests) -in light of George Lowther's proof that 3 -cannot be in a progression of length 4, George has shown that it is likely that $L_{\max}=3$ -but certification requires resolving an open problem. So I've added the open-problem tag. -Thanks for everyone's interest! - -REPLY [10 votes]: Despite the comments to the question (including mine), this is a bit easier than it seems at first sight. We can show that $L_{\max}=2$ or $3$. Almost certainly we have $L_\max=3$. However, determining which of these is actually the case seems to be beyond current technology, according to this MO answer "Are all primes in a PAP-3?". -Showing that, $L_{\max} < 4$ is easy. That is, not every odd prime is contained in an arithmetic progression of primes of length 4. More specifically, 3 is not contained in an arithmetic progression of length 4. Suppose that $\lbrace x, x+d, x+2d, x+3d \rbrace$ was such a progression for $d > 0$. Then $x\not=2$, otherwise we would have $x=2,d=1$, but $x+2d=4$ is not prime. So, $x=3$. But, then, $x+3d=3(1+d)$ is not prime. -Edit: Looking at $\tilde L_\max \equiv \max_A\liminf_i \vert A_i\vert$ might be more interesting. I expect that this is infinite but, again, showing that $\tilde L_\max > 2$ appears to be beyond current means.<|endoftext|> -TITLE: Characterizing forcings that don't add any dominating reals -QUESTION [7 upvotes]: Regarding reals as functions from $\omega$ to $\omega$, let's say a real $f$ eventually dominates $g$ iff $(\exists n)(\forall m > n)[ f(m) > g(m)]$. Let's say that a (non-trivial separative) forcing poset $P$ doesn't always add a dominating real iff there is a generic extension by $P$ which doesn't contains a real that eventually dominates every real from the ground model. Let's say that $P$ never adds a dominating real iff every generic extension by $P$ doesn't contain any real that eventually dominates all the ground model's reals. I'm interested in combinatorial/order-theoretic conditions which may be necessary or sufficient for either of these notions. - -$\omega$-closure implies you add no reals, hence you add no dominating reals; Cohen forcing is not $\omega$-closed but it never adds a dominating real -one can show that separability implies you never add a dominating real (by separability, I mean containing a countable dense subset); the Cohen forcing that adds uncountably many reals isn't separable but never adds a dominating real -Hechler forcing has size at most continuum but always adds a dominating real; the Cohen forcing that adds more than continuum many reals (where "continuum" is "continuum as computed in the ground model" obviously) has size greater than continuum but never adds a dominating real -Hechler forcing also has the countable chain condition yet adds a dominating real; the forcing that adds a function $\omega _1 \to \omega _1$ with countable partial functions doesn't have the ccc but it's $\omega$-closed hence adds no new reals and thus never adds any dominating reals. - -My question: -What are some combinatorial/order-theoretic conditions on a poset that are necessary and/or sufficient for the poset to never/not always add a dominating real? - -REPLY [7 votes]: Stefan's answer pointed me in the right direction, and then talking it over with prof. Leo Harrington we've got an answer: -A complete Boolean algebra $\mathbb{B}$ never adds a dominating real iff for any collection $\{ u _{m,k} : m, k \in \omega \} \subset \mathbb{B}^+$ the following weaker form of weak $(\omega ,\omega )$-distributivity holds: - -$\prod _{m \in \omega} \sum _{k \in \omega} u _{m,k} = \sum _{f \in \omega ^{\omega}} \prod _{n \in \omega} \sum _{n < m < \omega} \sum _{k < f(m)} u _{m,k}$ - -For the reverse implication, suppose "weak weak $(\omega ,\omega )$-distributivity" holds, and for contradiction let $u \neq 0$ be the Boolean value of the sentence "there exists a dominating real," and let $\dot{g}$ be a name witnessing this, i.e. the sentence "$\dot{g}$ is a dominating real" has Boolean value $u$. Define $u _{m,k} = || \dot{g} (m) = k ||$. Now if $G$ is any $\mathbb{B}$-generic filter containing $u$, noting that the left side of the distributivity identity is (at least) $u$, we know that the right side belongs to $G$. It's then not hard to see that: - -$(\exists f \in (\omega ^{\omega})^V )(\forall n \in \omega )(\exists m > n)(\exists k < f(m))(u _{m,k} \in G)$ - -which is to say that there's a real $f$ in the ground model such that: - -$(\forall n)(\exists m > n)(\dot{g}^G (m) < f(m))$ - -so $f$ is not dominated by $\dot{g}$, contradiction. -For the forward implication, it should suffice to show it in the case where for each $m$, the set $\{ u _{m,k} : k \in \omega \}$ is an antichain with least upper bound $u$ independent of $m$ (I haven't checked this detail personally). So let $\{ u _{m,k}\}$ be such a collection for which the identity fails. Consider the name: - -$\dot{g} = \{ (u _{m,k}, (m,k)) : m, k \in \omega \}$ - -It's not hard to see that the right side of the identity is at most $u$, so assuming the identity fails it's strictly less than $u$, so since $\mathbb{B}$ is separative there's a generic $G$ containing $u$ avoiding the right side of the identity. It's not hard to see from here that $\dot{g}^G$ will dominate all the ground model's reals. - -I should add that if we think of $u _{m,n}$ as saying "$\dot{g}(m) = n$" and replace the Boolean operations with the corresponding quantifiers, then the left side says "$\dot{g}$ is a real," and the right side says "$\dot{g}$ doesn't dominate every real in the ground model." This suggests how we can characterize forcings that don't add any unbounded reals, for example, namely the following identity holds: - -$\prod _{m \in \omega} \sum _{k \in \omega} u _{m,k} = \sum _{f \in \omega ^{\omega}} \prod _{m \in \omega} \sum _{k < f(m)} u _{m,k}$ - -Forcings that don't add any reals are precisely those that satisfy the following identity: - -$\prod _{m \in \omega} \sum _{k \in \omega} u _{m,k} = \sum _{f \in \omega ^{\omega}} \prod _{m \in \omega} u _{m,f(m)}$ - -You can easily generalize this to talking about functions $\kappa \to \lambda$; the above two results so generalized are precisely Theorem 15.38 and Lemma 15.39 in Jech, "Set Theory".<|endoftext|> -TITLE: Fundamental group of R^2 minus the (ir)rationals -QUESTION [10 upvotes]: Let -$$E =\{(x,0) \in \mathbb{R}^2 \colon x \in \mathbb{Q} \}$$ -$$F = \{(x,0) \in \mathbb{R}^2 \colon x \in \mathbb{R} \setminus \mathbb{Q}\}$$ -compute the fundamental group of $\mathbb R^2\setminus E$ and $\mathbb R^2\setminus F$. -How can I start? -(I don't know why the { symbols don't appear) - -REPLY [8 votes]: When you want to compute the fundamental group of a wild space very often the thing to do is identify it as the subgroup of an inverse limit of simpler fundamental groups (often the first shape group). A result of Fischer and Zastrow says that if $X\subseteq \mathbb{R}^{2}$, then the canonical homomorphism of $\pi_1(X,x)$ into the shape group $\check{\pi}_{1}(X,x)$ is injective for any $x\in X$. Of course, this homomorphism is not always injective (even for 2-dimensional compacta) but for a subset of the plane like you have this approach should work. This is, for instance, how you compute the fundamental group of the Hawaiian earring. I would begin by looking for some simple approximating spaces (probably with free fundamental groups) with projection maps and figuring out which elements of the inverse limit of the fundamental groups of these spaces are represented by loops. -Here is the paper I mentiond: -Fischer, Zastrow, The fundamental groups of subsets of closed surfaces inject into their first shape groups. Algebraic and Geometric Topology. Volume 5 (2005) 1655–1676.<|endoftext|> -TITLE: Asymptotic difference between a function and its "binomial average" -QUESTION [9 upvotes]: (I posted this question on Math.SE a few weeks ago. I got a few comments, but nothing definite, and so I thought I would try MO.) -The origin of this question is the identity -$$\sum_{k=0}^n \binom{n}{k} H_k = 2^n \left(H_n - \sum_{k=1}^n \frac{1}{k 2^k}\right),$$ -where $H_n$ is the $n$th harmonic number. -Dividing by $2^n$, we have -$$2^{-n} \sum_{k=0}^n \binom{n}{k} H_k = H_n - \sum_{k=1}^n \frac{1}{k 2^k}.$$ -The sum on the left can now be interpreted as a weighted average of the harmonic numbers through $H_n$ - where the weights, of course, are the binomial coefficients. Thus the difference between $H_n$ and its "binomial average" (I'm guessing there's no term for this) is -$$H_n - 2^{-n} \sum_{k=0}^n \binom{n}{k} H_k = \sum_{k=1}^n \frac{1}{k 2^k}.$$ -The sum on the right is known to converge to $\ln 2$ as $n \to \infty$. (Substitute $-\frac{1}{2}$ into the Maclaurin series for $\ln (1+x)$.) -This leads me to my question: - -Can we classify nonnegative functions $f(n)$ for which - $$\lim_{n \to \infty} \left(f(n) - 2^{-n} \sum_{k=0}^n \binom{n}{k} f(k) \right)$$ is finite and nonzero? - -It would seem that if $f$ increases sufficiently rapidly, then the limit would be $\infty$. This is the case with both $f(n) = a^n$ and $f(n) = n$. If $f$ decreases or is constant, then the limit is zero. If $f$ has basically logarithmic growth, then it seems the limit would behave as $H_n$. But can this be proved? And what about other sublinear, increasing functions? - -The two Math.SE responses were - -"I agree that logarithmic growth is what you need. The 'binomial average' of $f(n)$ should be about $f(n/2)$." (from Michael Lugo) -A reformulation of the problem in terms of exponential generating functions. (from Qiaochu Yuan) - -REPLY [5 votes]: The function $f(n)$ must be $\Theta (\log n)$. Update: As Didier Piau points out in the comments, we can say something stronger: $\frac{f(n)}{\log_2 n} \to L$ as $n \to \infty$. See the update at the end of the argument. -Suppose, for some positive $L$ (the negative case is similar), $$\lim_{n \to \infty} \left(f(n) - 2^{-n} \sum_{k=0}^n \binom{n}{k} f(k) \right) = L.$$ -Thus $$f(n) = L + r(n) + 2^{-n} \sum_{k=0}^n \binom{n}{k} f(k)$$ for some function $r(n)$ such that $r(n) \to 0$ as $n \to \infty$. This gives us a recurrence relation for $f(n)$. Since $r(n) \to 0$, $L + r(n) > 0$ for all sufficiently large $n$. So, for all sufficiently large $n$, there exist positive constants $C$ and $D$ such that $C < L + r(n) < D$. Since the initial terms in the function eventually become negligible in determining the value of $f(n)$ via the recurrence relation, there exist functions $g(n)$ and $h(n)$ such that for all sufficiently large $n$, $g(n) \leq f(n) \leq h(n)$ and $g(n)$ and $h(n)$ satisfy -$$g(n) = C + 2^{-n} \sum_{k=0}^n \binom{n}{k} g(k),$$ -$$h(n) = D + 2^{-n} \sum_{k=0}^n \binom{n}{k} h(k).$$ -So the problem reduces to showing that $g(n)$ and $h(n)$ are $\Theta (\log n)$. The argument is the same for both. -There are some different ways to do this; my favorite is to interpret the $g(n)$ recurrence probabilistically. Suppose we start at time $g(0)$, we flip a set of $n$ coins simultaneously each round, and it takes $C$ time units to do one round of flips. When a coin turns up heads for the first time, we cease flipping it. Let $T(n)$ be the time at which the last coin to achieve its first head does so. To find $E[T(n)]$, condition on the number of coins that achieve heads in the first round of flips. This yields $$E[T(n)] = C + 2^{-n} \sum_{k=0}^n \binom{n}{k} E[T(n-k)] = C + 2^{-n} \sum_{k=0}^n \binom{n}{k} E[T(k)].$$ Thus $g(n) = E[T(n)]$. -Another way to view $T(n)$ is that it is $g(0) + C M_n$, where $M_n = \max\{X_1, X_2, \ldots, X_n\}$ and the $X_i$'s are independent and identically distributed geometric $(1/2)$ random variables. (Each geometric random variable models the first time a head appears.) Thus $g(n) = g(0) + C E[M_n]$. It is known that $\frac{H_n}{\log 2} \leq E[M_n] \leq \frac{H_n}{\log 2} + 1$ and, more precisely, that $E[M_n]$ is logarithmically summable to $\frac{H_n}{\log 2} + \frac{1}{2}$. (See, for example, Bennett Eisenberg's paper "On the expectation of the maximum of IID geometric random variables," Statistics and Probability Letters 78 (2008) 135-143. See also this MO question, "What is the Expected Maximum out of a Sample (size $n$) from a Geometric Distribution?") -Thus $g(n) = \frac{C}{\log 2} \log n + O(1)$, which means that $h(n) = \frac{D}{\log 2} \log n + O(1)$ and $f(n) = \Theta (\log n)$. -Update: Given $\epsilon > 0$, if we take take $C > 0$ such that $L - \epsilon \leq C < L$ and $D = L + \epsilon$, this argument shows that -$$L - \epsilon + O\left(\frac{1}{\log n}\right) \leq \frac{f(n)}{\log_2 n} \leq L + \epsilon + O\left(\frac{1}{\log n}\right).$$ -Thus, as $n \to \infty$, $\frac{f(n)}{\log_2 n} \to L.$ -For some other ideas that pertain to this result, including what are effectively some alternative derivations, see Pradipta's recent MO question, "Coin Flipping and a Recurrence Relation". In fact, reading Pradipta's question and some of its answers gave me the ideas needed to construct this argument! So, thanks go to Pradipta, Didier Piau, Emil Jeřábek, and Louigi Addario-Berry.<|endoftext|> -TITLE: A learning roadmap for Additive combinatorics. -QUESTION [11 upvotes]: Hello, -I'd love to learn more about the field of additive combinatorics. From what I've understand, there's a book by Tao and Vu out on the subject, and it looks fun, but I think I lack the prerequisites. Right now, I've had basic real analyis (Rudin), read the first volume of Stanley's "Enumerative combinatorics", and some algebra (some graduate). I have no experience of probability theory whatsoever, or functional analysis or ergodic theory. So I'm curious, from my background, what would be needed to reach the level where I can read and understand the book of Tao and Vu? Are there any certain books to reach that level which you can recommend? -Best regards, -CM - -REPLY [9 votes]: Apart from Tao-Vu (which is a very useful resource) there aren't any obvious books from which to learn this subject at all comprehensively. The books Kevin recommends give an excellent survey of the state of the area in the late 1990s. -For more recent material, here are a few sets of course notes. Some of these are a bit more leisurely than Tao-Vu. -Jacques Verstraete: http://www.wix.com/annatar0/math262 -Andrew Granville: http://www.dms.umontreal.ca/~andrew/Courses/MAT6640.H10.html -Ben Green: http://www.dpmms.cam.ac.uk/~bjg23/add-combinatorics.html -Terry Tao: http://www.math.ucla.edu/~tao/254a.1.03w/ -For an overview of the content of Tao-Vu, you could consult my review of it in the Bulletin of the AMS: -http://www.ams.org/journals/bull/2009-46-03/S0273-0979-09-01231-2/home.html - -REPLY [4 votes]: Video lectures are the next best thing to seminars and discussions with others. Have a look at the videos of the workshop "Introduction to Ergodic Theory and Additive Combinatorics" from Fall 2008 at the MSRI.<|endoftext|> -TITLE: Is there a map of spectra implementing the Thom isomorphism? -QUESTION [23 upvotes]: A well known theorem in algebraic topology relates the (co)homology of the Thom space $X^\mu$ of a orientable vector bundle $\mu$ of dimension $n$ over a space $X$ to the (co)homology of $X$ itself: $H_\ast(X^\mu) \cong H_{\ast-n}(X)$ and $H^\ast(X^\mu) \cong H^{\ast-n}(X)$. -This isomorphism can be proven in many ways: Bott & Tu has an inductive proof using good covers for manifolds and I learned on MathOverflow that one can use a relative Serre spectral sequence. However, I believe that there should also be a proof using stable homotopy theory, in the case of homology by directly constructing a isomorphism of spectra $X^\mu \wedge H\mathbb{Z} \to X_+ \wedge \Sigma^{-n} H\mathbb{Z}$, where $X^\mu$ denotes the Thom spectrum, $H\mathbb{Z}$ the Eilenberg-Mac Lane spectrum for $\mathbb{Z}$ and $X_+$ the suspension spectrum of $X$ with a disjoint basepoint added. -Is there an explicit construction of such a map implementing the Thom isomorphism on the level of spectra? I am interested in such a construction for both homology and cohomology. If so, is there a similar construction for generalized (co)homology theories? I would also be interested in references. - -REPLY [3 votes]: I just noticed this question, so my apologies for a very belated answer. Proposition 20.5.5 of http://www.math.uchicago.edu/~may/EXTHEORY/MaySig.pdf states that a $k$-orientation of a spherical fibration $X$ over a base space $B$ specifies a $k$-trivialization of $X$. Here $k$ is any (homotopy) commutative ring spectrum, and the notion of a $k$-orientation is the standard one. However, a $k$-trivialization -is an equivalence $k\wedge X\simeq k\wedge S^n_B$ of parametrized spectra over $B$. Intuitively, after smashing with $k$ over $B$, $X$ thinks that it is the trivial fibration $B \times S^n \to B$.The Thom isomorphism follows directly.<|endoftext|> -TITLE: Simplicial "universal extensions", the hammock localization, and Ext -QUESTION [5 upvotes]: Let $M,B$ be $R$-modules, and suppose we're given an n-extension $E_1\to\dots\to E_n$ of $B$ by $M$, that is, an exact sequence $$0\to M\to E_1\to\dots\to E_n \to B\to 0.$$ -A morphism of $n$-extensions of $Y$ by $X$ is defined to be a hammock -$$\begin{matrix} -&&A_1&\to&A_2&\to&A_3&\to&\ldots&\to &A_{n-2}&\to &A_{n-1}&\to& A_{n}&&\\ -&\nearrow&\downarrow&&\downarrow&&\downarrow&&&&\downarrow&&\downarrow&&\downarrow&\searrow&\\ -X&&\downarrow&&\downarrow&&\downarrow&&&&\downarrow&&\downarrow&&\downarrow&&Y\\ -&\searrow&\downarrow&&\downarrow&&\downarrow&&&&\downarrow&&\downarrow&&\downarrow&\nearrow&\\ -&&B_1&\to&B_2&\to&B_3&\to&\ldots&\to &B_{n-2}&\to &B_{n-1}&\to& B_{n}&&\end{matrix}$$ -This detetermines a category $n\operatorname{-ext}(Y,X)$. Further, we're given an almost-monoidal sum on this category given by taking the direct sum of exact sequences, then composing the first and last maps with the diagonal and codiagonal respectively. Taking connected components, we're left with a set $ext^n(Y,X)$, and the sum reduces to an actual sum on the connected components, which turns $ext^n(Y,X)$ into an abelian group (and therefore an $R$-module). -It's well known that these functors, called Yoneda's Ext functors are isomorphic to the Ext functors $\pi_n(\underline{\operatorname{Hom}}(sM,sN))$ where $\underline{\operatorname{Hom}}(sM,sN)$ is the homotopy function complex between the constant simplicial $R$-modules $sM$ and $sN$ (obtained by means of cofibrant replacement of $sM$ in the projective model structure, fibrant replacement of $sN$ in the injective model structure, or by any form of Dwyer-Kan simplicial localization (specifically the hammock localization)). -In a recent answer, Charles Rezk mentioned that we can compute this in the case $n=1$ as $\pi_0(\underline{\operatorname{Hom}}(sM,sN[1]))$, where $sN[1]$ is the simplicial $R$-module with homotopy concentrated in degree $n$ equal to $N$. That is, these are exactly the maps $M\to N[1]$ in the derived category. -It was also mentioned that for the case $n=1$, there exists a universal exact sequence in the derived category: $$0\to N\to C\to N[1]\to 0$$ where $C$ is weakly contractible such that every extension of $N$ by $M$ arises as $$N\to M\times^h_{N[1]} C\to QM,$$ which is $\pi_0$-short exact. (And QM is a cofibrant replacement of $M$, for obvious reasons). -Questions: -Why can we get $\pi_1$ of the function complex by looking at maps into $N[1]$? At least on the face of it, it seems like we would want to look at maps into $N[-1]$, that is, look at maps into the "loop space", not the "suspension" (scare quotes because these are the loop and suspension functors in $sR\operatorname{-Mod}$, not in $sSet$. -What is this this "universal extension" $$0\to N\to C\to N[1]\to 0$$ at the level of simplicial modules? -Is there a similar "universal extension" for $n>1$? If so, what does it (one of its representatives) look like at the level of simplicial modules? -Given an $n$-extension $\sigma$ of $B$ by $M$, how can we produce a morphism in the derived category $B\to M[n]$ that generates an $n$-extension in the same connected component of $n\operatorname{-ext}(B,M)$? -Lastly, since Yoneda's construction of Ext looks suspiciously like Dwyer and Kan's hammock localization, and there is homotopy theory involved in the other construction, I was wondering if there was any connection between the two. That is, I was wondering if there is another construction of Ext using DK-localization directly that shows why Yoneda's construction works. - -REPLY [7 votes]: For your first question: - -Given an $n$-extension $\sigma$ of $B$ by $M$, how can we produce a morphism in the - derived category $M\to N[n]$ that generates an n-extension in the same connected - component - of $n{-}ext(B,M)$? - -Do you mean a morphism $B\to M[n]$? In which case, isn't this a the standard construction? Let $E$ be the extension of $B$ by $M$, let $P$ be a projective resolution of $B$, and cover the identity map of $B$ by a map of chain complexes $P\to E$. The resulting map $P_n\to M$ can be thought of as a morphism $B\to M[n]$ in the derived category, and it's the one that represents the corresponding component of $n{-}ext(B,M)$. -As for your second question, there's certainly some connection. What you've described is a category $C=n{-}ext(X,Y)$, whose objects are $n$-extensions of $Y$ by $X$. Let $W=C$, and perform the hammock construction $L_H(C,W)$ of the pair $(C,W)$; then $L_H(C,W)$ is a simplicially eniched category, which represents an $\infty$-groupoid since you inverted everything. The ext group is the set of equivalence classes of objects in $L_H(C,W)$. -Dwyer and Kan showed that if $C=W$, then the simplicial category $L_H(C,W)$ and the simplicial nerve $NC$ of $C$ represent the same $\infty$-groupoid. So $\mathrm{Ext}^n(X,Y)=\pi_0 (NC)$. -What is the homotopy type of $NC$? It is supposed to be equivalent to the space of maps $\mathrm{map}(X,Y[n])$ in the $\infty$-category associated to the derived category. -I'm unaware of a reference where this is proved, however. -But take a look at Stefan Schwede's very nice paper on An exact sequence interpretation of the Lie bracket in Hochschild cohomology, which shows that $\pi_1(NC)=\mathrm{Ext}^{n-1}(X,Y)$, and gets some interesting results from that. (He's looking at $A$-bimodule extensions of $A$ by $A$ (i.e., Hochschild cohomology), and he extracts the Gerstenhaber-algebra structure on the ext-groups by means of homotopical constructions involving the $NC$.<|endoftext|> -TITLE: Semisimplicity of Lie algebra in positive characteristic -QUESTION [10 upvotes]: Let $F$ be a field of characteristic $p > 0$. Let $\mathfrak{g}$ be a linear Lie algebra, that is $\mathfrak{g}\subset M_n(F)$ for some natural number $n$. Does there exist a condition involving $n$ and $p$ such that $\mathfrak{g}$ is semisimple if and only if its Killing form is non-degenerate? - -REPLY [9 votes]: If $g\subset M_n(F)$ and $n\le p-2$ then $g$ is semisimple if and only if the Killing form of g is non-degenerate. This statement is clear (and empty) if $p\in\{2,3\}$, while for $p>3$ one can use various results of the theory of modular Lie algebras. First one may assume that $g$ is semisimple as the converse statement is an easy exercise. One may also assume that $F$ is algebraically closed. Using Block's theorem one the structure of semisimple Lie algebras we then observe that $g$ is sandwiched between $\bigoplus_i S_i$ and $\bigoplus_i{\rm Der}(S_i)$ where the $S_i$'s are simple Lie algebras. As $n\le p-2$ the Classification Theorem implies that each $S_i$ is isomorphic to the Lie algebra of a simple algebraic $F$-group. Looking at the ranks of these groups more closely (and again using the inequality $n\le p-2$) we deduce that each $S_i$ has a non-degenerate Killing form and ${\rm Der}(S_i)=\mathrm{ad}\, S_i$ (this is discussed in Seligman's book quoted earlier). Hence the Killing form of $g\cong\bigoplus_i S_i$ is non-degenerate as well. The assumption on $p$ cannot be relaxed as for $p>3$ the Witt algebra $g={\rm Der}\,\mathcal{O}$ where $\mathcal{O}=F[X]/(X^p)$ is simple, has the zero Killing form, and acts irreducibly on the vector space $\mathcal{O}/F1$ of dimension $p-1$.<|endoftext|> -TITLE: Which curves have infinitely many rational points -QUESTION [14 upvotes]: Question: Assuming finiteness of the Tate-Shafarevich group, is there an algorithm to determine whether a curve $C$ defined over a number field $K$ has infinitely many $K$-rational points? - -I believe that this is (a) true and (b) sufficiently important that it has been carefully explained somewhere, but I don't know a reference. Any help from the MO community would be very much appreciated! -P.S. To make the question precise, $C$ is specified, say, by its function field (a simple extension of $K(x)$), and all abelian varieties over all number fields have finite Sha. (And if the algorithm takes $10^{10}$ years for C: y=3x+1 over the rationals, I don't care.) - -REPLY [7 votes]: Dear Tim: -How right you are to worry about computing the Jacobian of a genus one curve! Years ago, in the midst of many discussions with Bill McCallum, Alex Perlis, Nick Shepherd-Barron, and John Tate, I convinced myself this could be done, and it was eventually written up in Alex's Arizona thesis: -http://math.arizona.edu/~aprl/publications/dissertation/ -Unfortunately, my own argument had many waves of the hand, and I didn't actually read the final product. But hopefully, it's all worked out. Alex really was a pretty careful guy, with substantial computational saavy. -Possibly, you might also enjoy this unpublished note of mine: -http://www.ucl.ac.uk/~ucahmki/jockusch.pdf -where these issues are discussed somewhat informally. -Added: I only read the comments above just now. I'm not entirely sure that the algorithm described in Perlis' thesis is any more efficient than what Pete suggests. Let me know if it turns out to be so.<|endoftext|> -TITLE: On the polynomial $x^3 =ax^2 + (a + 3)x + 1, a \in \mathbb{Z}$ -QUESTION [7 upvotes]: I wonder if somebody can provide enlightenment on my question on the polynomial below. -The polynomial has the property that if you pick up one root, the other two can be obtained by successively acting certain fractional linear transformation on it. I wonder whether this phenomenon is ubiquitous (at least to a certain family of polynomials over rational numbers, say) or sporadic. -This fact is remarked by Shanks (an excerpt is attached below), but he gives no explanation as to why that is the case nor how he obtained the result. -Serre refers to this polynomial in one of his papers below, but he says nothing about the (fractional linear) group action. - -In this article, Shanks says: - -The cubic equation - $$(a) \qquad x^3 =ax^2 + (a + 3)x + 1$$ - has the discriminant - $$(b) \qquad D = (a^2 + 3a + 9)^2,$$ - and if $a^2 + 3a + 9$ is prime, (b) is obviously also the discriminant of the field $\mathbb{Q}(\rho)$ where $\rho$ is a root of (a). One may easily verify that the other two roots of (a) are $\rho_{2} = -l/(l+\rho)$ and $\rho_{3} = - 1/(1 +\rho_{2})$. - -Serre refers to this polynomial, saying "c'est même là une extension universelle" in Théorème 3 here. - -REPLY [18 votes]: This is true of every cubic polynomial with Galois group $C_3$ (in particular, in characteristic not equal to $2$ it is true of every separable irreducible cubic polynomial with discriminant a square). Let $f$ be such a cubic polynomial over a base field $k$ and let $L$ be the splitting field, hence $[L : k] = 3$. Let $r_1, r_2, r_3$ be the roots of $f$ in $L$. Then there is some nontrivial linear dependence among the elements $1, r_1, r_2, r_1 r_2$, hence a fractional linear transformation with coefficients in $k$ such that $r_2 = \frac{a r_1 + b}{c r_1 + d}$. This relation respects the action of the Galois group, so if we choose $r_1, r_2$ so that the Galois group acts by $r_1 \mapsto r_2, r_2 \mapsto r_3, r_3 \mapsto r_1$ we get $r_3 = \frac{a r_2 + b}{c r_2 + d}$ as desired. -On the other hand, this sufficient condition is necessary, since it implies that $[L : k] = 3$. -(Full disclosure: this was an exercise I did recently.)<|endoftext|> -TITLE: number of subdivisions needed to compare simplicial sets to simplicial complexes? -QUESTION [7 upvotes]: This is a question in elementary geometric topology, of which I know little. It has to do with the result that geometric realizations of simplicial sets and geometric realizations of abstract simplicial complexes coincide up to homeomorphism, and with how many subdivisions are needed to effect the argument. -Here is a sketch that the realization of a simplicial set is the realization of a simplicial complex. There are functors -$$Face: [\Delta^{op}, Set] \to Pos$$ -$$Nerve: Pos \to [\Delta^{op}, Set]$$ -where $Face(X)$ is the poset whose elements are simplices of $X$, ordered by $x \leq y$ if $x$ is a nondegenerate face of $y$. The composite $Nerve \circ Face$ is the barycentric subdivision of $X$. This subdivision is a regular simplicial set, whose realization is a regular CW complex. Associated to a regular CW complex is a simplicial complex, also called its subdivision, whose vertices are open cells in the CW decomposition, and where simplices are finite chains $e_1 < \ldots < e_n$ where each $e_i$ is a proper face of $e_{i+1}$ ($e_i$ is contained in the closure of $e_{i+1}$). There is a theorem that the realization of this simplicial complex is homeomorphic to the space obtained by gluing together the regular CW complex. -This line of argument almost feels like overkill to me: we subdivide once, realize, and then subdivide again to get to the simplicial complex. I'd like to simplify this if possible. Define a functor -$$Flag: Pos \to SimpComplex$$ -which takes a poset $P$ to the simplicial complex whose vertices are elements of $P$, and whose simplices are finite "flags" $x_1 < \ldots < x_n$. (In passing, I'll note that there is a fourth functor -$$U: SimpComplex \to Pos$$ -which takes a simplicial complex to the poset of simplices ordered by inclusion; the composite $Flag \circ U$ is the barycentric subdivision functor on simplicial complexes.) My question is: - -Let $X$ be any simplicial set. Is the realization of the simplicial complex $Flag \circ Face(X)$ homeomorphic to the realization of $X$? - -I am not hugely confident that the answer is "yes", and even suspect there are standard counterexamples, but it seems to be true for simple examples. -Edit: I have a feeling that I botched the description of the subdivision of simplicial sets, but I'll let the experts correct me if needed. - -REPLY [10 votes]: Let $X$ be a simplicial set. The partially ordered set $(X^{nd}, \leq)$ of non-degenerate simplices of $X$, with $x\le y$ if $x$ is a face of $y$, was considered by Barratt in a 1956 Princeton preprint. I write $B(X) = N(X^{nd}, \le)$ for its nerve, the Barratt nerve. -It is not quite clear to me if, in your definition of $Face \colon sSet \to Pos$, you only allow the non-degenerate simplices as elements. Let me assume that you make that restriction, so that $Nerve \circ Face(X) = B(X)$. -If $X$ is the simplicial set associated to an (ordered) simplicial complex, then $B(X)$ is -indeed the barycentric subdivision of that simplicial complex. In these cases, there is a homeomorphism $|B(X)| \cong |X|$. However, if $X = \Delta^n/\partial\Delta^n$ for $n\ge1$, -then $X^{nd}$ is isomorphic to $[1] = (0 < 1)$ and its nerve is contractible. So in these cases $|B(X)|$ is not homeomorphic (nor homotopy equivalent) to $|X|$. -From the definition of Kan's normal subdivision $Sd(X)$ as a coend, there is a canonical -natural map $b_X \colon Sd(X) \to B(X)$. It is an isomorphism if and only if $X$ is non-singular, meaning that the representing map $\bar x \colon \Delta^n \to X$ of each non-degenerate simplex of $X$ is a cofibration. Here $n$ is the dimension of $x$. -There exists a homeomorphism $h_X \colon |Sd(X)| \cong |X|$, due to Fritsch and Puppe (1967). However, it is not natural for most maps $X \to Y$ of simplicial sets. For instance, there is no way to fix $h_{\Delta^1}$ and $h_{\Delta^2}$ so that the homeomorphisms are compatible with both of the degeneracy maps $\sigma^j \colon \Delta^2 \to \Delta^1$. -The normal subdivision $Sd(X)$ of any simplicial set is a regular simplicial set, in the sense that each representing map $\bar x \colon \Delta^n \to X$ only makes identifications -along the last ($n$th) face. The geometric realization of a regular simplicial set -is a regular CW complex, i.e., if the closure of each $n$-cell is an $n$-ball and the boundary of the open $n$-cell in its closure is an $(n-1)$-sphere. Any regular CW complex admits a triangulation. See e.g. Fritsch and Piccinini (1990), sections 3.4 and 4.6 for proofs. -Combining the two previous paragraphs, the realization of any simplicial set can be -triangulated by a simplicial complex, but not in a natural way. Not every CW complex can be triangulated (Metzler, 1967), but the realizations of (regular) simplicial sets can. -In work of Waldhausen, Jahren and myself (Spaces of PL Manifolds and Categories of Simple Maps, -Annals of Maths. Studies, to appear in 2013) we consider finite simplicial sets $X$ -and prove in Proposition 2.5.8 that there are natural maps -$$ -Sd(Sd(X)) \to B(Sd(X)) \to Sd(X) \to X . -$$ -Each of these are simple, in the sense that their geometric realizations have contractible -point inverses. In particular, they are simple-homotopy equivalences. The term -$B(Sd(X))$ -is the nerve of a partially ordered set, hence an ordered simplicial complex. So there -is, indeed, a natural simple map -$$ -i_X \colon B(Sd(X)) \to X -$$ -from an ordered simplicial complex to $X$. -You mentioned geometric topology at the outset. If $X$ is a combinatorial manifold, -so that $M = |B(Sd(X))|$ and $N = |X|$ are PL manifolds, then Cohen (1970) proved that -any simple PL map $M \to N$ can be uniformly approximated by a PL homeomorphism (assuming -the manifolds are of dimension at least 5). Hence the natural map $|i_X| \colon |B(Sd(X))| \to |X|$ is a map from an ordered simplicial complex that is arbitrarily -close to a PL homeomorphism, but I have no reason to expect that this -approximation can be made natural.<|endoftext|> -TITLE: Advanced Math Jokes -QUESTION [24 upvotes]: I am looking for jokes which involve some serious mathematics. Sometimes, a totally absurd argument is surprisingly convincing and this makes you laugh. I am looking for jokes which make you laugh and think at the same time. -I know that a similar question was closed almost a year ago, but this went too much in the direction "$e^x$ was walking down the street ...". There is also the community wiki Jokes in the sense of Littlewood, but that is more about notational curiosities. In order to motivate you, let me give an example: - -The real numbers are countable. Indeed, let $r_1,r_2,r_3,\dots$ be a list of real numbers and suppose that there is a real number missing. Just add it to the list. - -If moderators or audience decide to close this question as off-topic or duplicate, I can fully understand. I just thought it could be interesting and entertaining to have this question open for at least some time. - -Added by joro Sat Apr 27 08:59:45 UTC 2019 There is chat room about general jokes and it appears close resistant. - -REPLY [33 votes]: This is from my blog, which I interestingly just posted today (at the time of this posting). -Several mathematicians are asked, "how do you put an elephant in a refrigerator?" -Real Analyst: Let $\epsilon\gt0$. Then for all such $\epsilon$, there exists a $\delta\gt0$ such that $$\left|\frac{\mathit{elephant}}{2^n}\right|\lt\epsilon$$ for all $n\gt\delta$. Therefore $$\lim_{n\to\infty} \frac{\mathit{elephant}}{2^n}=0.$$ Since $1/2^n \lt 1/n^2$ for $n\ge 5$, by comparison, we know that $$\sum_{n\ge 1}\frac{\mathit{elephant}}{2^n}$$ converges — in fact, identically to $\mathit{elephant}$. As such, cut the elephant in half, put it in the fridge, and repeat. -Differential Geometer: Differentiate it and put into the refrigerator. Then integrate it in the refrigerator. -Set Theoretic Geometer: Apply the Banach–Tarski theorem to form a refrigerator with more volume. -Measure Theorist: Let $E$ be the subset of $\mathbb{R}^3$ assumed by the elephant and $\Phi\in\mathbb{R}^3$ be that by the fridge. First, construct a partition $e_1,\ldots,e_i$ on $E$ for $1\le i \le N$. Since $\mu(E)=\mu(\Phi)$, and $$\mu(E)=\mu\left(\bigcup_{1\le i \le N}e_i\right)=\sum_{1\le i \le N}\mu(e_i),$$ we can just embed each partition of $E$ in $\Phi$ with no problem. -Number Theorist: You can always squeeze a bit more in. So if, for $i\ge 0$. you can fit $x_i$ in, then you can fit $x_i + x_{i-1}$ in. You can fit in a bit of the elephant $x_n$ for fixed $n$, so just use induction on $i$. -Algebraist: Show that parts of it can be put into the refrigerator. Then show that the refrigerator is closed under addition. -Topologist: The elephant is compact, so it can be put into a finite collection of refrigerators. That’s usually good enough. -Linear Algebraist: Let $F$ mean "put inside fridge". Since $F$ is linear — $F(x+y)=F(x)+F(y)$ — just put 10% of the elephant in, showing that $F\left(\frac{1}{10}\mathit{elephant}\right)$ exists. Then, by linearity, $F(\mathit{elephant})$ does too. -Affine Geometer: There exists an affine transformation $F:\mathbb{R}^3\to\mathbb{R}^3:\vec{p}\mapsto A\vec{p}+\vec{q}$ that will allow the elephant to be put into the refrigerator. Just make sure $\det A\neq 0$ so you can take the elephant back out, and $\det A \gt 0$ so you don't end up with a bloody mess. -Geometer: Create an axiomatic system in which "an elephant can be placed in a refrigerator" is an axiom. -Complex Analyst: Put the refrigerator at the origin and the elephant outside the unit circle. Then get the image under inversion. -Fourier Analyst: Will $\mathcal{F}^{-1}[\mathcal{F}(\mathit{elephant})\cdot\mathcal{F}(\mathit{fridge})]$ do? -Numerical Analyst: Eh, $\mathit{elephant}=\mathit{trunk}+\varepsilon$, and $$\mathrm{fridge}(\mathit{elephant})=\mathrm{fridge}(\mathit{trunk}+\varepsilon)=\mathrm{fridge}(\mathit{trunk})+O(\varepsilon),$$ so just put the trunk in for a good approximation. -Probabilist: Keep trying to push it in in random ways and eventually it will fit. -Combinatorist: Discretize the elephant, partition it, and find a suitable rearrangement. -Statistician: Put its tail in the refrigerator as a sample, and say, "done!" -Logician: I know it's possible, I just can't do it. -Category Theorist: Isn't this just a special case of Yoneda's lemma? -Theoretical Computer Scientist: I can't decide. -Experimental Mathematician: I think it'd be much more interesting to get the refrigerator inside the elephant. -Set Theorist: Force it. - -REPLY [15 votes]: Cosgrove's writings in the Mathematical Intelligencer about 20 or 30(?) years ago had lots of puns, many of which would be understood only by mathematicians. E.g. someone was even worse than an unprincipled infiltrator: he was a non-principal ultrafilter. The biographies of Victoria Cross (famous for Cross products and Cross-ratios, and also particular kinds of word puzzles and a certain style of country running), Montmorency Royce Sebastian Carlow (whose "methods" you've heard of), and Karl-Heinz Normal (Normal subgroups, the Normal distribution,....) were of that sort.<|endoftext|> -TITLE: Continuity in terms of lines -QUESTION [15 upvotes]: Let $f: \mathbb R^n \rightarrow \mathbb R^n$, where $n> 1$ be a bijective map such that the image of every line is a line. -Is $f$ continuous? -I think it is, but the proof isn't immediately obvious to me. -Related to this question on math.SE. -Feel free to retag. - -REPLY [22 votes]: This is called the fundamental theorem of affine geometry. Let $f : E \to E'$ be a map between affine spaces over a field $K$. Suppose that - -$f$ is bijective; -$\dim E=\dim E'\ge 2$; -If $a, b, c\in E$ are aligned, then so are $f(a), f(b), f(c)$. - -Then $f$ is semi-affine: fix some $a_0\in E$, then there exists a field automorphism $\sigma$ of $K$ such that the map $h: v\mapsto f(a_0+v)-f(a_0)$ (which goes from the vector space attached to ${E}$ to that attached to $E'$) is additive and $h(\lambda v)=\sigma(\lambda)h(v)$ for all $v$ and all $\lambda \in K$. I don't have an URL for this theorem, I find it in Jean Fresnel: Méthodes Modernes en Géométrie, Exercise 3.5.7. But I think it is in any standard textbook on affine geometry. -When $K=\mathbb R$, it is known that $K$ has no non-trivial field automorphism. So your $f$ is an affine function, hence continuous. If $K=\mathbb C$, as pointed out by Kevin in above comments, take any non-trivial automorphism of $\mathbb C$, then you get a semi-affine map $\mathbb C^n \to \mathbb C^n$ which will not be affine, even not continuous (if $\sigma$ is not the conjugation).<|endoftext|> -TITLE: How to generate all finite groups of order n? -QUESTION [9 upvotes]: I know how to generate all Abelian groups of order n, but how would I generate the others? I can't seem to find anything about this. -By "generate", I mean produce the Cayley tables for all groups of order n. - -REPLY [18 votes]: See -Hans~Ulrich Besche, Bettina Eick, and E.A. O'Brien. -A millennium project: constructing small groups. -Internat. J. Algebra Comput., 12:623-644, 2002. -for a description of the construction of groups of order up to 2000. (I believe they narrowly failed to achieve this before the end of the year 2000.) In fact they did not construct the groups of order 1024 individually, but it is known that there are $49\,487\,365\,422$ groups of that order. The remaining $423\,164\,062$ groups of order up to 2000 (of which $408\,641\,062$ have order 1536) are available as libraries in GAP and Magma. -I would guess that 2048 is the smallest number such that the exact number of groups of that order is unknown. It is known that, for $p$ prime, the number of groups of order $p^n$ grows as $p^{\frac{2}{27}n^3+O(n^{8/3})}$: see http://en.wikipedia.org/wiki/P-group.<|endoftext|> -TITLE: A free subgroup of GL(2,Z)? -QUESTION [7 upvotes]: Is the subgroup of $GL(2,\mathbb Z)$ generated by the matrices -$$ \left( \begin{array}{cc} -1 & 1 \\\ -1 & 0 \end{array} \right) \ \ \text{and} \ \ -\left( \begin{array}{cc} -2 & 1 \\\ -1 & 0 \end{array} \right) -$$ -free of exponential growth? More generally, how does one find all the relations between two matrices? -I am sure this is well known, so any relevant references will be appreciated. -My motivation comes from dynamical systems where these matrices specify two automorphisms of the 2-torus; I am interested in studying the orbits of their joint action. - -REPLY [13 votes]: I interpret the question "how does one find all the relations between the matrices" as "find a set of defining relations for the group generated by the two matrices". -To do that we need a presentation of $GL(2,\mathbb{Z})$. I found one in the paper: -T. Brady, Automatic structures on Aut$F_2)$, Arch. Math. 63, 97-102 (1994). -$\langle p,s,u \mid p^2=s^2=(sp)^4=(upsp)^2=(ups)^3=(us)^2=1 \rangle$ -where $p= \left(\begin{array}{cc}0&1\\1&0\end{array}\right)$, $s= \left(\begin{array}{cc}-1&0\\0&1\end{array}\right)$, $u= \left(\begin{array}{cc}1&1\\0&1\end{array}\right)$. -Denoting your two matrices by -$a= \left(\begin{array}{cc}1&1\\1&0\end{array}\right)$, -$b= \left(\begin{array}{cc}2&1\\1&0\end{array}\right)$ -we have $a=up$, $b=u^2p$. -Putting $H = \langle a,b \rangle$ and using coset enumeration in Magma, it turns out that $H={\rm GL}(2,\mathbb{Z})$. So your two matrices actually generate all of GL$(2,\mathbb{Z})$. -In fact, denoting $a^{-1}$ and $b^{-1}$ by $A$ and $B$, we have -$p=aBa$, $u=a^2Ba$, $s=abaBAbabA$. -Using the modified coset enumeration algorithm, we can compute a presentation of $H$, which came out as the not particularly enlightening -$H = \langle a,b \mid (aBa)^2, (AbaBA)^2, aBabABAbAbABAbAbABAbAbAB, (abABabaB)^3 \rangle$.<|endoftext|> -TITLE: What does the partially ordered class of cardinals look like in L(R)? -QUESTION [6 upvotes]: Assuming the existence of enough large cardinals (I'm not sure whether I mean in the original V or in L(R), do whatever is standard), is the partially ordered class of cardinals order-isomorphic to something simpler? - -If so, what is the weakest large cardinal assumption that gives this result? -Is L(R) ≠ L sufficient? -My best guess would be that it is order-isomorphic to $\omega \cup (\operatorname{Ord} \times \operatorname{Ord})$, with all elements of $\omega$ being smaller than all elements of $\operatorname{Ord} \times \operatorname{Ord}$, and $\operatorname{Ord} \times \operatorname{Ord}$ having the product partial order. - -REPLY [7 votes]: Ricky: -I assume you mean to ask your question in $L({\mathbb R})$. -In general, without choice, the ordering of cardinals tends to be rather pathological, although we do not yet know by how much. Here is an example: It is open whether ZF proves that, if there is no infinite set all of whose members are pairwise incomparable (cardinality-wise) then choice holds. -On the other hand, if choice fails, then for every finite $n$ there are $n$ pairwise incomparable sets. -Assuming enough large cardinals (in $V$ or, equivalently, determinacy in $L({\mathbb R})$), the ordering of cardinals in $L({\mathbb R})$ is much more complicated than you suggest. -To give you an idea of how little we know: We do not know yet whether there are infinitely many successors of $|{\mathbb R}|$. -For an example of the immense complexity present in the ordering, recall that $E_0$ is the equivalence relation on $2^{\mathbb N}$ given by $x E_0 y$ iff $\exists n\forall m\ge n (x(m)=y(m))$. Then: - -$|2^{\mathbb N}/E_0|$ is a successor of $|{\mathbb R}|$, and above it but still below $|{\mathcal P}({\mathbb R})|$, you can embed the partial order of Borel subsets of ${\mathbb R}$ under containment. - -In fact, you can realize these cardinals by taking suitable quotients of the reals by Borel equivalence relations. All these relations can be taken to be countable (i.e., each class has countably many members) and their definitions trace back to free measure preserving actions of $F_2$ (the free group in two generators) on Polish spaces. -I take this as a strong indication that there is no sense in which we may have a "reasonable" description of the whole partial ordering. But really, we are far from being able to say much. Ketchersid and I have some recent results at the very bottom of the ordering, see "A trichotomy theorem in natural models of $AD^+$," to appear in the Proceedings of Boise Extravaganza in Set Theory, and also our forthcoming paper on "$G_0$-dichotomies in natural models of $AD^+$." You may also want to take a look at a paper by Woodin that deals with a slightly more demanding version of determinacy, "The cardinals below $|[\omega_1]^{\lt\omega_1}|$," Annals of Pure and Applied Logic 140 (2006) 161–232. -Even if we restrict ourselves to better understood classes (specific collections of Borel sets), the ordering is rather elaborate and far from well understood. For a nice subclass for which there is a very decent picture, see Alex Andretta, Greg Hjorth, and Itay Neeman, "Effective cardinals of boldface pointclasses," J. of Mathematical Logic, vol. 7 (2007) 35–82. - -I didn't address this above, but I should probably say something: -The assumption that there are enough large cardinals ensures that the theory of $L({\mathbb R})$ is invariant under forcing, so it is in a sense as canonical as we can hope. -Under weaker assumptions, the partial ordering may vary greatly. For example, $L({\mathbb R})\ne L$ does not suffice to preclude choice in $L({\mathbb R})$. -Assuming you do not have choice, it is open (even for $L({\mathbb R})$) whether well-foundedness of the partial ordering of cardinalities must fail. This is believed to be the case, and it is certainly so in all reasonable cases I have checked. It is easy to give examples by forcing over $L$ where the $L({\mathbb R})$ of one extension and the $L({\mathbb R})$ of another have non-equivalent partial orderings of cardinalities. (For example, by replicating or excluding the behavior mentioned above of quotients by free actions.) -Sometimes we have some form of "control", for example, if $L({\mathbb R})$ is a kind of Solovay model. But it already takes effort to show that in "nice" situations there are no infinite Dedekind finite subsets of ${\mathbb R}$ in $L({\mathbb R})$. In my view, the "right" version of these questions is under large cardinals, so we have canonicity, but already without it there are many difficulties and possibilities that may be interesting to explore.<|endoftext|> -TITLE: Is the "Napkin conjecture" open? (origami) -QUESTION [52 upvotes]: The falsity of the following conjecture would be a nice counter-intuitive fact. -Given a square sheet of perimeter $P$, when folding it along origami moves, you end up with some polygonal flat figure with perimeter $P'$. -Napkin conjecture: You always have $P' \leq P$. -In other words, you cannot increase the perimeter using any finite sequence of origami folds. -Question 1: Intuition tells us it is true (how in hell can it increase?). Yet, I think I read somewhere that there was some weird folding (perhaps called "mountain urchin"?) which strictly increases the perimeter. Is this true? -Note 1: I am not even sure that the initial sheet's squareness is required. -I cannot find any reference on the internet. Maybe the name has changed; I heard about this 20 years ago. -The second question is about generalizing the conjecture. -Question 2: With the idea of generalizing the conjecture to continuous folds or bends (using some average shadow as a perimeter), I stumble on how you can mathematically define bending a sheet. Alternatively, how do you say "a sheet is untearable" in mathematical terms? -Note 2: It might also be a matter of physics about how much we idealize bending mathematically. - -REPLY [26 votes]: In addition to the answers above, here are some remarks from my paper in Russian; part of it used in the last lecture here. -(Sorry for self-advertisement.) -1. An other solution. -It is based on idea of Yashenko [Math. Intelligencer 20 (1998)]. -This way you can increase the perimeter just a bit, -but it is done by repeating one fold (which is very simple but not "simple" in the sense below). - -2. It is still not known if you can increase the perimeter by a sequence of natural folds; -i.e., folds like this: - -I just learned that this problem also appears in Pak's book, Problem 40.16b; -it is marked by [$*$] which means that the problem is open.<|endoftext|> -TITLE: Which properties of ultrafilters on countable sets hold for filters in general? -QUESTION [6 upvotes]: Background/motivation: I'm investigating the construction of models for a first-order modal system (S5) as products of classical models. Since ultraproducts are all classical models and I need non-classical ones as well, I need to look at reduced products where the filter is not an ultrafilter. This leads me to ask about filters in general: -J.L. Bell & A.B. Slomson, in Models and Ultraproducts (p. 116), state and prove: - -Lemma 1.17. Let I be a countable set. - Then the collections of non-principal, - $\omega$-incomplete, uniform, and - regular ultrafilters on I all - coincide. - -Suppose I alter their definitions slightly so the above properties are all defined for filters in general, then modify the lemma to assert that it holds for filters in general. Would that be true? Can anyone supply a reference to a proof or disproof? Thanks. - -REPLY [3 votes]: If you use literally the definitions in Bell and Slomson, only changing "ultrafilter" to "filter," and if, as in the lemma you cited, you're interested only in flters on a countable set, then I believe non-principal is equivalent to $\omega$-incomplete, while "regular" is strictly stronger and "uniform" is strictly weaker. Unfortunately, I don't have time right now to check this carefully, so I hope someone will object loudly if I've messed it up. -Now that I have a bit more time, let me add the counterexamples that justify "strictly". Partition $\omega$ into two infinite pieces $A$ and $B$. The principal filter $F_0$ generated by $A$ is uniform; that establishes the second "strictly" above. For the first, let $U$ be a nonprincipal ultrafilter that contains $B$, and let $F_1=U\cap F_0$. Then $F_1$ is nonprincipal (the intersection of all the sets in it is $A$, which isn't in it), but it is not regular. (A function $f$ as in Bell and Slomson's definition of "regular" on page 114 would have to send each $a\in A$ to a finite set $f(a)$ that contains all elements $j$ of $\omega$, a contradiction.)<|endoftext|> -TITLE: Intersections of irreducible components -QUESTION [5 upvotes]: Let $V$ be an algebraic variety (not irreducible) over $\mathbb{C}$, defined by an ideal $I = \{f_1,f_2,\dots, f_n\}$. $V$ is not necessarily pure dimensional. Suppose $V = R_1\cup R_2\cup\dots\cup R_k$ is a decomposition of $V$ into irreducible components. How can we characterize the set of points on $V$ that lies in at least two components? If this is hard to compute, is there a good approximation to this set (some bigger set that contains it)? -A second question related to the one above is: how can we find the equations that describe the singular loci (the difficulty lies in the fact that we don't know the dimension at the point we are interested in)? This loci will give an approximation to the set discussed above. - -REPLY [8 votes]: With regards to the first question, if the $R_i$ are all normal, then this is just the vanishing locus of the conductor ideal. This is not difficult to compute with something like Macaulay (via the method below). Without the normal hypothesis, the conductor gives a good approximation (which is better than the singular locus approximation you suggest in your question). -But the general question is easy too (and also easy to compute with Macaulay), consider the map -$$R_1 \coprod R_2 \coprod \dots \coprod R_k \to V$$ -corresponding to the inclusion -$$f : O_V \subseteq O_{R_1} \oplus O_{R_2} \oplus \dots \oplus O_{R_k}.$$ -Let $M$ be the cokernel of $f$ and simply compute $\text{Ann}_{\mathcal{O}_V} M$. The vanishing locus of this ideal will be exactly the points contained in at least two components. -EDIT: With regards to the singular locus question, compute the singular locus of each component individually and then union that set with the set computed above (which is also necessarily part of the singular locus).<|endoftext|> -TITLE: How to prove the connected sum of two closed aspherical n-manfolds (n >2) is not asperical? -QUESTION [9 upvotes]: The intuitive idea is that the sphere connected the two manifolds is not contractible, which implies the (n-1)th homotopy group is not zero. Another argument, which I am not totally understand, uses the fact that the universal covering of an aspherical manifold has only one end. I am wondering if someone here could clarify these for me or give me a new argument. - -REPLY [4 votes]: Darryl McCullough gives a very complete answer to your question in his "Connected Sums of Aspherical Manifolds" paper. Let $M$ be the connected sum of $g$ aspherical manifolds of dimension $d\geq 3$. Its universal cover $\tilde M$ is homotopy equivalent to $\bigvee_{\rho \in \pi_1(M)} \bigvee_{i=1}^{g-1}S^{d-1}$ (and in particular is not contractible). -I wonder if there is a short proof of this along the lines of Robert Bell's answer.<|endoftext|> -TITLE: Natural examples of sequences of adjoint functors -QUESTION [38 upvotes]: I am looking for examples of sequences of adjoint functors. That are (possibly bounded) sequences -$$(...,F_{-1}, F_{0}, F_1, F_2,...)$$ -such that each $F_n$ is left adjoint to $F_{n+1}$. We call such a sequence cyclic of order $k$ if for one $n$ (and hence for all) we have $F_{n} \cong F_{n+k}$. -It is relatively easy to prove that cyclic sequences of all orders and non-cyclic sequences of all possible length exist. This can e.g. be done using posets, see http://www.springerlink.com/content/pmj5074147116273/. -I am looing for more "natural" examples of such sequences that are as long as possible. By natural I mean that they grow out of "usual functors" (sorry for this vague statement...) -Let my give two short examples: -1) Let $U: Top \to Set$ be the forgetful functor from locally connected topological spaces to sets. This induces a sequence of length 4: -$$ (\pi_0 , Dis , U , CoDis) $$ -where $Dis$ and $CoDis$ are the functor that equip a set with the discrete and indiscrete topology. Then the sequence stops. Tons of examples of this type are induced by pullback functors in algebraic geometry. -2) a cyclic sequence of order 2: the Diagaonal functor $\Delta: A \to A \times A$ for any abelian category $A$ is left and right adjoint to the direct sum -$$ ( ...,\Delta,\oplus,\Delta,\oplus,...)$$ - -REPLY [7 votes]: There are some really nice answers here. Here's another contribution. -Let [n] denote the totally ordered (n+1)-element set, regarded as a category. For each positive integer n, we have the usual n+1 order-preserving injections from [n-1] to [n], and the usual n order-preserving surjections from [n] to [n-1]. (I mean the ones used all the time for simplicial anything.) When you regard them as functors, these injections and surjections interleave to form an adjoint chain of length 2n.<|endoftext|> -TITLE: Are there "real" vs. "quaternionic" conjugacy classes in finite groups? -QUESTION [38 upvotes]: The complex irreps of a finite group come in three types: self-dual by a -symmetric form, self-dual by a symplectic form, and not self-dual at all. -In the first two cases, the character is real-valued, and in the third -it is sometimes only complex-valued. The cases can be distinguished by -the value of the Schur indicator $\frac{1}{|G|} \sum_g \chi(g^2)$, -necessarily $1$, $-1$, or $0$. They correspond to the cases that -the representation is the complexification of a real one, the forgetful -version of a quaternionic representation, or neither. -A conjugacy class $[g]$ is called "real" if all characters take real -values on it, or equivalently, if $g\sim g^{-1}$. I vaguely recall the -number of real conjugacy classes being equal to the number of real irreps. - -Do I remember that correctly? -Can one split the real conjugacy classes into two types, -"symmetric" vs. "symplectic"? - -With #1 now granted, a criterion for a "good answer" would be that the number of symmetric real conjugacy classes should equal the number of symmetrically self-dual irreps. -(I don't have any application in mind; it's just bothered me off and on -for a long time.) - -REPLY [22 votes]: It's a great question! Disappointingly, I think the answer to (2) is No : -The only restriction on a `good' division into "symmetric" vs. "symplectic" conjugacy classes that I can see is that it should be intrinsic, depending only on $G$ and the class up to isomorphism. (You don't just want to split the self-dual classes randomly, right?) This means that the division must be preserved by all outer automorphisms of $G$, and this is what I'll use to construct a counterexample. Let me know if I got this wrong. -The group -My $G$ is $C_{11}\rtimes (C_4\times C_2\times C_2)$, with $C_2\times C_2\times C_2$ acting trivially on $C_{11}=\langle x\rangle$, and the generator of $C_4$ acting by $x\mapsto x^{-1}$. In Magma, this is G:=SmallGroup(176,35), and it has a huge group of outer automorphisms $C_5\times((C_2\times C_2\times C_2)\rtimes S_4)$, Magma's OuterFPGroup(AutomorphismGroup(G)). The reason for $C_5$ is that $x$ is only conjugate to $x,x^{-1}$ in $C_{11}\triangleleft G$, but there there are 5 pairs of possible generators like that in $C_{11}$, indistinguishable from each other; the other factor of $Out\ G$ is $Aut(C_2\times C_2\times C_4)$, all of these guys commute with the action. -The representations -The group has 28 orthogonal, 20 symplectic and 8 non-self-dual representations, according to Magma. -The conjugacy classes -There are 1+7+8+5+35=56 conjugacy classes, of elements of order 1,2,4,11,22 respectively. The elements of order 4 are (clearly) not conjugate to their inverses, so these 8 classes account for the 8 non-self-dual representations. We are interested in splitting the other 48 classes into two groups, 28 'orthogonal' and 20 'symplectic'. -The catch -The problem is that the way $Out\ G$ acts on the 35 classes of elements of order 22, it has two orbits according to Magma - one with 30 classes and one with 5. (I think I can see that these numbers must be multiples of 5 without Magma's help, but I don't see the full splitting at the moment; I can insert the Magma code if you guys want it.) Anyway, if I am correct, these 30 classes are indistinguishable from one another, so they must all be either 'orthogonal' or 'symplectic'. So a canonical splitting into 28 and 20 cannot exist. - -Edit: However, as Jack Schmidt points out (see comment below), it is possible to predict the number of symplectic representations for this group! - -REPLY [14 votes]: A standard strengthening of "real element" is "strongly real element". An element is strongly real if it is conjugate to its inverse by an involution, or equivalently, if it is a product of involutions (equivalently, it sits nicely inside a dihedral group). -The quaternion group of order 8 shows that this strengthening is distinct: the only strongly real element is the identity, but Q8 has 4 representations with Frobenius–Schur indicator +1. -However, in: - -Gow, R. "Real-valued and 2-rational group characters." - J. Algebra 61 (1979), no. 2, 388–413 - MR2222410 - DOI:10.1016/0021-8693(79)90288-6 - -some inequalities relating the two ideas are given as well as a few reasonably strong results. If the Sylow 2-subgroups are dihedral or large enough semi-dihedral, then a quaternionic representation exists iff a non-strongly real but real element of odd order exists (theorems are spread out in the paper). In the case of a 2-nilpotent group (the last section) more precise relationships are given, with the number of quaternionic reps being a mixture of the number of strongly and weakly real elements. -Beware of wanting to generalize the real case too broadly. An F-rational character is a C-irreducible character whose values are in F. An F-rational element is an element conjugate to the correct powers given the (cyclotomic) Galois group of C/F. In p-groups for odd p, the F-rational classes are 1–1 with F-rational characters, but not in general for 2-groups or groups of odd order.<|endoftext|> -TITLE: Zariski-closed subsemigroups of SL_n(C) are groups -QUESTION [6 upvotes]: I would like to show that any Zariski-closed subsemigroup of $SL_n(\mathbb{C})$ is a group. If I understand correctly, this is consequence 1.2.A of http://www.heldermann-verlag.de/jlt/jlt03/BOSLAT.PDF . -Is there a more elementary proof? For $SL_2(\mathbb{C})$, the result is quite easy to show directly, or using the Hilbert basis theorem, . - -REPLY [23 votes]: It is quite elementary. Let $S$ be the semi-group in question. Then for any $g \in S$, the set -$g^kS$ for $k=1,2, \dots$ is a decreasing sequence of closed sets, hence it has to stabilize. So, $g^kS=g^{k+1}S$ implies that $gS=S$. Hence $S$ is closed with respect to taking inverse, and therefore is a group. - -REPLY [3 votes]: I think it's a matter of basic linear algebra. -Generally, let $k$ be a field, and $L$ be some finite-dimensional $k$-algebra. (In our case, $k=\mathbb C$ and $L=\mathrm{M}_n\left(k\right)$.) If $A\in L$ is invertible, then $A^{-1}$ lies in the Zariski closure of the set $\left\lbrace 1,A,A^2,A^3,...\right\rbrace$. -Proof. Let $k\left[L\right]$ denote the algebra of all polynomial functions from $L$ to $k$ (where a "polynomial function" means a function that can be written as a polynomial in the coordinates). (If $k$ is infinite, this is isomorphic to the non-naive algebra of coordinate functions, i. e. the symmetric algebra $\mathrm{S}\left(L^{\ast}\right)$, but we don't care about this isomorphy and therefore we don't need $k$ to be infinite.) -Let $P\in k\left[L\right]$ be a polynomial such that $P\left(A^i\right)=0$ for every $i\in\mathbb N$. We must then prove that $P\left(A^{-1}\right)=0$ as well. -Define a $k$-algebra $U$ by $U=\bigoplus\limits_{i=0}^N L^{\otimes i}$ as a vector space, but with the multiplication being inherited from $L$ on each summand. So, as a vector space $U$ is a "cropped" tensor algebra over $L$, but as an algebra it is a direct product! -Let $N=\deg P$. Then the polynomial $P:L\to k$ can be written as $P=p\circ s$, where $U=\bigoplus\limits_{i=0}^N L^{\otimes i}$, where $s:L\to U$ is the canonical map given by -$s\left(B\right)=1\oplus B\oplus \left(B\otimes B\right)\oplus \left(B\otimes B\otimes B\right)\oplus ...\oplus B^{\otimes N}$, -and $p:U\to k$ is some $k$-linear map. (In fact, this follows from the properties of the tensor algebra, because here we are NOT using the algebra structure on our $U$, but we are only using the vector space structure on $U$, and as I said, as a vector space $U$ is just the tensor algebra of $L$ "cropped" at $N$, which is enough for linearlizing polynomial maps of degree $\leq N$.) -Now consider the element $s\left(A\right)\in U$. This element $s\left(A\right)$ is invertible (since $A$ is invertible, so that $A^{\otimes i}$ is invertible for every $i$, and since the multiplication on $U=\bigoplus\limits_{i=0}^N L^{\otimes i}$ is componentwise), and the algebra $U$ is finite-dimensional (although its dimension is usually quite large). Thus, $s\left(A\right)^{-1}$ lies in the $k$-linear span of the set $\left\lbrace 1,s\left(A\right),\left(s\left(A\right)\right)^2,\left(s\left(A\right)\right)^3,...\right\rbrace$ (because if $u$ is an invertible element of some finite-dimensional $k$-algebra, then $u^{-1}$ lies in the $k$-linear span of the set $\left\lbrace 1,u,u^2,u^3,...\right\rbrace$; this is easily proven using the fact that any element of a finite-dimensional $k$-algebra is algebraic over $k$). Since $s$ is a multiplicative map, we have $\left(s\left(A\right)\right)^i=s\left(A^i\right)$ for all $i$, so that this becomes: The element $s\left(A^{-1}\right)$ lies in the $k$-linear span of the set $\left\lbrace s\left(1\right),s\left(A\right),s\left(A^2\right),s\left(A^3\right),...\right\rbrace$. Since $p$ is a linear map, we can apply $p$ here and obtain: The element $p\left(s\left(A^{-1}\right)\right)$ lies in the $k$-linear span of the set $\left\lbrace p\left(s\left(1\right)\right),p\left(s\left(A\right)\right),p\left(s\left(A^2\right)\right),p\left(s\left(A^3\right)\right),...\right\rbrace$. Now $p\circ s=P$, so this becomes: The element $P\left(A^{-1}\right)$ lies in the $k$-linear span of the set $\left\lbrace P\left(1\right),P\left(A\right),P\left(A^2\right),P\left(A^3\right),...\right\rbrace$. So when $P\left(A^i\right)=0$ for all $i\in\mathbb N$, then $P\left(A^{-1}\right)=0$, qed.<|endoftext|> -TITLE: What is the second fundamental form of moduli space? -QUESTION [13 upvotes]: Away from the hyperelliptic locus, the moduli of curves immerses -in the moduli of principally polarized abelian varieties. The -ambient space has a riemannian metric, so one can ask about the -second fundamental form, the first-order deviation of the -submanifold from being totally geodesic. What is this second -fundamental form? Is anything known about it? -I think one could translate this into the language of algebraic -geometry by using the Serre-Tate formal coordinates, which exist -at each point of $A_g$. With respect to these coordinates, $M_g$ -is not linear; what is its quadratic approximation? One could -interpret this as a version of the Schottky problem, which -suggests that existing solutions to it might be applicable. - -REPLY [3 votes]: The following papers might be useful: -$(1)$ E. Colombo- G. Pirola- A. Tortora -"Hodge-Gaussian maps" -Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4) 30 (2001), no. 1, 125–146. -$(2)$ E. Colombo - P. Frediani -"Siegel metric and curvature of the moduli space of curves" -Trans. Amer. Math. Soc. 362 (2010), no. 3, 1231–1246.<|endoftext|> -TITLE: Automorphic form encoding the orders of $N$ modulo $p$. -QUESTION [5 upvotes]: Let $N$ be a nonzero rational number. For every prime number $p$ with $v_p(N)=0$, let $a_p$ denote the index in $\mathbb Z/p\mathbb Z$ of the subgroup generated by $N$ modulo $p$. So we have $a_p=1$ if and only if $N$ is a primitive root mod $p$. - -Is there an automorphic form which encodes the numbers $a_p$? - -Let me explain: To give $N$ is the same thing as to give a group homomorphism $\mathbb Z \to \mathbb G_m$ over $\mathrm{spec}\mathbb Q$. Such a morphism is also the same as an extension -$$0\to \mathbb Q(1) \to M \to \mathbb Q \to 0$$ -of motives over $\mathrm{spec}\mathbb Q$. The motive $M$ is an example of a "mixed Tate motive", and also an example of a 1--motive. Its weights are $0$ and $-2$. The condition $v_p(N)=0$ means then that $M$ has good reduction at the prime $p$, i.e. extends to a 1--motive over the local ring at $p$. The number $a_p$ is then given by -$$a_p = H^0(\mathbb F_p,M)$$ -where we interpret $M = [\mathbb Z \to \mathbb G_m]$ as a complex concentrated in degrees $-1$ and $0$. With $M$ are associated "realisations", in particular $M$ comes with integral $\ell$--adic representations. So to restate the question: - -Is there an automorphic form corresponding to $M$? - -We could look at the L-function associated with the $\ell$--adic representation of $M$ to get a hint. The problem with this is that the L-function does not see the extension structure, it only depends on the semisimplification of the representation. This is clear because it is constructed by taking traces of Frobenius elements -- the L-function is in fact $\zeta(s)\zeta(s-1)$, independently of $N$. -I don't know if morally the association Motives $\to$ Automorphic forms which is classically conjectured for pure motives should extend to mixed motives... also, I don't know what a mixed automorphic form is. Maybe these are simply automorphic forms for nonreductive groups? There is a theory of "mixed modular forms" by Min Ho Lee, but I don't think this is what I am looking for. - -REPLY [6 votes]: $\newcommand{\F}{\mathbf F}\newcommand{\Q}{\mathbf Q}$ -I want to say two things about this question. First is that I don't understand some of it: I don't know what you mean by "$a_p=H^0(\F_p,M)$" (number = group). Do you mean the number is the order of the group? If so, or indeed whatever you mean, I am a bit confused as to why this is the right idea. Consider for example an $\ell$-adic representation attached to a motive. Then $a_p$ is the trace of a Frobenius element at $p$ rather than the order of an $H^0$. So I'm missing something here. -But here is what I think the answer to a related question is. Toby Gee and I recently spent some time trying to fathom out as general a conjecture as we could, relating automorphic forms to Galois representations. The results of our labour are here: -http://arxiv.org/abs/1009.0785 -and the conjecture for $GL(2)$ boils down to something that was presumably in the literature long before us: we conjecture that to an algebraic automorphic representation on $GL(2)$ there should be an associated semi-simple Galois representation. In particular, we could not fathom out how to get automorphic gadgets corresponding to non-semisimple representations. I still vividly remember the train of thought, so let me explain it now. -Say we're looking for Galois representations of $Gal(\overline{\Q}/\Q)$ which are extensions of the trivial character by the cyclotomic character. Then on the automorphic side we are looking for automorphic representations on $GL(2)$ with certain Satake parameters which contradict the Weil bounds, so they can't be cuspidal. So we seek them in the Eisenstein series. And they exist! But here's the catch: they are parametrised by finite subsets of the places of $\Q$. Explicitly, if $S$ is a finite subset of the places, then construct $\pi$ as a tensor product of $\pi_v$ with $\pi_v=|\det|^{1/2}$ for $v$ not in $S$, and $\pi_v$ an appropriate twist of the Steinberg for $v\in S$ (the one with the same infinitesimal char as $|\det|^{1/2}$). These are what show up on the automorphic side. If $S$ is non-empty then $\pi$ does not show up discretely in the space of automorphic forms, and $\pi$ is not isobaric either (so in Clozel's work on Galois representations and automorphic forms he ignores these $\pi$s completely). -However, as you have observed, on the Galois side, more representations seem to show up. We couldn't fathom out how to match things up. So we decided to send every one of these automorphic reps to the direct sum of the trivial rep and the cyclo char. In other words, we had the same, or at least a very similar, problem to yours, and couldn't resolve it.<|endoftext|> -TITLE: Justifying the definition of arclength -QUESTION [14 upvotes]: Background -One of my friends told me the following story: A child must walk from his home at point A = (1,0) to his school at point B = (0,1). The laws in his country state that you can only walk parallel to the horizontal and vertical axis. No matter how he tries to get to school, he finds that he must walk at least 2 miles. He is very frustrated that he cannot walk diagonally. He doesn't want to get in trouble, so he puts up with this silly law until his senior year of high school. At this point he takes calculus, and learns about limits, and so he decides that each day he will walk a new "zig-zag" path around the line, in such a way that his sequence of paths approaches the line from A to B uniformly. On the 30th day he is pulled over by a policeman for walking diagonally. -The point of the story is to get you to think about the notion of the length of a curve. Here we have an instance of a sequence of polygonal paths which approach a curve uniformly, but the lengths of the polygonal approximations do not converge to the length of the curve. -Our usual definition of the arclength of a curve involves approximating it with polygonal secant segments. My question is why this is the "best" definition. Or to make this precise, how do we know that any other sequence of polygonal approximations to the curve will not have a shorter limit? EDIT: In the case of a straight line, this is clear, but my precise question below is about answering this for more general curves. -Precise Question: -Let $C:[0,1] \to \mathbb{R}^n$ be a rectifiable curve (or feel free to add as many smoothness requirements as you like), and let $P_n: [0,1] \to \mathbb{R}^n$ be a sequence of piecewise linear curves which converge uniformly to $C$. Is it true that limsup{Length($P_n$)} $\geq$ Arclength(C)? EDIT I goofed: I meant to ask whether or not liminf{Length($P_n$)} $\geq$ Arclength(C). -Hopefully this question is not too elementary; my analysis skills are almost pitifully weak. I strongly suspect that the answer to this precise question is "Yes", because otherwise I think that the usual definition of arc length is incorrect. I have enough faith in mathematics to believe that we have found the right definition, but I would still like to see a proof of this fact. - -REPLY [2 votes]: Here's a somewhat rigorous geometric argument (perhaps harking back to another of today's questions). Suppose that C is a continuous curve in $\mathbb{R}^n$ and that P is a polygonal path approximating C in that every point of C has a point of P within $\epsilon$ of it. I claim that it is approximately longer than some polygonal path of secants to C. -The idea is to modify P by performing rotations and translations (thus, isometries) on the segments to try to put them into secant position. Sometimes, we need to cut to finish the job, but (almost) never stretch. Using the parameterization of S, I will freely refer to the "beginning" and "end" of a segment on P without becoming confusing. -First, consider the first segment S of P, and denote by x the beginning of C. The beginning of S is within $\epsilon$ of some point of S by hypothesis, so for simplicity, let's assume it's within $\epsilon$ of x. (This assumption is harmless since we would otherwise have to add only up to the length of S, which goes to zero in the limit anyway.) Then by the triangle inequality we can move it to x without adding more than length $\epsilon$ to P. Assume that we have done so. -Now, assume inductively that we have a segment S of P and its beginning lies on C. We rotate S around this beginning until its end is also on C (this part requires continuity of C, if you are keeping track). In doing so, we keep the next segment attached to its end, allowing it to translate and rotate as necessary to keep its end within $\epsilon$ of C. -Suppose this latter goal is impossible: one cannot bring the end of the next segment T to within $\epsilon$ of C after transforming S. Then we can put some interior point y of T on C, and form a polygonal path P' of secants to C from x to y. Everything on P after y is extra length above that of P' (to within $\epsilon$, from the first step). -Conversely, if we can do this, then by adding at most another $\epsilon$ to the length of the last segment of P we can put its end at the end y of C and get P' again, which is at most $2\epsilon$ longer than P. -Either way, the liminf (as $\epsilon \to 0$) of lengths of all possible paths P is greater than or equal to that of paths P', as desired.<|endoftext|> -TITLE: Right-continuity of natural filtrations -QUESTION [9 upvotes]: My question: Is the natural filtration of a right-continuous process also right-continuous? -I would say yes, but don't know where to start proving it. -Thanks for your help/ideas! - -REPLY [9 votes]: Right continuity fails even for canonical continuous processes. -The natural filtration on $C([0,\infty))$ is not right continuous. -For example, the event $\{\omega: {d^+\over dt}\ \omega_t\mbox{ exists at }t=0\}$ belongs -to ${\cal F}_{0+}$ but not ${\cal F}_0$. In words, you can tell whether - the function $\omega_t$ has a right derivative at $t=0$ with an infinitesimal peek beyond time 0, -but you cannot tell just from the value of the function $\omega_t$ at time 0. -Right continuous filtrations are nicer to work with, and since it fails for the natural filtration, -we often use the right continuous version instead. Fortunately, many of the nice -properties of right continuous processes carry over even with this enlarged filtration. For example, -Brownian motion is still Markov with respect to ${\cal F}_{t+}$ which leads to interesting results like Blumenthal's zero-one law.<|endoftext|> -TITLE: What do singular, atomless invariant measures of $\times d$ look like? -QUESTION [7 upvotes]: Consider the circle map $\times d:x\mapsto dx \mod 1$. The lebesgue measure is the only absolutely continuous invariant probability measure, but this map has many other invariant measures. Of course, one can take barycentric combinations of invariant measures to get a new one, so let us restrict to the extremal points, namely ergodic measures. -One can consider a uniform measure on any periodic orbit. There are also singular, atomless invariant measures. For example, the "uniform" measure on the usual middle-third Cantor set is invariant under $\times 3$. All this is pretty explicit and I'm fine with it. But I also heard about a thermodynamical formalism that yield many invariant measures; Bowen's lecture note are on my desk but it does not seem to answer my question, which is the following: what do these measures look like? What is their support? I guess that we cannot answer this for all invariant measures, but maybe for some of them less trivial than the atomic and easy Cantor ones. -Other question: Cantor measures are easily constructed for all $\times d$, $d>2$, but I cannot really get one for $d=2$. Am I clumsy or is there something special to this case? - -REPLY [3 votes]: To supplement the previous answers (and perhaps further illustrate that there are a lot of invariant measures!) I thought I'd mention what "typical" invariant measures for the $d$-fold expanding map look like. Fix a $d$-fold expanding map $T$, and consider the set $\mathcal{M}$ of all $T$-invariant Borel probability measures on the circle. We can make $\mathcal{M}$ into a compact, metrisable topological space using the weak topology, which is characterised by the fact that a sequence of measures $(\mu_n)$ converges to $\mu$ if and only if the real sequence $(\int f d\mu_n)$ converges to $\int fd\mu$ for every continuous function $f$ from the circle to the reals. We can then ask what typical elements of $\mathcal{M}$ look like in the sense of Baire category. -It turns out that the answer is this: there is a dense $G_\delta$ subset of $\mathcal{M}$ in which every measure is fully supported, weak-mixing for $T$, but not strong-mixing for $T$. All of these measures are non-atomic and pairwise mutually singular with respect to one another. So there are enormously many fully supported invariant measures - so many that it's not easy to say anything about them at all which carries with very much generality. -This result is basically due to K. R. Parthasarathy in the 1961 paper "On the category of ergodic measures". In that paper it's proved for the two-sided full shift on $d$ symbols (or even infinitely many symbols) but the same proofs go through with little change.<|endoftext|> -TITLE: Gelfand theory Problem -QUESTION [7 upvotes]: I have 2 problems in Gelfand theory. I shall be thankful for any -answers. -1)What is the gelfand spectrum of l^1(N)? -A few of -the elements are evaluations of functions(defined below) on closed -unit disc. -For an element a = {a(n)} of l^1, corresponding function is -summation a(n)x^n over natural numbers. -Are there any others? -2)Can we find a non unital Banach Algebra with a compact Gelfand -spectrum? -Sincerely, -Madhuresh. - -REPLY [6 votes]: The Gel'fand spectrum of $\ell^1 \mathbb N$ is indeed the closed unit disc. After all, every functional to $\mathbb C$, must be given by sending the generator to some complex number. It is easy to see, that this works if and only if this complex number lies in the unit disc. -Your second question is rather nice. Anyhow, I think that there cannot be any commutative example (at least if it embeds into the algebra of continuous functions on the Gel'fand spectrum). As soon as there is the unit in the algebra of functions on the Gel'fand spectrum, then the Banach algebra contains at least an invertible element, and hence also the unit. -However, and now it is getting more interesting: there are non-unital Banach algebras whose universal $C^\star$-algebra has a unit. (Note that this is precisely the non-commutative analogue of your question in Gel'fand theory.) -Indeed, consider a group $\Gamma$ with Kazhdan's property (T), its $\ell^1$-algebra and the augmentation ideal $\omega(\Gamma) \subset \ell^1 \Gamma$. It is well-known that $C^\star(\Gamma)$ splits of the unit as a direct summand. The remaining summand is the universal $C^*$-algebra of $\omega(\Gamma)$ and it is unital.<|endoftext|> -TITLE: Proofs of the uncountability of the reals -QUESTION [92 upvotes]: Recently, I learnt in my analysis class the proof of the uncountability of the reals via the Nested Interval Theorem (Wayback Machine). At first, I was excited to see a variant proof (as it did not use the diagonal argument explicitly). However, as time passed, I began to see that the proof was just the old one veiled under new terminology. So, till now I believe that any proof of the uncountability of the reals must use Cantor's diagonal argument. -Is my belief justified? -Thank you. - -REPLY [4 votes]: Today I attended a lecture, and this question was asked, after the talk, I realized the following proof, that might be interesting, at least for set theorists!!!: - -Theorem (Cantor): The set of real numbers is uncountable. -Proof: Suppose not. Let $M$ be a countable transitive model of enough of ZFC such that $\mathbb R\in M$. -By our assumption $\mathbb R\subseteq M$. -Let $\mathbb P=\operatorname{Add}(\omega,1)$ be the Cohen forcing for adding a new real, let $G\subseteq\mathbb P$ be generic over $M$ and let $r=\bigcup G$. -Then: - -$r\notin M$, as otherwise $G=\{r \restriction n: n<\omega \}\in M$ which is not possible -$r\in\mathbb R\subseteq M$ - -A contradiction.<|endoftext|> -TITLE: Countable connected Hausdorff space -QUESTION [27 upvotes]: Let me start by reminding two constructions of topological spaces with such exotic combination of properties: -1) The elements are non-zero integers; base of topology are (infinite) arithmetic progressions with coprime first term and difference. -2) Take $\mathbb{R}^{\infty}\setminus \{0\}$ with product-topology and factorize by the relation $x\sim y \Leftrightarrow x=ty$ for some $t>0$ (infinite-dimensional sphere). Then consider only points with rational coordinates, all but finitely of them vanishing. -The first question is whether are these two examples homeomorphic or somehow related. -The second is an historical one. I've heard that the first example of such space belongs to P. S. Urysohn. What was his example? - -REPLY [27 votes]: First let us fix the terminology. -The space (1) is known in General Topology as the Golomb space. More precisely, the Golomb space $\mathbb G$ is the set $\mathbb N$ of positive integers, endowed with the topology generated by the base consisting of arithmetic progressions $a+b\mathbb N_0$ where $a,b$ are relatively prime natural numbers and $\mathbb N_0=\{0\}\cup\mathbb N$. -Let us call the space (2) the rational projective space and denote it by -$\mathbb QP^\infty$. -Both spaces $\mathbb G$ and $\mathbb QP^\infty$ are countable, connected and Hausdorff but they are not -homeomorphic. A topological property distinguishing these spaces will -be called the oo-regularity. -Definition. A topological space $X$ is called oo-regular -if for any non-empty disjoint open sets $U,V\subset X$ the subspace $X\setminus(\bar U\cap\bar V)$ of $X$ is regular. -Theorem. - -The rational projective space $\mathbb QP^\infty$ is oo-regular. -The Golomb space $\mathbb G$ is not oo-regular. - -Proof. The statement 1 is relatively easy, so is left to the interested reader. -The proof of 2. In the Golomb space $\mathbb G$ consider two basic open sets $U=1+5\mathbb N_0$ and $V=2+5\mathbb N_0$. -It can be shown that $\bar U=U\cup 5\mathbb N$ and $\bar V=V\cup 5\mathbb N$, so $\bar U\cap\bar V=5\mathbb N$. -We claim that the subspace -$X=\mathbb N\setminus (\bar U\cap\bar V)=\mathbb N\setminus 5\mathbb N$ of the Golomb space -is not regular. -Consider the point $x=1$ and its neighborhood $O_x=(1+4\mathbb N)\cap X$ in $X$. - Assuming that $X$ is regular, we can find a neighborhood $U_x$ of $x$ in $X$ such that -$\bar U_x\cap X\subset O_x$. -We can assume that $U_x$ is of basic form -$U_x=1+2^i5^jb\mathbb N_0$ for some $i\ge 2$, $j\ge 1$ and $b\in\mathbb N\setminus(2\mathbb N_0\cup 5\mathbb N_0)$. -Since the numbers $4$, $5^j$, and $b$ are relatively prime, by the Chinese remainder Theorem, the intersection $(1+5^j\mathbb N_0)\cap (2+4\mathbb N_0)\cap b\mathbb N_0$ -contains some point $y$. It is clear that $y\in X\setminus O_x$. -We claim that $y$ belongs to the closure of $U_x$ in $X$. We need to check that each basic neighborhood $O_y:=y+c\mathbb N_0$ of $y$ intersects the set $U_x$. Replacing $c$ by $5^jc$, we can assume that $c$ is divisible by $5^j$ and hence $c=5^jc'$ for some $c'\in\mathbb N_0$. -Observe that $O_y\cap U_x=(y+c\mathbb N_0)\cap(1+4^i5^jb\mathbb N_0)\ne\emptyset$ if and only if $y-1\in 4^i5^jb\mathbb N_0-5^jc'\mathbb N_0=5^j(4^ib\mathbb N_0-c'\mathbb N_0)$. The choice of $y\in 1+5^j\mathbb N_0$ guarantees that $y-1=5^jy'$. Since $y\in 2\mathbb N_0\cap b\mathbb N_0$ and $c$ is relatively prime with $y$, the number $c'=c/5^j$ is relatively prime with $4^ib$. So, by the Euclidean Algorithm, there are numbers $u,v\in\mathbb N_0$ such that $y'=4^ibu-c'v$. Then $y-1=5^jy'=5^j(4^ibu-c'v)$ and hence $1+4^i5^ju=y+5^jc'v\in (1+4^i5^jb\mathbb N_0)\cap(y+c\mathbb N_0)=U_x\cap U_y\ne\emptyset$. So, $y\in\bar U_x\setminus O_x$, which contradicts the choice of $U_x$. - -Remark. Another well-known example of a countable connected space is the Bing space $\mathbb B$. This is the rational half-plane $\mathbb B=\{(x,y)\in\mathbb Q\times \mathbb Q:y\ge 0\}$ endowed with the topology generated by the base consisting of sets $$U_{\varepsilon}(a,b)= -\{(a,b)\}\cup\{(x,0)\in\mathbb B:|x-(a-\sqrt{2}b)|<\varepsilon\}\cup -\{(x,0)\in\mathbb B:|x-(a+\sqrt{2}b)|<\varepsilon\}$$ where $(a,b)\in\mathbb B$ and $\varepsilon>0$. -It is easy to see that the Bing space $\mathbb B$ is not oo-regular, so it is not homeomorphic to the rational projective space $\mathbb QP^\infty$. -Problem 1. Is the Bing space homeomorphic to the Golomb space? -Remark. It is clear that the Bing space has many homeomorphisms, distinct from the identity. -So, the answer to Problem 1 would be negative if the answer to the following problem is affirmative. -Problem 2. Is the Golomb space $\mathbb G$ topologically rigid? -Problem 3. Is the Bing space topologically homogeneous? -Since the last two problems are quite interesting I will ask them as separate questions on MathOverFlow. -Added in an edit. Problem 1 has negative solution. The Golomb space and the Bing space are not homeomorphic since -1) For any non-empty open sets $U_1,\dots,U_n$ in the Golomb space (or in the rational projective space) the intersection $\bigcap_{i=1}^n\bar U_i$ is not empty. -2) The Bing space contain three non-empty open sets $U_1,U_2,U_3$ such that $\bigcap_{i=1}^3\bar U_i$ is empty. -Added in a next edit. Problem 2 has the affirmative answer: the Golomb space $\mathbb G$ is topologically rigid. This implies that $\mathbb G$ is not homeomorphic to the Bing space or the rational projective space (which are topologically homogeneous). -Problem 3 has an affirmative solution: the Bing space is topologically homogeneous. -Added in Edit made 14.03.2020. The rational projective space $\mathbb Q P^\infty$ admits a nice topological characterization: - -Theorem. A topological space $X$ is homeomorphic to $\mathbb Q P^\infty$ - if and only if $X$ is countable, first countable, and admits a decreasing sequence of nonempty closed sets $(X_n)_{n\in\omega}$ such that $X_0=X$, $\bigcap_{n\in\omega}X_n=\emptyset$, and for every $n\in\omega$, - (i) the complement $X\setminus X_n$ is a regular topological space, and - (ii) for every nonempty open set $U\subseteq X_n$ the closure $\overline{U}$ contains some set $X_m$.<|endoftext|> -TITLE: Brownian bridge interpreted as Brownian motion on the circle -QUESTION [14 upvotes]: Is it reasonable to view the Brownian bridge as a kind of Brownian motion indexed by points on the circle? -The Brownian bridge has some strange connections with the Riemann zeta function (see Williams' article http://www.statslab.cam.ac.uk/~grg/books/hammfest/22-dw.ps, for example). -I'm looking for a heuristic explanation of why this might be the case. If one could interpret the Brownian bridge as described above, then the heuristic would be that Brownian motion is naturally associated with heat flow, which goes hand in hand with theta functions, which goes some way toward explaining the appearance of the zeta function. -I don't know where to begin reading about Brownian motion indexed by anything other than $\mathbb{R}^{+}$. The standard stochastic analysis texts don't really address the idea. -Many thanks. - -REPLY [4 votes]: If you want an Ornstein-Uhlenbeck process (i.e. Brownian motion confined by a quadratic potential), there is no difficulty because this has a stationary distribution: start from that definition at time $0$, condition to be back at the same point at time $1$, and then extend by periodicity / wrap this on the circle. This gives a well-defined process indexed by the circle, and no point is special because we are always at equilibrium. -If you insist on Brownian motion, but do not want a special point on the circle, then there is a problem because there is no invariant probability measure. The right way to define things goes through white noise i.e. through the increments of Brownian motion. Start with a Brownian bridge $B$ on the interval $[0,1]$, continue it periodically / wrap it on the circle, and let $X_{s,t} = B_t-B_s$. This makes sense for any $s,t \in \mathbb R / \mathbb Z$, and it is easy to see that the distribution of $(X_{s,t})$ does not depend on the choice of the origin. So $(X_{s,t})$ is somehow "the" right object to look at. -From $(X_{s,t})$ you can recover $B_t$ if you want (because $B_t = X_{0,t}$), but then this depends on the choice of the origin. A better way is to say that you will identify two functions if they differ by a constant; then you will get the same equivalence class wherever you choose the origin, and that is as good a definition as any of Brownian motion on the circle. In particular, if what you want to do is stochastic calculus, the increments of the process are usually the relevant thing. -Another way to break the indeterminacy is to choose the path with average value $0$, which then coincides with the Fourier expansion version of Brownian motion (i.e., the Fourier series with independent Gaussian coefficients of appropriate variance).<|endoftext|> -TITLE: what is large compex structure limit of CY moduli space -QUESTION [7 upvotes]: What is the Large Complex Structure limit(LCL) of complex moduli space of a Calabi-Yau 3-fold and why do we need to consider LCL in Mirror symmetry. - -REPLY [4 votes]: The "large complex structure limit" is a family of CY manifolds -over a punctured disk having the maximal possible unipotent -(or, sometimes, quasiunipotent) monodromy. It seems that its existence is -proven in this paper http://arxiv.org/abs/math/0008061 -"Maximal Unipotent Monodromy for Complete Intersection CY Manifolds -Authors: Bong H. Lian, Andrey Todorov, Shing-Tung Yau" -for complete intersections. There are many claims of existence -of such family in other papers of Todorov, I am not sure how -much of them are correct. For simple examples (such as a K3) -it's not hard to find. -You can also complete this family adding a fiber in the center -and call this fiber "a large complex structure limit", but this is -(apparently) not as useful.<|endoftext|> -TITLE: What is the shortest route to Roth's theorem? -QUESTION [23 upvotes]: Roth first proved that any subset of the integers with positive density contains a three term arithmetic progression in 1953. Since then, many other proofs have emerged (I can think of eight off the top of my head). -A lot of attention has gone into the bounds in Roth's theorem, and in particular what kind of bounds different proofs get you (e.g. Fourier analysis gives log type bounds, regularity lemma arguments give Ackermann type bounds, etc.) -Also, some proofs are more amenable to generalisation (to longer arithmetic progressions) than others. -My question is - -If we are only concerned with brevity and directness (i.e. not with a deeper theoretical understanding or sharp quantitative bounds), what is the shortest proof of Roth's theorem? - -Running through the arguments I know, it seems like the shortest one may be Roth's original one (with a couple of simplifications): show there is a large Fourier coefficient and deduce some sort of density increment argument and iterate it - a good exposition is in Tao and Vu, or Ben Green's notes at http://people.maths.ox.ac.uk/greenbj/notes.html . With all details fleshed out, this could probably be done in 8 pages or an hour lecture. -The odd thing is that this proof also gives fairly good quantitative bounds; if $r_3(N)$ is the size of the largest subset of $\{1,...,N\}$ without three term arithmetic progressions, then even a crude version of this argument gives -$$ r_3(N)=O\left(\frac{N}{\log\log N^c}\right)$$ -whereas all Roth needs is $o(N)$. Hence my second question, - -Is it inevitable that the most direct and simple proofs would also lead to fairly good quantitative bounds? - -Finally, an exercise in Tao and Vu mentions that Behrend's example of a lower bound for $r_3(N)$ shows that simple pigeonhole type arguments couldn't be used to prove Roth's theorem. Hence, - -What other proof techniques wouldn't work with Roth's theorem? - -REPLY [8 votes]: On a slightly cheeky note, it seems that the shortest path to the best available bound, subject to peer review, is outlined in the following paper of Bloom-Sisak! Congratulations.<|endoftext|> -TITLE: Homologically nice commutative rings -QUESTION [11 upvotes]: Let $R$ be a commutative regular local ring. Is it true that for every $\mathfrak p \in \mathrm{Spec}(R)$ there is a finitely generated $R$-module $M$ such that $\mathrm{projdim}(M)=\mathrm{ht}(\mathfrak p)$ and $\mathrm{Ass}(M)=\{\mathfrak p\}$? -Or is there some family of commutative noetherian rings where is this true? - -I know that this holds if $R$ is commutative regular local of Krull dimension $\leq 4$ (up to dimension 3 it was easy, because factor rings $R/\mathfrak p$ where always CM, so we can take $M=R/\mathfrak p$. Dimension 4 was harder and in dimension 5 or more I don't know). -Is this an easy/hard/hopeless/already open/similar to something problem? -Thanks, -David - -REPLY [7 votes]: OK, so this is weaker than your last question, which is false but related to some open problems and also weaker than the conjecture in your first question which is another famous open problem. All of these sound like a nice project (-:. -The short anwser is this one is also open in general. I will show that your statement, if true for all regular local ring, would imply one of the outstanding open problem, the so-called Serre's Positivity conjecture. I will sketch the proof below. -Suppose your statement is true for any regular local ring $(R,m)$. Let $p,q$ be such that $\sqrt{p+q} =m$ and $\dim R/p + \dim R/q = \dim R$. By your statement, we can choose Cohen-Macaulay module $M,N$ such that $Ass(M)= \{p\}$ and $Ass(M)= \{q\}$. Let: -$$\chi(M,N) = \sum_{i\geq 0} (-1)^i\text{length}(\text{Tor}_i^R(M,N))$$ -the Serre's intersection multiplicity. -Then $M\otimes N$ has finite length and $M,N$ are Cohen-Macaulay, and a classical result by Serre (can be found in his book Local Algebra, V.6, Theorem 4, p 110 of the English version) says that $\text{Tor}_i^R(M,N)=0$ for $i>0$, so -$$\chi(M,N) =\text{length}(M\otimes N)>0$$ -(This is a nice generalization of Bezout theorem, since curves are Cohen-Macaulay) -But since $Ass(M)= \{p\}$ one has a prime filtration of $M$ by $a>0$ copies of $p$ and primes of smaller dimensions. Similarly $N$ has a filtration with $b>0$ copies of $q$ and primes of smaller dimensions. As $\chi$ is additive on short exact sequences, one has: -$$\chi(M,N) = ab\chi(R/p,R/q)$$ -(one needs to use the Vanishing part of Serre's conjectures, which is known, here) -So one can conclude that $\chi(R/p,R/q)>0$! But this has been open for about 50 years, so I doubt your statement is known (for all regular local rings).<|endoftext|> -TITLE: Uses of Divergent Series and their summation-values in mathematics? -QUESTION [8 upvotes]: This question was posed originally on MSE, I put it here because I didn't receive the answer(s) I wished to see. -Dear MO-Community, -When I was trying to find closed-form representations for odd zeta-values, I used -$$ \Gamma(z) = \frac{e^{-\gamma \cdot z}}{z} \prod_{n=1}^{\infty} \Big( 1 + \frac{z}{n} \Big)^{-1} e^{\frac{z}{n}} $$ -and rearranged it to -$$ \frac{\Gamma(z)}{e^{-\gamma \cdot z}} = \prod_{n=1}^{\infty} \Big( 1 + \frac{z}{n} \Big)^{-1} e^{z/n}. $$ -As we know that $$\prod_{n=1}^{\infty} e^{z/n} = e^{z + z/2 + z/3 + \cdots + z/n} = e^{\zeta(1) z},$$ -we can state that -$$\prod_{n=1}^{\infty} \Big( 1 + \frac{z}{n} \Big) = \frac{e^{z(\zeta(1) - \gamma)}}{z\Gamma(z)} \tag 1$$ -I then stumbled upon the Wikipedia page of Ramanujan Summation (see the bottom of the page), which I used to set $\zeta(1) = \gamma$ (which was, admittedly, a rather dangerous move. Remarkably, things went well eventually. Please don't stop reading). The $z^3$ -coefficient of both sides can now be obtained. Consider -\begin{align*} -(1-ax)(1-bx) &= 1 - (a+b)x + abx^2\\\ -&= 1-(a+b)x + (1/2)((a+b)^2-(a^2+b^2)) -\end{align*} -and -\begin{align*}(1-ax)(1-bx)(1-cx) &= 1 - (a + b + c)x\\\ -&\qquad + (1/2)\Bigl((a + b + c)^2 - (a^2 + b^2 + c^2)\Bigr)x^2\\\ -&\qquad -(abc)x^3. -\end{align*} -We can also set -\begin{align*} -(abc)x^3 &= (1/3)\Bigl((a^3 + b^3 + c^3) - (a + b + c)\Bigr)\\\ -&\qquad + (1/2)(a + b + c)^3 -(a + b + c)(a^2 + b^2 + c^2). -\end{align*} -It can be proved by induction that the x^3 term of $(1-ax)(1-bx)\cdots(1-nx)$ is equal to -$$\begin{array} {lll} -\phantom+ (1/3) \Bigl((a^3 + b^3 + c^3 +\cdots + n^3) - (a + b + c + \cdots + n)^3\Big) \\ -+ (1/2)\Big((a + b + c + \cdots + n)^3 \\ -\qquad\qquad -(a + b + c + \cdots + n)(a^2 + b^2 + c^2 + \cdots + n^2)\Big). -\end{array} \tag 2 $$ -On the right side of equation (1), the $z^3$-term can be found by looking at the $z^3$ term of the Taylor expansion series of $1/(z \Gamma(z))$, which is $(1/3)\zeta(3) + (1/2)\zeta(2) + (1/6)\gamma^3$. We then use (2) to obtain the equality -$$ (1/6)\gamma^3 - (1/2)\gamma \pi^2 - (1/6) \psi^{(2)}(1) = 1/3)\zeta(3) + (1/2)\zeta(2) + (1/6)\gamma^3$$ -to find that -$$\zeta(3) = - (1/2) \psi^{(2)} (1),\tag 3$$ -which is a true result that has been -known (known should be a hyperlink but it isn't for some reason) for quite a long time. The important thing here is that I used $\zeta(1) = \gamma$, which isn't really true. Ramanujan assigned a summation value to the harmonic series (again, see Ramanujan Summation), and apparently it can be used to verify results and perhaps to prove other conjectures/solve problems. - -My first question is: Is this a legitimate way to prove (3) ? - -Generalizing this question: - -When and how are divergent series and their summation values used in mathematics? - What are the 'rules' when dealing with summed divergent series and using them to (try to) find new results? - -Thanks, -Max -As I suspect someone (I was thinking of Qiaochu Yuan himself) will add this too, I will add this question for him/her, as it is somewhat related. - -REPLY [4 votes]: It seems likely to me that your derivation can be modified to just involve convergent series and products, but I have not tried it. -There is however a well-known easy derivation of your identity -$\zeta(3) = - (1/2) \psi^{(2)} (1)$ from the product formula, which is quite close to what you are doing. Note that $\psi=\Gamma'/\Gamma$. Take the logarithmic derivative of the product formula. Then the product becomes a series: -$$\frac{\Gamma'}{\Gamma}(z)=-\gamma-\frac 1z+\sum_{n=1}^\infty\left(\frac 1n-\frac 1{z+n}\right)=-\gamma+\sum_{n=1}^\infty\left(\frac 1n-\frac 1{z+n-1}\right). $$ -Now just differentiate twice and put $z=1$. All series and products involved converge nice enough so that termwise differentiation is justified.<|endoftext|> -TITLE: The flip graph of triangulations -QUESTION [14 upvotes]: A polygon $P_k$ divided by $k-2$ diagonals into triangles is called a polygonal triangulation. These are the vertices of the triangulation graph $\mathcal P_k$. Two vertices are connected by an edge if one triangulation is obtained from another by the diagonal flip, i.e. we take two triangles of the triangulation that share a side, and in their union (where that side is a diagonal), replace that diagonal by the other diagonal. Sleator, Tarjan, and Thurston proved that the diameter of the -triangulation graph ${\mathcal P}_k$ is bounded above by $2k-10$. Hence the problem of finding a shortest path in that graph between two triangulations is in NP. - Question 1. Is it in P? - Question 2. What is known about the complexity of finding the shortest path in the triangulation graph of other surfaces? - Update. I have posted a followup question. - -REPLY [9 votes]: Just to add a little to Joseph's nice answer, for part 1 of your question: although the problem of computing the flip distance in polynomial time is wide open for triangulations of convex polygons, it can be solved in polynomial time for triangulations of certain highly nonconvex point sets (such as the intersection of the integer lattice with a convex set): see my paper "Happy Endings for Flip Graphs", SoCG 2007 and JoCG 2010. - -REPLY [8 votes]: I believe your two questions are not directly related. As far as I know, Question 1 is open. -[Edit. I answered the 2nd question under the interpretation that "the triangulation graph of other surfaces" meant the graph of the surface triangulation, which, as Agol pointed out, -is likely not what Mark meant. Rather he meant the flip graph, which is the focus of the 1st -question. So the below does not answer the intended question. (The flip graph does not always -make sense on a surface.)] -Let me address Question 2. For a long time, the fastest algorithm for finding the shortest path -on a triangulated 2-manifold was the 1996 Chen and Han algorithm, which runs in $O(n^2)$ time for -a surface of $n$ vertices. Here is an example from an implemention of mine with students -of this algorithm, -showing the shortest paths from one point to all the vertices of a convex polyhedron: - - - -This and other algorithms are described in -Geometric Folding Algorithms: Linkages, Origami, Polyhedra, Section 24.2. -There was effort over many years toward improving the time complexity in the special case of surfaces of convex polyhedra, -which was finally cracked by -Schreiber and Sharir in their remarkable paper, - -Schreiber, Yevgeny, and Micha Sharir. "An optimal-time algorithm for shortest paths on a convex polytope in three dimensions." Discrete Comput. Geom., Springer, 39 (2008), 500-579; Twentieth Anniversary Volume, 2009. 1-80. - (Here is the earlier conference version.) - -They improved the speed to $O(n \log n)$; -the shortest paths are represented implicitly to avoid the $n^2$ complexity of explicit listing. -This time complexity has only been achieved to-date for subclasses of nonconvex surfaces.<|endoftext|> -TITLE: value of Theta in ZF+AD -QUESTION [5 upvotes]: Since I found out about it, I've always been interested in the Axiom of Determinacy rather than the Axiom of Choice. Along these lines, I've kept flipping back to http://en.wikipedia.org/wiki/%CE%98_%28set_theory%29, and occasionally looking on google, because I keep thinking ZF+AD should be able to prove non-obvious things about it, although I haven't found anything other than (I think) something saying it must be regular. So, I'm finally asking here. - -$\Theta := \operatorname{sup}(\{\alpha \in \operatorname{Ord} : (\exists f \in \alpha^\mathbb{R})(\operatorname{Range}(f) = \alpha)\})$ - -What is known about $\Theta$ in ZF+AD? In particular, how big is it? - -For example, is it known to be different from $\omega_2$? -Is anything more known in ZF + AD + V=L($\mathbb{R}$) ? - -REPLY [12 votes]: Ricky, -A good reference for this question specifically and determinacy in general is Kanamori's book on large cardinals, "The Higher infinite". The last chapter is devoted to determinacy. -We know a huge deal about $\Theta$. For example, it is much larger than $\omega_2$. It does not need to be regular, but it is regular if in addition we assume $V=L({\mathbb R})$. -The key fact to see that $\Theta$ is large is Moschovakis's coding lemma which, in its simplest version, says that: - -(Under AD) if there is a surjection $f:{\mathbb R}\to\alpha$, then there is a surjection $f:{\mathbb R}\to{\mathcal P}(\alpha)$. - -Harvey Friedman used this to prove that $\Theta$ is a limit cardinal. This is easy; the point is that there is a definable bijection between ${\mathcal P}(\tau)$ and ${\mathcal P}(\tau\times\tau)$ for any infinite ordinal $\tau$. But if $\tau$ is a cardinal, there is a surjection from ${\mathcal P}(\tau\times\tau)$ onto $\tau^+$: If $A\subseteq\tau\times\tau$ codes a well-ordering, send it to its order type. Else, to 0. -With a bit more effort, you can check that $\Theta=\aleph_\Theta$ and in fact it is limit of cardinals $\kappa$ such that $\kappa=\aleph_\kappa$, and it is a limit of limits of these cardinals, etc. -In $L({\mathbb R})$, $\Theta$ is regular (Solovay was first to prove this). In fact, if $V=L(S,{\mathbb R})$ for $S$ a set of ordinals, or $V=L(A,{\mathbb R})$ for $A\subseteq{\mathbb R}$, then $\Theta$ is regular. -(A technical aside: If AD holds, it holds in $L({\mathcal P}({\mathbb R}))$. Woodin defined a strengthening of AD that is now called $AD^+$. It is open whether $AD^+$ is strictly weaker than AD, since all known models of AD are models of $AD^+$ and any current technique that gives us a model of one gives us a model of the other. If $L({\mathcal P}({\mathbb R}))$ is a model of $AD^+$, then it is either of the form $L(A,{\mathbb R})$ for some $A\subseteq{\mathbb R}$, or else it is a model of $AD_{\mathbb R}$, the strengthening of AD where we allow reals (rather than integers) as moves of the games.) -However, ZF+AD does not suffice to prove that $\Theta$ is regular. If DC holds, the "obvious" diagonalization shows that ${\rm cf}(\Theta)>\omega$. But Solovay proved that $ZF+AD_{\mathbb R}+{\rm cf}(\Theta)>\omega$ implies the consistency of $ZF+AD_{\mathbb R}$, so by the incompleteness theorem, ZF+AD or even the stronger $ZF+AD_{\mathbb R}$ cannot prove that ${\rm cf}(\Theta)>\omega$. -Nowadays we know much more. For example, $\Theta$ is a Woodin cardinal in the HOD of $L({\mathbb R})$, and the computation of the large cardinal strength of $\Theta$ in the HOD of the models of AD is a guiding principle of what is now known as descriptive inner model theory. -You may be interested in the slides of recent talks by Grigor Sargsyan on the core model induction (which should be available somewher online, or from him by email). You will see there that the large cardinal strength of AD assumptions is calibrated by the large cardinal character of $\Theta$ inside HOD, and this is associated with the length of the so called Solovay sequence which keeps track of how difficult it is to define the surjections $$f:{\mathbb R}\to\alpha$$ as $\alpha$ increases. This difficulty is related to the Wadge degree of sets of reals present in the AD model. -(I can be much more detailed, but this will require me to get significantly more technical. Let me know.)<|endoftext|> -TITLE: Poincare conjecture and the graph of triangulations -QUESTION [6 upvotes]: This was an update to this question, but I decided to make it a separate question. The definition of the graph of triangulations can be found in the previous question. - Question. I was told a few years ago that some computational complexity problem for the triangulation graph (perhaps in dimension 3?) is related to the Poincare conjecture. Unfortunately I forgot what was the problem. Does anybody know? - -REPLY [2 votes]: Perhaps you are thinking of the recognition problem for the three-sphere? There are theorems of Aleksandar Mijatovic (Simplifying triangulations of $S^3$) and Simon King (How to make a triangulation of $S^3$ polytopal) that say: If $T$ is a triangulation of $S^3$ with at most $n$ tetrahedra then you need at most $f(n)$ "operations" to transform $T$ into a "standard" triangulation. The function $f$, the operations, and the definition of standard vary between the two papers. -I believe that the result is a very simple but very slow (doubly exponential) algorithm to recognize the three-sphere. The proofs, at the very end of the day, rely on normal surface theory and the Rubinstein/Thompson algorithm.<|endoftext|> -TITLE: Spaces that invert weak homotopy equivalences. -QUESTION [11 upvotes]: Are there any nontrivial spaces $Y$ so that for all weak homotopy equivalences -$A\to B$, the induced map $[B, Y]\to [A,Y]$ is bijective? -This would be a property of the homotopy type of $Y$, and -if $Y$ is homotopy equivalent to a space with -has some kind of local structure under which very close maps (probably of -compact spaces like $S^k$) are -necessarily homotopic, then it probably won't have this property. -My idea is to use the following construction: let $L^+ = \{ {1\over n} \mid n \geq 1\}$ -and let -$L = \{ 0\} \cup L^+$. -Then this hypothetical local property of $Y$ would ensure that -the restriction $L \times X \to L^+\times X$ -would induce an injection on homotopy sets. But - $L\times X$ is weakly equivalent to $\coprod_{0}^\infty X$, -and in the latter space we can have maps - which are $f$ on $X\times {n}$ for $n > 0$ and $g$ on $X\times 0$, -where $f\not \simeq g$. - -REPLY [10 votes]: The answer seems to be "no": only for contractible spaces Y (and Y=$\emptyset$) the functor [-,Y] inverts weak equivalences. As mentioned above I wrote an argument in https://arxiv.org/abs/1709.08734. It uses Jeff Strom and Tom Goodwillie's idea of considering a space whose path-components are the singletons. In this case its topology is similar to the cofinite topology.<|endoftext|> -TITLE: injectivity of torsion submodules of injectives -QUESTION [15 upvotes]: Local cohomology with respect to an ideal $\mathfrak{a}$ is often studied over a Noetherian ring $R$. However, the proof of a lot of basic results does not rely on noetherianity of $R$, but rather on the following two properties: - -(ITI) $\mathfrak{a}$-torsion submodules of injective modules are injective. -(ITR) $\mathfrak{a}$-torsion modules have injective resolutions whose components are $\mathfrak{a}$-torsion. - -If $R$ is Noetherian, then it has ITI with respect to every $\mathfrak{a}$. If $R$ has ITI with respect to $\mathfrak{a}$, then it has ITR with respect to $\mathfrak{a}$. -Is anything known about the converses of these implications? Does anybody know a ring that does not satisfy ITR? - -REPLY [3 votes]: 1) A ring has ITI with respect to an ideal $\mathfrak{a}$ if and only if it has ITR with respect to $\mathfrak{a}$. -2) A ring does not necessarily have ITI with respect to an ideal of finite type. (Note that the ideal in Quý's example is not of finite type.) -3) A ring that has ITI with respect to every ideal of finite type does not necessarily have ITI with respect to every ideal. -Proofs of the above, concrete examples, and further details on the ITI-property can be found in a joint work with P.H.Quý (who gave the accepted answer), available here. See also this question.<|endoftext|> -TITLE: What are some early examples of creation of lists / catalogues of (particularly) combinatorial objects? -QUESTION [17 upvotes]: A lot of effort in discrete maths / combinatorics is expended in the construction of lists, catalogues or census [sic] of combinatorial objects such as groups, graphs, designs etc. These catalogues are now a fundamental part of computer algebra systems. -Obviously nowadays most of this is done by computer, but a surprisingly large amount of this work predates (electronic) computers - for example, G.A. Miller worked on creating lists of "substitution groups" (permutation groups) in the late 1800s and early 1900s, while Ronald Foster created the "Foster Census" of cubic symmetric graphs in the 1930s. -I'd like to know some more examples of famous "cataloguers" of mathematical (well, particularly combinatorial) objects predating electronic computers. - -REPLY [3 votes]: A lot of people counted Latin squares, going back to Euler (1782) and Cayley and Frolov (independently, 1890). Many of those who tried got the wrong answer. A summary is in this paper: -B. D. McKay, A. Meynert and W. Myrvold, Small Latin squares, quasigroups and -loops, J. Combin. Designs, 15 (2007) 98-119. -A copy with a correction to Theorem 2 is here. -My favourite from the pre-computer age is: -P. N. Saxena, A simplified method of enumerating Latin squares by MacMahon’s -differential operators; II. The 7 × 7 Latin squares, J. Indian Soc. Agric. Statistics, 3 (1951) 24–79. -Saxena devoted 55 pages to the most intricate case-by-case calculations but amazingly got the right answer.<|endoftext|> -TITLE: Do proper polynomial mappings have a path-lifting property? -QUESTION [8 upvotes]: Suppose $f: \mathbb{C}^n \to \mathbb{C}^n$ is a proper polynomial mapping and $\gamma: [0,1] \to \mathbb{C}^n$ is a continuous path. Further, suppose $z_0 \in \mathbb{C}^n$ satisfies $f(z_0)=\gamma(0)$. Does there exist a (not necessarily unique) lift of $\gamma$ under $f$ based at $z_0$? Is there a published reference for this fact? - -REPLY [9 votes]: The answer is yes, here is a proof. -Let $k=2n$. The polynomial, regarded as a map $f:\mathbb R^k\to\mathbb R^k$, has the following properties: -(1) The pre-image of every point if finite, moreover the cardinality of a pre-image is uniformly bounded by some constant $N$ (by Bezout's theorem, see comments). -(2) The set $\Sigma$ of singularities of $f$ (i.e. points where $\det df=0$) is a union of finitely many smooth submanifolds of (real) codimension at least 2. -This (along with smoothness and properness) implies the path-lifting property. -First, observe the following facts: -(3) $f(\Sigma)$ is closed (because $\Sigma$ is closed and $f$ is proper). -(4) $\mathbb R^k\setminus f(\Sigma)$ is path connected. Moreover any path in $\mathbb R^k$ is a uniform limit of paths avoiding $f(\Sigma)$. This follows from the fact that $f(\Sigma)$ has Hausdorff dimension at most $k-2$. -(5) For every compact set $K\subset\mathbb R^k$ and every $\varepsilon>0$, there exists $\delta>0$ such that for every connected set $S\subset K$ of diameter greater than $\varepsilon$, the diameter of $f(S)$ is greater than $\delta$. Indeed, if this is not the case, there would be a sequence $S_i$ of connected subsets of $K$ with diameters at least some $\varepsilon_0>0$ and diameters of images going to zero. By choosing a subsequence, we may assume that the images converge to some $y_0\in\mathbb R^k$. -Since $S_i$ is connected and has diameter at least $\varepsilon_0$, it contains a finite subset $P_i$ of cardinality $N+1$ such that all points of $P_i$ are separated away from one another by distance at least $\varepsilon_0/2N$. A subsequence of $\{P_i\}$ converge to a set $P$ of cardinality $N+1$, and all points of $P$ are mapped by $f$ to $x_0$, contrary to (1). -Now, in order to lift a path $\gamma$, approximate it by paths $\gamma_i\subset\mathbb R^k\setminus f(\Sigma)$. The restriction of $f$ to the pre-image of $R^k\setminus f(\Sigma)$ is a covering map to $R^k\setminus f(\Sigma)$ because it is a proper local homeomorphism. So there are lifts $\tilde\gamma_i$ of $\gamma_i$. Now it suffices to find a converging subsequence of $\{\tilde\gamma_i\}$. To do this, it suffices to show that this sequence is equi-continuous. And this follows from (5): if you could find arbitrarily close pair of points on some $\gamma_i$ with lifts bounded away from each other, the lift of the segment of $\gamma_i$ between these points would have diameter bounded away from zero but the diameter of the image arbitrarily small, contrary to (5). -To make sure that the limit lift starts from $z_0$, choose $\gamma_i$'s so that their starting points have pre-images near $z_0$ and start $\tilde\gamma_i$'s from there.<|endoftext|> -TITLE: Does "antichain" mean something different in set-forcing than in lattice theory? -QUESTION [7 upvotes]: On page 3 of Introduction to Lattices and Order, Davey and Priestley define an antichain in a poset $\langle P,\leq\rangle$ as a set of pairwise incomparable elements: - -The ordered set P is an antichain if $x\leq y$ in P only if $x=y$ - -Gratzer's definition is equivalent, but stated in a manner which is difficult to excerpt. -On page 53 of Set Theory, an Introduction to Independence Proofs, Kunen defines an antichain in $\langle P,\leq\rangle$ as a set of pairwise incompatible elements, saying that two elements $p$ and $q$: - -are incompatible ($p\bot q$) iff $\neg\exists r\in P(r\leq p\wedge r\leq q)$. An antichain in $P$ is a subset $A\subset P$ such that $\forall p,q\in A(p\neq q\rightarrow p\bot q)$. - -So, given a three-element partially ordered set $\{0,a,b\}$ with $0\leq a$, $0\leq b$ the only (non-reflexive) related pairs in the partial order, it appears that $\{a,b\}$ is an antichain in the lattice sense but not in the forcing sense. -Question: is it in fact true that "antichain in a poset" means something different to set theorists than to lattice theorists? - -REPLY [9 votes]: Adam: -Yes, the notions are different, but I believe the ambiguity is older than forcing; doesn't Halmos use "antichain" for the forcing notion in his book on Boolean algebras? -Typically, when the need arises of distinguishing both notions, I've seen used (and used myself) "$A$ is a weak antichain" for "the elements of $A$ are pairwise incomparable", while "$A$ is a strong antichain" is reserved for the forcing version, "the elements of $A$ are pairwise incompatible." -Usually context suffices to know which version is used. In combinatorial contexts I would think using "antichain" for the "weak" version is more common. Certainly whenever forcing is used, it is the "strong", Boolean- (or forcing-)theoretic version that is used. In any paper where ambiguity could be an issue, I've seen at least a remark.<|endoftext|> -TITLE: Endomorphism of type III factor: can it satisfy $\phi\circ\phi = \phi\oplus\phi$? -QUESTION [13 upvotes]: I'm still trying to get some feeling about this question... -Given Jesse Peterson's answer to this question (he showed that $\phi\circ\phi\sim\phi$ is impossible), I suspect that the following is also impossible. But I'm unable to generalize his argument. - -Let $M$ be a type III factor, and let $\phi:M\to M$ be an irreducible endomorphism (the relative commutant of $\phi(M)$ in $M$ is trivial). -Let $v_1$, $v_2\in M$ be isometries with orthogonal ranges summing up to $1$ ($v_1^*v_1=v_2^*v_2=v_1v_1^*+v_2v_2^*=1$). Define $$\phi\oplus\phi:m\mapsto v_1\phi(m)v_1^*+v_2\phi(m)v_2^*.$$ - -Question: Is it possible to have $\phi\circ\phi$ conjugate to $\phi\oplus\phi$? $$ \phi ( \phi(m)) = u \Big( v_1 \phi (m) v_1^* + v_2 \phi (m) v_2^* \Big) u^* $$ - -REPLY [2 votes]: I think the situation you describe is impossible: Let $\bar{\phi}$ be the conjugate endomorphism to $\phi$. From the equation $d(\phi)^2 = 2d(\phi)$ we get $d(\phi) = 2$. Denote by $\langle \rho, \sigma \rangle$ the dimension of the intertwiner space between $\rho$ and $\sigma$. By Frobenius reciprocity and the irreducibility of $\phi$ we now have -$$ -\langle \bar{\phi} \circ \phi, \phi \rangle = \langle \phi, \phi \circ \phi \rangle = \langle \phi, \phi \oplus \phi \rangle = 2. -$$ -Thus, $\bar{\phi} \circ \phi$ contains two copies of $\phi$ and a copy of the identity. Therefore -$$ -4 = d(\phi)^2 = d(\phi)\cdot d(\bar{\phi}) = d(\bar{\phi} \circ \phi) \geq d(id \oplus \phi \oplus \phi) = 1 + 2d(\phi) = 5 -$$ -which is a contradiction. Note that if you drop the assumption that $\phi$ is irreducible, there should be examples: Suppose $M$ carries an involution $\alpha \colon M \to M$, i.e. an action of $\mathbb{Z} / 2\mathbb{Z}$. Consider $\phi = id \oplus \alpha$ with the definition of the sum similar to the one in your question. Then -$$ -[\phi \circ \phi] = [id \oplus \alpha] \circ [id \oplus \alpha] = [id \oplus \alpha \oplus \alpha \oplus \alpha^2] = [id \oplus \alpha \oplus id \oplus \alpha] = [\phi \oplus \phi] -$$ -where the brackets mean unitary equivalence classes of endomorphisms.<|endoftext|> -TITLE: Line bundles in abelian $\otimes$-categories -QUESTION [5 upvotes]: By an abelian $\otimes$-category I mean a symmetric monoidal category $(\mathcal{A},\otimes,\mathcal{O})$, such that $\mathcal{A}$ also is an abelian category and for every $M \in \mathcal{A}$ the functor $M \otimes - $ is cocontinuous (i.e. right exact and preserves coproducts; in particular additive). A line bundle is defined as an object $\mathcal{L}$ of $\mathcal{A}$ such that there is some object $\mathcal{L}'$ such that $\mathcal{L} \otimes \mathcal{L}' \cong \mathcal{O}$. An example is the category of (quasi-coherent) modules on a locally ringed space. The line bundles then coincide with the modules which are locally free of rank $1$ (see here). -Now I want to show that in general these line bundles have similar properties as in the case of the module category. For example it's not hard to show that if $\mathcal{L}$ is a line bundle, then it is flat in the sense that $\mathcal{L} \otimes -$ is exact (it is even an automorphism of $\mathcal{A}$ with inverse $\mathcal{L}^{-1} \otimes -$). The isomorphism classes of line bundles yield a group, which may be denoted as $\text{Pic}(\mathcal{A})$. -Question 1. Is there any literature about these abelian $\otimes$-categories which treats them systematically? Perhaps the "usual" definition differs a little from mine, this does not matter. -Question 2. Let $\mathcal{L}$ be a line bundle and $\phi : \mathcal{L} \to \mathcal{L}$ an epimorphism. Does it follow that $\phi$ is an isomorphism? -Question 3. Let $\mathcal{L}$ be a line bundle and assume $\phi : \mathcal{L} \to \mathcal{L}$ is an epimorphism. Does it follow that there is an epimorphism $\psi : \mathcal{L}^{-1} \to \mathcal{L}^{-1}$ such that $\phi \otimes \psi$ corresponds to the identity of $\mathcal{O}$ under the isomorphism $\mathcal{L} \otimes \mathcal{L}^{-1} \cong \mathcal{O}$? -Question 4. Assume we also have a $\lambda$-structure on $\mathcal{A}$ which is compatible with the given data. Is it possible to give a reasonable definition of a locally free object of rank $n$? See also this question. - -REPLY [3 votes]: Question 2: this does not really depend on line bundles. If $\phi:\mathcal {L\to L}$ is a non-invertible epimorphism, then $\phi\otimes 1:\mathcal{ L\otimes L'\to L\otimes L'}$ is epi, -since $-\otimes\mathcal L'$ is exact, and non-invertible, since otherwise -$\phi\otimes 1\otimes 1:\mathcal{L\otimes L'\otimes L\to L\otimes L'\otimes L}$ would be -invertible. Thus there is a non-invertible epimorphism $\mathcal {O\to O}$. -Question 3: this has the same answer as Question 2. If the answer to Q2 is yes, then we can take $\psi$ to be the identity and get a positive answer to Q3. If the answer to Q2 is no, then (as above) if $\phi:\mathcal{ L\to L}$ is a non-invertible epimorphism, also -$\phi\otimes 1:\mathcal{ L\otimes L'\to L\otimes L'}$ is a non-invertible epimorphism. -But now if $\psi:\mathcal {L'\to L'}$ is any map, then $\phi\otimes\psi=(1\otimes\psi)\circ(\phi\otimes 1)$ and if this is invertible then $\phi\otimes 1$ is split monic and so invertible (since it is already known to be epi).<|endoftext|> -TITLE: Exterior derivative on almost complex manifolds -QUESTION [6 upvotes]: Let $M$ be a complex manifold, and $\omega$ be a $(p,q)$-form. Then $d\omega$ is an element of $\Omega^{p+1,q}(M)\oplus\Omega^{p,q+1}(M)$, so that $d = \partial + \overline{\partial}$, where $\partial$ and $\overline{\partial}$ are the Dolbeault operators. -Now let $M$ be almost complex. It is commonly stated that $d = \partial + \overline{\partial}$ only holds for complex manifolds, and not for almost complex manifolds. But why is this? After extending $d$ to also be complex linear, if $\omega = \sum_i f(z)dz^i$ is a $(0,1)$-form, I'd say that we would have $ d\omega = \sum_i df\wedge dz_i -= \sum_{i,j} \frac{\partial f}{\partial z^j}dz^j\wedge dz^i + \frac{\partial f}{\partial \overline{z}^j}d\overline{z}^j\wedge dz^i,$ -which clearly does not have a $(0,2)$-part. Why is this wrong? -On the other hand, let $X, Y$ be antiholomorphic tangent vectors, then $d\omega(X,Y) = X(\omega(Y)) - Y(\omega(X)) - \omega([X,Y]) = -\omega([X,Y])$. Since $M$ is not nessecarily complex, $[X,Y]$ is not nessecarily also antiholomorphic, so that this term does not nessecarily vanish. But $d\omega$, being a 2-form, can only give a nonzero result if it has a $(0,2)$-part. So from this I can see that it has to have one, but I can't see why this contradicts the calculation of $d\omega$ above. - -REPLY [9 votes]: Just to follow up on Eric's correct answer: when you have an almost complex structure $J$, you can decompose $1$-forms into type $(1,0)$ and $(0,1)$. Locally, you can find a local basis $e^1, \ldots, e^n$ of $(1,0)$-forms, but these are not of the form $dz^1, \ldots, dz^n$. Indeed, as Eric mentioned, we do not have local holomorphic coordinates. Then $\bar e^1, \ldots, \bar e^n$ are a local basis of $(0,1)$ forms. Now if we compute $de^i$, it is a $2$-form, so it can be written in the form -\begin{equation*} -de^i = a^i_{jk} e^j \wedge e^k + b^i_{jk} e^j \wedge \bar e^k + c^i_{jk} \bar e^j \wedge \bar e^k. -\end{equation*} -The almost complex structure $J$ is integrable if and only if all the $c^i_{jk}$'s are zero.<|endoftext|> -TITLE: "Let $x \in A$", beginning a proof of "$\forall x \in A$ ...", if A were empty -QUESTION [13 upvotes]: I work at a four-year teaching school, where we pride ourselves on teaching pure math, proof, and a rather obsessive carefulness of work. Recently I have been criticized for saying that "Let $x \in A$" is often a good way to begin a proof of a statement about all elements of $A$. The criticism is based on the objection that $A$ could be empty, in which case there is no $x$ to be in $A$. The issue affects quite a lot of mathematical content, because more than half of our proofs are proofs of universal statements, and most of them begin this way. -I have objected that if $A$ is empty, then any universal statement $\forall x \in A ...$ is vacuously true, but people are telling me that this needs to be dealt with as a special case, or else the proof is technically incorrect, etc. -I have appealed to normal mathematical conventions, without success. Our department prides itself on being more careful than normal working mathematicians. Convention can do what it will, but we intend to be right! -I have appealed to serious logic, by talking about the underlying meaning of "Let $x \in A$." In my reading it plays a dual role of symbol introduction ("Use $x$ to represent a single thing") and assumption ("Assume $x \in A$"). But these arguments have no traction -- "our students can't be expected to understand clever subtleties of metalogic." -I fear the only option remaining is to appeal to authority -- some specific authority who says this act of "Let $x \in A$" has some sort of seal of approval. Maybe such an authority is here? -From this question you might thing I work with fools, but they're really very wonderful and intelligent people, and the sense of family here is unusually strong. Like family, they drive me out of my head sometimes. Probably it's mutual. -So I suppose I have two questions: 1. Am I right? 2. Is there any hope for me to persuade my colleagues that I'm right? -(3. But social advice would be welcome too.) -Thank you, -Anonymous Coward - -REPLY [6 votes]: I'm going to take a guess as to what your colleagues are worried about (though they are basically wrong to worry about it for the reasons that Andrej outlines). In classical first-order logic this sentence is a tautology: -$$ -\forall x. \phi(x) \implies \exists x. \phi(x), -$$ -but the same rule with bounded quantifiers is false: -$$ -\forall x \in A. \phi(x) \implies \exists x \in A. \phi(x). -$$ -This is only reason why I imagine they would be worried. I could imagine that the occasional student could start with "Suppose there exists an $x \in A$ such that $\phi(x)$..." and somehow slip and concludes that there exists an $x$ such that $\phi(x)$. -The reason for the difference between the the unbounded and bounded quantifiers is that classical first-order logic assumes that the universe of discourse is not empty. The correct rules of inference when the universe of discourse can be empty is known as free logic. It's discussed in great depth in this article at the Stanford Encyclopedia of Philosophy.<|endoftext|> -TITLE: Is every field the field of fractions of an integral domain? -QUESTION [19 upvotes]: Is every field the field of fractions of an integral domain which is not itself a field? -What about the field of real numbers? - -REPLY [30 votes]: Every field $F$ of characteristic zero or of prime characteristic -but not algebraic over its prime field -is the field of fractions of a proper subring of $F$. -But no algebraic extension of $\mathbb F_p$ is, since its only subrings are fields. -If $F$ is not an algebraic extension of some $\mathbb F_p$ -then $F$ contains a subring $A$ isomorphic to $\mathbb Z$ -or $\mathbb F_p[X]$. Each of these rings $A$ has a nontrivial -valuation $v$. The valuation $v$ can be prolonged to $F$. -Its valuation ring is a proper subring of $F$ whose quotient field -is $F$.<|endoftext|> -TITLE: Translation surfaces -QUESTION [9 upvotes]: I know that this definitely have some sort of reference out there, but I did not find any wikipidea page for it or any introductory Mathematical article about it . I just want definition and concrete examples of translation surfaces. For example, if I take four copies of equilateral triangle and glue them together side by side to form a tetrahedron, how do I put a Riemann surface structure on it ? I want some detailed explanations etc. Could you cite any reference(s) ? - -REPLY [7 votes]: People normally take the definition: translation surface = pair (X,w), where X is Riemann Surface and w is a non identically zero holomorphic form on X. Strictly, if you want to make a translation surface out of this pair you have to remove the zeros of w from X and then integrate w. -As a general reference take the book: -Flat surfaces (by Anton Zorich) -in collection "Frontiers in Number Theory, Physics and Geometry. Volume 1: -On random matrices, zeta functions and dynamical systems'', -P. Cartier; B. Julia; P. Moussa; P. Vanhove (Editors), -Springer-Verlag, Berlin, 2006, 439-586.<|endoftext|> -TITLE: Schemes (as in algebraic geometry) and first-order logic. -QUESTION [7 upvotes]: Affine schemes are simply the Zariski spectra of commutative rings, and commutative rings occurs as models of a first-order theory. -I would guess that general schemes do not naturally correspond to the models of any first-order theory and would like to know of any theorems that formalize this. -I would also like to know of interesting first-order theories that do capture more general schemes than merely the affine ones. -Even more interesting to me: first order theories that approximate/generalize scheme theory by allowing unintended models. For example, variables in the theory might range over partial sections of the structure sheaf, one relation might indicate when one section extends another, and another when two sections have the same domain. One would then formulate addition and multiplication as relations (with the various sections all required to have the same domain.) As above, I would guess that "locally affine" resists a first-order formulation in this language. Still one can consider the first-order theory of the models that have the form of schemes, then ask about systems of axioms for such a theory and what the non-scheme models look like. - -REPLY [2 votes]: One kind of answer is as follows. A commutative ring $R$ is local iff for all $a\in R$, either $a$ is invertible or $1-a$ is invertible. The logical form of this statement is obviously much simpler than the usual definition in terms of maximal ideals, so one can easily interpret it in a wide range of categories. The pair (sheaves on spec($R$), structure sheaf) is in some sense the free locally-ringed topos generated by the ringed topos (sets,$R$). If I remember correctly, this is explained in more detail in Johnstone's book on Stone Spaces. (He probably talks about it in his topos theory books as well.) More generally, the structure sheaf on a scheme is always a local ring object in the sense just discussed. Concretely, this means that for any section $a$ of the structure sheaf over an open set $U$ we can write $U=V\cup W$ with $V$ and $W$ open, such that $a|_V$ and $(1-a)|_W$ are invertible.<|endoftext|> -TITLE: Essential self-adjointness of differential operators on compact manifolds -QUESTION [22 upvotes]: Let $L$ be a linear differential operator (with smooth coefficients) on a compact differentiable manifold $M$ (without boundary). Suppose $M$ is endowed with a smooth volume form (actually, a smooth volume density, if one wishes to consider the non orientable case), so that we can speak about the Hilbert space $L^2(M)$. I regard $L$ as being a densely defined operator on $L^2(M)$ with domain $C^\infty(M)$. Assume that $L$ is symmetric. Is it true that $L$ is essentially self-adjoint? If $L$ is elliptic then the answer is yes (one possible proof: the domain of the adjoint $L^*$ is the set of those $f\in L^2(M)$ such that $L(f)$ --- understood in the distributional sense --- is in $L^2(M)$ and $L^*$ is the restriction of the extension of $L$ to distributions. Let $f$ be an eigenvector of $L^*$ with eigenvalue $\pm i$. Then $f$ is a weak solution of $L(f)=\pm if$ and, by elliptic regularity, $f$ is smooth and it is therefore an eigenvector of $L$ with eigenvalue $\pm i$, contradicting the symmetry of $L$). -Naively speaking, absence of essential self-adjointness is related to the existence of several possible "boundary conditions", which do not exist for compact manifolds. So, naively, the result seems plausible. But maybe I'm being too naive. -Edit: The result is false and the counterexample suggested by Terry Tao works. Let $M=S^1=\mathbb{R}/2\pi\mathbb{Z}$ and $L=\frac{d}{dx}\sin(x)\frac{d}{dx}$. The symmetric operator $L$ is not essentially self-adjoint in $C^\infty(S^1)$. A non zero solution of $(L^*+i)\psi=0$ is obtained using Fourier series. Here are the details: set $a_0=0$, $a_1=1$ and $a_{k+2}=\frac{k}{k+2}a_k+\frac{2}{(k+1)(k+2)}a_{k+1}$ for $k\ge0$. It is easily proven by induction that the sequence $a_k$ is $O(k^{-2/3})$ and hence it is square integrable. The function $\psi(x)=\sum_{k=0}^\infty a_ke^{ikx}$ is hence in $L^2(S^1)$ and it solves $(L^*+i)\psi=0$ (because it solves $(L+i)\psi=0$ in the distributional sense). - -REPLY [2 votes]: I would like to point out that, even if the classical dynamics is incomplete (with a large set of trajectories escaping in finite time), the quantum one might well be complete. -For example, consider on $M = \{(x,y) \in \mathbb{R}^2 \mid x >0 \}$ the Riemannian metric (Grushin metric) -$$ g = dx^2 + \frac{1}{x^2}dy^2 $$ -It is not hard to prove that the Laplace-Beltrami $\Delta$ of the above metric, with domain $C^\infty_c(M)$ is essentially self adjoint (see [2]). Notice that in the above example there is no external potential: the confinement is purely geometrical. -On the other hand, the principal symbol of $\Delta$ is -$$ 2H = p_x^2 + x^2p_y^2 $$ -and it can be prolonged smoothly on the whole $\mathbb{R}^2$, where it gives origin to a complete dynamic. In this case, essentially all trajectories starting at points with $x>0$ cross at some time the singular region $\{x=0\}$ (the only exception are trajectories with $p_y=0$ and $p_x>0$). More importantly, they do so without losing optimality (in the sense that all these trajectories are shortest paths for the Riemannian metric). -This is a particular instance of a more general fact (i.e. quantum completeness) for non-complete Riemannian structures satisfying suitable conditions at the metric boundary, as proved recently in [1]. -Let me mention that quantum completeness for two-dimensional almost-Riemannian structures (a class of which the Grushin metric above is part) has been proved originally in [2], using the normal forms for 2D almost-Riemannian structures. -A conjecture is still open for more singular situations (i.e. non-regular almost-Riemannian structures in the language of [1]).<|endoftext|> -TITLE: Are Penrose tilings universal? Do aperiodic universal tilings exist? -QUESTION [15 upvotes]: Consider a tiling of the plane using tiles of at least two types (e.g, a Penrose tiling such as that shown at the bottom of this question, which tiles the plane with two types of tiles). List the tile types as $t_1,\ldots,t_k$. Say that an animal using tiles $t_1,\ldots,t_k$ is a connected subset of the plane that can be obtained by gluing a finite number of tiles together along their edges; identify congruent subsets. If there is only one type, this is often called a polyomino; -here are some pictures of polyominoes in the square lattice (which has only one type of tile and is not in fact an interesting lattice from the point of view of this question). - (source) -Say that a tiling of the plane using (distinct) tiles $t_1,\ldots,t_k$ is universal if it contains every possible lattice animal using tiles $t_1,\ldots,t_k$. To explain what I mean by "possible", suppose that $k=2$, that $t_1$ is the "thin diamond" from the Penrose tiling and that $t_2$ is the "thick diamond. By gluing together four copies of $t_1$ one can obtain the following "animal". -http://www.math.mcgill.ca/louigi/images/penroseexample.jpg -This animal can't be contained within any tiling (Penrose or otherwise) using $t_1$ and $t_2$. So it makes sense to restrict to animals which, for example, are contained within some tiling of the plane with the given tiles. - -My question is: are there $k \geq 2$ for which (aperiodic -- adjective added in edit) universal tilings exist? -Edit: when I first posted the question I omitted the adjective aperiodic above. As pointed out in comments, in this case the answer is obviously yes, which is good to have had pointed out. - -We can also restrict the allowed animals. For example, we could restrict to animals which exhibit some form of symmetry. - -One could then ask: do aperiodic tilings exist which are universal for animals in a (non-trivial) restricted class? Are there any interesting results along these lines? Is the Penrose tiling itself known to be universal for some interesting class of animals? - -REPLY [5 votes]: A very nice reference on this family of questions is Grunbaum and Shephard's book Tilings and Patterns (now out in paperback). It has starred problems which are hard, and double-starred problems which are research projects. It is in this book that we find the problem of whether there is a tile which only tiles aperiodically. (There is such a beast for the 3-dimensional analog, but it's a bit of a cheat.) I contacted Grunbaum about this a few years ago, and he said the problem was still open.<|endoftext|> -TITLE: "Uniform probability" on a set of naturals -QUESTION [16 upvotes]: It's an obvious and well-known fact that there is no uniform probability measure on a set of natural numbers (i.e. the one that gives the same probability to each singleton). -On a recent probability seminar one professor mentioned that this nonexistance is one of the main objections to measure-theoretic formulation of probability and that even Kolmogorov admitted the problem in it. -What are possible solutions (i.e. alternative definitions of probability) to this problem? -I'm looking either for an informative answer or a good reference. - -REPLY [6 votes]: I think the usual way is to define a sequence $f_n$ of probability measures whose limit has whichever property you consider to be the essence of uniformity (invariance under translation, the evens have probability 1/2, finite additivitiy, et cetera). -For example, let $f_n(A) = |A \cap \{1,2,\dots,n\} | / n$. Each $f_n$ is a probability measure, and the limit $\lim f_n(A)$ (if it exists) is called the natural density of $A$. Another example, let $f_n(A) = \zeta(1+1/n)^{-1} \sum_{k\in A} k^{-1-1/n}$. The limit $\lim f_n(A)$ (if it exists) is the logarithmic density of $A$. -Another approach is to take an unlimited hyperinteger (as in, nonstandard analysis) $N$, and to take the measure $f(A)$ to be the standard part of $|{}^\ast A\cap \{1,2,\dots,N\}|/N$. This has a tendency to be meaningless, however, unless you can prove a result that works for all unlimited hyperintegers. Essentially, this is the same difficulty that arises in Denis Serre's answer.<|endoftext|> -TITLE: Are Kato's zeta elements integral? -QUESTION [11 upvotes]: Let $E$ be an elliptic curve over $\mathbb{Q}$ and $T$ the $p$-adic Tate module of $E$. Kato's Euler system, constructed in the paper "P-adic Hodge theory and values of zeta functions of modular forms" (Asterisque 295, 2004), gives rise to an element $\mathbf{z}_{\rm Kato}$ lying in the Iwasawa cohomology $ H^1_{\mathrm{Iw}}(\mathbb{Q}_p, T)[\frac{1}{p}]$. In Theorem 12.5(4) of the paper, Kato shows that if the image of $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ in $\mathrm{Aut}(T)$ contains $\mathrm{SL}_2(\mathbb{Z}_p)$, then in fact $\mathbf{z}_{\rm Kato} \in H^1_{\mathrm{Iw}}(\mathbb{Q}_p, T)$. -Is it known if there are weaker conditions that are sufficient to ensure that $\mathbf{z}_{\rm Kato}$ has this integrality property? Are there examples where it is genuinely non-integral, or is it conjectured that it should always be so? -I would be the last person to claim I understand Kato's argument, but it looks to me as if he only actually uses the weaker statement that the mod $p$ Galois representation $T/pT$ is irreducible. I'd be interested to know if this weaker condition is indeed sufficient, and whether anything is known in this direction if the weaker condition doesn't hold (i.e. if $E$ admits a $p$-isogeny). -(EDIT: Added more detail and references.) - -REPLY [9 votes]: There are two issues. Let $H=H^1_{\mathrm{Iw}}(\mathbb{Q},T)$ where $T=V_{\mathbb{Z}_p}(f)(1)$ and $f$ is the modular form associated to the isogeny class of $E$. -(1) What is $T$ ? -$T$ will correspond to the lattice $\Lambda$ in $\mathbb{C}$ generated by all modular symbols. This is because $V_{2,\mathbb{Z}}(f)$ is the image of $V_{2,\mathbb{Z}} (Y_1(N)) = H_1 \bigl( X_1(N)(\mathbb{C}),\{\text{cusps}\},\mathbb{Z}\bigr)$ inside $H_1(E(\mathbb{C}),\mathbb{Z})$. The lattice $\Lambda$ will contain the lattice $\Lambda_0$ of the $X_0$-optimal (strong Weil) curve $E_0$, but it may be strictly larger if $E_0$ has rational torsion points, e.g. for 11a. Note also that $\Lambda$ need not be the lattice of an elliptic curve in the isogeny class, but rather $\tfrac{1}{2}\Lambda_E$ for some elliptic curve $E$. in case $E_0[2] \subset E_0(\mathbb{Q})$, e.g. for 17a. But can only happen for $p=2$ and for $p$ of additive reduction. So if $p$ is a prime of odd semi-stable reduction, then $T$ is the Tate-module of an elliptic curve $E_*$ which is an étale quotient of $E_0$. So let $E$ be this curve for the rest of the answer. -(2) Are Kato's elements integral in $H$ ? -There are two kinds of them. The $z_{\gamma}$ and the ${c,d}$ $z_{m}$. (sorry I don't seem to be able to produce indices before the symbol in MathJax) The latter are in $H$, see 8.1 of Kato, but they depend on the choices of $c$ and $d$. They are useful for bounding the Selmer group as, for a fixed $c$ and $d$ they form an Euler system. -The $z_{\gamma}$ instead is linked to the $p$-adic $L$-function and they are independent of the choices. They are obtained by dividing by $\mu(c,d)$, page 229 of Kato. So they need not be integral anymore. The appendix A in Delbourgo's book "Elliptic curves and big Galois representations" discusses this in detail. Kato shows that they are in $H\otimes \mathbb{Q}_p$ -Kato shows that $z_{\gamma}$ is integral if $H$ is a free $\Lambda$-module of rank 1, e.g. as shown in 12.4.(3) if $T/pT$ is irreducible. In fact it is not hard to show that $H$ is free also if $E(\mathbb{Q})[p]$ is trivial. -Now if the curve admits an isogeny of degree $p$, one can show that for all curves $A$ in the isogeny class $H^1_{\text{Iw}}(\mathbb{Q},T_p A)$ is a free $\Lambda$-module of rank $1$, except for at most a single one of them (I mean up to non-$p$-isogenies of course). This exception - if present - will always be the minimal curve $E_{\text{min}}$ in the class. Moreover if $E_{\text{min}}$ is does not have a free $H^1_{\text{Iw}}$, then there is an embedding of it into $\Lambda$ with image equal to the maximal ideal of $\Lambda$. One can now conclude from the interpolation property of the $p$-adic $L$-function that even if $E_*=E_{\text{min}}$ then $z_{\gamma}$ will be in $H$. -Using this one can prove that the $p$-adic $L$-function is integral and that $\mu\geq 0$, but it would not say anything about Greenberg's $\mu=0$ conjecture. Furthermore one gets a proof of the divisibility in the main conjecture as in 12.5.(3), if the $T/pT$ is reducible. However this conclusion can not be extended at present to all odd semi-stable primes, because there may be primes for which the Galois representation is not surjective, yet $T/pT$ is irreducible; because the Euler system method requires and element $\binom{1\ 1}{0\ 1}$ in the image of the Galois representation, see Hyp($K_{\infty},T)$ in Rubin's book. So the integrality won't help, yet. -In summary, all zeta-elements of Kato are integral with respect to $T$ in the case of an elliptic curve. For Kato's divisibility on the other hand, the surjectivity of the representation to $\mathrm{GL}_(\mathbb{Z}_p)$ or its reducibility is still needed. -(edits: quite a few in the whole answer above, now that I konw the full answer to the question)<|endoftext|> -TITLE: Strength of Transfinite Induction on the Difference Hierarchy -QUESTION [7 upvotes]: I'm wondering if a particular theory of second order arithmetic has been studied or is known to be equivalent to some other theory. -Consider the formulas generated by $\Pi^1_1$ and $\Sigma^1_1$ formulas by propositional combinations (the title refers to this, rather informally, as the difference hierarchy, since these formulas are essentially the difference between two $\Pi^1_1$ formulas, the difference between two such, and so on). Let's call this class of formulas $\Delta$, for convenience. -My question is what the strength (either reverse mathematical or proof theoretic) of $\Delta-TI_0$, the theory which allows transfinite induction along any well-ordering for $\Delta$ formulas, is. This has proof theoretic strength strictly greater than $\Pi^1_2-TI_0$ (Rathjen and Weierman's proof that $\Pi^1_2-TI_0$ implies well-foundedness of $\psi\Omega^{\Omega^n}$ goes through, and furthermore this theory can carry out the key induction internally), and strictly less than $\Pi^1_1-CA_0$ (the standard argument that $\Pi^1_1-CA_0$ proves the existence of a $\beta$-model of $\Pi^1_\infty-TI_0$ suffices). - -REPLY [3 votes]: I hope it's not too tacky to answer my own question now that I've had a few days and train rides to think about it. -The answer is that the proof theoretic ordinal of $\Delta-TI_0$ is the Howard-Bachmann ordinal (the ordinal of $\Pi^1_1-CA_0^-$, $\Pi^1_\infty-TI_0$, $ID_1$, and $KP\omega$). -The upper bound is easy to see ($\Delta-TI_0$ is a subtheory of $\Pi^1_\infty-TI_0$). (In fact, each instance of $\Delta-TI_0$ is provable in $\Pi^1_1-CA_0^-$, modulo some work to handle parameters in the formulas in $\Delta$.) -For the lower bound, consider the easy embedding of $ID_1$ into the language of second order arithmetic, in which arithmetic formulas map to arithmetic formulas and the least fixed point in $ID_1$ is mapped to a $\Pi^1_1$ formula. Then all formulas of $ID_1$ map to $\Delta$ formulas, so every axiom of $ID_1$ becomes a theorem of $\Delta-TI_0$ (the point here is that the only instances of induction axioms appearing in $ID_1$ are induction on $\Delta$ formulas). In particular, the proof of well-foundedness below the Howard-Bachmann ordinal can be carried out, unchanged, in $\Delta-TI_0$.<|endoftext|> -TITLE: When is the projective line the seminaive projective line? -QUESTION [5 upvotes]: Excuse the possible naivete of this question. Since reading a nice survey article by Daniel Biss a few years ago, I'm always worried about what $P^1(R)$ is, for a ring $R$. -So that I stop worrying, I'm looking for an answer to the following question: For what (commutative, of course) rings $R$ is it true that $P^1(R)$ is naturally identifiable with the set of pairs $(a,b) \in R^2$ such that $(a,b)$ equals the unit ideal, modulo the natural action of $R^\times$? - -REPLY [10 votes]: This is equivalent to the property that every invertible (=rank-1 projective) $R$-module generated by two elements is free. Examples: semilocal rings, unique factorization domains, finite products of such rings.<|endoftext|> -TITLE: Quantitative bounds for multivariate central limit theorem -QUESTION [7 upvotes]: Hi, -For the univariate central limit theorem, the Berry-Esseen theorem gives a quantitative bound on the rate of convergence of distributions to the Normal distribution under Kolmogorov distance: -http://en.wikipedia.org/wiki/Berry%E2%80%93Esseen_theorem -Are similar statements known for the multivariate version of the central limit theorem, that use some standard distance measure? -http://en.wikipedia.org/wiki/Central_limit_theorem#Multidimensional_central_limit_theorem -This question is a re-post from -https://math.stackexchange.com/questions/11596/quantitative-bounds-for-multivariate-central-limit-theorem -Thanks, - -REPLY [7 votes]: There is a bunch of such statements which can be obtained by Stein's method. -You might be interested in the paper "On the Rate of Convergence in the Multivariate CLT" by Gotze, which is specifically devoted to Berry-Esseen theorems in the multidimensional setting. Have a look also at the very recent book Normal Approximation by Stein's Method by Chen, Goldstein and Shao.<|endoftext|> -TITLE: Automorphisms of constant sheaves -QUESTION [9 upvotes]: Let E be a Grothendieck topos, such as the category of sheaves of sets on a topological space. Then there is a unique geometric morphism $(\Delta \dashv \Gamma)\colon E\to \mathrm{Set}$, where $\Delta\colon \mathrm{Set}\to E$ constructs constant sheaves and its right adjoint Γ takes global sections. If E is locally connected (i.e. Δ has a further left adjoint), then Δ is a cartesian closed functor, i.e. $\Delta(B^A)\cong \Delta B^{\Delta A}$ for sets A,B. -Now in any cartesian closed category, we can define the "object of isomorphisms" Iso(X,Y) between any objects X,Y, as an equalizer of a pair of maps $X^Y \times Y^X \rightrightarrows X^X \times Y^Y$. In particular, when X=Y, we have the object Aut(X) of automorphisms of X. Since inverse image functors preserve finite limits, if E is locally connected then $\Delta(\operatorname{Aut}(X))\cong \operatorname{Aut}(\Delta X)$ for any set X. -My question is twofold: - -Is there a noticeably weaker condition on E than local connectedness which ensures that Δ preserves objects of automorphisms? -Can you give an explicit example of a topos for which Δ does not preserve objects of automorphisms? - -REPLY [5 votes]: Here is an answer to the second question. Consider the topos of sheaves on the one-point compactification of $\mathbb{N}$. Then for any set A, a global section of the constant sheaf ΔA is a function $\mathbb{N}\to A$ which is eventually constant. In particular, a global section of ΔAut(X) is an eventually constant sequence of automorphisms of X. -On the other hand, since ΔX is also the coproduct of X copies of the terminal object, a map ΔX → ΔX is just an X-indexed family of eventually constant functions $f_x\colon \mathbb{N}\to X$. In particular, an automorphism of ΔX is such a family such that for every n, the function $x \mapsto f_x(n)$ is an automorphism of X. So the difference between a global section of ΔAut(X) and an automorphism of ΔX is that the former is an eventually constant sequence of automorphisms, whereas the latter is a sequence of automorphisms such that for each $x\in X$, the sequence of images of x is eventually constant. Evidently the latter is strictly more general if X is infinite, so we cannot have $\Delta Aut(X) \cong Aut(\Delta X)$.<|endoftext|> -TITLE: Completeness of the club filter without AC -QUESTION [10 upvotes]: Let $\kappa$ be a regular cardinal. If I understand correctly, the proof that the intersection of $<\kappa$ many club subsets of $\kappa$ is a club does not require AC. However, the proof that the club filter on $\kappa$ is $\kappa$-complete does ostensibly require AC because given a sequence of sets in the club filter, we need to pick out a sequence of clubs that they contain in order to show that their intersection is in the club filter. So I'd like to know, in the absence of AC, how strong is the assertion that the club filter on $\kappa$ is $\kappa$-complete? Does it imply $AC_{\kappa}$, or is it weaker? If weaker, then how much weaker? -Is there any well-known model of ZF without AC in which there is a proper class or better of regular cardinals $\kappa$ such that the club filter on $\kappa$ is $\kappa$-complete? - -REPLY [6 votes]: If $M$ is an inner model of a forcing extension $V[G]$ of a model $V$ of Choice via a partial order of cardinality $\kappa$ in $V$, then for every regular cardinal $\lambda$ of $V$ greater than $\kappa$, every club subset of $\lambda$ in $V[G]$ will contain one from $V$. So, in $M$, the club filter on $\lambda$ will be $\lambda$-complete for any such $\lambda$, as $V[G]$ can carry out the usual argument with the sets from $M$, and then shrink the result to a set in $V$. A similar situation holds for models of ZF which have set-forcing extensions satisfying Choice. There are many such models of each type in which AC fails, for instance the Feferman-Levy model, in which $\omega_{1}$ is singular. -Andres's answer is more interesting, of course.<|endoftext|> -TITLE: nowhere vanishing vector field on a manifold -QUESTION [9 upvotes]: I am wondering if there are necessary and sufficient conditions under which an one-dimensional subbundle of $TM$ has a nowhere vanishing vector field. -More precisely let $M$ be a compact smooth manifold. -a. When dose there exist a one-dimensional (smooth or continuous) subbundle $L\subset TM$? -b. If $L\subset TM$ is a continuous/smooth line subbundle of $TM$, does there exist a nowhere vanishing continuous/smooth section $X:M\to L$? If so, the euler characteristic of $L$ should be zero. -This is related to the partially hyperbolic system $f:M\to M$ and $TM=E^s\oplus E^c\oplus E^u$. I am curious if there is a 'center flow' if $\dim E^c=1$. - -To Ryan: Am I right to say the following about your answers: - -If there exists an 1-dimensional subbundle $L$ of $TM$, then $\chi(M)=0$. This is independent of the case whether $M$ is orientiable or not. -If $M$ is orientable, then there always exists an orientable 1-dimensional subbundle $L$ of $TM$. - -Another question is, when is an 1-dimensional subbundle $L\subset TM$ orientable? Is it sufficient to assume that $M$ is orientable? - -Thank you all. I did not formulate some questions properly. What I really mean is: -I. For a given line bundle $L\subset TM$, what is the obstruction for $L$ beging orientable? (or equivalently trivial according to Georges) -For example let $L_{\mathbb{C}}$ be a complex line bundle over a complex manifold $M$, if the top Chern class $c_1(L_{\mathbb{C}})$ does not vanish, then $L_{\mathbb{C}}$ can not be trivial. Is there some similar results in the real case? -II. Is there an example such that $M$ is orientable and has a non-orientable line bundle $L\subset TM$? - -REPLY [5 votes]: A bundle is orientable if and only if its first Stiefel-Whitney class is 0 (one can see the first Stiefel-Whitney class as the function $w_1: H_1(M)\rightarrow \mathbb{Z}_2$ which associate to a loop the sign of the determinant of the monodromy). -As mentionned by Ryan, if a line bundle is non-orientable then there is a two sheeted cover of $M$ which orients $L$ (the covering correspond exactly to the index two subgroup $ker(w_1)$) this implies that if an oriented bundle admits a 1-dimensionnal sub-bundle the its Euler class has to be $0$ (regardless if the bundle is the tangent bundle or not). -Finally one can easily see that $T(T^2)$ is the sum of two non-trivial line bundle: -The canonical line bundle $\gamma$ on $\mathbb{R}P^1$ is non trivial but $\gamma\oplus\gamma^*$ is (it is oriented, 2 dimensionnal and admit a section given by the trace map). Pulling back this bundle by the projection $T^2\rightarrow S^1$ you get a trivial bundle (hence the tangent bundle) on $T^2$ written as a sum of two non trivial line bundle ($w_1$ is non zero on each summand). - -REPLY [2 votes]: To construct a non orientable line bundle on the torus $\mathbb{T}^2$ you can proceed as follows: -Consider a planar Reeb foliation of the anulus (see this drawing) and then glue the boundary circles obtaining a one dimensional non singular foliation of the two dimensional torus. Notice that it cannot be orientable since in one direction, when you follow the holonomy, you return with the wrong orientation (notice also, as explained above, that if you glue two such foliations, i.e. you consider the double cover, you get an orientable foliation).<|endoftext|> -TITLE: trivial subbundles -QUESTION [5 upvotes]: Let X be a projective variety (say, irreducible) and E a vector bundle on X or rank r. Is it true that there exists a codimension 2 closed subset Z in X such that restriction of E(n) (for n large enough) to U = X - Z has a trivial sub-bundle of rank (r-1)? Is this written somewhere? What happens when X varies in a flat family over a base S? -UPD: corrected the question - an ample twist of E is supposed to have a trivial sub-bundle. - -REPLY [6 votes]: Stated like this it can't be true. Indeed if $X$ is normal then by Hartogs lemma each embedding $O_{X-Z} \to E_{|X-Z}$ (i.e. a section of $E_{|X-Z}$) should extend to $O_X \to E$. But a priori $E$ can be without global sections. For example $E = O(-1) \oplus O(-1)$ on $P^2$. Maybe you want to have a subbundle which is trivial up to a twist? -UPD. If a twist is allowed then the answer is yes. Indeed, after an appropriate twist we can assume that $E$ is generated by a vector space $V$ of global sections, that is a map $V\otimes O_X \to E$ is surjective. Let $E' = Ker(V\otimes O_X \to E)$. It is a vector bundle on $X$. Consider the Grassmannian $Gr(r-1,V)$ of vector subspaces of rank $r-1$. Let $U$ denote the tautological subbundle. On the product $X\times Gr(r-1,V)$ we have a composition -$$ -U\boxtimes O \to V\otimes O\boxtimes O \to O\boxtimes E. -$$ -Let $Y$ be its discriminant locus (the scheme of points where the rank is $\le r-2$). -The fiber of $Y$ over a point $x \in X$ consists of all $(r-1)$-dimensional subspaces which intersect with the subspace $E'_x \subset V$ of codimension $r$. A simple parameter count shows that it has codimension 2 in the Grassmannian. Since this is true for all $x \in X$ we conclude that $codim Y = 2$. Hence for generic point $u \in Gr(r-1,V)$ the fiber $Y_u \subset X$ also has codimension 2. But since the corresponding map $O^{r-1} \to E$ is an embedding out of $Y_u$, we just take $Z = Y_u$ and restrict the above map to $X - Z$.<|endoftext|> -TITLE: Why is there no Borel function mapping every countable set of reals outside itself? -QUESTION [19 upvotes]: A choice function maps every set (in its domain) to an element of itself. This question concerns existence of an anti-choice function defined on the family of countable sets of reals. In an answer to a question about uncountability proofs it was suggested that while Cantor's diagonal method furnishes a Borel function mapping each countable sequence $S$ of real numbers to a number not in the seqence, the same is not true for countable sets of reals. This seems surprising (but believable) to me and an important insight, if true. But why is it true? Put more provocatively (if this is fair, if not explain it as stated) - -Let $\mathcal{C}$ be the family of countable subsets of $\mathbb{R}$ .` Is it the case that for every Borel function $f:\mathcal{C} \rightarrow \mathbb{R}$ there is an $X \in \mathcal{C}$ such that $f(X) \in X$? - -REPLY [4 votes]: There is no known forcing-free proof of the following strong version. (Here $E_{cntble}$ is yet another name for the equivalence relation that others call $T_{2}$ or $=^{+}$.) -Theorem Suppose that $E$ is a countable Borel equivalence relation on the standard Borel space $Z$ and that $f: {\mathbb{R}}^{\mathbb{N}} \rightarrow Z$ is a Borel homomorphism from $E_{cntble}$ to $E$. There there exists $z \in Z$ such that for every $r \in \mathbb{R}$, there exists $x \in {\mathbb{R}}^{\mathbb{N}}$ with $r \in \{ x_{n} \mid n \in \mathbb{N}\}$ such that $f(x) \mathbin{E} z$. -On the other hand, the analogous statement is false for Borel homomorphisms from $E_{cntble}^{+}$ to $E_{cntble}$. This suggests that there does not exist an easy forcing proof of the general form of Friedman's theorem.<|endoftext|> -TITLE: Non-mathematician submitting to top maths journal? -QUESTION [22 upvotes]: I am an amateur mathematician, and I had an idea which I worked out a bit and sent to an expert. He urged me to write it up for publication. So I did, and put it on arXiv. There were a couple of rounds of constructive criticism, after which he tells me he thinks it ought to go a "top" journal (his phrase, and he went on to name two). -His opinion is that my outsider status will have no effect on the reviewing process and my paper will be taken seriously. I am pleased and quite flattered to hear this, and my inclination is to do as he suggests. But I have to say it sounds too good to be true. Does this match other people's experience? I understand it's rare enough for undergraduates to publish anywhere, and I am not even an undergraduate! Surely a "non-mathematician submitting to top mathematical journal" must instantly rank high on the crackpot scale. How often does this actually occur successfully these days? Any advice? - -REPLY [34 votes]: Think of it this way: any graduate student who submits his first result for publication faces a similar issue: his paper will land on a referee's desk, who has never heard the name of the author before, so according to your logic, the better the result the more it must smell of crackpottery to the referee. Yet, PhD students and young postdocs do get their results published and in fact, they sometimes receive more patience from the referee, when it comes to the writing style, say, than experienced mathematicians. -Fact is, that an expert in the respective discipline can recognise 99% of the crackpots within less than 10 seconds. If your write-up passes this 10 second test and looks like a serious attempt at a mathematical problem, then the referee will read it. Of course, he might be extra sceptical about actual mistakes and extra diligent, but that can only be good for you. -At any rate, the threshold to receiving a careful reading is not nearly as high as you think. Key to this is a well written abstract and introduction with clear and precise statements of the results and, ideally, a clear indication of where the main difficulties were and how you overcame them, i.e. where the actual novelty lies. That's "all" it takes (I am not claiming that it's easy, but obtaining the result in the first place was likely harder). -And one last word: receiving rejections is part of the business. A mathematician who cannot handle them will lead a miserable life. None of us likes rejections, but we all must contend with the possibility of getting one. If you think that you will not be able to digest a rejection, then send the paper to a less prestigious journal. After all, your mental well-being is more important than this kind of prestige, especially since you are not relying on getting things published. - -REPLY [11 votes]: First of all, congratulations for making it as far as you have. I have a similar story, in 1986 I met with Stephen Wolfram and discussed with him my research. Wolfram was very supportive and even offered to edit and publish my research if I would just write it up. Then Wolfram left academia, forcing me to reach out to others to publish my work. Having only taken a few college level math classes and not knowing TeX, I was unprepared for how much work it would take to write a good paper. My first problem was that I was over-reaching in both the breath and depth of what I was trying to communicate. So my main advice is that small is beautiful. There are mathematicians of significance who's dissertations had a strong impact on mathematics, but if you engage in "empire building" as a novice, people are likely not to take you seriously. -In general no one is going to know who you are. As my knowledge of mathematics and my writing ability have grown I can now submit articles for publication and the publishers now refer to me as Doctor unless I correct them. So there is no barrier to a non-professional publishing except being unable to write at a professional level. But it took me fifteen years to get to this place by myself. -Know what you want and what you are willing to pay for it. I ended up going back to school in mathematics just so I could understand math better and I'm glad I did. -Enjoy the journey. It is now twenty five years later and I am still unpublished, but that is ok because I'm enjoying the journey. - -Note: I recommend when publishing in research journals that not only does your research need to be solid, but it should discuss the relevance of other important research in the field of your publication. You are not just adding another block of knowledge, you cementing it into place in the edifice of mathematics. Educational journals might be more accepting of stand alone material.<|endoftext|> -TITLE: Is projectivity local on the base? -QUESTION [9 upvotes]: Let $f:X\to Y$ be a morphism of schemes and assume that $Y$ has an open cover $\{U_i\}$ such that $f:f^{-1}U_i\to U_i$ is projective. Does it follow that $f$ is projective? - -REPLY [14 votes]: Unfortunately no. -Here is an example which is a variant on Hironaka's example of a non-projective smooth proper variety. -Let $g:Z\to Y$ be a smooth projective morphism and assume that $\dim Z\geq 3$ and that there exists two smooth curves $C_1, C_2\subset Z$ such that $C_1$ and $C_2$ intersect in exactly two (closed) points, say $P$ and $Q$. Assume that $g(P)\neq g(Q)$. This situation is easy to create. -Now, let $Z_1$ be the blow up of $Z\setminus g^{-1}(g(P))$ along $C_1\cap Z\setminus g^{-1}(g(P))$ first and then along the strict transform of $C_2$ and let $Z_2$ be the blow up of $Z\setminus g^{-1}(g(Q))$ along $C_2\cap Z\setminus g^{-1}(g(Q))$ first and then along the strict transform of $C_1$. Since $C_1$ and $C_2$ only intersect in $\{P,Q\}$, it follows that $Z_1$ and $Z_2$ are isomorphic over the open set $Y\setminus \{g(P),g(Q)\}$. Let $X$ be the scheme obtained by gluing $Z_1$ and $Z_2$ along the obvious open subset and $f:X\to Y$ the induced morphism. -Finally let $U_1=Y\setminus \{g(P)\}$ and $U_2=Y\setminus \{g(Q)\}$. Then $f^{-1}U_i=Z_i$ and -$f$ restricted to $Z_i$ is the combination of the original $g$ and a blow up, hence projective. -However, $f$ is not projective. The proof of this goes the same way as Hironaka's: take the cycles corresponding to the fibers of the blow-ups. One obtains that one of the irreducible components of the fiber over $P$ is numerically equivalent to the union of the irreducible components over $Q$, but similarly one of the irreducible components of the fiber over $Q$ is numerically equivalent to the union of the irreducible components over $P$. This means that an $f$-nef line bundle on $X$ has to be zero (acting with the appropriate power of its Chern class) on the "other" irreducible component of each fibers, but then it cannot be ample. -EDIT: Note that this $f$ is not projective in either the EGA or the Hartshorne sense. If in doubt, assume that $Y$ is projective over a field and then one only needs to show that $X$ is not, which is then the same according to either definition. -EDIT2: added condition $\dim Z\geq 3$ and added missing $g(\ )$'s for the images of the points $P$ and $Q$ following Qing Liu's comment. - -REPLY [6 votes]: No, there should be at least some noetherian hypothesis on $Y$. Take for example $Y$ to be an infinite disjoint union, say indexed by natural numbers, of points $x_i = \mathrm{spec}k$, and take for $X\to Y$ over each point $x_i$ the $i$--dimensional projective space. Then $f$ is not projective although "locally projective" in your sense.<|endoftext|> -TITLE: Universal homeomorphisms and the étale topology -QUESTION [16 upvotes]: Let $f:X\to S$ be a universal homeomorphism of schemes. Assume $X(S')\neq\emptyset$ for some étale surjective $S'\to S$. Does $f$ have a section? -The answer is yes if $S$ is reduced, by descent. Indeed, note that if $S_1$ is a reduced $S$-scheme then $X(S_1)$ has at most one element. Apply this to $S_1=S'\times_S S'$. -Interesting special case: if $S$ has prime characteristic $p$, let $G$ be a finite locally free $S$-group scheme with connected (i.e. "infinitesimal") fibers, such as $\alpha_p$ or $\mu_p$. Is $H^1_{\mathrm{et}}(S,G)$ trivial? - -REPLY [6 votes]: I think that the following might work. Let $X_0$ be a reduced scheme over a field $k$ of characteristic $p > 0$, and let $X$ be the product of $X_0$ with the ring of dual numbers $k[\epsilon]$. Then $\mathcal O_X = \mathcal O_{X_0} \oplus \epsilon\mathcal O_{X_0}$, and the $p^{\rm th}$ roots of 1 are those of the form $1 + \epsilon f$; hence the Zariski sheaf of $p^{\rm th}$ roots on 1 on $X$ is isomorphic to $\mathcal O_{X_0}$. Hence if $\mathrm H^1(X_0, \mathcal O_{X_0}) ≠ 0$ there is a non-trivial $\mu_p$-torsor on $X$ that is locally trivial in the Zariski topology, thus giving a counterexample.<|endoftext|> -TITLE: How To Present Mathematics To Non-Mathematicians? -QUESTION [151 upvotes]: (Added an epilogue) -I started a job as a TA, and it requires me to take a five sessions workshop about better teaching in which we have to present a 10 minutes lecture (micro-teaching). -In the last session the two people in charge of the workshop said that we should be able to "explain our research field to most people, or at least those with some academic background, in about three minutes". I argued that it might be possible to give a general idea of a specific field in psychology, history, maybe some engineering, and other fields that deal with concepts most people hear on a daily basis. However, I continued, in mathematics I have to explain to another mathematician a good 30 minutes explanation what is a large cardinal. -I don't see how I can just tell someone "Dealing with very big sizes of infinity that you can't prove their existence within the usual system". Most people are only familiar with one notion of infinity, and the very few (usually physicists and electrical engineering students) that might know there are more than one - will start wondering why it's even interesting. One of the three people who gave a presentation that session, and came from the field of education, asked me what do I study in math. I answered the above, and he said "Okay, so you're trying to describe some absolute sense of reality." to which I simply said "No". -Anyway, after this long and heartbreaking story comes the actual question. I was asked to give my presentation next week. I said I will talk about "What is mathematics" because most people think it's just solving huge and complicated equations all day. I want to give a different (and correct) look on the field in 10 minutes (including some open discussion with the class), and the crowd is beginner grad students from all over the academy (physics, engineering of all kinds, biology, education, et cetera...) -I have absolutely no idea how to proceed from asking them what is math in their opinion, and then telling them that it's [probably] not that. Any suggestions or references? -Addendum: -The due date was this morning, after reading carefully the answers given here, discussing the topic with my office-mates and other colleagues, my advisor and several other mathematicians in my department I have decided to go with the Hilbert's Hotel example after giving a quick opening about the bad PR mathematicians get as people who solve complicated equations filled with integrals and whatnot. I had a class of about 30 people staring at me vacantly most of the 10 minutes, as much as I tried to get them to follow closely. The feedback (after the micro-teaching session the class and the instructors give feedback) was very positive and it seemed that I managed to get the idea through - that our "regular" (read: pre-math education) intuition doesn't apply very well when dealing with infinite things. -I'd like to thank everyone that wrote an answer, a comment or a comment to an answer. I read them all and considered every bit of information that was provided to me, in hope that this question will serve others in the future and that I will be able to take from it more the next time I am asked to explain something non-trivial to the layman. - -REPLY [2 votes]: I'm also always struggling with these kinds of questions and I have essentially two answers for it. -If I carry a pen and paper and I feel like I have the time to explain stuff and want to go the less ridiculous (See my second answer) I talk about graphs and Eulerian paths. There is a childrens riddle known all across Germany called the "Haus vom Nikolaus"(House of Santa Claus) and it is the question whether some graph looking like a house has an eulerian path and almost everyone knows at least one Eulerian path. I then ask whether you can change the endpoint and the starting point. Afterwards I ask them whether the "Doppelhaus vom Nikolaus"(The double house of Santa Claus) i.e. two houses sharing a wall has an Eulerian path and if the audience is really interested I prove with them that this does not have an Eulerian path. -Nevertheless since I'm a topologists this is not really satisfying as it does not really represent my research interest. So at some point I came up with the following: -I take off my belt and close it again in front of the audience and declare this the "standard belt". I then twist the belt once to produce a Möbius band and ask whether we can play around with this to deform it to the standard belt. After playing around with it they are usually deeming this impossible but of course as a mathematician this is not the full answer so a proof has to be presented. Usually with a bit of help they count the boundary components or the sides and see that this is in fact not the standard belt. If I feel like the audience likes these kind of arguments I would then twist the belt twice and ask what happens now. So far I have never encountered a good answer to this, but after a while I explain to them how it is not possible in $\mathbb{R}^3$ and that the reason lies in the invariance of the winding of the normal bundle of the belt (explained in non-mathematical terms) and after all this I close with the final statement, that if we would live in $4$-dimensions we could untwist the two times twisted belt. If feel like this is quite an honest representation of the problems I'm actually interested in and people get an idea what I'm dealing with everyday.<|endoftext|> -TITLE: Origin of "versal"? -QUESTION [6 upvotes]: Any number of constructions guarantee the existence of maps $f$ without -guaranteeing their uniqueness. Some time ago, I was introduced to the terminology "versal" for such a construction. -I wonder: is this widespread? Is the origin of the terminology known? - -REPLY [5 votes]: coined in the 60's from "universal" by René Thom to describe "unfoldings" of singularities: -An $r$-unfolding of a function $f: \mathbb{R}^n \to \mathbb{R}$ is a function $F: \mathbb{R}^{n+r} \to \mathbb{R}$ such that $F(x_1,..., x_n, 0,..., 0) = f(x_1,..., x_n)$. An $r$-unfolding of $f$ is versal if all other unfoldings of $f$ can be induced from it. It is universal if $r$ is the smallest dimension for which a versal $r$-unfolding of $f$ exists. -Further comment (due credit to John Mather): -Going back to Thom's complete works, I came upon the following unpublished comment: -« En mathématique pure, mes propres résultats n'allèrent guère au-delà de développements limités de certaines singularités de potentiel. Il fallut la pertinence de mathématiciens américains (Milnor) ou européens (théorie du déploiement universel, Grauert, J. Martinet) pour sortir la théorie de son marasme initial. Mon seul apport à la théorie mathématique fut d'introduire la notion de « déploiement universel » - corrigé peu après en versel par les collègues algébristes (Mather). Il n'y a pas de doute que des mathématiciens américains (Mather,Milnor), puis soviétiques (Arnold) ont apporté à la théorie des singularités des progrès décisifs. La vision de ces mathématiciens m'a fait comprendre combien la théorie des singularités a des origines profondes en mathématiques. C'est la rencontre de mathématiciens soviétiques comme Arnold (souvent férocement critique de mes procédés rustres) qui m'a fait comprendre à quel point la théorie des singularités tire son origine de structures profondes (Polynômes de Dynkin, carquois de Gabriel, théorie des tresses, immeubles de Tits). L'intérêt de la T.C. est bien d'avoir attiré l'attention sur ces théories « profondes » dont la source reste (pour moi) bien mystérieuse.» -So, when Thom did introduce the terminology "d\'eploiement universel" (universal unfolding), John Mather is, according to Thom himself, responsible for the (relevant) alteration to "versal".<|endoftext|> -TITLE: Does a log-concave function on a convex set extend continuously to the boundary? -QUESTION [5 upvotes]: Let $U$ be an open convex set in a locally convex space $X$, and let $f : U \to [0,1]$ be a log-concave function on $U$ (i.e., bounded and real-valued). Under what conditions does $f$ have a continuous extension to the closure $\overline U$? - -REPLY [3 votes]: I do not see how $\log$-concavity should imply any form of continuity. For instance, if $\|\cdot\|$ is any semi-norm on the locally convex space $X$ then $f(x) = e^{-\|x\|}$ will be bounded and $\log$-concave but it will only be continuous if the semi-norm is continuous. -What you really want in order to be able to extend $f$ to the boundary of $U$ is uniform continuity and for that the fact that $f$ is $\log$-concave may help but is certainly not sufficient.<|endoftext|> -TITLE: When is the function of a median closer to the median of the function than the mean of the function is to the function of the mean? -QUESTION [18 upvotes]: Background -notation: RV= random variable, $\mu=$ mean $m=$ median -Jensen's Inequality considers the relationship between the mean of a function of an RV and the function of the mean of an RV. -If $f(x)$ strictly convex: -$$\mu (f(x)) > f(\mu (x))\mathrm{\hspace{20mm}(1)}$$ -Conversely if $-f(x)$ is strictly convex: -$$\mu (f(x)) < f(\mu (x))$$ -An analogous property of the median has been presented (Merkle et al 2005, pdf). -Motivation -I have a nonlinear function (pdf) of positive random variables, too complex to post here, not directly pertinent to this question; I am looking for a more general answer. It is worth noting that it is, however, neither strictly concave nor convex. -In practice, I find that the function of the medians provides a much better estimate of the median of the function than does the estimate of the mean of the function from the function of the means. I am interested in learning the conditions for which this is true. -Question -Under what conditions will the function of a median be closer to the median of a function than the mean of a function is to a function of the mean? -Specifically for what types of $f(x)$ and $x$ is -$$|\mu (f(x)) - f(\mu (x))| > |m (f(x)) - f(m (x))|$$ -I previously asked this on here on stats.stackexchange.com, but after not receiving an answer, was advised to post here on MO. -References -Merkle et al 2005 Jensen's inequality for medians. Statistics & Probability Letters, Volume 71, Issue 3, 1 March 2005, Pages 277-281 - -REPLY [4 votes]: If $f\ $ is a monotonic function then the median value of $f(X)$ is the same as the function applied to the median value of $X$. - -If there is no monotonicity (or approximate monotonicity) you wouldn't expect them to be even close: just think of $X$ being uniformly distributed on the interval and $f(x)=|x-1/2|$ say. - -On the other hand, a reasonable condition for the mean of $f(X)$ to lie close to $f$ applied to the mean of $X$ is for $f\ $ to be close to a linear function (the same example as above shows the mean of the function being very far from the function of the mean). Of course the mean is very sensitive to extreme values of $f(X)$ whereas the median is not - -In general for monotone or nearly monotone $f$, you can expect better closeness for the medians than the means. For seriously non-monotone $f$, I don't think there's anything useful you can say. - -By the way, you didn't ask this, but in case you are interested in the sample mean and sample median of $f(X_1),\ldots,f(X_n)$ as compared to the true mean and median of $f(X)$, the first typically errs by about std dev$(f(X))/\sqrt n$, whereas the latter errs by about $(1/\sqrt{8n})/\rho(m(f(X)))$ where $\rho$ is the density function of $f(X)$ (which I'm assuming exists). This means you can do a numerical test to see which of these will be closer.<|endoftext|> -TITLE: Is there such a thing as the sigma-completion of a Boolean algebra? -QUESTION [7 upvotes]: Hi all, -Suppose that $\mathcal{B}$ is a Boolean algebra. It there a way to extend $\mathcal{B}$ to a smallest Boolean algebra $\mathcal{B}'$ that contains an isomorphic copy of $\mathcal{B}$ and is countably complete, i.e. every countable subset of $\mathcal{B}'$ has a least upper bound in $\mathcal{B}'$? By "smallest" I mean that the inclusion $i: \mathcal{B} \hookrightarrow \mathcal{B}'$ has the obvious universal property, i.e. for every homomorphism $f$ from $\mathcal{B}$ to a countably complete Boolean algebra $\mathcal{C}$ there exists a unique homomorphism $g: \mathcal{B}' \to \mathcal{C}$ such that $g \circ i = f$ (it would be nice if $g$ turned out to commute with countable sups too). If no such $\mathcal{B}'$ exists, is there some other useful definition of "smallest" countably complete Boolean algebra containing $\mathcal{B}$? -If it makes any difference, I'm mostly interested in the special case where $\mathcal{B}$ is a direct limit of a sequence of finite Boolean algebras. -Edit: Thanks very much for the replies, it's a shame I can only mark one as the answer. It will take me a while to absorb the various references I've been given, so if I run into difficulty I'll bump the thread with an edit. -Edit 2: Bumping with followup question, please see my answer below. - -REPLY [7 votes]: In the course of a forthcoming research project with Asgar Jamneshan (in ergodic theory), we managed to discover a rather explicit answer to this question, which I am recording here if anyone is interested. -Let ${\mathcal B}$ be a Boolean algebra, let $\mathrm{Stone}({\mathcal B})$ be its Stone space (the space of Boolean homomorphisms $\alpha$ from ${\mathcal B}$ to $\{0,1\}$), let $\Sigma$ be the Baire $\sigma$-algebra of $\mathrm{Stone}({\mathcal B})$. Then there is a natural inclusion $\iota \colon {\mathcal B} \to \Sigma$ defined by sending each $E \in {\mathcal B}$ to the clopen set $\{ \alpha \in \mathrm{Stone}({\mathcal B}): \alpha(E) = 1 \}$. This is a Boolean homomorphism which is universal in the sense that for every Boolean homomorphism $f \colon {\mathcal B} \to {\mathcal X}$ into a $\sigma$-complete Boolean algebra ${\mathcal X}$ there is a unique lift $\tilde f: \Sigma \to {\mathcal X}$ such that $f = \tilde f \circ \iota$. - -Proof of uniqueness: the clopen sets in the Stone space $\mathrm{Stone}({\mathcal B})$ separate points, hence by Stone-Weierstrass the Baire algebra is generated by the clopen sets, hence $\Sigma$ is generated by $\iota({\mathcal B})$, giving uniqueness. -Proof of existence: Stone duality gives a continuous map from $\mathrm{Stone}({\mathcal X})$ to $\mathrm{Stone}({\mathcal B})$, which pulls back Baire sets to Baire sets, thus maps $\Sigma$ to Baire sets in $\mathrm{Stone}({\mathcal X})$, each one of which can be associated to an element of ${\mathcal X}$ by the Loomis-Sikorski theorem (every Baire set in $\mathrm{Stone}({\mathcal X})$ is equivalent up to meager sets to a unique clopen set). One easily verifies that this gives a map $\tilde f: \Sigma \to {\mathcal X}$ with the required properties. - -[EDIT: a previous version erronously assumed that the pullback of a meager set was meager, leading to an incorrect conclusion. Now corrected.]<|endoftext|> -TITLE: extension of $G$-bundles -QUESTION [12 upvotes]: Let $S$ be a smooth surface (let's say over an algebraically closed field) and let $D$ be a smooth divisor in $S$. Let also $G$ be a connected algebraic group. Assume that we are given a principal $G$-bundle ${\mathcal F}$ on $S\backslash D$. Under what conditions can we extend it to all of $S$? Do I understand correctly, that this is always the case when the derived -group $[G,G]$ is simply connected? - -REPLY [7 votes]: Nick Shepherd-Barron once asked me this question, and I think I can remember what was eventually concluded. -The short answer to the original question is negative for the additive group $\mathbb{G_a}$. Map $SL_2(\mathbb{C})$ to $X=\mathbb{C}^2-(0,0)$ by letting a matrix act on the vector $(1,0)^T$. This realizes the group as a torsor over $X$ for the group of unipotent matrices -\[ -\begin{pmatrix} 1 & a \\\\ 0& 1 \end{pmatrix}. -\] -We can trivialize it over the vectors $(v_1,v_2)^T$ with $v_1\neq 0$ with the section -\[ -\begin{pmatrix} v_1 & 0 \\\\ v_2& v_1^{-1} \end{pmatrix},\] -while it trivializes on the set $v_2\neq 0$ via the section -\[\begin{pmatrix} v_1 & -v_2^{-1} \\\\ v_2& 0 \end{pmatrix}.\] -The transition function on the overlap is easliy computed to be -\[\begin{pmatrix} 1 & (v_1v_2)^{-1}\\\\ 0& 1 \end{pmatrix}.\] -This represents the standard non-trivial generator of $H^1(O_X)$, and hence, the bundle is non-trivial. On the other hand, if you could extend it to $\mathbb{C}^2$, it would trivialize, since there are no non-trivial $\mathbb{G}_a$-bundles on an affine variety. -There is indeed a correspondence between principal bundles and tensor functors from representations to vector bundles. But if I recall correctly, the functor is required to be exact. In the case at hand, the extension of vector bundles is the direct image with respect to the inclusion $X\hookrightarrow \mathbb{C}^2$, which fails this. - -Added: OK, I see this was just an elaborate way to say: take any principal $\mathbb{G}_a$-bundle corresponding to a non-zero element of $H^1(O_X)$. Note, anyways, that the derived group is trivial in this example. Certainly the statement is false for general connected groups, contrary to some of the comments. - -Added, 25, November, 2011: -This question came back to me today while I was thinking about something unrelated. It occurred to me then to point out that for the example above, if we work in the analytic category, we have -$$H^1(X, \mathbb{G}_a)\simeq H^1(X, \mathbb{G}_m),$$ -via the exponential sequence. On the other hand, -$$H^1(\mathbb{C}^2, \mathbb{G}_a)=H^1(\mathbb{C}^2, \mathbb{G}_m)=0.$$ -So the desired extension property is false on analytic spaces even for reductive structure groups.<|endoftext|> -TITLE: Quantum cohomology rings as invariants -QUESTION [7 upvotes]: Let $X$ and $Y$ manifolds. What kind of relations between them (like homeomorphism, diffeomorphism, homotopy equivalence) gives an isomorphic quantum cohomology rings? - -REPLY [2 votes]: I am not an expert in the field, but I guess that there are examples of diffeomorphic symplectic manifolds with non-isomorphic quantum cohomologies (hence homotopy equivalence, homeomorphism, or even diffeomorphism alone is not enough to give an isomorphism on quantum cohomology). -Two simply connected smooth 4-manifolds having the same intersection form are homotopy equivalent by a theorem of Milnor. More than that, they are homeomorphic by a famous theorem of Freedman. However, very often these manifolds are not diffeomorphic. An example of homeomorphic but non-diffeomorphic pair is the Barlow surface and the blow-up of the complex plane in 8 points. Note that both manifolds admit a Kahler structure. -In the book of McDuff - Salamon J-holomorphic curves and quantum cohomology, they describe in Example 7.3.6 a construction due to Ruan of two diffeomorphic non-deformation equivalent 6-manifolds. The manifolds are the two 4-manifolds just described, both multiplied by $\mathbb{CP}^1$ (by a result of Wall, these two homeomorphic 4-manifolds become diffeomorphic after such a stabilization). -Ruan uses quantum cohmology to prove that these two diffeomorphic symplectic 6-manifolds are not deformation-equivalent, so my guess is that he uses that their quantum cohomologies are not isomorphic as a tool.<|endoftext|> -TITLE: Killing vectors and Ricci Tensor -QUESTION [5 upvotes]: Hi all, -We all know that the lie derivative of the metric tensor along a Killing Vector vanishes, by definition. I am trying to show that the Lie derivative of the Ricci tensor along a Killing vector also vanishes, and I am hoping to interpret it physically. -What might be a good direction to proceed? Thanks! - -REPLY [8 votes]: The Lie derivative of any tensor field $K$ with respect to a vector field $\xi$ is by definition -$$L_\xi(K)=-\frac d{dt}|_{t=0}(\phi_t)_*K,$$ -where $\phi_t$ is the local flow of $\xi$. Now, if $M$ has a Riemannian metric $g$ and $\xi$ is Killing with respect to $g$, each $\phi_t$ is a local isometry of $(M,g)$. From the uniqueness of the Levi-Civita connection, it follows that every isometry $\phi$ is affine, i.e. -$$\phi_*(\nabla_X Y)=\nabla_{\phi_*X}\phi_*Y.$$ -From here you get immediately $\phi_*R=R$ for the Riemannian curvature, and since the Ricci tensor of $R$ is obtained by a trace: -$$Ric(X,Y)=trace(V\mapsto R_{V,X}Y),$$ -one gets $\phi_*Ric=Ric$ for every isometry $\phi$. The first formula thus shows that $L_\xi Ric=0$ for every Killing vector field $\xi$. - -Edit: Here is another, purely tensorial, proof of the same statement. Let $\xi$ be Killing, in the sense that $g(\nabla_X\xi,Y)+g(X,\nabla_Y\xi)=0$ for all vector fields $X,Y$. After taking the covariant derivative wrt some vector field $Z$, and doing some standard manipulations, one gets the usual Kostant formula: -$$\nabla^2_{X,Y}\xi=R_{\xi,X}Y,\qquad\forall X,Y\in C^\infty(TM).$$ -This is just a rewriting of -$$L_\xi(\nabla_XY)=\nabla_{L_\xi X}Y+\nabla_X(L_\xi Y),$$ -i.e. some sort of Leibniz formula. -Applying this formula several times eventually yields the corresponding Leibniz formula for $R$: -$$L_\xi(R_{X,Y}Z)=R_{L_\xi X,Y}Z+R_{X,L_\xi Y}Z+R_{X,Y}(L_\xi Z),$$ -i.e. $L_\xi R=0$, and finally $L_\xi Ric=0$ after taking the trace. -Of course, this is just the infinitesimal version of the first proof... - -REPLY [5 votes]: Recall that the definition of the Lie derivative of a tensor field $T$ with respect to a vector field $X$ is given by "dragging" $T$ with respect to the one-parameter (quasi) group $\phi_t$ generated by $X$, i.e., computing $\phi_t^*(T)$, and differentiating wrt $t$ at $t = 0$. But to say that $X$ is a Killing field means that the $\phi_t$ are (partial) isometries, and so not only preserve the metric tensor but also the Riemann curvature tensor and its contraction the Ricci tensor or any other tensor field that is defined canonically from the metric tensor and so preserved by isometries. Thus any such tensor field is preserved by dragging, i.e., $\phi_t^*(T)$ is constant in $t$ and so has a zero derivative. -Regarding the physical interpretation, let me try to answer a slightly different question. Recall that the Ricci tensor comes up as the Euler-Lagrange expression for the Einstein-Hilbert functional, and that the latter is invariant under the group of ALL diffeomorphisms. So it is natural to ask what the Noether Theorem (connecting one-parameter groups that preserve a Lagrangian to constants of the motion of the corresponding Euler-Lagrange equations) leads to in this case. The answer is that it gives the contracted Bianchi identity for the Ricci tensor. Perhaps this is what your question about physical significance was aiming at.<|endoftext|> -TITLE: When is a local Artin C-algebra a subring of C[t]/t^n -QUESTION [15 upvotes]: Let $A$ be a local ring over $\mathbb{C}$, which moreover is a finite dimensional $\mathbb{C}$-vector space. - - -When is $A$ a subring of $\mathbb{C}[t]/t^n$? -What does the minimal such $n$ have to do with $A$? - - -Example: the ring $\mathbb{C}[x,y]/(x^3,y^2,xy)$ is a subring of $\mathbb{C}[t]/t^5$ by $x = t^2$ and $y = t^3$. - -REPLY [6 votes]: OK, this is not an elegant solution, but at least something... -So, let $A$ be a local $k$-algebra where $k$ is a field. I don't think that $k=\mathbb C$ makes a difference (and $k$ is easier to type than $\mathbb C$). -Anyway, the finite dimensionality condition implies that $A$ is Noetherian (in fact Artinian as the title says), so in fact it is a finitely generated $k$-algebra, in other words a surjective image of $k[x_1,\dots,x_p]$ for some $p\in \mathbb N$. -Let $I=\ker \left( k[x_1,\dots,x_p]\to A \right)$, i.e., $A\simeq k[x_1,\dots,x_p]/I$. -Now let $S\subset \mathbb N^p$ be the set of $p$-tuples $(\alpha_1,\dots,\alpha_p)$ for which $\prod x_i^{\alpha_i}\in I$, but $\prod x_i^{\alpha_i-\delta_{ij}}\not\in I$ for any $j\in\{1,\dots,p\}$ (where $\delta_{ii}=1$ and $\delta_{ij}=0$ for $i\neq j$). -Let us assume that there exists a homomorphism $\phi: k[x_1,\dots,x_p]\to k[t]/(t^n)$ such that $I=\ker \phi$. (Clearly this is equivalent to the existence of a homomorphism $A\hookrightarrow k[t]/(t^n)$.) -Let $r_i\in\mathbb N$ be such that $\phi(x_i)=u_it^{r_i}$ for some unit $u_i\in k[t]/(t^n)$. -Then it is necessary that the following inequalities are satisfied: -For any $(\alpha_1,\dots,\alpha_p)\in S$ and $j\in\{1,\dots,p\}$ , we must have -$$ n\leq \sum_i \alpha_i r_i < n + r_j $$ -This certainly gives a lot of restrictions. For example this shows that the ring -$$ k[x,y]/(x^my, xy^m, x^{2m}, y^{2m})$$ where $m\geq 2$ -cannot be embedded into a ring of the type $k[t]/(t^n)$. (The problem is that the first two terms imply that $mr_x+r_y\geq n$ and $r_x+mr_y\geq n$ has to hold, but this implies that then $(m+1)(r_x+r_y)\geq 2n$ which implies that then either $(m+1)r_x\geq n$ or $(m+1)r_y\geq n$ but then this contradicts $(2m-1)r_x< n$ (or the same with $r_y$) as soon as $m+1\leq 2m -1$, that is as soon as $m\geq 2$. -It seems to me that it is also a sufficient condition if the above system of inequalities has an integer solution $(r_1,\dots,r_p)$. I am not saying that it is necessarily easy to check, but at least with few variables and few relations it may not be so bad. If you want a more conceptual condition, I suppose one could associate a toric variety to the set of $S$ and there might be a nice condition of that variety that gives a condition. It will probably not be easier to check, but it might be interesting. Then again, you might not want to spend too much time on this. I think I already did. :) -EDIT: Corrected argument following Hailong's comment. -Remark: This argument is secretly about valuations...<|endoftext|> -TITLE: What's up with Wick's theorem? -QUESTION [24 upvotes]: Sorry about the dumb title. -I'd like to understand Wick's theorem. More specifically, I have seen it pop up in several different contexts and I am really puzzled by the different statements of it that I have seen. My own background/interest is in moduli of curves, if that helps. -The first version, that is also the only one I have seen more than one time, is in the context of infinite wedge space. Here Wick's theorem is a formula about how to decompose any product of the fermionic operators $\psi_k$ and their adjoints $\psi_k^\ast$ as a sum over normally ordered products. This is for instance how it is explained in Kac-Raina. -A second version is in "Graphs on surfaces" by Lando and Zvonkin as Theorem 3.2.5. Here it is a statement about how to integrate a polynomial against a Gaussian measure on the real line. If $\langle f \rangle$ denotes the integral $\frac{1}{\sqrt{2\pi}}\int_{\mathbb R} f(x) \exp(-x^2/2) dx$, then Wick's theorem states that if f is a product $f = f_1 f_2 \cdots f_{2k}$ of linear polynomials, then $\langle f \rangle$ can be written as an explicit sum of products of pairs $\langle f_if_j\rangle$. -Now I can somehow believe that the two theorems above are talking about the same thing or that the second is a special case of the first. But what really got me scratching my head was the following statement from page 2 of Getzler & Kapranov's paper on modular operads (sorry for the lengthy quote): -[...] -As a model for this calculation, take the formula for the enumeration of graphs known in mathematical physics as Wick’s Theorem. Consider the asymptotic expansion of the integral $W(\xi,\hbar) = \log \int \exp \frac 1 \hbar \left( x \xi - \frac{x^2}{2} + \sum_{2g-2+n > 0} \frac{a_{g,n}\hbar^g x^n}{n!}\right) \frac{dx}{\sqrt{2\pi\hbar}}$ -considered as a power series in $\xi$ and $\hbar$. (The asymptotic expansion is independent of the domain of integration, provided it contains 0.) Let $\Gamma((g,n))$ be the set of isomorphism classes of connected graphs $G$, with a map $g$ from the vertices Vert(G) of $G$ to $\{0,1,2,...\}$ and having exactly $n$ legs numbered from 1 to $n$, such that -$g = b_1 + \sum_{v\in \mathrm{Vert}(G)} g(v)$ where $b_1$ is the first Betti number of the graph. If $v$ is a vertex of $G$, denote by $n(g)$ its valence, and let |Aut(G)| be the cardinality of the automorphism group of $G$. Wick’s Theorem states that -$W \sim \frac 1 \hbar \left(\frac{\xi^2}{2} + \sum_{2g-2+n>0} \frac{\hbar^g\xi^n}{n!} \sum_{G\in \Gamma((g,n))} \frac{1}{|\mathrm{Aut}(G)|} \prod_{v\in \mathrm{Vert}(G)} a_{g(v),n(v)}\right)$. -I also heard a version of Wick's theorem at a talk of Rahul Pandharipandhe about two months ago, which I will not be able to state correctly here since I can't really make sense of my notes. In that version of Wick's theorem one studied an $n$-fold product of a variety with itself by interpreting it as a configuration space of $n$ "particles" moving on the variety. The goal was to simplify certain complicated products of cohomology classes given by diagonals (= particles coinciding) and Chern classes of the tangent/cotangent bundle at one of the "particles". This was all done pictorially, and one represented the diagonals as a line connecting the two points, which at least shows some connection with the Wick formalism since I think I have at one point seen these lines between particles also in the context of Feynman diagrams. -Can someone give a hint about how these Wick theorems fit together? - -REPLY [21 votes]: Let's take for granted the Gaussian integration formula, which holds for both bosonic and fermionic integrals, if they are properly interpreted: -Theoreom (Gauss, Wick): Let $X$ be a vector space with a chosen volume form ${\rm d}x$, $f: X \to \mathbb C$ a polynomial, and $a: X^{\vee 2} \to \mathbb C$ a symmetric bilinear form with inverse $a^{-1}\in X^{\vee 2}$ such that the Gaussian measure $\exp(-\frac12 a\cdot x^{\otimes 2}){\rm d}x$ is defined. (So for example $X$ can be bosonic finite-dimensional and $a$ can have positive-definite real part; or $a$ can be invertible pure-imaginary and all integrals can be taken as conditionally convergent; or $X$ can have even-dimensional fermionic parts and the integral can be defined a la Berezin.) Then we have -$$ \int_X \sum f^{(n)} \cdot \frac{x^{\otimes n}}{n!} \exp \left(-\frac12 a\cdot x^{\otimes 2}\right){\rm d}x = \sqrt{\det(2\pi a)} \sum f^{(2k)} \cdot \frac{(a^{-1})^{\otimes 2k}}{2^kk!} $$ -Or, anyway, when $X$ is finite-dimensional Bosonic this is correct. In the fermionic case, it's off by some $\sqrt{-2\pi}$s, and $\det$ is Berezin's superdeterminant. Here $f^{(n)} : X^{\vee n} \to \mathbb C$ is the $n$th Taylor coefficient of $f$ at $0$; it's a symmetric tensor. In the fermionic case, some care must be taken with words like "symmetric", but I will ignore this subtlety. -Proof: Integrate by parts. $\Box$ -Now the trick is to interpret the RHS combinatorially: you should draw each summand as a graph with one vertex, labeled $f^{(2k)}$, and $k$ self-loops, labeled $a^{-1}$; then if you count automorphisms correctly, the denominator of the summand is the number of automorphisms of the graph, and the numerator is the "evaluation" of the graph as a picture of tensor contractions. You can also draw the left hand summands combinatorially: a vertex with $n$ incoming strands corresponds to $f^{(n)}$, and the $n!$ counts the symmetries. -What would have happened if you had not just a single polynomial but a product? You can draw $\frac1{m!} f^{(m)} \otimes \frac1{n!} g^{(n)}$ as two vertices, one labeled $f$ with $m$ incoming strands and the other labeled $g$ with $n$ incoming strands, and the symmetries are correct, and using $\frac1{m!} f^{(m)} \otimes \frac1{n!} g^{(n)} = \binom{m+n}{m,n} (f^{(n)}\otimes g^{(m)})$ and the above theorem, the RHS now can be taken as a sum over all graphs (possibly disconnected) with two vertices, one labeled $f$ and the other labeled $g$, where the edges are still valued $a^{-1}$ and a labeled graph is weighted by its automorphism group. -Ok, so now let's try to calculate asymptotics of integrals, and I will henceforth ignore the "determinant" prefactors. My domain of integration will always be a "small" neighborhood of $0$. I'm interested in the $\hbar \to 0$ asymptotics of -$$ \int_X \exp \frac1\hbar\left( - a\cdot \frac{x^{\otimes 2}}2 + \sum_{n\geq 3} b^{(n)}\cdot \frac{x^{\otimes n}}{n!} \right) {\rm d}x $$ -First rescale $x \mapsto \sqrt\hbar x$; this just rescales the overall integral by $\hbar^{\dim X/2}$, and I'm dropping those terms. Then the integral is $\exp\bigl( - a\cdot \frac{x^{\otimes 2}}2 + O(\hbar) \bigr)$, so keep the $O(1)$ part in the exponent but expand the $O(\hbar)$ part out: -$$ = \int_X {\rm d}x \exp \left(-\frac12 a\cdot x^{\otimes 2}\right) \times \sum_{m\geq 0} \frac1{m!} \left( \sum_{n\geq 3} b^{(n)}\cdot \frac{x^{\otimes n}}{n!} \right)^m$$ -Expanding the sum further, the $b$s still look like vertices with $n\geq 3$ incoming strands (and $n!$ symmetries), but now I get to have $m$ many of them, weighted by the $m!$ symmetries from permuting the vertices. So the sum is over all collections of trivalent-and-higher vertices, counted with symmetry, and an $n$-valent vertex is valued $\hbar^{\frac n 2 - 1}b^{(n)}$. -Then we integrate by connecting them up. All together, we get: -$$ = \sum_{\text{graphs } \Gamma } \frac{ \hbar^{\#} \operatorname{ev} (\Gamma) }{ \lvert \operatorname{Aut} \Gamma \rvert } $$ -Graphs can be disconnected, empty, have parallel edges and self loops, etc. We compute $\operatorname{ev}(\Gamma)$ by assigning $a^{-1}$ to each edge, $b^{(n)}$ to an $n$-valent vertex, and contracting tensors. The power on $\hbar$ is $-1$ for each vertex, and $\frac12$ for each half-edge (each part of an edge that arrives at a vertex), i.e. it's $-1$ for each vertex, $+1$ for each edge, i.e. it's the negative of the Euler characteristic of the graph. -Finally, let me get to the version from Getzler-Kapranov. Above, I used the fact that if $\star$ is a sum of connected things (counted with symmetry), then $\exp(\star)$ is a sum (counted with symmetry) over all possible collections of disjoint copies of $\star$. We've ended up with a sum-with-symmetry over disjoint things. Taking $\log$ gives the sum of connected things. For a connect graph, the negative Euler characteristic is precisely one less than the first Betti number. -Exercise: Redo the above calculations with $O(\hbar)$ corrections to the exponent of the initial integral, to end up with the precise Getzler-Kapranov result. - -Finally, I should say one more thing about Feynman diagrams. Feynman and Dyson disagreed about the meaning of Feynman's diagrams: Feynman thought of them as pictures of particles interacting, and Dyson thought of them they way I do above: as graphs computing the asymptotics of integrals. -The point is the following. The most important integrals like those above that physicists care about come from particularly nice quantum field theories, where $X$ is a space of sections of some vector bundle (with some extra structure) on a Riemannian manifold, and $a$ is the Laplacian, and the $b^{(n)}$ are all "local". In this situation, $a^{-1}$ computes the "heat flow" for the Riemannian manifold, which is nothing more nor less than an integral over all paths connecting two endpoints of some "amplitude for a free particle to travel along this path". Then $\operatorname{ev}(\Gamma)$ can be interpreted as an integral over all embeddings of $\Gamma$ into your manifold of: the amplitude for your particle to travel along the edges, times an amplitude for an "interaction" at the vertices. -One good reference is the first chapter or two of Costello's recent book on QFT.<|endoftext|> -TITLE: how to think of monodromy transformations -QUESTION [9 upvotes]: I've come across the notion of Monodromy transformations while reading some aspects of variations of Hodge structures in context of Classical Mirror symmetry. I am having difficulty in grasping the concept of Monodromy transformations, probably due to lack of any good reference. My question is : -How to think of monodromy transformations, what are these intuitively and what are some examples of monodromy. -Please suggest any good references for Variations of Hodge structures and monodromy. - -REPLY [19 votes]: A local system is a sheaf of finite dimensional vector spaces that is locally isomorphic to the constant sheaf $k^n$. If $\gamma: [0,1] \to X$ is a continuous path in $X$, $\gamma^{-1}(L)$ is again local system on $[0,1]$ but one shows that any locally constant sheaf on $[0,1]$ is actually constant. So the fibers at 0 and 1 are canonically identified. This means that we have a map -$$ - \gamma_{*} : L_{\gamma(0)} \to L_{\gamma(1)} -$$ -Moreover this is - -linear: $\gamma_* (v + \lambda w) = \gamma_* (v) + \lambda \gamma_* (w)$ -invariant by homotopy: if $\gamma \sim \gamma'$, $\gamma_*v = \gamma'_*v$. -compatible with composition of homotopy classes of paths: $(\gamma')_*(\gamma_*x) = (\gamma'\gamma)_*x$. - -So to any local system $L$ corresponds a representation $\pi_1(X,x) \to GL(L_x)$ of the fundamental group at $x$. This is the monodromy representation. You can rebuild $L$ from it: this is the sheaf of sections of $(\tilde{X}\times V)/\pi_1(X,x) \to X$ where $\tilde{X}$ is the universal covering of $X$. We have sketched: -Theorem: If $X$ is connected, the functor "fiber at $x$" induces an equivalence of categories $LS(X) \to Rep(\pi_1(X,x))$. -This is all very abstract so let us look at an example. -Consider $\mathcal{K}$, the trivial rank 2 vector bundle $\cal{O}_{\mathbb{C}^\times}^2$ and connection -$$ - \nabla \begin{pmatrix} f_1 \cr f_2 \end{pmatrix} = - d\begin{pmatrix} f_1 \cr f_2 \end{pmatrix} - - \begin{pmatrix}0 & 0 \cr 1 &0 \end{pmatrix} \begin{pmatrix} f_1 \cr f_2 \end{pmatrix} \frac{dz}{z} - = \begin{pmatrix} df_1 \cr df_2 - f_1 \frac{dz}{z} \end{pmatrix} -$$ -Horizontal sections are the solutions of $\nabla f = 0$. This is a system of two first order linear differential equations. On any simply connected $U$ we can chose a determination of the logarithm and the solution can be written -$$ - \begin{pmatrix} f_1 \cr f_2 \end{pmatrix} = - \begin{pmatrix} A \cr A \log z + B \end{pmatrix} -$$ -where $\log$ is any determination of the logarithm function. This means that the sections -$$ - e_1 = \begin{pmatrix} 1 \cr \log z \end{pmatrix} \qquad - e_0 = \begin{pmatrix} 0 \cr 1\end{pmatrix} -$$ -trivialize the sheaf of solutions on $U$. Covering $\mathbb{C}^\times$ by simply connected open sets we see that the solutions form a local system $L$. -When we turn once around 0 following the orientation, our determination $\log z$ changes to $\log z + 2\pi i$. So if $\gamma(t) = xe^{2\pi i t}$ -$$ - v = \begin{pmatrix} A \cr A\log x + B \end{pmatrix} \mapsto - \begin{pmatrix} A \cr A(\log x + 2\pi i) + B \end{pmatrix} = \gamma_*(v) -$$ -The monodromy representation is -$$ - \pi_1(\mathbb{C}^\times,x) = \mathbb{Z} \to GL_2(\mathbb{C}) \qquad - 1 \mapsto \begin{pmatrix} 1 & 0 \cr 2\pi i & 1 \end{pmatrix} -$$ -It tells you everything there is to know about our differential equation (because it has regular singularities). For example, the space of global solutions is identified with the space of invariant of the representation: -$$ - \Gamma(\mathbb{C}^\times,L) = Hom(k_{\mathbb{C}^\times},L) = Hom_{\pi_1(X,x)}(k,L_x) = L_x^{\pi_1(X,x)} -$$ -This is the 1 dimensional space generated by $e_0$. -Another good example to work out is the equation $df = \alpha f\frac{dz}{z}$ (the monodromy can be quite different depending on $\alpha$). -Pierre Schapira's webpage has notes of a course on sheaves and algebraic topology focussing on local systems. Claire Voisin's book on Hodge theory is a good reference for variations of Hodge structures.<|endoftext|> -TITLE: subspaces of singular matrices -QUESTION [12 upvotes]: Let $A$, $B$ be square matrices over infinite field (we identify them with linear operators on the vector space of columns). It is given that for all scalars $a,b$ the matrix $aA+bB$ is singular. Does it follow that there exist matrices $P$, $Q$ such that rank$(P)$+rank$(Q) > n$ but $PAQ=PBQ=0$? -If yes, is the same true for arbitrary subspaces of singular matrices? Well, the answer is no for antisymmetric matrices $3\times 3$... But how can subspaces of singular matrices be described (if they can)? - -REPLY [2 votes]: `But how can subspaces of singular matrices be described (if they can)?' - -I doubt they can. For instance, there is the following counter-intuitive result: - -for infinitely many $n$, there exists -an 8-dimensional space of $n\times n$ matrices -(over any field of characteristic zero) that is maximal singular (i.e. inclusion-maximal subspace of singular matrices).<|endoftext|> -TITLE: Example for column rank $\neq$ row rank -QUESTION [30 upvotes]: The proof that column rank = row rank for matrices over a field relies on the fact that the elements of a field commute. I'm looking for an easy example of a matrix over a ring for which column rank $\neq$ row rank. i.e. can one find a $2 \times 3$-(block)matrix with real $2\times 2$-matrices as elements, which has different column and row ranks? - -REPLY [23 votes]: I will only consider matrices over division rings, since defining rank over more general rings, in particular those with zero divisors, involves difficulties rather independent of those posed by non-commutativity. -Contrary to what is implicit in the question, and was stated explicitly in Wikipedia,* and in seeming contradiction with the other answers given here (although what they say is correct), I maintain that there is only one natural notion of rank for a matrix over a division ring$~D$. That is, all natural definitions give the same notion of rank; in particular the row rank of a matrix always equals its column rank, if these notions are defined in a natural way. The point is that the inherent asymmetry of matrix multiplication makes that the set of column vectors naturally forms only a right $D$ vector space, while the set of row vectors forms a left $D$ vector space. Of course since the two sets are in bijection one can define an opposite vector space structure, but to do so in terms of matrix multiplication requires using an "unnatural" operation such as (1)$~$matrix transposition while (still) viewing the result as matrix over $D$, (2)$~$wrong matrix multiplication with the roles of rows and columns interchanged with respect to the usual definition, or (3)$~$reinterpreting a matrix over$~D$ as a matrix over the opposite division ring$~D^\mathrm{op}$ (the three transgressions can be seen to be more or less equivalent). For this reason it is natural to define the column rank of a matrix to be what some call the right column rank, and the row rank to be the left row rank (these are well known to be equal); the "switched" notions of left column rank and right row rank just work the wrong way, and would only be important in a setting where wrong matrix multiplication is treated on an equal footing with ordinary matrix multiplication, which fortunately nobody does. -So the rank of a $n\times m$ matrix $A$ over$~D$ can be defined in any one of the following ways, all giving the same value - -The dimension of the image of the linear map$~f$ of right $D$ vector spaces with matrix$~A$ (with respect to some bases); in coordinates, $f$ is given by left multiplication by$~A$ on column vectors. -The dimension of the image of the linear map$~f$ of left $D$ vector spaces with matrix$~A$ (with respect to some bases); in coordinates, $f$ is given by right multiplication by$~A$ on row vectors. -The dimension of the column space of $A$, the subspace of the right $D$ vector space $D^n$ spanned by the columns of $A$. -The dimension of the row space of $A$, the subspace of the left $D$ vector space $D^m$ spanned by the rows of $A$. -The decomposition rank of $A$: the smallest number$~r$ such that one can write $A=BC$ for some $n\times r$ matrix $B$ and $r\times m$ matrix $C$. -The largest number$~r$ such that there exist matrices $P,Q$ for which $PAQ=I_r$, the $r\times r$ identity matrix. - -*Until I modified the article, which was after I posted this answer. - -I'll add a word about the seeming contradiction which the other two answers. Rather than answering the stated question about row and columns ranks, they provide an example for the situation where the rank of the transpose of $A$ differs from that of$~A$. But they are implicitly interpreting $A^T$ as a matrix over$~D$, which is the first of the unnatural things I listed above. It is unnatural because one wants transposition the have some relation with composition; indeed one wants $(AB)^T=B^TA^T$ to hold, and this is only possible if $A^T$ is considered to be a matrix over$~D^\mathrm{op}$; doing so, transposition is a natural operation. But with that interpretation, the examples disappear, and the rank of $A^T$ is equal to the rank of$~A$.<|endoftext|> -TITLE: fundamental group and complete invariant of irreducible 3-manifolds -QUESTION [8 upvotes]: I heard that,by Perelman's work,we can get that the fundamental group is a -complete invariant of irreducible 3-manifolds (except for lens spaces). -can someone help explain this.Thank you! - -REPLY [2 votes]: An addendum to BM's answer: you might want to start with the Kneser-Milnor theorem which gives a unique decomposition of a 3-manifold as a connected sum and hence its fundamental group as a free product to reduce the question to prime (and then irreducible) 3-manifolds. The Poincare conjecture says you don't have any summands "hidden" from $\pi_1$. -Also, you have to be careful with what kind of 3-manifolds you consider, for example if $F_1$ is a 3 punctured sphere and $F_2$ a punctured torus, then $F_1\times S^1$ and $F_2\times S^1$ (or the complement of the square and granny knots for a more subtle example) have the same fundamental group but aren't homeomorphic. So perhaps you want to first consider closed 3-manifolds. The issues for general compact 3-manifolds can get technical, but they are understood. A good starting point is Hempel's book.<|endoftext|> -TITLE: The history of Proper Forcing -QUESTION [6 upvotes]: What were the initial motivations of the use of the proper forcing.? - -REPLY [11 votes]: Roslanowski once asked Shelah about this, and has kindly typed down the answer he got: -http://www.unomaha.edu/logic/papers/essay.pdf<|endoftext|> -TITLE: How much of ZFC does Quine's New Foundations prove? -QUESTION [20 upvotes]: Main Question: Does anyone know of a reference that can tell me which axioms of ZFC Quine's New Foundations prove, disprove, and leave undecided? -Secondary Question: I've read that diagonal arguments don't go through in NF and thus can't be used to prove that the reals are uncountable. Does NF manage to prove the uncountability of the reals by some other means or does that fact (normally rendered as "$P _1(\mathbb{N}) < P(\mathbb{N})$" in order to make sense in NF) turn out to be undecidable in NF? - -REPLY [6 votes]: I didn't see anyone specifically address the secondary question, so I thought I'd pipe in. In NF the natural numbers are provably Cantorian (see above for definition). Rosser's "Logic for Mathematicians" (available from Dover and a pretty nice resource) has a detailed proof on p.437. Basically, if $a \in n$ and $b \in m$ and $|a| = |\mathscr{P}_1(b)|$ then the relation pairing $n$ and $\{m\}$ turns out to be stratified and is a bijection between $\mathbb{N}$ and $\mathscr{P}_1(\mathbb{N})$. When you have such a bijection, Cantor's proof goes through in the usual fashion.<|endoftext|> -TITLE: Universities not listed on mathjobs -QUESTION [6 upvotes]: I am now applying for postdoc positions and faculty positions at teaching colleges, all through mathjobs. I have a location constraint, and so am applying to universities all within a single metropolitan area. There are many colleges and universities in this area, but not all of them have positions listed on mathjobs.org. -If a college has not listed a position, does this mean that they will not hire under any circumstances, or might they be able to hire if given the right opportunity? i.e. is it worth contacting department chairs at these colleges and inquiring? I am graduating from a top school and might perhaps be a "catch" for some of the smaller teaching colleges. - -REPLY [2 votes]: As the chair of a math department at a teaching institution (and Hailing is right, we want people who love to teach, but keep in mind we also expect people to be active researchers), I would highly recommend looking everywhere for job postings. Contacting chairs individually is great too, I can't imagine anyone would mind, and you might get lucky, things come up at the last minute sometimes. I would suggest email is by far the best way to do that. Good Luck!<|endoftext|> -TITLE: Don't the axioms of set theory implicitly assume numbers? -QUESTION [20 upvotes]: When one writes down the axioms of ZFC, or any other axiomatic theory for that matter, and making statements like "let x, y ..." doesn't this assume an understanding (and thus existence) of natural numbers implicitly? (Q1) -How is the reader to interpret statements such as existence of separate symbols, nevermind sets, without an intuitive notion of numbers? -Bourbaki talks about this within the framework of metamathematics, but then declares that the reader can read words, differentiate between different words etc. and that to assume otherwise is idiotic. -Is there an introduction to these circle of ideas & debates somewhere you'd recommend? (Q2) - -REPLY [2 votes]: If i may add (although this post is kind of old) -This circularity "problem" (if one wishes to see it as such), appears in (what is refered as) classical mathematics. -Now one should bear in mind that even these classical mathematics (continuing along the lines of aristotle, euclid, archimedes, leibniz, cantor, hilbert, russel, goedel, etc..), have been cutoff and formalised (or sterilised if you like) to a greater extened that originaly meant. -In any case this is not the main argument. -But i would like to draw attention to the intuitionistic flavor of mathematics (and especially of the LEJ Brouwer path) (see for example LEJ Brouwer, Cambridge Lectures on Intuitionism, most of first lecture plus the appendix on marxists.org). -There Brouwer, aware of the problem, explicitly takes on the issue of mathematics over language or syntax. -Excerpt: - -FIRST ACT OF INTUITIONISM -Completely separating mathematics from mathematical language and hence - from the phenomena of language described by theoretical logic, - recognising that intuitionistic mathematics is an essentially - languageless activity of the mind having its origin in the perception - of a move of time. This perception of a move of time may be described - as the falling apart of a life moment into two distinct things, one of - which gives way to the other, but is retained by memory. If the twoity - thus born is divested of all quality, it passes into the empty form of - the common substratum of all twoities. And it is this common - substratum, this empty form, which is the basic intuition of - mathematics.<|endoftext|> -TITLE: Connection between isomorphisms of algebraic topology and class field theory -QUESTION [16 upvotes]: I am considering the following two isomorphisms: -First, if $X$ is a reasonably nice topological space, then $X$ has a normal covering space which is maximal with respect to the property of having an abelian group of deck transformations. The group of deck transformations of this covering space is isomorphic to the abelianizatin of the fundamental group of $X$, which can be identified with the singular homology group $H_1(X,\mathbb{Z})$. -Second, if $K$ is a number field, class field theory gives an isomorphism between the Galois group of the maximal abelian unramified extension of $K$ (the Hilbert class field) and the ideal class group of $K$. The ideal class group can be identified with the sheaf cohomology group $H^1(\mathrm{Spec}(\mathcal{O}_K),\mathbf{G}_m)$. -Given the apparent similarity between these two theorems, is there some more general theorem which implies both of these results as special cases? - -REPLY [12 votes]: As BCnrd says, the theorem you want is geometric class field theory. One version says that the abelianization of the fundamental group of a curve over an algebraically closed field is the fundamental group of its Jacobian. One can use this to derive class field theory for curves over finite fields. Over the complex numbers, this is a topological statement, since the Jacobian of a Riemann surface can be constructed by topological methods, such as $J(C)=H_1(C;\mathbb R/\mathbb Z)$. Also, consider Weil's construction of the Jacobian by using Riemann-Roch to recognize a high symmetric power of a curve as a $\mathbb C\mathbb P^n$ bundle over the Jacobian. The projective space bundle is probably not a topological invariant, but the symmetric power is and it already has the right fundamental group. That has an extensive topological generalization, the Dold-Thom theorem, that the homology of a reasonable space is the homotopy groups of its infinite symmetric power. -The key unifying ingredient is, as BCnrd says, the Jacobian, even though it is missing from both of your statements.<|endoftext|> -TITLE: Does $SL_3(R)$ embed in $SL_2(R)$? -QUESTION [44 upvotes]: Is there any non-trivial ring such that $SL_{3}(R)$ is isomorphic to a subgroup of $SL_{2}(R)$? -$SL_{3}(\mathbb{Z})$ is not an amalgam, and has the wrong number of order $2$ elements to be a subgroup of $SL_{2}(\mathbb{Z})$. -Is there any non-trivial ring where this occurs? When can this definitely not occur? I am trying to understand if there is any sort of group-theoretically apparent notion of dimension here. - -REPLY [23 votes]: (Edit: I edited a bit. Now the answer should be clearer, simpler and even correct. Thank you Andrei Smolensky for repeating correcting my embarrassing mistakes here.) -I am surprised that this old question was not fully answered yet. -The answer is "No" and it is well known in some circles. -In fact, a far more general statement holds: -1) Let $R$ and $S$ be rings (commutative with 1). Then any group homomorphism $\text{SL}_3(R)\to\text{SL}_2(S)$ factors via the non-trivial quotient $\text{SK}_1(3,R):=\text{SL}_3(R)/\text{EL}_3(R)$, where $\text{EL}_3(R)$ is the subgroup generated by elementary matrices in $\text{SL}_3(R)$. -Note that $\text{SL}_3(R)$ contains an epimorphic image of $\text{SL}_3(\mathbb{Z})$ (induced by the map $\mathbb{Z}\to R$). It is well known (and easy) that $\mathrm{EL}_3(\mathbb{Z})\simeq \mathrm{SL}_3(\mathbb{Z})$, thus the image of $\text{SL}_3(\mathbb{Z})$ is contained in $\mathrm{EL}_3(R)$, and in fact $\mathrm{EL}_3(R)$ is generated by $\text{SL}_3(\mathbb{Z})$ as a normal subgroup (as you can observe by playing with commutation relation of elementary matrices). Thus (1) is equivalent to: -2) Let $S$ be a ring (commutative with 1). Then any group homomorphism $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(S)$ is trivial. -We now fix a homomorphism as in statement (2) and assume its image is non-trivial. Let $\mathfrak{n} -TITLE: Nice Classes of Non-Closable Operators -QUESTION [8 upvotes]: The only thing I know about non-closable operators can be summarised as "they exist, but they're nasty, so let's not talk about them!" This seems to be the case with everyone else I've talked to. I'd appreciate some references or ideas about the following: -What classes of non-closable operators have been studied (if any). For instance have the operators with non-empty, real, positive spectrum been studied? Are there weaker, but similar conditions to closable/symmetric which gaurentee nice properties (e.g. non-empty spectrum, adjoint with non-trivial domain, some form of polar decompostion)? -EDIT: As the comments suggest, the usual definition of spectrum doesn't make sense if the operator's not closable. Is there another way to define it - or is there a similar object which is useful? - -REPLY [3 votes]: Here's another simple example of a non-closable operator, from Reed & Simon Vol. I, Section VIII.4, Example 4 (p. 252): Let $D(T)$ consist of those $\psi\in L^2(\mathbf{R})$ such that $\int|f\psi|<\infty$, where $f$ is a fixed function not in $L^2(\mathbf{R})$, and set $T\psi=(f,\psi)\psi_0$ for some fixed $\psi_0 \in L^2(\mathbf{R})$. Then $D(T)$ is dense in $L^2(\mathbf{R})$; however, as they show by a short computation, $D(T^*)$ consists of the vectors orthogonal to $\psi_0$, so it's not dense.<|endoftext|> -TITLE: Difficult examples for Frankl's union-closed conjecture -QUESTION [46 upvotes]: Frankl's well-known union-closed conjecture states that if F is a finite family of sets that is closed under taking unions (that is, if A and B belong to the family then so does $A\cup B$), then there must be an element that belongs to at least half the sets. -I know that pretty well any naive approach one takes to this conjecture is known to fail. By "naive approach" I suppose I mean something like an observation that it would follow from such-and-such a stronger conjecture -- it seems that all sensible stronger conjectures one thinks of are false. A very simple example of a stronger conjecture would be that if you pick a random element then on average it will belong to at least half the sets. That is completely false: take the family that consists of the empty set, {1}, and {1,2,3,4,5,6,7,8,9,10}, for example. One can try to "correct" this strengthening by devices such as insisting that for any two elements there is a set that contains one and not the other (which WLOG is the case), but such corrections don't get one very far. -What I am asking for is examples, either small ones or ones that are constructed theoretically, of union-closed families that defeat more sophisticated strengthenings of the original conjecture. I'm fairly sure they are out there but I am not an expert on this problem so I don't know them myself. -Apologies in advance if this resembles an existing question (which it feels as thought it easily might). But I've looked and not found anything. - -REPLY [16 votes]: Here is a nice example due to Bjorn Poonen, which I have taken from this survey paper of Bruhn and Schaudt. It is motivated by the following observations. Let $\mathcal{A}$ be a union-closed family. Call an element $x$ abundant if $x$ is contained in at least half of the members of $\mathcal{A}$. -Observation. If $\mathcal{A}$ contains a singleton $\{x\}$, then $x$ is abundant. -Proof. Partition $\mathcal{A}$ as $\mathcal{A}_x$ and $\overline{A}_x$ where $\mathcal{A}_x$ consists of the members of $\mathcal{A}$ containing $x$ and $\overline{A}_x$ are the members of $\mathcal{A}$ not containing $x$. Since $\{x\} \in \mathcal{A}$ and $\mathcal{A}$ is union-closed, the map $S \mapsto S \cup \{x\}$ is an injection from $\overline{A}_x$ to $\mathcal{A}_x$. Thus, $|\overline{A}_x| \leq |\mathcal{A}_x|$, as required. $\square$ -Similarly, we also have the following. -Observation. If $\mathcal{A}$ contains a $2$-set $\{x,y\}$, then $x$ or $y$ is abundant. -This might lead one to conjecture the following false strengthening of Frankl's Union-closed Conjecture. -False Strengthening. Let $S$ be a non-empty member of $\mathcal{A}$ with the smallest number of elements. Then $S$ contains an abundant element. -Disproof. For two families $\mathcal{A}$ and $\mathcal{B}$ we let -$$ -\mathcal{A} \uplus \mathcal{B}=\{S \cup T: S \in \mathcal{A}, T \in \mathcal{B}\}. -$$ -Fix $k \geq 3$ and for each $i \in [k]$ define -$$ -\mathcal{B}^i=\{[k+1,3k] \setminus \{2i+k-1\}, [k+1,3k] \setminus \{2i+k\}\}. -$$ -Finally, define -$$ -\mathcal{A}^k=\{[k]\} \cup \bigcup_{i=1}^k(\{\emptyset, \{i\}, [k]\} \uplus \mathcal{B}^i) \cup (2^{[k]} \uplus [k+1,3k]). -$$ -It is easy to check that $\mathcal{A}^k$ is union-closed and that $[k]$ is the unique smallest set in $\mathcal{A}^k$. However, $|\mathcal{A}^k|=1+6k+2^k$ and each $i \in [k]$ is contained in only $1+(2k+2)+2^{k-1}$ members of $\mathcal{A}^k$. Thus, no element of $[k]$ is abundant. $\square$ -This example highlights one of the obstacles in proving the Union-closed Conjecture; it is difficult to predict where abundant elements will be.<|endoftext|> -TITLE: How many algebraic integers exist with degree $\leq k$ and bounds on the modulus of all Galois conjugates? -QUESTION [5 upvotes]: The precise question is the following: - -Question: Can one reasonably bound the number of algebraic integers $\alpha$ of degree at most $k$ - that means there exists a monic integer polynomial $p$ with $\deg(p) \leq k$ and $p(\alpha)=0$ - and such that $p(\beta)=0$ implies $|\beta| \leq n$, i.e. all Galois conjugates of $\alpha$ have a modulus bounded by $n$. - -Obviously, the number of relevant polynomials $p(t) = \sum_{i = 0}^k a_{k-i} t^i$ is bounded since $$|a_i| \leq {{k}\choose {i}} \cdot n^i$$ In particular, the number of such $\alpha$ is finite and one obtains a crude upper bound. One can also make a packing argument by observing that the distance between any two such algebraic integers cannot be too small. I am basically asking whether there are better bounds. - -REPLY [3 votes]: Schanuel, S. On heights in number fields. Bull. Amer. Math. Soc. 70 1964 262–263. -Masser, David; Vaaler, Jeffrey D., Counting algebraic numbers with large height. II. Trans. Amer. Math. Soc. 359 (2007), no. 1, 427–445 -Edit: As per Kevin's request, more context. The first reference is the basic result in the field and the second is the state of the art. The bound can be improved as expected. The second reference also has lots of additional references. Link to abstract and references (and also paper if your institution subscribes): -http://www.ams.org/journals/tran/2007-359-01/S0002-9947-06-04115-8/home.html<|endoftext|> -TITLE: CM for radical ideal -QUESTION [10 upvotes]: Let $R$ the polynomial ring in $n$ variables with complex coefficients and $I$ an ideal of $R$. Is it true that if $R/I$ is CM also $R/J$ is CM (where $J$ is the radical of $I$)? -Is there a relations between a resolution of $R/J$ and one of $R/I$? What if I suppose that $proj.dim(R/I)=2$? - -REPLY [11 votes]: It is not true, but the example is not easy to find $I = (x_2^2-x_4x_5,x_1x_3-x_3x_4, x_3x_4-x_1x_5)$!<|endoftext|> -TITLE: Is there an elementary proof that the Mertens function is not $O(x^\theta)$ if $\theta <1/2$? -QUESTION [15 upvotes]: The Mertens function is the partial sums of the Moebius function: -$M(x)=\sum_{n\leq x}\mu(n)$ -Since the zeta-function has a zero on the critical line it follows that $M(x)\ne O(x^\theta)$ for any $\theta<\frac 12$. -Does anyone know if there is an elementary proof of this statement? (By elementary I mean a proof which does not depend on complex analysis, in particular the existance of a zero of $\zeta$). even an elementary proof of $M(x)$ being unbounded would be interesting to me. - -REPLY [4 votes]: As far as I know, even an elementary proof that $M(x)$ is unbounded is not known.<|endoftext|> -TITLE: Can the simplicity of abelian varieities be implied by the reduction -QUESTION [6 upvotes]: A is an abelian variety over number field K, with simple good reduction at a finite field $\kappa$, can we deduce that $A$ itself is simple? - -REPLY [5 votes]: Here is a slightly different proof. We have the following facts: - - -(1) If $B, C$ are abelian varieties over $K$, then the Néron model of $B\times C$ is the product of the Néron models (this is a simple onsequence of the universal property of Néron models). In particular, if $B\times C$ has good reduction, then $B, C$ also have good reduction, and the reduction of $B\times C$ is the product of the reductions of $B$ and $C$. - - - -(2) If $A$ and $B$ are isogeneous abelian varieties over $K$ and if $A$ has good reduction, then $B$ has good reduction and any isogeny $A\to B$ induces an isogeny on the reductions. - - -Proof of (2): if $A\to B$ is an isogeny of degree $n$, then there exists an isogeny $B\to A$ such that the composition $A\to B\to A$ is the multiplication-by-$n$ map $[n]_A: A\to A$ on $A$. On the reductions we have the maps $A_k \to B_k^0 \to A_k$ whose composition is $[n]_{A_k}$. Consider a positive integer $\ell$ prime to $n$ and to $\mathrm{char}(k)$. As the restriction of $[n]_{A_k}$ to $A_k[\ell]$ is an isomorphism, $A_k[\ell]\to B^0_k[\ell]$ is injective, so $B^0_k[\ell]$ contains $(\mathbb Z/\ell \mathbb Z)^{2\dim B}$. Therefore $B_k^0$ is an abelian variety and $B$ has good reduction. Finally $A_k\to B_k=B_k^0$ is an isogeny because its kernel is contained in $A_k[n]$. -This is a special case of Néron-Ogg-Shafarevich criterion (Serre-Tate: Good reduction of abelian varieties, §1). -Now to answer the original quesiton, if $A$ was not simple, then it is isogeneous to a product of abelian varieties $B\times C$. Then $B\times C$, $B, C$ have good reduction by (2) and (1), and $A_k$ is isogeneous to $B_k\times C_k$. So $A_k$ would not be simple.<|endoftext|> -TITLE: Deformations of semisimple Lie algebras -QUESTION [13 upvotes]: In the questions Is "semisimple" a dense condition among Lie algebras? and What is the Zariski closure of the space of semisimple Lie algebras?, something equivalent to the following is mentioned: if you have a smoothly varying family of semisimple Lie algebras, all the Lie algebras in the family are isomorphic. e.g. the following quote: -"Because the Cartan classification of isomorphism classes of semisimples is discrete (no continuous families), connected components of the space of semisimples are always contained within isomorphism classes." -I can't see how this follows just from the discreteness of the classification. Can anyone explain why it's true or give a counterexample? -e.g. could you not have a $\mathbb{P}^1$ of semisimple Lie algebras which are generically isomorphic to $\mathfrak{d}_7 \oplus \mathfrak{a}_1$ say, but at one point you get $\mathfrak{e}_8$, or something similar? - -REPLY [6 votes]: I think that the explanation -"Because the Cartan classification of isomorphism classes of semisimples is discrete (no continuous families), connected components of the space of semisimples are always contained within isomorphism classes" -is a bit simplistic. The real reason, as many people have already mentioned, is Weyl's theorem on complete reducibility, which of course fails in charatcteristic $p$. And it shouldn't come as a surpise that over an algebraically closed field $K$ of characteristic $p>3$ one encounters situations where there are finitely many isoclasses of simple Lie algebras of dimension $N$ and, at the same time, there exist algebraic families of simple $N$-dimensional Lie algebras -{$\mathfrak{g}_t|\ t\in K$} over $K$ such that $\mathfrak{g}_t\cong L$ for all $t\ne 0$ and $\mathfrak{g}_0\not\cong L$ for some simple Lie algebra $L$. -Indeed, let $N=p^2-2$. Then it follows from the the classification theory that there are finitely many isoclasses of simple $N$-dimensional Lie algebras over $K$. Now consider the associative $K$-algebra $A$ generated by two elements $x,y$ such that $x^p=y^p=0$ and $[x,y]=1$. This is a fake modular version of the first Weyl algebra, and it is easy to see that it is simple and has dimension $p^2$. It has a finite increasing algebra filtration (with $x,y$ living in degree $1$) such that the corresponding graded algebra $P:={\rm gr}(A)$ is the truncated polinomial ring in $x,y$ with induced Poisson bracket satisfying {$x,y$} $=1.$ Then the Lie algebra $[A,A]/K1$ is isomorphic to $\mathfrak{psl}_p(K)$ whilst the Lie algebra {$P,P$}$/K1$ is nothing but the simple Cartan type Lie algebra $H(2;\underline{1})^{(2)}$. Both Lie algebras are simple of dimension $N$ and the latter is a contraction of the former. Moreover, they are not isomorphic when $p>3$.<|endoftext|> -TITLE: Cantor's argument revisited -QUESTION [19 upvotes]: This was inspired by this recent question. -In my answer there, I pointed out that, given $F:{\mathcal P}(X)\to X$, an argument dating back to Zermelo allows us to define a pair $(A,B)$ of distinct subsets of $X$ witnessing that $F$ is not injective. The pair is defined in terms of a well-ordering that is constructed using $F$. -Of course, the usual Cantor argument also shows that $F$ is not injective: One considers $$ A=\{F(Z)\mid F(Z)\notin Z\},$$ and argues that there must be a $B$ with $A\ne B$ and $F(A)=F(B)$. -My question is: - -Can we exhibit such a set $B$ (definably from $F$)? - -REPLY [7 votes]: It has been pointed out in the comments that Ewan's insightful solution shows that a negative answer to the question is consistent with ZF, since a positive answer implies AC. -But let me go one further. In fact, Ewan's solution shows that a negative answer to the main question is consistent with full ZFC. The reason is that a positive answer to Ewan's problem 2 actually implies the set-theoretic assertion $V=HOD$, that the universe consists of the hereditarily-ordinal-definable sets. To see this, supoose that $V\neq HOD$, then there is some cardinal $\kappa$ such that $Y=P(\kappa)$ has some non-ordinal definable elements. Let $Y'$ be the set of non-HOD subsets of $\kappa$. Both $Y$ and $Y'$ are ordinal definable, but $Y'$ has no ordinal definable elements. This contradicts any positive solution to problem 2. -Thus, any model of set theory having a positive answer to problem 2 must also satisfy $V=HOD$. And so any model of $ZFC+V\neq HOD$ is a model of a negative answer to the main question, with full ZFC. -Meanwhile, a positive answer to the main question is also consistent with ZFC, since there are models of ZFC in which every object is definable without parameters. For example, this is true in the minimal transitive model of set theory. Indeed, Reitz and Linetsky and I have recently proved that every countable model of ZFC and indeed of GBC can be extended to a pointwise definable model, in which every set and class whatsoever is definable without parameters. In such a model, we may definably find the desired B, since every B is definable. (But there is little uniformity in this definition.) -So the full answer to the main question is that it is independent of ZFC. Of course, the question of "definable" is not directly formalizable in set theory, so one should understand this assertion as the claim that if ZFC is consistent, then there are models of ZFC in which there is a positive solution, and models in which there is a negative solution, even when interpreted as the literal second-order claim. However, if one understands "definable" as "ordinal-definable", then the claim is formalizable in the language of set theory, and this claim is also independent of ZFC, for the same reasons.<|endoftext|> -TITLE: Positive polynomial having roots modulo any integer -QUESTION [14 upvotes]: Does there exist polynomial $p(x)$ with integer coefficients such that $p(x) > 0$ for all real values of $x$, but for any integer $n > 0$ there exists integer $k$ such that $n$ divides $p(k)$? The trick with multiplying quadratic polynomials (as here Diophantine equation with no integer solutions, but with solutions modulo every integer) does not work, if I am not mistaken. - -REPLY [13 votes]: $\newcommand{\Q}{\mathbf{Q}}\newcommand{\Z}{\mathbf{Z}}$ -I now suspect the answer might be no! This isn't a complete answer but it might be an idea that turns into one. -So let me assume that such $p$ exists and let me go for a counterexample. -First I claim that if such $p$ exists, then a monic $p$ exists. For if $p$ works, then so does $Np$ for a large positive integer $N$, and if $p=cx^d+\ldots$ and $N=c^{d-1}$ then $Np=q(cx)$ with $q$ monic with integer coefficients, and $q$ also works. So WLOG $p$ is monic. -Now say $p$ factors as $p_1p_2\ldots p_r$ in $\Q[x]$ with the $p_i$ monic. By Gauss' Lemma the $p_i$ are all in $\mathbf{Z}[x]$. Note -that $p$ must have positive degree so $r\geq 1$. Let $K$ be the splitting field of $p$ and let $K_i$ be the field $\Q[x]/(p_i)$. Now none of the $K_i$ have any real places. Let $c$ denote any complex conjugation in $Gal(K/\Q)$. By Cebotarev density, there's a prime number $\ell$, unramified in $K$, and such that $Frob_\ell$'s conjugacy class in $Gal(K/\Q)$ is that of $c$, and indeed there are infinitely many such $\ell$. -Certain primes cause me trouble, so let me get rid of them now: for each $K_i$ let $S_i$ be the index of $\Z[x]/(p_i)$ in the full ring of integers of $K_i$, and let $S$ be the product. -Now let $\ell$ be a prime as above, and such that $\ell$ doesn't divide $S$. -I claim that this $\ell$ is going to cause us problems. Say $p(k)$ is zero mod $\ell$. Then one of the $p_i(k)$ is zero mod $\ell$, so the factorization of $p_i$ mod $\ell$ -has a linear factor, and so $\ell$ factors in the integers of $K_i$ into a bunch of primes -one of which has degree 1. I now want to argue something like the following: the Frobenius element at $\ell$ corresponding to this prime is fixing a root of $p_i$ so complex conjugation is fixing a root of $p_i$ so $p$ has a real root! But the cricket has just started and I can't think straight. Can this be turned into a proof?<|endoftext|> -TITLE: How do I compute the compact cohomology of a hypersurface? -QUESTION [11 upvotes]: How do I compute the compact cohomology of a hypersurface? -For example, let $f$ be a Newton polynomial of a polytope in $\mathbb{R}^n$ and let $X = (f=0)$ -inside $(\mathbb{C}^\*)^n$ (maybe there is some dependency on the coefficients of $f\;$?). Can you tell me anything about $H^*_c(X)$? Perhaps I should know better, but I don't. -Thanks! - -REPLY [12 votes]: The classic reference is Danilov-Khovanskii's "Newton polyhedra and an algorithm for calculating Hodge-Deligne numbers". There is subsequent work by Cox, Batyrev, Malvyutov, etc. but they are mainly concerned with more general toric ambient spaces; if you want a hypersurface in the torus then this original paper should have all you need.<|endoftext|> -TITLE: mertens-function in the light of divergent summation - what summation method were best adapted -QUESTION [5 upvotes]: Just reading about the Mertens-function in the other thread -Mertens function I remember an earlier attempt to apply divergent summation -to the series which is constructed of the Moebius-function -at consecutive arguments, or in other words of which the Mertens-function-values represent -the partial sums. -Eulersummation, although relatively poorly adapted, suggested that the (divergent) sum should be -meaningfully evaluated to -2. But that sequence of partial sums (although seldom exceeding only the squareroot of its current index) seems to be a specific difficult case for -such common summation methods - the approximation is relatively poor even for 128 terms. -I tried Nörlund-means/Cesaro-sum, Euler-sums of different orders and also a selfmade matrix summation-method using the eulerian numbers (with which I could -on the other hand- well handle the even strongly -diverging $ 0!-1!+2!-3!+...$ -series), but I tried not yet for instance Abel and Borel. -Q1: What method would be most appropriate to sum the series -$ S = \sum _{k=1}^{\inf} moebius(k) $ -Evaluation of 128 terms (Euler,Cesaro) suggested the result $S = -2$ -Q2: And how could it be determined whether the Cesaro- and/or Euler-summation are at all capable to evaluate -that series to a final value? -Here is of the summation. - -REPLY [8 votes]: Well, -$$\sum_{n=1}^\infty\frac{\mu(n)}{n^s}=\frac1{\zeta(s)}$$ -for $s>1$, so setting $s=0$ should give -$$\sum_{n=1}^\infty\mu(n)=\frac1{\zeta(0)}=-2$$ -as $\zeta(0)=-1/2$. :-) -I should add that this is a trick often used in analytic -number theory (for instance in Eisenstein series). More generally -given a divergent sum -$$S=\sum_{i\in I}a_i$$ -then consider, for an appropriate choice of weights $b_i>0$ -the series -$$f(s)=\sum_{i\in I}\frac{a_i}{b_i^s}.$$ -We hope this converges in a suitable half-plane -and can be analytically continued to $0$. Then we "define" -$S=f(0)$.<|endoftext|> -TITLE: Classification of surfaces composed of circles -QUESTION [9 upvotes]: Define a circle as a geometric circle of positive, finite radius: -a set of points in $\mathbb{R}^3$ congruent to the -set $x^2 + y^2 = r^2$ in the $xy$-plane. [Edited as per BMann's comment.] -I am interested in classifying surfaces $S$ embedded in $\mathbb{R}^3$ in two categories: - - -Define $S$ to be a circled surface if every point of $S$ is contained in a -circle that lies in $S$. -(I choose the name "circled" in analogy with ruled surfaces.) -Thus, $S$ is a union of circles; it is covered by circles. - - -Define $S$ to be a hoop surface if it may be partitioned into circles, i.e., -it is a disjoint union of circles (hoops). In such a partition, every point of $S$ is contained in a -unique circle of the partition that lies in $S$. [Edited as per Charles Siegel's comment.] - - -A specialization would be to restrict the circles in either class to be congruent, -unit-radius circles. -Generalizations include: -(a) relaxing embedded to immersed; -(b) replacing circles by spheres in $\mathbb{R}^d$; -(c) replacing geometric circles with topological circles. -Setting aside these variants, let me list the hoop surfaces I know: - - - - - -Tubular surfaces, topological cylinders, either infinite in both directions, or with -a circular boundary at one or both ends. -These are sometimes called tubular splines in computer graphics. - - -Tubular surfaces capped at either or both ends but with one point missing from the cap. - - -Torus surfaces; genus 1. - - -Disk-like surfaces (composed of nested circles) with one interior point removed. -[Edited as per J.C. Ottem's comment.] - - -A surface homemorphic to the plane with one point removed. - - -[Due to Cristi Stoica, explained in his answer below.] A surface homeomorphic to two planes connected by a wormhole. - - -Is this list complete? -Every hoop surface is a circled surface, but planes and spheres are circled surfaces. -I cannot see how to form a circled surface of genus greater than 1. -Have these concepts been explored before, and if so, under what names? -My searches have been unsuccessful. Thanks for any ideas or literature pointers! -Addendum. The variant of immersed hoops surfaces partitioned -into topological circles has been thoroughly analyzed in -a paper by -Gábor Moussong and Nándor Simányi, -"Circle Decompositions of Surfaces": - -"The main result in this note is Corollary 3 ... stating that - any surface with a circle decomposition is homeomorphic to either - a torus, a Klein bottle, an annulus, a Möbius band, an open annulus, - a half-open annulus, or an open Möbius band." - -REPLY [14 votes]: Here's how to get circled surfaces of arbitrary genus: -First: Any open subset of $S^2$ is circlable. If the boundary is smooth, it is circlable with -circles of constant size: the size just needs to be less than the minimum distance of the boundary from the cut locus (the cut locus is the set of points with more than one shortest arc to the boundary). -Therefore, if you take any finite union of balls with no triple intersections and whose boundaries are not tangent, the boundary of the union is circlable by circles of constant (small) size. These surfaces can have arbitrary genus. -There are also $C^\infty$ smooth circlable genus $g$ surfaces, but they require circles of varying radii. Just take one of the surfaces above, and replace neighborhoods of the circles of intersection by short tubes that are smoothly tangent to the pair of intersecting spheres.<|endoftext|> -TITLE: function that sums to zero over cube vertices -QUESTION [6 upvotes]: Does anyone have an answer to the three-dimensional analogue of the 2009 Putnam Competition A1 problem, viz., if $f\colon \mathbb{R}^3 \rightarrow \mathbb{R}$ satisfies $\sum_{i=1}^8 f(a_i) = 0$ whenever $a_1, \ldots, a_8$ are the vertices of a cube, must $f$ be identically zero? -A few thoughts: Since the cube is not self-dual above dimension $2$, the solution (well, at least, my solution) to the $2$-dimensional problem doesn’t generalize.  To try to show that the answer to the $3$-dimensional problem is “no,” one might try letting $\Omega$ be the set of ordered pairs $(X, f_X)$ where $X \subseteq \mathbb{R}^3$ is a subset and $f_X\colon \mathbb{R}^3 \rightarrow \mathbb{R}$ is a function that sums to zero over any eight points of $X$ that form the vertices of a cube.  Ordering $\Omega$ in the usual way, we can invoke Zorn’s Lemma to obtain a maximal $(X, f_X) \in \Omega$.  If $X \neq \mathbb{R}^3$, then every $x \in \mathbb{R}^3 \setminus X$ must, in two “incompatible” ways, be the eighth vertex of a cube whose remaining vertices lie in $X$.  But I don't see how a contradiction arises from this (and of course one could make the same argument in two dimensions, where a contradiction does not arise). - -REPLY [7 votes]: First prove that the sum of the values of your function at the vertices of any regular tetrahedron is zero. You can do this by considering a $2\times 2\times 2$ cube made out of $8$ smaller cubes that you can color in a checkerboard fashion. Sum all the vertices of the black cubes and subtract the sum of the vertices of all the white cubes. -Second you can show that the sum of the values at the vertices of a $x\times x\sqrt{2}$ rectangle is always zero. You need to consider two tetrahedra sharing an edge for this. -From here it is very similar to the Putnam problem you mention. (In fact some solutions apply in the same exact way.) -P.S. This shows that it is enough to have the information only for cubes whose sides are parallel to the axis.<|endoftext|> -TITLE: Which groups have only real and quaternionic irreducible representations? -QUESTION [18 upvotes]: Consider a continuous irreducible representation of a compact Lie group on a finite-dimensional complex Hilbert space. There are three mutually exclusive options: -1) it's not isomorphic to its dual (in which case we call it 'complex') -2) it has a nondegenerate symmetric bilinear form (in which case we call it 'real') -3) it has a nondegenerate antisymmetric bilinear form (in which case we call it 'quaternionic') -It's 'real' in this sense iff it's the complexification of a representation on a real vector space, and it's 'quaternionic' in this sense iff it's the underlying complex representation of a representation on a quaternionic vector space. -Offhand, I know just four compact Lie groups whose continuous irreducible representations on complex vector spaces are all either real or quaternionic in the above sense: -1) the group Z/2 -2) the trivial group -3) the group SU(2) -4) the group SO(3) -Note that I'm so desperate for examples that I'm including 0-dimensional compact Lie groups, i.e. finite groups! -1) is the group of unit-norm real numbers, 2) is a group covered by that, 3) is the group of unit-norm quaternions, and 4) is a group covered by that. This probably explains why these are all the examples I know. For 1), 2) and 4), all the continuous irreducible representations are in fact real. -What are all the examples? - -REPLY [6 votes]: Torsten answered this question perfectly for the definition of real/complex/quaternionic in John's original question. But this usage of real/complex/quaternionic is foreign to my experience. Specifically, if you look at an irreducible real representation of a group, then its endomorphism ring is (by Schur and Frobenius) R, C, or H. And this seems to give a natural meaning of the terms "real", "complex" and "quaternionic" for irreps. This definition does not agree with John's, as you can see by considering the spin reps of Spin(7,1). -My definition is also what you find in Noah Snyder's answer here and in Wikipedia's definition of quaternionic representation. - -REPLY [4 votes]: This was a comment on Torsten's answer, but it got too long. -Suppose $G$ is connected and semisimple. Fixing a choice $\Phi^+$ of positive roots for $G$, we can describe $w_0$ as the unique element of the Weyl group of $G$ that takes $\Phi^+$ to the negative roots $\Phi^- = -\Phi^+$. Now, $-w_0$ is an involution of the Dynkin diagram of $G$. This involution is trivial when the components of the Dynkin diagram lack two-fold symmetry, and this happens precisely for components of type $A_1$, $B_n$, $C_n$, $D_{2n}$, $E_7$, $E_8$, $F_4$ and $G_2$, in which case $-w_0=1$. For type $A_n$ ($n>1$), the involution is given by $\alpha_i \leftrightarrow \alpha_{n-i+1}$, for $D_n$ it's given by $\alpha_i \leftrightarrow \alpha_{i-1}$, and for $E_6$ it's given by $\alpha_1 \leftrightarrow \alpha_6$ and $\alpha_2 \leftrightarrow \alpha_5$. -Now if $V$ is an irrep of highest weight $\lambda$, then $V^\ast$ has highest weight $-w_0\lambda$. So $V \cong V^\ast$ whenever $-w_0=1$, and the above discussion tells us when this happens. -Side note: There's a closely related MO question, which was asked not too long ago, whose answers might be helpful.<|endoftext|> -TITLE: Intersection Cohomology of Coordinate Hyperplanes -QUESTION [6 upvotes]: I'm trying to learn how to compute stalks of IC sheaves, and I was wondering about the following example: -Fix $n$. Let $X \subset \mathbb{C}^n$ be the variety cut out by the equation $x_1 \cdots x_n =0$, i.e. the coordinate hyperplanes. What are the stalks of $\mathrm{IC}(X)$ at the various points of $X$, in particular at the origin? -This seems like a natural toy example, but if the general answer is difficult, I'd be happy to know how to compute this for small $n$. - -REPLY [6 votes]: Let $Y$ denote the disjoint union of the coordinate hyperplanes in $\mathbb{C}^n,$ and let $f:Y \to X$ denote the corresponding resolution of singularities. -1) Show that $f_{\ast}\mathbb{C}_Y[n-1] \simeq IC_X$ (consider, for example, the support conditions and the fact that both sheaves are isomorphic to $\mathbb{C}_U[n-1]$ when restricted to the nonsingular open $U \subset X$). -Edit (some details added): Letting $U$ denote the complement of the set where any two coordinate planes intersect, $f$ is an isomorphism when restricted to $U.$ We therefore have that the restriction (i.e., pullback) of $f_{\ast}\mathbb{C}_Y[n-1]$ to $U$ coincides with $\mathbb{C}_U[n-1]$ (by proper base change if you like). -In order to conclude that $f_{\ast}\mathbb{C}_Y[n-1] \simeq IC_X,$ we now just need to check the support and cosupport conditions which uniquely define the intersection cohomology sheaf (together with the fact that its restriction to $U$ is the (shifted) constant sheaf). These conditions are similar to, but more restrictive than, the support and cosupport conditions for perverse sheaves. -I recommend looking at page 21 of the wonderful article by de Cataldo and Migliorini, which can be found at http://arxiv.org/abs/0712.0349 for a statement of these support and cosupport conditions (and figure 1 on page 25 for a visual illustration of the definition). -Since the fibers of $f$ consist of a finite number of points, the cohomology of the fibers is non-zero only in degree zero. This shows that the first condition (the support condition) is satisfied. -For the second condition (the cosupport condition), you can either derive it from the support condition using Verdier duality and the properness of $f,$ or you can simply note that an open ball in $\mathbb{C}^{n-1}$ has non-zero compactly supported cohomology only in degree $2n-2.$ -2) Now it's straightforward to compute any of the stalks since the fiber of $x \in X$ consists of anywhere between one point and n points, depending on how many hyperplanes $x$ lives inside of. -Alternatively, it is also possible to do this by using only basic definitions. To compute the stalk at $x,$ intersect a sufficiently small open ball around $x$ in $\mathbb{C}^n$ with $X$ and then calculate the intersection cohomology by considering intersection cochains (just like you would for singular cohomology, but now with a less restrictive notion of cochain).<|endoftext|> -TITLE: Brown representability beyond CW complexes -QUESTION [14 upvotes]: Brown representability states that any contravariant functor from the homotopy category $CW_*$ of pointed CW complexes to the category of pointed sets is representable if it turns coproducts into products and satisfies a type of Mayer-Vietoris gluability axiom, which I like to think of as a weak version of "the functor sends push-outs into pull-backs" as any representable functor must. The proof very much relies on the fact that CW complexes can be built up in a steady and predictable manner, as it uses Whitehead's theorem that a weak homotopy equivalence is automatically a homotopy equivalence. Namely, one shows that an element which is "universal" for the spheres is actually a "universal element" for this functor (in the sense of Yoneda's lemma). -Brown representability has many interesting consequences, e.g. that there is a "universal" principal $G$-bundle for pointed CW complexes (where $G$ is a topological group) or that the Eilenberg-Maclane spaces represent the cohomology functors. However, in the former case, it's actually true that the universal bundle exists for any topological space, not just CW complexes. I don't know whether the cohomology functors are representable on the category of all pointed topological spaces (even if one restricts to non-pathological ones: say Hausdorff, with nondegenerate basepoint), though I would imagine that a CW complex couldn't do it. This leads me to ask: - -Is there a version of Brown representability for arbitrary pointed topological spaces? - -There is a version of it on the nLab in more generality, but I don't know enough about categorical homotopy theory to understand anything. -Could someone perhaps translate some of that into the special case of topological spaces? - -REPLY [6 votes]: Let me try again. The original question - -Is there a version of Brown representability for arbitrary pointed topological spaces? - -is interesting, and I don't see how it has been fully answered. That Brown's theorem, as stated, does not hold beyond CW complexes doesn't necessarily mean that there is no reasonable version of this theorem for arbitrary (or almost arbitrary) spaces, generalizing an equivalent formulation of Brown's original theorem. On the other hand, versions coming with disclaimers like - -Then it is representable, but by construction rather than by any fancy theory. - -or - -any "generalization" you're going to get for general spaces is really an illusion - -are truly admirable for honesty of the disclaimers, but leave some doubt as to whether this is the end of the story. -More to the point. Given a homotopy invariant contravariant functor $h$ on the category of polyhedra, Watanabe shows that $h$ admits a unique (up to natural equivalence) extension $\check h$ to all topological spaces that is homotopy invariant and also continuous in the sense of satisfying two rather simple axioms (Theorems 5 and 6 here). This extension is also known as the Cech extension, and Lee and Raymond show that if $h$ is representable, then so is $\check h$ (See also Dold's Lectures in Algebraic Topology, Appendix, Example 3.12.) I understand that if $h$ is a (generalized) cohomology theory, then so is $\check h$. -Let me summarize: if we restrict our attention to functors that are cohomology theories and in addition are "continuous", then yes, we do have Brown's theorem for arbitrary topological spaces, by a certain fancy theory. (Assuming that I'm not missing any fine points concerning e.g. non-Hausdorff spaces. Note that in Dold's Appendix, some things work better for paracompact spaces.) -In case Watanabe's axioms of a "continuous" functor cut no ice with you, then there are the axioms mentioned in my previous reply in this thread (strong excision, strong homotopy invariance, additivity and multiplicativity) EDIT: which do suffice for locally compact separable metric spaces. Doing something similar for more general spaces could involve a fancier theory, which is probably not developed as yet. Certainly, doing something similar with homology (and other covariant functors) would involve a fancier theory. Going in the direction of such fancier theory are the following uniqueness theorems for generalized homology theories on metric compacta: Theorem 5.3 in Bauer and Corollary 2.17 in Kaminker-Schochet.<|endoftext|> -TITLE: Why are absolute values more natural than discrete valuations? -QUESTION [7 upvotes]: It is true that considering the archimedean places as well is more general, but that still doesn't explain why it is more natural. If we consider both the definitions of an absolute value and that of a discrete valuation, they both seem arbitrary. However, discrete valuations have the extra appeal of being related to geometric objects (they arise as local rings at non-singular codimension 1 subschemes of integral schemes, and in various other situations: local rings at non-singular codimension 1 subschemes of a scheme birational to the original scheme, or more abstract situations like Zariski-Riemann spaces). It seems (to me at least) that the archimedean places don't enjoy a similar geometric analogy. -This leads me to ask: why are the archimedean places natural? Why do they always feel like they're "missing points" of rings of integers? -I should note in the interest of full disclosure that I'm unfamiliar with the ways of Arakelov theory. - -REPLY [3 votes]: Perhaps the main reason why people care about archimedean places is that many results in algebraic number theory become much neater if you treat them on equal footing with finite primes, even before you get to Arakelov theory. E.g. zeta- and L-functions have contributions from infinite primes, Dirichlet's S-unit theorem is nicer to state, theorems in class field theory become nicer, and the whole adelic business becomes possible. As a concrete example, Kevin Buzzard's answer and especially Matt Emerton's comment in -Positive polynomial having roots modulo any integer -relies on the fact that you can treat complex conjugation like any other conjugacy class of Frobenius elements. -On the other hand, when there is no need to care about infinite places, many (most?) people will definitely state what they want in terms of discrete valuations rather than absolute values, because, as you say, they are nicer and more intuitive. I don't think many will say "because $4\in{\mathbb Z}[i]$ has $(1+i)$-adic absolute value $1/16$...", they will say "because $4\in{\mathbb Z}[i]$ has valuation $4$ at $(1+i)$".<|endoftext|> -TITLE: Spread polynomials -QUESTION [7 upvotes]: Norman Wildberger's "rational trigonometry" has been viewed by some mathematicians as a clever new take on an ancient topic. Wildberger's "spread polynomials" $S_n$ are characterized by the identity -$$ -\sin^2(n\theta) = S_n(\sin^2\theta) -$$ -(except that Wildberger refuses to refer explicitly to the sine function in the definition and does it by other means). In one sense these are trivially equivalent to the Chebyshev polynomials $T_n$ characterized by -$$ -\cos(n\theta) = T_n(\cos\theta). -$$ -Wildberger notes that $1 - 2S_n(s) = T_n(1 - 2s)$. -In thinking about whether this polynomial sequence is even worth mentioning after Chebyshev polynomials have been treated, three questions come to mind: - -Could it be that an essential difference justifying a separate treatment is the factorization of these polynomials? Wildberger factors the spread polynomials. Is there some important reason for doing that? -Is there a combinatorial interpretation of the coefficients? The coefficient of the $n$th-degree term in $S_n$ for $n=1,\dots,10$ is $n^2$ and the coefficient of the constant term is $4^{n-1}$. The first-degree term is $n/2$ times the constant term. -If there's no essential difference that justifies attending to the two sequences separately, the fact that the conventional way of viewing the sequence (Chebyshev polynomials) is chronologically first, doesn't mean it's necessarily better than Wildberger's way of viewing it (spread polynomials). Is it? - -REPLY [5 votes]: Factorization of the spread polynomials can be reduced to the factorization of the Chebyshev polynomials by observing that -$$ 1-T_{2n+2}(x)=2 U_n(x)^2 (1-x^2)$$ -and -$$ 1-T_{2n+3}(x)=(U_{n+1}(x)+U_n(x))^2 (1-x).$$ -Edit: -The following formulae show the connection between the factorization of spread polynomials and Chebyshev polynomials most clearly: -$$ S_{2n}(x^2)= (1-x^2)U_{2n-1}(x)^2 $$ -and -$$ S_{2n+1}(x^2)= T_{2n+1}(x)^2.$$<|endoftext|> -TITLE: When do we study maps into an object or from the object to another object? -QUESTION [5 upvotes]: In many Mathematical theories, to study an object, we usually consider the set of all maps from that object to some other object. For example, in differential geometry, we study the smooth maps from a manifold $M$ to $\mathbb{R}$. Or in Algebraic Geometry, we consider the structure sheaf, which is the set of maps from a variety to $\mathbb{A}^1$. -So, is there any heuristic idea about why we don't do the other way around, i.e. study the set of all maps from some object to the object we want to study (at least in the two examples above)? Would this give us any more information? And also, is there any subject in which we do that? -Edit: One more clarification that might make my question clearer. In algebraic geometry, when we write an $R-$scheme $\mathrm{Spec} A$, already implicitly, we are viewing $A$ as the ring of all $R-$functions from $\mathrm{Spec} A$ to $\mathrm{Spec} R[x]$. -Edit (based on Qiaochu Yuan's answer): maps in seem to give us local information while maps out gives us global one, at least in Differential Geometry and Algebraic geometry. For example, to learn about the tangent space at a point, we look at the map $I\to M$ (in differential geometry) and $\mathrm{Spec} k[x]/(x^2) \to X$ (in algebraic geometry). Is there any more example along these lines? - -REPLY [2 votes]: To me, it seems that your question is essentially "Why does contravariance occur more frequently than covariance?" -If one has a contravariant representable functor, then one is implicitly studying maps into a fixed (universal) object from the object one is interested in studying. -I think perhaps one reason why contravariance is more natural than covariance is what Qiaochu indicates in his answer: contravariant functors have more of an opportunity to be "local." For instance, let $X, Y$ be topological spaces, and $f: X \to Y$ a continuous map. Then a sheaf on $Y$ when pulled back to $X$ at a point $x$ depends only on the local nature near the single point $f(x)$. Things are not so nice when pushing forward a sheaf. Thus it happens that pull-backs preserve stalks, while push-forwards need not (unless you are working with a particularly nice map $f$). -In particular, if one has a bundle on $Y$, the local nature implies that it can be pulled back to $X$, while pushing a vector bundle forward will only give a sheaf, not necessarily a locally free one (i.e. a bundle). -In algebraic geometry, one of the first representable functors one encounters is the one that represents projective space. Namely, fix a field $k$ and an integer $n$; then a map from a $k$-scheme $X$ into $\mathbb{P}^n_k$ is given by a line bundle on $X$ and $n+1$ global sections generating it (up to isomorphism). This is contravariant because you can pull-back line bundles and the generating property of global sections. You can't push this forward in a reasonable manner. (This universal property generalizes to projective space bundles over any scheme.) -At an even more basic level, the definition of a sheaf itself is contravariant (it's a contravariant functor from the category of open sets with inclusions to the category of sets that satisfies unique gluing axioms).<|endoftext|> -TITLE: Uncountable preimage of every point -QUESTION [8 upvotes]: Let $f:[0,1]\to [0,1]$ be a continuous function. Must it have a point $x$ that $f^{-1}(x)$ is at most countable? -Added: Must it have a point $x$ that $dim_H(f^{-1}(x))=0$ ? ($dim_H$ means the Hausdorff dimension) - -REPLY [2 votes]: A simple example is to write any $x$ in $[0,1]$ as a binary sequence of $1$'s and $0$'s and then define continuous functions which takes odd (or even) values in the binary expansion of $x$. -That is, if $$x = a_1 a_2 a_3 a_4 a_5 \dots$$ -then $$f(x) = a_1 a_3 a_5 \dots$$ -This function is obviously continuous and has required property. -Also, the Peano curve could be defined in a similar fashion.<|endoftext|> -TITLE: What does it mean to extend a 2d (topological) conformal field theory to Deligne-Mumford space? -QUESTION [5 upvotes]: For 2D (topological) conformal field theory the corresponding moduli space is the space of Riemann surface with boundary, right? What does it mean by extending the theory to Deligne-Mumford space? How can we do that in terms of the methods we have by now? Thanks for any expository answers! - -REPLY [3 votes]: Cohomological Conformal Field theory (CCFT) introduced by M. Kontsevich, Yu. Manin as follows which is the extension of a 2D (topological) conformal field theory to Deligne-Mumford space -Let $(M,\omega)$ be a symplectic manifold with choice of complex structure $J$. -Define $ev:\overline{\mathcal M}_{g,k}(M,J,\beta)\to M^k$ and $d:\overline{\mathcal M}_{g,k}(M,J,\beta)\to\overline{\mathcal M}_{g,k}$ -Let $\Lambda_0$ be Novikov ring, now define the maps -$$I_{g,k}:H^*(\overline{\mathcal M}_{g,k})\otimes H^*(M)\to \Lambda_0$$ -as follows -$$<\alpha_1,\cdots\alpha_k>_{g,\beta}^\psi :=\sum_{\beta\in H_2(M)}\int_{\overline{\mathcal M}_{g,k}}d^*\psi\wedge ev^*(\alpha_1\otimes\cdots\otimes\alpha_k)T^{\omega(\beta)}\in\Lambda_0$$ -Moreover, we add the following axioms to introduce cohomological Conformal Field theory -1) $\overline{\mathcal M}_{g_1,k_1+1}\times \overline{\mathcal M}_{g_2,k_2+1}\to \overline{\mathcal M}_{g_1+g_2,k_1+k_2}$ -these induces maps, -2) $ H^*(\overline{\mathcal M}_{g_1+g_2,k_1+k_2})\to H^*(\overline{\mathcal M}_{g_1,k_1+1})\otimes H^*(\overline{\mathcal M}_{g_2,k_2+1})$ -and -3) $\overline{\mathcal M}_{g,k+2}\to \overline{\mathcal M}_{g+1,k}$ -As Kontsevich-Manin interpreted , the algebraic package of a genus zero Cohomological Field theory as follows -We can define a $\Lambda_0$-linear maps, -$$\tilde{I_k}:H_*(\overline {\mathcal M_{0,k+1}})\to Hom(H^*(M,\Lambda_0)^{\otimes k},H^*(M,\Lambda_0))$$ -via -$$<\tilde{I_k}(S)(\alpha_1,\cdots,\alpha_k),\alpha_{k+1}>:=<\alpha_1,\cdots,\alpha_{k+1}>_{g=0}^{P.D.(S)}$$ -where P.D. , here means Poincare dual.<|endoftext|> -TITLE: Bounding the growth of rational bivariate polynomials from below -QUESTION [6 upvotes]: The following question is an attempt to find a lower bound for the value of a polynomial at integer points. It is something that I originally thought about while trying to understand how it would be possible to approach this MO question about polynomials representing the nonnegative integers, but does seem to be very interesting in its own right. I guess this would fall under the study of Diophantine approximation/geometry, which I am not at all expert on. So it could even be a known conjecture or theorem, or obviously false. Maybe someone on MO will be able to say? -As is well known, the Thue-Siegel-Roth theorem says that, for an irrational algebraic number $\alpha$, there are only finitely many pairs of integers $p,q$ satisfying the inequality $\vert p/q-\alpha\vert\le q^{-2-\epsilon}$. Here, $\epsilon$ is any fixed positive real number. -This is easily seen to be equivalent to the following lower bound on the growth of homogeneous and irreducible polynomials $f\in\mathbb{Q}[X,Y]$ of degree $d > 1$. For all but at most finitely many integer pairs $x,y$, the inequality -$$ -\vert f(x,y)\vert\ge\vert y\vert^{d-2-\epsilon}\qquad\qquad{\rm(1)} -$$ -holds. The argument is very simple. We can decompose $f(x,y)$ as $y^d\prod_{i=1}^d(x/y-\alpha_i)$ for distinct irrational algebraic numbers $\alpha_i$. As $x/y-\alpha_i$ can only be made arbitrarily small for at most one $i$ at a time, (1) is equivalent to applying the Thue-Siegel-Roth theorem to $x/y-\alpha_i$. -Now for my question. Is there an extension of (1) to non-homogeneous polynomials $f$? Now, I realize that it cannot possibly carry directly across to the non-homogeneous case in the same form. For one thing, a simple change of variables allows us to replace $f$ by a polynomial of arbitrarily large degree, such as $\tilde f(x,y)=f(x+y^r,y)$, which would invalidate any inequality depending on the degree of $f$ and, similarly, the right hand side of (1) would change form under changes of variables. To guess how this can be fixed, we can look at Siegel's theorem, which says that $f(x,y)=a$ has only finitely many integer solutions in $x,y$ whenever $f-a$ defines a curve over $\mathbb{Q}$ of genus $g$ at least one. It seems reasonable then, that a generalization of (1) should involve the genus $g$ of the curve defined by $f-a$, for typical rationals $a$, and not the degree. -So, to be precise, my question is whether there is an increasing and unbounded function $\phi\colon\mathbb{N}\to\mathbb{R}$ with the following property: If $f\in\mathbb{Q}[X,Y]$ is such that $f-a$ defines a curve of genus $g$ (for all but finitely many $a$), there exists a nonconstant polynomial $h\in\mathbb{Q}[X,Y]$ such that the inequality -$$ -\vert f(x,y)\vert\ge \vert h(x,y)\vert^{\phi(g)-\epsilon}\qquad\qquad{\rm(2)} -$$ -holds on $\mathbb{Z}\times\mathbb{Z}$ outside of a finite set (*). We might even hope that $h(x,y)=y$ under a change of variables, but that seems like a bit much to ask. Comparing with the case where $f$ is homogeneous of degree d, the genus of $f-a$ for nonzero $a$ is given by $g=(d-1)(d-2)/2$ and we have $\phi(g)=(-1+\sqrt{1+8g})/2$, although I doubt that precise form would hold in the non-homogeneous case. -(*) Edit: It is easy to construct polynomials which vanish or degenerate into something simpler on a given finite set of curves. E.g., $f=\prod_{i=1}^n(X-a_i)$ is zero on the curves $x=a_i$. So I should say that (2) holds on $\mathbb{Z}\times\mathbb{Z}$ outside of a finite set of curves (of genus zero). - -REPLY [6 votes]: What you are asking is for a strong effective form of Siegel's theorem. There are a number of conjectures in this direction, all considered very hard. For example, if $f=x^3-y^2$, what you are asking is essentially Hall's conjecture: http://en.wikipedia.org/wiki/Hall%27s_conjecture . -In general, it follows from Vojta's conjectures (LNM 1239). Nothing close is known. For certain classes of polynomials (general cubics, $y^m - f(x)$, a few others) you can get a lower bound of the type $\log \log \log \max\{x,y\}$ give or take a log or two, from the theory of linear forms in logarithms.<|endoftext|> -TITLE: Computing the Chern-Simons invariant of $SO(3)$ -QUESTION [15 upvotes]: I recently tried asking this question on math.stackexchange.com but I have not yet received any feedback so I thought I should try asking here. I apologize if this question is too basic. The question is reproduced below: -I am an undergraduate learning about gauge theory and I have been tasked with working through the two examples given on pages 65 and 66 of "Characteristic forms and geometric invariants" by Chern and Simon. I will recount the examples and my progress at a solution. For ease here is the relevant text: - -Example 1. Let $M = \mathbb{R}P^3 = SO(3)$ together with the standard metric of constant curvature 1. Let $E_1, E_2, E_3$ be an orthonormal basis of left invariant fields on $M$, oriented positively. Then it is easily seen that $\nabla_{E_1}E_2 = E_3, \nabla_{E_1}E_3 = - E_2, \text{ and } \nabla_{E_2}E_3 = E_1$. Let $\chi : M \rightarrow F(M)$ be the cross-section determined by this frame. - $$\Phi(SO(3)) = \frac{1}{2}.$$ -Example 2. Again let $M = SO(3)$, but this time with left invariant metric $g_{\lambda}$, with respect to which $\lambda E_1, E_2, E_3$ is an orthonormal frame. Direct calculation shows - $$\Phi(SO(3),g_{\lambda}) = \frac{2\lambda^2 - 1}{2\lambda^4}.$$ - -For each of these examples I am expected to calculate -$$\Phi(M) = \int_{\chi(M)} \frac{1}{2} TP_1(\theta)$$ -which lies in $\mathbb{R}/\mathbb{Z}$. Previously in the paper they give an explicit formulation of $TP_1(\theta)$ in terms of the "component" forms of the connection $\theta$ and its curvature $\Omega$, -$$TP_1(\theta) = \frac{1}{4\pi^2}\left( \theta_{12}\wedge\theta_{13}\wedge\theta_{23} + \theta_{12}\wedge\Omega_{12} + \theta_{13}\wedge\Omega_{13} + \theta_{23}\wedge\Omega_{23}\right).$$ -I have verified this formula for myself given the information in the paper. Using the structural equation $\Omega = d\theta + \theta\wedge\theta$ I am able to reduce the expression for $TP_1(\theta)$ to -$$TP_1(\theta) = \frac{-1}{2\pi^2}\left( \theta_{12}\wedge\theta_{13}\wedge\theta_{23} \right).$$ -I don't believe I have assumed anything about the structure of $M$ during that reduction so I believe it should hold for both examples. I continue by claiming that since $E_1, E_2, E_3 \in so(3)$, the Lie algebra of $SO(3)$, I should be able to compute $\theta$ by considering -$$\nabla_{E_i}E_j := (\nabla E_j)(E_i) = \sum_k E_k \otimes \theta^{k}{}_{ij}(E_i)$$ -and comparing it with the given derivatives. -For example one this yielded for me $\theta_{12} = E^3, \theta_{13} = -E^2, \theta_{23} = E^1$ where $E^i$ are the 1-forms dual to the basis $E_i$. Then I think that $\chi^*$ should act trivially on $TP_1(\theta)$ as it is a horizontal form in $\Lambda^*(T^*F(M))$. Therefore I find that $\chi^*(TP_1(\theta)) = \frac{1}{2\pi^2}\omega$, where $\omega$ is the volume form of $M$, and when integrated this yields the correct answer of $\frac{1}{2}$ for the first example. -However, my approach fails completely for the second example. I assume that the set $\lambda E_1, E_2, E_3$ obeys the same derivate relationships as given in the first example, but this does not seem to give me enough factors of $\lambda$. I suspect that I am not handling the computation of the $\theta_{ij}$ forms or the application of $\chi^*$ correctly, however I am uncertain what my exact issue is. Is there a fundamental flaw in my understanding? I am hoping someone with more experience can point me in the right direction. - -REPLY [4 votes]: Not going into details, just would like to note that connection forms do depend on $\lambda$. -If I take $t=\lambda^{-1}$, then from Coshul formula -$$2(\nabla_{E_1}E_2,E_3) = E_1(E_2,E_3) + E_2(E_1,E_3) - E_3(E_1,E_3) - (E_1,[E_2,E_3]) - (E_2,[E_1,E_3]) + (E_3,[E_1,E_2])$$ -it follows $2(\nabla_{E_1}E_2,E_3) = -t^2 + 2$ (from $(E_1,E_1)=t^2$). -The double cover of $SO(3)$ with this metric $g_t$ is the well known "Berger sphere" which collapses to the base of the Hopf fibration $S^1 \to \pi: S^3 \to S^2$ when $t$ goes to $0$, and the fiber $S^1$ as t goes to $\infty$. You may find complete formulas for the connection $\nabla^t$ of $g_t$ by searching Berger sphere, I think.<|endoftext|> -TITLE: What makes four dimensions special? -QUESTION [137 upvotes]: Do you know properties which distinguish four-dimensional spaces among the others? - -What makes four-dimensional topological manifolds special? -What makes four-dimensional differentiable manifolds special? -What makes four-dimensional Lorentzian manifolds special? -What makes four-dimensional Riemannian manifolds special? -other contexts in which four dimensions or $3+1$ dimensions play a distinguishing role. - -If you feel there are many particularities, please list the most interesting from your personal viewpoint. They may be concerned with why spacetime has four dimensions, but they should not be limited to this. - -REPLY [2 votes]: 3 dimensions of space are special, because this is the lowest number of dimensions, where a random walk doesn't return to its origin with certainty (probability = 1), see -http://mathworld.wolfram.com/PolyasRandomWalkConstants.html -Similarly I think one time dimension is special, because less than one would mean no evolution at all, and more than one would lead to instabilities of all kinds.<|endoftext|> -TITLE: Random rotations in SO(3) and free group -QUESTION [10 upvotes]: Is it true that two random (w.r.t. Haar measure) rotations in $SO(3)$ generate a free group? - -REPLY [7 votes]: See also: Epstein - Almost all subgroups of a Lie group are free (MR).<|endoftext|> -TITLE: How to rearrange only negative part of a conditionally convergent series to get any sum greater then initial? -QUESTION [9 upvotes]: Suppose that $\sum^\infty_{n=1} u_n = s,$ where the series converges conditionally, and $s'>s$. How to prove the existence of such a permutation $\sigma,$ such that -1) $u_n\geq 0 \rightarrow \sigma(n)=n$ -2) $\sum^\infty_{n=1} u_{\sigma(n)} = s'?$ -I know how to rearrange positive divergent series to get speed of divergence as low as needed, but I don't know, how to apply this to mentioned problem - -REPLY [12 votes]: Georgii: -Let me start with some brief remarks. In a series of three papers: - -a. Wacław Sierpiński, "Contribution à la théorie des séries divergentes", Comp. Rend. Soc. Sci. Varsovie 3 (1910) 89–93 (in Polish). - - -b. Wacław Sierpiński, "Remarque sur la théorème de Riemann relatif aux séries semi-convergentes", Prac. Mat. Fiz. XXI (1910) 17–20 (in Polish). - - -c. Wacław Sierpiński, "Sur une propriété des séries qui ne sont pas absolument convergentes", Bull. Intern. Acad. Sci.: Cracovie A (1911) 149–158. - -Sierpiński proved three extensions of Riemann's theorem: -Let $\sum_{n\ge0}a_n=s$ be a conditionally convergent series. Then: - -For every $r\in{\mathbb R}\cup\{-\infty,+\infty\}$ there is a permutation $\pi:{\mathbb N}\to{\mathbb N}$ such that $\sum_{n=0}^\infty a_{\pi(n)}=r$ and for every $n$, $$ a_n<0\Longleftrightarrow a_{\pi(n)}<0. $$ -For every $r\le s$ there is a permutation $\pi$ with $\sum a_{\pi(n)}=r$ and $\pi(n)=n$ whenever $a_n<0$. -For every $r\ge s$ there is a permutation $\pi$ such that $\sum a_{\pi(n)}=r$ and $\pi(n)=n$ whenever $a_n>0$. - -The result asked in the original question is item 3 above. Note that items 1 and 2 follow easily from this. -There is a recent paper that addresses similar questions in a more general (descriptive set-theoretic) context, and I believe you may find interesting some of the results there. The paper is "Rearrangement of conditionally convergent series on a small set" by Rafał Filipów and Piotr Szuca, Journal of Mathematical Analysis and Applications, 362 (2010) 64–71. MR2557668 (2010i:40001). -In this paper, Riemann's theorem is extended in a spirit similar to Sierpiński's results: Now one considers an ideal ${\mathcal I}\subseteq{\mathcal P}({\mathbb N})$, and asks whether for every conditionally convergent series as above and any $r$ in the extended reals, there is a permutation $\pi$ of the natural numbers such that $\sum a_{\pi(n)}=r$ and $$\{n\mid\pi(n)\ne n\}\in{\mathcal I}.$$ If this is the case, one says that ${\mathcal I}$ has property (R). -(The standard intuition behind an ideal is that it defines a notion of smallness, so we are asking that only a small number of indices are changed.) The results in the paper relate ideals with property (R) to other well-known classes of ideals, studied by Farah and others. Section 4 of the paper specifically deals with Sierpiński-like results for ideals with property (R), see their Theorem 4.1. - -Here is a sketch of the proof of item 2. [Edit, Nov 17, 2014: My original argument for item 3 started with conditionally convergent series $\sum_i u_i=r$ and, for each $s\ge r$, produced a rearrangement fixing the positive terms, and whose partial sums have $\liminf s$, but to argue that their $\limsup$ is also $s$ requires additional work. Fixing this lead me to a proof very similar to Sierpiński's, so I decided to reorganize the answer and just present his argument, together with what I hope are some additional explanations. What follows is part of a post on rearrangements of series on my blog.] - -Theorem (Sierpiński). Let $\sum_i u_i$ be a conditionally convergent series of real numbers that converges to $U$. For any $V\le U$ there is a rearrangement $v_0,v_1,\dots$ of the $u_i$ such that $\sum_i v_i=V$, and the rearrangement leaves fixed all non-positive terms. -Similarly, for any $W\ge U$ there is a rearrangement $w_0,w_1,\dots$ of the $w_i$ such that $\sum_i w_i=W$, and the rearrangement leaves fixed all positive terms. -Moreover, for any extended real $\alpha$, there is a rearrangement $s_0,s_1,\dots$ of the $u_i$ such that $\sum_i s_i=\alpha$ and, for every $n$, $u_n>0$ iff $s_n>0$. - -Proof. Note first that it suffices to prove the result about rearrangements fixing non-positive terms, since the other results are consequence of it: simply apply the first result to the series $\sum_i -u_i$, noting that $-W\le -U$, to obtain the second result. For the first, apply first either the first or the second result if $\alpha\in\mathbb R$. The case where $\alpha=\pm\infty$ is handled easily. -Just as Riemann's theorem depended on a lemma about the series of positive and non-positive terms of the original sequence, Sierpiński's result depends on a lemma about the series of positive terms. Unlike Riemann's theorem, in this case, all the work goes into the proof of the lemma, and the theorem is more properly a corollary of it. -Lemma. Suppose $a_0,a_1,\dots$ is a sequence of positive numbers such that $\sum_i a_i=+\infty$ and $\lim_i a_i=0$ Let $A_0,A_1,\dots$ denote the sequence of partial sums. For any $L\ge0$ there is a rearrangement $c_0,c_1,\dots$ with sequence of partial sums $C_0,C_1,\dots$ such that $\lim_i (A_i-C_i)=L$. -To see that the lemma implies the result, suppose that $V\le U$, and let $L=U-V$. Let $a_0,a_1,\dots$ and $-b_0,-b_1,\dots$ be, respectively, the subsequences of positive and of non-positive terms of the $u_i$. Let $A_0,A_1,\dots$ and $B_0,B_1,\dots$ denote, respectively, the sequences of partial sums of the $a_i$ and of the $b_i$. For each $i$, let $p_i$ be the number of positive terms among $u_0,\dots,u_i$, and let $n_i$ be the number of non-positive terms, so that $p_i+n_i=i+1$ and, if $U_0,U_1,\dots$ denote the partial sums of the $u_i$, then $U_i=A_{p_i}-B_{n_i}$ for all $i$, and $p_i,n_i\to+\infty$. -Apply the lemma to the $a_i$ to obtain a rearrangement $c_0,c_1,\dots$ with partial sums $C_0,C_1,\dots$ such that $A_i-C_i\to L$. Consider the rearrangement $v_0,v_1,\dots$ of the $u_i$ that fixes the $-b_i$ and permutes the $a_i$ as indicated. If $V_0,V_1,\dots$ are the partial sums of the $v_i$, note that, by design, $V_i=C_{p_i}-B_{n_i}$, so that $U_i-V_i=A_{p_i}-C_{p_i}\to L$, and $V_i\to U-L=V$, as wanted. All that remains is to prove the lemma. -Proof. As in Riemann's theorem, we proceed by stages. Starting with $A_{-1}=C_{-1}=0$, at stage $n$ we examine $A_{n-1}-C_{n-1}$ to decide the value of $c_n$. As in that theorem, we want to arrange that the values $A_i-C_i$ increase if smaller than $L$, and decrease if larger, so that in the limit we obtain the desired value. Actually, we will need to modify slightly this strategy. To motivate the proof, consider first the case where the $a_i$ are monotonic, that is, $a_0\ge a_1\ge\dots$ In this case, the strategy works: At stage $n$, we consider two cases, according to whether $A_{n-1}-C_{n-1}\le L$ or $A_{n-1}-C_{n-1}>L$: - -If $A_{n-1}-C_{n-1}\le L$, then let $r$ be such that $a_rn-1$ is least such that $A_k-C_k>L$, then $A_k-C_k=(A_{k-1}-C_{k-1})+(a_k-c_k)\le L+a_k$. - -If $A_{n-1}-C_{n-1}>L$, then in particular $A_{n-1}\ne C_{n-1}$, so at least one of the $a_i$, $i(A_{k-1}-C_{k-1})-c_k>L-c_k.$$ -Note that our assumption that the $a_k$ are decreasing ensures that, in case 2, (with notation as above) $A_n-C_n=(A_{n-1}-C_{n-1})+(a_n-c_n)\le$ $A_{n-1}-C_{n-1}$, since $a_i\ge a_n$. Moreover, it cannot be that we have equality from some point on, since the $a_k$ converge to $0$. This means that the values $A_k-C_k$ decrease. -Since $a_i,c_i\to 0$, it follows that, if case 1 is also applied infinitely often, then $A_i-C_i\to L$.  Therefore, to conclude, it suffices to argue that we cannot stay in case 2 forever. To see this, argue by contradiction, and assume that from $n-1$ on, we always stay in case 2. Let $j_m$ be the number of indices $i\le m$ such that $a_i$ is not one of the $c_k$, $k\le m$. For any $m\ge n$, since $A_{m-1}-C_{m-1}>L$, then $j_m\le j_{m-1}$, with equality if and only if $a_m$ is not one of the $c_k$, $kL$. Also, once we enter case 2, we cannot stay there indefinitely, as the number of terms $a_i$ that are not some $c_k$, $k\le j$, decreases. The problem is that the sequence of $A_i-C_i$ is not necessarily decreasing. In fact, if we are at a stage where we define $c_n=a_i$ for some $ia_i$, then $A_n-C_n>A_{n-1}-C_{n-1}$. This means that, although we have ensured that $\liminf (A_i-C_i)=L$, and that there is a subsequence of the $A_i-C_i$ converging to $L$, it may well be the case that $\limsup (A_i-C_i)>L$. -Sierpiński deals with this situation in a clever fashion. He ensures that if we are in case 2, so that $c_n=a_i$ for some $iL$, now we consider two possibilities, according to whether or not the index $n$ was picked previously: If it was not, then we let $c_n=a_n$. Note that $A_n-C_n=A_{n-1}-C_{n-1}$ if this is the case, and that, since $C_{n-1}\ne A_{n-1}$, then necessarily some $c_i$ with $in$. This means that at some later stage (at most by stage $r$), we are no longer to be in this case. -If $A_{n-1}-C_{n-1}>L$, and the index $n$ was chosen previously, then we let $c_n$ be $a_i$, where $i$ is the first index less than $n$ not chosen yet. - -As before, cases 1 and 3 happen infinitely often, so we have indeed defined a rearrangement. Moreover, if we are in case 3, then $a_n=c_i$ for some $ii$ is if stage $i$ was by case 1, which means that $c_i<1/2^i$. -Fix $\epsilon>0$. Since cases 1 and 3 happen infinitely often, if we choose $n$ large enough, then we can ensure that all indices $m$ mentioned below are so large that $a_m,c_m<\epsilon$, and if $a_m=c_k$ for some $kN$ where $\sum_{i\ge N}1/2^i<\epsilon$. We want to ensure that $|(A_n-C_n)-L|<2\epsilon$. This proves that $A_n-C_n\to L$. -If stage $n$ is by case 1, and stage $mL$, then $A_{m+1}-C_{m+1}=(A_m-C_m)+(a_{m+1}-c_{m+1})>$ $L-c_{m+1}>L-\epsilon$. Since the sequence $A_i-C_i$ is increasing for $m+1\le i\le n$, we have $L\ge A_n-C_n>L-\epsilon$, as wanted. -Suppose then that stage $n$ is by case 3, and that stage $mN$, so -$$\sum_{i=m+2}^n(a_i-c_i)=\sum\{a_i-c_i\mid m+2\le i\le n,\mbox{ and stage }i\mbox{ was by case 3}\}\le\sum\{a_i\mid m+2\le i\le n,\mbox{ and stage }i\mbox{ was by case 3}\} \le\sum_{j>N}\frac1{2^j}<\epsilon.$$ -This means that $A_n-C_nL-c_n>L-\epsilon.$$ This completes the proof. $\Box$ -This concludes the proof of Sierpiński's theorem. $\Box$<|endoftext|> -TITLE: Alternative definition of monoidal categories -QUESTION [13 upvotes]: My question is about monoidal categories. To motivate it, let me first recall something about group objects. -Assume you define a group object in a category $C$ with products by an object $G$ together with morphisms $G \times G \to G, G \to G, * \to G$, so that the diagrams commute which correspond to the group axioms. Then you want to generalize some elementary properties from usual groups ($C=Set$) to these group objects, but it is quite hard to write down the relevant diagrams. For example when you want to prove that there is a actually at most one unit $* \to G$, a morphism $G \to G'$ between group objects which respects the multiplication already respects the inversion and the unit, left-inverse implies right-inverse, etc. But all this may be reduced to the case $C=Set$ by using the Yoneda-Lemma and the the definition of a group object as an object together with a factoriation of its hom-functor over the category of usual groups. Then these calculations are easy and you don't have to produce all these diagrams. -My question is: Is there a similar definition for a monoidal category? Specifically, I want to see a neat (diagram-free?) proof of Lemma 3.2.5 in this note (Pareigis' lectures on quantum groups), which is intuitive and does not come up with diagrams without motivating them. (For me, "We tried to prove it and finally this diagram worked" is no intuition.) The lemma implies for example that the endomorphism monoid of the unit object of a monoidal category is abelian, which is quite surprising (for me). This Lemma is just one example. It seems to me that monoidal categories are "weak monoids in the 2-category Cat", but I don't see yet if this description actually simplifies these diagrams. - -REPLY [9 votes]: The kind of thing you are looking for applies not just to monoidal categories, but to bicategories, and it is called the bicategorical Yoneda lemma. If $B$ is a small bicategory, one may form the strict 2-category $[B^{op}, Cat]$ consisting of weak 2-functors (aka homomorphisms), pseudonatural transformations, and modifications from $B^{op}$ to $Cat$. Then there is a Yoneda embedding -$$y: B \to [B^{op}, Cat]$$ -sending an object $b$ to $\hom(-, b)$, and the mapping of $B$ onto its image is a bi-equivalence. The image however is a strict 2-category, and hence we get from this a coherence theorem which assures us that all definable diagrams (in say the free monoidal category generated by a discrete category) commute. This line of thinking was introduced by Street in his paper Fibrations in Bicategories (for which there is also a correction). -This is really the modern point of view on coherence in monoidal categories. An exposition which focuses on monoidal categories can be found (I believe) in Braided Tensor Categories by Joyal and Street. -It is true that monoidal categories can be described as weak monoids, but this by itself doesn't solve coherence issues.<|endoftext|> -TITLE: What is the subgroup generated by involutions? -QUESTION [14 upvotes]: I was recently taking some notes on the Cartan-Dieudonné theorem: if $(V,q)$ is a nondegenerate quadratic space of finite dimension $n$ over a field of characteristic not $2$, then every element of the orthogonal group $O(q)$ is a product of at most $n$ reflections through nondegenerate hyperplanes. (Moreover -- this is not the hard part! -- $n$ is sharp, as one sees by considering $-1$.) -So that's how things go in $O(q)$. At some point my mind wandered and I started thinking about $\operatorname{GL}(V)$ instead. Hyperplane reflections don't seem to have the same significance here, so I started to think about involutions -- i.e., elements of order $2$ -- instead. What is the subgroup of $\operatorname{GL}(V)$ generated by all the involutions? -Well, I was smart enough to realize that every involution has determinant $\pm 1$, so the subgroup in question is definitely contained in the subgroup of all matrices with determinant $\pm 1$ -- let's call this $\operatorname{GL}(V)^{(2)}$ -- hence usually proper. But is it actually this whole group? I checked one example: $\operatorname{GL}_2(\mathbb{F}_3)$ is isomorphic to $S_4$ [not even close -- but still, one can see that my conclusion is true!], which is indeed generated by its involutions. -I didn't make much more progress than that, so I started googling. Eventually I came across the following paper: - -Gustafson, W. H.; Halmos, P. R.; Radjavi, H. -Products of involutions. -Collection of articles dedicated to Olga Taussky Todd. -Linear Algebra and Appl. 13 (1976), no. 1/2, 157–162. - -Apparently in this paper the authors prove, among other things, that for any finite-dimensional vector space over any field, any element of $\operatorname{GL}(V)^{(2)}$ is a product of at most $4$ involutions. This is a pretty striking result, so I have some questions. -Question 1: How do you prove it? (I can't immediately access the paper, and unfortunately the description in the MathSciNet review was not immediately illuminating to me.) -Question 2: Can this result really not have been known until 1976?? The Cartan-Dieudonné theorem was proved at the latest in $1945$, the first publication date given for Dieudonné's book on classical groups. No one wondered about general linear groups for another $30$ years?? -Question 3: In lots of generality we can take a group $G$ and ask what is the subgroup generated by its involutions. To fix ideas, suppose that $G$ is the group of $K$-rational points of a connected linear algebraic group over a field $K$. (Or choose whatever special case of this you like.) Can we say something about the subgroup generated by involutions? What about the least number $N$ of involutions so that every element of this subgroup is a product of at most $N$ involutions (if such an $N$ exists)? - -REPLY [2 votes]: Maybe I can comment on Question 2. To me the essential point is that this kind of result belongs to elementary linear algebra and basic group theory rather than to geometric algebra. Generation of special linear groups (or slightly larger groups) by involutions probably goes back a long way and may have multiple origins, though I don't have a definitive reference handy. (I'd probably try to ask someone like Gustafson or Djokovic rather than search the scattered literature.) -From a semi-modern viewpoint, for instance, generation of a general linear group over an arbitrary field focuses on the traditional building blocks: elementary, permutation, and diagonal matrices. Since a one-parameter group of elementary matrices along with a suitable permutation matrix will generate a copy of SL$_2$, generation of a special linear group just requires these two kinds of ingredients. Here a permutation matrix is obviously a product of involutions. On the other hand, SL$_2$ is almost simple, while its subgroup generated by involutions is obviously normal. (To include matrices of determinant -1 is a further step.) -Books like those by Dieudonne (and Artin) mix in further ideas arising from the geometry of a bilinear form, along with consideration of exact upper bounds on the number of involutions needed to generate any group element. Such upper bounds for products of involutions in general linear groups may have come along later. -P.S. I used SL$_2$ subgroups in the rough sketch above just to emphasize some elementary steps for showing that the big groups in question are generated by involutions; of course, simplicity of SL$_n$ modulo its center for most fields could be used directly here, plus ad hoc arguments for the few nonsimple cases. Then multiplying any matrix of det -1 by a diagonal involution reduces to the det 1 case. Anyway, none of this older theory deals with an explicit upper bound on number of involutions required. -FURTHER UPDATE: Sorry to have confused matters with my offhand remarks here. While my "essential point" is unchanged, a quick look at the paper quoted (along with other comments here) indicates that the statement of the theorem is true: Each matrix of determinant 1 or -1 is the product of at most four involutions (exactly four if you follow their convention that an involution is an element of square 1). I managed to make a copy of the bound journal article at the library this morning, where the methods used rely just on linear algebra. The authors are following in a tradition where products of 2 or 3 matrix involutions have been characterized. -The point seems to be that you work exclusively in the larger group of all matrices having determinant 1 or -1 (leaving special linear groups aside). Here the argument involves rational canonical forms and permutation matrices having either determinant. With this flexibility, the argument in the paper seems OK but is written down a bit loosely. It would help to state their theorem by giving the group in question a name (there seems not to be a standard one). Since I think first in terms of connected algebraic or Lie groups, I was too quick to interpret the theorem as applying to special linear groups. (Like Brian I am also tempted to consider other Lie types, relying on Bruhat -decomposition and the like.)<|endoftext|> -TITLE: orientations for zero-dimensional manifolds -QUESTION [14 upvotes]: I am teaching a course on manifolds, and soon I will have to prove the Stokes' theorem which, of course, involves defining oriented manifolds. There are many ways to define an oriented manifold. My favorite way is by the reduction of the structure group of the tangent bundle. But this definition and a couple of other that I know give just one orientation for the point: $GL(V)/GL^{+}(V)$ is -${\mathbb Z}/2{\mathbb Z}$ when $\dim V \ge 1$, but when $\dim V=0$ then $GL(V)$ has only one element. Of course, it is possible to define two orientations of a point by convention. I would like to know if is there any way to define the orientation for smooth manifolds in a uniform way that would also yield two orientations of a point. - -REPLY [10 votes]: Here is a definition that works for all dimensions, including zero. -Let $V$ be a real vector space equipped with a Riemannian metric. - -An orientation on $V$ is an isometric isomorphism between $\mathbb R$ and the top exterior power of $V$. - -If you want to get rid of the Riemannian metric, you can also say that an orientation is a ray inside the top exterior power of $V$. - -Here is a similarly well-behaved definition of spin structure. It also works all the way down to dimension zero. As above, -let $V$ be a real vector space equipped with a Riemannian metric. - -A spin structure on $V$ is a Morita equivalence between Cliff($\mathbb R^n$) and Cliff($V$). In other words, it is a Cliff($\mathbb R^n$)-Cliff($V$)-bimodule that has the property that it induces an equivalence between the category of Cliff($\mathbb R^n$)-modules and that of Cliff($V$)-modules - -To see the analogy with the above definition of orientation, -note that $\Lambda^{top}\mathbb R^n=\mathbb R$. And so requiring an isomorphism between -$\mathbb R=\Lambda^{top}\mathbb R^n$ and $\Lambda^{top}V$ is formally similar to requiring a Morita equivalence between Cliff($\mathbb R^n$) and Cliff($V$). You just replace the functor $\Lambda^{top}$ by the functor Cliff, and replace the 1-category of vector spaces with the 2-category of algerbas and bimodules.<|endoftext|> -TITLE: How to "globalize" the inverse function theorem? -QUESTION [7 upvotes]: Let $F: V \times W\rightarrow Z$, where $V,W,Z$ are finite-dimensional smooth (or analytic) manifolds and $F$ is smooth (or analytic). Assume that $\dim W=\dim Z$ and the usual inverse function theorem applies when we fix the first argument of $F$, so locally we have an inverse (w.r.t. the second argument of $F$) mapping $f: V \times Z \rightarrow W$ such that for any $v,w,z$ from suitably chosen small vicinities we have $F(v,f(v,z))\equiv z$ and $f$ is smooth (or analytic). Here comes the question: - -Is there any sensible way to glue together these $f$'s so that we can speak of them globally, and under which conditions? -More precisely, is there a way to treat this situation "as if" $f$ were defined globally, without doing the "restrict ourselves to suitably chosen small vicinities" thing all the time, even if we allow the Jacobian of $F$ w.r.t. $w$ to vanish at some points or even on some submanifolds of $V\times W$ of nonzero codimension? - -I do not quite expect that (under reasonably mild conditions) one can construct -a global $f$ defined on the whole $Q=Z\times V/Sing$ where $Sing=\lbrace(v,z)\in V\times Z|f \quad \mbox{can't be reasonably defined}\rbrace$ but perhaps one can have a sheaf of such $f$'s on $Q$ or something of the sort? I just started with learning the sheaf theory, so I don't quite (yet) know how to make things work on my own. -While in my particular application I would like $V$ to be a manifold, if this makes things easier, I can quite well make do with $W$ and $Z$ being just open domains in $\mathbb{R}^n$ or $\mathbb{C}^n$ (or whole $\mathbb{R}^n$ or $\mathbb{C}^n$ for that matter); that's the main reason why I wrote the domain of $F$ as $V\times W$ rather than just some general manifold $M$. -In fact, I am quite convinced that things like that should have been treated in the literature but I wasn't able to google up anything reasonable so far (maybe I just don't know the right keywords), so pointing out any suitable references is most welcome. - -REPLY [8 votes]: There exist several known global implicit function theorems. Those results tend to be tailored for specific applications. It seems to be rather difficult to state a universally useful one-size-fits-all version. -One result that I find particularly helpful goes back to Hadamard (see, e.g., Chapt. 6 of "The Implicit Function Theorem" by Krantz and Parks): - -Theorem. Let $M$ and $N$ be smooth, connected manifolds of dimension $d$ and let $f:M\to N$ be a $C^1$ mapping. If - -$f$ is proper (i.e. $f^{-1}(K)\subset M$ is compact whenever $K\subset N$ is compact), -the Jacobian of $f$ vanishes nowhere on $M$, and -$N$ is simply connected, - -then $f$ is a homeomorphism. - -You might be interested also in this paper by Rheinboldt. It contains some topological conditions on when the local solvability of the equation -$$F(x,f(x,z))=z$$ -leads to the global solvability.<|endoftext|> -TITLE: The Grothendieck plus construction for stacks of n-types -QUESTION [7 upvotes]: In Jacob Lurie's Higher Topos Theory, Section 6.5.3, he briefly mentions that to stackify a presheaf of $n$-groupoids, one needs to apply the "+"-construction $\left(n+2\right)$ times, and in general, for a presheaf of $\infty$-groupoids, one needs to apply a transfinite iteration. However, not much detail is given about this. Does anyone know where I can read more about this? Thanks. - -REPLY [3 votes]: This is discussed in section 3.4.3 of https://arxiv.org/abs/2004.00731 by Mathieu Anel and Chaitanya Leena Subramaniam.<|endoftext|> -TITLE: Does $\pi_1(Spec(\mathbb{Z}[1/p]))$ depend on p? -QUESTION [14 upvotes]: How difficult is it to know what $\pi_1(Spec(\mathbb{Z}[1/(p_1...p_r)]))$ is? Is it independent of the choice of $p_1,...,p_r$? When is it known, and what is known about it? - -REPLY [13 votes]: To add on to Pete's answer, let me comment that the differences are even more pronounced if we look at the maximal pro-$p$ quotient $\pi_1(\operatorname{Spec}(\mathbb{Z}\left[\frac{1}{p_1p_2\cdots p_r}\right])^{(p)}$ of this etale fundamental group. For example, if $r\geq 4$, then this group is infinite, in fact non-$p$-adic-analytic, if each $p_i\equiv 1\pmod{p}$ and is trivial if each $p_i\not\equiv1\pmod{p}$. The latter is basically for stupid reasons (only primes which are 1 mod p can ramify in a $p$-extension). But even ignoring stupid cases, there's a lot of fantastic arithmetic going on here. For example, $\pi_1(\operatorname{Spec}(\mathbb{Z}\left[\frac{1}{19\cdot 103}\right])^{(2)}$ is finite whereas $\pi_1(\operatorname{Spec}(\mathbb{Z}\left[\frac{1}{17\cdot 103}\right])^{(2)}$ is infinite, results which stem from simple quadratic residue calculations. Figuring out to generate these kinds of results more generally is an active and difficult area of research.<|endoftext|> -TITLE: Are there moves between Reidemeister moves? -QUESTION [22 upvotes]: Background -Knots are typically written in 2 dimensions as a loop in the plane with normal crossings. One then asks when two such diagrams describe the same knot. Two diagrams describe the same knot when one can be made into the other by a sequence of Reidemeister moves. These are three simple manipulations of a diagram which obviously don't change the underlying knot. -The Reidemeister Graph -One may consider the 'Reidemeister graph' of a knot diagram (probably not an official term), which consists of every diagram which is equivalent to the original, and an edge for every Reidemeister move between them. Since every Reidemeister move can be undone by the same move again, this is an undirected graph. -Two diagrams may be connected by a many different sequences of Reidemeister moves. It is not hard to find a sequence of moves which takes a knot to itself, and is not trivial in the sense that involves doing and immediately undoing the same move. As a consequence, the Reidemeister graph is infinite and not simply connected. -Question -I can think of a loop in the Reidemeister graph as a kind of 'relation between relations' (where the moves are the relations). I would like to find a finite list of relations between relations, such that every loop is built from these relations. -I'll be more specific. Define a higher Reidemeister move to be a locally-defined sequence of Reidemeister moves which relate a given diagram to itself. I would like a finite list of higher Reidemeister moves, such that if one fills in the corresponding loops in the Reidemeister graph with a 2-cell, the resulting space is simply connected. - -REPLY [7 votes]: Take a look at page 180 of Low dimensional topology by Tomasz Mrowka, Peter Steven Ozsvát, (following Ben Webster's comment about Movie Moves elucidated by Baez and Langford and their 30 basic movie moves, and by Carter and Saito who describe a 31st basic movie move.) A movie move is a sequence of frames of a braid (or subregion of a knot, I suppose). -Carter and Saito have a theorem that - -two movies represent the same tangle cobordism iff they can be related by a sequence of movie moves - -If you take the subset of Movie Moves where each movie is a composition of a sequence of Reidemeister moves, it seems like that would be similar or equivalent to what you are calling "Higher Reidemeister moves." Am I understanding you correctly? -I would point you out to the appropriate page on that wiki o' info, but "Movie moves" does not even show up on their search page.<|endoftext|> -TITLE: Gromov-Witten classes (as opposed to invariants)? -QUESTION [11 upvotes]: Let $\overline{M}_{g,n}(X,\beta)$ be the moduli of stable maps into $X$ of class $\beta \in H_2(X)$. We have the evaluation maps $\operatorname{ev}_i : \overline{M}\_{g,n}(X,\beta) \to X$. Given $\alpha_i \in H^\ast(X)$, the Gromov-Witten invariant corresponding to the tuple $(X,\beta,g,n,\alpha_i)$ is the integral $$\int_{[\overline{M}_{g,n}(X,\beta)]^\text{vir}}\bigwedge_i \operatorname{ev}_i^\ast(\alpha_i).$$ -There is also the "forgetful" map (or "stabilization" map) $F : \overline{M}_{g,n}(X,\beta) \to \overline{M}\_{g,n}$. I don't know if this is the standard terminology (is it?), but one can define the Gromov-Witten class corresponding to the tuple $(X,\beta,g,n,\alpha_i)$ to be the (virtual) pushforward $$F_\ast^\text{vir}\left(\bigwedge_i \operatorname{ev}_i^\ast(\alpha_i)\right) \in H^\ast(\overline{M}_{g,n}).$$ - -Question: Are there any nontrivial cases in which these Gromov-Witten classes have been identified explicitly, e.g., in terms of say tautological classes on $\overline{M}_{g,n}$? - -Just as a very naive motivation, note that in many situations Gromov-Witten invariants are zero simply because of "stupid reasons", like degree/dimension reasons (i.e. the degree of the integrand doesn't match the virtual dimension), and so provide no information. But the Gromov-Witten classes may still be nonzero and contain some information. - -REPLY [7 votes]: Consider $\overline{M}_{g+1}(C,1)$ where $C$ is a (fixed) genus g curve. Geometrically, the points in this moduli space correspond to maps $E\cup_p C \to C$ whose domain is the union of an elliptic curve $E$ attached to $C$ at a node $p$. The map is an isomorphism on $C$ and collapses the component $E $ to the point $p\in C$. Thus the moduli space is isomorphic to $\overline{M}_{1,1} \times C$ -which is of the expected dimension 2. Consequently, the virtual class is just equal to the usual class and the map -$ \overline{M}\_{1,1} \times C \to \overline{M}_{g+1}$ -is an inclusion and its image is clearly an explicit stratum in the boundary.<|endoftext|> -TITLE: Multiplicative Structures On Free Resolutions -QUESTION [5 upvotes]: Hello, -this question is related to Differential graded structures on free resolution?. -Given a regular local ring $S$ and $f\in{\mathfrak m}_S\setminus\{0\}$, I am interested in studying $R$-modules through their $S$-free resolutions. More precisely, given an $R$-module $M$, any $S$-free resolution $F^{\ast}$ of $M$ admits homotopies $s_n$ of respective degree $2n-1$ such that $s_0$ equals the differential, $s_1$ is a nullhomotopy for the multiplication with $f$ and for the higher $s_n$ we have -$s_0 s_n + s_1 s_{n-1} + ... + s_n s_0 = 0$ for $n\geq 2$. -If $s_1$ can be chosen in such a way that $s_1^2=0$, we can consider $F^{\ast}$ as a dg-module over the Koszul dg-S-algebra of $S/f$. -Now I have two questions: -(1) What happens if the $s_1$ can not be chosen such that $s_1^2=0$? Can we still write the datum of higher homotopies $s_n$ as a dg-module structure over some dg-resolution of $S/f$? A naive guess would be a free non-commutative dg-algebra generated by elements $s_n$ of degree $2n-1$ such that $\text{d}(s_n) = s_1 s_{n-1} + ... + s_{n-1} s_1$ for $n\geq 2$ and $\text{d}(s_1) = f\cdot 1$. Is this studied anywhere? -(2) Can we do all this somehow functorially in $M$? I'm thinking of something like the canonical functor from R-mod into the derived category, which after identification with the homotopy category of projectives turns "projective resolution" into a functor. However, in the concrete example I'm struggling with extending a morphism between R-modules to a morphism of free resolutions respecting the chosen homotopies. Perhaps this is just some diagram chase, but I don't see it. -Apart from technicalities, my goal is to study $S/f$-modules through dg-modules over an appropriate dg-$S$-algebra up to homotopy. If this is possible: do you know books or articles where it is treated? -Thank you very much! -Hanno - -REPLY [8 votes]: You can always find a free resolution $F$, of $M$ over $S$, such that $F$ is a dg-module over the Koszul complex. It may not be minimal, but in many cases that's not an issue. This is the path taken by Avramov and Buchweitz in their paper "Homological algebra modulo a regular sequence with special attention to codimension 2". In particular these resolutions can be taken functorially. -Also see Avramov's book "Infinite free resolutions" whose main themes are how to put dg-algebra and dg-module structures on free resolutions, and how this helps to study these resolutions. -More recently Dyckerhoff and Polishchuk/Vaintrob have used dg-categorical methods to study the situation you're interested in. See their recent papers on the Arxiv.<|endoftext|> -TITLE: Why not evaluate integrals using ODE-solvers? -QUESTION [25 upvotes]: Hello! -I have a question about the relationship between numerical integration and the solution of ordinary differential equations (ODE). Suppose I want to evaluate the integral -$I(x) = \int_{0}^{x} f(t) dt$, where $f$ is a continous function, for some fixed $x = x_{0}$. From the fundamental theorem of calculus I should be able to evaluate this integral by solving an initial value problem $\dot{I} = f(x)$, $I(0) = 0$, $x\in(0,x_{0}]$ right? However, I have seldom seen this method implemented, instead one uses specialized numerical integration codes. How come? -Best regards -Anne - -REPLY [5 votes]: Yes, integration of ODE involves extrapolation whereas integration of functions involves interpolation. -Take for example the Euler's method, we are projecting the slope at the initial point and assume that it applies throughout the time step (which is obviously not the case in general). Worse, the errors accumulate through the time steps. That is, the starting value for the 2nd time step is inaccurate to start with. -Integration of functions involves interpolation of a certain polynomial to fit the functional values and evaluating the area under that polynomial. -In fact, many researchers have tried to apply functional integration methods to integration of ODEs and not the other way round! We know the Gaussian quadrature is one of the most efficient and accurate numerical integration techniques around. Several ODE solvers are based on the Gauss method - e.g. the fully-implicit Gauss-Legendre method developed by Hollingsworth (1955) and later generalized by Butcher (1964) for arbitrary orders as the Gauss methods (implicit Runge-Kutta processes). -Still later, such implicit Runge-Kutta processes are shown to be identical to the collocation methods (Wright, 1970).<|endoftext|> -TITLE: Symmetric sequence of blow-ups for the Fulton-MacPherson compactification -QUESTION [7 upvotes]: Let $X$ be an algebraic variety and $X[n]$ be the Fulton-MacPherson compactification of the configuration space $F(X,n)$ introduced in the paper "A compactification of configuration spaces". -In this paper the authors give an explicit construction of the space $X[n]$ by a sequence of blow-ups, which is inductive. They assume that the space $X[n]$ is already constructed. They describe $X[n+1]$ as an explicit sequence of blow-ups of $X[n] \times X$. -As they mention in their paper their construction is not symmetric. For example, when $n=4$, if one starts by blowing up the small diagonal in $X^4$ and then blow-up proper transforms of the next larger diagonals, then the proper transform of succeeding diagonals will not be separated, so extra blow-ups are needed to get a smooth compactification. -I am interested in the case where $X$ is a smooth curve. I was wondering if there is a construction of the space $X[n]$ as an explicit and symmetric sequence of blow-ups of $X^n$. -I want to get a smooth space at each stage of the construction. -Question: Is there any such sequence? - -REPLY [3 votes]: Another thing worth knowing about $C[n]$, the Fulton-MacPherson compactification for a curve $C$ with $g>1$, is that it is given by the fiber of the forgetful map $\pi : \overline{M}_{g,n}\to \overline{M}_g$ over the point in $\overline{M}_g $ -corresponding to $ C$. -Since the usual construction of $\overline{M}_{g,n}$ is symmetric in $n$, this gives a symmetric construction of $C[n]$ (although not by blowups).<|endoftext|> -TITLE: Hypergraph Chromatic Number vs Degree, Clique-Size -QUESTION [5 upvotes]: For a hypergraph let $\chi$ be the least number of colours needed to colour the vertices, so that in each edge, each colour is used at most once (i.e., the strong chromatic number). Let $\Delta$ be the maximum number of hyperedges containing any vertex. Let $\omega$ be the maximum size of a clique, meaning a vertex set such that for every pair of vertices in the clique, some edge contains both. -Question: is there $\epsilon>0$ so that $\chi \le \Delta \omega / (1+\epsilon)$ in all hypergraphs? -Motive: let $R$ be the maximum edge size. A simple greedy algorithm for colouring can be used to establish that $\chi \le 1 + \Delta(R-1)$, and this bound cannot be improved in general. Note $\chi \ge \omega \ge R$; so I am essentially asking if $\omega$ approximates $\chi$ more closely than $R$. - -REPLY [6 votes]: Very nice question, Dave. I had a look at it with Matej Stehlik, here in Grenoble. We found a way to give a positive answer to your question, although it uses more heavy machinery than we would like. You would expect there is an easy argument, but if it exists, we haven't found it yet. -Given a hypergraph $H$, form a graph $G$ by replacing all hyperedges by cliques. Than $\chi(G)=\chi$, $\omega(G)=\omega$ and $\Delta(G)\le\Delta(R-1)$. Bruce Reed, in his paper "$\omega$, $\Delta$ and $\chi$", J. Graph Theory 27 (1998) 177-212, proves that if $a=1/140000000$, then for $\Delta(G)$ large enough, we have $\chi(G)\le a\omega(G)+(1-a)(\Delta(G)+1)$. So for the hypergraph we have -$$\chi\le a\omega+(1-a)(\Delta(R-1)+1)\le a\omega+(1-a)\Delta R\le -a\omega+(1-a)\Delta\omega=(1-a+a/\Delta)\cdot\Delta\omega.$$ -So if $\Delta\ge2$ (and $\Delta(R-1)$ large enough to allow Bruce's theorem), we get $\chi\le(1-a/2)\Delta\omega$. -I would be interested to hear about better bounds or more elementary approaches (Bruce's proof is not for the faint of heart).<|endoftext|> -TITLE: Matrix integral identity -QUESTION [8 upvotes]: 1) How to prove that $N\times N$ matrix integral over complex matrices $Z$ -$$ -\int d Z d Z^\dagger e^{-Tr Z Z^\dagger} \frac{x_1\det e^Z -x_2 \det e^{AZ^\dagger}}{\det(1-x_1e^Z)\det(1-x_2e^{AZ^\dagger})} -$$ -does not depend on the external Hermitian matrix $A$? $x_1$ and $x_2$ are numbers. The statement is trivial for $1\times1$ case. -2)The same for -$$ -\int d Z d Z^\dagger e^{-Tr Z Z^\dagger} \frac{x_1\det e^Z -x_2 \det e^{AZ^\dagger}}{\det(1-x_1e^Zg)\det(1-x_2e^{AZ^\dagger}g)} -$$ -where g - arbitrary $GL(N)$ matrix. - -REPLY [3 votes]: Let us first slightly generalize OP's first integral to -$$\tag{1} I(x,y,A,B) -~:=~ \int_{{\rm Mat}_{n\times n}(\mathbb{C})} \! dZ~dZ^{\dagger} -\frac{e^{-{\rm tr}(ZZ^{\dagger})}\left(xe^{{\rm tr}(ZA)}-ye^{{\rm tr}(BZ^{\dagger})}\right)}{\det \left({\bf 1}-xe^{ZA}\right)\det \left({\bf 1}-ye^{BZ^{\dagger}}\right)}, -$$ -where $A,B,Z\in {\rm Mat}_{n\times n}(\mathbb{C})$ and $x,y\in\mathbb{C}$. -I) Case $n=1$: As OP observes, it is straightforward to confirm that the integral (1) -$$I(x,y,a,b)~:=~ \int_{\mathbb{C}} \! dz~d\bar{z} -~e^{-|z|^2}\frac{xe^{az}-ye^{b\bar{z}}}{(1-xe^{az})(1-ye^{b\bar{z}})}$$ -$$\tag{2} ~=~\int_{\mathbb{C}} \! dz~~d\bar{z}~e^{-|z|^2}\left(\frac{1}{1-xe^{az}}-\frac{1}{1-ye^{b\bar{z}}}\right)~=~\frac{\pi}{1-x}-\frac{\pi}{1-y}$$ -is indeed independent of $a$ and $b$. -II) Case $n=2$: In two dimensions we have -$$\tag{3}\Delta(x,M)~:=~ \det \left({\bf 1}-xe^{M}\right) -~=~1-x{\rm tr}(e^{M})+x^2e^{{\rm tr}(M)},$$ -$$\tag{4}{\rm tr}(e^{M}) ~=~e^{{\rm tr}(MP_+)}+e^{{\rm tr}(MP_-)} -,\qquad M~\in~ {\rm Mat}_{2\times 2}(\mathbb{C}),$$ -where $P_{\pm}=\frac{1}{2}({\bf 1}_{2\times 2}\pm \sigma_{3})$ are projections. -The integral (1) now becomes -$$\tag{5} \int_{{\rm Mat}_{2\times 2}(\mathbb{C})} \! dZ~dZ^{\dagger}~ -e^{-{\rm tr}(ZZ^{\dagger})} -\frac{xe^{{\rm tr}(ZA)}-ye^{{\rm tr}(BZ^{\dagger})}}{\Delta(x,ZA)\Delta(y,BZ^{\dagger})}. -$$ -Let us view the integral (5) as a formal power series in the indeterminates $x$ and $y$. Let us consider just the coefficient in front of $x^2y$. It is$^1$ -$$ -\int_{{\rm Mat}_{2\times 2}(\mathbb{C})} \! dZ~dZ^{\dagger}~e^{-{\rm tr}(ZZ^{\dagger})} $$ -$$\times \left\{e^{{\rm tr}(ZA)} {\rm tr}(e^{ZA}){\rm tr}(e^{BZ^{\dagger}})- e^{{\rm tr}(BZ^{\dagger})}\left( \left({\rm tr}(e^{ZA})\right)^2- e^{{\rm tr}(ZA)} \right)\right\}, -$$ -$$ -~=~\int_{{\rm Mat}_{2\times 2}(\mathbb{C})} \! dZ~dZ^{\dagger}~e^{-{\rm tr}(ZZ^{\dagger})} $$ -$$\times \left\{ \left(e^{{\rm tr}(ZA(1+P_+))}+e^{{\rm tr}(ZA(1+P_-))}\right)\left(e^{{\rm tr}(P_+BZ^{\dagger})}+e^{{\rm tr}(P_-BZ^{\dagger})}\right)- e^{{\rm tr}(BZ^{\dagger})}\left( e^{2{\rm tr}(ZAP_+)}+e^{2{\rm tr}(ZAP_-)} +e^{{\rm tr}(ZA)} \right)\right\} -$$ -$$~\stackrel{(7)}{=}~ \pi^4 \left( {\rm tr}(e^{BA})-e^{{\rm tr}(BA)}\right), \tag{6} $$ -which depends on the matrices $A$ and $B$, even if $A={\bf 1}_{2\times 2}$ and $B$ is Hermitian as in OP's case. So OP's conjecture is false. --- -$^1$ Recall the Gaussian integral formula -$$\tag{7} I(A,B)~:=~ \int_{{\rm Mat}_{n\times n}(\mathbb{C})} \! dZ~dZ^{\dagger} ~e^{{\rm tr}(-ZZ^{\dagger}+ZA+BZ^{\dagger})}~=~~\pi^{n^2}e^{{\rm tr}(BA)}. $$<|endoftext|> -TITLE: two tetrahedra in $\mathbb R^4$ -QUESTION [45 upvotes]: It is relatively easy to show (see below) that if we have two equilateral triangles of side 1 in $\mathbb R^3$, -such that their union has diameter $1,$ then they must share a vertex. -I wonder whether we have an analog of this in higher dimensions. To start with $4$ dimensions, the question is whether the following statement is true: -If two regular tetrahedra of side length 1 are placed in $\mathbb R^4$ so that the diameter -of their union is $1,$ then the tetrahedra must share a vertex. -(here 'tetrahedron' is the convex hull of four points with equal pairwise distances and -'diameter' of a set is the maximum distance between two points of the set) -In general one is tempted to conjecture: -If two regular $(d-1)$-dimensional simplices of edge length 1 live in $\mathbb R^d$ so that their union has diameter $1,$ then they must share a vertex. -In fact, I believe they must share at least $d-2$ vertices, but let's start with just one vertex, probably it already contains the essence of the problem. -One may think of some even more general statements (what happens with $m$-simplex and $n$-simplex?), but I didn't come up with a reasonable conjecture. -It's easy to construct a tetrahedron and a triangle in $\mathbb R^4$ (with similar conditions) that do not share a vertex. Here I sketch it. I omit easy straightforward calculations, feel free to verify it on your own. Take a tetrahedron $abcd$ in $\mathbb R^4$ and take the midpoints $u$ and $v$ of the edges $ab$ and $cd$. We have $|uv|=1/\sqrt2$. Extend the segment $uv$ on both sides by an equal length to get a segment $xy$ of length $1.$ -Then the largest distance from $x$ and $y$ to a vertex of the tetrahedron is $\sqrt{\frac58+\frac{1}{2\sqrt2}}$ (e.g., the cosine law). Let the origin coincide with the center of the tetrahedron and let the tetrahedron lie in $\{x_4=0\}$. Translate the points $x$ and $y$ by vector $(0,0,0,\frac{\sqrt{3}}{2}-\sqrt{5/8})$ to get points $p$ and $q$ and let $r=(0,0,0,-\sqrt{5/8})$. Now $pqr$ is an equilateral triangle of side 1 and all the distances among the points $p,q,r,a,b,c,d$ are at most 1 (some of them are trivially at most $1,$ while for the others we use the Pythagorean theorem). -Now let me sketch the proof of the statement for $d=3$. It's enough to prove that one cannot have a triangle and a unit segment which lies on one side of the triangle's plane (without increasing the diameter). Suppose this is possible. First translate the segment by a vector orthogonal to the plane until one endpoint gets into the plane (the diameter hasn't increased). Now rotate the segment around the endpoint that is in the plane, so that the other endpoint also falls in the plane (and so that the diameter doesn't increase). Now we are in a $2$-dimensional space and it's easy to show that the segment and triangle must share a vertex. -A less elementary approach is to just apply the so called Dolnikov's theorem: any two odd cycles in a graph of diameters in $\mathbb R^3$ must share a vertex, but that's an overkill. - -REPLY [11 votes]: One can read the proof of the conjecture here: http://arxiv.org/abs/1402.3694 -Also in this paper there is the proof of Schur's conjecture. This conjecture states that any diameter graph in $\mathbb R^d$ on $n$ vertices may have at most $n$ cliques of size $d$.<|endoftext|> -TITLE: Is there a higher homotopical spinor theory? -QUESTION [16 upvotes]: Gabriele Vezzosi and I have been musing on the following. Consider the standard double cover $\mathop{\rm Pin}_{n} \to \mathrm{O}_{n}$, whose kernel is $\mathbb Z/2\mathbb Z$. This allows to associate with every $\mathrm{O}_{n}$-principal bundle $E$ over a space $S$ a class in $\mathrm H^{2}(X, \mathbb Z/2\mathbb Z)$, which is well known to be the second Stiefel--Whitney class of $E$. -Now, suppose that we have given a simplicial group $G$, with a central simplicial subgroup $Z$ such that the quotient $G/Z$ is homotopy equivalent to a topological group $H$. Assume that $Z$ is homotopy equivalent, as a simplicial group, to an Eilenberg--MacLane group $K(A, m)$, where $m$ is a non-negative integer and $A$ an abelian group. Then, if we are not mistaken, we should obtain a map from $H$-principal bundles on $S$ to $\mathrm H^{m+2}(S, A)$, and thus a characteristic class for $H$-principal bundles: the group $G$ gives a homomorphism from $H$ to the classifying simplicial group $B\,K(A,m) = K(A, m + 1)$, which yields a homorphism $B\,H \to B\,K(A,m+1) = K(A, m + 2)$, and this should give the characteristic class. - -Now, the question is: for each $m ≥ 0$, is there a simplicial group $\mathop{\rm Pin}_{n}^{(m)}$, with a surjective homomorphism $\mathop{\rm Pin}_{n}^{(m)} \to \mathrm{O}_{n}$, whose kernel is a central simplicial subgroup $K(\mathbb Z/2\mathbb Z, m)$, such that the associated characteristic class is the $(m+2)^{\rm nd}$ Stiefel--Whitney class? -If there isn't, is there some analogue in which $G$ is something less than a simplicial group (maybe an $H$-group, or something along that line)? We think this should be possible, but we'd be interested in an explicit construction, more than in an abstract existence theorem. - -Ultimately, we'd like to have an analogous construction for simplicial schemes over a field, in order to study higher Hasse--Witt classes, in the sense of Jardine; but we were wondering if such a construction, in the simpler case of simplicial sets, is known to topologists. - -REPLY [4 votes]: This seems to work for getting a central extension (even though it looks a little bit too simple). To be -specific I am using May's notation with $\newcommand{\hw}{\overline{W}}\hw$ for -the classifying space of a simplicial group and -$\newcommand{\kl}{\mathbf{G}}\kl$ for the Kan loop group. Start as Charles does -with the the map $\hw\mathbf{O}_n\rightarrow \hw\hw\mathrm{K}(\mathbb{Z}/2,m)$. -Applying the Kan loop construction to this map we get a map of simplicial groups -$\kl\hw\mathbf{O}_n\rightarrow \kl\hw\hw\mathrm{K}(\mathbb{Z}/2,m)$ and as $G$ -and $\hw$ are adjoint we have a map of simplicial groups -$\kl\hw\hw\mathrm{K}(\mathbb{Z}/2,m)\rightarrow\hw\mathrm{K}(\mathbb{Z}/2,m)$ -and composing we get a map of simplicial groups -$\kl\hw\mathbf{O}_n\rightarrow\hw\mathrm{K}(\mathbb{Z}/2,m)$. Now, there is an -exact sequence of abelian simplicial groups -$0\rightarrow\mathrm{K}(\mathbb{Z}/2,m)\rightarrow -H\rightarrow\hw\mathrm{K}(\mathbb{Z}/2,m)\rightarrow0$ with $H$ contractible. -Pulling back this along -$\kl\hw\mathbf{O}_n\rightarrow\hw\mathrm{K}(\mathbb{Z}/2,m)$ gives a central -extension $1\rightarrow\mathrm{K}(\mathbb{Z}/2,m)\rightarrow -H'\rightarrow\kl\hw\mathbf{O}_n\rightarrow1$ which has the right homotopy type. -Note that this goes more or less backwards in the following observations: If -$1\rightarrow Z\rightarrow G\rightarrow H\rightarrow1$ is a central extension, -then as multiplication $Z\times G\rightarrow G$ is a group homomorphism we get a -map $\hw Z\times\hw G\rightarrow\hw G$ which is an action of the simplicial -group $\hw Z$ on $\hw G$ making $\hw G\rightarrow\hw H$ a $\hw Z$-torsor giving -a classifying map $\hw H\rightarrow\hw\hw Z$.<|endoftext|> -TITLE: Proj for rings graded by different things then $\mathbb N$ ? -QUESTION [12 upvotes]: Given a commutative, $\mathbb N$-graded ring, one can associate to it a scheme via the $Proj$ construction. -What happens if one tries to copy this procedure but instead of $\mathbb N$ with another indexing gadget (say commutative monoid) ? -Some thoughts about this: -Considering projective varieties is roughly the same as studying affine varieties equivariant under the multiplicative group. -So I would guess that replacing $\mathbb N$ by something else corresponds to replacing the multiplicative group by something else. - -REPLY [13 votes]: Weighted projective spaces $\mathbb{P}(a_1,\ldots,a_n)$ are examples where a grading other than the standard grading is used. In general you can study gradings coming from any finitely generated abelian group, and this grading gives rise to a torus action on the ring. The Proj you speak of is then a GIT-quotient of $Spec R$ by this torus action (if you are familiar with GIT, the choice of linearization of the action corresponds to a choice of irrelevant ideal). This GIT-quotient construction is completely analogous to the usual construction -$$ -\mbox{Proj} k[x_0,\ldots,x_n]=\left(\mbox{Spec} k[x_0,\ldots,x_n]-V(x_0,\ldots,x_n)\right) // \mathbb{G}_m -$$ -The best reference I can give for this stuff is Chapter 1 of the book "Cox rings" by Arzhantsev, Derenthal, Hausen and Antonio Laface. Other nice references are -The Homogeneous Coordinate Ring of a Toric Variety by Cox, which deals with Toric varieties -Lectures on invariant theory by Dolgachev, which is a nice introduction to quotients in algebraic geometry.<|endoftext|> -TITLE: Centers of Semidirect Products -QUESTION [8 upvotes]: The following question is for my own curiosity as I take some time to get reacquainted with group theory. -Let G be a semi-direct product of the groups N and K with multiplication defined by the automorphism $\phi$ from K to Aut(N). Let Fix($\phi$) be the set of all elements of N that are mapped to themselves by all elements of the range of $\phi$. Clearly every element of Fix($\phi$) commutes with all elements of K and every element of the kernel of $\phi$ commutes with every element of N. -If Fix($\phi$) is the trivial group in N and Ker($\phi$) is the trivial group in K, does that imply that the center of G is trivial? If so, could someone point me to a reference or proof. If not, then a counter example. - -REPLY [7 votes]: Suppose that $z=xy$ is in the centre where $x\in N$ and $y\in K$. -Then for all $u\in K$, $uxy=xyu$. But $uxy=\phi(u)(x)uy$ so that -$x=\phi(u)(x)$ (and $uy=yu$). As this is true for all $u\in K$ -then by the assumption on Fix($\phi$), $x=1$. Therefore $z=y\in K$. -As $y$ commutes with all elements of $N$ then $y$ lies -in Ker($\phi$) and is trivial. So $z=1$ and the centre of $G$ -is trivial.<|endoftext|> -TITLE: Why the "W" in CGWH (compactly generated weakly Hausdorff spaces)? -QUESTION [44 upvotes]: In his 1967 paper A convenient category of topological spaces, -Norman Steenrod introduced the category CGH of compactly generated Hausdorff spaces -as a good replacement of the category Top topological spaces, in order to do homotopy theory. -The most important difference between CGH and Top is that in CGH there is a functorial homeomorphism $$\mathrm{map}(X,\mathrm{map}(Y,Z))\cong \mathrm{map}(X\times Y,Z),$$ -a fact that is only true in Top under the extra assumption that $Y$ is locally compact. - -But in more recent papers, I see that people use CGWH spaces instead of CGH spaces... -Why? -Could someone explain to me what goes wrong in CGH spaces -(please illustrate with an example), and explain how the "w" fixes everything? -Also (following Jeff's comment), to whom should the "w" be attributed? -One more wish: can someone give me an example of a CGWH space that isn't CGH? - -REPLY [6 votes]: Here is perhaps the simplest example of a CGWH space which fails to be Hausdorff. -Start with a countable metric space X so that with one exception x, each point is open, but so that at the exceptional point, X is not locally compact at x. -It is easy to find such a subspace of the real line. -( start with 0 and (1/n)+(1/(m+n)) -Now delete each 1/n). -Let Y be the one point compactification, adding to X a new point y, whose neighborhood complements are compact in X. In the new space Y, compact subsets are closed (and in particular Y is WH), but x and y are inseparable. -See for example, Example 99 from Counterexamples in Topology by Steen and Seebach.<|endoftext|> -TITLE: Does $S_4$ inject into $SL(2,R)$ for some commutative ring $R$? -QUESTION [29 upvotes]: $\newcommand{\Z}{\mathbf{Z}}$ -Given a nice infinite collection of groups, for example the symmetric groups, one can ask whether any finite group is a subgroup of one of them. Of course any finite group acts on itself, so any finite group is a subgroup of a symmetric group. Similarly any finite group acts linearly on its group ring over a finite field, so given a field $k$, any finite group embeds into $GL(n,k)$ for some sufficiently large $n$ (as permutation matrices). -Question 1: What if we do what I ask in the title, and consider the groups $SL(2,R)$ as $R$ ranges over all commutative rings. Given an arbitrary finite group, can I find a commutative ring $R$ (with a 1) such that this group is a subgroup of $SL(2,R)$? -Of course this is inspired by this pesky question which, at the time of typing, seems to remain unsolved. -Here is a more specific question: -Question 2: Is there a commutative ring $R$ (with a 1) such that the symmetric group $S_4$ injects into $SL(2,R)$? -I haven't thought much about question 1 at all. I'll tell you what I know about question 2. Let's consider first the case where $R$ is an algebraically closed field. If the characteristic is zero, or greater than 3, then by character theory any map from $S_4$ into $SL(2,R)$ must contain $A_4$ in its kernel (the map must give a semisimple representation and the irreducible 2-dimensional one has non-trivial determinant). -If the characteristic is 3 then considering the restriction of a map $S_4\to SL(2,R)$ to a Sylow 2-subgroup we see again by character theory that the kernel must contain the central element. But the kernel is a normal subgroup of $S_4$ so it must contain $V_4$ and hence factors through a map $S_3\to SL(2,R)$. Now the image of an element of order 2 must be central and it's not hard to deduce that the 3-cycles must again be in the kernel. -In characteristic 2 there are more possibilities. If I got it right, the kernel of a map $S_4\to SL(2,R)$ ($R$ alg closed char 2) is either $S_4$, $A_4$ or $V_4$ and of course the representation can be non-semisimple this time. -We conclude from this case-by-case analysis that if $R$ is any ring and $S_4\to SL(2,R)$ is any map then the image of $V_4$ is in $1+M_2(J)$, where $J$ is the intersection of all the prime ideals, that is, the nilpotent elements of $R$. -I now wanted to consider the case $J^2=0$ and check that $V_4$ must be killed mod $J^2$ and then go by induction, but I couldn't bash it out and wonder whether it's true. -It's clear that one could brute-force the argument if one could do a Groebner basis calculation over the integers. I have tried one of these in my life---when trying to solve the open problem of whether every finite flat group scheme of order 4 was killed by 4. That latter question seems to be beyond current computers, but perhaps the one I'm raising here might not be. The problem would be that one has to work over $\Z$ and this slows things down greatly. -I then looked for counterexamples, but convinced myself that $S_4$ was not a subgroup of either $SL(2,\Z/4\Z)$ or $SL(2,\Z/2\Z[\epsilon])$ with $\epsilon^2=0$ [edit: I am wrong; $SL(2,\Z/2\Z[\epsilon])$ does work, as pointed out by Tim Dokchitser]. I don't know how to get a computer algebra system to check $SL(2,\Z/2\Z[\epsilon,\delta])$ so I gave up and asked here. -I suspect I am missing some standard fact :-/ - -REPLY [16 votes]: For Question 2, The central extension $\tilde{S}_4$ is certainly a subgroup of $\mathrm{GL}_2(\mathbf{Z}[\sqrt{-2}]) \subset \mathrm{GL}_2(\mathbf{C})$. The image of the determinant is $\pm 1$. The image of $\tilde{S}_4$ in -$$\mathrm{GL}_2(\mathbf{Z}[\sqrt{-2}]/2) = \mathrm{GL}_2(\mathbf{F}_2[x]/x^2)$$ -is $S_4$, and all the elements have determinant one. It's easy to see that the central element -$$\left( \begin{matrix} -1 & 0 \\\ 0 & -1 \end{matrix} \right)$$ -lies in the kernel, so it suffices to note that nothing else does. Yet it's obvious that the map surjects onto $\mathrm{GL}_2(\mathbf{F}_2) = S_3$, and (from the character table) the image is larger than $S_3$, so the image is $S_4$. -For Question 1, if $G$ injects into $\mathrm{SL}_2(R)$ for some $R$ then it injects into such a ring where $R$ is Artinian. Here is the proof. -EDIT: Step 0. (This was in my head, but I forgot to mention it, as Kevin reminds me in the comments). One may replace $R$ by the subring generated by the images of the entries of $g-1$ for all $g \in G$, and hence assume that $R$ is finitely generated over $\mathbf{Z}$ and hence Noetherian. (The Krull intersection thm requires a Noetherian hypothesis.) -Step 1. If $x$ is a non-zero element of $R$, then there exists a maximal ideal $\mathfrak{m}$ of $R$ such that $x$ is non-zero in $R/\mathfrak{m}^k$ for some $k$. Proof: Let $\mathfrak{m}$ be some maximal ideal containing the annihilator of $x$. Then $x$ is non-zero in the localization $R_{\mathfrak{m}}$, and thus $x$ is non-zero in $R/{\mathfrak{m}^k}$ by the Krull intersection theorem. -Step 2. If $x_1, \ldots, x_n$ are non-zero elements of $R$, there exists an ideal $I$ such that each $x_i$ is non-zero in $R/I$ and $R/I$ is Artinian. Proof: Apply Step 1 to each $x_i$, and let $I = \bigcap \mathfrak{m}^{k_i}_i$. -Step 3. Suppose that $G$ has $n$ non-trivial elements. Let $x_1, \ldots, x_n$ denote -a non-zero entry in the matrix $g - 1$ for each element of $g$. Apply Step 2 to deduce that -$g$ is not the identity in $R/I$ for some Artinian quotient for all non-zero $g \in G$. -Remark: If $G$ is simple, then $G$ is actually a subgroup of $\mathrm{SL}(k)$ for some field $k$. Proof: Artinian rings are semi-local, so $G$ is a -subgroup of $\bigoplus_{i=1}^{n} \mathrm{SL}(A_i)$ for Artinian rings $A_i$. Since $G$ is simple, it must be a subgroup of $\mathrm{SL}(A)$ for one such $A$. This latter group is filtered by the groups $\mathrm{SL}(k)$ and copies of $M_0(k)$ (trace zero matrices). Since the latter is abelian and $G$ is simple, we are done. -It's easy to find examples of groups which are not subgroups of $\mathrm{SL}_n(k)$ for all fields $k$ and some fixed integer $n$.<|endoftext|> -TITLE: Spectrum of the Laplacian on G(n, p) and G(n, M) -QUESTION [8 upvotes]: A random graph in $G(n, p)$ model is a graph on $n$ vertices in which for each of the $n\choose{2}$ edges we independently flip a coin, then take the edge with probability $p$ or remove it with $1 - p$. -A random graph in $G(n, m)$ model is a graph on $n$ vertices in which a subset of edges of a fixed size $m$ is chosen at random. -We expect $G(n, p)$ and $G(n, m)$ for $m = p {n\choose{2}}$ to look asymptotically the same, because the number of edges in $G(n, p)$ is highly concentrated around the mean. -The (normalized) Laplacian $L$ on a graph is an operator (matrix) which has entries: -$L(v, v) = 1$ -$L(v, w) = - \frac{1}{\sqrt{deg(v)deg(w)}}$ -where $v \neq w$ are vertices of the the graph. -We are interested in $\lambda_{2}$, that is, the smallest nonzero eigenvalue of $L$. Suppose we know that, for $p = p(n)$ growing sufficiently fast (papers by Chung et al. show that, for example, $p \geq \frac{\log^2 n}{n}$), $\lambda_{2}$ for a random graph in $G(n, p)$ model is close to 1 with high probability (i. e. approaching 1). -Does it follow immediately that also in $G(n, m)$ we have $\lambda_{2} \to 1$ with high probability? It is known in random graph theory that such implications hold for monotone properties (that is, properties which still hold after adding an edge to the graph), however, the second eigenvalue is not monotone (although it is probably "monotone on average", so I expect the statement above is true for $G(n, m)$). - -REPLY [3 votes]: You might find this paper and the references therein useful: -Amin Coja-Oghlan, On the Laplacian eigenvalues of $G_{n,p}$, Combin. Probab. Comput., 16 (2007). MR 2008j:05212. -They discuss the classical model as well as Chung and Lu's model which is quite interesting.<|endoftext|> -TITLE: Weak and Strong Integration of vector-valued functions -QUESTION [23 upvotes]: This is probably an elementary question, but outside my area of expertise, and I was unable to find any suitable reference: -Suppose $f:X\to E$ is a continuous function from a compact spaces (endowed with a probability Borel measure $\mu$) to a locally convex, Hausdorff, complete topological vector space. We want to define a reasonable integral, i.e., $\int_Xfd\mu$. The question is : when exactly is that one needs weak integrals (e.g., the Gelfand-Pettis integral)? There is a strong integral called the Bochner integral which is typically defined for $E$ a Banach space; but it seems to me that its definition works at least for Frechet spaces. For a general locally convex, Hausdorff, complete TVS, when exactly does Bochner's approach, i.e., to define a strong integral by approximation of $f$ with simple functions (i.e., finite-valued measurable functions) fail? - -REPLY [4 votes]: Anton's answer prompted me to pull out Rudin's Functional Analysis. Exercise 23 of Chapter 3 is to exhibit the Gelfand-Pettis integral of a continuous function with values in a Frechet space as a strong limit of "Riemann sums" (looks slightly stronger than the Bochner integral, in fact). I couldn't make sense of how to use the hint, so I decided to see what I could come up with. It's been a long time since I've done this sort of thing, so I may have done something stupid. -We're going to assume $X$ is compact, $\mu$ is a probability Borel measure, $f$ is continuous, $E$ is locally convex and satisfies the technical criterion for the existence of Gelfand-Pettis integrals (closure of the convex hull of the image is compact). Then $f(X)$ is compact. If we also assume either $E$ is a metric space or $X$ is separable, then $f(X)$ is separable. Take a countable dense subset $e_i$ of $f(X)$. For a finite subset $B$ of seminorms and $\epsilon>0$, let $V(B,\epsilon,e_i)=\{e:p(e-e_i)<\epsilon\}$. These form a basis for the topology of $E$ (and they are convex and balanced). -First, we're going to construct a sequence of simple functions that converges to $f$, mimicking the example from measure theory. -Since $f(X)$ is compact, we can form a cover with finitely many of the $V(B,\epsilon,e_i)$, call them $A_i$. Let $E_i=A_i-\bigcup_{j=1}^{i-1}A_j$. Set $X_i=f^{-1}(E_i)$, so the $X_i$ are a disjoint partition of $X$. -Set $g_{B,\epsilon}(x)=\sum_i 1_{X_i}(x)e_i$ (where $i$ ranges over a finite set). -Consider $f(x)-g_{B,\epsilon}(x)$. If $x\in X_i$, $f(x)\in E_i$, so for all $p\in B$, $p\big(f(x)-g_{B,\epsilon}(x)\big)=p\big(f(x)-e_i\big)<\epsilon$. Convergence is easy from this. -Now we will show that $\int_X g_{B,\epsilon}\ d\mu=\sum_i \mu(X_i)e_i$ converges. We "cheat" by showing it converges to the Gelfand-Pettis integral of $f$ (instead of e.g. showing that it is Cauchy). So consider -$$\int_X f\ d\mu- \int_X g_{B,\epsilon}\ d\mu=\int_X f- g_{B,\epsilon}\ d\mu=\sum_i\int_{E_i}f-e_i\ d\mu$$ -On $E_i$, $f-e_i$ lies in $V(B,\epsilon)$, so $\int_{E_i}f-e_i\ d\mu$ is in the closure of $\mu(E_i)\cdot V(B,\epsilon)\subset \overline{V(B,\epsilon)}$ (since $V(B,\epsilon)$ is balanced) (this uses the estimate that the Gelfand-Pettis integral lies in the closure of the convex hull of the image, times the measure of the space). So $\sum_i\int_{E_i}f-e_i\ d\mu\subset \sum_i \overline{V(B,\epsilon)}\subset \overline{\sum_i V(B,\epsilon)}$. -By some basic tvs theorems, since $V(B,\epsilon)$ forms a basis for the topology, we can choose $V(B,\epsilon)$ so that $\overline{\sum_i V(B,\epsilon)}$ is inside any given neighborhood of $0$ (see p.10-11 of Rudin). -So Bochner integrals of continuous functions exist whenever Gelfand-Pettis integrals exist (plus a separability assumption), unless I made a mistake. Even though I left it in, I don't think that I used separability in an essential way (I could have just chosen an uncountable dense subset $e_i$ and proceeded). That requirement for strong measurability (almost-separably valued) may be an artifact of using sequences instead of nets.<|endoftext|> -TITLE: Ruelle Perron Frobenius Operator -QUESTION [10 upvotes]: Hello everybody! Recently in my research, I came across the Perron-Frobenius operator .. I would like to intuitive interpretation of this operator, ie, physical interpretations are possible, articles (can be physical). -I wonder how this operator originated. - -REPLY [9 votes]: I like to explain the Ruelle Perron Frobenius operator this way: -Suppose that $T$ is a map from some space $X$ to itself. Suppose also that $X$ has some distinguished ambient measure $\lambda$ on it (think of Lebesgue measure). One more assumption: the ambient measure is non-singular with respect to $T$ (i.e. if $A$ is a set of measure 0, then $T^{-1}A$ has measure 0 also). -Given a measure $\mu$ on $X$, the "push-forward" of $\mu$ is defined to be $\mu\circ T^{-1}$. Seems counter-intuitive, but in the case where $\mu$ is concentrated at a single point $a$ you can check that the push-forward is concentrated at $T(a)$. -The non-singularity condition implies that if $\mu$ is a measure absolutely continuous with respect to $\lambda$, then its push-forward is also absolutely continuous with respect to $\lambda$. By the Radon-Nikodym theorem, measures absolutely continuous with respect to $\lambda$ have essentially unique densities. -The Ruelle-Perron-Frobenius operator applied to $f$ gives the density of the push-forward of the measure whose density is $f$. -In more picturesque language: if $X$ is a random variable with density $f(x)$, then $T(X)$ is a random variable with density $L[f](x)$.<|endoftext|> -TITLE: Reference for the Hodge Bundle -QUESTION [9 upvotes]: For the purposes of this question, let the Hodge bundle $\lambda$ be the bundle on a fibration of abelian varieties $X\to B$ with fiber over $b\in B$ the space of 1-forms on $X_b$, or the pullback to $B$ along the zero section of the sheaf of relative differentials. The most interesting examples are when $B$ is $M_g$ or $A_g$, and the fibrations are the Jacobian fibration or the universal family of abelian varieties. -Is there a good reference for the properties of this bundle? And for the determinant line bundle on these spaces? Things like self-intersection numbers, cohomology (in particular global sections) are particularly of interest. - -REPLY [4 votes]: The list of references is way too long. Here are some classical texts containing both the setup and calculations: -1) P. Deligne, Le déterminant de la cohomologie, Current Trends in Arithmetical Algebraic Geometry, Contemp. Math., no. 67, AMS, Providence, 1987. -2) Gerd Faltings, Ching-Li Chai, Degenerations of abelian varieties, Springer-Verlag, 1990. -3) L. Moret-Bailly, Pinceaux de variétés abéliennes, Astérisque 129 (1985). -And here are a couple of more recent papers that deal with the self-intersection and cohomology calculations: -4) Alexis Kouvidakis, Theta line bundles and the determinant of the Hodge bundle, arXiv:alg-geom/9604017. -5) Alexander Polishchuk, Determinant bundles for abelian schemes, arXiv:alg-geom/9703021.<|endoftext|> -TITLE: Ellipse naturally associated with a polygon -QUESTION [25 upvotes]: My colleagues and I have stumbled onto a way to associate an ellipse, or equivalently a positive definite symmetric matrix, to a polygon that is different from other better known ways. We want to know if anyone has ever seen this before. Before describing this, I do want to note that there are other well-known ways of associating an ellipse to a polygon, including the matrix of second moments of the uniform distribution supported on the interior of the polygon and the so-called John ellipses, which are the ellipse of largest volume inscribed in the polygon and the one of smallest volume circumscribing the polygon. -Here is the ellipse we found: Given a polygon $P$ that contains the origin in its interior, let $\ell_1, \dots, \ell_n$ denote the lines that contain the sides of $P$. For each $i$, let $n_i$ denote the unit vector orthogonal to $\ell_i$, $h_i$ be the distance from the origin to the line $\ell_i$, and $s_i$ be the length of the side lying in $\ell_i$. For each $v \in R^2$, define -$ q(v) = \sum_{i=1}^n (v\cdot n_i)^2\frac{s_i}{h_i}. $ -Let $E_P = \{ q(v) \le 1\}.$ The ellipse $E_P$ does not behave particularly nicely under translations of $P$. It however, behaves nicely under linear transformations. In particular, if $A$ is an invertible linear transformation with determinant $1$, then $E_{AP} = AE_P$. It is scale-invariant in that $E_{tP} = E_P$ for any $t > 0$. -The ellipse $E_P$ can in fact be defined without using the inner product and is a linear invariant of the polygon $P$. Its definition can be extended to any body $P$ containing the origin with a sufficiently regular boundary. It is an example of a matrix-valued valuation on the set of all convex bodies that contain the origin in their interior. Its volume is maximized if and only if the body $P$ is itself an ellipsoid centered at the origin. -Does anyone recognize this association of an ellipse to a polygon? I would be grateful for any information or references. -EDIT: Vladimir's comments below, especially the first sentence, indicate clearly that I omitted something crucial. The definition of the ellipsoid depends on the choice of the volume form in the ambient vector space (and the dual volume form in the dual vector space). Changing the volume form will rescale the ellipsoid by a factor. However, if you fix the volume form, then the ellipsoid is invariant under rescaling of the convex body. -This observation, however, makes the existence of a scale-invariant ellipsoid associated with the polygon much less surprising. - -REPLY [8 votes]: Maybe it would have been more convincing if you had given the definition of $q_K$ in a unimodularly equivariant way from the beginning. That's not at all hard to do, and it's also easy to see how to create many more such (and how to create many more in all dimensions). -For example, suppose $K\subset V$ has vertices $v_0,v_1,\ldots,v_n=v_0$ (cyclically ordered counterclockwise, say). Let $\alpha_i\in V^\ast$ be the unique element that satisfies $\alpha_i(v_i)=\alpha_i(v_{i+1})=1$. Then -$$ -q_K = \sum_{i=0}^{n-1} \Omega(v_i,v_{i+1})\ {\alpha_i}^2 -$$ -(where $\Omega$ is the area form.) This is clearly unimodularly equivariant with respect to $K$ and, since, for $K' = tK$ (with $t>0$), one has $v'_i = tv_i$ and $\alpha_i' = (1/t)\alpha_i$, it follows that $q_{tK} = q_K$. Of course, anything like this would have worked. For example, you could have taken -$$ -\tilde q_K = Area_\Omega(K)\left( \sum_{i=0}^{n-1} {\alpha_i}^2\right), -$$ -and this would also have had the same equivariance property. -In dimension $n$, I think that the right formula would be to define, for each face $F$ of $K$, the element $\alpha_F\in V^\ast$ to be the linear function that equals $1$ on $F$, let $\Omega(F)$ denote the volume of the cone with vertex $0\in V$ whose base is $F$, and then set -$$ -q_K = \sum_{F\in\mathcal{F}(K)} \Omega(F)\ {\alpha_F}^2. -$$ -If you want it to be invariant under scaling, you should take -$$ -q_K = \sum_{F\in\mathcal{F}(K)} \Omega(F)^{2/n}\ {\alpha_F}^2. -$$ -Perhaps, better, though, would be to take -$$ -q_K = Vol_\Omega(K)^{(2-n)/n}\ \left(\sum_{F\in\mathcal{F}(K)} \Omega(F)\ {\alpha_F}^2\right), -$$ -since this is also invariant under subdivision of the faces of $K$.<|endoftext|> -TITLE: Freyd-Mitchell's embedding theorem -QUESTION [31 upvotes]: Freyd–Mitchell's embedding theorem states that: if $A$ is a small abelian category, then there exists a ring R and a full, faithful and exact functor $F\colon A \to R\text{-}\mathrm{Mod}$. -I have been trying to find a proof which does not rely on so many technicalities as the ones I have found. I have leafed through: - -Freyd's Abelian Categories says that the text, excepting the exercises, tries to be a geodesic leading to the theorem. If you take out the exercises, probably the text is 120 pages long... - -Mitchell's Theory of Categories (pdf) is very hard to read, and also to prove the theorem you have tons of definitions and propositions and lemmas to prove. For example, the study of AB-5 categories (C3 in Mitchell's notation) is fairly tedious. - -Weibel's An Introduction to Homological Algebra (pdf) redirects me to Swan, The Theory of Sheaves, a book which is unavailable in my university's library. I've leafed through Swan's Algebraic K-Theory: the theorem is proved, but it is also long, hard and painful to read, and assumes a lot of knowledge I don't have (I had never seen a weakly effaceable functor, or a Serre subcategory; and it certainly is not well known to me that the category of additive functors from a small abelian category to the category of abelian groups is well-powered, right complete, and has injective envelopes!) - - -Maybe there are more modern proofs which require less heavy machinery and technicalities? - -REPLY [54 votes]: $\DeclareMathOperator{\Hom}{Hom}\newcommand{\amod}{\mathscr{A}\text{-}{\bf Mod}}\newcommand{\scrA}{\mathscr{A}}\newcommand{\scrE}{\mathscr{E}}\newcommand{\Ab}{\mathbf{Ab}}\DeclareMathOperator{\Lex}{\mathbf{Lex}}\DeclareMathOperator{\coker}{Coker}$I only know one proof of the embedding theorem—the expositions differ heavily in terminology but the approaches all are equivalent, as far as I can tell. I think the proof in Swan's book on K-theory makes the relation between the Freyd-Mitchell approach and Gabriel's approach pretty clear. Let me just say that there is no cheap way of getting the Freyd-Mitchell embedding theorem since there is a considerable amount of work you need to invest in order to get all the details straight. On the other hand, if you manage to feel comfortable with the details, you will have learned quite a good amount of standard tools of homological algebra, so I think it's well worth the effort. -Think of the category of functors $\scrA \to \Ab$ as $\scrA$-modules, that's why the notation $\amod$ is fairly common. The Yoneda embedding $A \mapsto \Hom(A,{-})$ even yields a fully faithful contravariant functor $y\colon \scrA\to\amod$. -The category $\amod$ inherits a bunch of nice properties of the category $\Ab$ of abelian groups: - -It is abelian. -It is complete and cocomplete ((co-)limits can be computed pointwise on objects) -The functors $\Hom(A,{-})$ are injective and $\prod_{A \in\scrA} \Hom{(A,{-})}$ is an injective cogenerator. -Since there is an injective cogenerator, the category $\amod$ is well-powered. -etc. - -However, the Yoneda embedding is not exact: If $0 \to A' \to A \to A'' \to 0$ is a short exact sequence, we only have an exact sequence -$$0 \to \Hom(A',{-}) \to \Hom(A,{-}) \to \Hom(A'',{-})$$ -in $\amod$. -It turns out that the functor $Q = \coker{(\Hom(A,{-}) \to \Hom(A'',{-}))}$ is “weakly effaceable”, so we would want it to be zero in order to get an exact functor. How can we achieve this? Well, just force them to be zero: say a morphism $f\colon F \to G$ in $\amod$ is an isomorphism if both its kernel and its cokernel are weakly effaceable. If this works then a weakly effaceable functor $E$ is isomorphic to zero because $E \to 0$ has $E$ as kernel. Now the full subcategory $\scrE$ of weakly effaceable functors is a Serre subcategory, so we may form the Gabriel quotient $\amod/\scrE$. By its construction, isomorphisms in the Gabriel quotient have precisely the description above. -On the other hand, the category $\Lex(\scrA,\Ab)$ of left exact functors $\scrA \to \Ab$ is abelian. This is far from obvious when you start from the definitions. However, $\Lex(\scrA,\Ab)$ sits comfortably inside the abelian category $\amod$. The inclusion has an exact left adjoint (!) (= “sheafification”), so again ${\bf Lex}({\scr A}, \Ab)$ inherits many useful properties from $\amod$. Moreover, the kernel of the left adjoint can be identified with the weakly effaceable functors, and that's why $\Lex{(\scrA, \Ab)} = \amod/\scrE$. -All this work shows that $A \mapsto \Hom{(A,{-})}$ is a fully faithful and exact embedding of $\scrA$ into ${\Lex}{(\scrA, \Ab)}$, so it remains to show that the latter can be embedded into a category of modules. This is well described in Weibel's or Swan's books, so I won't elaborate on that point and content myself by saying that you simply need to look at the endomorphism ring of an injective cogenerator. -As for references, I think you can't do much better than Freyd's book. Don't be too intimidated by Swan's exposition in his K-theory book. If you're really interested in understanding this proof, I think it's worth reading the two expositions (first Freyd, then Swan). There also is a proof in volume 2 of Borceux's Handbook of categorical algebra with a more “hands on” approach.<|endoftext|> -TITLE: Cohomology of the Moduli of G-bundles on a Curve -QUESTION [5 upvotes]: For a simple complex group G and Riemann surface X, are the (integral, if possible) cohomology groups of the moduli of holomorphic G-bundles on X written down somewhere, either explicitly or implicitly? If not, are there some specific cases where these groups have been calculated? - -REPLY [3 votes]: See Atiyah-Bott, "Yang-Mills Equations over Riemann Surfaces" for the case of stable holomorphic vector bundles. See in particular section 9 and Theorem 9.11. -They compute the Betti numbers of $N(n,k)$, the moduli space of stable holomorphic vector bundles of rank $n$ and first Chern class $k$, when $n$ and $k$ are relatively prime. They also find generators for the cohomology ring. These "Atiyah-Bott generators" are the Künneth components of the Chern classes of the universal bundle. In this paper they don't compute the relations for these generators. -The main tool in this paper is Morse theory. -I think the reason for restricting to stable bundles is to get a moduli space (rather than some kind of stack or whatever). The reason for fixing the first Chern class is just because different Chern classes correspond to different connected components. I think the reason for the condition that $n$ and $k$ are relatively prime is just to ensure that the moduli space is a smooth manifold. -Since these moduli spaces are also algebraic varieties, one can also compute their Betti numbers using the machinery of the Weil conjectures... There are some comments about this, as well as references, in the introduction of the paper and in section 11. -EDIT: I just did a Google search and found this paper of Heinloth and Schmitt, which claims to compute the cohomology ring of the entire moduli stack: http://staff.science.uva.nl/~heinloth/HS_v4.pdf -This paper says that the Atiyah-Bott generators are in fact free generators...<|endoftext|> -TITLE: Pointed Hurewicz model structure -QUESTION [15 upvotes]: In Strøm's (no relation) paper "The Homotopy Category is a Homotopy Category" he proves -that the category of unpointed topological spaces, with Hurewicz fibrations and ordinary cofibrations and homotopy equivalences, is a model category. -Then he has a fairly long section on the pointed case, and his results are not as good: he restricts to well-pointed spaces, and doesn't get all the model category axioms. -But if $\mathcal{M}$ is a model category, then for any -object $A\in \mathcal{M}$, the category $A \downarrow \mathcal{M}$ inherits a model category structure in perfectly straightforward way. Since -$\mathbf{Top}_* = * \downarrow \mathbf{Top}$, we get a model structure right away. -So: what is going on here? Did he simply miss an easy extension to the pointed case? -Is it possible that the natural model structure on $* \downarrow \mathbf{Top}$ is not the one -he wants (i.e., the received cofibrations, fibrations or weak equivalences differ somehow from pointed cofibrations, fibrations and pointed homotopy equivalences)? - -REPLY [10 votes]: You must allow the weak equivalences to be unpointed homotopy equivalences. These become honest pointed homotopy equivalences between fibrant-cofibrant objects by the generalized whitehead theorem. Strøm's mistake, I think, was that he didn't realize that unbased $Top$ with his model structure has the very special property that all objects are fibrant-cofibrant. This is not the case for based spaces, however (all objects are fibrant but not cofibrant). The cofibrant objects are precisely the nondegenerately based spaces. -By restricting to nondegenerately based spaces, he restricted himself to working in the category of cofibrant objects of the actual model category $*\downarrow Top$ equipped with the relative Strøm model structure. It shouldn't be too surprising that this subcategory does not admit a (compatible) model structure. -(Proof that all objects in Strøm's model structure on $Top$ are fibrant-cofibrant): -Lemma: -A Hurewicz fibration $A\to B$ has the RLP with respect to every inclusion of a space $X$ into its cylinder $X\times I$. When $B$ is the terminal object, let $X\times I\to X$ be the map killing the cylinder. Then we get a lift to $X\times I$ of any map $X\to A$ by composing $X\times I\to X\to A$. -Lemma: -A closed Hurewicz cofibration is a closed inclusion $A\hookrightarrow B$ such that $A\times I \coprod_{A\times \{0\}} B\times \{0\}\hookrightarrow B\times I$ has the LLP with respect to any map $Y\to *$. When $A$ is empty, this reduces to finding an extension of a map $B\to Y$ to a homotopy extending this, but again, this is immediately possible by composition with the trivial homotopy $B\times I\to B$. -The theorem of Strøm is that there exists a model structure on Top with - -$C$ = closed hurewicz cofibrations -$W$ = homotopy equivalences -$F$ = hurewicz fibrations - -From which it follows that all objects are fibrant-cofibrant. -(Source of definitions: Dwyer-Spalinski example 3.6)<|endoftext|> -TITLE: String theory "computation" for math undergrad audience -QUESTION [23 upvotes]: I am giving a talk on String theory to a math undergraduate audience. I am looking for a nice and suprising mathematical computation, maybe just a surprising series expansion, which is motivated by string theory and which can be motivated and explained relatively easily. Examples of what I have in mind are the results in Dijkgraaf's "Mirror symmetry and the elliptic curve", or the "genus expansion" of the MacMahon function (aka DT/GW for affine three-space), but I am not sure I can fit either into the time I have. Any thoughts? - -REPLY [2 votes]: Derive the Casimir Energy in Bosonic String Theory. -You start with the $\hat L_0$ operator and get rid of the non-vacuum $\displaystyle\frac{\alpha_0^2}{2}+\sum_{n=1}^\infty\alpha_{-n}\cdot\alpha_n$, then you use a Ramanujam sum to do $\zeta$-function renormalisation, from which you find out that the vacuum energy denoted by $\varepsilon_0$ is -$$\varepsilon_0=-\frac{d-2}{24}$$ -However, the most interesting part comes when you go around deriving the critical dimension of Bosonic String Theory. -After which, the expression surprisingly simplifyies to a $-1$. -For a more detailed derivation of the above stuff, see these lecture notes/. (Section 4) (Equation 4.5-4.10)<|endoftext|> -TITLE: Derived functor -QUESTION [6 upvotes]: Let $F:A\longrightarrow B$ be a left exact functor of Abelian categories. My question is about the derived functor $RF: D(A)\longrightarrow D(B)$. -Let $X$ be an object of $A$. If $0\longrightarrow X\longrightarrow I^0 \longrightarrow I^1\longrightarrow \cdots$ -is an injective resolution of $X$, we know $RF(X)=F(I^n)$. When $X$ is a complex in $A$, can we write $RF(X)$ explicitly? The abstract definition of $RF$ is not easy to understand. - -REPLY [15 votes]: It's my opinion that derived functors are better understood through a universal property. On derived categories it is possible to express the condition of derivability in a concise way due to the additivity of everything in sight. My impression is that you do not need the full machinery of homotopical algebra unless you are in a non-additive situation (which is often the case, I should add). -Let me spell it out. Let $F\colon A\to B$ be a left exact functor between Abelian categories. Being additive, this induces a (Delta) functor between their homotopy categories that by abuse of notation we keep calling it $F\colon K(A)\to K(B)$. Recall that $K(A)$ is the category of (possibly unbounded) complexes of objects of $A$ with maps homotopy classes of chain maps. Now if we localize a homotopy category making invertible the quasi-isomorphisms (i.e. those maps that induce isomorphisms in homology) we obtain the derived category $D(A)$. If we call the canonical localization functor $Q \colon K(A) \to D(A)$ and similarly for $B$, the problem of deriving $F$ is to extend "in an optimal way" $Q \circ F$ to $D(A)$. The solution, if it exists, it is expressed by the following universal property: -The (Delta) functor $RF \colon D(A)\to D(B)$ is the derived functor of $F$ if there is a natural transformation $Q \circ F \to RF \circ Q$ such that for any other (Delta) functor $G \colon D(A)\to D(B)$ the natural transformation induces a bijection -$$ Hom(RF, G) \longrightarrow Hom(Q \circ F, G \circ Q)$$ -I suggest you draw a diagram (It is difficult to draw a diagram here, but very instructive if you do). -It turns out that if there are unbounded injective resolutions on $K(A)$ they provide a right adjoint to $Q \colon K(A) \to D(A)$ that automatically gives a way to express $RF$. This is related to the Bousfield localization philosophy in the context of derived categories. -Yet another question is the description of the adjoint functor i.e. the unbounded injective resolution. If the complex is bounded below, then the resolution can be constructed step by step. In the case that $A$ is the category of modules over a ring, one can use the exactness of products and use countable homotopy limits as in Bökstedt-Neeman. In more general cases such as Grothendieck categories, one needs more sophisticated tools. - -REPLY [4 votes]: The "correct" abstract definition of a derived functor involves the theory of "homotopical categories". An abelian category $\cal A$ does not have enough structure alone to consider homotopical properties. To this end, we usually look at $Ch^+(\cal A)$ or $s\cal A$ of chain complexes in $\cal A$ or simplicial objects in $\cal A$ respectively. We equip these larger categories with the structure of a homotopical category by fixing the lluf subcategories consisting of quasi-isomorphisms or weak homotopy equivalences respectively (by the Dold-Kan correspondence, these are essentially identical). In the terminology of Dwyer-Hirschhorn-Kan-Smith, left-exact functors are deformable in the sense that they preserve weak equivalences (that is to say, they are homotopical) on a deformation retract of our original category. So in truth, a derived functor is simply the evaluation of a functor on a suitable approximation of our original object by a homotopically equivalent one on which the functor is well-behaved. -When $\cal A$ has enough structure, we can upgrade the homotopical structure on our homotopical version of $\cal A$ (that is, chain complexes or simplicial objects) to a model structure, which makes it substantially easier to compute derived functors, since the existence of approximations is guaranteed, and further, such approximations can usually be taken to be functorial. For Quillen functors between model categories (these are functors that become homotopical on the deformation retracts consisting of cofibrant or fibrant objects), we can, in the presence of functorial factorization, define the derived functors to be the composition of the functorial approximation and the original functor. These approximations have the property that they descend to the usual Kan extensions at the level of the derived categories (for a proof, see Dwyer-Hirschhorn-Kan-Smith Homotopy Limit Functors on Model Categories and Homotopical Categories). -That is, in your case, we approximate the chain complex $X$ (concentrated in degree zero, so it is constant) by an injective approximation (fibrant for the injective model structure) and evaluate the functor on this new complex. In fact, if injective approximations exist for all chain complexes, we see that when we compose with an appropriate functorial injective approximation, our functor becomes homotopical (it preserves quasi-isomorphisms) and therefore descends uniquely by the universal property of localization to a total derived functor between derived categories of chain complexes.<|endoftext|> -TITLE: failure of $\square(\kappa)$ at an inaccessible $\kappa$ -QUESTION [8 upvotes]: How can we force the failure of $\square(\kappa)$ at an inaccessible $\kappa$, where -$\square(\kappa)$ is defined as follows: There is a sequence $(C_i:i< \kappa)$ such that: -(1) $C_{i+1} = \{i\}$ and $C_i$ is closed and cofinal in $i$ if $i$ is a limit -ordinal. -(2) If $i$ is a limit point of $C_j$, then $C_i = C_j \cap i$. -(3) There is no club $C$ (a subset of $\kappa$) such that for all limit points $i$ in -$C$ the equality $C_i= C \cap i$ holds. -Update. By a recent result of Magidor and Vaananen ( On Löwenheim-Skolem-Tarski numbers for extensions of first order logic), it is consistent, relative to the existence of a supercompact cardinal, that $\square(\kappa)$ fails at the least strongly inaccessible cardinal. -Is it possible to reduce the consistency strength to a weakly compact cardinal or even a measurable cardinal? - -REPLY [7 votes]: The consistency strength of the failure of $\square(\kappa)$ for non-weakly Mahlo inaccessible cardinal $\kappa$ (in particular, for the first strongly inaccessible) is higher than weakly compact. For example it implies that $0^\#$ exists. -http://www.jstor.org/stable/27590333 -We know that global square holds in $L$, i.e. there is a sequence $\langle C_\alpha | \alpha \text{ is singular limit ordinal} \rangle$ with the properties: $\forall \alpha,\, \text{otp } C_\alpha < \alpha$ and for every accumulation point $\beta$ of $C_\alpha$, $C_\alpha \cap \beta = C_\beta$. -In $V$, there is a club $D\subset \kappa$ through the singular cardinals below $\kappa$. If $0^\#$ doesn't exists, the covering theorem will imply that those cardinals are singular in $L$ as well, and therefore $C_\alpha$ is defined for every $\alpha \in D$. -Now we can define a $\square(\kappa)$ sequence, $\langle E_\alpha | \alpha < \kappa \rangle$ in $V$ as follow: - -For $\alpha \in \text{acc } D$, $E_\alpha = C_\alpha \cap D$ (unless this set is bounded below $\alpha$. In this case, $\text{cf }\alpha = \omega$ and we can choose $E_\alpha$ to be arbitrary $\omega$ sequence cofinal in $\alpha$). -For $\alpha \notin \text{acc }D$, $E_\alpha = \alpha \setminus (\max D\cap \alpha)$ - -It's easy to see that $\langle E_\alpha | \alpha < \kappa\rangle$ is coherent (since the global square is coherent) and thread-less (since every thread meets $D$ club many times, but $\text{otp }E_\alpha < \alpha$ for every $\alpha \in D$). -Edit: As Mohammad noted, we can get better lower bound by using more appropriate inner model. We need to violate global square on the singular cardinals. By $\square$ On the singular cardinals (Theorem 1), we need at least many measurable cardinals of uncountable Mitchell order in order to get a model without global square on the singular cardinals.<|endoftext|> -TITLE: Can a curve intersect a given curve only at given points? -QUESTION [15 upvotes]: Clearly the question in the title has a positive answer for analytic (or smooth, or continuous ...) curves, but what about the algebraic category? More specifically, given an irreducible polynomial curve $X \subseteq \mathbb{CP}^2$ and points $P_1, \ldots, P_k$ on $X$, when can we find another curve $Y$ (defined by a polynomial) such that the $Y$ intersects $X$ only at $P_1, \ldots, P_k$? -I find the question to be nontrivial even for $k = 1$. Here are some observations for $k = 1$ case: - -If $P$ is a point on $X$ with multiplicity $\deg X - 1$, then a tangent of $X$ through $P$ intersects $X$ only at $P$ (by Bezout's theorem). -If $X$ is a rational curve and $X \setminus \{P\} \cong \mathbb{C}$, then there is a curve $Y$ such that $X \cap Y = \{P\}$. -Let $X$ be a non-singular cubic. Give it a group structure such that the origin is an inflection point. Then for all $P \in X$, there exists $Y$ such that $Y \cap X = \{P\}$ iff $P$ is a torsion point in the group. - -If $X$ (of degree $d$) is non-singular at $P$, then the most direct approach for finding a $Y$ of degree $e$ intersecting $X$ only at $P$ seemed to blow it up $de$ times and look for the conditions under which $Y$ goes through each of the points on $X$ in the $i$-th infinitesimal neighborhood of $P$, $0 \leq i \leq de - 1$. But the conditions on the coefficients of the polynomial defining $Y$ did not appear very tractable. - Edit: I would like to make a correction to observation 3. This is what I know about a non-singular cubic curve $X$: If $P$ is an inflection point, then there is a curve $Y$ such that $Y \cap X = P$ (take $Y$ to be the tangent of $X$ at $P$). If $P$ is a non-torsion point (for the group structure on $X$ for which the origin is an inflection point), then there is no such $Y$. I don't know what happens for torsion points. - -REPLY [2 votes]: This is actually an answer (a bit belated!) to one of the questions Juan posed in his answer. I did not want to put it as just a comment, since I thought the question was good, and the answer is cute enough to be interesting to some people. -The answer to Juan's first question is affirmative: "Every finite set of points in the projective plane is a set-theoretic complete intersection." Here are the arguments: we may assume number of points is $d + 2$, $d \geq 0$. Choose $\mathbb{C}^2 \subseteq \mathbb{P}^2$ and coordinates $(x,y)$ on $\mathbb{C}^2$ such that - -one of the points is on the intersection of $y$-axis and the line at infinity, -The others are in $\mathbb{C}^2$, -The finite points have mutually distinct $x$-coordinates. - -Now by Lagrange interpolation we can find a polynomial curve $C$ with equation of the form $y = f(x)$ which passes through each of the finite points. We may (and will) assume $\deg(f) \geq 2$, so that $C$ passes also through the other point which is at infinity. Let $a_1, \ldots, a_{d+1}$ be the $x$-coordinates of the finite points. Let $g_1 := z\prod_{i=1}^{d+1}(x - a_1z)$ and $g_2(x,y,z)$ be the homogenization of $y - f(x)$ (with respect to $z$). Then $g_1 = 0$ and $g_2 = 0$ intersects precisely at the given points - answering Juan's question. -Note that $g_2$ is irreducible. Replacing $g_1$ by an element of the form $g_1^p + \lambda g_2^q$ for suitable $\lambda$, $p$ and $q$, we may ensure that $g_1$ is also irreducible. Is it possible to ensure that the curves are non-singular?<|endoftext|> -TITLE: Basis for the Algebraic numbers over the rationals -QUESTION [9 upvotes]: Is there an explicit basis for the algebraic numbers as a vector space over the rationals? - -REPLY [12 votes]: Every computable field which is an algebraic extension of the rationals $\mathbb{Q}$ has a computable basis (as a vector space over $\mathbb{Q}$). The idea is to build up this basis by recursion: let $F_0 = \mathbb{Q}$, with basis $B_0=$ {$1$}, and, given a basis for $F_s$ over $\mathbb{Q}$, find the least element $x$ of $F$ whose minimal polynomial over $F_s$ has degree $d > 1$. This can be done because $F_s$ has a splitting algorithm, enabling one to recognize irreducible polynomials in $F_s[X]$, uniformly in $s$. Then $C_s=${$ 1, x, x^2, x^3,...x^{d-1}$} is a basis for $F_{s+1}=F_s[x]$ over $F_s$, and a basis $B_{s+1}$ for $F_{s+1}$ over $\mathbb{Q}$ is given by all products of one element of $C_s$ with one element of the basis $B_s$ for $F_s$ over $\mathbb{Q}$. Since $1$ lies in $C_s$, we have $B_s \subset B_{s+1}$, and the union (over $s$) of all these $B_s$ will be a basis $B$ for $F$ over $\mathbb{Q}$. The same construction would work with any finite field in place of $\mathbb{Q}$, and indeed over every c.e. ground field which has a splitting algorithm. -However, when $F$ is transcendental over $\mathbb{Q}$ (especially when it has infinite transcendence degree), this can no longer be done. In this case a computable vector-space basis follows if there exists a computable transcendence basis for $F$ over $\mathbb{Q}$, but not all computable field extensions of $\mathbb{Q}$ have computable transcendence bases. Metakides and Nerode produced a computable field of infinite transcendence degree over $\mathbb{Q}$ which has no computable vector-space basis over $\mathbb{Q}$. One could probably also build a computable field $\mathbb{Q}$ with computable vector-space basis over $\mathbb{Q}$, but with no computable transcendence basis over $\mathbb{Q}$. -A discussion of splitting algorithms, over $\mathbb{Q}$ and over finitely generated extensions, appears in Harold Edwards' book Galois Theory. Fried & Jarden is another good source.<|endoftext|> -TITLE: Set of maximal subfields not containing particular elements. -QUESTION [9 upvotes]: Instead of extending a field, by adjoining a new element, consider what happens if we remove an element or elements. -This started as a question on math.SE Field reductions where Pete L. Clark explained that there isn't a unique subfield that could be called a "field reduction", rather there is a set of maximal subfields not containing the element. -Let $\mathbb{R}(\setminus a)$ be the set of maximal subfields of $\mathbb{R}$ that don't contain $a$. -Question: What is the cardinality of $\mathbb{R}(\setminus a)$ ? -Pete L. Clark suggested that this sounds "ultrafiltery" so I thought it would be an appropriate question for Mathoverflow. -Let $\mathbb{A}'$ be the set of non-rational algebraic numbers. -In Field reductions. part two Arturo Magidin showed that the cardinality of $\mathbb{R}(\setminus\mathbb{A}')$ is $2^\mathfrak{c}$. - -REPLY [5 votes]: I might be missing something but I think Arturo´s idea can also be used to show that $\mathbb{R}(\setminus a)$ has cardinality $2^\mathfrak{c}$: -Let $T$ be any set of reals algebraically independent over $\mathbb{Q}(a)$ with $|T|=\mathfrak{c}$. For any $X \subseteq T$ the set $T_X:=aX \cup (T \setminus X)$ is still algebraically independent over $\mathbb{Q}(a)$ and by Zorn´s lemma we can find $F_X \in \mathbb{R}(\setminus a)$ such that $\mathbb{Q}(T_X) \subseteq F_X$. If $X \neq Y$ then $F_X \neq F_Y$ because otherwise there would be a $t \in T$ such that both $t$ and $at$ (and hence $a$) are elements of $F_X$, a contradiction.<|endoftext|> -TITLE: Explicit injective resolutions of (Laurent) polynomial rings -QUESTION [5 upvotes]: Hi, -Despite being nothing but the dual notion of projective resolution, injective resolutions seem to be harder to grasp. For example, the general form of Poincaré-Lefschetz duality given in Iversen's Cohomology of sheaves (p. 298) uses an injective resolution of the coefficient ring k (which is assumed to be Noetherian) as a k-module, a notion whose projective equivalent is rather meaningless. -My question is: do we have explicit injective resolutions of some simple (but not principal) rings (as modules over themselves) like the polynomial ring $\mathbb{C}[X,Y]$ or its Laurent counterpart $\mathbb{C}[X,Y,X^{-1},Y^{-1}]$? By general dimension arguments, some short resolutions exist, but I'm unable to find them explicitly. -Thanks! - -REPLY [5 votes]: $\newcommand{\C}{\mathbb C} -$I think this is OK. -The first step is the inclusion of $\C[X,Y]$ into its fraction field which is -$\C(X,Y)$. For each irreducible polynomial $f$ (normalised so that the top degree -monomial for some ordering is $1$) we map $\C(X,Y)$ to $\C(X,Y)/\C[X,Y]_{(f)}$ -and then we map $\C(X,Y)\rightarrow\bigoplus_f\C(X,Y))/\C[X,Y]_{(f)}$ which is -the next step in an injective resolution, the kernel of this map is clearly -$\C[X,Y]$. Finally, the cokernel of this map is injective (as the global -dimension of $\C[X,Y]$ is $2$). -Addendum: A systematic way of getting this resolution as well as identifying the last term is to note that the Cousin complex of $\C[X,Y]$ is an injective resolution (Hartshorne: Residues and duality, SLN 20, p. 239) which in degree $p$ is the sum of the injective hulls of the residue fields of points of dimension $p$.<|endoftext|> -TITLE: Grothendieck and Non-commutative Geometry? -QUESTION [22 upvotes]: When Grothendieck and his followers were working on their profound progress of algebraic geometry, did they ever consider non-commutative rings? Is there anyway evidence that Grothendieck foresaw the developments that would later come in non-commutative geometry or quantum group theory? - -REPLY [29 votes]: No and yes, depending on the level of understanding. The consideration of noncommutative rings telling about geometry is almost nonexistent in Grothendieck's published opus. One of the exceptions is that he considered cohomologies for the possibly noncommutative sheaves of $\mathcal{O}$-algebras for commutative $\mathcal{O}$ (the latter is used in Semiquantum geometry). On the other hand, Grothendieck has been pioneer on abandoning the points of spaces as primary objects and promoting the category of sheaves over the space as defining the space. This is the point of view of topos theory which he invented; he noticed that the topological properties do not depend on a site but only on the associated topos of sheaves, and proposed a topos as a natural generalization of a topological space. Manin took Grothendieck's advice that one should consider the topos of sheaves as replacing the space, together with Serre's theorem that the category of quasicoherent modules determines a projective variety, as a motivation to his approach to noncommutative geometry and quantum groups. The modern view of noncommutative geometry is that it is about the presentation of space via the structures consisting of all possible objects of some kind living on a space (algebra of functions, some structures consisting of cocycles, like category of vector bundles, category of sheaves, higher category of higher stacks). -In late 1960s W. Lawvere, with help from Tierney, extended the Grothendieck topoi to the theory of elementary topoi. This was not the only contribution of Lawvere in the 1960s. Lawvere promoted also the duality between spaces and dual objects which he calls quantity (cf. space and quantity). While Lawvere's impact has been deep, I object to the terminology: in physics a quantity is normally a single observable; physicist do not consider the algebra of all observables a quantity, but rather a field of quantities, or algebra of quantities. But never mind the terminology, Lawvere went on very deeply in presenting this point of view, which is really generalized noncommutative geometry. Of course, neither Grothendieck nor Lawvere did not pay that particular attention to reconstructing the differential geometry and measure theory from the study of operator algebras, what is the huge contribution of Connes, or from the study of noncommutative rings, which was implicit in Gabriel 1961 and more explicit with works of J. S. Golan, van Oystaeyen (and P. M. Cohn with his affine spectrum) and others in mid 1970s, working with spectra of noncommutative rings and noncommutative localization theory as a noncommutative analogue of Zariski topology. One should mention that sporadic appearance of operator algebras from the noncommutative geometry point of view is present to some extent in 1970s book of Semadeni on Banach spaces of continuous functions (MR296671), where he studies, among other topics, the noncommutative analogues of many topological properties of topological spaces; in less explicit form there are also works of Irving Segal which had a similar motivation. -Grothendieck says in his memoirs that the concept of abelian category as he promoted it in Tohoku is part of the same philosophy -- abelian categories, possibly with additional axioms like AB5 are sort of categories of sheaves of modules, and should be viewed as an idea which is sort of abelian/stable version of Grothendieck topoi. More precisely, in this line, there is a recent Nikolai Durov's concept of a vectoid. Pierre Gabriel, who was close to Grothendieck's school in his early days, had in his prophetic work of 1961 reconstruction theorem for schemes and study of subcategories and localizations in abelian categories which represent open or closed subschemes and so on. Gabriel's work is in fact the first big work in noncommutative algebraic geometry and his reconstruction theorem is really the basic motivation in algebraic flavour of the theorem. In a sense, Gabriel's work is an abelian version of some Grothendieck's basic ideas of topos theory (cf. noncommutative scheme for one of the modern ideas along that line of thought) and Grothendieck was well aware of the abelian direction of this thinking from the Tohoku times. -For a general vista, I recommend - -Pierre Cartier, A mad day's work: from Grothendieck to Connes and Kontsevich The evolution of concepts of space and symmetry, Bull. Amer. Math. Soc. 38 (2001), 389-408, pdf.<|endoftext|> -TITLE: When can you reverse the orientation of a complex manifold and still get a complex manifold? -QUESTION [17 upvotes]: I'm told that $\overline{\mathbb{C}P^2}$, i.e. $\mathbb{C}P^2$ with reverse orientation, is not a complex manifold. But for example, $\overline{\mathbb{C}}$ is still a complex manifold and biholomorphic to $\mathbb{C}$. -This makes me wonder, if $X$ is complex manifold is there a general criterion for when $\overline{X}$ also has a complex structure? For example, it seems that if $X$ is an affine variety than simply replacing $i$ with $-i$ gives $\overline{X}$ a complex structure and $X, \overline{X}$ are biholomorphic. -EDIT: the last claim is wrong; see BCnrd's comments below and Dmitri's example. Also, as explained by Dmitri and BCnrd, $X$ should be taken to have even complex dimension. -Another question: if $X$ and $\overline{X}$ both have complex structures, are they necessarily biholomorphic? -Edit: No per Dmitri's answer below. - -REPLY [11 votes]: If you take an odd dimensional complex manifold $X$ with holomorphic structure $J$ then $-J$ defines on $X$ a holomorphic structure as well. And, of course, $J$ and $-J$ induce on $X$ opposite orientations. In general it is not true that these two complex manifolds are biholomorphic. Indeed, if $X$ is a complex curve, then $(X,J)$ is biholomorphic to $(X,-J)$ only if $X$ admits an anti-holomorphic involution (this will be the case for example if $X$ is given by an equation with real coefficients). -Starting from this example on can construct a (singular) affine variety $Y$ of dimension $3$, such that $(Y,J)$ is not biholomorphic to $(Y,-J)$. Namely, let $C$ be a compact complex curve that does not admit an anti-holomorphic involution say of genus $g=2$. Consider the rank two bundle over it, equal to the sum $TC\oplus TC$ ($TC$ is the tangent bundle to $C$). Contract the zero section of the total space of this bundle, this gives you desired singular $Y$.<|endoftext|> -TITLE: Generalized geometry and spin structures -QUESTION [25 upvotes]: Let $(M,g)$ be a $d$-dimensional, oriented pseudo-Riemannian manifold, and $V$ the subbundle of $E=TM\oplus T^*M$ given by the graph of the musical linear isomorphism $g^\flat:TM\rightarrow T^*M$ associated to the metric $g$. The nondegeneracy of $g$ entails that $E$ decomposes as the Whitney sum $E=V\oplus V'$, where $V'$ is the graph of $-g^\flat=(-g)^\flat$. Indeed, the projections -$P_\pm(X\oplus\xi)=\frac{1}{2}(X\pm g^\sharp(\xi))\oplus(\xi\pm g^\flat(X))$ -satisfy $P_-=\mathbb{1}-P_+$, $P_+(E)=V$ and $P_-(E)=V'$, where -$g^\sharp:T^*M\rightarrow TM$ is the musical linear isomorphism associated to $g^{-1}$. Moreover, $E$ carries a canonical, pseudo-Riemannian metric $h$ of signature $0$ -$h(X_1\oplus\xi_1,X_2\oplus\xi_2)=\frac{1}{2}(\xi_1(X_2)+\xi_2(X_1))$ -such that $(\mathbb{1}\oplus g^\flat)^*(h|_V)=g$. It can be shown that $E$ admits a spin structure associated to $h$ - for example, the (space of sections of the) exterior algebra bundle $\Lambda^*T^*M$ is a Clifford module, and its tensor product $\Lambda^*T^*M\otimes(\Lambda^dT^*M)^{1/2}$ with the real line bundle of half-densities over $M$ is a spinor bundle. Moreover, the space of spin structures on $(E,h)$ is an affine space modelled on the group $H^1(M,\mathbb{Z}_2)$ of real line bundles over $M$, which maps the corresponding spinor bundles onto each other by tensoring (the above preliminary results can be found on Chapter 2 of Marco Gualtieri's PhD thesis on generalized complex geometry, arXiv:math.DG/0401221). - -Question(s): if $(M,g)$ admits a spin structure, does a choice of spin structure on - $(TM\oplus T^*M,h)$ descend by restriction to $V$ to a choice of spin structure on - $(M,g)$? Does this establish a one-to-one correspondence between both sets of spin - structures? If so, how does this generalize to, say, generalized Riemannian metrics (i.e. - rank-$d$ subbundles $W$ of a twisting of $TM\oplus T^*M$ by a Cech 1-cocycle $B$ with - values at closed 2-forms, such that the restriction of $h$ to $W$ is positive definite)? - -In other words, I want to know if, in the above convext, there is a specific converse to the well-known property that, given two pseudo-Riemannian vector bundles $V,V'$ and their Whitney sum $E=V\oplus V'$, a choice of spin structure for any two of these bundles uniquely determines a spin structure on the remaining one. In the case $V$ and $V'$ are Riemannian, this is Proposition 2.1.15, pp. 84-85 of H. B. Lawson and M.-L.Michelsohn, "Spin Geometry" (Princeton, 1989). See also M. Karoubi, "Algèbres de Clifford et K-Théorie". Ann. Sci. Éc. Norm. Sup. 1 (1968) 161-270. -More precisely, here we use the fact that there is a canonical isomorphism between $Spin(p,q)$ and $Spin(q,p)$ which covers the canonical isomorphism between $SO(p,q)$ and $SO(q,p)$ (see M. Karoubi, ibid.) to establish a canonical one-to-one correspondence between the set of spin structures on $(M,g)$ and the set of spin structures on $(M,-g)$. This correspondence, on its turn, is used to induce spin structures on $V$ and $V'$ from that of $(M,g)$ together with the orientation-preserving, isometric bundle isomorphisms $\mathbb{1}\oplus(\pm g^\flat)$. What I want to know is if, among the pairs of spin structures on $V$ and $V'$ which determine the given spin structure on $E$ in the above fashion, there is (only?) one pair which, once pulled back to $M$, is related by the above one-to-one correspondence between $Spin(p,q)$- and $Spin(q,p)$-structures. - -REPLY [4 votes]: OK, I am making an assumption: I can re-interpret the problem (using the musical isomorphism) as $V$ being the diagonal embedding of $TM$ inside $TM\oplus TM$ and studying spin structures on them. -Actually, it turns out (see comments) that this "re-interpretation" is slightly different from the original construction. But the main bulk still goes through: -I will restrict to $\dim M=3$, in which case our (closed oriented) manifold is always spinnable. A spin structure $\mathfrak{s}$ on $M$ induces a canonical spin structure $\mathfrak{S}_0=\mathfrak{s}\oplus\mathfrak{s}$ on $TM\oplus TM$, and this is actually independent of the choice of $\mathfrak{s}$ (these appear in the notion of a 2-framing on 3-manifolds, which Atiyah and Witten have used for some of their QFT studies). As a result, the "restriction" $\mathfrak{S}_0|_V$ on $M$ is ill-defined. -[proof of claim of canonical spin structure (learned from conversation with Rob Kirby): the spin structure fixes a trivialization over the 1-skeleton, and over circles there are two trivializations, so changing a trivialization of $\mathfrak{s}$ is doubled in $\mathfrak{s}\oplus\mathfrak{s}$ which modulo-2 is no change.] -This also implies that the "restriction" to each $TM$-summand is ill-defined. (What I know that works: the collar-neighborhood theorem does allow an induced spin structure on $T(\partial X$) from a spin structure on $TX$ thanks to the splitting $TX|_\partial=T(\partial X)\oplus\underline{\mathbb{R}}$ near the boundary.)<|endoftext|> -TITLE: How manifold-like is Aut(C^n) in the holomorphic category? -QUESTION [10 upvotes]: This question is similar to, but not the same as this one. Take the space of automorphisms of $\mathbb{C}^n$ in the holomorphic category, with the compact-open topology. For $n=1$ this is just $\mathbb{C} \times \mathbb{C}^*$, but for larger $n$ is very complicated. It clearly contains $GL(n,\mathbb{C})$, and translations. -To give a taste of how big $Aut(\mathbb{C}^n)$ is, there is a theorem that given any two countable dense subsets $X,Y \subset \mathbb{C}^n$, $n >1$, there is a volume preserving automorphism taking $X$ to $Y$. But I have no idea about what this space is like. Is it some sort of infinite dimensional manifold? Analytic space? Does it contain an infinite dimensional Lie group as a (closed) subgroup? - -REPLY [3 votes]: It is a group generated by flows of holomorphic vector fields. Its Lie algebra is a set of all holomorphic vector fields. As I can for now remember there were a computation of its group of cohomologies (by Feigin and Fuchs: http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=rm&paperid=3301&option_lang=eng -http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=intf&paperid=93&option_lang=eng -) -Maybe this book can help if you have no translation of these articles from Russian: -Fuks D.B. Cohomology of infinite-dimensional Lie algebras, Consultants Bureau, 1986 -I think that you have much to learn from this book. -Also there is an article: http://arxiv.org/abs/0708.3398 - it may be rather useful<|endoftext|> -TITLE: Examples of Mixed Hodge Structures -QUESTION [32 upvotes]: Does anyone know a user-friendly, example-laden introduction to mixed Hodge structures? I get from Wikipedia how to calculate for a punctured and pinched curve (http://en.wikipedia.org/wiki/Hodge_structure#Mixed_Hodge_structures), but I want more! I want tables and numbers and everything explicit and spoonfed. Thanks! - -REPLY [17 votes]: I just noticed the question. The references mentioned so far are good. So I'll just do -the example that I normally do on a blackboard when anyone asks me. -Take a smooth complex projective variety $X$ with a smooth divisor $D$. Set $U=X-D$. There's -a long exact sequence -$$ \ldots H^i(X)\to H^i(U)\to H^{i-1}(D)\to H^{i+1}(X)\ldots$$ -say with rational or complex coefficients. What are the maps? -The first is restriction, the second using $\mathbb{C}$ coefficients is a residue map, -and the third is the Gysin map which is of type $(1,1)$ (or you want to want to get -fancy you need a Tate twist here). The mixed Hodge theory of $U$ can be read off from this. -For example, for the Hodge numbers -$$\dim H^{pq}(U)= \dim im[H^{p-1,q-1}(D)\to H^{pq}(X)]+\dim ker[H^{pq}(D)\to H^{p+1,q+1}(X)]$$ -and so on. (By the way, $H^{pq}$ is taken to be the $(p,q)$ part of the $p+q$ -weight graded quotient.) -OK, let me make it more concrete. Let $X$ be a surface with irregularity $q=0$, perhaps $\mathbb{P}^2$, then $D \subset X$ is a curve of say genus $g$. -Then from above, the interesting Hodge numbers are -$$h^{20}(U)=h^{02}(U)=h^{20}(X)$$ -$$h^{11}(U)=h^{11}(X)-1$$ -$$h^{12}(U)=h^{21}(U)=g$$ - Maybe that's enough for now. -I can't resist squeezing in one more example. Suppose $D$ on the above surface $X$ -can be contracted to a point in a normal surface $Y$. So for example, $Y$ might be a cone over a plane curve, and $X$ the blow up of the vertex. Using duality and the standard exact sequence for compactly supported cohomology, we can conclude -$$H^i(Y)=H_c^i(U)= H^{4-i}(U)^*(-2),\quad i>0$$ -As far as Hodge numbers are concerned, the dual means $(p,q)\mapsto (-p,-q)$ and $(-2)$ means shift by $(2,2)$.<|endoftext|> -TITLE: Radicals of binomial ideals -QUESTION [6 upvotes]: Let $R=k[x_1,x_2,...,x_n]$ be the polynomial ring in $n$ indeterminates over a field $k$. An ideal (that can be) generated by monomials is called a monomial ideal. For the monomial ideal $M=(m_1,m_2,...,m_t)$, the radical of $M$ itself is monomial and can be written as, $Rad(M)=(\sigma(m_1),\sigma(m_2),...,\sigma(m_t))$ where $\sigma(x_1^{a_1}x_2^{a_2}...x_n^{a_n})$ is the product of indeterminates $x_i$ s.t. $a_i\geq 1$. -A binomial ideal in $R$ is generated by binomials. I was wondering if we have similar theorems for the case of binomial ideals where we can write down a generating set for the radical by just knowing a generating set of the ideal. Eisenbud and Sturmfels, in their monumental paper on binomial ideals, showed that the radical itself is binomial. I am especially interested in finding generators for radical of binomial ideals in the case where char$(k)=0$ (or even when $k=\mathbb{C}$) and what kind of binomials generate radical binomial ideals. -Becker, Grobe and Niermann discuss the case of zero dimensional binomial ideals. Ojeda and Sanchez prove some results for radicals of lattice (binomial) ideals. I have also seen some results in positive characteristic, but they are not relevant to my research. - -REPLY [2 votes]: The minimal primes (and sometimes also their intersection) can be computed relatively quickly (compared to primary decomposition) using Algorithm 4 of http://arxiv.org/pdf/0906.4873v3. I've looked at binomial ideals for some time and I doubt that there is an easy way to see the generators of the radical.<|endoftext|> -TITLE: "Noncommutative heat equation" -- a strange generalization of Killing vectors for a flat metric -QUESTION [8 upvotes]: Let $(M,g)$ be a smooth (pseudo)Riemannian manifold with a flat metric $g$, and $X$, $Y$ be vector fields on $M$ such that -$$ -L_X^2 (g)=L_Y(g). \hspace{70mm} \mbox{(1)} -$$ -where $L_Z$ is the Lie derivative along $Z$ and $L_X^2(g)\equiv L_X(L_X(g))$. -If $X=0$ then $Y$ is just a Killing vector for $g$ but - -is there any (geometric?) interpretation for $X$ and $Y$ in the general case? - -In fact I wonder whether Eq.(1), be it for fixed $g$ and unknown $X,Y$ or conversely for fixed $X,Y$ and unknown $g$, was studied systematically at all: I failed to find any relevant references. -Motivation: Equation (1) follows from the last formula in Proposition 5 from a paper on classification of compatible Hamiltonian structures that I have recently come across. It looks a bit like the heat equation for the metric, hence the title. - -REPLY [8 votes]: To understand the local geometry of this equation, I think one should apply the Calabi resolution of the Killing equation. (See E. Calabi, On compact, Riemannian manifolds with constant curvature. I, in Differential Geometry, C. B. Allendoerfer, ed., vol. 3 of Proceedings of Symposia in Pure Mathematics, pp. 155–180. AMS, Providence, RI, 1961. For a useful modern exposition, see http://arxiv.org/pdf/1409.7212.pdf.) -The basic point is that, when $g$ is flat (or, more generally, has constant sectional curvature) the Killing operator $X\mapsto D_1(X) = L_X(g)$ as a map from $C^\infty(TM)$ to $C^\infty(S^2(T^*M))$ can be embedded into a locally exact sequence of sheaves -$$ -0 \longrightarrow \mathscr{K}_g\longrightarrow C^\infty(TM)\longrightarrow C^\infty(S^2(T^*M))\longrightarrow C^\infty\bigl(C^{(2,2)}(TM)\bigr)\longrightarrow\cdots\tag1 -$$ -that resolves the sheaf $\mathscr{K}_g$ of $g$-Killing vector fields on $M$ and where the sheaf mappings are given by known linear differential operators. An element $h$ of $C^\infty(S^2(T^*M))$ can be written (locally) in the form $h = L_Y(g)$ for some vector field $Y$ if and only if $D_2(h) = 0$. -This shows that if $L_X(L_X(g)) = L_Y(g)$, then $X$ must satisfy the system of differential equations -$$ -D_2\bigl(L_X(L_X(g))\bigr) = 0.\tag2 -$$ -Conversely, if a vector field $X$ satisfies this equation, then, at least locally (and globally if $M$ is simply-connected), there will exist a vector field $Y$ such that $L_X(L_X(g)) = L_Y(g)$, and this $Y$ will be unique up to the addition of a Killing vector field. -The differential equation (2) for $X$ is generally overdetermined (and nonlinear) when the dimension $n$ of $M$ is greater than $2$, and one expects the space of its local solutions to be finite dimensional. Thus, one expects that, even locally, the space of solutions of the original system for $(X,Y)$ will be finite dimensional when $n>2$. To check this, one should compute the characteristic variety of (2). If there is interest, I can do this and report the result.<|endoftext|> -TITLE: Functions which form continuous curve with its own iterations -QUESTION [7 upvotes]: The following function -$$f(x)=-2 \cos \left(\sqrt{2} \arccos \left(\frac{x-1}{2}\right)\right)+1$$ -has interesting property to form a continuous curve with its own integer iterations. The following image illustrates this property: -      (source) -Here blue is $f(x)$, red is $f(f(x))$, yellow is $f(f(f(x)))$ and green is $f(f(f(f(x))))$. It seems that all these functions form a continuous, and, probably, smooth curve. -The question is what is the general criterion for a function to have such property. Can you point some more examples of functions with such property? -P.S. If to use the following function -$$f(x)=-2 \cos \left(\sqrt{2} \arccos \left(\frac{x-1}{2}\right)\right)+1.1$$ -the curve becomes as below with one continuous branch and numerous closed circular branches. -      (source) -For function -$$f(x)=-2 \cos \left(\sqrt{2} \arccos \left(\frac{x-1}{2}\right)\right)+0.9$$ -the curve is completely continuous and seems to be smooth. -      (source) - -REPLY [2 votes]: Playing with the first example, it appears that the answer is negative. Let $x_0$ be the fixed point of $f(x)=-\cos\left(\sqrt{2} \arccos\left(\frac{x-1}{2}\right) \right)+1$. Looking at the point $-1< x_1<0$ where $f(f(f(x_1)))=x_0$, we see that $f(f(x_1))=-1$ and the point where -$f(x)=-1$ is $x=3$, so it's easy to see that $x_1=1+2\cos\left(\frac{\sqrt{2}\pi}{2}\right)$. -So, if I am understanding the question correctly, we want to know whether $\frac{df^{[4]}}{dx}(x_1^{-}) = \frac{df^{[3]}}{dx} (x_1^{+})$. I find that $\frac{df^{[4]}}{dx}(x_1^{-})=-\frac{8\sin(\sqrt{2}\pi)}{\sqrt{2-2\cos(\sqrt{2}\pi)}}$ whereas $\frac{df^{[3]}}{dx} (x_1^{+})=-\frac{4\sqrt{2}\sin(\sqrt{2}\pi)}{\sqrt{2-2\cos(\sqrt{2}\pi)}}$. -(Here I've used $f^{[n]}$ for the nth iterate of $f$ and $f(x^{\pm})$ for the right/left hand limit of $f$.) -It seems to me the graphs are illustrations of two facts: 1.) if $f^{[n]}(X)=x_0$, then $f^{[n+1]}(X)=x_0$, and 2.) since $f$ is decreasing near the fixed point $x_0$, if $f^{[n]}(x)$ is increasing approaching an $X$ such that $f^{[n]}(X)=x_0$, then $f^{[n+1]}(x)$ will be decreasing near $X$ (and vice versa). But the matching of derivatives is not guaranteed.<|endoftext|> -TITLE: Number of integral points inside a small sphere -QUESTION [5 upvotes]: Is there a good asymptotic estimation for the number of integral points -inside a $n$-sphere of a small radius ( less than $\sqrt{n}$)? -It looks like a problem which has been studied, however, I cannot -find any reference. Thanks for help! - -REPLY [11 votes]: Your question is treated in "Lattice points in high dimensional spheres" by J.E. Mazo and A.M. Odlyzko. -The article starts with - -A general principle, dating back to Gauss, that is widely used in estimating the number of integer lattice points in nice sets $S$ in $\mathbb{R}^n$ is that this number equals the volume of $S$ with a small error term [4,5,13]. This approach is very useful, and can be proved to be rigorous, for example, if one considers the number of lattice points in sets $rT$, where the dimension $n$ is fixed, $T$ is a given nice set, and $r \to \infty$. However, we will show below that this principle fails completely if one considers the dimension $n \to \infty$, and the sets $S$ to be spheres of radii proportional to $\sqrt{n}$. That this general principle cannot be justified rigorously in this case is not surprising, since (i) the surface area of the sphere of radius $\sqrt{\alpha n }$ is larger than the volume by a factor of $\sqrt{n/\alpha}$, and (ii) the diameter of the unit cube is comparable to the diameter of the sphere. However, it is rather interesting that it is not just the proof, but the principle itself, that fails. - -Now, I know that this is probably not the most recent paper on this problem but maybe you can search through the articles that reference it. See also Fedja's answer to this previous question.<|endoftext|> -TITLE: Does a triangulated category that possesses a subcategory $B$ of generators with no extensions of non-zero degree between them have to be isomorphic to $K^b(B)$? -QUESTION [10 upvotes]: Suppose that a triangulated category $C$ contains a full additive subcategory $B$ of (strong) generators (i.e. there does not exist a proper strict triangulated subcategory $C'\subset C$ that contains $B$) such that: there are no non-zero $C$-morphisms between $B_1$ -and $B_2[i]$ for any $B_1,B_2\in Obj B$ and $i\neq 0$. -Is it true that $C\cong K^b(B)$? Is anything known about this question (in general)? -Upd. I know how to prove this statement for any 'algebraic' triangulated $C$ (i.e. if $C$ admits a differential graded enhancement); this includes all derived categories of sheaves. So, one can reformulate my question as follows: is such a $C$ necessarily algebraic? I know the proof of this fact when $C$ is an $f$-category (in the sense of Beilinson). - -REPLY [3 votes]: Mikhail, if you assume more generally that $C$ is topological (in an appropriate sense) then your claim is also true. As Matthias suggests above, this is a tilting-like theorem, Theorem 5.1.1 in 'Stable model categories are categories of modules' by Schwede and Shipley. This assumption may comprise all examples of interest to you. So far, the very few known examples of exotic model categories do not satisfy your assumptions.<|endoftext|> -TITLE: Suggestions for sonifying math -QUESTION [12 upvotes]: Let me apologize first as I see this may be way off topic. Still it is a really fun question I've been meaning to ask a few fellow grads/faculty members, and so I think it's worth a shot here. -I'm interested in suggestions for using math formulas or concepts in coding algorithmic music. -In Stephen Cope's Workshop in Algorithmic Computer Music in 2004 I was introduced to the art of algorithmic composition, through coding LISP to generate midi compositions, frequently using markov chains to weight transitions, from the large scale harmonic progression and rhythmic structure to the individual notes and their time values. -We played a bit with simple math functions for generating simple pieces. One of these I wrote "sonified" the towers of Hanoi. The movement of the kth largest disk generated a bleep of frequency N(2/3)^k, for N some (high) starting frequency. Since 2/3 is roughly the ratio to get the next lower 5th, the I was able to stay roughly in the 12 tone equal temperament, while superimposing the same pulse at (2/3)^k the (tempo and wave) frequency. The piece wasn't particularly interesting musically, but conceptually fun. -In the workshop many other math themes are explored such as cellular automata, genetic algorithms, Brownian motion. I've been thinking since about interesting curves on the orbifold $T^n/\Sigma_n$ ($n$ continuous voices modulo the octave and modulo their labeling), and also about energy functions which give harmonic progressions as geodesics. (Perhaps harmonic functions would be applicable here, after all!) -I wonder what specific examples others have for making interesting pieces of music (art), or vague examples for that matter. -I'm happy to close this off, too, if no one is interested. Sorry for the softy. - -REPLY [3 votes]: I messed around with this when I was a C++ TA. I came up with "musical sorting algorithms", "musical Gauss Seidel" and a terrible sounding FFT. Details: http://www.math.ucla.edu/~rcompton/art.html<|endoftext|> -TITLE: Characterizations of Euclidean space -QUESTION [9 upvotes]: I posted this question at math.stackexchange.com but didn't get an answer. Is it a dumb question, eventually? -There are three ways of characterizing the abstract Euclidean space $E^n$ that are quite different in spirit: - -axiomatically (with axioms concerning dimension) -by the abstract Euclidean group $E(n)$ (as its symmetry group, determining $E^n$ uniquely) -by presupposing a metric and requiring that the space is a maximal one with respect to the property that the $(n+1)$-dimensional Cayley-Menger determinant vanishes for all $k$-tuples of points for $k \geq n+2$ and does not vanish for all $k$-tuples of points "in general position" for $k < n+2$. - - -Question 1: Is it correct, actually, that $E^n$ is uniquely - determined by 2 and 3? -Question 2: What are still other ways of - characterizing $E^n$ "different in spirit"? - -REPLY [9 votes]: Herbert Busemann provided many metric characterizations of the elementary spaces in his 1955 book The Geometry of Geodesics. To characterize $E^n$, we can then further restrict to cases with zero curvature, or to non-compact cases with non-negative curvature, or probably in several other ways. -For a first pass at these characterizations, we can think of them as being about complete connected Riemannian manifolds; the results actually hold for the more general G-spaces instead. In any case, here are five of the characterizations in that book. -Theorem 15.4, via Desargues's theorem (p. 87): In a Riemannian G-space, if the geodesic through two points is unique, and any three points lie in a plane, then the space is either Euclidean, hyperbolic, or spherical. -Theorem 47.4, via bisectors (p 331): If each bisector $B(a,a')$ (i.e. the locus $xa=xa'$) of a G-space contains with any two points $x,y$ at least one geodesic segment between them, then the space is Euclidean, hyperbolic, or spherical of dimension greater than 1. -Theorem 48.8, via motions of three points (p. 337): If a G-space possesses for any four points $a,a',b,c$ with $ab=a'b$ and $ac=a'c$ a motion which leaves $b$ and $c$ fixed and carries $a$ in to $a'$, then the space is Euclidean, hyperbolic, or spherical. -Theorem 49.7, via reflections (p. 347): A G-space which can be reflected in each lineal element is Euclidean, hyperbolic, spherical or elliptic. -Theorem 55.3, via doubly transitive motions (p. 395): A G-space whose dimension is finite and odd (or two) and which possesses a pairwise transitive group of motions is Euclidean, hyperbolic, spherical or elliptic. -There are still more characterizations in section 24, via the parallel postulate, and in chapter VI.<|endoftext|> -TITLE: trace of the atiyah class equals chern class -QUESTION [7 upvotes]: In several textbooks ("The Geometry of Moduli Spaces of Sheaves" by Huybrechts and Lehn, "Calcul differentiel et classes caracteristiques..." by Angeniol and Lejeune-Jalabert) it is mentioned that the trace of the p-th atiyah class equals the p-th chern class or the p-th component of the chern character. I could not find a reference where this statement is proven. Thanks for any help. - -REPLY [6 votes]: I might be wrong but it seems to me that the $p$-th Atiyah class does not have any reason to agree with the usual $p$-th Chern class unless the manifold under consideration is Kahler. -Namely, if $X$ is not Kahler then for a holomorphic vector bundle $E\to X$, $c_p(E)\in H^{2p}(X)$ and $at_p(E)\in H^{p}(X,\Omega^p_X)$ live in different spaces. -The point is that $c_p(E)$ can be defined as the class of $tr(R^p)$, where $R$ is the curvature of an hermitian connection on $E$, while $at_p(E)$ can be defined as the class of $tr(R_{1,1}^p)$, where $R_{1,1}$ is the $(1,1)$-part of the curvature of a $(1,0)$-connection on $E$. -The point is that if $X$ is Kahler then there exists an Hermitian $(1,0)$-connection with curvature being of type $(1,1)$. The relation between the Atiyah classes and the Chern classes can be made through the Hodge-to-de Rham spectral sequence. - -So, I think that the Chern classes you are talking about are not the usual (i.e. topological) ones, but the Chern classes in Hodge cohomology. Then they coincide with the Atiyah classes almost by definition (by the way, there is a very nice paper of Grothendieck on Chern classes in Hodge cohomology).<|endoftext|> -TITLE: Holomorphic functions in almost-complex geometry -QUESTION [6 upvotes]: Maximum principle implies that every holomorphic function on a compact complex manifold is constant. Is this still true if the manifold is only almost complex? - -REPLY [9 votes]: This is still true, although as Francesco says in his comment above, it is trivially so in general : in complex dimension 2 and more, a generic almost complex structure has only constant holomorphic functions, even locally. -Proof : if $f:(V,J)\to\mathbb{C}$ is such a function, namely $df\circ J=i\\,df$, then (obviously) $d(df\circ J)=0$. -But the second order operator $f\mapsto (d(df\circ J))^{1,1}$ from functions to $(1,1)$-forms has the "same" principal symbol at each point as in the integrable case (the "plurisubharmonic Hessian", so to speak, perhaps up to some $-2i$ factor). -In particular you can compose it with contraction by a positive smooth $(1,1)$ form (given by any hermitian metric) to obtain a "Laplace operator", which satisfies the maximum principle. EDIT (after comment by OP): it is important to observe that the operator vanishes on constants to derive the maximum principle -- locally it writes $\sum g_{jk}(x) \partial_j\partial_k +\sum b_i(x) \partial_i$, with $g_{jk}$ symmetric positive definite.<|endoftext|> -TITLE: Varieties where every non-zero effective divisor is ample -QUESTION [20 upvotes]: The following question seems very intuitive, but I haven't been able to find any proof (or counterexample). - -Let $X$ be a non-singular projective variety of $\dim X\ge 2$ and let $NS^1(X)$ be its Neron-Severi group. If every non-zero effective divisor on $X$ is ample, does it follow that $X$ has Picard number one, i.e., $\rho=$ rank $NS^1(X)=1$? - -Motivation: -1) In the case of Fano varieties the result is true (the proof is an easy application of Riemann-Roch). In fact this result was a key ingredient in Mori's proof of Hartshorne's conjecture for projective 3-space (i.e., any 3-fold with ample tangent bundle is isomorphic to $\mathbb{P}^3$). See Mori's original article for the details. -2) In this Mathoverflow question Charles Staats asks for a surface with the property that any two curves on the surface have nontrivial intersection. In his comment, BCnrd considered a K3 surface with Picard number one, which satisfies the condition precisely because any effective divisor is ample. A natural question is whether any such surface has Picard number one. -I am mostly interested in the case where $X$ is a complex projective variety. In the case the result does not hold, I'd also be interested in seeing a concrete counterexample and other examples of varieties where the result holds. - -REPLY [7 votes]: It is a general fact that on any simple abelian variety, an effective divisor is ample. The following result underlies the usual algebraic proof of the projectivity of abelian varieties (defined initially only as complete group varieties). It is therefore rather standard; the first reference which comes to mind is Lemma 8.5.6 on page 253 in the abelian varieties chapter of the book Heights in diophantine geometry by Bombieri and Gubler, from which I quote literally. -Let A be an abelian variety and $D$ an effective divisor such that the subgroup -$$ -Z_D : \hspace{3cm} \{ a \in A \mid a + D = D \} -$$ -is finite. Then $D$ is ample on $A$. -If the abelian variety $A$ is simple, and $D$ is non-zero, then the $Z_D$ is a fortiori finite, since it is a proper algebraic subgroup of $A$; and it follows from the quoted Lemma 8.5.6 that $D$ is ample. -An application. A surjective morphism $f: A \to X$ from a simple abelian variety onto a positive-dimensional projective variety $X$ is finite. -Proof. Choose $H \subset X$ an ample divisor. The divisor $f^*H$ is effective on $A$, hence it is ample. This is equivalent to $f$ being finite.<|endoftext|> -TITLE: Fibered/cofibered higher categories, relative model structures, slicing, and (∞,2)-category theory -QUESTION [8 upvotes]: Jacob Lurie defined a model structure on the category of marked simplicial sets sliced over a fixed simplicial set $S$ called the cartesian model structure. (For a definition, see here or HTT Ch.3.1). How does this model structure behave with respect to slicing over objects? That is, is the natural model structure on the slice over a simplicial set $S$ the same as the cartesian model structure on the slice over $S$? -Do any of the other flexible models of higher categories (specifically complete segal spaces and segal categories) have relative forms? If so, do these model categories give the right results under slicing? Whether they do or not, are there appropriate analogues of the straightening and unstraightening constructions? If so, do they give a powerful enough theory of fibered and cofibered categories (which are extremely important in ordinary 2-category theory (these give the "right" slice bicategories)). - -REPLY [15 votes]: It seems that the first question only makes sense for marked simplicial sets $X$ over $S$ -where every edge of $X$ is marked (otherwise, the slice category is not equivalent to marked simplicial sets over $X$). Under this assumption, the answer is yes at least if $X$ is fibrant (so that the underlying map of simplicial sets $X \rightarrow S$ is a right fibration). -If you take any model category ${\mathbf A}$ for higher category theory and take the slice -category ${\mathbf A}_{/X}$ for some fibrant object $X$, it will be a model for higher categories $Y$ with an arbitrary functor $p: Y \rightarrow X$. If you want to enforce the requirement that $p$ should be a Grothendieck fibration, you need to modify the definitions somehow. In the quasicategory model, this is what the markings are for.<|endoftext|> -TITLE: Utility of virtual knot theory? -QUESTION [23 upvotes]: Virtual knot theory is an interesting generalization of knot theory in which ``virtual" crossings are allowed. See Kauffman's Virtual Knot Theory for an introduction. Greg Kuperberg gave a nice topological interpretation of virtual knots in this paper. One reason to be interested in virtual knots and links is that many knot and link invariants generalize to the virtual setting. For example, my naive understanding is that virtual knots are a more natural domain for Vassiliev invariants than knots are. My question is whether anyone knows of examples that demonstrate the utility of virtual knot theory? For example, are there any interesting theorems outside of virtual knot theory that can be most easily proven using virtual knot theory? The papers I have seen seem to pursue VKT for its own sake, but my sense is that such a natural area must be of much wider use. -Added: I just ran across a paper by Rourke that explains a geometric interpretation of "welded links" which are like virtual links but an additional move is allowed. This is a beautiful little paper which explains how welded knot theory corresponds to a theory of certain embedded tori in $\mathbb R^4$. It's really amazing how the Reidemeister moves, both virtual and classical, correspond to isotopies of what Rourke calls toric links. This is a part of Bar-Natan's program mentioned by Theo Johnson-Freyd and Daniel Moskovich below. Dror calls the toric links "flying rings". - -REPLY [2 votes]: This paper by Chrisman and Manturov is as close as possible to an answer to my original question. From their introduction: - -By classical knot theory we mean the study of knots and links in the $3$-sphere. By virtual knot theory we mean the study knots and and links in thickened surfaces $\Sigma\times I$ modulo stabilization, where $\Sigma$ is compact orientable surface (not necessarily closed), and I is the closed unit interval. The goal of the present paper is to study classical knots using the methods of virtual knot theory. To do this, we introduce the concept of a virtual cover of a classical knot.<|endoftext|> -TITLE: Occurrence of semi-spin groups -QUESTION [9 upvotes]: In the classification of simple Lie algebras one has the familiar picture of 4 families, $A_n$, $B_n$, $C_n$ and $D_n$, and 5 exceptional groups, $F_4,$ $G_2,$ $E_6$, $E_7$ and $E_8$. The $D_n$ family has the unique feature that it contains, among all the corresponding Lie groups, groups whose center is non-cyclic: for $Spin(4n+2)$ the center is $\mathbb{Z}_4$, but one has $$Z(Spin(4n))=\mathbb{Z}_2\times \mathbb{Z}_2.$$ One can take the quotient of $Spin(4n)$ by any of the 3 subgroups of its center isomorphic to $\mathbb{Z}_2$ - one of these quotients is $SO(4n)$, the other are two are isomorphic to each other and are sometimes referred to as the half-spin or semi-spin groups, denoted by $SSpin(4n)$ (the notation may not be completely standard). They are a little bit the forgotten Lie groups - simple groups that are neither exceptional nor a quotient or covering of a classical group. -My question is: other than occuring in the classification, are there any places where these half-spin groups show up (naturally so to speak)? - -REPLY [7 votes]: $SSpin(32)$ appears in two superficially unrelated places that I know of. - -There are two heterotic string theories: the one based on the root lattice of $E_8 \oplus E_8$ and the one based on the weight lattice of $SSpin(32)$. This is often (but incorrectly!) called the $SO(32)$ heterotic string theory. -Milnor gave the first example of isospectral manifolds which are not isometric: the flat tori obtained by quotienting $\mathbb{R}^{16}$ by the above two lattices. - -Another factoid is the following. Although they are isomorphic, it is convenient for this to distinguish between, say, $SSpin(4n)$ and $S'Spin(4n)$, which are the quotients of $Spin(4n)$ by different $\mathbb{Z}_2$ subgroups of the centre, which are not $SO(4n)$. Then $SSpin(8n)$ and $S'Spin(8n)$ are self-dual (under Langlands duality), whereas $SSpin(8n+4)$ is the Langlands (a.k.a. magnetic) dual of $S'Spin(8n+4)$.<|endoftext|> -TITLE: Relationship between determinants. -QUESTION [13 upvotes]: Given an orthogonal matrix $O$ with dimensions $4n \times 4n$ and $\det O = -1$, how to prove that -$\det[O_{11} - O_{22} + i (O_{12} + O_{21})] = 0$? -Here $O$ is a block matrix $[[O_{11}, O_{12}], [O_{21}, O_{22}]]$, and all blocks have equal size. -An equivalent statement is the following: if $\det O = -1$, then $\det(\Omega - O \Omega O^T)=0$, with $\Omega = [[0, I],[-I, 0]]$. The reason why I believe this is correct is due to the physical meaning of it (it's about presence of bound states in a certain system), and numerical checks. I am not sure why odd $n$ doesn't work, but it makes sense for physics reasons. -Example of a matrix for which it works is below: -$$\left(\begin{array}{cccc} -0.44090815& -0.71206204& -0.44576549& 0.31600755 \\\ -0.12767731& 0.35584235& 0.19884606& 0.90417641 \\\ -0.88038152& 0.229104 & 0.30519084& -0.28159952 \\\ -0.11927654& 0.56022784& -0.81768693& -0.05749749 -\end{array}\right)$$ -Edit(Will Jagy):In particular, this fails for $(4n - 2) \times (4n - 2),$ as in the $ 2 \times 2$ -$$ O \; \; = \; \; -\left( \begin{array}{cc} - \cos t & \sin t\\\ - \sin t & - \cos t -\end{array} - \right) , $$ -with the relevant $ 1 \times 1$ matrix of complex numbers -$$ \left( \cos t - ( - \cos t) + i ( \sin t + \sin t )\right) = \left( 2 e^{it} \right) . $$ -Edit(Anton Akhmerov): -Yet another statement which could solve the problem is the following: -Prove that any $4n\times 4n$ orthogonal matrix $O$ can be brought to a block-diagonal form by a transformation $O\rightarrow S_1 O S_2$ with $S_1$ and $S_2$ symplectic matrices. If this was correct, the rest follows immediately. - -REPLY [13 votes]: I denote your matrix $\Omega$ by $W$ for the sake of brevity. Note that $W$ is both skew-symmetric and orthogonal, i. e. it satisfies $W=-W^T$ and $W^2=-I$. (And this is all I am going to use about $W$.) -The only thing I am going to use about the matrix $O$ is that $O^TO=I$. The assumption that $O$ is a real matrix will not be needed (it could be from any field of characteristic $\neq 2$). -We are going to use notion of the Pfaffian of a skew-symmetric matrix. -Lemma 1. Let $R$ be a commutative ring with $1$. Let $A\in R^{2n\times 2n}$ be a skew-symmetric matrix, and $B\in R^{2n\times 2n}$ be an arbitrary matrix. Then, the matrix $B^TAB$ is skew-symmetric as well, and $\mathrm{Pf}\left(B^TAB\right)=\det B\cdot\mathrm{Pf} A$. -Proof of Lemma 1. Clearly, the matrix $B^TAB$ is skew-symmetric. It remains to prove that $\mathrm{Pf}\left(B^TAB\right)=\det B\cdot\mathrm{Pf} A$. Let us WLOG assume that $R=\mathbb Q$ (this is WLOG indeed because we are proving a polynomial identity). Since the square of the Pfaffian of a skew-symmetric matrix is the determinant of this matrix, we have -$\left(\mathrm{Pf}\left(B^TAB\right)\right)^2=\det\left(B^TAB\right)=\underbrace{\det B^T}_{=\det B}\cdot\underbrace{\det A}_{=\left(\mathrm{Pf}A\right)^2}\cdot \det B = \left(\det B\cdot\mathrm{Pf} A\right)^2$.` -Thus, FOR EVERY $A$ and $B$, we have either $\mathrm{Pf}\left(B^TAB\right)=\det B\cdot\mathrm{Pf} A$ OR $\mathrm{Pf}\left(B^TAB\right)=-\det B\cdot\mathrm{Pf} A$. By a Zariski-topological argument, we can interchange the words "for every" with the words "or" here, so we obtain: (FOR EVERY $A$ and $B$, we have $\mathrm{Pf}\left(B^TAB\right)=\det B\cdot\mathrm{Pf} A$) OR (FOR EVERY $A$ and $B$, we have $\mathrm{Pf}\left(B^TAB\right)=-\det B\cdot\mathrm{Pf} A$). Since (FOR EVERY $A$ and $B$, we have $\mathrm{Pf}\left(B^TAB\right)=-\det B\cdot\mathrm{Pf} A$) is wrong (take $A=W$ and $B=\mathrm{id}$), we thus must have (FOR EVERY $A$ and $B$, we have $\mathrm{Pf}\left(B^TAB\right)=\det B\cdot\mathrm{Pf} A$), and Lemma 1 is proven. -Lemma 2. Let $R$ be a commutative ring with $1$. Let $B\in R^{2n\times 2n}$ and $C\in R^{2n\times 2n}$ be arbitrary matrices. Then, $\det\left(W-BWC\right)=\det\left(W-CWB\right)$. -Proof of Lemma 2. We have $W-BWC=W\cdot\left(I+WBWC\right)$ (because $-I=W^2$), and thus -$\det\left(W-BWC\right)=\det\left(W\cdot\left(I+WBWC\right)\right)=\det W\cdot\det\left(I+WBWC\right)$ -and similarly -$\det\left(W-CWB\right)=\det W\cdot\det\left(I+WCWB\right)$. -Thus, in order to prove Lemma 2, it remains to show that $\det\left(I+WBWC\right)=\det\left(I+WCWB\right)$. This follows from the general fact that if $U$ and $V$ are two square matrices of the same size, then $\det\left(I+UV\right)=\det\left(I+VU\right)$ (for a proof of this fact, apply Corollary 2 in MathLinks post #1491761 to $X=-1$). Thus, Lemma 2 is proven. -Lemma 3. Let $R$ be a commutative ring with $1$. Let $B\in R^{2n\times 2n}$ be an arbitrary matrix. Then, $\mathrm{Pf}\left(W-BWB^T\right)=\mathrm{Pf}\left(W-B^TWB\right)$. -Proof of Lemma 3. Let us WLOG assume that $R=\mathbb Q$ (this is WLOG indeed because we are proving a polynomial identity). Applying Lemma 2 to $C=B^T$, we obtain $\det\left(W-BWB^T\right)=\det\left(W-B^TWB\right)$. Since the determinant of a skew-symmetric matrix is the square of its Pfaffian, this rewrites as $\left(\mathrm{Pf}\left(W-BWB^T\right)\right)^2=\left(\mathrm{Pf}\left(W-B^TWB\right)\right)^2$. Again, a Zariski density argument like in the proof of Lemma 1 shows us that we must have $\mathrm{Pf}\left(W-BWB^T\right)=\mathrm{Pf}\left(W-B^TWB\right)$ (because we cannot have $\mathrm{Pf}\left(W-BWB^T\right)=-\mathrm{Pf}\left(W-B^TWB\right)$ for all $B$, since this would fail for $B=0$). This proves Lemma 3. -Now let us solve our problem: -We have $\mathrm{Pf}U=\mathrm{Pf}\left( -U\right) $ for every skew-symmetric $4n\times 4n$ matrix $U$ (because $\mathrm{Pf}$ is a homogeneous polynomial of degree $2n$). Now, -$\mathrm{Pf}\left( W-OWO^{T}\right) =\mathrm{Pf}\left( W-O^{T}WO\right) $ (after Lemma 3, applied to $B=O$) -$=\mathrm{Pf}\left( -\left( W-O^{T}WO\right) \right) $ (since $\mathrm{Pf}U=\mathrm{Pf}\left( -U\right) $ for every skew-symmetric $4n\times4n$ matrix $U$) -$=\mathrm{Pf}\left( O^{T}WO-W\right) $ -$ =\mathrm{Pf}\left( O^{T} WO-O^{T}OWO^{T}O\right) $ (since $O^{T}O=I$ yields $W=O^{T}OWO^{T}O$) -$=\mathrm{Pf}\left( O^{T}\left( W-OWO^{T}\right) O\right) $ -$=\underbrace{\det O}_{=-1}\cdot\mathrm{Pf}\left( W-OWO^{T}\right) $ (by Lemma 1, applied to $B=O$ and $A=W-OW^{T}O^{T}$) -$=-\mathrm{Pf}\left( W-OWO^{T}\right) $, -so that $2\mathrm{Pf}\left( W-OWO^{T}\right) =0$ and thus $\mathrm{Pf}\left( W-OWO^{T}\right) = 0$, so that $\det\left( W-OWO^{T}\right) =\left(\mathrm{Pf}\left( W-OWO^{T}\right) \right)^2=0^2=0$, qed.<|endoftext|> -TITLE: Is Riemannian distance function equivalent to Euclidean one? -QUESTION [5 upvotes]: Consider the Riemannian manifold $\mathbb{R}^n$ and a smooth Riemannian metric $G:\mathbb{R}^n\rightarrow\mathbb{R}^{n\times{n}}$. We know that if $w_1I_n\leq{G}(x)\leq{w_2}I_n$ for some $w_1,w_2\in\mathbb{R}^+$, any $x\in\mathbb{R}^n$, and identity matrix $I_n$, then we have $\sqrt{w_1}\Vert{x_1}-x_2\Vert_2\leq\mathbf{d}_G(x_1,x_2)\leq\sqrt{w_2}\Vert{x_1}-x_2\Vert_2$, where $\mathbf{d}_G$ is the Riemannian distance function with respect to the metric $G$. Now, I am interested in the following equivalency between $\mathbf{d}_G$ and the usual Euclidean distance: $\alpha_1(\Vert{x_1}-x_2\Vert_2)\leq\mathbf{d}_G(x_1,x_2)\leq\alpha_2(\Vert{x_1}-x_2\Vert_2)$, where $\alpha_i:\mathbb{R}^+_0\rightarrow\mathbb{R}^+_0$ are continuous, strictly increasing functions, $\alpha_i(0)=0$ and $\alpha_i(r)\rightarrow\infty$ as $r\rightarrow\infty$, for $i=1,2$. I tried to find a characterization on the metric $G$ to reach to this version of equivalency, but I could not find anything. I am just wondering if there is any nice result for this kind of equivalency. - -REPLY [4 votes]: This equivalence between distances (even on metric space more in general) is exactly the uniform equivalence: the distance $d$ and $d'$ (in your case the Euclidean distance) induce the same uniform structure. In other words, the identity map on $\mathbb{R}^n$ is uniformly continuous from $d$ to $d'$ and from $d'$ to $d$, with modulus of continuity respectively $\alpha_1^{-1}$ and $\alpha_2.$ The first equivalence you mentioned is indeed a special case, with linear moduli of continuity: it's the "Lipschitz equivalence", meaning that the identity map is bi-lipschitz; you may define analogously an Hoelder equivalence etc). -Also, it is not difficult to see that in your case $d_G$ may be uniformly, but not Lipschitz equivalent to the Euclidean distance (that is, the latter notion is really weaker than the former). It is possible to give simple sufficient conditions in terms of point-wise inequalities on $G(x)$, that ensure the uniform equivalence to the Euclidean distance. However, I don't see an easy characterization of this, the reason being that the distance $d_G$ is defined in terms of minimization on paths, and there may be uniform equivalence even if $G(x)$ has very high bumps at some points (say, a sequence with $G(x_k)\geq kI_n$).<|endoftext|> -TITLE: Nonnegative to Positive Curvature. -QUESTION [22 upvotes]: This questions asks for your intuition and insight as I'm surprised by how little is known about the difference between nonnegative and positive curvature. I don't want to be completely vague, so I could ask: What are the difficulties and currently blocked paths to solving the Hopf Conjecture? (Does $S^2\times S^2$ support a metric of positive curvature?). But in general, I would like to know what others might know on why it's difficult to determine if a given closed simply-connected space of nonnegative curvature can also admit positive curvature. As far as I know, there are no obstructions, how come? The amount of examples of nonnegative curvature compared to that of examples of positive curvature seem to suggest there should be something distinguishing the two. - -REPLY [12 votes]: Yau asked in 1982 if there is any compact simply connected manifold with nonnegative curvature for which one can prove that it does not admit a metric of positive curvature. This question opens his list of unsolved problems in geometry (see "Seminar on Differential Geometry", p. 670.) -Let me quote from "A Panoramic View of Riemannian Geometry" by Berger (Springer 2003, p. 579): - -It is not surprising that many people tried to address Yau’s remark, starting - with the Hopf conjecture on $S^2 × S^2$, by trying to deform such a metric - with $K ≥ 0$ into one with $K >0$. This means considering some one parameter - family $g(t)$ of metrics and computing the various derivatives at $t = 0$ of the - sectional curvature. Technically it is very easy to compute such a derivative - for a given tangent plane, but what is difficult is to find a variation for which - all the derivatives would be positive. Today this approach still does not work. - -A major difficulty is that it is not clear how to find the critical set of the sectional curvature (as a function on the set of tangent planes). -The earlier short survey by Bourguignon contains a discussion of the reasons why some seemingly natural approaches fail.<|endoftext|> -TITLE: Narratives in Modular Curves -QUESTION [25 upvotes]: I've tried several times to read up on modular curves, and I've despaired every time. It seems that there are several competing narratives, that all get enfused and conspire to befuddle me. There are modular forms, automorphic forms, Hecke theory, L-functions, Galois representations, hyperbolic space, primitive forms, Fuchsian groups, the Langlands program and so forth. -It is somewhat of an introductory question, but I would be very grateful if one of you could put things in some sort of order, and separate the narratives. (What generalizes what, what is parrallel to what, and what implies what) - -REPLY [5 votes]: I recommend reading - -Modular points, modular curves, modular surfaces and modular forms - -by Don Zagier.<|endoftext|> -TITLE: Deform a compact Kähler manifold to a non Kähler one -QUESTION [9 upvotes]: Could you give me an example of a compact Kähler manifold which analytically deforms to a non Kähler one? -For example, there is no hope to find a complex structure on a Hopf manifold in order to make it Kähler because of topological obstructions (the second Betti number is zero). -For instance, I think that the Iwasawa manifold should not have topological obstructions. -Of course, the algebraic counterpart of my question has an affirmative answer: complex tori of dimension greater than one give examples of manifolds that can be analytically deformed from an algebraic one to a non algebraic one (but still Kähler). -Thanks in advance! - -REPLY [7 votes]: (Just so this question has an answer. All manifolds are compact.) -In dimension one every deformation of Kähler manifolds is Kähler because every Riemann surface is Kähler. -In dimension two the same is true but for less trivial reasons. A two-dimensional complex manifold is Kähler if and only if its first Betti number is even, which is purely a topological condition. The result then follows from the fact that the fibres of a deformation are diffeomorphic. -In dimension three, the result is no longer true; that is, there is a deformation of Kähler manifolds such that the central fibre is not Kähler. As Gunnar Magnusson pointed out in the comments, Hironaka gave an example of such a deformation in his paper An Example of a Non-Kählerian Complex-Analytic Deformation of Kählerian Complex Structures. In fact, the construction used gives rise to a whole host of interesting phenomena in dimension three including the existence of a Moishezon manifold which is not projective. See the associated page on Wikipedia for more details.<|endoftext|> -TITLE: A-infinity tensor categories -QUESTION [11 upvotes]: My question is rather simple: -What is the correct notion of a monoidal A-infinity category C? -Or is there any reference where such a notion is explained? - -REPLY [5 votes]: One way to define a monoidal $A_\infty$-category (which is what Chris Brav suggests in the comments - and of course presumably to Kevin and Scott's) is as an associative algebra object in the monoidal $(\infty,1)$-category of $A_\infty$ categories (which is the same as that of dg categories). In other words, to define a monoidal structure we first need to ask "what do $A_\infty$-categories form?" and then just take the notion of associative multiplication in that world. Now of course to say that sentence I need to know there's a theory of monoidal $(\infty,1)$ -categories, for which we have Lurie's DAG II. But the advantage is that with this definition we can immediately perform all the operations of ordinary algebra as if we were just dealing with an associative ring -- there's automatically a theory of module categories eg, inner homs, bimodules and tensors, we can even talk about Hochschild homology and each module category defines a character object in this Hochschild homology, and of course many many other things. So I strongly believe it's worth the investment. -As to how to define this monoidal $(\infty,1)$-category, there are many ways, but I'll take the cheapest (again modulo the worthwhile investment which is DAG): I would define them as modules for dg-Vect, the monoidal $\infty$-category of chain complexes of $k$-vector spaces. This category itself is given as modules for the commutative ring k, hence its monoidal structure. (Again modulo DAG 2 and now 3 for commutative rings and their tensor products of modules). -[Let me attempt to preempt the obvious complaint with all this: yes it's not very explicit and close to the ground, i.e. to models, and for calculations you very well may want a more concrete objectwise definition in terms of higher homotopies. This approach is however a very powerful one for proving formal properties -- ie if the questions you're asking have the feeling "this would be easy/formal/follow for abstract general reasons if I were living in the toy model world of just plain monoidal categories, or even associative algebras, but seem hard in this homotopical world" then this is the approach for you, otherwise it's not.]<|endoftext|> -TITLE: comparison of de Rham cohomology and etale cohomology -QUESTION [18 upvotes]: I have a basic question concerning comparison of different cohomology theories. Let $X$ be a projective smooth (or just proper smooth) variety over a separably closed field $k$ of characteristic $p,$ which is not liftable to characteristic zero. -Is there any relation between the de Rham cohomology $H^n(X,\Omega^{\bullet}_X)$ and the $\ell$-adic cohomology? For example, do they have the same dimension (over $k$ and $Q_l$ resp.)? - -REPLY [27 votes]: Often, yes. What always has the same dimension as $H^n_{et}(X,Q_l)$ is the rational crystalline cohomology $H^n_{cr}(X)\otimes K$ with coefficients in the fraction field $K$ of the Witt vectors $W$ of $k$. $H^n_{cr}(X)$ itself will have coefficients in $W$, and of course, have rank equal to the dimension of $H^n_{cr}(X)\otimes K$. But it might have torsion in general. On the other hand, there is an exact sequence -$$0\rightarrow H^n_{cr}(X)\otimes_W k\rightarrow H^n(X,\Omega_X^{\cdot})\rightarrow H^{n+1}_{cr}(X)[p]\rightarrow 0$$ -as in the universal coefficient theorem. This is because crystalline cohomology can be taken with any of the torsion coefficients $W/p^n$, and when you take it with coefficients in $W/p=k$, you get exactly De Rham cohomology. (One of the most important things to learn at the beginning about crystalline cohomology with $W/p^n$ coefficients is that it can be computed using the divided power De Rham complex associated to a smooth embedding over $W/p^n$, which reduces to the De Rham complex of $X$ itself when the coefficients are $W/p$.) -So you will get the same dimensions you want if enough of crystalline cohomology is torsion-free. All this is explained in introductory books, such as the one by Berthelot and Ogus, except the comparison with \'etale cohomology. That is perhaps explained in a paper by Katz and Messing from the 70's. - -REPLY [7 votes]: I believe the answer is no, that these two spaces need not have the same vector space dimension. Grothendieck here cites an example of Serre in a footnote on the last page; unfortunately, I don't have access to Serre's original paper at the moment. -http://www.numdam.org/item?id=PMIHES_1966_29_95_0<|endoftext|> -TITLE: Proving interesting theorems about S_n using its character table. -QUESTION [6 upvotes]: Hi, -i wonder if there are interesting proofs about $S_n$ (group theoretic or not) using its character table. -Using the Murnaghan-Nakayama rule you can for example prove that for $n>4$ $A_n$ is the only normal subgroup of $S_n$ because there are no nonlinear characters $x$ and $g$(not 1) in $S_n$ -with $ x(g)=x(1)$, since $x(1)>x(g) $ . -Do you know any other nontrivial theorems about $S_n$ with a proof using its charactertable ? - -REPLY [4 votes]: In an answer to an earlier question, I showed how to prove that the square root counting function $r_2: S_n\rightarrow \mathbb{N},\;g\mapsto \#\{h\in S_n|h^2=g\}$ assumes its maximum at the identity, using the representation theory of $S_n$. Admittedly, you need to know slightly more than the character table. You need to be able to compute the Frobenius-Schur indicators of the characters, so you need to know how the conjugacy classes multiply. Alternatively, you just need to know that all representations are defined over $\mathbb{R}$, which you prove along to way to computing the character table anyway. In a comment to my answer, Richard Stanley remarks that, also using the representation theory of $S_n$, you can generalise this to the $k$-th root counting function for any positive integer $k$. In an answer to the same question, Alon Amit remarks on possible generalisations to solving other polynomial equations in the elements of $S_n$. - -REPLY [3 votes]: The following is not strictly speaking something that can be read off from the character table. However, it is an elementary combinatorial identity about partitions which one can deduce from understanding the character theory of symmetric groups well enough, and looking at the character table does play a central role: -For $\lambda \vdash n$ a partition of $n$ (i.e., $n = 1 \lambda_1 + 2 \lambda_2 + \cdots + n \lambda_n$) define -$$ A(\lambda) = \prod_{i=1}^{n} n^{\lambda_n}, \qquad B(\lambda) = \prod_{i=1}^{n} (\lambda_n)! $$ -Claim: $$\prod_{\lambda \vdash n} A(\lambda) = \prod_{\lambda \vdash n} B(\lambda) $$ -The character-theoretic proof proceeds as follows: - -For an element in the conjugacy class of $S_n$ indexed by the partition $\lambda$, it's centralizer has cardinality $A(\lambda) B(\lambda)$, i.e., the number of elements in the conjugacy class is $$\frac{n!}{A(\lambda) B(\lambda)}$$ -Take the character matrix $M$. The orthogonality relations tells us that a suitable rescaling of the character matrix is orthogonal, so has $\det = \pm 1$. From this, together with 1 to find the scaling factors for the columns, we obtain -$$ (\det M)^2 = \prod_{\lambda} A(\lambda) B(\lambda) $$ -$M$ relates two bases for the spaces of class-functions: The characters of irreps (indexed in the Schur ordering by partitions), and the delta functions on conjugacy classes (indexed obviously by partitions). For symmetric groups, there is a third nice basis: For $\lambda \vdash n$, let $$ S_\lambda = \prod_{i=1}^{n} S_i^{\lambda_i} $$ and consider the characters of the induced reps $Ind_{S_\lambda}^{S_n} \mathbb{C}$. -Consider the change of basis matrix relating characters of induced reps and the delta functions on conjugacy classes: Easy character theory shows that it is triangular with diagonal entries equal to $B(\lambda)$. -Consider the change of basis matrix relating characters of induces reps and characters of irreps: Knowing how character theory for symmetric groups works over $\mathbb{Z}$ (i.e., that both span integrally), we can show that it has determinant $\pm 1$. More precisely, knowing the character theory well enough we can show that the change of basis matrix between them is upper triangular with ones on the diagonal. - -Putting together 2, 4, 5 we obtain -$$ \det(M)^2 = \prod_{\lambda \vdash n} A(\lambda) B(\lambda) = \left(\prod_{\lambda \vdash n} B(\lambda)\right)^2 $$ and so the claimed identity.<|endoftext|> -TITLE: Is P=NP relevant to finding proofs of everyday mathematical propositions? -QUESTION [33 upvotes]: Disclaimer: I don't know a whole lot about complexity theory beyond, say, a good undergrad class. -With increasing frequency I seem to be encountering claims by complexity theorists that, in the unlikely event that P=NP were proved and an algorithm with reasonable constants found, mathematicians wouldn't bother trying to prove things anymore beause we could just use our P-time algorithm to search for proofs. Usually this is part of an argument for why all mathematicians and logicians should care a lot about P=?=NP. -I think most of these claims are exaggerations of the first full paragraph on page 8 of the Cook's problem description for the Clay Institute (which itself is stated in a completely reasonable and unexaggerated manner). -However, it's quite clear from the Clay Institute description that P=NP is relevant only to classes of problems, parameterized by some integer $n$, for which we have already proved all three of the following: - -the question is not independent of our chosen axioms ($T\vdash \phi\vee T\vdash \neg\phi$) -any proof of the proposition must have size at most polynomial in $n$ -any proof of the negation of the proposition must have size at most polynomial in $n$ - -This way we know there's a proof of either the proposition or its negation, and the search problem for the one that does exist falls inside NP, so we can dovetail the two searches and stop when one of them succeeds. -This puzzles me. Most of the propositions mathematicians care about don't come in integer-parameterized classes, let alone classes with known proof-size bounds. Usually they come in classes of size 1 with no knowledge of proof-size. Is there some trick for turning the sorts of results mathematicians care about into these integer-parameterized-polynomially-bounded classes? -Example: how would you do this for the question of whether or not CH is independent of ZFC? -Cook and Reckhow's JSL article The Relative Efficiency of Propositional Proof Systems (which seems to be the starting point for the literature) actually mentions that if you take the problem class to consist of all propositions in some proof system (such as first-order predicate calculus), take the length of the proposition as the parameter, and take the question to be "is it entailed by the axioms", then at the time the paper was published (1979) no real-world proof system was known to have the desired property, and a few were known not to have the desired property. -I suppose I am being slightly lazy here, since the study of which problems have this property is a whole subfield with plenty of literature I could read, but really I'm only interested in whether or not that subfield's positive-results-to-date justify the claims I've been hearing lately. A reference to a paper containing the "trick" above would be fine as an answer. - -REPLY [4 votes]: Others have already given good answers, but I'd like to point out that the constraint of "reasonable constants" is potentially much more stringent than most people think. For example, in the real world, space tends to be more expensive than time. So suppose that there is some proof of the Riemann hypothesis that is just barely small enough that we can just barely implement a computer program that, when run using all the world's computer resources, can just barely step through the entire proof and verify it within an acceptable human time-frame. Chances are, we would not have enough resources to write down such a proof explicitly; the computers would probably be constantly reusing space and erasing stuff that had been used in checking earlier stages of the proof but that would no longer be needed. The plan that Andreas Blass outlined, of writing down explicitly the length of the proof in unary and running our magic algorithm on this input, would not be feasible in this scenario. -Anyway, the point is that in today's world, one should not think of a proof as something that takes just a few million characters to write down explicitly. Although it is an important observation that it would be revolutionary to discover an algorithm for searching for proofs that is as efficient as verifying them, we should recognize that the claim that a proof of P = NP would achieve this is a slight exaggeration.<|endoftext|> -TITLE: (un)decidability in matrix groups -QUESTION [14 upvotes]: Given a collection of matrices $S=\{M_1, \dots, M_k\}$ in (say) $SL(n, Z), \ n>2$ does $S$ generate $SL(n, Z)?$ -Similar are questions are undecidable for $n\geq 4$ (eg, given a set $S$ as above, is a given matrix in the subgroup generated by $S$) but I cannot find any reference for the above. -For $n=2$ all questions of this sort are reasonably efficiently decidable. -EDIT (in response to @Misha's interesting comments). -It is not clear that Mihailova tells you that the generation problem is undecidable. I believe that it IS a result of Baumslag, Miller, Short that this is undecidable for some word-hyperbolic groups (see MR1246477 (94i:20053) -Baumslag, G.(1-CCNY); Miller, C. F., III(5-MELB); Short, H.(1-CCNY) -Unsolvable problems about small cancellation and word hyperbolic groups. (English summary) -Bull. London Math. Soc. 26 (1994), no. 1, 97–101. -20F10 (20F06) ) [they use the Rips construction @Misha alludes to]. -For $n\geq 4,$ there are the non-free Zariski-dense examples of Margulis-Soifer (1979). I haven't read their paper in detail, but it seems that their technique does not work in $SL(3, \mathbb{Z}).$ However, there is the nice result of Stephen Wang: -Wang, Stephen(1-HAV) -Representations of surface groups and right-angled Artin groups in higher rank. (English summary) -Algebr. Geom. Topol. 7 (2007), 1099–1117. -20F36 (20F65 57M25) -Which can presumably be generalized to other RAAGs. -Geometric finiteness: I think the action of $SL(n)$ on the positive semidefinite cone was studied first by Minkowski (for $SL(2)$ the PSD cone is just the light cone in the usual Minkowski space), and I had actually implemented this. The program usually run forever. -AND ALSO Mihailova's counterexample for generalized word problem (AKA membership problem) uses SEVEN generators. Undoubtedly she had tried to get it down lower, but apparently failed. It turns out that in many applications, we have two generators, in which case it seems that even the generalized word problem is open even for $F_2 \times F_2.$ - -REPLY [3 votes]: I do not know direct answer to your problem, but I will think about this -as it is close to the area of problems which we are currently working on. -Although I would like to comment that not all questions are "reasonably efficiently decidable" for SL(2,Z) as most of known to me problems for -SL(2,Z) are at least NP-hard, meaning that none of these problems have efficient (polynomial time) solutions unless P=NP. -The membership for Identity Matrix is NP-hard for a finitely generated matrix semigroup from SL(2,Z) and there is no even simple NP brute-force algorithm as there is a class of semigroups where the shortest identity could require exponential length of products to reach the identity. -See: Paul C. Bell, Igor Potapov: On the Computational Complexity of Matrix Semigroup Problems. Fundam. Inform. 116(1-4): 1-13 (2012) -Similar results exist for Mortality problem (membership of the Zero matrix) Paul C. Bell, Mika Hirvensalo, Igor Potapov: Mortality for 2×2 Matrices Is NP-Hard. MFCS 2012: 148-159 -Recently we proved that the Vector Reachability for SL(2, Z) is decidable (in general it is harder to check than membership as there is an infinite set of matrices that can connect two points and decidability of the membership is not lead to vector reachability problem decidability). -Potapov, Pavel Semukhin: Vector Reachability Problem in SL(2, Z). MFCS 2016: 84:1-84:14 http://drops.dagstuhl.de/opus/volltexte/2016/6492/ -The other fresh 2016 result is the proof that the membership for nonsingular matrix semigroup from $Z^{2x2}$ is decidable. -http://arxiv.org/abs/1604.02303 -In relation to the membership in SL(2,Z) we recently managed to improve -the complexity for the membership problem from PSPACE to NP, the main idea of the paper that we are able to operate effectively with compressed versions of matrices from SL(2,Z) by application of number of new results about structural properties of matrix products. - -For undecidability side the membership problem for a semigroup generated by matrices from SL(2,H) or even for a semigroup of double quaternions is undecidable http://www.sciencedirect.com/science/article/pii/S0890540108000771 -Moreover the membership of the identity matrix for semigroup generated by matrices from SL(4,Z) is undecidable: -Paul C. Bell, Igor Potapov: On the Undecidability of the Identity Correspondence Problem and its Applications for Word and Matrix -Semigroups. Int. J. Found. Comput. Sci. 21(6): 963-978 (2010) -Archive vestion: arxiv.org/abs/0902.1975 -See Problem 10.3 http://press.princeton.edu/math/blondel/solutions.html -but the same problem for 3x3 is sill open.<|endoftext|> -TITLE: Interpreting Witten's Asymptotic Expansion of the WRT invariant. -QUESTION [8 upvotes]: Witten's asymptotic expansion conjecture as described in "Problems on invariants of -knots and 3-manifolds" in Geometry and Topology Monographs, Volume 4 states that -$$Z_r^{SU(2)}(M)\sim_{r\rightarrow \infty} e^{-3\pi \bf{i}(1+b^1(M))/4}\times -\sum_{[A]}e^{2\pi \bf{i}CS(A)} r^{(h^1_A-h^0_A)} e^{-2\pi \bf{i}(I_A/4+h^0_A/8)}\tau_M(A)^{1/2},$$ -where $b^1(M)$ is the first betti number of $M$, $CS(A)$ is the chern-simons invariant of $A$, $h^i_A$ is the rank of the $i$th cohomology of $M$ with coefficients in the $su(2)$ bundle twisted by the adjoint action of the monodromy of $A$ and $I_A$ is the spectral flow of the -Laplace operator along a path connecting $A$ to the trivial flat connection. Finally, -$\tau_M(A)$ is the Reidemeister torsion. -Since $h^1$ is often the dimension of the Zariski tangent space of the space of gauge equivalence classes of flat connections, if $h^1>0$ in this formula, you are admitting -the possibility that the moduli space of flat connections has positive dimension. In this case -the sum is not really a sum, but an integral. The torsion in this case defines a volume -form on $H_0(M,adA)\oplus H^1(M;adA)\oplus H_2(M,ad A)\oplus H^3(M,adA)$ ,where -I mean the complex with twisted homology as appears above. Clearly what is -meant by $\tau_M(A)^{1/2}$ must be a volume form on the moduli -space of gauge equivalence classes of flat connections for the integral to work. That means $\tau_M(A)^{1/2}$ is a top dimensional -form on $H^1(M;adA)$. -What is the explicit formula for the volume form? In other words, how do you go about cutting down -the torsion and taking its square root as as a differential form? - -REPLY [7 votes]: In the conjecture, which should hold in some way for any other simple, simply-connected, compact Lie group $G$, $M$ is closed and $A$ is a flat connection in a trivialized $G$-bundle over $M$. It is easier to think of Reidemeister torsion as a density on $H^0(M,adA)^* \oplus H^1(M,adA) \oplus H^2(M,adA)^* \oplus H^3(M,adA)$: It is a norm on the determinant line of the cohomology, or equivalently, an element of $|\det H^0(M,adA) \otimes \det H^1(M,adA)^* \otimes \det H^2(M,adA) \otimes \det H^3(M,adA)^*|.$ I believe, it was first mentioned by Jeffrey, how to interpret the conjecture using Reidemeister torsion as a volume form or density. -We need to fix an inner product on the Lie algebra $\frak{g}$ of $G$, e.g. the Killing form, so that we can use Poincaré duality to identify $H^0(M,ad A)$ with $H^3(M,ad A)^* $ and $H^2(M,ad A)$ with $H^1(M,ad A)^* $. We denote the identification on the level of densities by $PD$. If we also assume that $A$ is irreducible, then $H^0(M,adA) = H^3(M,adA)^* = 0$ and we can write -$$ -\tau_M(A)=\omega_A \otimes PD^{-1}(\omega_A) -$$ -for some $\omega_A \in |\det H^1(M,adA)^*|$. Then $\sqrt{\tau_M} = \omega$ is a density on the moduli space of gauge-equivalence classes irreducible flat connections, which we can integrate. -If $H^0(M,adA) \neq 0$, then an orthonormal basis of $H^0(M,adA) \subset \frak{g}$ with respect to the chosen inner product on $\frak{g}$ gives a canonical element $\eta_A \in |\det H^0(M,adA)|$. Then -$$ -\tau_M(A) = \eta_A \otimes \omega_A \otimes PD^{-1}(\omega_A) \otimes PD(\eta_A). -$$ -The density on the moduli space of flat connections is therefore $\omega$, but integration needs to be made sense of. The conjecture depends on a suitable choice of inner product on $\frak{g}$, if the cohomology is not trivial. -I should add, that the conjecture you wrote is only a conjecture for the leading order term, not for the full asymptotic expansion. In my paper with Joergen Andersen we give a more precise statement of the same conjecture and prove it for finite-order mapping tori. The version using spectral flow is pre-eminent in the literature, but it becomes more compact if you use the Rho invariant $\rho$ instead: -$$Z_{k}^{G}(M)\sim \sum_{c\in C} \frac{1}{|Z(G)|}\int_{A \in c} \sqrt{\tau_{M}(A)} e^{2\pi i CS(A)k} e^{\frac{\pi -i}{4} \rho_A(M)} k^{d_c},$$ -where $C$ is the set of all connected components of the moduli space of gauge-equivalence classes of flat connections on $M$, and $$d_c = -\frac{1}{2} \max_{A \in c} \left( \dim(H^1(M,d_A)) - -\dim(H^0(M,d_A)) \right),$$ where $\max$ is the maximum attained on a Zariski open subset of $c$. The Rho invariant is related to the spectral flow and the Chern-Simons invariant via the APS index theorem, which gives $r = k + h$ in your statement (where the $r$ is missing next to the Chern-Simons invariant), where $h$ is the dual Coxeter number. Note that this formula is not really the leading order term, but only the leading order term at each flat connection. Also note that my signs are a little different because I use different conventions for the spectral flow as well the normalization of the Chern-Simons invariant.<|endoftext|> -TITLE: Artin's conjecture for $n=2$ -QUESTION [7 upvotes]: I am interested in the following question: -Is it known that $2$ is a primitive root modulo $p$ for infinitely many primes $p$? -There is some information about Artin's conjecture in Wikipedia. -I need to know if it is up-to-date and if one can say something about the case $n=2$. - -REPLY [3 votes]: It follows from GRH. Not known on its own.<|endoftext|> -TITLE: Ricci flow with surgery in dimension 2 -QUESTION [9 upvotes]: Is it possible to define the Ricci flow with surgery in dimension 2 and use it to classify the surfaces? -I know this is overkill, there are simpler ways to classify surfaces, but I would like to understand the Ricci flow with surgery in dimension 3 and perhaps that this is simpler in dimension 2. - -REPLY [13 votes]: Regarding (3) in Dan Lee's answer, you don't need my work in the Ricci flow approach to the differential geometric version of the uniformization theorem. The reason is as follows. Take any metric $h$ on a closed orientable surface $M$ with $\chi (M) > 0$. Let $r$ denote the average scalar curvature of $h$, which is positive since $\chi$ is. Solve the equation $\Delta_h u = R_h - r$, where $R_h$ denotes the scalar curvature of $h$. This is possible since the integral of $R_h - r$ with respect to the measure induced by $h$ is zero. Define the pointwise conformally equivalent metric $g=e^uh$. Then we have $R_g=e^{-u}(-\Delta_h u + R_h) = e^{-u}r>0$. We can now use $g$ as an initial data for the normalized Ricci flow and apply Hamilton's theorem to obtain exponential convergence in any $C^k$-norm to a constant curvature metric in the same conformal class as $h$. -By now there are many approaches to the Ricci flow on closed surfaces, such as the Aleksandrov reflection method employed by Bartz--Struwe--Ye (inspired by Schoen's work on Yamabe (constant scalar curvature) metrics; in PDE see Serrin and Gidas--Ni--Nirenberg, etc.), Hamilton's isoperimetric monotonicity, application of Perelman's entropy formula, Andrews--Bryan's isoperimetric profile monotonicity, etc. -Regarding Igor Rivin's answer, it's been a long time since I've looked at this, but this is what I remember (please correct me if I am wrong). Osgood--Phillips--Sarnak looked at the Polykov action from the spectral point of view and also from the variational point of view. Aside: Ray--Singer's work on analytic torsion predates their work (I should also mention McKean, etal.). However they did not explicitly make the connection to Ricci flow, although they may have known this. They also did not reprove the differential geometric version of the uniformization in the positive Euler characteristic case since their proof of sequential convergence assumed conformality to the standard $S^2$. In my 2-sphere paper, I actually used the Polykov energy and the fact that it is bounded from below (I believe due to Onofri), which I learned from Osgood--Phillips--Sarnak (I am indebted to Richard Melrose for asking that I read this paper when I first arrived at MIT). I used the Polykov energy to control Hamilton' entropy in the variable signed curvature case. However, later in my entropy on 2-orbifolds paper, I found a way to avoid Polykov entropy to control the modified Hamilton's entropy. Yet another proof of the entropy bound, adapting the original Hamilton's contradiction argument, was in my paper with Lang-Fang Wu on 2-orbifolds with variable signed curvature. -Regarding some sort of convexity of the functional, a fancy way to interpret the energy functional is in terms of Bott--Chern secondary characteristic classes and this was originally used in Donaldson's work on Hermitian-Einstein metrics/Hermitian Yang-Mills connections on (semi-)stable vector bundles over algebraic surfaces (and later algebraic manifolds); Uhlenbeck--Yau had a different approach. In the case of closed Riemannian surfaces, the formula is: -$$\ln \det \Delta_g - \ln \det \Delta_h = -\frac{1}{48\pi} E_h(g),$$ -where the relative energy of two pointwise conformal metrics is defined by: -$$E_h(g) = \int_M \ln (g/h)(R_g d\mu_g + R_h d\mu_h).$$ -The Bott--Chern class is in effect the term $\ln (g/h)$ since $\partial \bar{\partial}$ of it is essentially the difference in the first Chern classes (Gauss--Bonnet integrands in this case) of $g$ and $h$.<|endoftext|> -TITLE: Where can I learn about Formal Schemes? -QUESTION [18 upvotes]: I am trying to learn formal schemes. I tried to read the section in Hartshorne but I don't get very far from there since things are not done quite explicitly enough, at least in my opinion. I cannot read French, so EGA is out of the question. I would really appreciate it if you could tell me a good introduction to this topic. - -REPLY [4 votes]: If you are interested in the infinitesimal structure of formal schemes (with emphasis on non necessarily adic maps) I suggest to look at my papers with Jeremías and Pérez "Infinitesimal Lifting and Jacobi Criterion for Smoothness on Formal Schemes" (Comm. Alg.) and "Local structure theorems for smooth maps of formal schemes" (JPAA). Also for the deformation theory of formal schemes look at M. Pérez "Basic deformation theory of smooth formal schemes" J. Pure Appl. Algebra 212 (2008), pp. 2381–2388 (MR 2009h:14006). I apologize for this self-promotion...<|endoftext|> -TITLE: Commuting supremum and expectation -QUESTION [7 upvotes]: Given a one-parametric random function on a probability space $(\Omega,\mathcal F,\mathbb P)$: -$X:U\times\Omega\to \mathbb R \text{ and } (a,w)\mapsto X(a,w), \text{ with } \sigma(X(a))\subseteq \mathcal F\quad\quad\forall a\in U\subseteq\mathbb R, -$ -Then the following holds: -$E\left[\sup\limits_{a\in U}X(a)\right]=\sup\Bigr\lbrace E\left[X(A)\right]\Bigr|\sigma(A)\subseteq{\mathcal F},\;A(\omega)\in U\Bigr\rbrace$ -and also -$E\left[\sup\limits_{a\in U}X(a)\right]=\sup\Bigr\lbrace E\left[X(A)\right]\Bigr|\sigma(A)\subseteq{\bigcup\limits_{a\in U}\sigma(X(a))},\;A(\omega)\in U\Bigr\rbrace$ -Proof: -The following holds trivially: -$E[X(A)]\le E[\sup_{a\in U} X(a)]$ -it remains to show the other direction. This is done by applying zhoraster's answer: -Clearly, $M(\omega) = \sup_{a\in U} X(a,w)$ is $\mathcal F$-measurable. -Define for $\delta>0$ -$\mathfrak A_\delta = \lbrace(a,\omega)\in U\times \Omega\mid X(a,w) >M(\omega)-\delta\rbrace$ -This set is in $\mathcal B(\mathbb R)\otimes \mathcal F_t$, and it has a full projection onto $\Omega$. By a measurable selection theorem (which I think one can find in Bogachev Measure Theory) there is an $\mathcal F$-measurable $A_\delta$ such that $(A_\delta(\omega),\omega)\in\mathfrak A_\delta$ almost surely. Hence $E[X(A_\delta)]≥E[M(\omega)]−\delta$. We get the desired statement by letting $\delta\to 0$. -(One can also use Kuratowski--Ryll-Nardzewski theorem to prove the existence of a measurable $A_\delta$.) - -After a very good answer of zhoraster, I realized, that my initial question was a mixup of several different things. Thats why I changed it community wiki and clearified the problem. - -REPLY [4 votes]: Clearly, $M(\omega) = \sup_{a\in U} g(a,S_t)$ is $\mathcal F_t$-measurable. -Define for $\delta>0$ -$$ -\mathfrak A_\delta = \{(a,\omega)\in U\times \Omega\mid g(a,\omega)>M(\omega)-\delta\} -$$ -This set is in $\mathcal B(\mathbb R)\otimes \mathcal F_t$, and it has a full projection onto $\Omega$. By a measurable selection theorem (which I think one can find in Bogachev Measure Theory) there is an $\mathcal F_t$-measurable $A_\delta$ such that $(A_\delta(\omega),\omega)\in \mathfrak A_\delta$ almost surely. Hence $E[g(A_\delta,S_t)]\ge E[M(\omega)]-\delta$. We get the desired statement by letting $\delta\to 0$. -(One can also use Kuratowski--Ryll-Nardzewski theorem to prove the existence of a measurable $A_\delta$.) -UPD: This is perhaps wrong (I just noticed Did's comment).<|endoftext|> -TITLE: Is every ring the direct limit of Noetherian rings? -QUESTION [12 upvotes]: Are there any examples of commutative rings that do not occur as direct limits of Noetherian rings? - -REPLY [28 votes]: Every commutative ring is the directed colimit of its subrings that are finitely generated as $\mathbb{Z}$-algebras. The Hilbert Basis Theorem implies that these subrings are noetherian. Actually this method is used in EGA IV, §8.9 to generalize some theorems from noetherian schemes to more general schemes.<|endoftext|> -TITLE: Cardinality of $\eta$-bush; on a Lemma from Wolff's paper. -QUESTION [6 upvotes]: The Question -This question is about Lemma 1.2 on the fifth page of Thomas Wolff's paper, "A sharp bilinear cone restriction estimate", Annals of Mathematics, 153 (2001), 661--698. The Lemma states (the definitions to be given after) - -If $x\in Q(1)$ is a smooth $\mu$-fold point for the white tubes with $\mu > \delta^B$ then $x$ is a base-point for an $\eta$-bush of white tubes with cardinality $\gtrsim (\log \frac1\delta)^{-1} \mu (\frac\eta\delta)^M$ for some $\eta \leq \delta^{1-\epsilon}$. Conversely... - -(I understand the converse part, and can sketch the proof, so I omit that here.) The Lemma is given without proof in the paper, and left deliberately as an exercise. The problem is I am not even sure if I understand the Lemma correctly! (In particular, I don't see why the restriction $x\in Q(1)$ is necessary at all: the statement seems to be translation invariant.) -Question Can someone supply a sketch of the proof for this Lemma (beyond the one-liner in the paper)? In particular, where does the logarithm loss come from? The converse statement does not require a logarithm. Thanks. -The Definitions -Now, the definitions to make sense of the Lemma. (Do let me know if I missed anything.) - -$Q(1)\subset \mathbb{R}^d$ is the unit cube centered at the origin. -A small constant $\epsilon$ is fixed throughout, and a large constant $B$ depending on dimension $d$ is fixed throughout. -For the purpose of this lemma, a white light-ray is a line in $\mathbb{R}^d$ transversal, and making an angle of 45 degrees, with the plane $\{x_d = 0\}$ -Let $\mathcal{W}$ denote a (finite or countable) collection of white light rays. Fix $\delta > 0$. For an element $W\in \mathcal{W}$, denote by $w$ the set of points $\{ |x - W| \leq \delta\}$. For $w$ the function $\phi_w$ is defined: $ \phi_w(x) = \min (1, \frac{\delta}{|x-w|})^M $ for some fixed large constant $M$ depending on $\epsilon$. $|x-w|$ denotes the Euclidean distance from the point $x$ to the set $w$. -$\mathcal{W}$ is assumed to be $\delta$ separated: for each $W$ we can associate a direction in real projective space corresponding to the axis. Then for any $D$ disc in projective space of radius $\delta$, let $\mathcal{W}_D$ denote the subset of light rays whose direction points in $D$, the $\delta$-separation assumption requires that the set $\cap_{W\in \mathcal{W}_D} w$ be a bounded set. In particular this implies two parallel light rays cannot be closer than $2\delta$ apart. -We say that $x$ is a smooth $\mu$-fold point for the white tubes if $\sum_{W\in\mathcal{W}} \phi_w(x) \geq \mu$. Roughly speaking this means around $\mu$ light rays get close to $x$. (As far as I can tell, this "roughly speaking" is the content of the Lemma described above.) -$x$ is said to be a base-point for an $\eta$-bush $P\subset \mathcal{W}$ if $\sup_P |x-w| \leq \eta$. That is, we have a family of light-rays that all pass within $\delta + \eta$ of $x$. - -REPLY [3 votes]: Ah, with Terry's comment it turns out the situation was a lot simpler than I thought. -Step 1: Lack of contribution from distant pieces -Divide the region $\{|y-x| > \delta : y \in \mathbb{R}^d\}$ into concentric shells of thickness $\delta$. Consider the contribution to $\sum \phi_w(x)$ from $W\in\mathcal{W}$ such that $|x-w| \in (k\delta, k+1\delta]$, for $k\in \mathbb{N}$. We can divide the shell up into $\sim k^{d-1}\delta^{1-d}$ balls of size $\delta$. Write $y(W)$ for the point of closest approach on $W$ to $x$. In each ball $V$ of size $\delta$, by the $\delta$-separation assumption, there exists sum large constant $B'$ such that -$$ \left|\{ W\in\mathcal{W} : y(W) \in V \}\right| \leq B \delta^{-B'} $$ -So -$$ \sum_{k > \delta^{-\epsilon}} \sum_{k\delta < |y(W)-x| - \delta \leq (k+1)\delta} \phi_w(x) \lesssim \sum_{k > \delta^{-\epsilon}} k^{d-1}\delta^{1-d-B'} k^{-M} $$ -For sufficiently large $M$ (and if we fix $\delta < \delta_0 < 1$), the RHS can be bounded $\leq \delta^B$. So we can neglect the contribution from far away pieces up to a $\delta^B$ term. -Step 2: Pigeonhole the inside -In view of Step 1, it suffices to show that there exists a constant $C$ such that if for any $0 < k < \delta^{-\epsilon}$ ($k$ should be though of as $\eta/\delta$) we have -$$ \left| \{ W\in\mathcal{W} : |x-w| < k\delta \} \right| \leq C\frac{1}{\log \frac1\delta} \mu k^M $$ -we must have $\sum \phi_w(x) < \mu$. This follows by noting that the above means that there are no rays within distance $\delta$ of $x$ ($k = 0$), and thus we only need to sum over $k > 0$ -$$ \sum_{k = 1}^{\delta^{-\epsilon}}\phi_w(x) \lesssim \sum_k \left| \{ W\in\mathcal{W} : |x-w| < k\delta \} \right| \cdot k^{-(M+1)} $$ -using that $\phi_w(x) \sim k^{-M}$. Plugging in the hypothesis, we have that the expression is controlled by -$$ C\frac{1}{\log \frac1\delta} \mu \sum_{k = 1}^{\delta^{-\epsilon}} k^{-1} \lesssim C \mu \log_\delta(\delta^{-\epsilon}) = C\epsilon \mu$$ -So for sufficiently small $C$, we can bound -$$ \sum_{k = 1}^{\delta^{-\epsilon}}\phi_w(x) \leq \frac12 \mu $$ -which gives the desired contradiction.<|endoftext|> -TITLE: Rational powers of ideals in Noetherian rings -QUESTION [6 upvotes]: Let $R$ be a Noetherian ring, and let $I$ be an ideal of $R$. Fix a rational number $a=\frac{p}{q}$ with $p, q\in \mathbb{Z_\geq 0}$ $q\neq 0$. We define $I_a = \{x \in R: x^q\in \overline{I^p}\}$, where overline denotes integral closure. -Huneke-Swanson show, (in their book on Integral Closure), that $I_a$ is well defined, is indeed an ideal and is integrally closed. If $a\leq b$, $I_b\subseteq I_a$ and if $n$ is a positive integer, then, $I_n=\overline{I^n}$. -I was curious to know what else is known about these rational powers. Are there any computations done in special cases (e.g. monomial ideals)? Do these ideals find any use other than computing the integral closure of the extended Rees ring in the Laurent polynomial ring? I would appreciate if anyone has any references on this subject other than Huneke-Swanson book. -So far I have only found a paper by David Rush "Rees valuations and asymptotic primes of rational powers in Noetherian rings and lattices" -PS: I am not sure why, but I am unable to typeset curly braces (I am using backslash followed by brace sign, but it does not show up) and the "less than" (causes text after the symbol to vanish) symbol. - -REPLY [4 votes]: I give some hints about computing them in my paper Balanced normal cones and Fulton-MacPherson's intersection theory, section 3. (Also I compute 14 examples.) -I find it kind of astounding that this easy and beautiful construction has sat essentially unused in the literature for over 50 years now. There is Rees' book on asymptotic properties of ideals, and the Huneke-Swanson book, but that's almost the entire literature.<|endoftext|> -TITLE: Hodge-Tate decomposition for formal groups -QUESTION [5 upvotes]: Hi All, -Where can I find a proof of the Hodge-Tate decomposition for Lubin-Tate formal -groups? -Thanks! - -REPLY [7 votes]: Dear jjj, -I recommend reading Tate's original paper, which proves the Hodge--Tate decomposition -for all $p$-divisible groups. If you are nervous about $p$-divisible groups, rather -than formal groups, it would not be difficult to restrict to just this case while -reading the paper. (And the paper includes an entire section devoted to relating -formal groups to the $p$-divisible groups picture, which is valuable in its own right.)<|endoftext|> -TITLE: Unique equilibrium states for systems without specification -QUESTION [6 upvotes]: Let $X$ be a compact metric space and let $f\colon X\to X$ be a continuous expansive map. Let $\mathcal{V}$ denote the space of H��lder continuous potential functions $\phi\colon X\to \mathbb{R}$, and let $\mathcal{W}$ denote the set of potential functions (not necessarily Hölder continuous) that have a unique equilibrium state. -It is well known that if $f$ satisfies the specification property, then every $\phi\in \mathcal{V}$ has a unique equilibrium state, and hence $\mathcal{V} \subset \mathcal{W}$ (Rufus Bowen, Some systems with unique equilibrium states, Math. Systems Theory 8 (1974/75), no. 3, 193–202). -There are many systems without the specification property for which something is known about the set of potentials $\mathcal{W}$. For example, intrinsic ergodicity (existence of a unique measure of maximal entropy), which is equivalent to the statement that $\mathcal{W}$ contains the constant functions, has been studied in a number of recent works (Buzzi-Fisher, Bufetov-Gurevich, Climenhaga-Thompson), and there are stronger results for particular examples, such as $\beta$-shifts, for which $\mathcal{W}$ is known to contain the space of Lipschitz functions (Peter Walters, Equilibrium states for $\beta$-transformations and related transformations, Math. Z. 159 (1978), no. 1, 65–88). -However, I do not know of any examples of expansive maps without specification for which the inclusion $\mathcal{V} \subset \mathcal{W}$ is known. Does anybody know of such an example? -(Note that it should be not too difficult to adapt the answers to this question to obtain a system for which $\mathcal{W}$ contains the constant functions, but does not contain all of $\mathcal{V}$.) -Edit: Of course if $f$ is uniquely ergodic then $\mathcal{W}$ contains all potential functions. The most obvious examples of uniquely ergodic systems, irrational circle rotations (or rather, their symbolic counterparts, which are expansive) possess a weak version of the specification property, but I don't know if this weak specification holds for every uniquely ergodic system. -What I'd really like to know is if there is an expansive map that is not uniquely ergodic and does not have the specification property for which $\mathcal{V} \subset \mathcal{W}$. I'd also be interested in knowing whether unique ergodicity implies weak specification. -(By "weak specification" I mean that orbits can be consecutively shadowed with uniformly bounded gaps, as in the usual specification property, but that we do not require the shadowing orbit to be periodic, and we allow the length of the gaps to vary.) - -REPLY [2 votes]: In the time since I asked this question, Dan Thompson and I have posted a preprint showing, among other things, that every Hölder continuous potential function on a β-shift has a unique equilibrium state. Since β-shifts do not have specification for most β>1, this gives an example of the sort I was after.<|endoftext|> -TITLE: Is intersection homology the usual homology of something else? -QUESTION [15 upvotes]: Let $X$ be the sort of topological space for which it makes sense to talk about the intersection homology. Fix a perversity $p$, or just take $p= 1/2$ if you like. - - -Is there some naturally defined $X'$ such that ${}^p IH_* (X) = H_* (X')$ ? - -REPLY [9 votes]: Such a space $X'$ cannot possibly depend functorially on $X$: suppose that there is a functor $F$ from the category of topological spaces (that have reasonable stratifications) and continuous maps to itself such that IH(X)=H(F(X)). Then IH would be functorial, but considering the example -$$(S^1 \hookrightarrow X \rightarrow S^1)=id_{S^1}$$ where $X$ is the pinched torus and $S^1$ is the non-collapsed circle (see pages 55 and 56 of Kirwan-Woolf "Introduction to Intersection Homology") would imply that the intersection homology of $S^1$ (=ordinary homology of $S^1$) imbeds in the intersection homology of $X$. But the first IH group of $X$ (middle perversity) is $0$.<|endoftext|> -TITLE: does this group have a name? -QUESTION [8 upvotes]: In my work this week I came across a group with presentation with two generators $a$ and $b$ subject to the relations $baba=1$, $a^2b=ba^2$, and $ab^{-n}ab^n=b^nab^{-n}a$. This group looks like the lamplighter group or something to me, but I couldn't get a sequence of Tietze transformations from this group to the standard presentation for the lamplighter. Does anyone know what this group is? thanks. - -REPLY [11 votes]: All relations of the form $ab^{-n}ab^n=b^nab^{-n}a$ follow from $baba=1$, $a^2b=ba^2$ (exercise). So the group is isomorphic to $G=\langle a,b\mid baba=1, a^2b=ba^2\rangle$. The later splits as a central extension $1\to \mathbb Z\to G \to D_{\infty }\to 1$. I do not think the group has a name. - -REPLY [4 votes]: The first two relations alone give a polycyclic group of Hirsch length 2 ($a^2$ is central, quotienting by it gives the infinite dihedral group $D_\infty$), which, thanks to Denis Osin's answer, is already the whole group. Even without that knowledge, it is still a quotient of this group, and so polycyclic of Hirsch length $\leq 2$. In particular, it is far too small to be the lamplighter.<|endoftext|> -TITLE: A metric for Grassmannians -QUESTION [15 upvotes]: I'm reading an article by Ricardo Mañé, "The Hausdorff dimension of horseshoes of diffeomorphisms of surfaces" (https://doi.org/10.1007/BF02585431). I'm having a technical problem. Sorry for my ignorance, but I would like a reference which explains how to equip the Grassmann manifolds with a metric. - -REPLY [6 votes]: My "answer" is just a (still another) reformulation of Ian's. -Given two subspaces $V$, $W \subset \mathbb{R}^n$ with the same dimension, define their distance as: -$$ -d(V,W):= \inf_J \| J - i_V\| -$$ -where $i_V : V \to \mathbb{R}^n$ denotes the inclusion, $\|\mathord{\cdot}\|$ is the Euclidian operator norm, and $J$ runs over the linear maps $V \to \mathbb{R}^n$ whose such that $J(V) \subset W$. It's not too difficult to prove that the the infimum is attained at map $J=P_W|V$ (i.e. the orthogonal projection onto $W$ restricted to $V$). Trivially, -$$ -\|P_W|V - i_V\| = \sup_{v\in V, \|v\|=1} \inf_{w\in W} \|w-v\|, -$$ -so my formula is equivalent to Ian's. -As an easy consequence, we have the following interesting fact: - -Fact: Let $T$ be a linear automorphism of $\mathbb{R}^n$. Then the induced trasformation of the grassmannian $G(k,n)$ is (bi-)Lipschitz with constant $\|T\|\,\|T^{-1}\|$. - -Proof: -$$ -d(TV,TW) = \inf_{K(TV)\subset TW} \|K-i_{TV}\|=\inf_{J(V)\subset W}\big\|T(J-i_V)T^{-1}|_{TV}\big\| \le \|T\| \, \|T^{-1}\| \, d(V,W). -$$ -(The proofs of the "fact" using other characterizations of $d$ become messier, I think.)<|endoftext|> -TITLE: A Game of Knights and Queens -QUESTION [9 upvotes]: Let $m,n,u,v \in \mathbb{N}$ be parameters with $m,n \geq 3$. Suppose two players play a game on a $m \times n$ chess board and we denote the squares of the board by the set of points $ (i,j) $ such that $1 \leq i \leq m, 1 \leq j \leq n $. Player 1 has $u$ queens, and player 2 has $v$ knights. The initial configuration is some given subset of the board where no piece is placed at a position that can be immediately attacked by another piece of the opposing player, and the player with at least one piece when the game terminates is the winner. Both players must make a move when it is their turn. The game terminates when either player captures all of the pieces of the opposing player. If Player 1 goes first, under what circumstances does either player have a winning strategy? What can be said about the probability as a function of the parameters given with both players playing optimally, that the game will terminate in finitely many steps? -I have obtained the result for $m = 3$ and arbitrary $n$ with $u= v = 1$. Ed Dean proved that Player 1 always has a winning strategy for arbitrary $m,n$ and $u = v = 1$ (see below), and he sketched a winning strategy for Player 1 in the $u = 1, v = 2$ case. He also gave an argument for the lack of a winning strategy for Player 1 for the $u = 1, v = 3$ case. All other cases are still unknown as of yet. -Edit: Previous suggestion on a way to show in the $u = v = 1$ case that the knight can avoid capture for sufficiently large m,n is removed. -Edit 2: Thanks to Ed Dean for resolving some cases, as indicated above. -Edit 3: Included a new question regarding a related probability distribution. -Edit 4: Moved edit 3 to a new question: A random variable in a game of knights and queens - -REPLY [6 votes]: Here's how White (my new name for Player 1) wins in the $u=v=1$ case. The idea is of course for White to force the knight to an edge, where it can then be summarily captured. WLOG let's force the knight to the lower edge (in my coordinate system, that'll be the edge given by $n=1$). It's enough to show that whenever the knight stands at a point $(a,b)$, we can force a situation where its next move will either allow it to be captured, or will place it on a square whose $n$ coordinate is $ -TITLE: A problem on sums of arctangents of rationals -QUESTION [5 upvotes]: Let $S$ be a set of rational numbers. -For "special" sets $S$, we can ask if $\pi$ can be written as a $\mathbb{Q}$-linear or $\mathbb{Z}$-linear combination of elements from '$\{\tan^{-1}(x): x \in S\}$'. -In particular, let $\hat S$ be the closure of $S$ under both negation and the binary operation $p*q=\frac{(p+q)}{(1-pq)}$. - -For any natural number $b \geq 2$, $S_b =\{ \frac{1}{b^k}: k \geq 1 \}$. Prove that $0 \notin \hat {S_b}$. -Note: $1 \notin \hat S$ since any rational number $P/Q$ in $\hat {S_b}$ in non-reduced form satisfies $(P,Q) \in \{(0,1), (0,-1),(1,0),(-1,0)\}$ (mod $b$). - -$tan^{-1}(1/2) + tan^{-1}(1/3) = \pi/4$. -Let $b_1,b_2 \geq 2$ be two natural numbers. For which pairs does 1 belong to the closure of $S_{b_1} \cup S_{b_2}$? For which pairs does zero belong to the closure? Are there pairs for which $\pi$ can be written as a $\mathbb{Q}$-linear combination of the arctangents of their negative powers but not as $\mathbb{Z}$-linear combinations? - -REPLY [3 votes]: In addition to Paul's comment let me point out that the structure of the underlying group was investigated in - -Sam Northshield, Associativity of the secant method, Amer. Math. Mon. 109 (2002), 246-257, - -and that the relation with the arithmetic of ${\mathbb Z}[i]$ was observed by - -John Todd, A problem on arc tangent relations, Amer. Math. Mon. 56 (1949), 517-528 - -The associativity of the secant method is, by the way, a consequence of the usual geometric interpretation of the group law on conics.<|endoftext|> -TITLE: turbulence as an unsolved problem of classical mechanics -QUESTION [5 upvotes]: Why is it that turbulence is considered to be an unsolved problem of classical mechanics? What is meant by "unsolved"? Don't the Navier-Stokes equations apply to turbulent flows? It's difficult to grasp these concepts because there doesn't seem to be a mathematically precise definition of turbulence. - -REPLY [2 votes]: I think we don't need a mathematically precise definition of turbulence to make sense of "turbulence as an unsolved problem of classical mechanics". -The question of singularities in Navier-Stokes solutions is a challenging math problem but almost unrelated to "turbulence as an unsolved problem"; -which in my opinion amounts to: what, if anything, is universal in the statistics of actual fluid flows at (extremely) high Reynolds numbers? -No answer yet, so unsolved seems right... (Well, no full answer: Kolmogorov's 4/5th law is a universal feature, but there should be more, for instance a probability measure on a space of velocity fields having suitable invariance properties with respect to Navier-Stokes or Euler dynamics).<|endoftext|> -TITLE: Generalize the Proj construction? -QUESTION [5 upvotes]: I'm wondering if there is a generalization of the Proj construction used in algebraic geometry. Given a graded ring R, which is a monoid homomorphism $R\to \mathbb{N}$, we can form the scheme Proj(R), which is the union of Spec$(R_f)_0$. -I'm wondering if there is a way to extend this construction to general homomorphisms from a ring to a (sharp) monoid. -In particular, I want to see if there is a ``Proj'' construction which gives the product $\mathbb{P}^m\times\mathbb{P}^n$ from the bidegree $k[x_0,\ldots,x_m,y_0,\ldots,y_n]\to\mathbb{N}^2$. -(I know this can be seen via two consecutive Proj, but want to see if this generalizes, as I want to see what happens if the degree map take value in a general toric monoid.) - -REPLY [5 votes]: The monoid and the sharpness condition feel misplaced to me. After all, there's no reason a ring must be positively graded in order to take Proj of it. -A grading on $R$ is "really" a $\mathbb G_m$-action on $R$, which happens to be the same as a monoid homomorphism from $R$ to the character group of $\mathbb G_m$. $Proj(R)$ is obtained by cutting the $\mathbb G_m$-fixed locus out of $Spec(R)$ and quotienting what's left by the action of $\mathbb G_m$. So it feels like the more natural generalization to shoot for is "Proj" of a ring with group action. -But maybe not. I think a productive case to think about is $R=k[x,xy,x^2y]$ with the bi-grading inherited from $k[x,y]$. It seems to me unlikely that any reasonable generalized $Proj(R)$ would depend on which specific monoid you take, whether its $\mathbb N\times \mathbb N$, or something bigger, or something smaller. It may as well be $\mathbb Z\times \mathbb Z$.<|endoftext|> -TITLE: Sublattices of Young's Lattice -QUESTION [14 upvotes]: Young's Lattice is the lattice of inclusions of Young tableaux, or integer partitions. -In R. Stanley's Enumerative Combinatorics Vol. 1, Proposition 1.4.4., there is a generalization of integer partitions where you limit the number of times each positive number may occur in the partition (and this limit may be zero). -Apparently, this gives a sublattice of Young's lattice -- where can I find a discussion of such sublattices in the literature? They may be discussed somewhere in Stanley's books but I can't seem to find them. -The particular case which i'm interested in is when only a given finite set of positive integers may occur, each of which at most a specified number of times. This produces a finite self-dual sublattice with many nice properties... do these lattices have a name? -For example, if 2 is allowed at most once and 1 is allowed at most twice, then we have the lattice - (2,1,1) - | - (2,1) - / \ - (2) (1,1) - \ / - (1) - | - ( ) - -Another way of thinking of it would be as a lattice of subsequences of a finite sequence of weakly decreasing positive integers. -Update: such a lattice is a finite distributive lattice, so by the Birkhoff representation theorem, it is isomorphic to the lattice of lower sets of the poset of its join-irreducible elements. The posets in question are sub-posets of a certain infinite poset, but I am hoping for a description of this class of lattices which goes beyond simply identifying their Birkhoff representation. - -REPLY [3 votes]: It appears these lattices can be described as a kind of twisted product of the simple lattice $[n]=\{0,1,2,\ldots, n-1\}$ with the usual order. To construct the lattice of subsequences of the sequence of weakly decreasing positive integers $(k_1,\ldots, k_N)$, let $m(k)$ be the multiplicity of the integer $k$ in the given sequence and form the product $$[d(1)]\times \cdots \times [d(N)],$$ -equipped with the order $(a_1,\ldots, a_N)\leq (b_1,\ldots, b_N)$ iff $$a_N\leq b_N$$ $$a_N+a_{N-1}\leq b_N + b_{N-1}$$ $$\ldots$$ $$a_N+\cdots + a_1\leq b_N+\cdots + b_1.$$ -For example, for the sequence $(3,2,1,1)$, we have $d(1)=2$ and $d(2)=d(3)=1$, so that the lattice is given by $[2]\times [1]\times [1]$, with the order given above. -This must be a standard construction in poset/lattice theory, please let me know if you've seen it somewhere.<|endoftext|> -TITLE: is the closure of a totally convex set again totally convex? -QUESTION [5 upvotes]: Recall that a totally convex subset $C$ of a complete Riemannian manifold $M$ is a set which contains with any two points $p,q$ also all the geodesics between them. -We know that there is a totally geodesic, totally convex submanifold $N\subset M$ such that $N\subset C \subset \bar N$. So the question is: Is $\bar N$ totally convex? - -REPLY [6 votes]: No. Define a Riemannian metric tensor $g$ on $\mathbb R^2=\{(x,y)\}$ by -$$ - g(x,y) = \begin{pmatrix} 1 & 0 \\ 0 & f^2(x) \end{pmatrix} -$$ -where $f:\mathbb R\to\mathbb R$ is a positive smooth even function such that -$f(x) = \cos x$ for $|x|\le 1$ and $f''(x)/f(x)$ increases after $x=1$. Let $N$ be an open segment of length $\pi$ in the $y$-axis, e.g. the one between points $A=(0,0)$ and $B=(0,\pi)$. -(The plane with this metric is isometric to the universal cover of a surface of revolution that looks like a unit sphere with two infinite tubes attached near a pair of opposite points. Note that the Gaussian curvature $K$ is given by $K=-f''/f$, so $K\le 1$ everywhere.) -The strip $\{|x|\le 1\}$ is isometric to the universal cover of a neighbourhood of the equator of the standard sphere, so there are plenty of geodesics between $A$ and $B$. On the other hand, using Clairot integral and the fact that the Gaussian curvature does not exceed 1, it is easy to see that no geodesic can intersect the $y$-axis at two points with distance less than $\pi$ between them. Hence $N$ is totally convex but its closure is not.<|endoftext|> -TITLE: A good reference for the wave front set -QUESTION [6 upvotes]: Hello, -I am wondering whether anyone know some good references for the theory of wave front set, microlocal analysis? I have some basic knowledge of distribution theory at the level of the Rudin's functional analysis (the first part). As for PDE theory, I learned this topic mainly by Folland's "Introduction to Partial Differential Equations". -When I learned the distribution theory, the book by Strichartz gave me many intuition and helped me -a lot. I am wondering whether there is a similar book to introduce the theory of wave front set, microlocal analysis? -Thank you very much! - -REPLY [8 votes]: There are many references at various levels of difficulty; it also depends on what aspects are you interested in. I cite out of memory, so beware of inaccuracies (which can be corrected according to your needs). -A very good reference is Hormander I (The theory of linear partial diff. op), chapter VIII. The emphasis there is on the $C^\infty$ theory, with Hormander's own definition of WF as a limit. You cutoff the function near a point, Fourier transform it, then examine in which directions the Fourier transform decays fast and in which ones it does not. These last directions stabilize as the cutoff support tends to the point, and what remains is the WF set at the point. Further results, written in an even denser style, are contained in Michael Taylor's book on Pseudodifferential Operators. More recently, the concept has been generalized to include directions of Sobolev regularity and has found applications in nonlinear equations; there results are scattered in a number of papers (JM Bony wrote some papers on this). -There are other points of view; an important one is the analytic wave front set. Here the accent is more on the algebraic aspects, since the set of all solutions to an equation or system of PDEs is studied as a whole. Here the best references are japanese, a good starting point being Akira Kaneko's book on hyperfunction theory, and continuing with the works of Kashiwara (including a book), Sato, Schapira. -EDIT: I understand that an easier introduction would be helpful. You should try with section II.B of Alinhac-Gerard Operateurs pseudodifferentiels et Theoreme de Nash-Moser. It's very readable (assuming you read french :)<|endoftext|> -TITLE: Abelian subvarieties of abelian varieties --- reference request -QUESTION [6 upvotes]: This question may be too naive, in which case I apologise in -advance. Anyway, it is a well-known fact (see e.g. Milne's notes) -that any abelian variety A has only finitely many direct factors -up to automorphisms of A. (Here a direct factor of A is an -abelian subvariety B for which there exists another abelian -subvariety C of A such that $A \cong B \times C$.) -My question is: how much is known about the corresponding -question for arbitrary abelian subvarieties, rather than -direct factors? That is, is it known whether every abelian -variety A has finitely many abelian subvarieties, up to -automorphisms of A? If not, what's the best known result in this -direction? -I've asked a couple of people about this, and their opinion seems -to be that it's "more or less" known. But I would like -something a little more concrete, if possible. Any relevant -references would be appreciated! - -REPLY [5 votes]: For the benefit of others who might look at this question, let me mention that I found the following reference proving exactly what I wanted. (More precisely, I was told about it by David Ploog.) -Lenstra, H; Oort, F; Zarhin, Yu. Abelian subvarieties. J. Algebra 180 (1996), no. 2, 513–516.<|endoftext|> -TITLE: Representability of matroids over $\mathbb R$ -QUESTION [12 upvotes]: Let $M$ be a matroid, for example viewed as being given by a finite set $X$ and a rank function $d : P(X) \to {\mathbb N}$ such that -1) $d(\varnothing)=0$, $d(\lbrace x \rbrace)=1$, for all $x \in X$, -2) $A \subset B$ implies $d(A) \leq d(B)$, and -3) $d(A \cap B) + d(A \cup B) \leq d(A) + d(B)$ for all $A,B \in P(X)$. -A matroid is said to be representable over a field $k$, if there exists a collection of vectors $\lbrace \xi_x \in V \mid x \in X \rbrace$ of some $k$-vectorspace $V$, such that -$$d(A) = \dim {\rm span}_k \lbrace \xi_x \mid x \in A \rbrace \quad \forall A \in P(X).$$ -It is well-known by results of Tutte, that representability of $M$ over $GF(2)$ and representability over all fields is characterized by certain finite lists of excluded minors that $M$ should not contain. At the same time Vámos has shown that there is no such finite list of excluded minors which characterizes representability over $\mathbb R$. - -Question: What are sufficient conditions for representability of $M$ over $\mathbb R$? - -By Tutte's result, $M$ is representable over any field if $M$ does not contain $U_{24}$, $F_7$ and $F^\ast_7$ as minors. Here, $U_{24}$ denotes the matroid of four points on a line, $F_7$ is the Fano plane and $F^\ast_7$ its dual. The question is whether there is a general result, that describes a larger class of matroids which are representable over $\mathbb R$. - -REPLY [2 votes]: There is a negative answer in terms of excluded minors (this has been somehow hinted in the existing answers): "for any infinite field $\mathbb{F}$, there are infinitely many excluded minors for $\mathbb{F}$-representability." -This is mentioned at the end of section 3 of the survey What is a matroid? by James Oxley, and more precisely stated in theorem 5.9, where an infinite family of forbidden minors for representability over $\mathbb{Q}$, $\mathbb{R}$ or $\mathbb{C}$ is presented.<|endoftext|> -TITLE: A synopsis of Adyan’s solution to the general Burnside problem? -QUESTION [17 upvotes]: Where can I find a high-level overview - of Adyan’s original proof of the existence of finitely generated infinite groups with finite exponent? - -Additionally: - -If possible, would an expert please write a short - synopsis of this proof here on MO? - -Many years ago at Illinois, Ken Appel took notes for lectures on combinatorial group theory given by Adyan. He told me once that Adyan had said that the key thing missing from Britton’s approach to the general Burnside problem was the notion of a cascade. I’d like to know, for example, why this is the case. How, precisely, is Adyan's proof related to Britton's attempt? Is the former a perturbation of the latter in which cascades resolve the inconsistency? If so, can one succinctly say how? -In an English translation of S. I. Adian, “An axiomatic method of constructing groups with given properties”, Uspekhi Mat. Nauk, 32:1(193) (1977), 3–15, we find, regarding the interlocking facts to be proved by simultaneous induction on the rank: -"...as the reader accumulates experience and gradually forms intuitive pictures of the relevant concepts, our proofs become less formal..." -What are these intuitive pictures? (Of course, I don't expect an answer for this.) -Also, - -Is Adyan's notion of cascade used anywhere else in group theory today? (Or has this idea been totally abandoned for Olshanskii's diagrammatic method.) - -I apologize for not including a definition of cascade, but the ability to do so concisely is implicitly part of the question! - -REPLY [28 votes]: A few things: - -Britton did not have a proof at all. -The notion of cascades is central to Novikov-Adyan's proof. That concept was missing in the original announcement of Novikov, and that is why the period between the announcement and the actual proof was so long. Approximately it means the following. You start with a word W and apply relations $u^n=1$ for relatively small $u$. The word changes, of course, but certain part of it stays almost the same. So cascades reflect the stable part of the process. Without them, there is no control on what can happen when you start applying relations. The notion of cascades is not used now, as far as I know (except for the later papers by Adyan and his students): it is too connected with the Burnside problem. -There is no "short description" of Novikov-Adian work. One of the things simplifying reading his text is that you really need to understand the base of induction, the step of induction is much easier (the proof consists of proving hundreds of statements simultaneously using induction on some parameter). A relatively short proof is in Olshanskii's book "Geometry of defining relations in groups", and the new proof of Gromov and Delzant show much more explicitly what is going on. -Olshanskii's method was based on van Kampen diagrams and the key concept (as key as cascades in Novikov-Adian's proof) was the concept of contiguity subdiagram. Contiguity subdiagrams are not related to cascades, but play similar role. They help replacing the classic small cancelation condition. In Gromov-Delzant proof the key concepts are "very small cancelation", "mesoscopic curvature", and "Cartan-Hadamard"-type theorems. These are "big guns" which make proof conceptually easier. -If you really need more explanation, you should contact Igor Lysenok. He knows all three proofs very well. - - Update. I can probably add a little bit more information. The proofs of Adyan and Olshanskii are based on the idea that if $W$ is a six-power-free word (without subwords of the form $u^6$) in the generators of the Burnside group of sufficiently large odd exponent, then $W\ne 1$ in the group. Since there are infinitely many cube-free words in two generators by Thue, there are infinitely many different six-power-free words in the free Burnside group. To prove thatfact, you start inserting words $u^n$ to $W$. Then parts of the powers $u^n$ will possibly cancel but the original word should not be completely destroyed. This simple idea turned out to be very hard to implement. As I said, in order to control what happens to $W$ when we insert the relations, Adyan and Novikov introduced the cascades. -Olshanskii's idea was to consider a possible van Kampen diagram over the presentation of the Burnside group with boundary label $W$. If that presentation satisfied a small cancelation condition, then the boundary of one of the cells would have a large intersection with the boundary of the diagram by the Greendlinger lemma, and $W$ would contain large powers, a contradiction. Unfortunately, the presentation does not satisfy the standard small cancelation condition. So Olshanskii basically introduced his own condition. Although a cell may not be attached to the boundary of the diagram, there exists a subdiagram (called the contiguity subdiagram) which has a form of a rectangle with two long sides and two short sides: one long side on the boundary of the cell, another long side on the boundary of the diagram. He proves that this subdiagram consists of cells that correspond to shorter relations. This alows him to use induction. As in the case of Adyan-Novikov, in order to prove the statement by induction, one needs to first make it stronger. In the case of Adyan-Novikov, the resulting statement consists of many parts which are proved by simultaneous induction, in the case of Olshanskii, there are very few parts. -Update (4/23/2018) A relatively short explanation (a road map) of Olshanskii's proof can be found in my book "Combinatorial algebra: syntax and semantics", Chapter 5.<|endoftext|> -TITLE: The "girthwidth" of a graph -QUESTION [7 upvotes]: Abstractly, tree/path decompositions of a graph $G$ can be thought of as doing the following: - -fix a "skeleton" class of graphs (tree or path) -Pick a member of this class $H$. Associate with each node of $H$ a subset of vertices of $G$ such that (1) For each edge $e = (u,v)$ in $G$, there exists a node of $H$ containing both $u$ and $v$ and (2) the set of nodes of $H$ containing a fixed vertex of $G$ form a connected subgraph -minimize the maximum cardinality of the set associated with a node of $H$, over all such associations and skeletons. - -Suppose that rather than choosing a tree or a path, we are allowed to pick any graph of girth at least $g$, for some fixed parameter $g$. Is there anything known about the resulting "girthwidth" of a graph ? - -REPLY [3 votes]: The (finite) "girthwidth" is always 1 so this is not going to be that useful (as defined here anyway). First some notation: Let a $g$-decomposition of a graph $G$ consist of a graph $H$ with girth at least $g$ and an assignment to each node of $H$ of a set of vertices as above, the width of such a decomposition be one less than the maximum number of vertices assigned to any node of $H$, and $w_g(G)$ be the minimum width of any $g$-decomposition of $G$. Then $w_3(G)$ is the minimum width where $H$ is completely unrestricted. So $w_3(G)=1$ as we can take $H$ to be a complete graph and put the endpoints of each edge in their own node. Since the girth of a tree is infinite a tree decomposition is an $\infty$-decomposition and the tree width is $w_{\infty}$. Actually these are the only widths that matter since $w_3=w_b \le w_{\infty}$ for any finite $3 \le b$: If $a \lt b \le \infty$ then $w_a \le w_b$ simply by the definitions. However if $a \lt b \lt \infty$ then also $w_b \le w_a$ since we can turn an $a$-decomposition into a $2a$-decomposition by putting a new node in the middle of each edge and assigning it the same vertex set as one of its neighbors. -Maybe one could restrict to $H$ being a planar graph although $w_3(G)$ would still be 1 for planar $G$. I wonder when the "cycle-width" is less than the path-width. If one demands that no two nodes get the same vertex set (which is no restriction for tree width) it would get harder (and sometimes impossible if we forbid empty nodes) -A use of tree-width is that many hard (NP complete, say) problems become easy (say polynomial time) when restricted to bounded tree-width. A variant width could be justified merely by its own intrinsic appeal, but would be more motivated if it had similar applications.<|endoftext|> -TITLE: Compactness properties of plurisubharmonic functions -QUESTION [7 upvotes]: I'm quite interested in this topic, but the main text on Several Complex Variables say little of nothing about it. Here are my questions, and I'd be grateful of any reference or information. -Let $\Omega$ be an open subset of $\mathbb{C}^n$ and let $F$ be a family of plurisubharmonic functions on $\Omega.$ We may assume that these functions are continuous or smooth if we wish; also, I'm particularly interested in $n=2.$ -Q1. In what convergence can we get a converging subsequence to a plurisubharmonic function? -Q2. Are there further special conditions on the family $F$ that ensure compactness in stronger sense? - -REPLY [3 votes]: Besides the above-mentioned results in various topologies, the following is quite useful: -Let $\Omega$ be an open subset of $\mathbb{C}^n$. If $\Omega$ is connected and $u_j, j \in \mathbb{N}$ is a decreasing sequence of PSH functions, then $u = \lim_{j \to \infty}u_j \in PSH(\Omega)$ or $u \equiv -\infty$. This is Theorem 2.9.14 (ii) in: M. Klimek: Pluripotential Theory, Clarendon Press 1991. It also appears as Proposition I.3 (iii) in the Appendix I by Russakovskii and Favorov to the Russian translation of P. Lelong, L. Gruman: Entire functions of several complex variables, Springer-Verlag 1986. -The result mentioned by Alex Eremenko can also be found in the main body of Lelong and Gruman's book, as Theorem 1.27 (in a slightly more general form, for upper-semicontinuous regularization of the upper limit (in the sense of filters) of a locally bounded from above family of PSH functions). -The conspiracy is vast indeed, but it seems to be limited to authors of texts in complex analysis.<|endoftext|> -TITLE: Is a smooth action of a semi-simple Lie group linearizable near a stationary point? -QUESTION [17 upvotes]: Suppose that $G$ is a semi-simple Lie group that acts smoothly (i.e., $C^\infty$) on a smooth, finite dimensional manifold $M$. Does it follow that the action of $G$ is linearizable near any stationary point of the action? This was conjectured by Steve Smale and myself in 1965, and was proved for the case that $M$ and the action were analytic by Bob Herman and Guillemin and Sternberg in two papers from long ago: -Hermann, R.: The formal linearization of a semi-simple Lie algebra of vector fields about a singular point. Trans. Am. Math. Soc. 130, 105-109 (1968) -Guillemin, V., Sternberg, S.: Remarks on a paper of Hermann. Trans. Am. Math. Soc. 130,110-116 -(1968) -I have not heard whether any progress has been made since then and I would be interested to hear from anyone who has heard of a proof or a counter-example. The reason is not just idle curiousity; this is the missing step in a proof that what I call The Principle of Symmetric Criticality is valid for smooth finite dimensional actions of a semi-simple group: see (particularly page 29 of) the paper downloadable here: -http://www.springerlink.com/content/wur75t1t65371812/ -for more details on this principle and why it is important, particularly in mathematical physics. - -REPLY [16 votes]: There are smooth counter-examples by Cairns and Ghys [Ens. Math. 43, 1997], for instance a smooth non-linearizable action of $SL(2,\mathbb{R})$ on $\mathbb{R}^3$ (fixing the origin) or of $SL(3,\mathbb{R})$ on $\mathbb{R}^8$. -By contrast, they show that any $C^k$ action of $SL(n,\mathbb{R})$ on $\mathbb{R}^n$ (same $n$, fixing the origin) is $C^k$-linearizable. Here is a link to their paper.<|endoftext|> -TITLE: Re: Mordell's Equation $y^2 = x^3 + k$ and Perfect Numbers -QUESTION [5 upvotes]: I have already tried a somewhat exhaustive search of the literature, but couldn't find anything close to the problem that I am working on. -My question is: When does Mordell's Equation -$$y^2 = x^3 + K$$ -have only FINITELY many solutions over the field of rational numbers, if we allow $K$ itself to be a rational number? -I've seen a "criterion" (i.e. a set of sufficient conditions) related to the class number of the (real/imaginary) quadratic field $\mathbb{Q}(\sqrt{K})$, but it is limited only to $K$ being either 1 or 2 modulo 4. -[The actual "criterion" (as stated in the Japanese[?] paper that I allude to) is: -Mordell's equation $y^2 = x^3 + K$ has finitely many solutions in $\mathbb{Q}$ if -(1) $-K$ is not of the form $3t^2 + 1$ or $3t^2 - 1$; AND -(2) $K \equiv 1 (mod 4)$ or $K \equiv 2 (mod 4)$; AND -(3) $3$ does not divide the class number of the (real/imaginary) quadratic field $Q(\sqrt{K})$.] -Edit: Please refer to this hyperlink for more information as to the context of the previous "criterion". These have since been refuted by Kevin Buzzard (@Kevin - thank YOU!). -Thanks to Kevin for pointing out some of the subtle errors in my original post, particularly in the third condition. I was considering the case $K > 0$ (i.e. for real quadratic fields). -Now for the context: -Let -$$Y = W + Z$$ -and -$$X = WZ$$ -where $W$ and $Z$ are defined as: -$$W = I(p^k) = \frac{\sigma_{1}(p^k)}{p^k}$$ -$$Z = I(m^2) = \frac{\sigma_{1}(m^2)}{m^2}$$ -Let $$N = {p^k}{m^2}$$ be a perfect number. (At this point, we don't have to distinguish between even or odd $N$ because the Euclid-Euler model for perfect numbers fits both cases. For more details regarding this, please refer to this link.) -We "know" that the exponent $k$ allows us to distinguish between even and odd $N$ in the sense that: -(1) If $k$ = 1, then $N$ is even. -(2) If $k$ > 1, then $N$ is odd. (Again, refer to the link for more details. There is also a related MathOverflow post here.) -Thus, a (possibly) feasible and modern approach to the OPN problem (i.e. determining nonexistence or otherwise) will be to try establishing a finiteness result first (for particular values of $K$). -In other words, checking for finiteness of OPNs amounts to checking for finiteness of solutions for Mordell's equation -$$Y^2 = X^3 + K$$ -for particular values of $K$. -And you will only have to check for values of $K$ in the range $[50, 399]$ (for a total of 350 elliptic curves), per the previous answer to this MathOverflow question. -$K$ falls in that range because the sum -$$Y = W + Z$$ -is known to lie in the open interval $(57/20, 3)$. -Of course, the "juicy" implication is that: If you will be able to find a condition (e.g. equation, inequality, etc.) relating $k$ to $K$ and you are also able to FURTHER show that the number of solutions to the corresponding Mordell equation $Y^2 = X^3 + K$ is finite FOR ALL SUCH $K$, then it would follow that there are only finitely many perfect numbers (odd AND even). -Disclaimer: This is a "naive" approach based on my current understanding of elliptic curve theory. I am well-aware, of course, that the rationals are dense over the real numbers. [Edit: In addition, the abundancy indices and the abundancy outlaws are both dense over the rationals.] Which is why I was kinda surprised that there is NO need to assume ("strict") rationality (i.e. $K \in \mathbb{Q}$ but not in $\mathbb{Z}$) for $K$ when checking for finiteness of solutions to Mordell's equation. - -REPLY [6 votes]: Here's another reason why there's unlikely to be a simple answer, unless you can find a simple answer to another question that appears to be very deep. You can try to do a descent to compute the $p$-Selmer group for some small prime $p$, and if that succeeds, you will have computed the rank. But in computing the $2$-Selmer group, you'll need to know the $3$-part of the ideal class group of $\mathbb{Q}(\sqrt{-k})$. Similarly, in computing the $3$-Selmer group, you'll need to know the $2$-part of the ideal class group of $\mathbb{Q}(\sqrt[3]{k})$. I'm not aware of any elementary conditions on $k$, for example, congruence conditions, which would let you determine the ranks of those parts of those ideal class groups. The situation is very different for the curves $E_k:y^2=x^3+kx$, since here there is a rational $2$-torsion point, so the $2$-Selmer group can be computed without working in an extension of $\mathbb{Q}$. That's why, for example, there are some special cases such as $k$ prime and congruent to 7 or 11 modulo 16 for which one knows that the Mordell-Weil rank is zero, so $E_k(\mathbb{Q})$ is finite.<|endoftext|> -TITLE: Applications of knot theory -QUESTION [43 upvotes]: An answer of André Henriques' inspired the following closely related CW question. Parts of the following is extracted from his answer and my comments. - -I regularly teach a knot theory class. Every time, students ask about applications. What should I say? - -I have two off-the-cuff replies when students ask. The first is that knot theory is a treasure chest of examples for several different branches of topology, geometric group theory, and certain flavours of algebra. The second is a list of engineering and scientific applications: untangling DNA, mixing liquids, and the structure of the Sun's corona. I'm interested hearing about other applications. I am also interested in hearing your take on the pedagogical issues involved. Thank you! - -REPLY [10 votes]: It's still a bit of a far cry, but literary scholars are looking for more and more visualization tools to help them compare and contrast large sets of self-similar stories (like the countless variations on the Grail Quest, or the ~350 existing Cinderella stories). One tool I'm trying to develop makes use of braid theory, a subbranch of knot theory, to chart the history of speech-acts between characters -- if strand A passes over strand B, it means character A speaks to character B, and vice versa. The result is often, and expectedly so, messy. But, applied on a single story for all of the speaking characters, it can prove valuable by exposing patterns of speech-acts, moments of conversational dominance (where one thread supersedes all the others), and a number of other visually obvious phenomena that might otherwise have been overlooked. Plays and oral stories prove particularly good subjects. Applied on a slough of stories, a whole data set, especially stories that are supposed to tell the same tale likes the ones mentioned above, it can prove pretty useful. -Check out the linked TEDx talk for more details: -http://www.tedxsmu.org/talks/arnaud-zimmern-braiding-red-riding-hood-tedxsmu-spring-2014/<|endoftext|> -TITLE: Parametric vs Non-parametric Estimation of Quantiles -QUESTION [7 upvotes]: Motivation -Suppose that we need to estimate the median from a normal distribution with known variance. One non-parametric approach is to use the sample median as an estimator. However, this does not take advantage of our distributional assumptions. Exploiting, the symmetry of the normal distribution, we can, instead, use the sample mean as an estimator of the median. This new estimator is more efficient than the sample median. In fact, it is known that the median is asymptotic inefficient in normal sampling, and the relative efficiency is given by $\frac \pi 2$. -The previous paragraph suggests a similar approach for any given quantile: we estimate the mean, fit a normal distribution, and analytically compute the quantile for the estimated distribution. It is not hard to show that the relative efficiency of the non-parametric estimator is worse than $\frac \pi 2$. -Does the previous result holds in a more general sense? -Problem -Consider a sample $X_1,\ldots, X_n$ drawn independently from a population with p.d.f. $f(x|\theta)$, where $\theta$ is a scalar unknown parameter to be estimated. We impose the following regularity conditions for the distribution: (i) the p.d.f. is continuously differentiable in $\theta$, (ii) the c.d.f. $F(x|\theta)$ is strictly monotonic, (iii) the distributions $f(x|\theta)$ have common support. -We are interested in estimating the $p-$quantile, which is given $q=F^{-1}(p|\theta)$. We are going to consider two estimators. The first one is the sample quantile. The second is going to be our parametric estimator, and is constructed as follows. First, we obtain the maximum likelihood estimator of our unknown parameter $\theta$. Then, we compute the quantile analytically using our fitted distribution. -Non-parametric estimator. Our non-parametric estimator is the sample $p-$quantile, which we denote by $\hat{q}_n$. From Bahadurs’ representation, we know that our estimator is asymptotically, normal and verifies that -$$ - \sqrt {n} (\hat{q}_m - q) \Rightarrow \mathcal{N}(0,u^\prime(\theta)), -$$ -where the variance is $u^\prime(\theta) = \frac {p(1 - p)} {f(q|\theta)^2}$. -Parametric estimator. Next, let $\hat{\theta}_n^{\rm mle}$ be the maximum likelihood estimator of the unknown parameter. The parametric estimator is given by $\hat{q}_n^{\rm mle}=F^{-1}(p|\hat{\theta}_n^{\rm mle})$. In turn, by the invariance property of the MLE, it is the case that $\hat{q}_n^{\rm mle}$ is the maximum likelihood estimator of the quantile $q$. As a consequence, under some regularity conditions, we have that our new estimator is asymptotically normal -$$ - \sqrt{n} (\hat{q}_n^{\rm mle} - q) \Rightarrow \mathcal{N}(0,u(\theta)), -$$ -where the $u(\theta)$ is the Cramer-Rao lower bound on the variance of any unbiased estimator. That is, $\hat{q}_n^{\rm mle}$ is a consistent, and asymptotically efficient estimator of $q$. The Cramer-Rao lower bound is $u(\theta) = \left( \frac {\partial F^{-1}} {\partial \theta} (p | \theta) \right)^2 I(\theta)^{-1}$, where $I(\theta)$ is the Fisher information of parameter $\theta$. -Analysis. Both the parametric and the non-parametric estimators converge, as the number of samples increases, to the true quantile. We want to study the efficiency, as given by the variance, of these estimators. The relative efficiency of the non-parametric estimator to the MLE, denoted by $\varepsilon(\theta)$, is -$$ - \varepsilon(\theta) = \frac {u^\prime(\theta)} {u(\theta)} = p (1 - p) I(\theta) \left( \frac {\partial F} {\partial \theta} (v | \theta) \right)^{-2}, -$$ -where we have used the fact that $\frac {\partial F^{-1}} {\partial \theta} (p | \theta) = - \frac {\partial F} {\partial \theta} (q | \theta) / f(q|\theta)$. In view of Cramer-Rao lower bound, we have that $\varepsilon(\theta) \ge 1$. -An example. To fix ideas we consider a simple example. Suppose that $X \sim \exp(\theta)$. The maximum likelihood estimator is given by $\hat{\theta}_n^{\rm mle} = \left( \frac 1 n \sum_{i=1}^n X_i \right)^{-1}$, and the Fisher information is $I(\theta) = \theta^{-2}$. The true quantile is $v = - \theta^{-1} \ln (1-p)$. Hence, the efficiency is $\varepsilon(\theta) = \frac {p} {(1-p) \ln^2(1-p)}$. In this case, the efficiency is lower bounded by $\varepsilon(\theta) \ge 1.544$. The lower bound is tight, and attained at $p \approx 0.80$. - -Questions - -What is it known of the parametric estimation of quantiles? -In experiments with some common distributions we have seen that $\varepsilon(\theta) \ge \frac 3 2$. Is it possible to prove, under some assumptions, a more tighter bound for the relative efficiency than $\varepsilon(\theta) \ge 1$? - -REPLY [2 votes]: Just in case someone is following, I want to post a somewhat negative answer to my second question. I found an example that satisfies the assumptions, and achieves an efficiency arbitrarily close to 1. -The example is inspired on Laplace distribution with an unknown location parameter $\theta$, and p.d.f. $f(x|\theta) = \frac 1 2 e^{-|x-\theta|}$. In this case, when $p=\frac 1 2$, both estimators coincide and the efficiency is 1. This is due to the fact that (i) the MLE of the location parameter of a Laplace distribution is the median, and (i) the distribution is symmetric. -The problem with Laplace distribution is that it does not satisfy our assumptions: its log likelihood is not differentiable because of the absolute value in the exponent. The trick is to replace the absolute value by an analytic approximation, such as $\frac 1 k \ln(\cosh(k x))$, which converges point-wise to the absolute value as $k\rightarrow\infty$. Indeed, the sequence of distributions given by -$$ -f_k(x|\theta) = \frac {a_k} 2 \cosh(k (x-\theta))^{- 1 / k}, -$$ -where $a_k$ is a normalization constant, achieve an efficiency of 1 as $k$ goes to infinity. -Bummer.<|endoftext|> -TITLE: Relationship between the cohomology of a group and the cohomology of its associated Lie algebra -QUESTION [20 upvotes]: Let $G$ be a group and let $k$ be a field (characteristic 0 if you want). Let $L$ be the graded Lie ring associated to the lower central series of $G$, that is, $L$, as a graded abelian group is $\oplus_{i \geq 1} G_{i}/G_{i+1}$ where $G_1 = G$, $G_i = (G_{i-1}, G)$, and the Lie bracket is induced by the commutator $(a,b)$ on $G$. Tensor $L$ with $_{\mathbb Z}k$ to get a Lie algebra $\hat L$ over $k$. -Is there any relationship between the cohomology algebra of $G$, $H^{*}(G, k)$, and the cohomology of the (graded) Lie algebra $\hat L$, $H^{*}(\hat L, k)$? -I realize I am not being precise about what category these cohomologies are being computed in, as in graded versus ungraded. I am hoping someone can tell me there is a relationship or that there is no connection. Perhaps there are conditions on $G$ that ensure these cohomology algebras are isomorphic? References? - -REPLY [3 votes]: I just happened to stumble upon this old question. The checked answer -refers to a 1999 paper for a spectral sequence that was studied in a -1994 paper that addresses the question at hand: -Annetta Bajer. The May spectral sequence for a finite $p$-group stops. -J. Algebra 167 (1994), no. 2, 448–459. -Here is the Math Review of that paper: ``Let $k$ be a field of characteristic $p>0$, let $G$ be a finite $p$-group, and let $gr kG$ be the graded $k$-algebra associated to the filtration of $kG$ by the powers of its augmentation ideal. The May spectral sequence of the title is a spectral sequence whose $E_2$-page is $Ext_{gr kG}(k,k)$ which converges to a filtration of $H^{*}(G,k)=Ext_{kG}(k,k)$. The author shows that this spectral sequence has only finitely many nonzero differentials, and deduces a similar result for a related spectral sequence converging to a filtration of $H^*(G,M)$ for $M$ a finitely generated $kG$-module.'' -The published reference for the MSS in general goes back to 1966: The cohomology of restricted Lie algebras and of Hopf algebras. -J. Algebra (1966), 123--146. ([3] on my web page).<|endoftext|> -TITLE: Given a cardinal k, what's the biggest dense linear order with a dense subset of size k? -QUESTION [8 upvotes]: It's not hard to show that for any cardinal $\kappa$, there is no dense linear order without endpoints (DLO) of size greater than $2^{\kappa}$ that has a dense subset of size $\kappa$. But one can show that if $\mu$ is the least cardinal such that $\kappa^{\mu} > \kappa$, then there's a DLO of size $\kappa^{\mu}$ with a dense subset of size $\kappa$, so in particular there's a dense linear order of size greater than $\kappa$ with a dense subset of size $\kappa$. Thus in models of GCH, the answer to the question in the title is "as big as possible." Also from ZFC alone, the case $\kappa = \omega$ also has the answer "as big as possible": just consider $\mathbb{Q}$ and $\mathbb{R}$. -So the question is: Is it consistent that for some $\kappa$, there is no DLO of size $2^{\kappa}$ with a dense subset of size $\kappa$? -Apparently the answer's supposed to be "yes," and the relevant notion is that of a "Dedekind number," but searches for "Dedekind number" haven't yielded anything relevant, so if someone could point me to a relevent reference, that'd be great too. - -Unless I made a mistake, I've shown that a DLO $(L,\leq)$ is a DLO with a dense subset of size $\kappa$ iff it is (up to isomorphism) some set $X \subset \mathcal{P}(\kappa)$ ordered by inclusion such that $(X,\subseteq)$ satisfies the following property: - -$(*)\ \ \forall \alpha < \kappa$, the set $\{ x \in X : \alpha \notin x \}$ has a least upper bound, let's call it $x _{\alpha}$, in $(X,\subseteq)$ - -Furthermore, I've shown that given $X \subset \mathcal{P}(\kappa)$ which is a DLO under inclusion (and $\cup X = \kappa$, and $|X| > \kappa$), it can be modified to give $X' \subset \mathcal{P}(\kappa)$ which is also a DLO under inclusion, has the same size as $X$, and has a dense subset of size $\kappa$: First, obtain $X_0$ from $X$ by removing, for each $\alpha < \kappa$, the greatest lower bound in X (if it exists) of the collection $\{x \in X : \alpha \in x\}$; Next obtain $X'$ from $X_0$ by adding, for each $\alpha < \kappa$, the set $x _{\alpha} = \cup \{x \in X_0 : \alpha \notin x\}$. -So an equivalent reformulation of the question is: Is it consistent that for some $\kappa$, there is no $X \subset \mathcal{P}(\kappa)$ of size $2^{\kappa}$ which forms a DLO under inclusion? - -REPLY [6 votes]: This is very late, but in 2012, Artem Chernikov, Itay Kaplan and Saharon Shelah posted on arXiv a paper titled "On non-forking spectra" ( http://arxiv.org/abs/1205.3101 ). They claim that it is consistent that $Ded(\kappa)< Ded(\kappa)^\omega$. In particular, it is consistent that $Ded(\kappa)<2^\kappa$. -It is still open whether both inequalities can be made strict: $Ded(\kappa) -TITLE: How should I publish my "story"? i.e. strategies and advice on chopping up a thesis -QUESTION [12 upvotes]: Background -In my thesis I look at same problem from a couple of different angles. To state it roughly, in each chapter I use a different technique or area of mathematics to try and gain further insight into a hard case that was shown to have a "negative" result in the early 90s. I believe my main contribution is a few cute insights/tricks and building connections between a couple of different areas/problems in mathematics. -Chopping it -I think I have about about 4-5 publishable chapters in my thesis but they all consider the same problem. - -Do I send them all to the same journal and let them take their pick? or different journals? but then how do I choose which chapter to which journal? - -My topic is 'probabilistic' and I think my results are novel. Unfortunately, I don't really know where to submit. Some related problems and theory have previously appeared in the Annals of Probability and Journal of Functional Analysis. Although novel, I don't really know if my work is of that technical calibre though. What other journals should I consider? Should I post on ArXiv first? -My favourite result is a very short chapter where I derive some cute estimates for a special case by "bare hands" without any big machinery. It is totally non-standard estimate but shows a new type of question one might pose about my problem (and also other problems). The paper would only be about 8-10 pages long, is this worthy to submit? -Timing -I think my ideas have a lot of potential and would like to submit to a top journal, but due to timing constraints and working by myself, I still think my work is a bit "shallow" (i.e. easy to extend). Ideally, I would like to have a good co-author to refine and deepen the ideas and turn it into a great publication but every time I mention my example problem it seems like I inspire them and shortly after they produce a paper with their co-authors on a similar or more abstract result. Very frustrating as I don't have the time to do these extensions myself at the moment! -I'm aiming to submit my thesis in about a month, then I will spend a couple of weeks putting the chapters into a publication format for submission to arXiv or a journal. - -Should I submit to arXiv or journals before I receive my thesis referee reports back? or wait until I receive their comments? - -Thanks in advance for your advice :) - -REPLY [7 votes]: Caveat: I just finished my PhD, and I have spent the bulk of this semester converting my dissertation to papers. The best advice is Pete Clark's: Ask your thesis advisor for advice. Nonetheless, I'll try to answer your questions to the best of my ability. -Should I submit to arXiv or journals before I receive my thesis referee reports back? or wait until I receive their comments? -It can't hurt to stick the entire thesis on the arXiv right now, then update it when you've completely finished the thesis. If you're worried about any copyright issues, this is the place to put it. -Do I send them all to the same journal and let them take their pick? or different journals? but then how do I choose which chapter to which journal? -Absolutely not. You shouldn't publish your thesis, per se. You should write and publish good papers, which will be based on the work in your thesis. -You write a thesis because it is a formal document certifying you have earned your degree. You write papers in order to communicate ideas and methods to other mathematicians. -How do you choose which journal to submit something to? Ask your thesis advisor for advice. -If you think your work is "shallow", then it's not appropriate for a top journal. That's okay: there's no shame in submitting it to a lower-tier journal. The important thing is to get it out there ASAP. You can then spend your time and energy on doing something totally new and deep then submit that to a top journal. -My favourite result is a very short chapter where I derive some cute estimates for a special case by "bare hands" without any big machinery. It is totally non-standard estimate but shows a new type of question one might pose about my problem (and also other problems). The paper would only be about 8-10 pages long, is this worthy to submit?It is absolutely worthy to submit. Is it worthy to be published? Well, that's between you, the referee and the editor of the journal you submit it to. -Since you have 5 chapters, here's my suggestion. Turn your favorite chapter into a publication first. Then condense three of the other chapters into a second publication. Don't publish the last chapter outright. Instead of answering the same question a fifth time, use the method you developed in the last chapter to answer a new question, then publish that.<|endoftext|> -TITLE: Recursive Non-Well-Orders that are Sneaky, but not THAT Sneaky. -QUESTION [6 upvotes]: This is a variant on - Sneaky Recursive Non-Well-Orders -where it was asked - -Is there a recursive function $f$ such that whenever $a\in\mathcal{O}$, $f(a)$ is a Turing index for a linear non-well-order with no $H_a$ -computable descending chain? - -The answer to the original question gave a single non-well-order with no hyperarithmetic descending chain at all. Instead can $f(a)$ be a non-well-order with an $H_a$-computable descending chain but no $H_b$-computable descending chain for $b <_{\mathcal{O}} a$ ? - -REPLY [3 votes]: Here is something close to what you ask for. Hopefully others can improve it. -Step 1. For each $a$ there is a tree $T_a\subseteq \mathbb{N}^{<\mathbb{N}}$ which has a unique path $P$, and $P\equiv_T H_a$. -To construct $T_a$, use the fact that $H_a$ is a $\Pi^0_2$ singleton, that is $$X = H_a \iff \forall n \exists k R(X\upharpoonright k, n, a),$$ where $R$ is a recursive predicate. (For details see Higher Recursion Theory by Sacks, Chapter II, page 37.) Define \begin{align*} T_a = \{ (\sigma_1, \dots, \sigma_r) : \sigma_1 \preccurlyeq \dots \preccurlyeq \sigma_r \text{ and } \forall i\leq r (\exists k \leq |\sigma_i| R(\sigma_i\upharpoonright k, i) \text{ and no } \tau\prec\sigma_i \\ \text{ with } \sigma_{i-1} \preccurlyeq \tau \text{ has this property)}\}.\end{align*} Then $T_a$ has exactly one path $P$, and $\cup P = H_a$. One can check that $P\equiv_T H_a$. This construction is recursive in $a$. -Step 2. Define $f(a)$ to be the Kleene-Brouwer ordering on the tree $T_a$. -Then $H_a$ computes a descending sequence because it computes the unique path $P$. I suspect a weaker oracle could compute a descending sequence, but if $a$ is a successor notation, $a=2^b$, then I don't know whether $H_b$ does. If $D = \overline{\sigma}^1, \overline{\sigma}^2, \dots$ is a descending sequence from $T_a$ in the KB ordering, then $\lim_{i\rightarrow \infty} \overline{\sigma}^i = P$, so $D' \geq_T H_a$.<|endoftext|> -TITLE: If $f:R^n \to R$ is a smooth real-valued function such that $\nabla f : R^n \to R^n$ is a diffeomorphism, what can one conclude about the behavior of $f(x)$ at infinity? -QUESTION [20 upvotes]: This question may seem peculiar, so let me preface it by saying that it arose while I was trying to understand Legendre transformations better, and in that context it is fairly natural. Anyway, suppose that $f$ is a smooth real-valued function on $R^n$ such that the gradient map, $\nabla f: - p \mapsto {\partial f \over \partial x_i}(p)$, - is a diffeomorphism of $R^n$ with itself. Of course two necessary conditions for this are that: - (1) the hessian matrix ${\partial^2 f \over \partial x_i \partial x_j}(p)$ - is everywhere non-singular, and - (2) $\nabla f$ is a proper map, i.e., - if $M > 0$ the the set where $ ||\nabla f|| \le M $ is compact. Moreover, it - is not hard to show that these two conditions are together sufficient for the - gradient map to be a diffeomorphism. Now my question is this: if $f$ is such a function does it follow that $f$ is also proper, i.e., that - $\lim_{||x||\to \infty} |f(x)| = \infty$ ? That's clearly so if $n = 1$, but that is a very special case. For general $n$ I hope someone can show me a simple proof (but I also wouldn't be too surprised if I were shown a simple counter-example). -Added in response to Theo's simple and very nice counter-example: Suppose that the hessian is not only everywhere non-singular, but even everywhere positive definite. Can one then deduce that $f$ is proper? - -REPLY [27 votes]: The map $f(x,y) = xy$ has $\nabla f(x,y) = \left(\begin{array}{c} y \\\ x\end{array}\right)$ but $f(x,0) \equiv 0$, so $f$ is not proper. - -Edit in response to the modified question: -Positive definiteness of the Hessian implies strict convexity of $f$ and this indeed implies properness of $f$ as follows: -Since you assume that $\nabla f: {\mathbb R}^{n} \to {\mathbb R}^{n}$ is a diffeomorphism, $f$ has a unique minimum, namely the point $x_{0}$ where $\nabla f(x_0) = 0$. If $f$ were not proper, there would be a constant $M$ such that the closed convex set $C = \{x\, :\, f(x) \leq M\}$ is not compact. But then you would find a direction $y \in \mathbb{R}^{n} \smallsetminus \{0\}$ such that $x_{0} + t y \in C$ for all $t \geq 0$. The function $t \mapsto f(x_{0} + t y)$ is bounded and convex on $\mathbb R_{+}$ and it assumes its minimum. Thus it must be constant, in contradiction to the fact that $x_0$ is the unique minimum.<|endoftext|> -TITLE: Are innermost reductions perpetual in untyped $\lambda$-calculus? -QUESTION [6 upvotes]: Background -In the untyped lambda calculus, a term may contain many redexes, and -different choices about which one to reduce may produce wildly -different results (e.g. $(\lambda x.y)((\lambda x.xx)\lambda x.xx)$ -which in one step ($\beta$-)reduces either to $y$ or to itself). -Different (sequences of) choices of where to reduce are called -reduction strategies. A term $t$ is said to be normalising if -there exists a reduction strategy which brings $t$ to normal form. -A term $t$ is said to be strongly normalising if every reduction -strategy brings $t$ to normal form. (I'm not worried about which, but -confluence guarantees there can't be more than one possibility.) -A reduction strategy is said to be normalising (and is in some sense -best possible) if whenever $t$ has a normal form, then that's where -we'll end up. The leftmost-outermost strategy is normalising. -At the other end of the spectrum, a reduction strategy is said to be -perpetual (and is in some sense worst possible) if whenever there -is an infinite reduction sequence from a term $t$, then the strategy -finds such a sequence - in other words, we could possibly fail to -normalise, then we will. -I know of the perpetual reduction strategies $F_\infty$ and $F_{bk}$ -given respectively by: -\begin{array}{ll} - F_{bk}(C[(\lambda x.s)t]) = C[s[t/x]] & \text{if $t$ is strongly normalising}\\\\ - F_{bk}(C[(\lambda x.s)t]) = C[(\lambda x.s)F_{bk}(t)] &\text{otherwise} -\end{array} -and -\begin{array}{ll} - F_\infty(C[(\lambda x.s)t]) = C[s[t/x]] &\text{if $x$ occurs in $s$, or if $t$ is on normal form}\\\\ - F_\infty(C[(\lambda x.s)t]) = C[(\lambda x.s)F_\infty(t)] &\text{otherwise} -\end{array} -(In both cases, the indicated $\beta$-redex is the leftmost one in -the term $C[(\lambda x.s)t]$ - and on normal forms, reduction -strategies are necessarily identity.) The strategy $F_\infty$ is even -maximal - if it normalises a term, then it has used a longest -possible reduction sequence to do so. (See e.g. 13.4 in Barendregt's -book.) -Consider now the leftmost-innermost reduction strategy. Informally, -it will only reduce a $\beta$-redex which contains no other redexes. -More formally, it is defined by -\begin{array}{ll} - L(t) = t - &\text{if $t$ on normal form}\\\\ - L(\lambda x.s) = \lambda x. L(s) - &\text{for $s$ not on normal form}\\\\ - L(st) = L(s)t - &\text{for $s$ not on normal form}\\\\ - L(st) = s L(t) - &\text{if $s$, but not $t$ is on normal form}\\\\ - L((\lambda x. s)t) = s[t/x] - &\text{if $s$, $t$ both on normal form} -\end{array} - -The natural intuition for leftmost-innermost reduction is that it will -do all the work - no redex can be lost, and so it ought to be -perpetual. Since the corresponding strategy is perpetual for -(untyped) combinatory logic (innermost reductions are perpetual for -all orthogonal TRWs), this doesn't feel like completely unfettered -blue-eyed optimism... - -Is leftmost-innermost reduction a perpetual strategy for the untyped $\lambda$-calculus? - -If the answer turns out to be 'no', a pointer to a counterexample -would be very interesting too. - -REPLY [3 votes]: Over at cstheory.stackexchange.com, Radu GRIGore provided the following very nice counterexample: -Let $t$ be the term $t=(\lambda x. (\lambda y.z) (x x))$. Then the term $tt$ normalises under $L$ (in fact, $L(tt) = (\lambda x.z)t$), but it is clearly not strongly normalising, since $F_\infty(tt) = (\lambda y.z)(tt)$.<|endoftext|> -TITLE: Quotient of two Laplace integrals -QUESTION [6 upvotes]: Let $\varphi(x)$ and $\psi(x)$ be two -complex-valued continuous functions on $[a,b]$, and let $f(x)$ be -a complex-valued continuously differentiable function on $[a,b]$. -Suppose that $|f(x)|$ has an absolute maximum at an interior point, -say $\xi$, of the interval. Prove or disprove -\begin{equation}\label{eq3} -\lim_{n\to\infty}\frac{\int_a^b\varphi(x)[f(x)]^ndx}{\int_a^b\psi(x)[f(x)]^ndx}=\frac{\varphi(\xi)}{\psi(\xi)}. -\end{equation} -Remark 1: This is true for $f(x)\in C^2$, by Laplace's method. -Remark 2: Micheal has given a counter example, which makes the use of the fact $f'(\xi)\neq0$. This is a good example. Now, if we further assume that $f'(\xi)=0$, and also we assume that $\phi\neq0$ and $\psi(\xi)\neq0$. how about now? I believe this will be more difficult. -Remark 3: Sorry for my question style. Well, this is not just a home work. It's an open problem, when one try to prove a theorem of Chung and Erd\"os 1951. That theorem is essentially said that the ratio of two coefficients in Fourier series of $f(x)^n$ will tend to 1. Where the assumption on $f$ could be translated as $f\in C^1[-\pi,\pi]$, $f(0)=1$, $f'(0)=0$ and $|f(x)|<1$ for $x\neq0$. This theorem will be a corollary if the limit is true here. So I make a further assumption in Remark 2 that $f'(\xi)\neq0$. -Thank you. - -REPLY [7 votes]: This isn't necessarily true. Let $\phi(x) = \cos(x)$ and $\psi(x) = \cos(2x)$. Let $f(x)$ be a complex-valued function such that $|f(x)|$ has its absolute maximum at some $\xi$ for which the (complex-valued) derivative $f'(x)$ is nonzero at $\xi$. Then one can integrate by parts to get -$$\int_0^{\pi} \cos(x)f(x)^n dx = n\int_0^{\pi}\sin(x)f'(x)f(x)^{n-1} dx$$ -$$\int_0^{\pi} \cos(2x)f(x)^n dx = {n \over 2}\int_0^{\pi}\sin(2x)f'(x) f(x)^{n-1}dx$$ -If your statement were true, then by looking at the ratio of the left-hand sides and taking limits as $n$ goes to infinity one should get ${\cos(\xi) \over \cos(2\xi)}$, while looking at the ratio of the right-hand sides and taking limits as $n$ goes to infinity one should get ${2 \sin(\xi) \over \sin(2\xi)}$, which is generally not the same.<|endoftext|> -TITLE: Associativity with infinite nesting -QUESTION [7 upvotes]: I was trying to understand the Eilenberg-Mazur swindle (which I learned about here) especially as it could be used to show that if $A, B$ are compact (topological) $n$-manifolds whose connect sum is $S^n$ (i,e. $A \# B= S^n$) then $A=B=S^n$. -The trick seems to make use of the fact that: -(1) The infinite connect sum (when properly defined) of spheres is a non-compact manifold with one end (indeed is euclidean space). -(2) The infinite connect sum is associative in that any placement of parentheses that are not infinitely nested does not change the infinite connect sum. -I was confused for a bit because if one allows for infinitely nested parentheses one seems to be able to get a non-compact manifold with an arbitrary number of ends. The idea here is that with infinitely nested parentheses one could add spheres "on alternating sides" and so produce a cylinder i.e. a manifold with two ends. -On the other hand, if one is looking at absolutely convergent sums of real numbers then it would appear that one can allow infinite nesting of parentheses without issue. Though I'm not clear on what happens for conditionally convergent sums. -My question is whether my intuition about what can go wrong with infinite nesting justified? (I am coming at it from a very geometric point of view which may be a problem). What is the right framework to think about when one is asking whether infinite nesting is possible or note? I could imagine there is some sort of abstract algebra/category theoretic context in which such things are studied. -I apologize if this is sort a trivial question but I haven't thought much about algebra in a long time. (P.S I didn't know how to categorize this question so made a rough stab at it). - -REPLY [3 votes]: First, it's important that the infinite connect sum $A\# B\#A\#\cdots$ is not the limit of the finite connect sums $A,A\#B, A\#B\# A,\dots$; in fact, I'm sure the binary connect sum is as wrong a notation for the infinite connect sum as $+$ is the wrong notation for $\sum$. Similarly, a conditionally-convergent infinite sum of numbers is the limit of a particular sequence: the initial finite sums; whereas an absolutely convergent sum of real numbers $x_i$ is a (sum of) limit(s) for much bigger diagrams: -$$ \sum x_i = \sup_{W\ \mbox{finite}} \sum_{i\in W} x_i + \inf_{V\ \mbox{finite}} \sum_{j\in V} x_j$$ -Looking for a better description of our infinite connect sum, let's first note that $A\# B$ means a particular colimit -$$\begin{array}{c} - {} & & S^n & & \\\\ -&\swarrow & & \searrow & \\\\ -Q & & & & R \\\\ -& \searrow & & \swarrow & \\\\ -& & Q\sqcup_{S^n} R & & -\end{array}$$ -Note that $A$ and $B$ are also colimits, $A= Q \sqcup_{S^n} D$ and $B=D\sqcup_{S^n} R$. In fact, it's best to keep things open at both ends: for $X$ and two maps $S^n\rightrightarrows X$, let $Q = D\sqcup_{S^n}X$ and similarly define $R$ in terms of $Y$ and two maps. -Instead of the infiite connect sum, we have an infinite diagram -$$\begin{array}{c} - {} & & S^n & & & & S^n & & \\\\ -&\swarrow & & \searrow & &\swarrow & & \searrow & & \swarrow \cdots \\\\ -X & & & & Y & & & & X \\\\ -\end{array}$$ -and its colimit. -The two-ended construction you describe is (if I understand you correctly) a colimit for a different diagram. The fact that you'll get the homeomorphic spaces in the colimit spot for these two diagrams is a nontrivial fact, and is where you need to use the hypothesis $X\sqcup_{S^n} Y \simeq I\times S^n$. -In summary, to test your intuition, pay attention to what limit (or limits) you're considering, and then see if additional hypotheses let you decide on or against equivalence.<|endoftext|> -TITLE: More open problems -QUESTION [79 upvotes]: Open Problem Garden and Wikipedia are good resources for more or less famous open problems. But many mathematicians will be happy with more specialized problems. They may want to find a research theme, e.g. for their PhD thesis, or they may have one, and want to connect their work to other problems, to find applications. Or they may simply want to check if something is already done about a particular question. -So, I would like to ask you (I apologize in advance if the question is not appropriate for MO): - -can you suggest some links to other compilations of open problems, even if they are very specialized and not as famous? - -Please note that I would like it to be large and specialized, not to contain only famous conjectures. I think each subject in mathematics has many such open problems. It would be useful to be maintained such a list, containing problems classified on subject. It would be useful to be able to check there if new progress was made, who is working to those particular problems, and how important are these problems (e.g. rated as in Open Problem Garden). -Thank you. - -REPLY [4 votes]: Igor Shparlinski maintains a large list of open problems in exponential and character sums. You can find it on his website: http://web.science.mq.edu.au/~igor/ - -REPLY [3 votes]: Miller has a list of set theory problems here (upper right). -Schindler also has a list of open problems specifically in inner model theory here.<|endoftext|> -TITLE: Proper classes and their consequences -QUESTION [9 upvotes]: I have two main questions: - -What is a proper class? I've read that it's collection of objects that's "too big" to be a set, but in what sense is such a collection "too big"? Since I'd like this post to be accessible to people who aren't necessarily with the intricacies of symbolic logic, it'd be much appreciated if any symbolic logic expression be accompanied by an explanation of what it means and maybe a quick example of how it works (e.g. "One consequence of this axiom is that objects such as x={x}, which isn't disjoint from any of its elements, can't be a set. However, it [is/is not] a class because..."). -Beyond the implications for philosophy of mathematics, would most mathematicians ever need to worry about proper classes vs. sets? If so, when and why? If not, what types of mathematicians would need to worry about this distinction? - -REPLY [2 votes]: Let me restrict and better formalize this question to ask, "If we have some first-order formula $\varphi(x)$ expressible (with parameters) in the language of set theory, when will $\lnot \exists x \varphi(x)$ be true?" In other words, when will we have an object that cannot exist as a set when sufficiently described? For example, the question of whether the class of all sets is proper is equivalent to the question of whether $\lnot \exists x \varphi(x)$ where $\varphi(x) \equiv \forall y(y \in x)$. Since this statement is easily provable in ZFC, we say that the class of all sets is indeed proper. As another example, we can consider the existence of a generic filter for some partial order as mentioned by Adam. In this case, we could fix a partial order $\mathbb{P}$ and have $\varphi(x)$ be the statement "$x$ is a filter on $\mathbb{P}$ intersecting all dense subsets of $\mathbb{P}$.'' There may be many such $x$ external to the universe that satisfy these criteria, but if $\mathbb{P}$ is a nontrivial forcing notion such as the forcing to add a Cohen Real, we will be able to prove $\lnot \exists x \varphi(x)$ so that any such filter $x$ cannot be a set in the universe. Thus any such generic object must be a proper class. -In this sense, I would characterize a class as being proper when its existence as an object in the universe would cause set theory to be inconsistent. Since we consider a wide variety of models in set theory today (some which may even be sets themselves), I would argue that "too big" is a somewhat outdated intuition for characterizing all proper classes that can arise. For example, if we consider a countable model $M$ of ZFC, we can force to add a wide variety of sets to it that are proper classes of $M$. Also, if $0^{\sharp}$ (a Real number encoding truth in $L$) exists, then this will be a proper class of the constructible universe $L$ containing all ordinals. I therefore claim that a proper class is merely an object that reveals too much about the universe in question to be a set in this universe. -In light of these considerations, I would always consider proper classes in terms of specific models of a theory. In this sense, proper classes are ubiquitous. For example, the set of all primes in any model of Peano arithmetic would be a proper class in the context of this model. Pure mathematicians may need to consider questions about proper classes of models in proofs. For example, when Andrew Wiles proved Fermat's Last Theorem, he appealed to the existence of a Grothendieck universe, which is a proper class of any set-theoretic universe without an inaccessible cardinal. Since we cannot even prove the consistency of the existence of such a cardinal in ZFC, a mathematician should be aware that he/she may be appealing to a stronger theory than set theory alone when assuming that Fermat's Last Theorem is true (although the discussion on Inaccessible cardinals and Andrew Wiles's proof seems to suggest this large cardinal assumption was inessential). As Greg pointed out, it may be sufficient to only consider universes with an inaccessible cardinal for most mathematics thereby eliminating the need to consider proper classes. However, one never knows when a new theorem may rely on the use of large cardinals with even greater consistency strength. For example, there is a result in category theory appealing to the existence of a cardinal with very large consistency strength (Vopěnka's Principle). There are also proper class considerations to be made in determining whether it even makes sense to talk about a conjecture. (i.e., is it formalizable in the language?) -Finally, as a point of clarification to Greg's post, $\aleph_0$ is technically not an inaccessible cardinal under the standard definition I'm accustomed to, which requires the cardinal to be uncountable. The reason for this is because inaccessibility is considered a large cardinal notion, whose consistency should always be independent of ZFC. Since the (consistency of the) existence of $\aleph_0$ is provable from ZFC, it should not be characterized as a large cardinal. However, as noted, it could be considered a large cardinal in the absence of the axiom of infinity.<|endoftext|> -TITLE: Ranicki symmetric L-groups of finite fields? -QUESTION [10 upvotes]: Can anyone tell me what the Ranicki symmetric L-groups $L^*(F)$ are when $F$ is a finite field? (and maybe provide a reference?) Thanks! - -REPLY [10 votes]: The symmetric $L$-group $L^*(F)$ of a field $F$ are 4-periodic, -$$L^n(F)=L^{n+4}(F)$$ -by Proposition 7.1 of -http://www.maths.ed.ac.uk/~aar/papers/ats1.pdf -$L^{2i}(F)$ is the Witt group of $(-)^i$-symmetric forms: -see Milnor and Husemoller! -$L^{2i+1}(F)=0$, see -http://www.maths.ed.ac.uk/~aar/papers/simple.pdf -(my shortest paper).<|endoftext|> -TITLE: The Dirichlet heat semigroup, $L^1_\delta$, and the dimension shift phenomenon -QUESTION [8 upvotes]: In relation to the question on the Hardy inequality and the answer by Terry Tao, I've always been curious about the following: -Let $U \subset \mathbb{R}^n$ be a bounded domain of class $C^2$, $(e^{-t A})_{t \ge 0}$ be the Dirichlet heat semigroup(s) on $L^p(U)$, $1 \le p \le \infty$. $A$ is the Dirichlet Laplacian (i.e. zero boundary conditions). Compare the following (where $\lesssim$ hides a constant dependent on $p,q,U$): - -For $\varphi \in L^p(U)$ and $1\le p \le q < \infty$, we have $$\| e^{-t A} \varphi\|_{q} \lesssim \|\varphi\|_p t^{-\frac{n}{2}(\frac{1}{p}-\frac{1}{q})}, \quad t >0.$$ - -and now on the weighted spaces $L^p(U,\delta)$ where $\delta(x):=\text{dist}(x,\partial U)$, $1\le p \le \infty$. - -For $\varphi \in L^p_\delta(U)$ and $1\le p \le q < \infty$, we have $$\| e^{-t A} \varphi\|_{q,\delta} \lesssim \|\varphi\|_{p,\delta} t^{-\frac{n+1}{2}(\frac{1}{p}-\frac{1}{q})}, \quad t >0.$$ - -Quittner and Souplet call this the dimension shift phenomenon: the weighted space estimates are similar to those in standard $L^p$-spaces in $n+1$ dimensions. - -Question 1: Is there something subtle and interesting happening here? - -It seems to be based on the following (sketched) observations: from the estimate$$|e^{-tA} \varphi(x)| \lesssim \|\phi\|_{\infty} \frac{\delta(x)}{\sqrt{t}},\quad x \in U, t > 0,\quad \varphi \in L^\infty(U),$$ -and as $e^{-tA}$ is self-adjoint in $(L^2,(\cdot,\cdot))$, -$$\|e^{-tA} \varphi\|_1 = (e^{-tA} \varphi, \chi_{U}) = (\varphi, e^{-tA} \chi_U) \lesssim t^{-1/2}(\varphi,\delta)$$ -so -$$\|e^{-tA} \varphi\|_{\infty} = \|e^{-(t/2)A}(e^{(t/2)A)} \varphi)\|_\infty \lesssim t^{-n/2} \|e^{-(t/2)A} \varphi\|_1 \lesssim t^{-(n+1)/2}\|\varphi\|_{1,\delta}$$ -and the weighted estimate is obtained by Holder's inequality and some additional rigour (see the book by Quittner/Souplet for details). - -Question 2: The weight $\delta$ seems very special. What can be said in the case $\delta^\alpha$ where $\alpha > 1$? It seems a different argument is needed. - -I would love to hear any insightful or interesting remarks about the above. Thanks. - -REPLY [6 votes]: It seems dimensional analysis already reveals the exponent behaviour. If we use $m$ (say) to denote the unit of length, then an unweighted $L^p$ norm has units $m^{n/p}$, while a weighted $L^p$ norm has units $m^{(n+1)/p}$. The Laplacian $A$ has units $m^{-2}$, so time should have units $m^2$ in order for the exponent in $e^{tA}$ to be dimensionless. This soon predicts the right exponents for both estimates. -A bit more directly; a ball of radius $r$ has unweighted volume comparable to $r^n$, but has weighted volume comparable to $r^{n+1}$ if it is near the boundary (and the boundary is where all the "action" takes place). This already largely explains the dimension shifting phenomenon (noting also that at time t, a heat flow will have spread things out at the spatial scale of $r \sim t^{1/2}$.) -One can make the above dimensional analysis arguments rigorous by a scaling argument, at least in the model case when U is a half-space. These arguments do not actually prove the above estimates, but they show that the specified powers of $t$ are the only possible choices for such an estimate to be true.<|endoftext|> -TITLE: graded ring associated to a line bundle in a tensor category -QUESTION [5 upvotes]: Let $\mathcal{A}$ be an abelian tensor category with unit $\mathcal{O}$. An object $\mathcal{L}$ is called invertible or a line bundle if there is some $\mathcal{L}^{-1}$ such that $\mathcal{L} \otimes \mathcal{L}^{-1} \cong \mathcal{L}^{-1} \otimes \mathcal{L} \cong \mathcal{O}$. Equivalently, $\mathcal{L} \otimes -$ is an equivalence of categories. Now define a graded ring $\Gamma_*(\mathcal{L})$ as follows: -As an abelian group, take the direct sum of the $\text{Hom}(\mathcal{O},\mathcal{L}^{\otimes n})$, where $n \geq 0$. The product of homogenuous elements $s : \mathcal{O} \to \mathcal{L}^{\otimes n}, t : \mathcal{O} \to \mathcal{L}^{\otimes m}$ is defined by $s \otimes t$, where we identify $\mathcal{O} \otimes \mathcal{O} \cong \mathcal{O}$ and $\mathcal{L}^{\otimes n} \otimes \mathcal{L}^{\otimes m} \cong \mathcal{L}^{\otimes (n+m)}$. -Question Is $\Gamma_\*(\mathcal{L})$ commutative? Note that this is known in degree $0$ since $\text{End}(\mathcal{O})$ is commutative, even if we do not assume that $\mathcal{A}$ is symmetric. Actually it's not hard to see that $\text{End}(\mathcal{O})$ is central in $\Gamma_*(\mathcal{L})$. Remark that all this is well-known in the case of $\mathcal{A} = \text{Qcoh}(X)$ for a scheme $X$. - -REPLY [3 votes]: The following arose originally as a comment above and is being moved to an answer (per suggestion). -Let $R = R^*$ be any graded ring which is graded-commutative in the sense of homological algebra, i.e. for homogeneous elements $x$ and $y$ we have $xy = (-1)^{|x| |y|} yx$. Consider the category of graded left $R$-modules. This has a tensor structure as follows. Any left $R$-module $M$ inherits a right action by $R$ via the formula $m\cdot r=(−1)^{∣m∣∣r∣}rm$. Using this, we can define a monoidal structure on left $R$-modules using the graded tensor product $M \otimes_R N$. -(Note that all of this really comes because graded abelian groups form a symmetric monoidal category under tensor product, using twist isomorphism $\tau(x \otimes y) = (-1)^{|x| |y|} y \otimes x$. In this category, $R$ is a commutative monoid object and the tensor is just defined by a standard coequalizer on modules.) -Now let $\mathcal{L} = R[1]$, by which I mean a shifted copy of $R$ so that the degree $n$ group $R[1]^n$ is $R^{n+1}$ (grading cohomologically in order to align with the delicate sensibilities of the ag.algebraic-geometry tag). As a left $R$-module, it is free on a generator $e$ with $|e|=-1$. Then this object is invertible, and tensor powers $\mathcal{L}^{\otimes n} = R[n]$ are free on generators $e^{n}$ for $n \in \mathbb{Z}$. -At this point, one should verify for themselves that the ring $\Gamma_*(\mathcal L)$ is isomorphic to $R$ as a graded ring. (Seriously, you should check this, especially if you usually take the attitude that "the signs just work themselves out". There may be a clever perspective that avoids sign issues, but the straightforward perspective is not so.) -However, if you choose not to verify this: -One then gets an identification $Hom_{gr. R-mod}(R, \mathcal{L}^{\otimes n}) \cong R[n]^{0} = R^n \cdot e_n$, and the multiplicative structure is given by $$r e^{|r|} \cdot s e^{|s|} = (-1)^{|r| |s|} (rs) e^{|rs|}.$$ In particular, this graded ring is noncommutative precisely when $R$ is noncommutative, (which is most of the time). -ADDED: You can verify that there is an isomporphism between $R$ and this ring, given by the formula: -$$ -\phi(r) = (-1)^{\binom{|r|}{2}} r \cdot e^r -$$ -There is no canonical sign switch if you use $\mathbb{Z}/2$-graded objects rather than $\mathbb{Z}$-graded objects (although mod-4 gradings are fine).<|endoftext|> -TITLE: Minimal subset of axioms for ZFC -QUESTION [11 upvotes]: Hello all, one may look for "minimal system of axioms" for ZFC (or any other -theory) in the following (unusual) sense : say that a subset S of ZFC is -"sufficient" if there is an explicit procedure that -constructs a model of ZFC from any model of S. -Thus, for example, ZF is sufficient since inside ZF we can construct Godel's -universe L which is a model for ZFC. - My questions : are minimal sufficient subsets of ZFC known? -Is extensionality+infinity+(power set)+(separation scheme) sufficient? - -REPLY [12 votes]: I pondered this question some time ago. As pointed out by Stefan Geschke, infinity and power set are indispensable. Moreover: - -The replacement or collection schema is indispensable: $V_{\omega+\omega}$ is a model of all other axioms of ZFC (including separation). -The sum set axiom is indispensable: $H_{\beth_\omega}$ is a model of other axioms of ZFC. -The collection schema (if included in your axiomatization of ZFC) or extensionality is indispensable. One can build a model of other axioms by essentially taking $V_{\omega+\omega}$, and blowing it up to have a lot of automorphisms. Namely, put $M_0=\varnothing$, $M_{\alpha+1}=2\times\mathcal P(M_\alpha)$, and $M_\gamma=\bigcup_{\alpha<\gamma}M_\alpha$ for limit $\gamma$. Then take the model $\langle M,\in^M\rangle$, where $M=M_{\omega+\omega}$, and $\in^M$ is defined as follows: $x\in^M\langle i,y\rangle$ iff $x\in y$. The only nontrivial axiom to check is the replacement schema. It follows from the following property: if $M\models\exists!x\,\phi(x,\vec a)$, where $\vec a\in M_\alpha$, $\alpha<\omega+\omega$, then the unique witness for $x$ also belongs to $M_\alpha$. This in turn holds because there exists an automorphism $f$ of $M$ which is identical on $M_\alpha$, but which moves all elements of $M-M_\alpha$, namely -$f(\langle i,x\rangle)=\begin{cases}\langle i,x\rangle&\text{if }\langle i,x\rangle\in M_\alpha,\\\\ \langle 1-i,x\rangle&\text{otherwise.}\end{cases}$ - -On the other hand: - -Infinity, sum set, power set, extensionality and replacement interpret ZFC. That's just ZF without foundation: it interprets ZF in the well-founded kernel $WF=\bigcup_\alpha V_\alpha$, and ZF interprets ZFC in $L$. -Infinity, sum set, power set, separation (or replacement) and collection interpret ZFC. This requires some work, but here's at least the definition: the domain of interpretation consists of triples $\langle a,e,x\rangle$ such that $e$ is a well-founded extensional binary relation on $a$, and $x\in a$. Equality is interpreted by the relation $\langle a,e,x\rangle =^*\langle a',e',x'\rangle$ defined to hold iff there exists a partial embedding $f$ of the relational structure $\langle a,e\rangle$ to $\langle a',e'\rangle$ with $e$-transitive domain, $e'$-transitive range, and such that $f(x)=x'$. The elementhood predicate is interpreted by $\langle a,e,x\rangle\in^*\langle a',e',x'\rangle$ iff there exists $y\in a'$ such that $\langle y,x'\rangle\in e'$ and $\langle a,e,x\rangle=^*\langle a',e',y\rangle$. One can check that this is an interpretation of all axioms of ZF without foundation, which in turn interprets ZFC by the previous point. - -EDIT: I realized that I'm using here a nonstandard definition of $H_\kappa$ which is equivalent to the usual one for regular $\kappa$, but not for singular. The $H_{\beth_\omega}$ above is meant to denote the set of all sets $x$ such that every set in the transitive closure of $\{x\}$ has cardinality (strictly) less than $\beth_\omega$ (but the transitive closure itself can have cardinality $\beth_\omega$).<|endoftext|> -TITLE: Harmonic forms on Ricci-flat Kahler manifolds -QUESTION [6 upvotes]: Let $X$ be a compact Kahler manifold with $c_1(X) = 0$. Any Kahler metric $\omega$ on $X$ gives a Laplacian $\Delta_\omega$ and the $(1,1)$-form $\omega$ is harmonic with respect to this Laplacian. - -Take $\omega$ to be Ricci-flat. Is any other Ricci-flat metric $\alpha$ harmonic with respect to $\omega$? -If this is not true in general, and $X$ is a manifold on which this holds, then is $X$ a quotient of a torus? - -The first question has a positive response in the case of tori, which is where it comes from in the first place. There the Ricci-flat metrics are exactly those Kahler metrics that have constant coefficients, and the $(1,1)$-forms which are harmonic with respect to a such a metric have constant coefficients. So a Ricci-flat metric on a torus is harmonic with respect to any other Ricci-flat metric. - -REPLY [5 votes]: First let me give some cheap examples of CY manifolds that satisfy condition of 1) while they are not tori quotients. -Example. Let $X$ be any CY manifold with $h^{1,1}(X)=1$, for example $X$ can be a quintic in $\mathbb CP^4$. Then the conclusion of 1) holds, because all Kahler Ricci flat metric on $X$ are proportional. Now, to get an example with $h^{1,1}(X)=2$ one can just take a direct product $X_1\times X_2$ of CY manifolds satisfying $h^{1,1}(X_i)=1$. -So the list of manifolds with $c_1=0$ satisfying 2) is larger than tori quotients. But maybe all these examples can be understood. For example, it should be easy to show that a $K3$ surface does not satisfy 1). I'll give a brief proof in the case of a generic K3 -that does not have $-2$ curves. -EDDITED. I put more details here so that it becomes clear that was said previously is correct. -The good thing about harmonic forms is that the sum of two harmonic forms is harmonic. -By Nakai-Moishenzon (for Kahler surfaces), the Kahler cone -of a K3 without $-2$ curves coincides with a connected component -of $(1,1)$ classes with positive square in $H^{1,1}$. -Now, by Yau in any such class there is a unique Ricci-flat metric. And also obviously -there exists a unique harmonic metric. Suppose that assumption 1) holds for our K3. -Then it is obvious that this means that each harmonic form $w$ with $\int_{K3}w^2>0$ -in the correct component of the cone -is Ricci flat. Let me get a contradiction from this -So let $w_1$ and $w_2$ be two Kahler Ricci-flat forms harmonic with respect to the metric defined by $w$ on a $K3$ surface, and moreover that $aw_1+bw_2$ is Ricci flat again. Then we know that $(aw_1+aw_2)^2$ is proportional to $\Omega\wedge \bar\Omega$, where $\Omega$ is the complex volume form on $K3$. Consider now the family of forms $w_1-tw_2$, $t>0$. Take the first moment $t_0$ when we have $\int_{K3}(w_1-t_0w_2)^2=0$. But since $(w_1-tw_2)^2=c_t\Omega \wedge \bar\Omega$ for some $c_t>0$ for all $t$ less than $t_0$ (since this form is Ricci flat) -we conclude that $(w_1-t_0w_2)^2$ is equal to zero point-wise. So the kernel of $w_1-tw_2$ should define a holomorphic folitation on $K3$. Since for $K3$ it holds $h^{1,1}=20$, we should get a tremendous amount of holomorphic foliations on it, but this is clearly impossible. -This is still not a 100% complete reasoning, but I am sure it can be completed. So it seems to me that the complete list of manifolds satisfying 1) are all manifolds that are finite quorients of a torus times a collection of CY manfiolds with $h^{1,1}=1$. The idea is simple: whenever you have two Ricci-flat forms such that $w_1+tw_2$ is Ricci flat for small $t$, this should lead to a local metric splitting of the manifold into direct product. Then we should just use De-Rahm decomposition theorem.<|endoftext|> -TITLE: Social Reading Platform for Math or LaTeX texts -QUESTION [11 upvotes]: Social reading is considered to be one of the big trends that could be catalysing learning by reading. Features could include: - - Highlighting or annotating paragraphs or single steps in a proof for yourself (ok, this is not yet social reading), e.g. in a proof you could add to the sentence "Therefore we have ..." of the original work a more detailled explanation, thus "zooming in". - Publishing those of your own annotations you consider helpful for others - Reading all or a selection of the other reader's published annotations via a "show helpful annotations" button - Thumbing up other's annotations. - Asking questions to specific paragraphs or steps in a proof. - Answering questions asked by others and thumbing up questions and answers. - The original text would always keep the same, only being annotated - -Thus all the insights or questions you have while reading a paper or a textbook could be shared with others and learning could be far more efficient. Authors of textbooks could take into account the annotations/questions/answers, thus optimizing their text in future editions (or developing different versions or an annotated version). -Questions: a) Is there any tool that can provide these features, e.g. LaTeX-tools or pdf-tools? b) Would the Math overflow be a good starting point to build such a platform? c) Does anybody know of such a platform that is already in place? - -(Edit: Changed to community wiki) - -REPLY [3 votes]: Your "social reading platform" looks like what HTML and the WWW=World-Wide-Web was supposed to be when Tim Berners-Lee first set up a web-server and web-browser platform applications at CERN on a NeXT machine using Objective-C (I think he programmed it in objective C). Researchers were supposed to have their web pages listing and highligthing their research with hyperlinks pointing to the publications and datasets. If you look at the majority of academic webpages, the pulication and research interests are listed in that way. It's just that the majority of the internet world has gone into walled gardens such as the social media pages, with the cost of entry usually being the loss of any privacy or control over what can be done with your user-provided content. Look at the issues discussed on the meta website here at mathoverflow about why there hasn't been an upgrade to the StackExchange 2.0 software. -Wiki pages (not just that encyclopedic site that everyone uses, but a wiki page and wiki server which you can set up for yourself) allow for multiple users to modify a text using html or internal markup language. The requirement that $\LaTeX$ be usable in the markup language may require the use of MathML, or MathJax, or the jsMath package. -I think the correct answer is most likely an internal wiki server, with password-accessed accounts for modifying the wiki-pages. The problem is going to lie in placing a full copy of possibly copy-righted material, particularly in the case of wanting to do an "annotated version" of a research paper, or of a book chapter. If the author of a particular paper or book chapter, or the full book itself, wanted to do the experiment and set up their own wiki for the paper or book, and allow either free-for-all access or password-required gateway granted access to allow modifications and annotations, I would be very interested in taking part in that collaborative effort.<|endoftext|> -TITLE: Failure of Theorem of the Cube? -QUESTION [10 upvotes]: I am trying to understand the theory of cubical structures and am interested in knowing if a disconnected commutative group variety whose identity component is a semi-abelian variety satisfies the theorem of the cube. -Recall that if $X$ is an abelian variety and $L$ is a line bundle on $X$ that is rigidified along the identity sectionn, then the -Theorem of the Cube implies that the line bundle -$$ -\Theta(L) := m_{123}^{*}(L) \otimes m_{12}^{*}(L^{-1}) \otimes m_{13}^{*}(L^{-1}) m_{23}^{*}(L^{-1}) \otimes m_{1}^{*}(L) \otimes m_{2}^{*}(L) \otimes m_{3}^{*}(L) -$$ -admits a unique trivialization that makes -$$ -\Lambda(L) := m^{*}(L) \otimes p_1^{*}(L^{-1}) \otimes p_{2}^{*}(L^{-1}) -$$ -into a symmetric biextension of $X \times X$ by $\mathbb{G}_{m}$. -Here $m$ denotes the multiplication map, $p_i$ the projection maps, and -$m_{\underline{i}}$ the morphism $X \times X \times X \to X$ given by summing the -coordinates whose indices are in $\underline{i}$. -More generally, Breen proved this fact remains true when $X$ is a semi-abelian variety. -Does this statement remain true if we allow $X$ to have non-trivial component group? -If not, what is a example of a rigidified line bundle that does not have canonical cubical structure. -Does the theorem remain valid you rigidify along $1$ fixed point on every component (rather along the identity element)? -Added As BCnrd notes, over a more general base the formulas should be modified by adding the term $0^{*}(L^{\otimes \pm 1})$, which should be thought of as $m_{\emptyset}^{*}(L^{\otimes \pm 1})$. This (rigidified) line bundle is trivial when the base if a field, but not in general. - -REPLY [7 votes]: Uniqueness may fail if $X$ is a finite group: any two cube structures "differ" by a "quadratic" map $X\to \mathbb{G}_m$. If, say, $X=\mathbb{Z}/n\mathbb{Z}$ and $\zeta$ is an $n$-th root of unity, then $m\mapsto \zeta^{m^2}$ is such a map. -This of course implies non-uniqueness whenever the component group of $X$ admits nontrivial quadratic maps to $\mathbb{G}_m$.<|endoftext|> -TITLE: Homotopical descent information contained in the Dwyer-Kan function complexes of a presheaf category? -QUESTION [9 upvotes]: Recall that the category of sheaves on some site $C$ equipped with a grothendieck topology $\tau$ is equivalent to the localization of the category of presheaves $W^{-1}Psh(C)$ at $W$ where $W$ is the system of local isomorphisms generated by the Grothendieck topology (one way to describe these is as the morphisms that become isomorphisms under sheafification, but there are other, more direct methods to deduce what they are). -Recall that given any pair $(C,W)$ of a category and a class of morphisms, we can apply any of the methods of Dwyer-Kan simplicial localization to enrich $C$ to a simplicial category such that $\pi_0$ of the function complexes are exactly the hom-sets in the classical localization of the category. -Hammock localization is probably the best choice in this case, since local isomorphisms arising from a Grothendieck topology admit a calculus of left fractions, which means that the resulting function complexes are exactly the nerves of certain cospan categories. -Anyhow, is there any interesting descent-related information contained in the higher homotopical data of the function complexes? Is there any relation between this sort of thing and cohomological descent? - -REPLY [8 votes]: The function complexes will have no higher homotopy: they will be weakly equivalent to discrete sets, and so the simplicial localization will be DK-equivalent, as a simplicial category, to the category of sheaves regarded as a locally discrete simplicial category. This is because - -Every presheaf is locally isomorphic to a sheaf, so the simplicial localization will be DK-equivalent to its full subcategory on the sheaves, and -Between sheaves, all local isomorphisms are already isomorphisms, so there is no further localization to do.<|endoftext|> -TITLE: Are there any abelian 2 groups with Complete Holomorphs Other than $C^2_2$ and $C^4_2$? -QUESTION [7 upvotes]: The title says it. I'm reviewing some group theory concepts I haven't touched in quite a while, since I don't teach abstract algebra in my current position, and could not find the answer to this question. I've searched on google and found some papers that discuss other types of groups, but not 2-groups. I know that the holomorph of all non-trivial finite abelian groups of odd order are complete groups, due to a theorem by Miller(1908), where complete means trivial center and all automorphisms are inner. Also, since the holomorph of a finite abelian group is the direct product of the holomorphs of its Sylow subgroups, again Miller(1903?), does that imply the following: -Let G be an even ordered abelian group. If the holomoprh of the Sylow 2-subgroup of G is complete then the holomorph of G is complete. -As an added note, the automorphism group of $C^n_2$ is isomorphic to $PSL(n,2)$ since $C^n_2$ can be thought of as an $n$ dimensional vector space over the finite field $Z_{2}$. In the case where $n$ = 4, $PSL(4,2)$ is isomorphic to $A_{8}$. Is this connected to the holomorph being complete? Is the answer known for $C^n_2$, if not for all 2-groups? -Thanks - -REPLY [5 votes]: The completeness of ${\rm AGL}(n,2)$ for $n \ne 3$ follows easily from the following two facts: - -${\rm GL}(n,2)$ is complete for all $n \ge 1$. - -CORRECTION: Sorry - that was very careless of me! As Greg and Jack have pointed out, ${\rm GL}(n,2)$ is NOT complete for $n \ge 3$. The inverse-transpose automorphism which, in terms of groups of Lie type is the graph automorphism of groups of Lie-type $A_n$, is an outer automorphism of ${\rm GL}(n,2)$. However, as Jack pointed out, this does not induce an automorphism of ${\rm AGL}(n,2)$, because it interchanges subspaces of $V$ of dimension $r$ with subspaces of dimension $n-r$, and so does not act on $V$ itself. - -The cohomology group $H^1({\rm GL}(n,2), V)$ (with $V$ the natural module) is trivial for $n \ne 3$. This is equivalent to saying that for $n \ne 3$ all complements of the elementary abelian normal subgroup of order $2^n$ in ${\rm AGL}(n,2)$ are conjugate in ${\rm AGL}(n,2)$. - -I believe that the second of these was first proved in: -D.G. Higman, Flag-transitive collineation groups of finite projective spaces. Illinois J. Math. 6 (1962), 434-466. -I don't know when the first statement was first proved, but it follows from then general theory of automorphism groups of finite groups of Lie type. -Note that, for $n=3$, there are two conjugacy classes of complements of the $2^3$ normal subgroup, and they are interchanged by an outer automorphism of ${\rm AGL}(3,2)$.<|endoftext|> -TITLE: Uncountable nonstandard models of PA -QUESTION [10 upvotes]: Standard techniques (no pun intended) can be used to show that countable nonstandard models of Peano Arithmetic are order isomorphic to $\mathbb{N} + \mathbb{Z} \cdot \mathbb{Q}$. Once we have used the compactness theorem to verify that nonstandard models of PA exist, we can appeal to the Löwenheim–Skolem theorem to ensure that we have countable ones. We can then verify that these models form unbounded dense linear orders of $\mathbb{Z}$ blocks beyond the standard portion and use a back and forth argument to confirm that the nonstandard parts must always be order isomorphic. -My questions concern the structure of uncountable nonstandard models of PA. The Löwenheim–Skolem theorem can be used to show that we have nonstandard models of arbitrarily large cardinality so: -(1) Is there an example of a set $M$ of the form $\mathbb{N} + \mathbb{Z} \cdot D$ where $D$ is a dense linear order such that $M$ is not a model of PA? -(2) Can we find arbitrarily large $\kappa$ and nonstandard models of PA of size $\kappa$ such that there will be a an unbounded well-orderable subset of size $\kappa$ respecting the ordering relation of the nonstandard model (i.e. an order isomorphism between an ordinal and a subset of the model)? -(3) If the answer to (2) is no, what types of related results can we get? - -REPLY [14 votes]: Question 2 is an immediate consequence of the compactness theorem. Let $\kappa$ be fixed, and for each $\alpha < \kappa$ add a constant symbol $c_\alpha$ to the language. Add axioms of the form $c_\alpha < c_\beta$ for every $\alpha < \beta < \kappa$. The new theory $T$ is finitely satisfiable (every finite fragment is interpretable in the standard model of PA). So $T$ it is satisfiable, and the map $\alpha \mapsto c_\alpha$ is the desired embedding. -For question 1, it is known that $D$ cannot have the order type of the reals. There is a nice survey by Bovykin and Kaye [1] that includes this fact and has a lot more information about order types of models of PA. In the comments below, Andrey Bovykin suggests downloading his thesis from his homepage [2]; the paper is a summary of parts of the thesis. -1: Bovykin, Andrey; Kaye, Richard. Order-types of models of Peano arithmetic. (2002) Logic and algebra, 275--285, Contemp. Math., 302; MR1928396, DOI: 10.1090/conm/302/05055, http://www.maths.bris.ac.uk/~maaib/orders.ps -2: https://logic.pdmi.ras.ru/~andrey/research.html<|endoftext|> -TITLE: Names of finite groups -QUESTION [41 upvotes]: Question: If you have a finite group, how do you name it? - -If, for whatever reason, you have to list all subgroups of $GL_2({\mathbb F}_5)$ up to isomorphism in a paper, you are likely to write something along the lines of -$$ -C_1, C_2, C_2, C_3, -C_{2,2}, C_4, -C_5, -C_6, S_3, -Q_8, C_8, C_{2,4}, D_4, -$$ -$$ -C_{10}, D_5, -D_6, C_{12}, C_3\rtimes C_4, -C_{2,4}\rtimes C_2, OMC_{16}, C_{4,4}, -$$ -$$ -C_{20}, D_{10}, G_{20}, C_5\rtimes C_4, -SL_2(F_3), C_4\times S_3, C_3\rtimes C_8, C_{24}, -$$ -$$ -Q_8\rtimes C_4, -C_2\times G_{20}, C_2\times G_{20}, C_4\times D_5, -(C_{2,4}\rtimes C_2)\rtimes C_3, C_3\rtimes OMC_{16}, -$$ -$$ -C_4\times G_{20}, C_2.A_5, -SL_2(F_3)\rtimes C_4, -(C_2.A_5)\rtimes C_2, -GL_2(F_5). -$$ -Computer algebra packages tend to produce a human-unfriendly output of -generators and relations or generating permutations in $S_n$. How do -you convert from one to the other and decide how to name complicated groups? -I am looking for standard names, standard constructions, conventions and -notations. For me a good notation is informative, -human friendly, short and is generally as close as possible to what you would -use in a paper. I am also looking for any kind of canonical conventions: e.g. $(C_5\times C_5)\rtimes C_4$ or $(C_5\rtimes C_4)\times C_5$? - -(The reason I am asking is that I seem to have to work with funny groups all -the time recently. I have a Magma function for personal use that analyzes and -names finite groups; e.g. it produces the list -above for $GL_2({\mathbb F}_5)$, and I personally find this really useful. -Currently it knows various standard groups: cyclic, abelian, dihedral, -alternating, symmetric, special $p$-groups (semi-dihedral, generalized -quaternion, "other maximal cyclic", Heisenberg), simple groups, linear groups (SL, GL, O, -SP) and eventually their projective versions; it tries to recognize -direct, semidirect (and eventually wreath) products if the group is not too -large, and reverts to chief series if everything else fails. -Recently sufficiently many people asked me to share the code that I'll make -it public domain. But before that I'd very much like to get suggestions from the MO -community how to make it as useful as possible for most people.) - -Edit (6 years later): The names are finally in public domain (groupnames.org), and comments and suggestions are still very much welcome. - -REPLY [6 votes]: For transitive permutation groups the first paper in Journal of Computation & Mathematics -by Conway, Hulpke, & McKay lists the smaller degrees with "respectable names".<|endoftext|> -TITLE: Recurrent sequences and Bernoulli-like numbers -QUESTION [13 upvotes]: Consider the Fibonacci polynomials defined by $$F_n(s)=F_{n-1}(s)+sF_{n-2}(s)$$ with initial values $F_0(s)=0$ and $F_1(s)=1$ and define a linear functional $L$ on the polynomials in $s$ by $$L(F_{2n})=\delta_{n,1}.$$ Then $$L(F_{2n+1})=(2n+1)B_n,$$ where $B_n$ are the Bernoulli numbers defined by $B_n={\sum{n\choose k} B_k\}$ for $n\ge2$ and $B_0=1.$ -Choosing the linear functional $M$ defined by $$M(F_{2n+1})=\delta_{n,0},$$ gives $$M(F_{2n})=(-1)^n G_{2n},$$ where $G_{2n}$ are the Genocchi numbers $G_{2n}=(-1)^n 2 (1-4^n) B_{2n}.$ -Finally let $H_n$ be a variant of the Hermite polynomials defined by $$H_n(s)=H_{n-1}(s)-(n-1)s H_{n-2}(s)$$ and the linear functional $N$ defined by $$N(H_{2n})=\delta_{n,0},$$ then we get $$N(H_{2n-1})=(-1)^{n-1} T_{2n-1},$$ where $T_{2n-1}=(-1)^{n-1} \frac{4^n (4^n-1)}{2n} B_{2n}$ are the tangent numbers. -My question is: Are these isolated results or special cases of a more general theorem? Does anyone know other such examples? -Edit. To make my question somewhat more precise: Define the Fibonacci polynomials by $F_n(x,s)=xF_{n-1}(x,s)+sF_{n-2}(x,s)$ and the Hermite polynomials by $H_n(x,s)=xH_{n-1}(x,s)-(n-1)s H_{n-2}(x,s).$ -The above results follow from the identities -$$(e^{xz} + 1)\sum {\frac{{F_{2n} (x,s)}}{{(2n)!}}z^{2n} =(e^{x z}-1) \sum {\frac{{F_{2n + 1} (x,s)}}{{(2n + 1)!}}z^{2n + 1} } } $$ and -$$(e^{2xz} + 1)\sum {\frac{{H_{2n + 1} (x,s)}}{{(2n + 1)!}}z^{2n + 1} = (e^{2xz} - 1)\sum {\frac{{H_{2n} (x,s)}}{{(2n)!}}z^{2n} } }. $$ -Thus a more precise question would be: Are there polynomial sequences which satisfy similar identities? -Further edit. A more precise question: Are there "naturally occurring sequences" $A_n(x,s)$ satisfying $A_n(x,s)=xA_{n-1}(x,s)+c(n,s)A_{n-2}(x,s)$ such that -$$\sum {\frac{{A_{2n} (x,s)}}{{(2n)!}}z^{2n} =b(z,x) \sum {\frac{{A_{2n + 1} (x,s)}}{{(2n + 1)!}}z^{2n + 1} } }, $$ where $b(z,x)$ does not depend on $s?$ -The only examples I know besides the Fibonacci and Hermite polynomials are the Lucas polynomials $L_n(x,s)$ defined by $L_n(x,s)=xL_{n-1}(x,s)+sL_{n-2}(x,s)$ with initial values $L_0(x,s)=2$ and $L_1(x,s)=x.$ - -REPLY [4 votes]: The question is a bit vague (not that that is all bad) so my answer will be as well. There are (infinitely) many ways to define sequences and matrices of numbers and polynomials. Some of the simpler ones lead to sequences with special names. Simple transformations will send some of these nice ones to others. A sequence of polynomials gives an array of coefficients . Diagonals can give something else. For example the binomial coefficients have their recurrence $\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$. If we change it to $B(n,k)=s\,B(n-1,k-1)+B(n-1,k)$ (with appropriate initial conditions) then we just have $B(n,k)=\binom{n}{k}s^k$ and each row can be viewed as a polynomial $(1+s)^n$. It is well known that the Fibonacci numbers can be viewed as sums of certain diagonals of Pascals triangle. If we use instead the B(n,k) then we get the Fibonacci polynomials which start your question. $1,1,s+1,2s+1,s^2+3s+1,3s^2+4s+1\dots$ Splitting these into even and odd terms gives us $1,s+1,s^2+3s+1,\dots$ and also $1,2s+1,3s^2+4s+1,\dots$ either is a basis of the space of polynomials as is $1,s,s^2,\dots$ sending any one of these to an integer sequence (starting with 1) sends the others to another. See what the linear transformations in your examples do to $1,s,s^2,s^3,\dots$ -Here is another example (which I was happy to discover) My personal favorite recurrence is that for convergents to $\sqrt2$, $1/1,3/2,7/5,17/12,41/29,\dots$ The numerators $1,3,7,17,41,\dots$ and the denominators 1,2,5,12,29,... both satisfy the recurrence $a_n=2a_{n-1}+a_{n-2}$ just with different initial conditions. They could be called the Pell and Pell-Lucas numbers. If we say $A_n=2A_{n-1}+sA_{n-2}$ then we get a sequence $1,1,s+2,3s+4,s^2+8s+8,5s^2+20s+16,\dots$ These are closely related to Chebyshev polynomials of the first kind.(replace s by -s^2 and multiply every other one by s). -Avoiding many digressions, Consider the two sequences -$$1,s+2,s^2+8s+8,s^3+18s^2+48s+32,\dots$$ -$$1,3s+4,5s^2+20s+16,7s^3+56s^2+112s+64,\dots$$ -If the first is mapped to 1,0,0,0,0...then the last is mapped to 1, 2/3, -8/15, 32/21, -128/15, 2560/33, -1415168/1365, 57344/3, -118521856/255 -which is intriguingly close to the negatives of the cosecant numbers 1, -1/3, 7/15, -31/21, 127/15, -2555/33, 1414477/1365, -57337/3, 118518239/255 -If the last one is mapped to 1,0,0,0,0 then the first is mapped to -2,16,-272,7936,... which are the tangent numbers (but with alternating signs). -later: The OEIS and associated Journal of Integer Sequences are full of sequences and transformations taking one to another. The example above was an illustration (the only one I tried) of what I am sure would generate many examples (probably some nicer than the one I gave). Start with an integer recurrence of the form $a_0=0$ $a_1=1$ and $a_m=u(m)a_{m-1}+v(m)a_{m-2}$. Replace this with $A_0=1$ $A_1=1$ and $A_m=u(m)a_{m-1}+sv(m)a_{m-2}$. Then (in general) $A_{2t}$ and $A_{2t+1}$ are both polynomials of degree $t$ so one has two bases for the additive vector-space of polynomials. Take the linear transformation which sends one to $1,0,0,0,\dots$ and see what it does to the other (or some other basis of that space). Look up the associated sequence (perhaps the numerators and denominators if it is a rational sequence) in the OEIS and see what you got. Examples (which I have not investigated) include powers of 2 ($a_n=a_{n-1}+2a_{n-2}$) and factorials ($a_n=(n-1)a_{n-1}+(n-1) a_{n-2}$). Those two are shifted, it would be nicer to start with $a_0=1$ and $a_1=1$ making appropriate adjustments. Starting the second example with $a_0=1$ and $a_1=0$ gives derangements. One could consider other initial conditions $A_0=c_0$ and $A_1=c_1$ -There are many sequences of orthogonal polynomials like the Chebshev polynomials I mentioned above, (Legendre, Laguerre, Hermite, Jacobi,etc.) which are alternately even and odd. Dividing the odd ones by $x$ and then replacing $x$ with $\sqrt{x}$ (maybe taking absolute value of coefficients) gives more examples. Exponential generating functions might be more appropriate in some cases. -All that we are using of the polynomial structure is the (additive) book-keeping for the coordinates of a vector-space. Of course for famous sequences of polynomials that is still interesting. In general we could instead look at infinite lower triangular matrices. Then I think that your linear transformations correspond to the first column of the inverse matrix and the image on the companion sequence is simply the first column of a certain matrix product. Other columns might be of interest as well. -even later I did not say that it would give you what you wanted, only that I thought it might. Taking up my idea for powers of 2 gives two sequences of polynomials $1,2s+2,4s^2+10s+2,8s^3+36s^2+18s+2,\dots$ and $2,6s+2,16s^2+14s+2,40s^3+64s^2+22s+2,\dots$. Sending the first to $1,0,0,0,\dots$ sends the powers of $s$ to $1,-1,2,-7,38,-295,3098,\dots$ which look to be the absolute values of Genocchi numbers of second kind divided by 2^(n-1). This sends the other sequence to $2,-4,20,-172,2228,-40300,\dots$. The OEIS does not immediately identify that. Dropping the first term, still nothing. Since the remaining terms are multiples of 4 one divides that out to get $-1, 5, -43, 557, -10075, 241949, -7437547,\dots$ which does get identified as values B(2,n)/4 of Gandhi polynomials along with a reference to an article on combinatorial interpretations of Genochi numbers. I did not identify anything happening if we try to send the second sequence to $2,0,0,0,\dots$. I did not try for a functional equation.<|endoftext|> -TITLE: A set that can be covered by arbitrarily small intervals -QUESTION [26 upvotes]: Let $X$ be a subset of the real line and $S=\{s_i\}$ an infinite sequence of positive numbers. Let me say that $X$ is $S$-small if there is a collection $\{I_i\}$ of intervals such that the length of each $I_i$ equals $s_i$ and the union $\bigcup I_i$ contains $X$. And $X$ is said to be small if it is $S$-small for any sequence $S$. -Obviously every countable set is small. Are there uncountable small sets? -Some observations: - -A set of positive Hausdorff dimension cannot be small. -Moreover, a small set cannot contain an uncountable compact subset. - -REPLY [11 votes]: The problem was also studied by Besicovitch from the geometric measure-theoretic point of view in the 1930s. In particular, Besicovitch was motivated by the problem of determining the sets of reals on which the variation of any continuous monotone function vanishes. He proved that assuming CH there exists an uncountable set of reals which has strong measure zero (see "Concentrated and rarified sets of points", Acta Mathematica (1934), Vol. 62, pp. 289-300); doi: 10.1007/BF02393607. -Besicovitch constructed what he called a concentrated set (an uncountable set of reals $E$ is said to be concentrated on a countable set $H$ iff for any open set $U$ if $H \subset U$, then $E \setminus U$ is countable). The Besicovitch concentrated sets can be viewed as a weaker measure-theoretic analogue of Lusin sets. -The earlier stages of the theory of strong measure zero sets are summarized in Sierpinski's monograph Hypothèse du continu. Sierpinski refers to these objects as sets satisfying Property C (see the definition on p. 37). -[EDIT. Somewhat surprisingly the work of Besicovitch and Sierpinski on strong measure zero sets is not mentioned at all in the book recommended by Andres. The historical development of the theory is discussed in detail in the fundamental survey article "History of the Continuum in the 20th Century" by Steprans.] (Wayback Machine)<|endoftext|> -TITLE: Analytic lower bounds on the first sign change of pi(x) - li(x)? -QUESTION [12 upvotes]: There have been many results on the first sign change of $\pi(x)-{\mathrm{li}}(x)$: among others, Lehman, te Riele, Bays & Hudson, Demichael, Chao & Plymen, and most recently Saouter & Demichel. These provide upper bounds (as well as lower bounds on some region in which $\pi(x)>{\mathrm{li}}(x)$). -Could these methods be used to generate a lower bound? That is, a region [2, y] in which $\pi(x)<{\mathrm{li}}(x)$. (Or even a region [x, y] where y is smaller than the best known upper bound, such that, in principle at least, direct calculations could yield a lower bound.) -It seems unlikely that direct searches like [2] will ever be able to resolve the exact value of the first crossing. - -[1] C. Bays and R. H. Hudson. "A new bound for the smallest x with $\pi(x)>{\mathrm{li}}(x)$" Mathematics of Computation 69 (2000), pp. 1285–1296. -[2] T. Kotnik. "The prime-counting function and its analytic approximations", Advances in Computational Mathematics 29 (2008), pp. 55-70. -[2] R. Sherman Lehman. "On the difference $\pi(x)-{\mathrm{li}}(x)$" Acta Arithmetica 11 (1965), pp. 397–410. -[3] H. J. J. te Riele. "On the sign of the difference $\pi(x)-{\mathrm{li}}(x)$" Mathematics of Computation 48 (1987), pp. 323–328. -[4] Yannick Saouter and Patrick Demichel. "A sharp region where $\pi(x)-{\mathrm{li}}(x)$ is positive" Mathematics of Computation 79 (2010), pp. 2395-2405. - -REPLY [4 votes]: There are some explicit results in the work S. B. Stechkin, A. Yu. Popov, “The asymptotic distribution of prime numbers on the average”, Uspekhi Mat. Nauk, 51:6(312) (1996), 21–88<|endoftext|> -TITLE: Never appeared forthcoming papers -QUESTION [47 upvotes]: This has been inspired by this MO question: Harmonic maps into compact Lie groups -Just for joking: which is your favourite never appeared forthcoming paper? -(do not hesitate to close this question if unappropriate) - -REPLY [8 votes]: There is a result by Oesterle, that proves that you can find the first non residue quadratic modulo a prime in no more than $70\log(p)^2$ step assuming the GRH, this result was then improved by Bach who replaced the constant $70$ by $2$. The result of Oesterle was never published and when I asked him why, he told me because the laptop containing the proof was stolen from his car. However I think he exposed his proof to the mathematical community, so it is widely recognized. - -REPLY [6 votes]: Kurt Gödel referred to part II (Der wahre Grund für die Unvollständigkeit, welche allen formalen Systemen der Mathematik anhaftet, liegt, wie im II. Teil dieser Abhandlung gezeigt werden wird, darin, daß die Bildung immer höherer Typen sich ins Transfinite fortsetzen läßt) in his seminal paper Über formal unentscheidbare Sätze der Principia Mathematica und verwandter Systeme I, Monatshefte für Mathematik und Physik 38 (1931) p. 191. This part never appeared.<|endoftext|> -TITLE: How does the mixing time of a geodesic flow on a surface vary with the genus? -QUESTION [9 upvotes]: I have been looking at the numerical behavior of a particular quantity (of no direct importance here, though if you must know the gory details start with figure 17 here) associated to the geodesic flow on a surface of constant negative curvature and genus $g$. The behavior is quantitatively similar for $g = 2,3,4$ and physical intuition based on this quantity suggests that some "intrinsic" timescale--prime candidates are the mixing or relaxation time--should therefore depend on $g$ only weakly or even not at all. - -So: what is known about the behavior - of the mixing and relaxation (or - similar) times associated to these flows as $g$ - varies? - -REPLY [4 votes]: Here is another thought that struck me on the way home, that I should have realized earlier. Suppose that $S$ is a closed hyperbolic surface, of genus $g$. Then the area of $S$ is $-2\pi\chi(S) = 2\pi(2g - 2)$. Since the area of a disk in the hyperbolic plane is exponential in its radius, it follows that the diameter of $S$ is at least logarithmic in $g$. The mixing time of a space has to be at least the diameter, right? So this gives a uniform lower bound on the mixing time. -Thurston's comment is pointing out that there is no uniform upper bound. To see this: The injectivity radius is one-half the systole (the length of the shortest closed geodesic). For a hyperbolic surface, the collar lemma implies that as the injectivity radius goes to zero the diameter goes to infinity (this is the previously mentioned "long thin tube"). Thus the mixing time also has to grow, by the previous paragraph. -I roughly expect the mixing time can be estimated from the logarithm of the genus and the inverse of the injectivity radius. One reference for the geometric facts above is Peter Buser's book "Geometry and spectra of compact Riemann surfaces".<|endoftext|> -TITLE: Group Extensions and Line Bundles on $BG$ -QUESTION [8 upvotes]: I am sure the answer to this question is well-known, but -It is well known that the group cohomology $H^2(G,\mathbb Z)$ classifies group extensions $0\to \mathbb Z\to E\to G\to 1$ and that for a topological space $X$ elements of $H^2(X,\mathbb Z)$ are in natural bijection with complex line bundles on $X$. -My question is thus: -What is the direct correspondence between extensions of $G$ and line bundles on $BG$? -That is, given an explicit line bundle $L$ how does one construct an explicit group extension $E$ such that the two give the same cohomology class and vice versa? - -REPLY [5 votes]: Your line bundle $L$ over $BG$ can be seen as a $G$-equivariant line bundle over a point. That is, up to isomorphism, just a continuous group homomorphism $f:G \to \mathbb{C}^{\times}$. Try to lift $f$ along $\exp: \mathbb{C} \to \mathbb{C}^{\times}$ on an open cover of $G$. The error is a $\mathbb{Z}$-valued Cech-1-cocycle on $X$, and $E$ is the total space of the associated $\mathbb{Z}$-bundle over $X$.<|endoftext|> -TITLE: Are complements of non-isomorphic affine hypersurfaces necessarily non-isomorphic? -QUESTION [10 upvotes]: Fix $n$ and let $H_1$ and $H_2$ be two hypersurfaces in ${\bf A}^n$ (not necessarily smooth or irreducible, but we'll assume reduced). If the complements $U_i = {\bf A}^n \setminus H_i$ are isomorphic as schemes, does this imply that the $H_i$ are isomorphic? -From asking other people, the answer is yes via Euler characteristic arguments when the $H_i$ are smooth, but I have in mind singular examples. -As for the ground field, the polynomials I have in mind are defined over ${\bf Z}$, so choose whatever field you like. Probably I'd prefer an answer for the complex numbers though. - -REPLY [8 votes]: The paper of Jeremy Blanc, "The correspondence between a plane curve and its complement" (Crelle 2009) shows that two irreducible plane curves can have isomorphic complements in P^2 without being isomorphic. I would look at his argument and see if you can arrange for the affine complements to be isomorphic as well (if indeed you really care about the distinction between A^n and P^n here.)<|endoftext|> -TITLE: Finite type vs. finite dimensional cohomology? -QUESTION [9 upvotes]: I am not quite sure about the terminology, but let's call a $n$-dimensional manifold of finite type if it has a finite open cover $U_1,\ldots,U_k$ such that all intersections are either empty or diffeomorphic to ${\mathbb R}^n$. Every compact manifold is of finite type (endow it with a Riemannian metric and use geodesically convex neighbourhoods), while $\mathbb{R}^2\setminus \mathbb Z$ isn't. -A straightforward argument using the Mayer-Vietoris sequence shows that the cohomology of a manifold of finite type is finite dimensional. My question is about the converse statement: -If the cohomology algebra of a smooth manifold $M$ is finite dimensional, does this imply that $M$ is of finite type? -Edit: From Tom and Charles' examples, it is clear that the answer is "no" if one uses de Rham or even singular cohomology. As Tom suggests, there is some hope for a positive answer under an appropriate assumption on cohomology with twisted coefficients, but not being an expert on these matters I will not take the risk of further conjectures. - -REPLY [13 votes]: How about the unbounded complement $X$ of the Alexander horned sphere? Its cohomology is certainly finite dimensional (by the usual generalization of the Jordan curve theorem). The fundamental group is a direct limit $G=\mathrm{colim} F_{2^n}$ of of an increasing sequence free groups, where a generator in $F_{2^n}$ is sent to the commutator of a pair of generators of $F_{2^{n+1}}$; in particular, the group $G$ is -not finitely generated. This means that $X$ cannot have a nice finite cover of the type you describe (by application of Van Kampen's theorem, a space with such a finite cover must have finitely generated fundamental group in each path component.) - -REPLY [12 votes]: It does matter what kind of cohomology you use: -First think of the connected sum of an infinite number of real projective spaces end to end. De Rham cohomology cannot tell that these were not spheres, but mod 2 cohomology shows that the thing is not of finite type. -Now take it further. Take a (necessarily non simply connected) closed manifold whose integral homology is the same as that of a sphere (for example Poincare's 3-dimensional example whose fundamental group is the binary icosahedral group). Just as this looks like sphere to integral cohomology, the connected sum of infinitely many of these appears is indistinguishable from the connected sum of infinitely many spheres. -If you allow cohomology with twisted coefficients, then things get more hopeful. I would be inclined to separate the problem into two parts: (1) if a smooth manifold has the homotopy type of a finite CW complex, does this imply that it is of finite type in your sense? (2) What finiteness conditions on cohomology suffice to guarantee that a CW complex (space) has the (weak) homotopy type of a finite CW complex? (2) has the expected affirmative answer if you confine yourself to simply connected spaces, but the general case is more subtle. -EDIT What I really mean is: First assume that there are finitely many components, and that each of these has a finite presented fundamental group. Then use Wall's finiteness obstruction (which has to do with modules over the group ring of the fund gp -- it's a long story). -RE-EDIT Some more details: There are finite groups $G$ for which the ring $\mathbb ZG$ has some projective modules that are not stably free. For any such $G$ there are examples of finite CW complexes $K$ and $L$ with fundamental group $G$ such that $L$ is finite and $K$ is a retract of $L$ up to homotopy (i.e. there are maps $K\to L\to K$ whose composition is homotopic to the identity) but $K$ is not homotopy equivalent to any finite complex. The other composition $p:L\to L$ is then such that $p\circ p$ is homotopic to $p$, and $K$ is equivalent to the mapping telescope or homotopy colimit of $L\to L\to L\to\dots$. Take any such example and realize it by manifolds: Choose $L$ to be a smooth compact parallelizable manifold with boundary and (by taking the dimension big enough) take $p$ to be a smooth embedding of $L$ in the interior of $L$, and take $K$ to be the increasing union of infinitely many nested copies of $L$. So $K$ is a noncompact manifold without boundary whose (co)homology is in any sense no bigger than that of $L$, but which is not homotopy equivalent to any finite complex. In particular it is not the interior of any compact manifold with boundary. I can't see why it cannot be of finite type in the sense of the problem, but it looks unlikely. -It just struck me that this example can be improved: By doing surgery on a finite number of embedded $1$-spheres, one can arrange for the noncompact manifold $K$ to be simply connected. In this case it is homotopy equivalent to a finite complex (basically because $\mathbb Z$-modules are (stably) free), but it still is not the interior of any smooth compact manifold (because outside a compact set it's still the same example). -Background to this (but I am not using the full strength of it) is Wall's finiteness obstruction for CW complexes and Siebenmann's thesis in which Wall's obstruction is adapted to give an obstruction to putting a boundary on an isolated end of a manifold.<|endoftext|> -TITLE: Partitions of partial orders -QUESTION [7 upvotes]: (I am assuming choice.) -Suppose that ${\mathbb P}=(P,{\lt})$ is a partially ordered set (a poset), and that $\kappa\le\lambda$ are ordinals. The notation $$ {\mathbb P}\to(\kappa)^1_\lambda $$ means that whenever $f:P\to\lambda$, we can find some $i\in\lambda$ and some subset $H$ of $P$ such that $(H,{\lt})$ is order-isomorphic to $\kappa$ and $f(a)=i$ for all $a\in H$. - -Theorem. Suppose that $P$ is a poset and $P\to(\kappa)^1_\kappa$, where $\kappa$ is an infinite cardinal. Then $P\to(\alpha)^1_\kappa$ for all $\alpha\lt\kappa^+$. - -This is due to Galvin (unpublished). A nice combinatorial argument is presented in Stevo Todorcevic, "Partition relations for partially ordered sets", Acta Math. 155 (1985), no. 1-2, 1-25. -In fact, Galvin result is that: - -For any poset $P$, $$ P\to(\kappa)^1_\lambda $$ implies $$ P\to(\alpha)^1_\lambda $$ - whenever $\kappa\le\lambda$ are infinite cardinals and $\alpha\lt\kappa^+$. - -This can be proved by collapsing $\lambda$ to $\kappa$ with a $\kappa$-closed forcing, noting that in the extension $P\to(\kappa)^1_\kappa$, so (by the theorem) $P\to(\alpha)^1_\kappa$, and using the closure of the forcing to find such a homogeneous set of type $\alpha$ in the ground model. -My question: - -How can we prove Galvin's result without appealing to a forcing argument? - - -Very briefly, Stevo's proof of the theorem proceeds as follows: Given a poset $P$, let $\sigma'P$ be the collection of injective sequences $\tau$ whose domain is a successor ordinal and whose range is strictly increasing in the ordering of $P$. This is a poset under the "initial segment" ordering of sequences. Stevo proves two results: - -If $P\to(\kappa)^1_\kappa$ holds, then $\sigma' P\to(\kappa)^1_\kappa$ holds. -If $\sigma' P\to(\alpha)^1_\gamma$ holds, then $P\to(\alpha)^1_\gamma$ holds. - -Item 2 is straightforward, and a more general result holds. Item 1 uses a delicate argument and I do not know of a more general statement. The general version of 2 says that many partition relations that hold for $\sigma' P$ must hold for $P$ as well. -The combination of 1 and 2 is very powerful: It says that to prove partition results for posets $P$ satisfying $P\to(\kappa)^1_\kappa$ it suffices to prove the result for trees, for which the combinatorics tends to be much better understood than for arbitrary posets. -For example, for trees $T$ it is essentially obvious that if $T\to(\kappa)^1_\kappa$, then $T\to(\alpha)^1_\kappa$ for all $\alpha\lt\kappa^+$, and the theorem follows. - -REPLY [7 votes]: Assume that $P\rightarrow(\kappa)^1_\lambda$, yet for some $\alpha<\kappa^+$, $P\not\rightarrow(\alpha)^1_\lambda$. Accordingly, there is a decomposition $P=\bigcup_{\xi<\lambda}P_\xi$ such that no $(P_\xi,<)$ does contain a chain of type $\alpha$. For some $\xi<\lambda$ we must have $P_\xi\rightarrow (\kappa)^1_\lambda$ as otherwise all of them would be the union of $\lambda$ subsets, each missing a chain of type $\kappa$, and then, takigk union, the same would hold for $P$. Applying Galvin's original theorem, $P_\xi$ contains a chain of type $\alpha$, contradiction.<|endoftext|> -TITLE: Forcing over an arbitrary model of ZFC -QUESTION [14 upvotes]: I’m learning set theory and forcing in the (french) book from Jean-Louis Krivine “Théorie des ensembles”. -Given a countable transitive model $\mathscr{M}$ of ZFC together with a poset $P$, he constructs the model $\mathscr{M}[G]$ where $G$ is a $P$-generic in the ambient universe $\mathscr{U}$. The countability of $\mathscr{M}$ is essential in order for such a generic $G$ to exist. -But I saw several times in answers on MO that forcing could be defined over any model of ZFC and that, for example, CH and ~CH can both be forced over any model of ZFC. -My questions are: - -How does this kind of forcing work? I guess that we cannot anymore suppose that there exists a generic $G$, so how is the new universe constructed? And what does the truth lemma becomes? -When do we need to use this kind of forcing? For example if I want to prove that some proposition $P$ is independent of ZFC, I can always assume that my initial model of ZFC is countable and “usual” forcing will probably be sufficient. - -Any reference about this will be more than welcome. - -REPLY [6 votes]: $\def\zfc{\mathit{ZFC}}$No one seems to have mentioned another way to think about forcing, that can be sometimes helpful. -Let $\zfc_B$ denote $\zfc$ extended with an additional constant $B$ and an axiom that $B$ is a complete Boolean algebra. Let $\zfc_{B,G}$ be a theory in the language $\langle\in,B,G,U\rangle$ where $G$ is a constant, $U$ a unary predicate, with axioms stating that $U$ defines a transitive class, $\zfc_B$ holds in $U$, $\zfc$ (including replacement for all the new stuff) holds in the whole universe $V$, $G$ is a $U$-generic filter on $B$, and $V=U[G]$. -Any method of making sense of forcing (such as in the answers above) shows that $\zfc_{B,G}$ is a conservative extension of $\zfc_B$, in the sense that if $\phi$ is a $\langle\in,B\rangle$-sentence such that $\zfc_{B,G}\vdash\phi^U$, then $\zfc_B\vdash\phi$. Note that this can even be shown purely syntactically in a very weak metatheory. -In model-theoretic terms, this means that if $M\models\zfc$ and $M$ thinks that $B\in M$ is a complete Boolean algebra, then $M$ has an elementary extension $M'$ which has a generic extension $M'[G]$ with a generic for $B$. -However, the main point is that if you are now working inside $\zfc$ and you come upon a complete Boolean algebra $B$, you can safely pretend that your universe $V$ is included in a generic extension $V[G]$, even though you cannot actually define it from inside.<|endoftext|> -TITLE: How additive is Lebesgue measure in ZF+AD ? -QUESTION [7 upvotes]: What is known about the additivity of Lebesgue measure under the Axiom of Determinacy? - -That is, for what cardinals $\kappa$ do we have -with $|I| = \kappa$, for all functions $f : I \to 2^\mathbb{R}$, $\; \lambda(\displaystyle\bigcup_{i\in I}\ f(i)) = \operatorname{sup}(\{\lambda(\displaystyle\bigcup_{i\in J}\ f(i)) : J\subseteq I \land |J|<\infty\})$. - -(This is equivalent to the usual definition when $\kappa \leq \aleph_0$.) - -For example: - -Does it hold for $\aleph_1$? - -If so, what alephs does it hold for? - -Does it hold nontrivially for any infinite cardinals that are not alephs? - -(nontrivially meaning sets of that cardinality can be mapped onto non-well-orderable subsets of $2^\mathbb{R}$) - -REPLY [3 votes]: If $\kappa$ is a well orderable cardinal (i.e. an $\aleph_\alpha$) I don't think there is a need to involve determinacy. If all sets of reals are Lebesgue measurable (and DC holds), then Lebesgue measure is $\aleph_\alpha$ additive for every $\alpha$. -This is sufficient (at least granting DC) since AD certainly implies all sets of reals are Lebesgue measurable. -First argue that if $\kappa$ is a well orderable cardinal (i.e. an $\aleph_\alpha$) and Lebesgue measure is not $\kappa$ additive, then this can be witnessed in such a way that the sets in the union are all null sets (there is a countable subcollection whose union has measure equal to the sup in your equation). Without loss of generality, we may arrange that -$$ -\lambda ( \bigcup_{\xi \in \alpha} f(\xi) ) = 0 -$$ -for all $\alpha \in \kappa$. Define $Y = \bigcup_{\alpha \in \kappa} f(\alpha)$. -Now define -$$ -X = \{(x,y) \in Y^2 : \alpha_x < \alpha_y\} -$$ -where $\alpha_x = \min\{\alpha \in \kappa : x \in f(\alpha)\}$. -The set $X$ now violates Fubini's theorem: -the set of all horizontal sections are the union of fewer than $\kappa$ many of the sets -$f(\xi)$ and the vertical sections have their complement (in $Y$) being the union of fewer than -$\kappa$ sets of the form $f(\xi)$. -Thus by integrating in one direction, $X$ has measure 0, while in the other direction, -its measure is $\lambda(Y)^2$.<|endoftext|> -TITLE: Branching rule from symmetric group $S_{2n}$ to hyperoctahedral group $H_n$ -QUESTION [14 upvotes]: Embed the hyperoctahedral group $H_n$ into the symmetric group $S_{2n}$ as the centralizer of the involution $(1, 2) (3, 4) \cdots (2n-1, 2n)$ (cycle notation). Label representations of $S_{2n}$ by partitions of $2n$ and label representations of $H_n$ by pairs of partitions whose sizes add up to $n$ in the standard way. I am looking for a combinatorial description of the branching rule from $S_{2n}$ to $H_n$. This should be in the literature, but I couldn't find it. -Second small question: Is it possible to have a different embedding of $H_n$ into $S_{2n}$ so that the rule changes? I'd guess probably, but didn't think hard about it. - -REPLY [8 votes]: Okay I found it in the article by Koike and Terada, "Littlewood's formulas and their application to representations of classical Weyl groups". Here is the theorem. Define $d^\nu_{\lambda, \mu}$ as the coefficients in the product of plethysms: -$\displaystyle (s_\lambda \circ h_2)(s_\mu \circ e_2) = \sum_\nu d^\nu_{\lambda, \mu} s_\nu$ -where $s_\lambda$ is a Schur function, $h_2$ is the complete homogeneous symmetric function of degree 2, and $e_2$ is the elementary symmetric function of degree 2. Then the restriction of the representation $\chi(\nu)$ of $S_{2n}$ to $H_n$ is $\sum_{\lambda, \mu} d^\nu_{\lambda, \mu} \chi(\lambda, \mu)$. -Richard Stanley has informed me that a more general statement appears as Theorem A2.8 of Enumerative Combinatorics Vol. 2 (set $k=2$). This formula gives the coefficients for representations of $H_n$ indexed by bipartitions where one of the partitions is the empty set. The general case comes from the fact that $\chi(\lambda, \mu)$ is induced from a parabolic subgroup $H_k \times H_{n-k}$ of $H_n$ and that the map ch takes induction to product of characters.<|endoftext|> -TITLE: Primes that must occur in every composition series for a given module -QUESTION [15 upvotes]: Let $M$ be a finitely generated module over the commutative noetherian ring $R$. Let ${\cal C}(M)$ be the set of all primes $P$ in $R$ such that $R/P$ appears as a quotient in every composition series for $M$. Clearly ${\cal C}(M)$ includes all the primes associated to $M$. But it can contain other primes as well. For example, let $R$ be the local ring of a singular point on an irreducible curve and let $M$ be the maximal ideal of $R$. Then $R/M$ appears as a quotient in every composition series for $M$, but $M$ (as an ideal) is not associated to $M$ (as a module). -So: Is there a nice characterization of the set ${\cal C}(M)$? - -REPLY [10 votes]: I think this is a subtle question. The best result I am aware of is the following paper "Filtrations of Modules, the Chow Group, and the Grothendieck Group", by Jean Chan. She proved the following: Let $\mathcal F$ be any composition series of $M$. Let $c_i(\mathcal F)$ be the formal sum of primes of height $i$. Then $c_0(\mathcal F), c_1(\mathcal F)$, as elements in the Chow group of $R$, does not depend on the composition $\mathcal F$, so we can talk about $c_0(M), c_1(M)$. -So, for example, any such series will always contain the same minimal primes. As for height one primes, we only know that the sum of them is constant up to rational equivalences. However, if there is only one height one prime (your example of local ring of curves), then you can deduce its presence in any series by showing $c_1(M) \neq 0$ in the Chow group of $R$. -I will also note that in Eisenbud's "Commutative Algebra..." after Proposition 3.7, the author remarks that modules which always have a filtration consisting only the associated primes are called clean. Clearly for clean modules one has $C(M) = Ass(M)$. There is no clean(!) criterion for cleanliness, as far as I know, but see this paper for some partial results!<|endoftext|> -TITLE: Bounding the modular discriminant of an elliptic curve in the j-invariant -QUESTION [9 upvotes]: Consider an elliptic curve $X=\mathbf{C}/ (\mathbf{Z}+\tau \mathbf{Z})$, where $\tau$ is an element in the complex upper half plane. We define $$\Vert \Delta\Vert(X) = (\Im \tau)^6 \vert q\prod_{k=1}^\infty (1-q^k)^{24}\vert,$$ where we write $q=\exp(2\pi i \tau)$ as usual. This is called the modular discriminant of $X$. -Assume $j$ is algebraic, i.e., $X$ can be defined over a number field. -Question. Can the function $\log\Vert \Delta \Vert(-)$ (on the moduli space of elliptic curves over $\mathbf{C}$) be bounded (from above or below) in terms of the (height of the) $j$-invariant? -Firstly, one should be able to answer this question ineffectively. That is, to give a yes or no answer to the question. An effective bound (if it exists) might be a bit harder to obtain. -I heavily edited this old question. Therefore, the first four comments below might not make sense anymore. - -REPLY [11 votes]: The new $\| \Delta \|$, defined as $\mathop{\rm Im}(\tau)^6$ times the absolute value of the usual modular form $\Delta$, is invariant under the full modular group $\Gamma = {\rm PSL}_2({\bf Z})$ acting on the upper half-plane $H$. This $\| \Delta \|$ is nonzero and continuous on the quotient $H / \Gamma$, and approaches zero exponentially as $\tau$ approaches the one cusp of $H / \Gamma$. Hence $\|\Delta\|$ is uniformly bounded above, without any hypothesis on $j$; and $\|\Delta\|$ is bounded below if we have an upper bound on $|j|$. The latter bound is completely effective, namely -$$ -\| \Delta \| \gg (\log|j|)^6 / |j| -{\rm\ \ \ \ as\ \ \ \ } -|j| \rightarrow \infty, -$$ -and indeed $\| \Delta \| \sim C (\log|j|)^6 / |j|$ for some universal constant $C$, which is $(2\pi)^{-6}$ if I did this right. Now if you bound the height of $j$ from above then you impose an upper bound on the absolute value of any conjugate of $j$, and thus on $\| \Delta \|$. -Whether and how this lower bound depends on the height of $j$ then hinges on which flavor of height you're using, i.e. whether you normalize according to the degree $[{\bf Q}(j) : {\bf Q}]$, and whether you take logarithms. There is no such bound in the other direction: large height of $j$ does not force small $\| \Delta \|$ because it does not force $j$ to have a large conjugate (e.g. $j$ could be $1 / 10^{100}$).<|endoftext|> -TITLE: A collection of intervals that can cover any measure zero set -QUESTION [20 upvotes]: This is a follow-up to this question (in fact, this is what originally motivated me to ask that one.) -Let's say that a sequence $\{s_i\}$ of positive reals covers a set $X\subset\mathbb R$ if there is a collection if intervals $\{I_i\}$ such that $X\subset\bigcup I_i$ and the length of each $I_i$ equals $s_i$. -Does there exist a sequence $\{s_i\}$ such that $\sum s_i<\infty$ and $\{s_i\}$ covers any set of Lebesgue measure zero? -For example, simple things like geometric progressions do not work: they cannot cover a union of infinitely many copies of a compact set of positive Hausdorff dimension, separated by a distance at least $\max_i \{s_i\}$ from one another. -(Sorry for the strange collection of tags. It is hard to see in advance which area this question really belongs to.) - -REPLY [19 votes]: No. If you can cover every set of measure $0$ by your sequence of intervals, you can certainly scale (shrink all intervals some number of times) and still have covering (just cover the expanded set by the original sequence) . If $\sum s_j<+\infty$, then $\sum H(s_j)<+\infty$ for some measuring function $H$ with $H(x)/x\to+\infty$ as $x\to 0$. Thus, every set of measure $0$ would have the Hausdorff measure associated with $H$ zero, which can be ruled out by the standard Cantor construction.<|endoftext|> -TITLE: Advantages of Diffeological Spaces over General Sheaves -QUESTION [31 upvotes]: I have been playing with/thinking about diffeological spaces a bit recently, and I would like understand something rather crucial before going further. First a little background: -Diffeological spaces are a Cartesian-closed, complete, and cocomplete category containing all infinite dimensional manifolds, and in fact even form a quasi-topos. -Diffeological spaces, concisely, are nothing more than concrete sheaves on the site of Cartesian manifolds (manifolds of the form $\mathbb{R}^n$): -http://ncatlab.org/nlab/show/concrete+sheaf -However, the category of ALL sheaves on Cartesian manifolds, categorically is even nicer, since it is a genuine topos. -$\textbf{My question is:}$ -What can you do with diffeological spaces that you cannot do with general sheaves? Or, more generally, what are the advantages of diffeological spaces over general sheaves? -All of the generalizations of differential geometry concepts to diffeological spaces I have seen so far, actually carry over to genuine topos of sheaves (though sometimes with a little more work). -I'm aware that you gain the ability to work with a set with extra structure and talk about its points etc, but, what does this gain you? It seems that you can always use Grothendieck's functor of points approach instead. -Is it that limits and colimits are more like their counterparts for manifolds? - -REPLY [5 votes]: I'm not used to read this website, so I post my remark very late, after having randomly googled this page. -I think that one of the reason why diffeological spaces are better than general sheaves is the possibility to consider infinitesimal Fermat extensions: - -Giordano P. " Fermat reals: nilpotent infinitesimals and infinite dimensional spaces" . Book in preparation, see http://arxiv.org/abs/0907.1872, July 2009. -Giordano P. "The ring of Fermat reals", Advances in Mathematics 225 (2010), pp. 2050-2075. https://doi.org/10.1016/j.aim.2010.04.010 -Giordano P. "Infinitesimals without logic", Russian Journal of Mathematical Physics, 17(2), pp.159-191, 2010 https://doi.org/10.1134/S1061920810020032 -Giordano P., Kunzinger M. "Topological and algebraic structures on the ring of Fermat reals". Submitted to Israel Journal of Mathematics on April 2011. See http://arxiv.org/abs/1104.1492 -Giordano P. "Fermat-Reyes method in the ring of Fermat reals". To appear in Advances in Mathematics, 2011. https://doi.org/10.1016/j.aim.2011.06.008 -Giordano P. "Infinite dimensional spaces and cartesian closedness". Journal of Mathematical Physics, Analysis, Geometry, 2011. http://mi.mathnet.ru/eng/jmag/v7/i3/p225 - -This is a new theory, and I post this answer also because I think it is not known. -The basis of the theory is a surprisingly simple extension of the real field containing nilpotent infinitesimals. We start from the class of little-oh polynomials, i.e. functions $x:\mathbb{R}_{\ge 0}\rightarrow \mathbb{R}$ that can be written as $x(t)=r+\sum_{i=1}^{k}\alpha_{i}\cdot t^{a_{i}}+o(t)$ as $t\to 0^+$, where all the coefficients and powers are reals. Then, we introduce the equivalence relation between little-oh polynomials $x\sim y$ iff $x(t)=y(t)+o(t) \text{ as }t \to 0^{+}$. The ring of Fermat reals $ {}^\bullet\mathbb{R}$ is the corresponding quotient set. -The theory of Fermat reals has been developed trying always to obtain a good dialectic between formal mathematics and intuitive interpretation. Even if there are several theories of infinitesimals, only a couple of them always have this intuitive interpretation, and this contradicts the idea that (rigorous) infinitesimals are a strong support to guess some mathematical truths. Of course, Fermat reals take strong inspiration from smooth infinitesimal analysis, even if, at the end, it is a radically different theory. In fact, in the corresponding ring of scalars, which extends the classical reals, we have nilpotent infinitesimals of every order, infinitesimal Taylor's formulas (analogous of the Kock-Lawvere axiom), powers, roots of (nilpotent!) infinitesimals, logarithms, a total order relation, and the ring is also geometrically representable, so that we can finally state that infinitesimals are no longer ghosts of departed quantities. -It is also very interesting to note that its mathematical definition uses only elementary analysis and Landau's little-oh notation, without requiring a background in mathematical logic. In particular, the model is so simple that can be studied directly in classical logic without any need to switch to intuitionistic logic. On the other hand, this extension of the real field is generalizable both to finite and infinite dimensional manifolds (more generally to diffeological spaces). The extension ${}^\bullet(-): \mathcal{C}^\infty \rightarrow {}^\bullet\mathcal{C}^\infty$ (here $\mathcal{C}^\infty$ is the category of diffeological spaces and $ {}^\bullet\mathcal{C}^\infty$ is the category of Fermat spaces, which are defined similarly to diffeological spaces) is functorial and has very good preservation properties: a full transfer theorem for intuitionistically valid sentences is indeed provable (the "true" logic of smooth spaces is always intuitionistic!). -Several applications to differential geometry has been already developed: e.g. tangent vectors to any diffeological space $X \in \mathcal{C}^\infty$ can be defined, similarly to SDG, as smooth functions of the form $t:D\rightarrow {}^\bullet X$, where $ D :=\{h\in {}^\bullet\mathbb{R}|h^2=0\}$ is the ideal of first order infinitesimals and where ${}^\bullet X\in {}^\bullet\mathcal{C}^\infty$ is the Fermat space obtained extending $X$ with new infinitesimally closed points. -At present, we are developing several notions of differential geometry in this framework and are trying to extend the theory so as to include infinities and generalized functions (distributions).<|endoftext|> -TITLE: Borel set plus a closed set = Borel -QUESTION [14 upvotes]: Hi, -Let $R$ be equipped with the usual Borel structure. Let $F$ be a Borel subset and $E$ be a closed subset of $R$. Then $F+E=(f+e: f\in F, e \in E \)$ is Borel? If yes, is it true for any locally compact topological group? Thanks in advance. - -REPLY [20 votes]: No. Erdös and Stone showed that the sum of two subsets $E$, $F\subset\mathbb R$ may not be Borel even if one of them is compact and the other is $G_\delta$ (see "On the Sum of Two Borel Sets", Proc. Am. Math. Soc., Vol. 25, (1970), pp. 304-306). -Their argument works for every connected locally compact (or abelian) topological group with a complete metric.<|endoftext|> -TITLE: Simple examples of equivariant homology and bordism -QUESTION [6 upvotes]: I'm looking for simple examples of calculations of equivariant homology and of equivariant bordism. -I have a finite group G acting on an CW-complex X. I would like to calculate the equivariant homology $H_\ast^G(X)= H_\ast(X\times_G EG)$. In all of the examples I know, either the calculation is degenerate, for example because action of $G$ is taken to be free in which case $H_\ast^G(X)\simeq H_\ast(X/G)$, or else the setting is quite complex (string topology for instance) and I have trouble following. What I need is a good "second example". -I am looking first for a useful example of a simple non-degenerate equivariant homology calculation. One example I am specifically interested in is to calculate $H_\ast^{C_p}(C_q)$, where $C_p\ltimes C_q$ is a metacyclic group, so that $C_p$ is a cyclic group of order p acting on $C_q$ (with addition as the group operation) by multiplication by a $p$th root of unity in $C_q$. For example, $C_2$ would act on $C_7$ by multiplication by $6$, and $C_3$ would act on $C_7$ by multiplication by $2$. The action is not free (it fixes $0\in C_q$) so this example is not degenerate, yet the groups are small and finite. -I have the same question with regard to equivariant bordism. In the above example for instance, consider smooth maps from CW-complexes to the Eilenberg-Maclane space $K(C_q,1)$, where two such maps are considered equivalent if their image differs by a smooth $C_p$-action on $K(C_q,1)$. How to calculate $\Omega_\ast^{C_p}(C_q)$? - -REPLY [2 votes]: Sorry this got too long for a comment, so I post it as an answer. -A CW-complex equipped with a $G$- action is not the right thing to consider. A $G$-CW complex is a space, that can be built using blocks of the form $G/H \times (D^n,S^{n-1})$ (in the same way you build a CW-complex). A $G$-CW complex has better properties than a CW-complex, for example the fixed point set of any subgroup is a subcomplex. However most CW-complexes with a $G$ action can also be given the structure of a CW-complex. -In the example of $G=\{1,t\}=\mathbb{Z}/2$ acting on $S^1\times S^1$ by flipping the components, one could for example take the following $G$-CW- structure with 1 $0$-cell $P$ of type $G/G$, two 1-cells $A,B$ of type $G/1,G/G$ and 1 2-cell $C$ of type $G/1$: -image of G-CW structure http://www.freeimagehosting.net/uploads/d0acae4eb9.jpg -Then the cellular chain complex may be considered as $\mathbb{Z}[G]$ chain complex. Here it is: -$\mathbb{Z}[P]\leftarrow\mathbb{Z}[A,tA,B] \leftarrow \mathbb{Z}[C]$ -, where the differentials are given by $A,B\mapsto 0, C\mapsto A+tA+B$. Forgetting the names, we can write the chain complex as -$\mathbb{Z}[G/G]\leftarrow\mathbb{Z}[G/1]\oplus \mathbb{Z}[G/G] \leftarrow \mathbb{Z}[G/1]$ -These maps are $G$-equivariant. If one would apply $\otimes_{\mathbb{Z}[G]}\mathbb{Z}$ to this chain complex and take the homology, one would just get the cellular homology of the quotient $H_*(X/G)$. -I think, that if one takes a projective resolution of this chain complex first, one should get $H_*(X\times_GEG)$. (The cellular chain complex of $X\times EG$ is a free resolution of the cellular chain complex of $X$). -EDIT:I am not sure, whether I understand the second paragraph correctly. Group homology can be defined as the homology opf the classifying space, which reads $H_*(G):=H_*(BG)$. So this causes confusion, if it is not clear, whether one considers the group homology or the homology of the discrete space given by forgetting the group structure. -If the second was the case: Then the answer would be: write the discrete space as a union of orbits $\amalg_{i\in I}G/H_i$ and the result would be (like above) $\bigoplus_{i\in I} H_*(H_i)$ (denoting the group homology). -Or (more likely) it asks for some sort of "equivariant group homology", which I would read as -$H_*^C_p(BC_q)$. Using a nice functorial construction for $BC_q$, the $C_p$ action on the group also gives a $C_p$ action on the classifying space $BC_q$ and also on $EC_q$. However in this case I would guess, that $EC_q\times EC_p$ is a free,contractible $C_p\ltimes C_q$-CW-complex. Then it would be a model for $EC_p\ltimes C_q$ and hence: -$(BC_q \times EC_p)/C_p=(EC_q/C_q\times EC_p)/C_p=(EC_q\times EC_p)/(C_p\ltimes C_q)$. -Then $H_*^{C_p}(C_q)$ would just be the group homology of $C_p\ltimes C_q$.<|endoftext|> -TITLE: examples of finitely generated semigroups that are not residually finite -QUESTION [5 upvotes]: Does anybody know of any finitely generated semigroups that are not residually finite and whose group of units (if there is an identity) is trivial? Basically, I'm looking for finitely generated semigroups that are not residually finite, but I don't just want to punt and look at groups (like Thompson's groups or something). Thanks! - -REPLY [3 votes]: To generalise @BenjaminSteinberg's (by now rather old) answer, Lallement showed in 1971 that a one-relation monoid $M = \text{Mon}\langle A \mid w= w'\rangle$ where $w$ is a left and right factor of $w'$ (that is, there exist $w_1, w_2 \in A^\ast$ with $w' \equiv ww_1 \equiv w_2 w$, where $\equiv$ denotes graphical equality of words) is not residually finite if the presentation relation of the left special monoid $L(M)$ associated to $M$ is not the presentation relation of a group. In this case, it cannot be embedded into a compact semigroup, just as the bicyclic monoid. -The left special monoid comes about through so-called compression, which is a little too long to get into here, but one particular corollary is that if one takes $w \equiv \varepsilon$, then $L(M) = M$. Hence $\text{Mon}\langle A \mid w' = 1\rangle$ is not residually finite if it is not the free product of a free monoid by a group. So this includes the bicyclic monoid $\text{Mon}\langle b,c \mid bc = 1\rangle$, but also more general examples such as $\text{Mon}\langle a,b \mid ababb = 1\rangle$ and $\text{Mon}\langle a,b,c \mid abacac = 1\rangle$, both of which have trivial group of units. -One can easily check whether or not a one-relation special monoid $\text{Mon}\langle A \mid w' = 1\rangle$ is the free product of a free monoid by a group by Adian's algorithm.<|endoftext|> -TITLE: Cliques, Paley graphs and quadratic residues -QUESTION [35 upvotes]: A question I've thought about, on and off for a long time, is how to improve the best bounds that (seem to be) known for the clique numbers of Paley graphs. -If p=1 mod 4 is a prime, we can define the Paley graph, call it $P(p)$, of order p as follows. This graph has vertex set {0, 1, 2, ...., p-1} with two vertices i and j joined by an edge if and only if i - j is a quadratic residue modulo p. If p=1 mod 4 then -1 is a quadratic residue modulo p so this is a bona fide undirected graph. Say $c(p)$ is the clique number, or size of the largest complete subgraph of $P(p)$. -We can ignore graphs, etc. and just phrase it in terms of finding a maximum set, A, of quadratic residues modulo p (p still is 1 modulo 4) such that the difference of any two distinct elements of A is still a quadratic residue. -I'll make several comments here at this point: - -There are the "standard bounds" which are of the form $c(p) > (1/2-o(1)) lg(p)$ and -$c(p) < sqrt(p)$ where $lg(p)$ is the base 2 logarithm. -One or both of these bounds have been rediscovered every 5-10 years or so, going all the way back to Erdos and Newman(?) back in the 1950's. Sometime soon, I'll collect a comprehensive list and make it available online. :::::grin:::::: Another interesting observation is that both bounds can be proved either using combinatorial tools (Ramsey's theorem, Lovasz's theta function, SDP and vertex transitive graphs, etc.) OR using number theoretic tools (Gaussian sums, Weil's theorem, etc.) -We can look at generalizations involving quadratic nonresidues, nonprime finite fields/rings, etc. which I won't get into, though I can at least refer to Ernie Croot's problem list in arithmetic combinatorics and work of Gasarch and Ruzsa. I'm not interested ( in this post at least:::grin::::) in this. - -Finally my question: - Has anyone been able to materially improve either of these bounds? -What I do know about related work is: - -Maistrelli and Penman give a discussion of these bounds along with some computational work, in Discrete Mathematics in 2006. -Fan Chung, Friedlander, Iwaniec and others have studied related problems in character sums and applications in combinatorics but haven't seemed to show (yet?) any improvement in the bounds for $c(p)$. Or, have I missed something obvious here? -Andrew Thomason, Chung-Graham-Wilson, etc. have related work on "pseudo-random graphs" -which I will assume is known, or at least accessible to all the readers here. Thomason, in one article, makes some interesting assertions which are plausible but which I need to check. -There's the work of a number of people from Alon to Wigderson, exploring related questions and their applications to problems in theoretical computer science. -Finally, we have estimates for $n(p)$, the least quadratic nonresidue modulo p, following Chowla, Salie, Graham-Ringrose and others. The connection with $c(p)$ is obvious. -What else, of a substantial nature, is there? -I think, in particular, that using work of Granville and Soundarajan, the 1/2 in the first bound can be improved and using a combination of number theoretic methods (Burgess, etc.) and combinatorial methods (Ruzsa, Chang, Green, etc.) the second (square root) bound can be improved. I'm going to stop here and not go into any more specifics, either attempts, propositions, conjectures or computational evidence. - -REPLY [2 votes]: Actually if you use multi-dimensional character sum estimates instead of the single variable Hasse-Weil estimate then you should be able to obtain $c(p)>\log_2(p)$. In fact I have done this assuming a reasonably plausible estimate for the character sums required. (In fact I did this for k-colourings of graphs but the case k=2 is what is needed here). It is a very involved and difficult calculation. -The problem is the required estimates are not obviously derivable from the standard estimates of Deligne, Katz, Sperber for example and although I am reasonably sure they can be derived by someone with knowledge in this area I personally am unable to do this! -Montgomery showed that the least non-residue must sometimes have value $\epsilon \ln(p)\ln\ln(p)$ which shows that definitely $c(p)=O(\ln(p))$ is wrong. However it is possibly true infinitely often and if so this would keep alive the hope that the Paley graphs provide explicit constructions of graph colourings with no monochromatic clique larger than $c\log_2(n)$, where $n$ is the number of vertices, for infinitely many n.<|endoftext|> -TITLE: Motivic cohomology with finite coefficients for singular varieties -QUESTION [16 upvotes]: Let $X$ be a smooth variety over a field $K$ whose characteristic does not divide a positive integer $m$. Then the motivic cohomology of $X$ with coefficients in $\mathbb Z/m(j)$ can be computed in terms of the etale and Zariski topologies by the familiar rule -$$ - H_M^i(X,\mathbb Z/m(j)) = H_{Zar}^i(X,\tau_{\le j}R\pi_*\mu_m^{\otimes j}), -$$ -where $\pi:Et\to Zar$ is the natural map between the (big or small) etale and Zariski sites, $\mu_m$ is the etale sheaf of $m$-roots of unity, and $\tau$ denotes the canonical truncation of complexes of Zariski sheaves. In fact, this result is a combination of two: the Beilinson-Lichtenbaum etale descent -$$ - \mathbb Z(j) = \tau_{\le j}R\pi_*\pi^*\mathbb Z(j) = \tau_{\le j+1}R\pi_*\pi^*\mathbb Z(j) -$$ -and (a version of) the Suslin rigidity theorem -$$ - \pi^*\mathbb Z/m(j) = \mu_m^{\otimes j}. -$$ -These results also remain true if one replaces the Zariski topology with the Nisnevich one. -For singular varieties, the above formula for motivic cohomology no longer holds. It suffices to consider the example when $X$ is the affine line with two points glued together. Then one has $H^1_M(X,\mathbb Z/m(0))=\mathbb Z/m$, though the formula would give $0$ as the answer for this motivic cohomology group. It appears that in this particular case ($X$ as above and $j=0$) the rigidity assertion still holds, so it must be the etale descent that breaks down. -Is there any way to make any or all of the above assertions true for singular varieties by changing the topologies one works with? E.g., replace the etale topology with the h topology and the Zariski/Nisnevich with cdh? If there is more than one way to do this, I would be greatly interested to hear about each and every of them. - -REPLY [7 votes]: The problem is the definition of the motivic complex: it is a priori only defined for smooth schemes. Voevodsky defines motivic cohomology of non-smooth schemes by pulling the complex back to the category of all (separated, finite type) schemes and taking cdh-cohomology (etale cohomology has cdh-descent by the proper base-change theorem). -If you do this, your counterexample disappears , and I think the conjecture holds for non-smooth schemes as well.<|endoftext|> -TITLE: Proving that a map is a morphism -QUESTION [5 upvotes]: Example : Consider the (open, not compactified) moduli space of stable maps $ \mathcal M_g(1,d)$ of maps of smooth curves of genus $ g$ to $\mathbb P^1$. To each map -we associate its branch divisor, which is an element of $ Sym^r(\mathbb P^1)$. Then we should have a morphism from $ \mathcal M_g(1,d)$ to $ Sym^r(\mathbb P^1)$. How do -we prove this? -In general, if we have a family of objects $A$ and for each $ a \in A$ we can -choose an element $ b \in B$ that "depends continuously" on $ a$. How do we prove -that we have established a morphism $A \to B$? What is the general method to do this? -Note about the example: Extending the morphism to the Kontsevich compactification of the moduli space -was the main objective of a paper by Fantechi-Pandharipande. But I couldn't filter out -the proof that the map is a morphism, or the proof wasn't there. - -REPLY [6 votes]: Emerton has already answered, but let me summarize all the steps: -1) to a perfect torsion complex in the derived category associate a Cartier divisor; -2) to every sequence of morphisms $C\to X\to S$ satisfying suitable assumptions associate a perfect torsion complex, hence a Cartier divisor; -3) when $C\to X\to S$ is a family of stable maps to a smooth curve $Y$ (thus $X=Y\times S$), the divisor obtained on $X$ is effective and commutes with base change. -4) since $M_g(X,d)$ is the stack of stable maps, i.e. it represents a pseudofunctor (or, if you prefer, its coarse moduli space corepresents a functor), to define a morphism from it to a scheme $T$ means to define for every family over $S$ a mor $S\to T$ commuting with base change. -5) as Emerton wrote, $Hilb^rY=Sym^rY$. Hence an effective Cartier divisor on $X=S\times Y$ which is of degree $r$ on every fiber of $X\to S$ defines a morphism $S\to Sym^rY$.<|endoftext|> -TITLE: How many finite simple groups of order $p+1$? -QUESTION [6 upvotes]: I'm looking at finite simple groups of order $p+1$ where $p$ is a prime number. -But they don't seem to fall into any classification - have these all been determined? Is the number of them even finite? - -REPLY [5 votes]: Standard heuristics (together with orders from the list of finite simple groups ) suggest that by far the most common orders of the form $p+1$ for $p$ prime will come from $A_1(q) = PSL_2(\mathbb{F}_q)$, of order $\frac{q^3-q}{2}$, as $q$ ranges over odd primes (or prime powers, if you want an additional small contribution). In particular, for large $N$, one should expect roughly $\frac{\sqrt[3]{N}}{(\log N)^\alpha}$ satisfactory numbers $p+1$ less than $N$, for some fixed positive number $\alpha$, and this sequence of numbers certainly grows without bound. -As others have remarked, the question of proving that the set of suitable primes satisfies the rough asymptotics I gave above, or even proving that the set is infinite, seems to be beyond current technology. For example, we still don't know if there are infinitely many primes of the form $n^2+1$ for $n$ an integer.<|endoftext|> -TITLE: Has the mathematical content of Grothendieck's "Récoltes et Semailles" been used? -QUESTION [69 upvotes]: This question is partly motivated by Never appeared forthcoming papers. -Motivation -Grothendieck's "Récoltes et Semailles" has been cited on various occasions on this forum. See for instance the answers to Good papers/books/essays about the thought process behind mathematical research or Which mathematicians have influenced you the most?. However, these citations reflect only one aspect of "Récoltes et Semailles", namely the nontechnical reflexion about Mathematics and mathematical activity. Putting aside the wonderful "Clef du Yin et du Yang", which is a great reading almost unrelated to Mathematics, I remember reading in "Récoltes et Semailles" a bunch of technical mathematical reflexions, almost all of which were above my head due to my having but a smattering of algebraic geometry. However, I recall for instance reading Grothendieck's opinion that standard conjectures were false, and claiming he had in mind a few related conjectures (which he doesn't state precisely) which might turn out to be the right ones. I still don't even know what the standard conjectures state and thus didn't understand anything, but I know many people are working hard to prove these conjectures. I've thus often wondered what was the value of Grothendieck's mathematical statements (which are not limited to standard conjectures) in "Récoltes et Semailles". -The questions I'd like to ask here are the following: - -Have the mathematical parts of "Récoltes et Semailles" proved influential? If so, is there any written evidence of it, or any account of the development of the mathematical ideas that Grothendieck has expressed in this text? If the answer to the first question is negative, what are the difficulties involved in implementing Grothendieck's ideas? - -Idle thoughts -In the latter case, I could come up with some possible explanations: - -Those who could have developed and spread these ideas didn't read "Récoltes et Semailles" seriously and thus nobody was aware of their existence. -Those people took the mathematical content seriously but it was beyond anyone's reach to understand what Grothendieck was trying to get at because of the idiosyncratic writing style. -Should one of these two suppositions be backed by evidence, I'd appreciate a factual answer. -The ideas were already outdated or have been proven wrong. -If this is the case, I'd appreciate a reference. - -Epanorthosis -Given that "Pursuing Stacks" and "Les Dérivateurs" were written approximately in 1983 and 1990 respectively and have proved influential (see Maltsiniotis's page for the latter text, somewhat less known), I would be surprised should Grothendieck's mathematical ideas expressed around 1985 be worthless. - -REPLY [49 votes]: Begging your pardon for indulging in some personal history (perhaps personal propaganda), I will explain how I ended up -applying R'ecollte et Semaille. I do apologize in advance for interpreting the question in such a self-centered fashion! -I didn't come anywhere near to reading the whole thing, but I did spend many hours -dipping into various portions while I was a graduate student. Serge Lang had put his copy into the -mathematics library at Yale, a very cozy place then for hiding among the shelves and getting lost -in thoughts or words. Even the bits I read of course were hard going. However, one thing was clear -even to my superficial understanding: Grothendieck, at that point, was dissatisfied with motives. Even though I wasn't knowledgeable enough to have an opinion about the social commentary in the book, I did wonder quite -a bit if some of the discontent could have a purely mathematical source. -A clue came shortly afterwards, when I heard from -Faltings Grothendieck's ideas on anabelian geometry. I still recall my initial reaction to the section conjecture: `Surely there -are more splittings than points!' to which Faltings replied with a characteristically brief question:' Why?' -Now I don't remember if it's in R&S as well, but I did read somewhere or hear from someone that -Grothendieck had been somewhat pleased that the proof of the Mordell conjecture came from outside -of the French school. Again, I have no opinion about the social aspect of such a sentiment (assuming the story true), but it is -interesting to speculate on the mathematical context. -There were in Orsay and Paris some tremendously powerful -people in arithmetic geometry. -Szpiro, meanwhile, had a very lively interest in the Mordell conjecture, as you can see from his -writings and seminars in the late 70's and early 80's. But somehow, the whole thing didn't come together. -One suspects that the habits of the Grothendieck school, -whereby the six operations had to be established first in every situation -where a problem seemed worth solving, could be enormously helpful in some situations, and limiting in some others. In fact, my impression is that Grothendieck's discussion of the operations in R&S has an ironical tinge. [This could well be a misunderstanding due to faulty French or faulty memory.] -Years later, I had an informative conversation with Jim McClure at Purdue on the demise of sheaf theory in topology. [The situation has -changed since then.] But already in the 80's, I did come to realize that the motivic machinery didn't fit in very well -with homotopy theory. -To summarize, I'm suggesting that the mathematical -content of Grothendieck's strong objection to motives was inextricably linked with his ideas on homotopy theory as appeared in 'Pursuing Stacks' and the anabelian letter to Faltings, and catalyzed by his realization that the motivic philosophy had been of limited use (maybe even a bit of -an obstruction) -in the proof of the Mordell conjecture. More precisely, motives were inadequate for the study of points (the most basic maps between schemes!) in any non-abelian setting, but Faltings' pragmatic approach using all kinds of Archimedean techniques may not have been quite Grothendieck's style either. Hence, arithmetic homotopy theory. -Correct or not, this overall impression was what I came away with -from the reading of R&S and my conversations with Faltings, and it became quite natural to start thinking about a workable approach to -Diophantine geometry that used homotopy groups. Since I'm rather afraid of extremes, it was pleasant to find out eventually that -one had to go back and find some middle ground between the anabelian and motivic philosophies to get definite results. -This is perhaps mostly a story about inspiration and inference, but I can't help feeling like I did apply R&S in some small way. (For a bit of an update, see my paper with Coates here.) - -Added, 14 December: I've thought about this question on and off since posting, and now I'm quite curious about the bit of R&S I was referring to, but I no longer have access to the book. So I wonder if someone knowledgeable could be troubled to give a brief summary of what it is Grothendieck really says there about the six operations. I do remember there was a lot, and this is a question of mathematical interest.<|endoftext|> -TITLE: Algebraic surfaces and their (intrinsic) geometry -QUESTION [14 upvotes]: Recently I began to consider algebraic surfaces, that is, the zero set of a polynomial in 3 (or more variables). My algebraic geometry background is poor, and I'm more used to differential and Riemannian geometry. Therefore, I'm looking for the relations between the two areas. I should also mention, that I'm interested in the realm of real surfaces, i.e. subsets of $\mathbb{R}^n$. -On my desk you could find the following books: Algebraic Geometry by Hartshorne, Ideals, Varieties, and Algorithms by Cox & Little & O'Shea, Algorithms in Real Algebraic Geometry by Basu & Pollack & Roy and A SINGULAR Introduction to Commutative Algebra by Greuel & Pfister. Unfortunately, neither of them introduced notions and ideas I'm looking for. -If I get it right, please correct me if I'm wrong, locally, around non-singular points, an algebraic surface behaves very nicely, for example, it is smooth. Here's the first question: is it locally (about non-singular point) a smooth manifold? Is it a Riemannian manifold, having, for instance, the metric induced from the Euclidean space? -Further questions I have are, for example: - -Can I define geodesics (either in the sense of length minimizer or straight curves) in the non-singular areas of the surface? Can they pass singularities? -How about curvature? Is it defined for these objects? -Can we talk about convexity of subsets of the algebraic surface? -What other tools and term can be imported from differential/Riemannian geometry? - -I will be grateful for any hint, tip and lead in the form of either answers to my questions, or references to books/papers which can be helpful, or any other sort of help. - -REPLY [8 votes]: It seems to me that your interest is not in algebraic geometry, but in the differential geometry of spaces defined by algebraic equations. An algebraic variety defined over $\mathbb R$ or $\mathbb C$ is a manifold away from the singularities. The singular set is is a proper closed subset, where closed means defined by some algebraic equations, so this set is actually lower dimensional than the original object, so you have a very nice big open subset where you have a manifold. The question of extending differential geometric constructions to singularities is in general a difficult one and is the focus of a lot of research. You might be able to get a better idea by looking at complex analytic geometry. In any case, you can obviously do any of those you ask on the smooth part, but it will not be algebraic geometry. But that's OK. -One possibility you could try is indeed looking at the resolution of singularities, do your Riemannian magic there and try to get bring the results back to the original space. I suspect that you don't know what a resolution of singularities is since it is actually an very specifically algebraic geometric notion. It is the following: Let $X$ be your starting object. A resolution of singularities is a morphism $\pi:\widetilde X\to X$ such that it is an isomorphism outside a smaller dimensional subspace of $X$. You can read more about these in Lectures on resolution of singularities by János Kollár. The difficulty will be in taking whatever you do on the resolution back to the original, but the good news is that it is differential geometry, so my suggestion would be the following: -For now assume that such a $\pi$ exists and see if what you want to do you can on $\widetilde X$. If so, try to see if you can "push-forward" some of those results to $X$. Perhaps you will realize that "if only $\pi$ satisfied property $P$, then I could do this" and it is possible that $\pi$ does. SO, if you get to that point, then look at Kollár's book or come back to MO and ask more specific questions.<|endoftext|> -TITLE: Why are this operator's primes the Sophie Germain primes? -QUESTION [20 upvotes]: I was seeking a binary operator on natural -numbers that is intermediate between -the sum and the product, and explored this natural -candidate: -$$x \star y = \lceil (x y + x + y)/2 \rceil \;.$$ -Then I wondered which numbers are prime with respect to $\star$, -i.e., only have one factoring. -For example, -$11 = 1 \star 10$ is prime but -$13 = 1 \star 12 = 2 \star 8$ -is not. -Computing these $\star$-primes, I found they begin: -$$ 2,3,5,11,23,29,41,53,83,89,113,131,173,179,191, \ldots $$ -I tried to prove there were an infinite number of -$\star$-primes, but then -discovered my primes are precisely the Sophie Germain primes -(primes $p$ such that $2p+1$ is also prime), -and it is unknown if there are an infinite number of them. -Two questions: - -Q1. Why are the $\star$-primes as I defined - them precisely the Sophie Germain primes? - -I see that factoring $xy + x + y$ to $(x+1)(y+1)-1$ reveals the connection, but -my argument for iff is not precise. -(Incidentally, the ceiling cannot be ignored: -replacing ceiling by floor results in different -"primes.") - -Q2. Is it possible to express the number of $\star$-divisors - of $n$ in terms of - a mixture of the number of - divisors $\tau(n)$ and the number of partitions $p(n)$? - -For example, -here are the factors of $n=40$: -$$(1 \star 39), (2 \star 26), (3 \star 19), (4 \star 15), (7 \star 9), (8 \star 8) \;,$$ -And so 40 has 11 $\star$-divisors: -$1, 2, 3, 4, 7, 8, 9, 15, 19, 26, 39 \;.$ -I label this recreational because I'm sure this is like eating -candy for many of you! Enjoy the snack! -Addendum. Incidentally, if $\star$ is defined using floor rather than -ceiling, then the $\lfloor \star \rfloor$-primes $>3$ are even numbers $n$ -such that $n+1$ and $2n+1$ are (conventionally) prime. I don't know if these primes -have been named, or if it is known whether there are an infinite supply. - -REPLY [19 votes]: Q1. $p$ is $\star$-prime iff equation $xy+x+y=2p$ has no solution and $xy+x+y=2p-1$ has exactly one solution, i.e. $(x+1)(y+1)=2p+1$ has no solution (which is iff $2p+1$ is prime) and $(x+1)(y+1)=2p$ has only one solution $\{x,y\}=\{1,p-1\}$. This last holds iff $p$ is prime. -Q2. Why partitions?! The number of $\star$-divisors of $n$ equals $\tau(2n+1)+\tau(2n)-4$. - -REPLY [10 votes]: In answer to your first question: -As you hinted at, it simplifies things to make the change of variable $x \mapsto x+1$. -Then the product becomes $x \star y = \lceil\frac{xy+1}{2}\rceil$. And we want to find $z$ that can't be expresed as $x \star y$ where $x,y > 2$. -Well that's equivalent to $z$ such that you can't solve $2z=xy+1$ with $x,y>2$, and also you can't solve $2z-1=xy+1$ with $x,y>2$. -The first condition is equivalent to $2z-1$ being prime, since the oddness of $2z-1$ makes the constraint on $x$ and $y$ irrelevant. -The second condition is equivalent to $z-1$ being prime, since it's saying that $2z-2=2(z-1)$ has no non-trivial factorizations beyond the obvious two involving the factor $2$. -Hence overall, we see that $z$ prime in the new sense $\Leftrightarrow$ $z-1$ Sophie Germain prime. Changing variables back, we get the answer to the original question.<|endoftext|> -TITLE: Can we deduce that two rings $R_1$ and $R_2$ are isomorphic if their polynomial ring are isomorphic? -QUESTION [7 upvotes]: Given two rings $R_1$ and $R_2$ (with or without identity: it's not specified). If $R_1[x]$ is isomorphic to $R_2[y]$ (No such requirement that the isomorphism sends the constant terms to constant terms), can we deduce that $R_1 \cong R_2$? -I feel there might be a counterexample but it's quite hard to find one. - -REPLY [4 votes]: Let X be an affine variety with two non-isomorphic vector bundles V and W that become isomorphic after adding a trivial line bundle to each. Then the coordinate rings of the total spaces of V and W should yield a counterexample. (Though you might have to do some extra work to verify that the total spaces of V and W are non-isomorphic as varieties.) -The counterexample cited by Tobias takes X to be the 2-sphere, V the tangent bundle to X, and W the trivial plane bundle over X.<|endoftext|> -TITLE: Is Deligne's central extension sofic? -QUESTION [6 upvotes]: In P. Deligne. Extensions centrales non résiduellement finies de groupes -arithmétiques. CR Acad. Sci. Paris, série A-B, 287, 203–208, 1978. Deligne proves the existence of a certain central extension of a residually finite group. -See the section Deligne's central extension in Cornulier-Guyot-Pitsch's paper for a quick discussion of the group. -A countable, discrete group $\Gamma$ is sofic if for every $\epsilon>0$ and finite subset $F$ of $\Gamma$ there exists an $(\epsilon,F)$-almost action of $\Gamma$. See, for example Theorem 3.5 of the nice survey of Pestov. Gromov asked whether all countable discrete groups are sofic. It is now widely believed that there should be a counterexample to this. - -Is Deligne's central extension sofic? - -This question is related to the one here, but is not sharpened enough to be an answer. (In fact, in its original form not even to be a question! Thank you Henry and Andreas.) - -REPLY [5 votes]: Deligne's central extension is certainly an interesting group to consider. I can say something about the question whether this group is hyperlinear. For hyperlinearity, you ask for approximation of the group law by unitary matrices instead of permutations. This is (as you of course know) weaker than being sofic, but being sofic implies that the group is hyperlinear. Hence, it is a neccesary condition and natural generalization. -In the paper Examples of hyperlinear groups without factorization property, Groups Geom. Dyn. 4 (2010), no. 1, 195–208 I obverved that a central extension of a group $G$ by an abelian group $A$ is hyperlinear if and only if the twisted group von Neumann algebra $L_{\phi \circ \alpha} G$ is embeddable, where -$$\alpha \colon G \times G \to A$$ is the 2-cocycle which classifies the central extension, and $\phi$ belongs to a dense set in the Pontrjagin dual of $A$. This is the same as asking for a approximation of the group laws of $G$ with unitaries on a finite-dimensional Hilbert space, twisting the multiplication with the cocycle $\phi \circ \alpha$. The twisted setup is natural anyway and I propose to call a $S^1$-valued 2-cocycle on a group $G$ hyperlinear if you can find such an approximation by unitaries. It seems natural to consider the possibility that there are $S^1$-valued 2-cocycles even on residually finite groups which are not hyperlinear. However, I do not have any examples. -On the other side, if $G$ is residually finite and you can at the same time approximate $\phi \circ \alpha$ for sufficiently many $\phi$'s by suitable almost 2-cocycles defined on the finite quotients, you are in business. This would show that the central extension is at least hyperlinear. -I do not know about sofic in place of hyperlinear, since in the combinatorial setup, it is not possible to disintegrate the central extension over the Pontrjagin dual of $A$.<|endoftext|> -TITLE: When is a connected sum of torus knots a slice knot? -QUESTION [6 upvotes]: This question is about the beaviour of 4-genus of knots with respect to connected sum. -Let us indicate with $T(k)$ a Torus knot of type $(2,k)$, $k$ is an odd integer. -Fix an orientation for every $T(k)$ so that $T(-k)$ represents the same knot with reversed orientation. -$T(k)\sharp T(-k)$ is a slice knot. More generally if $K^* $ denotes the mirror of $K$ then $K\sharp K^*$ is slice (infact ribbon). -My question is: - -Is it true that a knot of the form $T(a_1)\sharp\dots\sharp T(a_n)$ is slice if and only -if n is even, say $n=2k$, and we can arrange the coefficients so that for every $i\leq n$ -we have $a_{k+i}=-a_i$? - -Note that for any pair of knots $K_1$ and $K_2$, if both $K_1$ and $K_1\sharp K_2$ are slice then $K_2$ is also slice. Therefore one only needs to show that there exists -$i$ and $j$ such that $a_i=-a_j$. -Of course we can generalize this problem: - -Which connected sums of torus knots are slice? - -Here are some links for definitions: -Torus knot: http://en.wikipedia.org/wiki/Torus_knot -Slice knot: http://en.wikipedia.org/wiki/Slice_knot -Slice genus: http://en.wikipedia.org/wiki/Slice_genus - -REPLY [10 votes]: I believe the answer to your last two questions is yes, and it follows from Litherland's (1979) computation of the Tristram-Levine invariants of torus knots. See Kearton's survey here: -http://www.maths.ed.ac.uk/~aar/slides/durham.pdf -i.e. a connect sum of torus knots is slice if and only if the prime summands appear in balancing mirror-reflected pairs.<|endoftext|> -TITLE: Spectral theory of pseudo-differential operators -QUESTION [8 upvotes]: Consider a finite rank complex bundle $E$ over $S^1$ with connection $\nabla$. Let $Q_0, Q_1 \in C^\infty(S^1, E)$ be pseudo-differential operators. $Q_0$ is defined by the symbol $\sigma_0(x, \xi) = i \xi$, and $Q_1$ by $\sigma_1(x,\xi) = 2H(\xi)-1$ where $H$ is the Heaviside step function. Both of these have discrete spectra. -Now take $Q_t$ to be defined by $\sigma_t = t \sigma_0 + (1-t) \sigma_1$. Does $Q_t$ have a discrete spectrum? -Less rigorously, is there a ``continuous" way to connect $Q_0$ and $Q_1$ through operators with discrete spectra? Also, are there well known theorems about when such operator have discrete spectra, or well known classes of operators with discrete spectra? (Compact operators do, but that is insufficient for my purposes). Apologies for being vague, I'm a physicist. -Thanks, -Lukasz Fidkowski. - -REPLY [3 votes]: You should have a look at p. 267, 23.35.2 of Dieudonné's Treatise on Analysis, part 7, as well as Lawson-Michelson, Spin Geometry, p. 196, Thm. 5.8, and check if your symbols are elliptic (they seem to be) and notice that your base manifold is compact. This might help.<|endoftext|> -TITLE: Geometric motivation for the Stanley-Reisner correspondence -QUESTION [10 upvotes]: The Stanley-Reisner ring of an abstract simplicial complex $\Delta$ on the vertex set $\{1,...,n\}$ is the $k$-algebra -$$ -k[X_1,...,X_n]/I_\Delta -$$ -where $I_\Delta$ is the ideal generated by the $X_{i_1}...X_{i_r}$ with ${i_1,...,i_r}\notin \Delta$. -Somebody told me that this construction helps to study varieties $k[X_1,...,X_n]/I$ for an arbitrary ideal $I$ as follows (if I am not missing something): Let $X_1<...< X_n$ be a monomial order and consider the initial ideal $I':=in_{<}(I)$ of $I$. The passage from $k[X_1,...,X_n]/I$ to $k[X_1,...,X_n]/I'$ is called 'flat deformation' and this term makes sense if I draw pictures of the varieties. Many properties of $I$ (like dimension) are directly related to properties of $I'$. -The aim is to find a to $I'$ related ideal $I_\Delta$ for an abstract simplicial complex $\Delta$. -I was told that the problem that $I'$ has a generator like $X_1^2$ could be resolved by introducing a new variable $X_1'$, replacing $X_1X_1$ by $X_1X_1'$ and mod out $X_1-X_1'$ of $k[X_1,...,X_n,X_1']$. First, I don't understand why this should be closer to the form $I_\Delta$ because $I_\Delta$ is generated by monomials. -My main question is: - -Can one see in a concrete affine - example how the geometry of $\Delta$ - relates to the initial variety $I$? - -If you take for example the simplicial complex $\Delta$ (I apologize that I can not typeset the brackets) $\emptyset,X, Y, Z, XY, XZ, YZ$ which looks like a one sphere, the associated variety is the union of the $XY$, the $XZ$ and the $YZ$ hyperplane in $\mathbb{R}[X,Y,Z]$. Why is this reasonable? On the other hand, I would like to know how the simplicial complex (after the transformation indicated above) corresponding to the circle $I=(X^2+Y^2-1)$ in $\mathbb{R}[X,Y,Z]$ looks like. It would be nice if this is associated to the $\Delta$. -Please tell me, if this question is completely unreasonable or doesn't make any sense. I am an absolute beginner in algebraic geometry. - -REPLY [7 votes]: I think you're asking if there's a direct geometric relationship between an algebraic variety $X=V(I)$ (i.e., the zero set of an ideal $I$) and the Stanley-Reisner complex $\Delta_I$ or $\Delta_{in(I)}$ --- in other words, does the variety look like the simplicial complex? In general, I think the answer can be quite subtle, and is probably best approached using the algebra as an intermediary.. For example, if $\Delta$ is a simplicial sphere, then the number $d$ of its facets will correspond to the degree of $X$, but you will not necessarily be able to ``see'' directly from $X$ how those facets fit together combinatorially in $\Delta$ (e.g., if $d=20$, is $\Delta$ a decagonal bipyramid or an icosahedron?). On the other hand, you can calculate $d$ and similar invariants fairly easily from the ideal $I$ (in this case, compute the Hilbert series as a rational function in $q$, and plug $q=1$ into the numerator). -For "flat deformation", here's the example I always keep in mind (very similar to Torsten's): take the hyperbola $xy=c$ and let $c\to0$. In the limit, the hyperbola degenerates into a pair of lines $x=0$ and $y=0$. This corresponds to replacing the ideal $(xy-c)$ (or $(xy-cz^2)$, if you want to think projectively) with the monomial ideal $(xy)$, which is its initial ideal under the right term ordering on $\Bbbk[x,y,z]$. The general principle is that some invariants (e.g., Hilbert series) don't change when you degenerate, and some (e.g., singularities) can only get \emph{worse}, so if you can prove that the degeneration is, say, Cohen-Macaulay then so was the original thing. -As the operation of replacing, e.g., $X_i^2\in I$ by $X_iX_i'$, it is called ``polarization'' and has similarly mild effects. See, e.g., \S3.2 of Miller and Sturmfels \emph{Combinatorial Commutative Algebra}.<|endoftext|> -TITLE: What does an etale topos classify? -QUESTION [24 upvotes]: Any Grothendieck topos E is the "classifying topos" of some geometric theory, in the sense that geometric morphisms F→E can be identified with "models of that theory" internal to the topos F. For the topos of sheaves on a site C, the corresonding theory may tautologically be taken to be "the theory of cover-preserving flat functors on C." However, for some naturally arising toposes of interest, the classified theory has a different, more intuitive expression. For instance, the topos of simplicial sets classifies linear orders with distinct endpoints, and the "Zariski topos" classifies local rings. -My question is: if X is a scheme—say affine for simplicity—then what theory does its (petit) etale topos $Sh(X_{et})$ classify? Can it be expressed in a nice intuitive way, better than "cover-preserving flat functors on the etale site"? I hope/suspect that it should have something to do with "geometric points of X" but I'm not sure how to formulate that as a geometric theory. - -REPLY [14 votes]: For a general scheme $(X,\mathbb{O}_X)$, the étale topos over $X$ classify the strict henselisation (also called the separable closure) of the locale ring $\mathbb{O}_X$. -The result is clearly due to Monique Hakim in her thesis ("topos annelés et schémas relatifs") but as she did not use at all the concept of internal logic there is no description of the precise geometric theory involved. So if you look for a full and definite answer to this question, you should look in : -G. C. Wraith - Generic Galois theory of local rings -which indeed give a clear description of the theory involved.<|endoftext|> -TITLE: Organizing principles of mathematics -QUESTION [5 upvotes]: In his famous paper "The two cultures of Mathematics" T. Gowers gives examples of organizing principles in combinatorics. -(i) Obviously if events $E1, \cdots,E_n$ are independent and have non-zero probability, then -with non-zero probability they all happen at once. In fact, this can be usefully true -even if there is a very limited dependence. [EL,J] -(ii) All graphs are basically made out of a few random-like pieces, and we know how -those behave. [Sze] -(iii) If one is counting solutions, inside a given set, to a linear equation, then it -is enough, and usually easier, to estimate Fourier coefficients of the characteristic -function of the set. -(iv) Many of the properties associated with random graphs are equivalent, and can -therefore be taken as sensible definitions of pseudo-random graphs. [CGW,T] -(v) Sometimes, the set of all eventually zero sequences of zeros and ones is a good -model for separable Banach spaces, or at least allows one to generate interesting -hypotheses. -(vi) Concentration of Measure -More examples (by Tao and other) you can see at -http://ncatlab.org/davidcorfield/show/Two+Cultures -Do you know another examples in various areas? I mean, for example, globalization techniques in topology (structure functor in Hirsh, Differential Topology, $\S 2.11$ and Mayer–Vietoris sequence, in Bott & Tu, Differential Forms in Algebraic Topology $\S 5$). -So, many proofs look like "prove the local version of theorem and globalize". -Do you know such principles? It should be more specific than undergraduate course but it should be common used in your branch and be situated in "common wisdom" of mathematics. - -REPLY [2 votes]: The Choquet theory in convex analysis / functional analysis / whatever you want to call it. An element of a convex set should be some kind of "average" of extreme points. This has the status of a theorem for compact sets in normed linear spaces but is a useful guiding principle for not-necessarily-compact sets in not-necessarily-normed linear spaces. Chapter 14 in Lax's Functional Analysis book gives good examples of the wide array of applications of the same simple idea.<|endoftext|> -TITLE: Smooth structures on the connected sum of a manifold with an Exotic sphere -QUESTION [16 upvotes]: What can we say about the connected sum of a manifold $M^n$ with an Exotic sphere? Is is possible some of them are still diffemorphic to $M^n$. Is it possible to classifying all the exotic smooth structures for a given $M^n$? - -REPLY [8 votes]: This is more a long comment on a special case of the first two questions than a complete answer, but I am surprised no one else mentioned this. -One can, even without a proof or refutation of the smooth 4-dimensional Poincare conjecture, state something about the case $M^4 = \sharp m S^2 \times S^2$, where $m$ is greater than or equal to a particular positive integer $k$. -Assume there exists an exotic 4-dimensional sphere $\textbf{S}^4$ homeomorphic but not diffeomorphic to a standard $S^4$. Then, by a theorem of Wall, there exists a positive integer $k$ such that $S^4 \sharp k S^2 \times S^2$ is diffeomorphic to $\textbf{S}^4 \sharp k S^2 \times S^2$. Thus, $S^4 \sharp m S^2 \times S^2$ is diffeomorphic to $\textbf{S}^4 \sharp m S^2 \times S^2$ for $m \ge k$. Trivially, $S^4 \sharp m S^2 \times S^2$ is diffeomorphic to $\sharp m S^2 \times S^2$. Thus, $\textbf{S}^4 \sharp m S^2 \times S^2$ is diffeomorphic to $\sharp m S^2 \times S^2$ for $m \ge k$.<|endoftext|> -TITLE: Characterize a continental divide -QUESTION [7 upvotes]: Here is something I've wondered about from time to time: The continental divide in North America is commonly described as the geographic line curve seperating points where a drop of water would drain to the Atlantic from those where it would drain to the Pacific. My question is how to characterize such a curve mathematically given a "reasonable" height function described over a region of the plane. I am not concerned with applied topography but also not interested in exotic pathologies. I'll propose a crude model now but feel free to propose a better one. -MODEL: The domain is the unit disk. A pre-mountain with peak at $(h,k,p)$ is a function $M=M(x,y)=\frac{p}{1+s((x-h)^2+(y-k)^2)}$ where $s>>0$ controls how steep it is and $p>>0$ how high. (note that a sum M_1+M_2 will have local maxima somewhat higher than $p_1$ and $p_2$ and somewhat displaced from $(h_i,k_i)$) The surface will be $b(x,y)(M_1+M_2+\cdots+M_n)$ where the $M_i$ are a large but finite number of pre-mountains and b(x,y) is a function such as $1-x^2$ or $1-x^2-\frac{y^2}{2}$ which is positive except at (-1,0) and (1,0) where it is 0. From each initial point the path of steepest gradient leads somewhere, usually (one might suppose) to $(1,0)$ or $(-1,0).$ - -Using the crude model as above, or a better one (describe it!) characterize the boundry between the basin of attraction of $(1,0)$ and that of $(-1,0)$ - -Comments: Of course a ring of mountains could create a pit with a sink in the middle, but that can be ignored or the problem can be changed to "characterize the boundries of the various basins of attraction". At a peak or saddle point the gradient is 0 but usually any direction one goes leads to the same sink. I imagine that there are (useful) applied approximate solutions starting from a grid of sample points with edges joining nearest neighbors. But I'd like some kind of minimax description like the solution of a continuous linear programing problem. - -REPLY [2 votes]: As Thierry and Gerry mentioned, Brian Hayes wrote an article "Dividing the Continent" -in American Scientist (Volume 88, Number 6, page 481), -reprinted in his book, Group Theory in the Bedroom -and Other Mathematical Diversions. His focus is algorithms to compute -the continental divide, and so does not shed much light on -the thrust of your question. But he does mention two interesting connections, which I will -mention in the hope that it triggers further associations. -First, there is considerable algorithmic work by those interested in watersheds. -For example, he cites the work of Luc Vincent and Pierre Soille, likely this paper: -"Watersheds in Digital Spaces: An Efficient Algorithm Based on Immersion Simulations," -(IEEE Transactions on Pattern Analysis and Machine Intelligence, -Volume 13 Issue 6, June 1991). Generally the algorithms are variants on flooding the surface from -minima, preventing the merging of water from different sources. -In image processing, this is called the watershed transformation. -E.g., see the images here. The watershed transformation is apparently available in MatLab. -Second, the problem was studied in some form by James Clerk Maxwell, although Hayes does not -give enough information (in the article—I don't have the book here) for me to locate a precise reference. Perhaps it is related to what -I know (from the work of Bob Connelly) as Maxwell-Cremona lifts? I would be interested to learn -if anyone knows. (See citation in comments.) Here is what Hayes says: - -Maxwell relates the number of topographic peaks, pits and saddles on a surface. In the case of a sphere, the formula is $p+q–s=2$, where $p$ is the number of peaks, $q$ the number of pits and $s$ the number of saddles. Maxwell also outlines a procedure for dividing the landscape into watershed regions.<|endoftext|> -TITLE: Escher, Conway, Kali, etc. -QUESTION [7 upvotes]: One can express the symmetry types of, say, Escher's "Circle Limit" prints using -Conway's orbifold notation, best known in the context of symmetries of Euclidean -plane patterns. -For example, Circle Limit III has symmetry type $433$ (with Euler characteristic $-1/12$). -Where can I find an explicit algorithm that produces generators for some appropriate -subgroup of the isometries of the Poincaré model of the hyperbolic plane given a suitable Conway notation? Only certain notation give rise to rigid orbifolds, so I'd also like -to know how to read off the number of moduli from the Conway notation. Absent rigidity, -I'd really like a parametric family of generating sets realizing all the distinct forms of -the underlying orbifold. -The popular program Kali facilitates drawing symmetric Euclidean patterns? Does anyone -distribute some appropriate hyperbolic counterpart? - -REPLY [2 votes]: Can't speak for the Conway -> generators, but for drawing, there is this -http://www.plunk.org/~hatch/HyperbolicApplet/ -I am not sure why "rummaging" was necessary for the D. Huson program, since there is a link to it on the "OrbifoldNotation" wikipedia page.<|endoftext|> -TITLE: Is every compact topological ring a profinite ring? -QUESTION [20 upvotes]: There are a lot of compact (Hausdorff) groups, whereas every compact field is finite. What about rings? Is there a classification theorem for compact rings? If you take a cofiltered limit of finite rings, you get a compact ring; for example, the $p$-adic integers $\mathbb{Z}_p$ are obtained as a limit of -$$ -\cdots \twoheadrightarrow \mathbb{Z}/p^{n+1}\mathbb{Z} \twoheadrightarrow \mathbb{Z}/p^n\mathbb{Z}\twoheadrightarrow \cdots \twoheadrightarrow \mathbb{Z}/p\mathbb{Z}\twoheadrightarrow 0. -$$ -Can every compact ring be obtained as a cofiltered limit of finite rings? -For a counterexample, a compact ring that is not totally disconnected would suffice. In the other direction, proving that such a ring has to be totally disconnected wouldn't suffice a priori: It would show the the additive group is profinite, but not that the ring is a cofiltered limit of rings. -Remark: By "compact," I consistently mean "compact Hausdorff." - -REPLY [6 votes]: The earliest reference I could find to the fact that compact Hausdorff rings are profinite (objects in the category of topological rings) is in Johnstone's "Stone Spaces" (VI.4.11 on page 266); in the bibliographical notes on page 269 it is attributed to Kaplansky, "Topological Rings" (Amer. J. Math. 69 (1947) 153-183). Interestingly, Kaplansky himself refers to Otobe ("On quasi-evaluations of compact rings," Proceedings of the Imperial Academy of Tokyo, 20 (1944), pp. 278-282) who in turn refers to Anzai ("On compact topological rings", same Proceedings, 19 (1943), 613-616); and Anzai says he owes the proof of the profiniteness part to Nakayama. -Decided to mention all this since none of the references in answers or comments seem to contain these attributions, except for the comment by Gjergji Zaimi "...it appears that the result is old and possibly due to Kaplansky, so I'd like to research that a little more." -PS Concerning units - actually Anzai does not require it, the only thing needed is that the zero is the sole element annihilating the whole ring.<|endoftext|> -TITLE: homology and cohomology of a quotient manifold -QUESTION [7 upvotes]: suppose $M$ is a manifold , $G$ is a lie group (may be finite) ,then let $G$ act on $M$ freely , $N=M/G$ is then a manifold ,so my question is what relations may be between the homology and cohomology of $M$ and $N$ ? - In the surface case ,if $G$ is a finite group ,then we can get no differences between the cohomolgy and homology of $M$ and $N$ then the euler number will be the same what will mean that the order of the group must be one ,and so there are no actions of finite groups freely acting on a surface and it is a very beautiful claim. -Does there any effective way to compute the cohomology of a quotient space when the action is not free? -So let us consider more about the question raised above , references about this question are also welcomed!! - -REPLY [2 votes]: To elaborate a little on Kostya's answer, for the case where $G$ acts freely you may want to look up the Cartan-Leray spectral sequence (eg Chapter 7 of the book by Brown). Also, transfer homomorphisms are a useful elementary tool (see section 3.G of Hatcher's book). -When $G$ doesn't act freely you're into the realm of orbifolds, about which I have much to learn!<|endoftext|> -TITLE: Geodesic rays and horofunctions -QUESTION [5 upvotes]: Let $(X,d)$ be a metric space. -Let $(x_n)_{n\in\mathbb N}$ be a geodesic ray: $d(x_n,x_m)=\vert n-m\vert$. - -Is it true that, for all $y\in X$, the sequence $(d(x_{n+1},y)-d(x_n,y))$ converges to $1$ as $n$ goes to infinity ? - -I am particularly interested in the case of $\delta$-hyperbolic spaces. A positive answer to the above question would imply that any geodesic ray converges to a Busemann function (or horofunction). -More generally, is anything known about Busemann functions on hyperbolic spaces ? In particular, how do the Busemann compactification relates to the visual boundary ? These two boundaries are the same for CAT(0) spaces, but need not be in general, as shown by the example $\mathbb Z\times \mathbb Z/2\mathbb Z$, with obvious generating set. - -REPLY [9 votes]: Yes. Let $a_n:=d(y,x_n)-|n|$. By the triangle inequality, we see that $(a_n)_{n\ge0}$ is non-increasing and bounded. Therefore it has a finite limit $a$. -Now if $n\ge0$, -$d(y,x_{n+1})-d(y,x_n)=a_{n+1}-a_n+1\rightarrow a-a+1=1$.<|endoftext|> -TITLE: Relation between holonomic D-modules and perverse sheaves -QUESTION [9 upvotes]: Given a smooth complex algebraic variety, the Riemann-Hilbert-correspondence tells us, that the category of perverse sheaves is equivalent to the category of regular, holonomic D-modules. -However not every interesting holonomic D-module is regular. For example the solution sheaves of all the $D_{\mathbb A^1}$-modules $\mathbb C[x]e^{\chi x}$ are isomorphic to the constant sheaf and only for $\chi=0$ our module $\mathbb C[x]e^{\chi x}$ is regular. -So my question is, is there an analogue of the Riemann-Hilbert correspondence if we replace regular by something else (and perhaps also perverse sheaves by something else)? -For example in the above example one could do the following: -One could fix a $\chi$ and tensor first with $\mathbb C[x] e^{-\chi x} $, -before applying the deRham functor. This gives an equivalence between perverse sheaves and holonomic modules with "$e^{-\chi x}$-like " singularities. - -REPLY [6 votes]: The answer is yes but it's not easy. You need additional data to describe the irregular part of your connexion. These are known as Stokes structures. Very loosely it's a filtration of your sheaf of solutions according to their growth in a given sector. Very recently, Claude Sabbah has written lecture notes on the subject (arXiv:0912.2762).<|endoftext|> -TITLE: Decomposing an arbitrary unitary representation of a connected nilpotent Lie group in terms of its irreps -QUESTION [7 upvotes]: For a locally compact (Hausdorff) abelian group $G$ we have following theorem (see e.g. Folland): -"For every (strongly continuous) unitary representation $(\pi,\mathcal{H_{\pi}})$ of $G$, there exists a unique regular $\mathcal{H}_{\pi}$-projection-valued measure $P$ on $\hat{G}$ such that $\pi$ decomposes as: -$\pi (g)=\int_{\hat{G}}\left\langle g,\chi\right\rangle dP\left(\chi\right)$ for every $g \in G$." -To which extent is this theorem true for nilpotent Lie groups (say, connected and simply connected)? That is, do we have a canonical decomposition of a unitary representation of such a group in terms of its irreducible unireps and some sort of measure on the unitary dual? -The proof of the above theorem has two major ingredients: the identification of the spectrum of $L^1 (G)$ with $\hat{G}$ when $G$ is abelian and the spectral theory of commutative Banach algebras. It is not clear to me whether any of these ingredients has a suitable analogue in the nilpotent case. Furthermore, in this case $\hat{G}$ is not a group or even a Hausdorff space, plus one would have to integrate an operator-valued function which assumes operators acting on different Hilbert spaces as its values. Thus I am not so sure if the standard theory of projection-valued measures can be so easily applied in this case. - -REPLY [2 votes]: Brad Currey has actually obtained a pretty explicit description of the Plancherel measure for general exponential solvable Lie groups which is used to decompose the representation into direct integral of irreducibles. Notice that nilpotent Lie groups are part of this larger class of exponential Lie groups. He uses the technique of jump indexes and orbit method. Here is his paper -http://mathcs.slu.edu/~currey/030401-Currey-1.pdf -Vignon S. Oussa<|endoftext|> -TITLE: Minimal dimension of maximal abelian subalgebras -QUESTION [9 upvotes]: Let $\mathfrak{g}$ be a semi-simple finite dimensional Lie algebra over $\mathbb C$. -Is it true that -$$ -\dim \mathfrak h \ge \mathop{rank} \mathfrak g -$$ -for any maximal abelian subalgebra $\mathfrak h \subset \mathfrak g$ ? Here by maximal I mean that $\mathfrak h$ is not properly contained in any other abelian subalgebra of $\mathfrak g$. - -REPLY [3 votes]: Some of my previous comments were unfortunately too casual and unfocused, but I still suspect that the answer to the original question is yes. At the same time, I can't document precisely enough what is known about the exceptional simple Lie algebras over $\mathbb{C}$. I'm not aware of any unified Lie-theoretic arguments, in any case. -For the classical simple types (including the trace zero matrices?), the older papers are sometimes hard to read but have many concrete details. The main tool used is linear algebra, inspired in part by a long tradition in the study of (associative) matrix algebras going back to Schur, Kravchuk, and Mal'cev. In particular, there is a series of papers by Patera, Winternitz, and Zassenhaus on classical real or complex Lie algebras, some published in the journal Linear Algebra and its Applications which is in our university library but not available online to us. One article I can access online is typical: -MR713527 (84i:17006) 17B05 (22E10), -Patera, J. (3-MTRL-R); Winternitz, P. (3-MTRL-R); Zassenhaus, H. (1-OHS), -Maximal abelian subalgebras of real and complex symplectic Lie algebras. -J. Math. Phys. 24 (1983), no. 8, 1973–1985. -Their emphasis is on classifying the maximal ones up to conjugacy under the adjoint group, with emphasis on those like the Cartan subalgebras and the contrasting ones consisting of nilpotent matrices. Dimension considerations seem secondary. But as far as I can see, over $\mathbb{C}$ they always find that the minimum dimension of a maximal abelian subalgebra is equal to the rank, while the maximum dimension tends to be larger. -Concerning jvp's own references to work of Gerstenhaber and others on the full matrix algebra, it has to be kept in mind that the simple linear Lie algebras in question aren't usually closed under ordinary matrix multiplication. This makes comparisons of commutative subalgebras and abelian Lie algebras tricky, I think. However, I haven't gone far enough into the literature to feel confident about exactly what is true. It would help to have a modern survey of the entire problem area for both matrix algebras and semisimple Lie algebras. -ADDED: Though I haven't yet seen the Courter paper or his pre-computer proof method, the description here and jvp's reported computer verification suggests that for type A this Lie subalgebra of the special linear algebra consisting of nilpotent upper triangular matrices is maximal and of dimension less than the rank. On the other hand, the published results for classical simple Lie algebras suggest that the rank is a lower bound in those types. I'm uncertain how much is known about the exceptional types, so I guess I'd emphasize the last line of my earlier answer and the related question whether a Lie-theoretic understanding of the problem exists (beyond linear algebra). But I'm happy to drop my original guess that the rank is always the minimum dimension if Courter's proof says otherwise.<|endoftext|> -TITLE: Proofs that require fundamentally new ways of thinking -QUESTION [212 upvotes]: I do not know exactly how to characterize the class of proofs that interests me, so let me give some examples and say why I would be interested in more. Perhaps what the examples have in common is that a powerful and unexpected technique is introduced that comes to seem very natural once you are used to it. -Example 1. Euler's proof that there are infinitely many primes. -If you haven't seen anything like it before, the idea that you could use analysis to prove that there are infinitely many primes is completely unexpected. Once you've seen how it works, that's a different matter, and you are ready to contemplate trying to do all sorts of other things by developing the method. -Example 2. The use of complex analysis to establish the prime number theorem. -Even when you've seen Euler's argument, it still takes a leap to look at the complex numbers. (I'm not saying it can't be made to seem natural: with the help of Fourier analysis it can. Nevertheless, it is a good example of the introduction of a whole new way of thinking about certain questions.) -Example 3. Variational methods. -You can pick your favourite problem here: one good one is determining the shape of a heavy chain in equilibrium. -Example 4. Erdős's lower bound for Ramsey numbers. -One of the very first results (Shannon's bound for the size of a separated subset of the discrete cube being another very early one) in probabilistic combinatorics. -Example 5. Roth's proof that a dense set of integers contains an arithmetic progression of length 3. -Historically this was by no means the first use of Fourier analysis in number theory. But it was the first application of Fourier analysis to number theory that I personally properly understood, and that completely changed my outlook on mathematics. So I count it as an example (because there exists a plausible fictional history of mathematics where it was the first use of Fourier analysis in number theory). -Example 6. Use of homotopy/homology to prove fixed-point theorems. -Once again, if you mount a direct attack on, say, the Brouwer fixed point theorem, you probably won't invent homology or homotopy (though you might do if you then spent a long time reflecting on your proof). - -The reason these proofs interest me is that they are the kinds of arguments where it is tempting to say that human intelligence was necessary for them to have been discovered. It would probably be possible in principle, if technically difficult, to teach a computer how to apply standard techniques, the familiar argument goes, but it takes a human to invent those techniques in the first place. -Now I don't buy that argument. I think that it is possible in principle, though technically difficult, for a computer to come up with radically new techniques. Indeed, I think I can give reasonably good Just So Stories for some of the examples above. So I'm looking for more examples. The best examples would be ones where a technique just seems to spring from nowhere -- ones where you're tempted to say, "A computer could never have come up with that." -Edit: I agree with the first two comments below, and was slightly worried about that when I posted the question. Let me have a go at it though. The difficulty with, say, proving Fermat's last theorem was of course partly that a new insight was needed. But that wasn't the only difficulty at all. Indeed, in that case a succession of new insights was needed, and not just that but a knowledge of all the different already existing ingredients that had to be put together. So I suppose what I'm after is problems where essentially the only difficulty is the need for the clever and unexpected idea. I.e., I'm looking for problems that are very good challenge problems for working out how a computer might do mathematics. In particular, I want the main difficulty to be fundamental (coming up with a new idea) and not technical (having to know a lot, having to do difficult but not radically new calculations, etc.). Also, it's not quite fair to say that the solution of an arbitrary hard problem fits the bill. For example, my impression (which could be wrong, but that doesn't affect the general point I'm making) is that the recent breakthrough by Nets Katz and Larry Guth in which they solved the Erdős distinct distances problem was a very clever realization that techniques that were already out there could be combined to solve the problem. One could imagine a computer finding the proof by being patient enough to look at lots of different combinations of techniques until it found one that worked. Now their realization itself was amazing and probably opens up new possibilities, but there is a sense in which their breakthrough was not a good example of what I am asking for. -While I'm at it, here's another attempt to make the question more precise. Many many new proofs are variants of old proofs. These variants are often hard to come by, but at least one starts out with the feeling that there is something out there that's worth searching for. So that doesn't really constitute an entirely new way of thinking. (An example close to my heart: the Polymath proof of the density Hales-Jewett theorem was a bit like that. It was a new and surprising argument, but one could see exactly how it was found since it was modelled on a proof of a related theorem. So that is a counterexample to Kevin's assertion that any solution of a hard problem fits the bill.) I am looking for proofs that seem to come out of nowhere and seem not to be modelled on anything. -Further edit. I'm not so keen on random massive breakthroughs. So perhaps I should narrow it down further -- to proofs that are easy to understand and remember once seen, but seemingly hard to come up with in the first place. - -REPLY [4 votes]: Having read his name among the commenters, I thought I ought to mention his theorem: Monsky's theorem is proved using valuations from algebra.<|endoftext|> -TITLE: Rank of a free module without the axiom of choice -QUESTION [9 upvotes]: Perhaps my question is really naive. I teach this semester in algebra. I am embarassed about the proof that a free module over an integral domain has a well-defined rank. It is based on the theorem that two bases have the same cardinal. But the proof I know of this statement uses a reduction modulo a maximal ideal (then we are brought back to the theory of vector spaces). It seems to me that the existence of a principal ideal needs in general the axiom of choice (not for concrete domains, of course). Whence my question: - -Do we really need the axiom of choice to prove in full generality that the bases of a free module have the same cardinal ? - -One day later. I did not realize that the question even makes sense in the context of vector spaces. I'd like being able to accept a comment. - -REPLY [6 votes]: Hans Läuchli [Auswahlaxiom in der Algebra, Comment. Math. Helv. 37, MR0143705, DOI:10.5169/seals-28602] has constructed a model of ZFA wherein a vector space has two bases of different size. (See also Jech, The Axiom of Choice, §10.3, problem 5.) The Jech–Sochor Embedding Theorem allows to transfer the result to ZF.<|endoftext|> -TITLE: Non finitely-generated subalgebra of a finitely-generated algebra -QUESTION [28 upvotes]: Ok, I feel a little bit ashamed by my question. -This afternoon in the train, I looked for a counter-example: -— $k$ a field -— $A$ a finitely generated $k$-algebra -— $B$ a $k$-subalgebra of $A$ that is not finitely generated -Finally, I have found this: -— $k$ any field -— $A=k[x,y]$ -— $B=k[xy, xy^2, xy^3, \dots]$ -(proof : exercise) -My questions are: -1) What is your usual counter-example ? -2) Under which conditions can we conclude that $B$ is f.g. ? -3) How would you interpret geometrically this counter-example ? - -REPLY [12 votes]: Dear Nicojo, since you now have many counter-examples, let me give you a situation where $B$ is finitely generated, in line with your question 2). I am going to adopt your notations with the important caveat that $k$ is a ring which needn't be a field . -Theorem of Artin-Tate Consider the inclusions of rings $k \subset B \subset A$ . Suppose that $k$ is Noetherian, that $A$ is a finitely generated algebra over $k$ and that $A$ is a finitely generated module over $B$. Then $B$ is a finitely generated algebra over $k$. -You might interpret this as saying that when $B$ is sufficiently close to $A$, finite generation is preserved. -You can find the proof in Atiyah-Macdonald, Proposition 7.8, page 81. -From this theorem you can then prove Zariski's result that an extension of fields that is finitely generated as an algebra is actually a finite-dimensional extension (Proposition 7.9 page 82 loc.cit.) and then Hilbert's Nullstellensatz is literally an exercise: exercise 14, page 85 . So this result of Artin-Tate is really basic in commutative algebra and algebraic geometry, not surprisingly if you consider the authors (the Artin here is Emil, Mike's father.)<|endoftext|> -TITLE: Relationships between the roots of an entire function and the roots of its derivative -QUESTION [7 upvotes]: Hey everyone, -I would like to know if anybody could help me find references for the following. -Take a suitably well defined entire function $f(x)$ and it's derivative $\tilde{f}(x)$ to which the roots $x_n$ and $\tilde{x}_n$ are associated. The function $f(x)$ may have infinitely many roots, though naturally one needs to be more careful in this case. Let's discount these subtleties for the time being. Define -$$Z(s)=\sum_n\frac{1}{x_n^s} \hspace{8mm} \textrm{and} \hspace{8mm} \tilde{Z}(s)=\sum_n\frac{1}{\tilde{x}_n^s}.$$ -One of the identities which I have proven which relates these is -$$\tilde{Z}(3)=Z(3)-\left(\frac{Z(2)}{Z(1)}\right)^3+3\left(\frac{Z(3)Z(2)}{(Z(1))^2}-\frac{Z(4)}{Z(1)}\right).$$ -I also have proven other identities (some much more simple) of this form. I have searched the internet, journals and every analysis book in my University library and found nothing of the sort. Also the main applications which I would expect to find curiously also do not appear in any of the aforementioned sources. -Edit -Just to say thanks to those who replied. The question has been answered :) - -REPLY [3 votes]: Let $f(x) = (1 - r_1 x)...(1 - r_n x)$ be a polynomial. Then $f(x) = 1 - e_1 x^1 + e_2 x^2 \mp ... $ where the $e_i$ are the elementary symmetric functions in the $r_i$. We define also $p_k = \sum_i r_i^k$, the power symmetric functions in the $r_i$. Then Newton's identities state that -$$ke_k = \sum_{i=1}^{k} (-1)^{i-1} e_{k-i} p_i.$$ -This identity is equivalent to the generating function identity -$$\frac{f'(x)}{f(x)} = \sum_{i=1}^{n} \frac{r_i}{1 - r_i x} = \sum_{k \ge 0} p_{k+1} x^k$$ -which follows from taking logarithmic derivatives on both sides. Now, you want to relate the functions $p_k$ to the functions $\tilde{p}_k$, the power symmetric functions of the reciprocals of the roots of the derivative $f'(x)$. Applying Newton's identities to $f'(x) = -e_1 + 2e_2 x - 3e_3 x^2 \pm ...$ gives -$$k(k+1) \frac{e_{k+1}}{e_1} = \sum_{i=1}^k (-1)^{i-1} \frac{(k+1-i) e_{k+1-i}}{e_1} \tilde{p}_i.$$ -This pair of identities should get you your results (at least formally, letting $n \to \infty$). You may or may not find it useful to write Newton's identities in the equivalent form -$$e_n = \frac{1}{n!} \sum_{\sigma \in S_n} \text{sgn}(\sigma) p_{\sigma}$$ -where $p_{\sigma} = p_{\lambda_1} ... p_{\lambda_i}$, where $\sigma$ has cycle type $(\lambda_1, ... \lambda_i)$. This gives -$$\frac{(n+1) e_{n+1}}{e_1} = \frac{1}{n!} \sum_{\sigma \in S_n} \text{sgn}(\sigma) \tilde{p}_{\sigma}.$$<|endoftext|> -TITLE: Why need the morphisms to form a set ? -QUESTION [7 upvotes]: For a category $C$ it is required that the morphisms of any two objects of $C$ form a set (c.f. Lang: Algebra, or Weibel: An introduction to homological algebra). -What's the point about this requirement ? Would there be any disadvantages / logical deficiencies if one allows the morphisms to form a proper class ? - -REPLY [4 votes]: As David mentioned, category theorists generally tend to shy away from such `size issues', because in a sense they do not touch the heart of the matter, as the original question rightfully suggests. Apart from such foundational issues, I can think of two practical reasons where it is important that homs be sets. -First, there is Freyd's celebrated Adjoint functor theorem. It gives conditions that characterize precisely when a given functor has an adjoint. Crucially, one of the conditions, called the `solution set condition', is that a certain class is in fact a set. This shows that size issues do play a fundamental role in category theory, which came as quite a surprise to most people. -Second, one can think of Enriched categories. It turns out that a lot of category theory goes through if homs are not necessarily objects of the category Set of sets and functions, but objects in an arbitrary monoidal category, with composition being a morphism of that category. For example, relating to Kevin and Qiaochu's comments above, a 2-category can be seen as a category enriched in Cat, the category of categories. But this also gives some surprising examples. For example, a metric space can be seen as a category enriched in $\mathbb{R}^+$, i.e. the poset $[0,\infty]$ with monoidal structure given by addition. And of course a locally small category is just a Set-enriched category. This is not an argument against `large categories' per se, but does indicate that a lot of murky waters can be avoided by only considering locally small categories.<|endoftext|> -TITLE: on common fixed points of commuting polynomials (and rational functions) -QUESTION [11 upvotes]: By the Ritt's classification, for any pair of commuting polynomials (i.e. $f(g(z))=g(f(z))$) over $\mathbb C$ there is a common fixed point of them. My questions are: - -Is that true that this can be obtained rather simpler than with Ritt's classification? -Is that true for any pair of commuting rational functions? - -REPLY [3 votes]: I don't know the answers. Consider this as an extended comment. -Let $P$ be the set of fixed points of $g$. Then $f$ maps $P$ into itself. -Indeed let $x\in P$, so $g(x)=x$. Then -$$f(x)=f(g(x))=g(f(x)),$$ -which means that $f(x)\in P$. Now $P$ is finite, so $f$ must have -a periodic point in $P$. So $f^m$ and $g$ have a common fixed point. -Of course one can interchange $f$ and $g$ here. -With little more care, one can find a common fixed point of $f^m$ and $g^n$ -with some $m,n$, which is REPELLING for both $F=f^m$ and $G=g^n$. -Let this common repelling fixed point be $a$. -Evidently $F$ commutes with $G$. If two communing functions share a -repelling fixed -point, then they have the same Poincare function at this point. -Poincare function $\phi$ is the ``linearizer'', that is the solution -of the functional equation -$$\phi(\lambda z)=F(\phi(z)), \quad \phi(0)=a,\quad\phi'(0)=1, -$$ -where $\lambda=F'(a)$. -Now if $F'(a)=G'(a)$ we easily conclude that $F=G$. -Which means that $f$ and $g$ have a common iterate. -If $F'(a)\neq G'(a)$ there is only a very restricted set of possibilities -which were completely described and classified by Ritt. -Thus we come to a question: - -Suppose that $f$ and $g$ have a common iterate. - Must they have a common fixed point? - -Or, perhaps those pairs which do not -permit some explicit classification? -The question about rational functions that have a common iterate is interesting in itself -and I am posting it separately, Rational functions with a common iterate<|endoftext|> -TITLE: Is there a functorial proof that Eilenberg-MacLane spaces are unique up to homotopy equivalence? -QUESTION [12 upvotes]: Let $(G, n)$ be a pair where $G$ is an abelian group and $n \in \mathbb{N}$. Recall that an Eilenberg-MacLane space is a connected CW complex $X$ such that $\pi_r(X) = G$ if $r=n$ and $0$ otherwise. An Eilenberg-MacLane space is unique up to homotopy equivalence. A direct argument is given in Hatcher's book based on the Whitehead theorem (that a weak equivalence of CW complexes is a homotopy equivalence): one shows that there is always an Eilenberg-MacLane space of a given form (that starts with a wedge of $r$-spheres and then goes up). Then, one explicitly checks that given any Eilenberg-MacLane space, one can define a map from this one into the other one which induces an isomorphism on the homotopy group in dimension $n$, though. This relies on a particular construction of the Eilenberg-MacLane space, though. -Is there a more functorial way of seeing this? -One half of an approach is the following. We know that the Eilenberg-MacLane spaces represent the cohomology functors (with coefficients in $G$) on $CW_*$, the homotopy category of pointed CW complexes. This implies uniqueness. -But I don't think this is logically correct given the above constraint. Namely, the fact that any Eilenberg-MacLane space represents the cohomology functor relies, I think, on the fact that any two are homotopy equivalent. Namely, the proof I know of this fits the various $\{ K(G,n) \}$ into an $\Omega$-spectrum and then argues that they represent a cohomology theory which satisfies the dimension axiom, so is singular cohomology with coefficients. I don't see how it is obvious that any $K(G,n)$ can be fit into an $\Omega$-spectrum, though, without using the homotopy equivalence above. - -REPLY [3 votes]: If you accept the Whitehead theorem, then simple obstruction theory (without cohomology) determines the homotopy type of spaces with homotopy groups in only one dimension. That is, -you can easily show that if the only nonzero homotopy group of the CW complex $Y$ -is in dimension $n$, then for any $(n-1)$-connected CW complex -$$ -[X, Y] \cong \mathrm{Hom}( \pi_n(X), \pi_n(Y) ) . -$$ -Then if the homotopy groups of $X$ are also concentrated in dimension $n$, -an isomorphism $\pi_n(X) \xrightarrow{\cong} \pi_n(Y)$ gives rise to -a homotopy equivalence $X\xrightarrow{\simeq} Y$. -But what if we don't want to use Whitehead? I think you are stuck, because basic -(model category theoretic) arguments can be used to prove Whitehead (at least for -CW complexes with finitely many finitely generated homotopy groups?) from the uniqueness -of Eilenberg-Mac Lane spaces. -(Prove it by induction for CW complexes with at most $n$ nonzero homotopy groups; this requires us to convert maps to fibrations, and we don't want to end up with non-CW complexes, which is why I said we might need finitely generated homotopy groups.) -CLARIFICATION: When I said "model-theoretic" I meant the basic facts about cofibrations, -fibrations, induced maps, etc. The finite generation comes in when we convert a map to a fibration: this is done with a path space and a pullback, and these things can easily take us away from CW complexes, but my plan was to impose conditions that would guarantee we can apply Milnor's theorem (on loops of CW complexes and so on) to the CW complexes in question.<|endoftext|> -TITLE: Non-negative quadratic maximization -QUESTION [6 upvotes]: For a given symmetric and positive semidefinite $n \times n$ matrix $A$, we want to solve the problem $$\max_{||x|| = 1, \ x\geq 0} x^T A x.$$ -Here, $x\geq 0$ indicates that $x$ must be component-wise non-negative. Without the $x\geq 0$ constraint, the solution $x$ must be the eigenvector corresponding to the largest eigenvalue (from Ky Fan). Is there a well-known solution for the case when $x\geq 0$? -If not, are there any good relaxations (or randomized algorithms) to find $x$? For instance, are there any approximation bounds on how far from the optimum is the "largest eigenvector" of $A$? - -REPLY [11 votes]: First, note that the condition that $A$ be positive semidefinite (PSD) doesn't buy you anything. Replacing $A$ by $A+kI$ changes the objective value of any feasible solution by $k$, so if we could solve the given problem when the matrix in the objective is PSD, we could just choose some $k$ large enough to make $A+kI$ PSD, solve the resulting problem with $A+kI$, and subtract $k$ to get the answer to the problem with $A$ instead. -A matrix is called copositive if $x^T A x\geq 0$ for all $x\geq 0$. Checking copositivity of $A$ is the same as checking whether the optimal value of your problem on $-A$ is nonpositive. As shown by Murty and Kabadi ("Some NP-Complete Problems in Quadratic and Nonlinear Programming"), this problem is co-NP-complete. Thus your given problem is NP-hard. -That said, it is a quadratically constrained quadratic program ("QCQP"), which is a well-studied kind of problem, so I would suggest looking into these. For certain classes of them there are well-performing approximation algorithms. In general there are SDP relaxations, although proving performance guarantees is often difficult. -One last thing to note: at least when $A$ is PSD, your formulation is equivalent to one in which the equality constraint is replaced by $\leq$, and then the feasible set will be convex. So you may see the problem in this form. -EDIT: I'm not sure if this is the kind of thing you are looking for, but there is a sequence of SDP relaxations which will give you upper bounds on the objective function. To construct these, we will need the completely positive matrices, which form a convex cone dual to the copositive matrices above. The completely positive matrices are convex combinations of outer products $xx^T$ where $x\geq 0$. -Your problem can be rewritten as maximizing $Tr(AX)$ subject to the conditions that $Tr(X) = 1$ and $X = xx^T$ for $x\geq 0$. Since the objective is now linear, we can go ahead and convexify the feasible set to get the equivalent problem of maximizing $Tr(AX)$ subject to $Tr(X)=1$ and $X$ is completely positive. -Now, being dual to copositivity, complete positivity is also hard to test. But there are a nice sequence of SDP relaxations which give tighter and tighter outer approximations to the completely positive cone; these are just dual to the inner approximations of the copositive cone given by Parrilo ("Semidefinite Programming Based Tests for Matrix Copositivity"). For example, the first relaxation is that $X$ be elementwise nonnegative and PSD, two obvious necessary conditions for complete positivity (in fact these are sufficient for $4\times 4$ and smaller matrices, in which case the relaxation is exact). -Substituting in any such relaxation will give you an SDP which upper bounds the value of your problem of interest.<|endoftext|> -TITLE: Nonessential use of large cardinals -QUESTION [16 upvotes]: In Awfully sophisticated proof for simple facts, we are asked for examples of complex proofs of simple results. To quote from the questioner's post, we are asked for proofs that are akin to "nuking mosquitos." In set theory, a natural "nuke" with respect to a certain result is a large cardinal axiom with unnecessarily high consistency strength (i.e. applying to a much stronger collection of axioms than is required to provide a proof of the possibility of the result in question). -A research focus in set theory is a search for large cardinal axioms with the weakest consistency strength that can be used to prove the possibility of a certain result. My question is of an opposing nature: -Can you think of results that can be proven in a different manner by appealing to a large cardinal axiom with unnecessarily large consistency strength? -There are plenty such examples where the proofs become less technical (e.g., using a $\kappa^{++}$-supercompact cardinal $\kappa$ to show that the GCH can fail at a measurable cardinal is much more than is required), but I'm thinking of examples where the original proof was accomplished without such a strong large cardinal hypothesis or any large cardinal hypothesis at all. For example (from my post to the aforementioned question): -Theorem (ZFC + "There exists a supercompact cardinal."): There is no largest cardinal. -Proof: Let $\kappa$ be a supercompact cardinal, and suppose that there were a largest cardinal $\lambda$. Since $\kappa$ is a cardinal, $\lambda \geq \kappa$. By the $\lambda$-supercompactness of $\kappa$, let $j: V \rightarrow M$ be an elementary embedding into an inner model $M$ with critical point $\kappa$ such that $M^{\lambda} \subseteq M$ and $j(\kappa) > \lambda$. By elementarity, $M$ thinks that $j(\lambda) \geq j(\kappa) > \lambda$ is a cardinal. Then since $\lambda$ is the largest cardinal, $j(\lambda)$ must have size $\lambda$ in $V$. But then since $M$ is closed under $\lambda$ sequences, it also thinks that $j(\lambda)$ has size $\lambda$. This contradicts the fact that $M$ thinks that $j(\lambda)$, which is strictly greater than $\lambda$, is a cardinal. - -For the people who are unfamiliar with large cardinal embeddings, let me mention that the critical point of an embedding $j$ is the first ordinal $\kappa$ that is moved (i.e., $j(\alpha) = \alpha$ for all $\alpha$ less than the critical point $\kappa$ and $j(\kappa) > \kappa$.) A cardinal $\kappa$ is $\theta$-supercompact if there exists an elementary embedding $j: V \rightarrow M$ into a transitive (proper class) $M$ with critical point $\kappa$ such that $M^{\theta} \subseteq M$ and $j(\kappa) > \theta$. A cardinal is supercompact if it is $\theta$-supercompact for all $\theta$. - -REPLY [7 votes]: When dealing with the singular cardinals hypothesis ($SCH$), one may face with many such examples, let me say a few: -$\star_1:$ The consistency of the failure of $SCH$ was proved first by silver using supercompact cardinals. Later, Woodin reduced it to large cardinals up to strong cardinals, and finally Gitik showed that a measurable cardinal with $o(\kappa)=\kappa^{++}$ is suffices (which is also necessary). -$\star_2:$ Magidor first proved the consistency of $GCH$ below $\aleph_\omega$ with $2^{\aleph_\omega}=\aleph_{\omega+2}$ from a supercompact cardinal and a huge cardinal above it. Later Woodin reduced it to the level of strong cardinals, and finally it turned out that a measurable cardinal with $o(\kappa)=\kappa^{++}$ is suffices. -$\star_3:$ Foreman and Woodin proved the consistency of the total failure of $GCH$ from a supercompact cardinal and infinitely many inaccessibles above it. Later, it turned out that a $(\kappa+3)$-strong cardinal is suffices (and even less is needed).<|endoftext|> -TITLE: Landau's constant -QUESTION [12 upvotes]: (Hi. This is my first question here.) -A well known result in complex analysis says that there is an $\varepsilon\gt 0$ such that if $f$ is holomorphic in (a neighborhood of) the closed disk ${\mathbb D}$ of radius 1, and $f'(0)=1$, then $f({\mathbb D})$ contains a disk of radius $\varepsilon$. -This is due to Landau and, accordingly, the largest possible $\varepsilon$ is called Landau's constant. The standard proof (see, for example, the book Complex Variables by Berenstein-Gay) gives $\varepsilon\ge1/16$. -As far as I understand, the best known bounds are -$$ \frac 12\lt \varepsilon\le\frac{\Gamma(\frac13)\Gamma(\frac56)}{\Gamma(\frac16)}=0.54325\dots $$ -However, I have been unable to locate any proofs of the first inequality, or any updated treatments of the second one (due to Rademacher). For all I know, current bounds may be better, or there may be a standard source to read about this. - -Could you please give me some suggestions on where to look, or ideas on how to improve the $1/16$ bound, even if shorter of $1/2$? - -(Many thanks!) - -REPLY [4 votes]: Is the following reference helpful? -Current status of Bloch Constant and Landau Constant bounds<|endoftext|> -TITLE: Galois theory: Generalization of Abel’s Theorem? (Better version!) -QUESTION [15 upvotes]: (Unintentionally I have previously asked a similar and perhaps in itself not uninteresting -question -Galois theory: Generalization of Abel's Theorem? -but this is what I originally had in mind.) - -Let $L$ stand for the smallest extension of ${\Bbb Q}$ closed under the operation of adjoining all roots of all polynomials of the form $x^n+ax+b,a,b∈L$. -What polynomials $p$ don't split over $L$? In particular, how low -can one make the degree of such a $p$? (This -http://en.wikipedia.org/wiki/Bring%E2%80%93Jerrard_form#Bring.E2.80.93Jerrard_normal_form -would seem to guarantee degree($p$) $> 5$.) -Classically, $S_n$ occurs as a Galois group for -certain $x^n+ax+b$, $n\geq 5$. That means that -obstructions for $p$ splitting over such $L$ -must reflect information beyond the Galois group -of $p$. So absent a full answer to my question, -what candidates does one have for such an obstruction? -For example, does the form of the polynomial single -out particular representations of $S_n$? -Again, absent a full answer, does the literature contain theorems about -polynomials not splitting over similar large extension of ${\Bbb Q}$? - -REPLY [5 votes]: A minor note: it suffices to limit oneself to closure under roots of polynomials of form $x^n + b$ or $x^n + x + b$, since any other polynomial of form $x^n + ax + b$ can be transformed into the latter by the change of variables $x = a^{1/(n-1)}y$ (and $a^{1/(n-1)}$ is 'available' by virtue of the former).<|endoftext|> -TITLE: If a map restricts to an isomorphism on a closed subscheme and its open complement, must it be an isomorphism? -QUESTION [14 upvotes]: Suppose $f:X\to Y$ is a morphism of smooth connected schemes (over some base). Say $Z\subseteq Y$ is a closed subscheme with complement $U$ so that $f$ pulls back (restricts) to isomorphisms on $Z$ and $U$. Does it follow that $f$ is an isomorphism? - -If we drop the condition that $X$ and $Y$ have to be smooth, the normalization of a cusp is a counterexample. If we remove the hypothesis that $X$ and $Y$ must be connected, taking $X=U\sqcup Z$ gives a counterexample. - -REPLY [5 votes]: This is essentially the argument in Bhargav's comment. Matt Satriano showed me the separatedness argument. -We first apply the valuative criterion for separatedness to show that $f:X\to Y$ is separated. In fact, this shows that any morphism which is injective on topological spaces must be separated. Suppose $\Delta$ is the spectrum of a valuation ring with closed point $\ast$. Given any map $\Delta\to Y$, there is a unique set theoretic lift to $X$. Therefore, any failure of the valuative criterion will be witnessed on the induced map from an affine open neighborhood of $\ast$ in $X$ to an affine open neighborhood of $\ast$ in $Y$. Since any morphism of affine schemes is separated, the valuative criterion holds, so $f$ is separated. -Since $f$ induces isomorphisms on geometric points, it is quasi-finite. Supposing $Y$ is normal, integral and locally noetherian and $f$ is finite type, Zariski's Main Theorem† tells us that $f$ is an open immersion. Since $f$ is surjective, it is an isomorphism. -† I'm always surprised at how everything seems to be ZMT, so here's a precise reference: EGA III, Corollary 4.4.9. If $Y$ is normal, integral, and locally noetherian, $f:X\to Y$ is separated, birational, finite type, and quasi-finite, then $f$ is an open immersion.<|endoftext|> -TITLE: Can the I-fold direct product be free? -QUESTION [7 upvotes]: Let $A$ be a non-zero commutative ring with unit, $I$ a infinite set. - -Can $\prod_{i\in I}A$ be free as an $A$-module? - -I found when $A$ is a field or is isomophic to $\mathbb{Z}/m\mathbb{Z}$, then it is free. -But even when $A=\mathbb{Z}$, it is not free. (Baer Specker group) -It seems $\prod_{i\in I}A$ is always not free when $A$ is a domain with dim$\geq1$? -But I've found it difficult to prove. So I want to prove that when $A$ is PID, $\prod_{i\in I}A$ is always not free. -It is suffice to prove that when $I=\mathbb{N}$, $\prod_{i\in \mathbb{N}}A$ is not free. -we have already checked that when $A$ is not DVR,then $\prod_{i\in \mathbb{N}}A$ is not free. (similar to $\mathbb{Z}$) -we remain DVR need to check. -My question is : - -Is any other $A$ such that $\prod_{i\in I}A$ is free as an $A$-module ? -Is $\prod_{i\in I}A$ always NOT free when $A$ is an integral domain with dim$\geq1$? - -REPLY [11 votes]: If $A$ is a noetherian domain and not a field then the infinite product $M=A\times A\times \dots$ is not free. Suppose there is a basis. For $x\in M$ define its support to be the finite set of basis elements for which the coefficient is not zero. Note that if the supports of $x$ and $y$ are disjoint then their union is the support of $x+y$. Choose $\pi\in A$ neither zero nor invertible. Define the $n$-support of $x$ to consist of those basis elements for which the coefficient is not divisible by $\pi^n$. Note that $n$-support is contained in $(n+1)$-support is contained in support. -Choose an infinite sequence of nonzero elements $m_1,m_2,\dots $ of $M$ such that -(1) $m_n$ projects to zero in the first $n-1$ factors of the infinite product, -(2) the $m_n$ have pairwise disjoint support. -To get $m_n$ when all the previous $m_k$ are given, you just have to know that the kernel of a certain map from $M$ to a finite product of copies of $A$ is nontrivial (project the product on the first $n-1$ factors and project the free module on the span of a finite subset of basis). -Then divide each $m_n$ by as high a power of $\pi$ as possible; this preserves 1 and 2 while also arranging -(3) $m_n$ is nonzero mod $\pi$. -Now let $s_n=\pi m_1+\pi^2 m_2+\dots +\pi^nm_n$ and let $s$ be the limit of $s_n$ (defined because of 1). -The contradiction is that the support of $s$ must contain arbitrarily large finite sets $S_n$: Let $S_n$ be the $(n+1)$-support of $s_n$. Then the support of $s$ contains the $(n+1)$-support of $s$, which equals $S_n$. And $S_n$ properly contains $S_{n-1}$ because it is the disjoint union of the $(n+1)$-support of $s_{n-1}$ and the $(n+1)$-support of $\pi^nm_n$, this last being the (by 3 nonempty) $1$-support of $m_n$. -EDIT This implies that if $A$ is noetherian and has dimension $>0$ then the infinite product is not free, because $(A/P)\otimes \prod A=\prod (A/P)$ if $P$ is a finitely generated ideal -- choose $P$ to be a non-maximal prime. Also, the argument above proves more than I said: for a noetherian domain the infinite product is not even a submodule of a free module.<|endoftext|> -TITLE: Is the Invariant Subspace Problem interesting? -QUESTION [29 upvotes]: There's an amusing comment in Peter Lax's Functional Analysis book. After a brief description of the Invariant Subspace Problem, he says (paraphrasing) "...this question is still open. It is also an open question whether or not this question is interesting." -To avoid lengthy discussions involving subjective views about what makes math interesting, I'd simply like to know if there are examples of math papers out there that begin with something like, "Suppose the invariant subspace problem has a positive answer..." -Of course, papers that are about the ISP itself don't count! - -REPLY [4 votes]: If the invariant subspace problem has a positive answer then every bounded operator $A \in B(H)$ can be put in upper triangular form, in the sense that there is a maximal chain $(E_\lambda)$ of closed subspaces of $H$ such that every $E_\lambda$ is invariant for $A$. -In $\mathbb{C}^n$, a maximal chain of subspaces looks like $$\{0\} = E_0 \subset E_1 \subset \cdots \subset E_n = \mathbb{C}^n,$$ where the dimension of $E_i$ is $i$, and any operator for which all the $E_i$ are invariant is literally upper triangular for an orthonormal basis whose first $i$ elements belong to $E_i$, for all $i$. The infinite dimensional version is a natural generalization and seems to say rather a lot about the structure of $A$. -No doubt this result would be considered "known" by experts, but I could not find it explicitly stated anywhere, so I included it near the end of this paper. Also, Matt Kennedy answered this question with a reference to a result that easily implies it.<|endoftext|> -TITLE: Noether-Deuring for injections and surjections? -QUESTION [13 upvotes]: Noether-Deuring theorem (not in the strongest form, but in the one I usually need): -Let $L\diagup K$ be a field extension. Let $A$ be a $K$-algebra which is finite-dimensional as a vector space over $K$. Let $U$ and $V$ be two left $A$-modules which are finite-dimensional as vector spaces over $K$. If $U\otimes_K L \cong V\otimes_K L$ as $A\otimes_K L$-modules, then $U\cong V$ as $A$-modules. -This was nicely proven by Torsten Ekedahl in Hilbert 90 for algebras . The same proof shows something more general: If $U\otimes_K L$ is isomorphic to a direct summand of $V\otimes_K L$ as $A\otimes_K L$-modules, then $U$ is isomorphic to a direct summand of $V$ as $A$-modules. (Actually I have seen this also being called the Noether-Deuring theorem. Yes, this is stronger than the original Noether-Deuring theorem.) -But what about this: If there exists an $A\otimes_K L$-linear map $U\otimes_K L\to V\otimes_K L$ of some $L$-rank $r$, must it follow that there exists an $A$-linear map $U\to V$ of $K$-rank at least $r$ ? Again, the only hard case is when $K$ and $L$ are both finite fields, but Krull-Remak-Schmidt doesn't come to our help now... -Of course, the interesting cases are when $r=\dim_K U$ and when $r=\dim_K V$, so we are talking about the existence of injective resp. surjective maps. -Remarks: -(1) The condition that $A$ be finite-dimensional as a vector space over $K$ can be removed: we can always replace $A$ by the image of $A$ in $\mathrm{End}_K U \times \mathrm{End}_K V$, and that image is finite-dimensional if $U$ and $V$ are so. -(2) When $A$ is the polynomial $K$-algebra $K\left[X\right]$, then this question generalizes the conjugacy rank question. - -REPLY [6 votes]: Hendrik W. Lenstra has just informed me that the answer to my question is "no", both in the case of $r = \dim U$ (so we are looking at injective $A$-linear maps) and in the case of $r = \dim V$ (so we are looking at surjective $A$-linear maps). Counterexamples can be found in Exercise 21 (c) and Exercise 22 of his notes The unit theorem for finite-dimensional algebras (arXiv:1703.07273v1). -Thanks, Hendrik, for resolving this nagging question!<|endoftext|> -TITLE: Smooth functions for which $f(x)$ is rational if and only if $x$ is rational -QUESTION [43 upvotes]: A friend of mine introduced me to the following question: Does there exist a smooth function $f: \mathbb{R} \to \mathbb{R}$, ($f \in C^\infty$), such that $f$ maps rationals to rationals and irrationals to irrationals and is nonlinear? -I posed this question earlier in math.stackexchange.com (link to the question) where it received considerable interest. There hasn't been an answer so far, but one commenter suggested to bring it here. -Related results - -The friend who told me the problem -has been able to prove that no -polynomial satisfies the required -conditions. -If we required just that $f \in C^1$, then we can cut and paste the function -$x \mapsto \frac{1}{x}$ to provide a nonlinear example: -$$f(x) = \begin{cases}\frac{1}{x-1} + 1, & x \le 0 \\\\ \frac{1}{x+1} - 1, & x \ge 0\end{cases}$$ - -REPLY [35 votes]: There are such functions. Moreover any diffeomorphism $f_0:\mathbb R\to\mathbb R$ can be approximated by such $f$. For the sake of simplicity I assume that $f_0'\ge 2$ everywhere. -Enumerate the rationals: $\mathbb Q=\{r_1,r_2,\dots\}$, and construct a sequence $f_0,f_1,f_2,\dots$ of self-diffeomorphisms of $\mathbb R$ satisfying the following: - -$f_{2k-1}(r_k)\in\mathbb Q$, and $f_n(r_k)$ is the same for all $n\ge 2k-1$ -$f_{2k}^{-1}(r_k)\in\mathbb Q$, and $f_n^{-1}(r_k)$ is the same for all $n\ge 2k$. -The first $k$ derivatives of the difference $f_k-f_{k-1}$ are bounded by $2^{-k}$ everywhere on $\mathbb R$. - -Such a sequence has a limit $f$ in $C^\infty$, and this limit is a diffeomorphism satisfying $f(\mathbb Q)\subset\mathbb Q$ and $f^{-1}(\mathbb Q)\subset\mathbb Q$. -The sequence $\{f_i\}$ can be constructed by induction. To construct $f_{2k-1}$ from $g:=f_{2k-2}$, consider $g(r_k)$. If it is rational, let $f_{2k-1}=g$. If not, let $I$ be an open interval containing $r_k$ and not containing any of the points $r_i$ and $g^{-1}(r_i)$ for $i\le k-1$. (Note that $r_k$ is different from these points due to the fact that $g(r_k)\notin\mathbb Q$). Then define $f_{2k-1}=g+\varepsilon\cdot h$ where $h$ is your favorite smooth function with support contained in $I$ and such that $h(r_k)\ne 0$, -$\varepsilon$ is so small that the above derivative estimates hold and is chosen so that $f_{2k-1}(r_k)\in\mathbb Q$. To construct $f_{2k}$ from $f_{2k-1}$, do a similar perturbation near the pre-image of $r_k$, assuming it is not yet rational.<|endoftext|> -TITLE: Reference for tensor products of fields -QUESTION [11 upvotes]: Does anybody know a reference for basic properties of tensor products of (finite) algebraic extensions of fields? -Ideally, I would like a description of $L \otimes_k K$ for arbitrary finite extensions $L, K$ of $k$ but I would settle for a reference for results such as -1) If $K / k$ is Galois with group $G$ then $K \otimes_k K \cong \oplus_{g \in G} K$. -2) If $K / k$ is purely inseparable then $K \otimes_k K$ is local with residue field $K$ and length $[K : k]$. -3) If $K / k$, $L / k$ are separable then $K \otimes_k L$ has no nilpotent elements. - -REPLY [2 votes]: If $K/k$ is separable, then $K=k[\alpha]\approx k[X]/(f(X))$ where $f(X)$ is -the minimal polynomial of $\alpha$. Let $L$ be a field containing $k$. Then -$f(X)=f_1(X)...f_r(X)$ in $L[X]$ with the $f_i$ irreducible and distinct (because $K/k$ is separable). Therefore, -$L\otimes_kK\approx L[X]/(f(X))\approx \prod L[X]/(f_i(X))$ -by the Chinese remainder theorem. This describes $K\otimes L$ completely as a product of fields when $K/k$ is separable. For example, if which $f(X)$ splits in $L$, say $f(X)=(X-\alpha_{1})\cdots(X-\alpha_{n})$, then -$L\otimes_{k}K\approx L[X]/(f(X))\approx\prod_{i}L[X]/(X-\alpha_{i})\approx\prod L_{i}$ -with $L_{i}=L$. The map $L\otimes_{k}K\rightarrow L_{i}$ sends $a\otimes g(\alpha)$ -to $ag(\alpha_{i})$. This takes care of 1) and 3). -As for 2), if $K=k[\alpha]$ -with $\alpha^{p}\in k$, then $K\otimes_{k}K=K[\epsilon]$ where $\epsilon -=\alpha\otimes1-1\otimes\alpha$ and $\epsilon^{p}=\alpha^{p}\otimes -1-1\otimes\alpha^{p}=0$. That gets you started on 2).<|endoftext|> -TITLE: uneven spaced time series -QUESTION [8 upvotes]: Let $(t_k), k \in \mathbb{N}$, be an increasing sequence of real numbers ($t_{k-1} < t_k$) and $(X_{t_k}$) be a sequence of real numbers indexed by $(t_k)$. Such a sequence is sometimes called a time series. -The idea is that this series represents a sequence of measurements of some sort, like, for example, the average temperature of some location at time $t_k$. -The analysis of time series is an established area of statistics. In concrete applications, for example in climate science, there are two common problems when applying statistical algorithms to time series: - -The time series are finite, which produces artefacts in statistical algorithms that are designed for infinite time series. This problem is well known and there exist several approches to handle it. -The times series are uneven spaced, that is $t_k - t_{k-1}$ is not independent of $k$. - -I don't know of any textbook, algorithm or paper that explicitly addresses the latter problem. My question is therefore: Is this not a problem, is the solution trivial or, if not, are there any treatments? -Of course it is possible to interpolate missing values to generate a time series with an even time spacing $\min_k (t_k - t_{k-1})$, but it seems to me that this is not a solution, because algorithms like the fast fourier transform, nonlinear regression analysis or wavelet transforms would produce artefacts that depend on the kind of interpolation (linear, qubic splines, whatever). And therefore an explicit explanation of why the kind of interpolation one uses does not produce any artefacts in the analysis of the time series seems to be warranted to me, but I have never seen one in the literature. - -REPLY [2 votes]: Haven't used it yet, but I did happen to stumble upon this Python library - https://github.com/datascopeanalytics/traces (A Python library for unevenly-spaced time series analysis). Docs here - https://traces.readthedocs.io/en/latest/. -Might or might not be relevant.<|endoftext|> -TITLE: Sweep-segment bot: Will this random walk sweep the plane? -QUESTION [17 upvotes]: This model is inspired by the random behavior of the -Roomba sweeping robot. -Let a unit segment $ab$ in the plane be placed -initially with $a=(0,0)$ and $b=(1,0)$. -The segment is first rotated a random angle $\theta_1$ about $a$, -$\theta_1 \in (-\pi,+\pi)$, sweeping out a sector -of angle $\theta_1$ centered on $a$. -Then it is rotated a random angle $\theta_2$ about $b$, -sweeping out a sector of angle $\theta_2$ centered on $b$. -Odd steps rotate about $a$, even about $b$, -each with $\theta_i \in (-\pi,+\pi)$; each sector sweeps $< \pi$. -An example is shown below. The first, darker sector sweeps -$\theta_1=114^\circ$ (ccw), the second sweeps -$\theta_2=-140^\circ$ (cw), -and the sequence continues -$$\theta_3,\ldots = 104^\circ,-112^\circ,-93^\circ,-156^\circ,15^\circ,-97^\circ,-37^\circ,72^\circ \;.$$ - - - -Q. Does the described random walk of this segment eventually - sweep every point of the plane with probability 1? - That is, will the union of - the sectors cover the plane? - -My guess is: 'Yes,' because of the analogy with -Pólya's recurrence theorem for random walks on a 2D lattice. -Indeed if the random angle is discretized to -$\theta_i \in \{ -\pi, -\pi/2, +\pi/2, +\pi \}$, -then the segment midpoint executes a random walk on a lattice. -On the other hand, it seems remarkable that this process -would actually sweep the whole plane. -Here is the same example extended to 10,000 steps: - - - -Despite the similarities, I don't see how to directly reduce this model to a -conventional random walk to answer the posed question. -If anyone does, I would appreciate suggestions. Even just your intuitions are welcomed. -Thanks! -Addendum. -The answer to my question is a clear Yes!, -as shown by -Didier, Hugh, George, Anthony, and fedja. -Although I find Pólya's recurrence surprising, -that this segment-walk sweeps the plane is -somehow more remarkable, even if it relies -(as fedja's exposition shows) on roughly the same logic. - -REPLY [9 votes]: Once Didier touched it, let me comment that we do not need the full $[-\pi,\pi]$ swing here. Any fixed non-degenerate sectors for $A$ and $B$ rotations would do the job perfectly well. -As Didier proposed, we consider any "reinforced random walk with translation and circular symmetry" that operates like that: at each moment, we have a point $x_k$ (with $x_0=0$, for definiteness) and a rotation $R_k$ ($R_0$ can be anything). Each step consists of adding to $x_k$ the vector $R_k w_k$ and multiplying $R_k$ by $S_k$ to get $x_{k+1}$ and $R_{k+1}$ respectively where $(w_k,S_k)$ are i.i.d. pairs with some joint distribution satisfying some natural assumptions (we do not need much here: to know that $w_k$ are bounded and that the joint density is bounded is more than enough and our process falls under those assumptions if we agree to view 10 steps as one) -The proof starts the same way as Polya's recurrence argument: if the probability fof $x$ to return to some neighborhood of the origin is less than $1$, then, for any bounded region $U$, the probability for $x$ to visit $U$ more than $k$ times decays exponentially with $k$, so all moments of the number of visits are finite, etc. -Now, cover the plane by unit disks in some nice way. Two cases are possible: -A) The probability to ever visit the disk $Q$ decays to $0$ as the distance from $Q$ to the origin tends to $0$. -B) There are arbitrarily far disks $Q$ (depending on $R_0$, of course) that are hit with probability at least $q$ for some fixed $q>0$. -We'll show that both (A) and (B) imply recurrence. -In case (A), notice that the expectation $E|x_k|^2\le Cn$ for $k\le n$ (just estimate scalar products of increments $(R_kw_k,R_mw_m)$. they decay exponentially with $|k-m|$) , so if we denote by $v(Q)$ the number of visits to $Q$ after $n$ steps, we'll have $E\sum_Q v(Q)d(Q)^2\le Cn^2$ where $d(Q)$ is the distance from $Q$ to the origin. Thus, the sum of $v(Q)$ over the disks with $d(Q) -TITLE: Two Equal Series? -QUESTION [5 upvotes]: Hey all, -I'd like to know if anyone has a link to a proof of the following? -Take two infinite sequences $a_n$ and $b_n$ such that -$$\sum_{n=1}^\infty a_n^s=\sum_{n=1}^\infty b_n^s=finite$$ -for all $s\in\mathbb{N}$. Then, after some rearrangement, $a_i=b_i$ for all i? I have my own proof which I am pretty sure is correct and I'd very much like to know if such a proof has been given before (to see if it is any different) -Thanks in advanced for any help :) - -REPLY [6 votes]: Here is a link to a counterexample for conditionally convergent complex series.<|endoftext|> -TITLE: Smallest n for which G embeds in $S_n$? -QUESTION [33 upvotes]: Question: Given a finite group $G$, how do I find the smallest $n$ for which $G$ embeds in $S_n$? - -Equivalently, what is the smallest set $X$ on which $G$ acts faithfully by permutations? -This looks like a basic question, but I seem not to be able to find answers or even this question in the literature. If this is known to be hard, is there at least a good strategy that would give a small (if not the smallest) $n$ for many groups? -Note: I do not care whether $G$ acts transitively on $X$, so for example for $G=C_6$ the answer is $n=5$ (mapping the generator to (123)(45)), not $n=6$ (regular action). -Edit: If this is not specific enough, is there a method that could find the smallest $n$ (or one close to the smallest one) for any group of size $\le 10^7$ in 5 seconds on some computer algebra system? - -REPLY [5 votes]: Let $\mu(G) = \min\{n \mid G \text{ embeds in } S_n\}$. Here are some results on $\mu(G)$ from this paper by O. Becker: - -$\mu(G)$ is known for abelian groups. -It is known eactly when $\mu(G) = |G|$. If $\mu(G) < |G|$, then $\mu(G) \le \frac{5}{6}|G|$. -The identity $\mu(G\times H) = \mu(G) + \mu(H)$ holds for a wide family of groups, for instance - for all $G,H$ with central socle.<|endoftext|> -TITLE: An example where $Pic(X) = H^0(k,Pic(\overline{X}))$? -QUESTION [6 upvotes]: Let $X$ be a geometrically integral smooth projective variety over a number field $k$. Then if $X$ is everywhere locally soluble, we have $Pic(X) = H^0(k,Pic (\overline{X}))$, where $\overline{X}=X \times_k \overline{k}$. This can be proved using the Hochschild-Serre spectral sequence and the fundamental exact sequence of class field theory. -My question is about the converse: If the canonical injection $Pic(X) \hookrightarrow H^0(k,Pic (\overline{X}))$ is an isomorphism, then is $X$ everywhere locally soluble? -An example of a class of varieties for which this holds is for Brauer-Severi varieties, however my intuition tells me that this is not the case in general. Does anybody have a good counterexample? - -REPLY [6 votes]: No, it is possible for $X$ not to have points everywhere locally and still have every $k$-rational divisor class be represented by a $k$-rational divisor (this is the "Picard equality" you want). -For instance, the Picard equality holds if $X$ has points everywhere locally except at a single place. Many examples of genus one curves satisfying this property -- over any number field $k$ -- are given in this paper of mine. -There should be no lack of other examples as well. In some sense the simplest family is as follows: it is enough to find hyperelliptic quartic curves $C: y^2 = P_4(x)$ which do not have points everywhere locally and for which the Jacobian elliptic curve $E$ has $E(k)$ finite of odd order. The point here is that then the "elliptic Kummer sequence" -$0 \rightarrow E(k)/2E(k) \rightarrow H^1(k,E[2]) \rightarrow H^1(k,E)[2] \rightarrow 0$ -gives an isomorphism $H^1(k,E[2]) \rightarrow H^1(k,E)[2]$. This means that for each torsor $C \in H^1(k,E)[2]$ there is a unique rational divisor class of degree $2$, so $C$ comes from a hyperelliptic quartic curve iff this divisor class is represented by a rational divisor. -Actually, a simpler argument can be given if we assume a little more, that $E(k) = \{0\}$. (By a recent theorem of Mazur and Rubin, such curves exist over every number field $k$.) Then the degree zero part of $H^0(\operatorname{Pic} C) = E(k) = 0$, so the equality of $H^0(\operatorname{Pic} C) = \operatorname{Pic}(C)$ for torsors is equivalent to having period equals index, so if you can find a hyperelliptic quartic curve $C$ with Jacobian having trivial Mordell-Weil group and failing to have points over more than one completion, this will give an example of what you want. -In practice, it should be easy to write down such curves $C$ over $\mathbb{Q}$, say. -Note that the situation is different for genus one curves over a $p$-adic field $k$: by a theorem of Roquette-Lichtenbaum, the order of $C$ in $H^1(k,\operatorname{Jac} C)$ is equal to the order of $H^0(\operatorname{Pic} C)/ \operatorname{Pic}(C)$.<|endoftext|> -TITLE: What are some natural and attractive open problems in Jones's theory of planar algebras? -QUESTION [7 upvotes]: I'm hoping to learn something about planar algebras while attacking a planar algebra question with an undergrad research student. I'm thinking about reading this paper, as Kuperberg's program seems like the sort of thing I'm looking for, but maybe there are better ideas: - -What are some open questions in planar - algebra theory that are self-contained - within, and need minimal motivation outside, - the planar algebra formalism? - -If there are such questions that are well-known in the community, I'd appreciate being pointed to some of them. Short answers and references would be fine. -Some supplemental questions that would also be helpful, if answered: -What are the main open problems in the theory of planar algebras, proper? (This is kind of silly, because these are certainly the ones corresponding to the important open questions in subfactor theory. Perhaps, though, an answer to a question in planar algebras would resolve deep things in several areas where planar algebras appear. Such a question I'd consider a question in planar algebras, proper.) -Which of the main problems driving this subject are interesting even if they are not directly traced back to their implications in subfactor theory? -Of course, the theory of subfactors is the clear motivation for using the formalism. However, the diverse examples of planar algebras give evidence that we should study them in their own right. -Ideally, I am looking for problems that can be stated in the planar algebra formalism and do not require strong, direct reference back to the "subfactor world" to motivate. - -REPLY [8 votes]: One important kind of question in planar algebras is given generators and relations for a planar algebra can you: - -Find an algorithm which takes an arbitrary closed diagram and evaluates it to give a number -Show that any two ways of evaluating a closed diagram gives the same answer -Find an explicit basis for every box space - -Here's some examples. These are motivated by questions outside of planar algebras as such, but the motivations don't come from subfactor theory, and the questions are not hard to state in purely elementary terms. - -Given the generators and relations defining the Yamada polynomial (this gives quantum so(3)) at circle value a real number 4 or bigger, can you find a manifestly positive evaluation algorithm? (This is the 4-color theorem.) -Write down the natural generators and relations for the planar algebra coming from the adjoint representation of Deligne's (conjectural!) exceptional Lie algebra. (These are antisymmetry, the Jacobi relation, and Vogel's relation for simplifying a square.) Answer each of the above three questions. This would give a construction of the exceptional Lie algebra. -Same as the last question but for the adjoint representation of Vogel's (conjectural) universal Lie algebra. Here you just have anti-symmetry, Jacobi, and killing negligibles. - -Some other interesting questions not of the above type: - -Find a continuous family of planar algebras which isn't one of: HOMFLY polynomial, Kauffman polynomial, the symmetric group S_t, Fuss-Catalan, or one of a few other known constructions. -Consider the planar algebra of planar graphs with a bipartite shadings of both their vertices and faces. Find explicit quotients of this planar algebra with finite dimensional box spaces. (Not counting the easy ones which come from relations simplifying 2-valent vertices.) This question sounds topological, but is secretly about quadrilaterals of factors. -Find a direct (i.e. non-number theoretic) proof that there exists a closed diagram which evaluates to two different numbers for each of the "further extended Haagerups" (see our paper with Bigelow-Morrison-Peters). Here the "jellyfish algorithm" gives an explicit way to evaluate closed diagrams (depending on some choices), and you just want to show that making different choices gives different answers.<|endoftext|> -TITLE: Asymptotic expansion of an integral -QUESTION [5 upvotes]: I am looking for an asymptotic expansion of J(n) -$J(n)=\frac {2} {\pi} \int_{0}^{\pi/n} \prod_{k=1}^n \frac {\sin kx} {\sin x} dx$, $n=2,3,4,\dots$ -The first approximation is managed to get -$F_1(n)=\frac {n!} {\sqrt{\pi A}}$, $A=n(n-1)(2n+5)/36$ -Is a general expansion known for this? - -REPLY [5 votes]: This is probably an overkill way to a solution, but applying Euler reflection formula -$$ -\sin(z)=\frac{\pi}{\Gamma(1-\frac{z}{\pi})\Gamma(\frac{z}{\pi})}, -$$ -one gets products of gamma functions both in the numerator and denominator. From this point on, I think, your integral is computable/representable via a generalized Hypergeometric Function of Fox type: http://en.wikipedia.org/wiki/Fox_H-function (I am mentioning the wiki link mainly because of the references there), i.e. Barnes-Integral with a kernel of fraction of products of Gamma functions. If my memory serves me right, this type of integral has been treated also in the book of Bleistein and Handelsman. If it does not, then the relevant paper in this case is "Asymptotic expansions and analytic continuations for a class of Barnes-integrals" by Braaksma, Compositio Math. 15,1964, p. 239–341.<|endoftext|> -TITLE: In a contact manifold, is every tranverse 1-foliation given by some Reeb vector field? -QUESTION [9 upvotes]: Let $M$ be a contact manifold, and let $F$ be an oriented 1-dimensional foliation that is transverse to the contact structure. -Is there a contact form $\alpha$ whose associated Reeb vector field generates the foliation $F$? - -REPLY [3 votes]: If there is already a contact structure given say $\eta$ one might try to see if there exists a function $f$ so that Reeb vector field $R_f$ of $\eta_f = f\eta$ lies inside the given line bundle $\ell$. Assume that you can give $\ell$ as the kernel of two independent differential 1-forms $\eta_1,\eta_2$. Then we want $\eta_1(R_f)=\eta_2(R_f)=0$ which is if and only if $d\eta_f = \theta_1 \wedge \theta_2$. Let $E_i = ker(\theta_i)$ and $\Delta = ker(\eta)$ which is iff $d\eta_f|_{E_i}=0$ for both $i$. This condition could be written as a system of PDE. -Take $X_i \subset E_i \cap \Delta$ a vector field. Since $\ell$ is transverse to $\Delta$, $X_1$ and $X_2$ are distinct. And take $Z_i$ any other vector fields independent from $X_i$ inside $E_i$. Then the conditions above are true iff -$$(fd\eta + df \wedge \eta)(X_i,Z_i)=0$$ -which is iff -$$X_i(ln(f)) = d\eta(X_i,\frac{Z_i}{\eta(Z_i)}) = \eta([X_i,\frac{Z_i}{\eta(Z_i)}])$$ -or in more suggestive notation -$$\mathcal{L}_{X_i}(ln(f)) = \eta(\mathcal{L}_{X_i}W_i)$$ -where $W_i=\frac{Z_i}{\eta(Z_i)}$. Hence $f$ must be a solution to this system of PDE.<|endoftext|> -TITLE: Is there a name for the involution on Laurent polynomials? -QUESTION [6 upvotes]: This is a simple terminology question: I want to know if the involution $z \mapsto z^{-1}$ on Laurent polynomials (over some ring, I happen to be working over $\mathbb{Z}$ but that's not important) has a special name. -My motivation is perhaps a little unusual for this site. I'm doing some computations that involve manipulating Laurent polynomials and, being a lazy sort of fellow, I'm letting the computer do it. Being extra lazy, I don't particularly want to learn a new programming language to do this so I'm using Perl as it's the only one that I know. However, there my laziness stops as whilst there's a Perl module for ordinary polynomials there isn't one for Laurent polynomials. Still, it wasn't hard to adapt it to Laurent polynomials so I did and the program is chugging away churning out these computations to its heart's content. In writing the methods (meaning, things you can do to a Laurent polynomial), most already have obvious names (add, subtract - actually called sub_, mul(tiplication), and so forth) but I don't know one for the obvious involution $z \mapsto z^{-1}$. inv sounds a little to easy to mistake for inverse. -So, is there a name for this? If not, would anyone like to suggest one (preferably with an unambiguous shortening - I've already gotten fed up of typing monomial every time)? - -REPLY [7 votes]: I would call it the antipode. If your base ring is commutative, then the Laurent polynomials are the coordinate ring of the multiplicative group, and the antipode gives you the inversion on the group scheme.<|endoftext|> -TITLE: Does the everywhere unramified extension of Q(mu_37) of degree 37 grow into a Z_37-extension? -QUESTION [23 upvotes]: Let $p=37$. Since $p$ divides the numerator of $B_{32}$, by Ribet's proof of the converse of Herbrand's theorem, we know that the class group of ${\bf Q}(\mu_p)$ has size divisible by $p$. More specifically, the $p$-part of the class group decomposes under the action of $\Delta = {\rm Gal}({\bf Q}(\mu_p)/{\bf Q})$, and we know exactly which eigenspace is non-trivial; in this case, the $\omega^5$-eigenspace is non-trivial where $\omega$ is the mod $p$ cyclotomic character. -Now let $X$ denote the $p$-Hilbert class field of ${\bf Q}(\mu_p)$. As above, $X$ decomposes under the action of $\Delta$, and we have that $X^{(\omega^5)}$ is 1-dimensional over ${\bf F}_p$. Thus, there is some abelian everywhere unramified extension $H_5$ of ${\bf Q}(\mu_p)$ -such that $\Delta$ acts on ${\rm Gal}(H_5/{\bf Q}(\mu_p))$ by $\omega^5$. -Let $Y$ denote the Galois group of the maximal abelian $p$-extension of ${\bf Q}(\mu_p)$ unramified outside of $p$. By Leopoldt, the ${\bf Z}_p$-rank of $Y$ (modulo its torsion) equals $(p+1)/2$. We can also decomposing $Y$ under the action of $\Delta$; I presume what happens is that eigenspace of $\omega^0$ is rank 1 over ${\bf Z}_p$ (corresponding to the cyclotomic ${\bf Z}_p$-extension), and the eigenspace of $\omega^i$ is rank 1 over ${\bf Z}_p$ for $i$ odd. In particular, there is some ${\bf Z}_p$-extension $M_i$ over ${\bf Q}(\mu_p)$ such that $\Delta$ acts on ${\rm Gal}(M_i/{\bf Q}(\mu_p))$ by $\omega^i$ for each odd $i$. -My question: is $H_5$ the first layer of the ${\bf Z}_p$-extension $M_5/{\bf Q}(\mu_p)$? - -REPLY [22 votes]: $\newcommand{\L}{\mathcal{L}}$ -$\newcommand{\Q}{\mathbf{Q}}$ -$\newcommand{\Z}{\mathbf{Z}}$ -$\newcommand{\F}{\mathbf{F}}$ -The answer is yes. It suffices (as Brian mentions) to show that the corresponding space -of extensions is one dimensional. -Let $p$ be an odd prime, let $k$ be an integer, and let $\omega$ be the mod-$p$ cyclotomic character. The extensions we are interested in correspond to elements of the Selmer group $H^1_{\L}(\Q,\omega^k)$, where - $\L$ is defined by the following local conditions: unramified away from $p$, and -unrestricted at $p$ (so $\L_p = H^1(\Q_p,\omega^k)$). The dual Selmer group $\L^*$ -corresponds to classes which are unramified outside $p$, and totally split at $p$. -Let us further suppose that $k \not\equiv 0,1 \mod p-1$, so neither $\omega^k$ nor -$\omega^{1-k}$ have invariants over $\Q_p$, and so $|\L_p| = p$. A formula of Wiles relating the Selmer group to the dual Selmer group implies (under the condition on $k$) that -$$ \frac{|H^1_{\L}(\Q,\omega^k)|}{|H^1_{\L^*}(\Q,\omega^{1-k})|} -= \frac{|\L_p|}{|H^0(G_{\infty},\omega^k)|} = -\begin{cases} p, & k \equiv 1 \mod \ 2, \\\ -1, & k \equiv 0 \mod \ 2. \\ \end{cases}$$ -(This result also follows more classicaly from so-called mirror theorems, and, phrased slightly differently, occurs in Washington's book on cyclotomic fields.) -In your situation, $k$ is odd, and so the space -$H^1_{\L}(\Q,\omega^k)$ is one dimensional -if and only if $H^1_{\L^*}(\Q,\omega^{1-k})$ vanishes. - Since $1-k$ is even, however, the vanishing of this latter group is essentially the same as Vandivier's conjecture (or rather, the $\omega^{1-k}$-part of Vandivier's conjecture). Since Vandiver's conjecture is true for $p = 37$, everything is ok in this case. -Update: Brian reconciles this answer with his previous comments in the comments to this answer below. -(If you don't assume Vandivier's conjecture, then it isn't clear what question to ask for general $p$, since -$H^1_{\L^*}(\Q,\omega^k)$ could have dimension $\ge 2$.) - -Dear Rob, -For your second question (in the comments below), let $\Gamma$ denote the $\omega^{k}$-part of the maximal -extension of $\Q(\zeta_p)$ unramified outside $p$. Then there -is an exact sequence: -$$0 \rightarrow \Z_p \rightarrow \Gamma \rightarrow C \rightarrow 0,$$ -where $C$ is the $\omega^k$-part of the class group tensor $\Z_p$. You ask -whether the $\Z_p$-extension is totally ramified. This -boils -down to the following question: Is the image of $\Z_p$ saturated -in $\Gamma$? -This is equivalent to asking -that the sequence above remains exact after tensoring -with $\F_p$. Equivalently, it is the same as asking that -$$\Gamma/p \Gamma \simeq \F_p \oplus C/pC \simeq -\F_p \oplus H^1_{\L^*}(\Q,\omega^k).$$ -Yet $\Gamma/p \Gamma \simeq H^{1}_{\L^*}(\Q,\omega^k)$. Thus, using -Wiles' formula again, this is the same as asking that: -$$p = \frac{|H^1_{\L}(\Q,\omega^k)|}{|H^1_{\L^*}(\Q,\omega^k)|} -= \frac{p |H^1_{\L^*}(\Q,\omega^{1-k})|}{|H^1_{\L}(\Q,\omega^{1-k}|},$$ -or equivalently: -The $\Z_p$-extension is totally ramified if and only if -there does not exist a $\omega^{1-k}$ extension of $\Q(\zeta_p)$ which -is unramified outside $p$ but ramified at $p$. -Note that: - -If $p$ is regular, then there is no such extension, and so the -$\Z_p$-extension is totally ramified. (This is obvious more directly.) -If $p$ is irregular, then mirror theorems imply that there does exist -a $\omega^{1-k}$-extension unramified outside $p$; if Vandiver's conjecture holds this extension is ramified, and so the -$\Z_p$-extension is not totally ramified (this also -follows from the first answer). -If $p$ is irregular, and the $\omega^k$ part $C$ of the class group is cyclic -(a consequence of Vandiver's conjecture), then -$H^1_{\L}(\Q,\omega^{1-k})$ has order $p$. Thus, in this case, -the $\Z_p$-extension is not totally ramified if and only if Vandiver's -conjecture holds (for the $\omega^{1-k}$ part of the class group). -This example shows that one can't really give a particularly clean -answer, since cyclicity of the class group is seen as a strictly weaker -property than Vandiver. This shows that you can't really expect a cleaner answer than I gave above (since proving Vandiver from cyclicity sounds very hard). - -BTW, there is a little check mark button next to this answer, if you press it, it goes green, which lights up the dopamine receptors in my brain.<|endoftext|> -TITLE: Does $\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$ carry an almost complex structure? -QUESTION [22 upvotes]: Recall the list of irreducible simply connected inner symmetric spaces of compact type in dimension $4k+2$: - -Hermitian symmetric spaces (one can write them down explicitly); -Grassmannians of oriented real $p$-planes in $\mathbb{R}^{p+q}$, with $pq=4k+2$; -The 70-dimensional exceptional symmetric space -$\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$. - -Hermitian symmetric spaces are of course complex manifolds, and real Grassmannians carrying almost complex structures were classified by P. Sankaran (PAMS, 1991) and Z. Tang (PAMS, 1994). Thus my question: - -Does - $\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$ - carry an almost complex structure? - -Note that in dimension $4k$ it is known that an irreducible simply connected inner symmetric space of compact type carries an almost complex structure iff it is Hermitian symmetric. -Edit concerning the dimensions multiple of 4: It is on purpose that I did not mention our paper http://fr.arxiv.org/abs/1003.5172 when I first asked the question. Indeed, the method used in that paper to treat the $4k$-dimensional situation (an index-theoretical obstruction) seems uneffective in the $4k+2$-dimensional cases so I expected other methods to tackle this problem. -Let me nevertheless give a short account on how things work in dimension $4k$. The idea is to use the following criterion: - -If a compact Riemannian manifold - $(M,g)$ carries a self-dual vector - bundle $E$ such that the index of the - twisted Dirac operator on $\Sigma M \otimes E\otimes -> TM^{\mathbb{C}}$ is odd, then $M$ has - no almost complex structure (in fact $TM$ is not even stably isomorphic to a complex bundle). - -Note that the complex spin bundle $\Sigma M$ and $E$ can be locally defined, only their tensor product has to be globally defined. On then checks that this criterion applies to all irreducible inner symmetric spaces of compact type which are neither Hermitian symmetric nor spheres. -Returning to the dimensions $4k+2$, it is easy to check that there exists a priori no bundle satisfying the above criterion. Nevertheless, one can show real Grassmannians and $\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$ do carry -such a bundle $E$ which satisfies all conditions in the criterion except that it is not self-dual. I don't know if this can lead anywhere... - -REPLY [7 votes]: I'm sorry if this is obvious to everyone, but I thought that it was worth mentioning: -I don't know the answer to the question asked, but the answer to the easier question, "Does the space $\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$ carry an $\mathrm{E}_7$-invariant almost complex structure?", is 'no'. -To see this, consider the orthogonal complement of the subalgebra $\mathfrak{su}(8)$ in $\mathfrak{e}_7$, say $V = \mathfrak{su}(8)^\perp$. By inspecting the root diagram of $\mathfrak{e}_7$ (as described by Cartan's representation, not the usual one you see in most books), one sees that this real vector space of dimension $70$ has the property that its complexification is the irreducible complex representation $\Lambda^4(\mathbb{C}^8)$ of $\mathrm{SU}(8)$. Since the complexification of $V$ is irreducible under $\mathrm{SU}(8)$, the action of $\mathrm{SU}(8)$ does not preserve any complex structure on $V$. -Thus, the space $\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$ does not have an $\mathrm{E}_7$-invariant almost complex structure.<|endoftext|> -TITLE: Griffiths and Harris reference -QUESTION [19 upvotes]: Trying to read the section on Poincare duality from Griffiths and Harris is a nightmare. I want to know if there is a place where Poincare duality and intersection theory are done cleanly and rigorously in the order that GH do them (usually, one proves Poincare duality for singular cohomology and then defines the intersection pairing by cup product and proves that (using the Thom isomorphism) that indeed the intersection pairing counts the number of points taken with sign (and convert everything to forms using De Rham's theorem). GH on the other hand define intersection pairing and then proceed further. However, one has to wave their hands at relativistic speeds to make some things work here). - -REPLY [5 votes]: I ran into the same problem with G&H treatment of Poincare duality. It seems incomplete, for example they refer to "cells" without defining that term. Following the suggestion from one of the comments above I turned to Seifert and Threlfall and I did find it to be a good source to fill in the missing pieces of the G&H approach. Having studied homology from the singular point of view (Massey) I found the simplicial approach of S&T enlightening.<|endoftext|> -TITLE: What is a Cheeger deformation? -QUESTION [8 upvotes]: I'm currently at a Differential Geometry meeting and there is a mini-course on positively curved Riemannian manifolds. There, we were told that a technique to construct such manifolds is a Cheeger deformation, which (if I understood correctly) is a generalization of a one-parameter family of surfaces of revolution given by $\frac{f}{\lambda f + 1}$, where $f$ is the curve that generates a surface of revolution in Euclidean space and $\lambda$ is a positive parameter that varies over $[0,\infty)$. Can anyone tell me what is the concrete definition of a Cheeger deformation and how are they used to construct manifolds of positive (or non-negative, perhaps; I don't remember) curvature? Thanks a lot. - -REPLY [15 votes]: Cheeger deformations were first introduced in Jeff Cheeger's 1973 paper Some examples of manifolds of nonnegative curvature, and were inspired by the Berger sphere construction. The basic idea is to consider a Lie group acting by isometries on a Riemannian manifold, and then shrink the lengths of vectors tangent to the orbits of the action while keeping those perpendicular to the orbits fixed. Here is a formal description: - -Let $(M,g_M)$ be a Riemannian manifold with $G$, a closed subgroup of the isometry group of $M$, acting on the left. Choose any bi-invariant metric $g_G$ on $G$, and consider the one-parameter family of product metrics $\{l^2g_G + g_M\}_{l>0}$ on the product manifold $G\times M$. Then $G$ acts on $G \times M$ via the map -$$y\cdot(x,p) = (xy^{-1}, y\cdot p).$$ -This action is free, so taking the quotient gives a manifold $(G\times M)/G$, and it is diffeomorphic to $M$. Thus by imposing that the quotient map -$$q_l:(G \times M, l^2g_G + g_M)\to (G\times M)/G\cong M$$ -be a Riemannian submersion, we obtain a one-parameter family of metrics $g_l$ on $M$. The family of Riemannian manifolds $\{(M,g_l)\}_{l>0}$ is called a Cheeger deformation of the original manifold $(M,g_M)$. - -As $l\to \infty$, the Cheeger deformed metrics $g_l$ converge to the original metric $g_M$, and as $l\to 0$, the manifolds $(M,g_l)$ converge in the Gromov-Hausdorff metric to the quotient space $M/G$. Moreover, if the original manifold $(M,g_M)$ is one of nonnegative (resp. positive) sectional curvature, then the Cheeger deformed manifolds $(M,g_l)$ will have nonnegative (resp. positive) sectional curvature for all values of $l$. -The survey by Wolfgang Ziller mentioned by @Dan in the comments is a great reference for general information about these deformations and how they affect curvature. See my blog post for an animation of the Cheeger deformed plane under the circle action given by rotation about the origin.<|endoftext|> -TITLE: Complex root systems -QUESTION [5 upvotes]: This question is twofold. -1) What is the best reference on root systems? -2) Do complex root systems exist? - -REPLY [19 votes]: To supplement what Pete says, I'd emphasize that "root systems" have been defined in a variety of ways for a variety of purposes related to Lie theory or to some type of "reflection" group. (For instance, Vinay Deodhar proposed a general notion in the setting of Coxeter groups, which I used in my 1990 book.) The traditional notion arose in the study of semisimple Lie groups and their complexified Lie algebras, where you have finite real reflection groups (Weyl groups = finite crystallographic Coxeter groups). Witt built an early bridge between Coxeter's viewpoint and this kind of Lie theory. Bourbaki's axiomatic treatment encodes all of this efficiently and also includes affine Weyl groups. Eventually many of these notions have been carried over to $p$-adic groups, Kac-Moody theory, etc. -Concerning "complex root systems", these arise in the separate but overlapping theory of finite complex (= unitary) reflection groups. After the classification by Shephard-Todd, Arjeh Cohen systematized somewhat the notion of root system in a 1976 paper "Finite complex reflection groups", freely available here. More recently Michel Broue and his collaborators have used complex reflection groups and associated diagrams (more elaborate versions of Dynkin diagrams) extensively in their study of representations of finite groups of Lie type. There is a useful 2009 Cambridge text by Lehrer and Taylor, Unitary Reflection Groups, with a suitable notion of root system defined at the end of the first chapter.<|endoftext|> -TITLE: Did Grothendieck write about modular forms? -QUESTION [16 upvotes]: This question might be astoundingly naive, because my understanding of modular forms is so meek. It occurred to me that the reason I was never able to penetrate into the field of modular forms, automorphic forms, the Langland's program and so forth was because my appeal is to things that have the feel of SGA1, and those things do not. -I was wondering, therefore, if Grothendieck had devoted thought to this, and if so where it can be found, and how it is treated in the field at the moment. - -REPLY [4 votes]: Grothendieck was aware (at least) of modular forms and their relations to motives by J P Serre in a letter dated December 31, 1986. - -Grothendieck (also) develop on "La longue Marche" a theory of "topos modulaire" (now completely available online!).<|endoftext|> -TITLE: Nontrivial criteria for polynomials to have no common zeros? -QUESTION [6 upvotes]: When we work in $C[x_1,x_2,...,x_n]$,here $C$ denotes the complex field, we know that when polynomials $f_1,f_2,...,f_k$ have no common zeros, then there exists polynomials $g_1,...,g_k$ , such that the sum of $g_i\cdot f_i$ is $1$, (Nullstellensatz). -However, this theorem doesn't seem to have any practical value, consider the following problem: -Is there any nontrivial criterion such that we can determine whether some polynomials (chosen randomly) have common zeros or not by using it? Of course, I don't want to use the Grobner base theory, what I need is just a criterion involving information on the algebraic properties of the given polynomials, something just like Eisensten's criterion on irreducibility. - -REPLY [4 votes]: Not known. Your problem is $NP_\mathbf{C}$-complete in the Blum-Shub-Smale complexity model. So anything significantly simpler than Groebner basis would yield significant progress on the $P=NP$ question over $\mathbf{C}$ compared to the current state of the art.<|endoftext|> -TITLE: mirror symmetry with algebraic geometry? -QUESTION [17 upvotes]: Why is it that mirror symmetry has many relations with algebraic geometry, rather than with complex geometry or differential geometry? (In other words, how is it that solutions to polynomials become relevant, given that these do not appear in the physics which motivates mirror symmetry?) I would especially appreciate nontechnical answers. - -REPLY [13 votes]: Here are a few scattered observations: - -Our ability to construct examples (e.g. of CY manifolds) is limited, and the tools of algebraic geometry are perfectly suited to doing so (as has been noted). -Toric varieties are a source of many examples -- Batyrev-Borisov pairs -- and they are even "more" than algebraic, they're combinatorial. In fact, the whole business is really about integers in the end, so combinatorics reigns supreme. -The fuzziness of $A_\infty$ structures is more suited to algebraic topology rather than geometry. -Continuity of certain structures (which are created from counting problems) across walls, scores some points for analysis over combinatorics and algebra. -Elliptic curves are only "kinda" algebraic, and the mirror phenomenon there is certainly transcendental. -Physics indeed does not care too much about how the spaces are constructed, but (as has been noted) even the non-topological version of mirror symmetry is an equivalence of a very algebraic structure (which includes representations of superconformal algebras). - -I was hoping to unify these idle thoughts into a coherent response, but I don't think I can. Maybe the algebraic geometric aspects just grew faster because the mathematics is "easier" (or at least better understood by more mathematicians): witness the slow uptake of BCOV and its antiholomorphicity within mathematics. -To respond personally: these days, I try to transfer the algebraic and symplectic structures to combinatorics so that I can hold them in my hand and try to understand them better.<|endoftext|> -TITLE: Can the graph of a continuous function be a rotation of the graph of a discontinuous function? -QUESTION [5 upvotes]: Can there exist two functions $f,g: \mathbb R \to \mathbb R$ so that $f$ is continuous, $g$ is discontinuous, and their graphs $\Gamma_f, \Gamma_g \subseteq \mathbb R^2$ are related by an isometry? (I think you can assume the isometry is a rotation.) -The graph of a continuous function must be path connected, so a natural intermediate question is, "can a discontinuous function have path connected graph?" - -REPLY [6 votes]: I think the question has an easy answer, and it's essentially a 1st course in analysis type question that reduces pretty quickly to an intermediate value theorem application. Here's a more general statement. Let $f : \mathbb R \to \mathbb R$ be continuous and $h : \mathbb R^2 \to \mathbb R^2$ a homeomorphism. Then if $h(graph(f))$ is the graph of a function $g : \mathbb R \to \mathbb R$, then $g$ is continuous. -Sketch: consider the function $P : \mathbb R \to \mathbb R$ given by $P(x) = \pi_1 \circ h(x,f(x))$, where $\pi_1(x,y)=x$. This is a continuous monotone function by the assumptions, so it is an open map by the intermediate value theorem, so it's inverse exists and is continuous. The claimed function $g$ is then $g(x)=\pi_2 h(P^{-1}(x), f(P^{-1}(x))$ where $\pi_2(x,y)=y$. -The question gives me the feeling it's a homework problem, but I've never seen it before. This was solved in meta, with a correction by Anton. I wish I had known about this question yesterday -- I would have put it on my analysis class final exam!<|endoftext|> -TITLE: Strange real functions -QUESTION [8 upvotes]: I know there are a lot of strange functions $f~:~\mathbb R \to \mathbb R$. -I'm looking for an "elementary but complete" exposition of a result discovered by W. Sierpi\'nski and A. Zygmund in "Sur une fonction qui est discontinue sur tout ensemble de puissance du continu." Fund. Math., vol. 4, pp.316–318, 1923 stating (in simple form) that there exist a function $f~:~\mathbb R \to \mathbb R$ such that for every non empty open interval $I$, $f(I)=\mathbb R$ . -And also, a probably well-known fact for experts in real analysis: -If an arbitrary continuous function $f~:~ [0,1] \to \mathbb R$ is given, is it true that $f_{|D}$ is monotonic on some dense set $D\subset [0,1]$ or some set $D$ of positive measure? -It seems to be true if I replace "dense set" by "perfect subset" (according to Jack Brown). -If this is not true an example is highly appreciated. - -REPLY [5 votes]: Although you asked about continuous functions, here is an example of a discontinuous -function $f:[0,1]\to\mathbb{R}$ -which is not monotone on any measurable set with positive measure. -Let $V\subset [0,1]$ be the usual Vitali set, selecting one -element from each equivalence class under translation by -the rationals. Thus, $V$ is not measurable, and the -translates $V+q$ (working modulo 1) for rational $q$ are -disjoint and cover $[0,1]$. It follows that none of the -translates $V+q$ contains a measurable set of positive measure. Enumerate -the rationals $\mathbb{Q}=\{ q_n \mid n\in\mathbb{N}\}$, -and let $f(x)=n$ for $x\in V+q_n$. Thus, $f$ is constant on -each $V+q$, and the range of $f$ involves only natural -number values. Suppose that $f$ is monotone on a measurable -set $A\subset [0,1]$. If $f|A$ is constant or has only -finitely many values, then $A$ will be contained in the -union of finitely many $V+q$, and hence not have positive -measure. Otherwise, $A$ must contain points from infinitely -many $V+q$, and since the range is contained in -$\mathbb{N}$, it must be that $f$ is nondecreasing on $A$. -For any $a\in A$, note that if $f(a)=n$, then $A\cap [0,a]$ -is contained in the union of $V+q_m$ for $m\leq n$, a -finite number of translations of $V$. Thus $A\cap [0,a]$ -has measure $0$ for any $a\in A$, and it follows that $A$ -has measure $0$ altogether.<|endoftext|> -TITLE: Rainbow matchings (in random graphs) -QUESTION [11 upvotes]: Suppose we have an $(n,n)$-bipartite graph with edges colored with $k$ colors. Is anything known about the existence of rainbow matchings (i.e. a matching that uses each color exactly once, for $k=n$) for a random bipartite graph (e.g. that for $k$ colors and more than $f(k,n)$ edges we get a rainbow matching with $p \rightarrow 1$)? -In the noncolored case, Hall's theorem makes proving this kind of results relatively simple, since we are interested in the non-existence of "no matching possible" witness (i.e. a subset that violates Hall condition) and we can use union bound to bound the probability from above (for $A_k$ = "k-th subset is a witness" give bound to $\mathbb{P}(\cup A_k)$). However, there is no simple condition of this kind equivalent to the existence of a rainbow matching. - -REPLY [3 votes]: Isn't this very much related to the problem of a transversal in a Latin square? Suppose we have an $(n,n)$ bipartite graph with $n$ edge colors, such that every vertex has one edge of each color. This is equivalent to an $n\times n$ Latin square. A rainbow matching is a transversal of the Latin square. There is a conjecture (due to Ryser) that every Latin square with $n$ odd has a transversal, that is, a perfect rainbow matching. For even $n$, the conjecture (due to Brualdi) is that it has a partial transversal of length $n-1$ (i.e., a rainbow matching of cardinality $n-1$). To indulge in a little self-promotion, the best known result is that there exists a partial transversal of length $n -O(\log^2 n)$. There are also a number of results about transversals and partial transversals in near-Latin squares, which will probably be relevant to rainbow matching questions. -I guess the relevant Latin square question would be: does a random Latin square have a transversal with high probability? I know extensive calculations have been done which suggest that the answer is yes. I don't know whether anybody has proven this.<|endoftext|> -TITLE: Is pattern recognition NP-complete? -QUESTION [7 upvotes]: Hello, -is the problem of pattern recognition (for a given sequence of n numbers, find the shortest Turing machine with an alphabet of 42 elements that will output these n numbers in, say, 5*n^3 time) an NP-complete problem? -To me, it seems practically impossible that the problem is in P: If you had the level of understanding of algorithms required to show quickly that all algorithms smaller than this-or-that size will produce a different output or require more than 5*n^3 time, what reasons could there be that prevent you from having the level of understanding required to solve the Halting problem? (in this case, actually, the problem of determining whether a program will run in less than 5*n^3 steps) But I have no clue how you could possibly prove this. On the one hand, this seems to be for exactly the same reason: If you use the method "cleverly encode your SAT problem as a pattern; then prove that if the SAT instance is satisfiable, you could use a Turing machine of form A; if it is not satisfiable, you must use a longer Turing machine of form B" (or vice versa), how do you want to show that there exists no Turing machine C that is smaller than both of them? And: If you had the understanding to prove that, how could you possibly not be able to solve the Halting problem? -Even if you left that issue aside, I can't even think of approaches that could create such an A-B dichotomy. -Is the problem possibly NP-intermediate? - -REPLY [5 votes]: The key phrase you are looking for is "resource-bounded Kolmogorov complexity". This paper by Allender, et. al. may be a good starting point. Also, this PhD thesis might provide some helpful background. -Edited to add: -According to the first article, Ko, K.-I. "On the complexity of learning minimum time-bounded Turing machines", SIAM Journal on Computing, 20 (1991) may be even more relevant. In that paper, Ko demonstrates that determining whether or not computing time-bounded Kolmogorov complexity is NP-hard requires non-relativizing techniques.<|endoftext|> -TITLE: Should a published paper with a published correction be replaced on arXiv? -QUESTION [18 upvotes]: I submitted a paper to a journal. It was accepted and published. -I had also posted a version of the paper on arXiv. The arXiv version of the paper and the journal's version are not identical, because the journal's version uses their house LaTeX style. However, the two versions are otherwise the same, and share the same theorem numbering. -Then someone pointed out an error in the paper. So I wrote a short correction, and submitted it to the journal. It has been accepted and published as a separate paper. -My question is: what should I do with the arXiv version of the paper? Here are some possibilities. - -Replace the version on arXiv with a new version, which incorporates the necessary changes into the paper. Doing so would probably change the theorem numbering, so that it does not agree with the original paper. -Post the correction to the arXiv as a separate paper, and leave the old arXiv version of the original paper alone (so that it still agrees with the published version). -Post the correction to arXiv as a separate paper, and to replace the original paper on arXiv with a new version, which incorporates footnotes pointing out the erroneous statements, and referencing the correction (but otherwise not changing theorem numbering). - -The third choice seems like the right one to me. But I'd like to know what the accepted practice is here (if there is one). -Added. Several answers suggest a fourth choice: -4. Replace the original paper on arXiv with a revised version, but make sure the revision does not change theorem numbering, or at least carefully indicate the changes in numbering. -Added. The consensus seems to be for choice 4. This is what I'll do, since it looks like I'll be able to keep the numbering the same. - -REPLY [2 votes]: Probably there are also ways to force LaTeX to renumber the new theorems in such a way as it doesn't conflict with the old numbering. (Theorem 6.3B or something similar, or perhaps have the added theorems with their own numbering system using a different counter, roman numerals or something?)<|endoftext|> -TITLE: Individual mathematical objects whose study amounts to a (sub)discipline? -QUESTION [13 upvotes]: Certain mathematical objects have a theory so rich that their study -alone arguably constitutes a distinct (sub)discipline. My own list -would begin with -1) the absolute Galois group of the rationals; -2) the Mandelbrot set; -3) the Stone-Cech compactification of the integers; -4) the three-dimensional Cremona group; -5) the Riemann $\zeta$ -function; -6) the hyperfinite type $II_1$ factor; -7) the set of rational prime numbers; -8) $SL_2({\mathbb R})$; -9) the 27 lines on a cubic surface; -etc. -I suppose one might add "the real line," "the Euclidean plane," "the axioms -of ZFC," but I'm looking for objects that have emerged out of research and whose -richness itself might carry an element of surprise, rather than objects -purpose-built for their universal or foundational character. -I think a survey of such objects would make a lovely text for an undergraduate -capstone course, so I'm asking for your favorite examples. -My question has a sociological underpinning - there actually exist communities -of mathematicians who would recognize the objects I've listed as central to their -focus. I'm not allergic to suggestions of objects that should enjoy that level -of attention, but for whatever reason, don't yet. -In the same spirit, I recognize that all the objects mentioned belong to broad -categories, and could thus abstractly could be deemed mere examples, and certainly -then studied in a broader context. But de facto, these objects enjoy a distinctive critical level of attention in relative isolation. For example, each makes an appropriate subject for a monographic treatment. But please don't hesitate to make a suggestion because your favorite object doesn't have a monograph yet! - -REPLY [3 votes]: The Erdos-Renyi random graph model $G(n,p)$ - a single, concrete model that more or less created the field of random graph theory and is still studied.<|endoftext|> -TITLE: How is the longest increasing subsequence a matrix integral? -QUESTION [8 upvotes]: In "Random Matrices Random Permutations", the longest increasing subsequence of a permutation is related to an expectation over Hermitian matrices. -$$ \frac{1}{2^{|k|} n^{|k|/2}} \left\langle \prod_{j=1}^s \mathrm{tr}H^{k_j} \right\rangle$$ -Can anyone clarify this relation? I vaguely remember this coming from a paper of Gessel. -In general, I wonder is there a "gadget" turning permutation statistics (such as inversion number, or number of cycles) into integrals over unitary matrices? - -REPLY [3 votes]: There is a roundabout way of putting this. -A discrete analogue of random matrix spectra is random partitions (=Young diagrams). There clearly is a 'gadget' relating random permutations with random Young diagrams - the celebrated Robinson-Schensted correspondence. -On the other hand, the passage from random Young diagrams to random matrices is rather well understood (mainly, from the algebraic point of view). In particular, many random matrix ensembles arise from random Young diagrams via certain degenerations. And of course, in various limit regimes the asymptotic distributions of random partition and random matrix ensembles are precisely the same (the Airy ensembles, etc.). -On the other hand, maybe a look on some generalizations of your identity (arXiv:math/9905083) can help?<|endoftext|> -TITLE: Is the generic deformation of a symplectic variety affine? -QUESTION [11 upvotes]: Kaledin and Verbitsky have shown that symplectic varities have a remarkably nice deformation theory as symplectic varieties. -Let $X$ be a symplectic variety (a smooth quasi-projective variety over $\mathbb{C}$ equipped with a nondegenerate closed algebraic 2-form $\Omega$); then there is a universal formal deformation $\tilde X$ over $H^2(X;\mathbb{C})$ completed at the class $[\Omega]$. -In many examples, it seems that this deformation is generically affine; that is, its fiber at the generic point is an affine scheme. For example, $T^*G/B$ deforms to a generic coadjoint orbit in $\mathfrak{g}^*$, and if $X$ is a hyperkähler quotient, like a quiver variety, then the deformation comes from varying the complex moment map, and the deformation is also generically affine (since a GIT quotient by a free action of a reductive group is always affine). -Of course, I don't think this always happens; the product of two elliptic curves is symplectic, but doing this deformation should just change the $j$-functions of the curves simultaneously. So, clearly one needs some kind of extra condition. I've opted for "resolution of singularities of its affinization." -So my question is: - -Let $X$ be a symplectic variety which is a resolution of singularities of its affinization. Is $\tilde X$ generically affine? - -EDIT: As was pointed out in Misha Verbitksy's answer below, the deformation is not canonically algebraic. I believe though that if you assume that $X$ has a $\mathbb{C}^*$-action which is dilating (Definition 1.7 of this paper), then $\tilde X$ will also have a $\mathbb{C}^*$-action which acts by dilation on $H^2(X;\mathbb{C})$ and there will be a unique algebraic structure for which weight vectors of the $\mathbb{C}^*$-action are algebraic functions. It is this algebraic structure I want. -EDIT: Since it's buried a little bit in comments, let me just put here that it's true and proven by Kaledin in this paper. - -REPLY [7 votes]: Being "affine" in this case does not make much sense, -because the hyperkaehler deformation is a complex manifold, without -a fixed algebraic structure. Simpson produced an example of a -hyperkaehler deformation of a space of flat bundles -admitting several algebraic structures, both -inducing the same Stein complex structure; one of them -is affine, another has no global algebraic functions. -In fact, the space F of flat line bundles on elliptic curve -(with an appropriate algebraic structure, defined by -Simpson) is an example of such a manifold, -it is biholomorphic to $C^*\times C^*$, but -this biholomorphic equivalence is not algebraic, -and F has no global algebraic functions. -However, you can show that a hyperkaehler deformation -of a resolution of something affine has no non-trivial complex -subvarieties (arXiv:math/0312520), except, possibly, some -hyperkaehler subvarieties The latter don't exist, because -the holomorphic symplectic form $\Omega$ on such a manifold -is is lifted from the base, which is affine, hence $\Omega$ -vanishes on all complex subvarieties. -Therefore, a typical fiber of such a deformation is Stein. -Indeed, a hyperkaehler deformation of a -resolution of something affine remains holomorphically convex. To see this -if you produce a function which is strictly plurisubharmonic outside of -a compact set (we have such a function, because we started from something -affine), and apply the Remmert reduction.<|endoftext|> -TITLE: Labelled spanning trees without edge crossings -QUESTION [10 upvotes]: Draw the complete graph $K_n$ on a plane in general position with every edge a straight line and randomly label the edges $0$ or $1$. Does this graph always have a spanning tree with no edges crossing and edge-labels either all $0$ or all $1$? - -REPLY [9 votes]: This is a theorem of Gyula Károlyi, János Pach and Géza Tóth: Ramsey-type results for geometric graphs. I. ACM Symposium on Computational Geometry (Philadelphia, PA, 1996). -Discrete Comput. Geom. 18 (1997), no. 3, 247–255. Link to preprint -In this paper they indeed give an induction proof, but IMHO not an easy one. - -REPLY [2 votes]: Here is a proof under the assumption that the vertices $V$ are the vertices of a convex polygon. Label these as $p_1, \dots, p_n$ in cyclic order (the subscripts should be read modulo $n$). If all the edges $p_ip_{i+1}$ are red, then we are done. Otherwise, we may assume that $p_1p_2$ is red, and $p_n p_1$ is blue. By induction, we have that $V-p_1$ has a spanning red or blue plane tree $T$. In either case, we can extend $T$ to a spanning monochromatic plane tree of $V$.<|endoftext|> -TITLE: Teichmuller modular forms and number theory -QUESTION [16 upvotes]: Do higher genus Teichmuller modular forms have, or are they expected to have, implications for number theory that generalize the sorts of results that flow from the study of classical modular forms? - -REPLY [12 votes]: $$\hbox{Vector-valued Teichmueller Modular forms}$$ -Vector-valued Siegel -modular forms are the natural generalization of the classical elliptic -modular forms as seen by studying the cohomology of the universal abelian variety. -In spite of their relevance they have been studied essentially for genus $g=2$, where -correspond to suitable commutators of Siegel modular forms. -In the case $g=2$ and $g=3$ Ichikawa introduced the concept of Teichmueller modular forms. -It turns out that the Mumford forms for $g>3$ lead to the concept of vector-valued Teichmueller modular forms. -The main steps are the following. -For each fixed positive integers -$g,n$, define $$M_n(g)=M_n:={g+n-1\choose n}\ ,\; -N_n(g)=N_n:=(2n-1)(g-1)\ ,\quad K_n:=M_n-N_n\ , $$ -so that, for a curve $C$ of genus $g\ge -2$, $M_n$ and $N_n$ are the dimensions of ${\rm Sym}^n H^0(K_C)$ and -$H^0(K_C^n)$, respectively. -Let ${\frak H}_g:=\{Z\in M_g({\Bbb C})\mid {}^tZ=Z,\mathop{\rm Im} Z>0\}$, be the Siegel -upper half-space. Let -$\{\alpha_1,\ldots,\alpha_g,\beta_1,\ldots,\beta_g\}$ be a -symplectic basis of $H_1(C,{\Bbb Z})$. Denote by -$\omega_1,\ldots,\omega_g$ the basis of $H^0(K_C)$ satisfying the -standard normalization condition -$\oint_{\alpha_i}\omega_j=\delta_{ij}$, and by -$\tau_{ij}:=\oint_{\beta_i}\omega_j$ the Riemann period matrix, -$i,j=1,\ldots,g$. -Denote by ${\cal I}_g$ the closure of the locus of Riemann -period matrices in ${\frak H}_g$ and by ${\cal M}_g$ the moduli space of curves of genus $g$. -Consider the case $g\ge 2$ and a given symplectic basis for $H_1(C,{\Bbb Z})$. For each positive integer $n$, consider the basis $\tilde\omega_1^{(n)},\ldots,\tilde\omega_{M_n}^{(n)}$ of ${\rm Sym}^n H^0(K_C)$ whose elements are symmetrized tensor products of $n$-tuples of vectors of the basis $\omega_1,\ldots,\omega_g$, taken with respect to an arbitrary ordering chosen once and for all. Denote by $\omega_i^{(n)}$, $i=1,\ldots, M_n$, the image of $\tilde\omega_i^{(n)}$ under the natural map $\psi:{\rm Sym}^n H^0(K_C)\to H^0(K_C^n)$. It is well known that such a map is surjective if and only if $g=2$ or $C$ is non-hyperelliptic of genus $g>2$. For $n=2$, $g=2$ and $g=3$ non-hyperelliptic, this map is an isomorphism. -Consider the Thetanullwerte -$\chi_k(Z):=\prod_{\delta\hbox{ even}} \theta\[\delta\](0,Z)$, -$Z\in{\frak H}_g$, with $k=2^{g-2}(2^g+1)$. -Set -$$F_g:=2^g -\sum_{\delta\hbox{ -even}}\theta^{16}\[\delta\](0,Z)-\bigl(\sum_{\delta\hbox{ -even}}\theta^{8}\[\delta\](0,Z)\bigr)^2 \ . -$$ -It turns out that $F_4$, the Schottky-Igusa form, vanishes only on the Jacobian. Furthermore, there is a nice relation -between $F_g$ and the theta series $\Theta_\Lambda$ -corresponding to the even unimodular lattices $\Lambda=E_8$ and -$\Lambda=D_{16}^+$: -$$ -F_g=2^{-2g}(\Theta_{D_{16}^+}-\Theta_{E_8}^2) \ . -$$ -Let $\{\phi^n_i\}_{1\le i\le N_n}$ be a basis of -$H^0(K_C^n)$, $n\geq2$. -The Mumford form is, up to a universal constant -$$ -\mu_{g,n}={\kappa[\omega]^{(2n-1)^2}\over -\kappa[\phi^n]}{\phi^n_1\wedge\cdots\wedge\phi^n_{N_n}\over -(\omega_1\wedge\cdots\wedge\omega_g)^{c_n}} \ , -$$ -where $\kappa[\omega]$ is a constant that depends only on the choice of the homological basis whereas $\kappa[\phi^n]$ also depends on the choice of the basis $\phi^n$ (see Prop.1.2). In the case $n=2$ and $g<4$ one may choose the natural basis -${\rm Sym}^2 H^0(K_C)$ for $H^0(K_C^2)$, and for $g=2$ gets -$${\kappa[\omega]^{9}\over -\kappa[\omega^{(2)}]} ={1\over \pi^{12}\chi_{5}^2(\tau)}\ ,$$ -whereas for $g=3$ -$${\kappa[\omega]^{9}\over -\kappa[\omega^{(2)}]} ={1\over 2^6\pi^{18}\chi_{18}^{1/2}(\tau)}\ . $$ -For $g>3$ one has $g(g+1)/2-(3g-3)>0$, so apparently there is no a natural extension. Nevertheless one may continue to take $3g-3$ elements of ${\rm Sym}^2 H^0(K_C)$, or, more generally $N_n:=(2n-1)(g-1)$ elements of ${\rm Sym}^n H^0(K_C)$. Doing this leads to some surprise. - -To simplify notation, denote here by $\omega^{(n)}$ the basis $\{\omega^{(n)}_k\}$ with -$k=i_1,\ldots,i_{N_n}\in\{1,\ldots,M_{n}\}$. -$$[i_{N_n+1},\ldots,i_{M_n}|\tau]=\epsilon_{i_1,\ldots,i_{M_n}} -{\kappa[\omega^{(n)}]\over\kappa[\omega]^{(2n-1)^2}}\ ,$$ -are vector-valued Teichmuller modular forms without poles on ${\cal M}_g$ and vanishing on the hyperelliptic locus, of weight -$$ -d_n:=6n^2-6n+1-{g+n-1\choose n-1} \ . -$$ -Note that the vector-valued nature is just a consequence of the inequality $M_n-N_n>0$ for some $g$. For example, for $n=2$ one has $g(g+1)/2-(3g-3)>0$ satisfied for $g>3$. This implies that there are free indeces: the $i_{N_n+1},\ldots,i_{M_n}$, -a nice hint that the theory of vector-valued Teichmueller modular forms is a key tool to investigate the Schottky problem, see below for the case of genus 4 (presumably here should also appear some interesting Number Theoretical structures). -For each integer $n\geq 2$ and for all $i_{2},\ldots,i_{K_n}\in\{1,\ldots,M_n\}$ one has -$$\sum_{i=1}^{M_n}[i,i_{2},\ldots,i_{K_n}|\tau]\omega^{(n)}_{i}(x) -=0\ .$$ In particular, for $n=2$ these are all the quadrics characterizing the canonical curve in projective space. -Remarkably, one finds that at $g=4$, $[(ij)|\tau]\equiv [i|\tau]$ (see the paper for the indexing) is proportional to -$$S_{4ij}(Z):={1+\delta_{ij}\over 2}{\partial F_4(Z)\over \partial Z_{ij}} \ .$$ -For $g=4$ the discriminant of the quadrics is proportional to the square root of -$\chi_{68}$, the $g=4$ Thetanullwerte -$$ \det S_4(\tau)=d\chi_{68}(\tau)^{1/2}\ , $$ -with $d$ a constant. -A key step here is the following lemma. -Let $C$ be either a non-hyperelliptic Riemann surface of genus $g=4$ -or a non-trigonal surface of $g=5$. Then, the canonical model of $C$ -is contained in a quadric of rank $3$ if and only if -$\prod_{\delta\hbox{ even}}\theta[\delta]=0$. -The $g=4$, $n=2$ Mumford form is -$$\mu_{4,2}=\pm{1\over c -S_{4ij}}{\omega_1\omega_1\wedge\cdots\wedge -\widehat{\omega_i\omega_j}\wedge\cdots\wedge \omega_4\omega_4\over -(\omega_1\wedge\cdots\wedge\omega_4)^{13}} \ , -$$ -with $c$ a constant. - -Note that $S_{4ij}(Z)$ -trasforms with affine terms proportional to $F_4$, so that it is a vector-valued modular form only when $F_4=0$, that is in the Jacobian. This motivates the name vector-valued Teichmueller modular forms. Note that also the square root of $\chi_{68}(\tau)$ exists only in ${\cal I}_4$. It follows that $\det S_4$ and $\chi_{68}(\tau)^{1/2}$ are modular forms (of weight $34$) only when restricted to ${\cal I}_4$. -This clearly shows that the vector-valued Teichmueller modular forms -$[i_{N_n+1},\ldots,i_{M_n}|\tau]$ -are deeply related to the geometry of the Jacobian and to the Schottky's problem. The structure of the vector-valued Teichmueller modular at any $g$, generated by Mumford's forms, and their properties, such as the one of generating canonical curves, that is Petri's relations, strongly support Mumford's suggestion -that Petri's relations are -fundamental and should have basic applications: see pg.241, D. Mumford, The Red Book of Varieties and Schemes, Springer Lecture Notes in Math. 1358, 1999. Actually, it seems Mumford was right, one has just to use his forms. -A suggestion for the literature: the papers by John Fay are excellent and not very well-known as they should. The one in the Memoirs of the AMS is a masterpiece.<|endoftext|> -TITLE: When does the direct image functor commute with tensor products? -QUESTION [6 upvotes]: Let $i : U \to X$ be a quasi-compact open immersion of schemes. Under which conditions is the natural map -$i_* M \otimes i_* N \to i_* (M \otimes N)$ -for all $M,N \in \text{Qcoh}(U)$ an isomorphism? We may assume that $X=\text{Spec}(A)$ is affine. If $U$ is affine, then we may reduce to the case $M=N=\mathcal{O}_U$ (using presentations) and use that $i^{\#}$ is a flat epimiorphism to get an affirmative answer. If there are counterexamples for general $U$, what conditions are sufficient? - -REPLY [15 votes]: This is an addition to the Laurent's answer. First, it should be said that if one derives all the functors, one will get an isomorphim --- $Ri_*M \otimes^L Ri_*N \cong Ri_* (M \otimes^L N)$. Indeed, it is a simple corollary of the projection formula: -$$ -Ri_*M \otimes^L Ri_*N \cong -Ri_*(M \otimes^L i^*Ri_*N) \cong -Ri_*(M \otimes^L N) -$$ -(the second isomorphism is by the flat base change). What goes wrong with the underived version is that $M$ and $N$ have higher direct images which then have $Tor$'s all of which eventually get canceled. In the particular example of Laurent one has -$$ -R^1i_*O_{Y'} = y_1^{-1}y_2^{-1}k[y_1^{-1},y_2^{-1}], -\quad -R^1i_*O_{Z'} = z_1^{-1}z_2^{-1}k[z_1^{-1},z_2^{-1}], -$$ -where are $y_1,y_2$ are coordinates on $Y$ and $z_1,z_2$ are coordinates on $Z$. -In addition to $O_Y\otimes O_Z = k$ we have -$$ -Tor_2(O_Y,R^1i_*O_{Z'}) = Tor_2(R^1i_*O_{Y'},O_Z) = k, -$$ -$$ -Tor_4(R^1i_*O_{Y'},R^1i_*O_{Z'}) = k, -$$ -and it is easy to see that all this cancels in the spectral sequence calculating $Ri_*O_{Y'} \otimes^L Ri_*O_{Z'}$.<|endoftext|> -TITLE: Kähler small resolution of 3-dim ADE singularities -QUESTION [5 upvotes]: In Tristan Hübschs book on Calabi-Yau manifolds it is stated that 3-dimensional ADE singularities admit a crepant resolution but this resolution does not need to be Kähler. -Are there some results on when one of these singularities admit a small resolution that is Kähler? -Best, -Peter - -REPLY [2 votes]: An analysis of when this type of singularities admits (Kähler) small resolutions is carried out in this paper by Sheldon Katz, see also this question and the answer to -this one.<|endoftext|> -TITLE: Most intricate and most beautiful structures in mathematics -QUESTION [62 upvotes]: In the December 2010 issue of Scientific American, an article "A Geometric Theory of -Everything" by A. G. Lisi and J. O. Weatherall states "... what is arguably the most -intricate structure known to mathematics, the exceptional Lie group E8." Elsewhere in the -article it says "... what is perhaps the most beautiful structure in all of -mathematics, the largest simple exceptional Lie group. E8." Are these sensible -statements? What are some other candidates for the most intricate structure and for the -most beautiful structure in all of mathematics? I think the discussion should be confined -to "single objects," and not such general "structures" as modern algebraic geometry. -Question asked by Richard Stanley - -Here are the top candidates so far: -1) The absolute Galois group of the rationals -2) The natural numbers (and variations) -4) Homotopy groups of spheres -5) The Mandelbrot set -6) The Littlewood Richardson coefficients (representations of $S_n$ etc.) -7) The class of ordinals -8) The monster vertex algebra -9) Classical Hopf fibration -10) Exotic Lie groups -11) The Cantor set -12) The 24 dimensional packing of unit spheres with kissing number 196560 (related to 8). -13) The simplicial symmetric sphere spectrum -14) F_un (whatever it is) -15) The Grothendiek-Teichmuller tower. -16) Riemann's zeta function -17) Schwartz space of functions -And there are a few more... - -REPLY [2 votes]: I want to mention the small category $\bf{\Delta}$, where its objects are $[n]$ for a natural number $n$ and its morphism are all maps preserve orders. -I've been told $\bf{\Delta}$ is magic for a long time. Topologists use it to say what a space (simplicial set) is, and algebraic-geometors also use it. We even need it to define the higher categories. -I'm still in learning about this category, and I wish I could know some deep reason why it works so well.<|endoftext|> -TITLE: What is an example of a regular realization of $C_5$ over $\mathbb{Q}(x)$? -QUESTION [7 upvotes]: It's known that all abelian groups are regularly realizable over $\mathbb{Q}(x)$, but it occurred to me that I don't even have an example of a cyclic regular extension of $\mathbb{Q}(x)$ handy. -So: what is an example of a regular realization of $C_5$ over $\mathbb{Q}(x)$? - -REPLY [11 votes]: Emma Lehmer's quintic -\begin{align*} -y^5 +& x^2y^4 - (2x^3 + 6x^2 + 10x + 10)y^3 +\\ - &(x^4 + 5x^3 + 11x^2 + 15x + 5)y^2 + (x^3 + 4x^2 +10x + 10)y + 1 -\end{align*} -has Galois group $C_5$ over ${\mathbf Q}(x)$ and the splitting field over ${\mathbf Q}(x)$ is a regular extension. Her paper is "Connection between Gaussian periods and cyclic units" Math. Comp. 50 (1988), 535--541. -Typing "Lehmer quintic" in Google produces several hits which will give more information.<|endoftext|> -TITLE: What's an example of a locally presentable category "in nature" that's not $\aleph_0$-locally presentable? -QUESTION [16 upvotes]: Recall the notion of locally presentable category (nLab): $\DeclareMathOperator{\Hom}{Hom}$ -Definition: Fix a regular cardinal $\kappa$; a set is $\kappa$-small if its cardinality is strictly less than $\kappa$. A $\kappa$-directed category is a poset in which every $\kappa$-small set has an upper bound. A $\kappa$-directed colimit is the colimit of a diagram for which the indexing category $\kappa$-directed. An object $a$ of a category is $\kappa$-small if $\Hom(a,-)$ preserves $\kappa$-directed colimits. A category is $\kappa$-locally presentable if it is (locally small and) cocomplete and there exists a set of objects, all of which are $\kappa$-small, such that the cocompletion of (the full subcategory on) this set in the category is the entire category. -It is a fact that every $\kappa$-locally presentable category is also $\lambda$-locally presentable for every $\lambda > \kappa$. -In a current research project, we have some constructions that work naturally for $\kappa$-locally presentable categories for arbitrary regular cardinals $\kappa$. But all of our applications seem to be to $\aleph_0$-locally presentable categories. For example, for any ring $R$, I'm pretty sure that the category of $R$-modules is $\aleph_0$-locally presentable. (Every module has a presentation; any particular element or equation in the module is determined by some finite subpresentation.) The category of groups is $\aleph_0$-locally presentable (by the same argument). I'm told that every topos is locally presentable for some $\kappa$; is a topos necessarily $\aleph_0$-locally presentable? -Indeed, although I know of some categories that are not $\aleph_0$-locally presentable but are $\kappa$-locally presentable for some larger $\kappa$; the example, apparently, is the poset of ordinals strictly less than $\kappa$ for some large regular cardinal $\kappa$. But this is not a category I have ever encountered "in nature"; it's more of a zoo specimen. Hence my somewhat ill-defined question: - -Question: Do there exist "in nature" (or, "used by working mathematicians") categories that are $\kappa$-locally presentable for some $\kappa > \aleph_0$ but that are not $\aleph_0$-locally presentable? - -Put another way, is there any use to having a construction that works for all $\kappa$? -So far, I haven't thought much about general topoi, and we do want our project to include those, so I would certainly accept as an answer "yes, this particular topos". But there might be other "representation theoretic" categories, or other things. - -REPLY [5 votes]: The category of $\Delta$-generated spaces (the full category of the category of general topological spaces which are colimits of simplices) is locally $\lambda$-presentable for a regular cardinal $\lambda \geq 2^{\aleph_0}$. See http://www.tac.mta.ca/tac/volumes/21/1/21-01abs.html. This category contains all realizations of simplicial sets, all CW-complexes, all spheres, etc... i.e. all topological spaces used in algebraic topology. It is moreover cartesian closed.<|endoftext|> -TITLE: Is every saft category cocomplete? -QUESTION [14 upvotes]: Here is a word that I think should be adopted by the category theorists. (If there is another synonymous word already in existence, please let me know.) -Definition: A category $C$ is saft if every cocontinuous functor $C \to D$ has a right adjoint. -The word "saft" is an abbreviation for "special adjoint functor theorem (SAFT)", of which there are many, because SAFTs always take the form "If a category $C$ is XYZ, then it is saft." For example: it suffices for a category to be some cocompletion of some small subcategory (F. Ulmer, The adjoint functor theorem and the Yoneda embedding, Illinois Journal of Mathematics, 1971 vol. 15 (3) pp. 355-361). -It is an easy exercise that a category $C$ is saft if and only if every continuous functor $C^{\rm op} \to \text{Set}$ is representable. In particular any saft category is complete, because any putative limit corresponds to a continuous functor, and hence necessarily representable. -On the other hand, I don't see any particular reason why a saft category is necessarily cocomplete, except that every example I know is, and every SAFT uses cocompleteness as one of the conditions. - -Question: Is a saft category necessarily cocomplete? - -REPLY [5 votes]: This is just an extension of Todd's answer to summarize Adamek's example, which is a bit convoluted. By a graph we mean a set equipped with a binary relation, call it ∼. If A is a graph, let $A^{(3)}$ be the set of triples (x,y,z) such that x∼y∼z. And let F(A) be the power set $P(A^{(3)})$, equipped with the relation defined by ∅∼X for all nonempty X. Adamek's category is the category of algebras for the endofunctor F of the category of graphs, i.e. of graphs A equipped with a graph-morphism F(A)→A. He proves that the forgetful functor from this category to Graphs has a left adjoint, namely $A\mapsto A \sqcup F(A)$, and is monadic. But he gives the following example of a pair of parallel morphisms of F-algebras that have no coequalizer. -Let A be the set {p,q} with the empty relation ∼, and let B be the set {s,t} with s∼t only. Then F(A) and F(B) are both the graph {∅} with the empty relation, and we make A and B into F-algebras by sending ∅ to p and s, respectively. Now let f:A→B send p to s and q to t, while g:A→B sends p and q both to s. Adamek goes on to prove that if f and g had a coequalizer in F-algebras, then one could construct from this a weakly initial P-algebra, where P is the powerset endofunctor on Set; from this one could then construct an initial P-algebra, hence a fixed point of P, which contradicts Cantor's diagonal argument. -I don't have time to summarize the construction of the weakly initial P-algebra from a coequalizer of f and g, but I'm making this CW, so anyone else who wants to add it, feel free.<|endoftext|> -TITLE: What is a convenient shorthand notation for a category -QUESTION [10 upvotes]: Set theory has a very convenient and well established curly brace notation to specify a set by its elements: $\{2,3,4,6\}$ or $\{\text{finite subgroups of }SU(2)\}$ are simple examples. -There should be a similar convenient notation for specifying a category by its objects and morphisms. Such a notation should easily accommodate categorical constructions such as slice categories. For example a double slash notation to separate objects and morphisms would define a slice category by something like (I am making this up!) -$$ -\mathcal{C}\downarrow X= [ Y\to^f X : [Y//f]\in \mathcal C\quad // \quad (Y\to^f X)\to^h (Z\to^g X): [Y,Z//h] \in \mathcal C, g\circ h=f] -$$ -(a commutative diagram in the second part of the specification would be more convenient here, but it should also be possible to typeset the notation inline). -Do such notations already exist? Whether they do or not, what notations would contributors recommend or suggest? -Update. Many thanks for comments made here. So far I most like the observation that clearly describing the morphisms makes the objects implicit. Still, I think beginners need the objects too, and have been experimenting with a notation like the one above, but using "staples" instead of square brackets, and introducing morphisms after objects by a vertical rectangular block (a bit like a closed staple). 2-morphisms could then be introduced in a similar way by a double block (a block with a vertical line through it). While the answers convince me that such notation is often unnecessary and maybe unhelpful sometimes, I'm not convinced such notation would be worthless. - -REPLY [4 votes]: One thing I often do is work with set-builder notation $\{blah\in thing| conditions\}$ where either or both of the $blah$ and $conditions$ are allowed to be (collections of) (2-)commuting diagrams. I also work with objects and arrows separately. (Edit: by which I mean I write $Obj(C) := \ldots$ and $Mor(C) := \ldots$ or similar)<|endoftext|> -TITLE: What formal properties should resolution of singularities have? -QUESTION [15 upvotes]: If I were going to propose a new construction as a "replacement for resolution of singularities", what properties would my replacement have to have? [I am going to do no such thing -- this is purely speculative.] Is there a shortish list of theorems such that any construction verifying the properties on the list would thereby deserve to be called a resolution of singularities? - -REPLY [6 votes]: It should exist in all characteristics, even mixed characteristic! - -This sounds somewhat cheeky, but I was fairly serious. To algebraists, and the OP is one last time I met him, a purely ring-theoretic statement which can largely only be verified in char. $0$ feels more incomplete than to a geometer. -This is a main reason why the use of resolution of singularity is restricted in attacking some of the open questions in commutative algebra: often the hardest case is mixed characteristic, and if you are extremely lucky and smart you can reduce it to a statement in char. $p>0$, then you are dead! -Because of the above reason, I would mention that de Jong's alteration has found some spectacular success in commutative algebra (so one can replace birational by surjective and generically finite). A main example is Gabber's proof of the non-negativity part of Serre's conjectures on intersection multiplicities (see number 1 here for an expository account) .<|endoftext|> -TITLE: What are some examples of journals that will accept undergraduate student research? -QUESTION [9 upvotes]: I am currently doing a research project with a professor and 3 other students in an area that is usually seen as a "recreational" math topic; that of change-ringing and its relation to group theory. The papers regarding change-ringing that I have managed to find on JSTOR are generally of an expository nature, and since we do have one or two somewhat original results (we were trying to see what subgroups of Sn can be enumerated using transformations allowed under the restrictions of change-ringing), I think we might have something at least worth trying to submit to a small journal. -Does anyone have any recommendations? - -REPLY [3 votes]: I haven't seen mentioned the AMS undergraduate mathematics page, in particular the "Clubs, Conferences, Events, Online Journals" section. The section mentions the following journals targeting undergraduates: - - Involve, mentioned in another answer. - The Rose-Hulman Institute of Technology Undergraduate Mathematics Journal, which requires an established mathematician to sponsor your submission and recommend referees. - The Harvard College Mathematics Review, focuses on expository articles by undergraduates. It also sadly seems to be semi-defunct, the website promises an issue in March but one has not yet appeared. - -There is also a new undergraduate journal The Waterloo Mathematics Review from the University of Waterloo (full disclosure: I am one of the editors) in a similar style to the HCMR, though it also accepts original research. We are currently accepting submissions for our second issue, while this answer may find you too late I hope you consider submitting.<|endoftext|> -TITLE: Who colored in my Dynkin diagrams? -QUESTION [12 upvotes]: Many of you will recognize these as the ADE diagrams, festively colored for the holidays! -     (source) - - -Does anyone know a mathematical interpretation for these diagrams, when colored like this? - - -Edit: the answers below seem to treat the set of red nodes and the set of green nodes as essentially equivalent. However in the combinatorics relevant to me, the red and green ones play slightly different roles; does this also happen in the Coxeter group story? - -REPLY [3 votes]: The fact that Dynkin diagrams are bipartite allows one to define the box product $X\square Y$ of two Dynkin diagrams (or rather, of two quivers whose underlying graphs are Dynkin diagrams). This is very relevant in the study of Zamolodchikov perodicity; see e.g. https://arxiv.org/abs/1001.1531 and https://arxiv.org/abs/1506.05378.<|endoftext|> -TITLE: Classes of graphs for which isospectrum implies isomorphism? -QUESTION [8 upvotes]: The spectrum of a graph is the (multi)set of eigenvalues of its adjacency matrix (or Laplacian, depending on what you're interested in). In general, two non-isomorphic graphs might have the same spectrum. -Prompted in part by this discussion on reverse engineering a graph from its spectrum, I was wondering: - -Are there interesting classes of - graphs for which isospectrality - implies isomorphism ? - -REPLY [4 votes]: If you consider a class of graphs which is closed under taking covering spaces, then this is likely not to work, since one may then apply Sunada's method to construct isospectral pairs.<|endoftext|> -TITLE: comparing diffusions -QUESTION [7 upvotes]: Consider a probability distribution $\pi$ on the real axis that has a density (w.r.t Lebesgue) proportional to $e^{-V(x)}$, where $V(\cdot)$ is a potential function. For any reasonable volatility function $\sigma:\mathbb{R} \to (0:+\infty)$ the diffusion -$$ dX^{\sigma}_t = [ -\frac{1}{2} \sigma(X_t^{\sigma})^2 V'(X_t^{\sigma}) + \sigma(X_t^{\sigma}) \sigma'(X_t^{\sigma}) ] dt + \sigma(X_t^{\sigma}) \, dW_t $$ -has $\pi$ as unique invariant distribution. -Question: -Given two volatility functions $\sigma_1, \sigma_2$, are there tractable ways of comparing the speed of convergence to equilibrium of the two associated diffusions? -For example, if $\sigma_2(x) = \alpha \cdot \sigma_1(x)$, the diffusion $X^{\sigma_2}$ is just $X^{\sigma_1}$ slowed down by a factor $\alpha$: any ways of comparing the two diffusions should say that if $\alpha > 1$ then $X^{\sigma_2}$ converges 'faster' than $X^{\sigma_1}$. Spectral Gaps work but are not very tractable when comparing two non-proportional diffusions. Is it hopeless ? -Motivations: -I consider several MCMC algorithms with target density $\pi$: each one of them, after some time-rescaling, looks like a diffusion $X^{\sigma}$. Which algorithm is the best $i.e.$ what diffusion $X^{\sigma}$ mixes the fastest ? - -REPLY [4 votes]: A nice quantitative and very general tool to study the speed to convergence of symmetric Markov processes to equilibrium is the Bakry-Emery criterion. More precisely, let $(X_t)_{t \ge 0}$ be a diffusion Markov process with generator $L$, semigroup $P_t$ and symmetric and invariant probability measure $\pi$. Define the carre du champ by -$\Gamma(f,g)=\frac{1}{2} (L(fg) -fLg-gLf)$ -and the iterated carre du champ by -$\Gamma_2(f,f)=\frac{1}{2} (L\Gamma(f,f)- 2\Gamma(f,Lf))$ - -Assume that $\Gamma_2(f,f) \ge \rho \Gamma(f,f)$ for some positive constant $\rho$, then for every $t \ge 0$ - $\int (P_t f -\int f d\pi)^2 d\pi \le e^{-2\rho t} \int ( f -\int f d\pi)^2d\pi$ - -As a consequence, we get a convenient criterion for exponential speed to equilibrium. In your one -dimensional case, $\Gamma$ and $\Gamma_2$ are easy to explicitly compute so the criterion is easy to check. -Further details on the Bakry-Emery method may be found in these Lecture Notes by Dominique Bakry.<|endoftext|> -TITLE: Give an example of monoid with property $m^2 = m^3$ -QUESTION [8 upvotes]: Give an example of finitely generated, infinite monoid $M$ with property that for all $m \in M$ we've got $m^2 = m^3$. -This question comes from the problem I was given during algebraic languages theory class at CS department. I've got construction that is using methods outlined during that class but the structure of the monoid is not very clear. -I thought someone could propose more direct construction that would give better insight into methods of constructing such algebraic structures. -In a case there's no better solution I'm planning to share my own with brief explanation of methods used in that construction. - -REPLY [19 votes]: This is a classic result of Morse and Hedlund (they actually attribute it to Dilworth). Take the alphabet $\{a,b,c\}$ and an infinite word $W$ in that alphabet which does not contain subwords of the form $uu$ (such an infinite word was first constructed by Thue, search Google for Thue sequence, then by Morse-Hedlund, then by many others, all done independently). Now let $S$ be the set of all finite subwords of $W$ (including the empty word) and symbol $\{0\}$. The product of two words in $S$ is either their concatenation: $u\cdot v=uv$ if $uv\in S$ or 0 otherwise. That is a monoid satisfying the law $x^2=0$ (for all $x\ne 1$), hence -$$x^2=x^3.$$<|endoftext|> -TITLE: When is the product of two ideals equal to their intersection? -QUESTION [54 upvotes]: Consider a ring $A$ and an affine scheme $X=\operatorname{Spec}A$ . Given two ideals $I$ and $J$ and their associated subschemes $V(I)$ and $V(J)$, we know that the intersection $I\cap J$ corresponds to the union $V(I\cap J)=V(I)\cup V(J)$. But a product $I.J$ gives a new subscheme $V(I.J)$ which has same support as the union but can be bigger in an infinitesimal sense. For example if $I=J$ you get a scheme $V(I^2)$ which is equal to "double" $V(I)$. -Vague Question : What is geometric interpretation of $V(I.J)$ in general? -Precise question : When is $I\cap J=I.J$? Everybody knows the case $I+J=A$ but this is absolutely not necessary. For example if $A$ is UFD and $f,g$ are relatively prime then $(f).(g)=(f)\cap(g) $ but in general $(f)+(g)\neq A$ (e.g. $f=X, g=Y \in k[X, Y]$) -Thank you very much. - -REPLY [12 votes]: I apologize in advance for resurrecting such an old question but I absolutely could not resist the urge of sharing a precise characterization for $I \cap J=IJ$, that I read recently in a beautiful short paper, when $I,J$ are monomial ideals in a polynomial ring. -First some terminology: Let $k$ be a field and $R=k[x_1,...,x_n]$ . Every monomial ideal $I$ of $R$ has a unique minimal monomial set of generators, usually denoted by $G(I)$ . For a set of monomials $T$ in $R$, let $\newcommand{\Supp}{\operatorname{Supp}}\Supp (T) :=\{i | x_i $ divides $m$ for some $m \in T \}$ . -With this, we can state the characterization: Let $I,J$ be monomial ideals in $k[x_1,...,x_n]$ , then $I \cap J=IJ$ if and only if $\Supp (G(I)) \cap \Supp (G(J))$ is empty. -This is Theorem 2.2 in https://link.springer.com/article/10.1007/s12044-019-0509-5 -Here's the arxiv link https://arxiv.org/abs/1705.00488<|endoftext|> -TITLE: intersection of ideals in a commutative ring vs their product -QUESTION [10 upvotes]: This question was inspired by this one. Given two ideals $A,B$ in a finitely generated commutative ring $R$. Is it possible to decide whether $A\cap B=AB$? Here $R$ is given by generators and relations, i.e. as a factor-ring of the free commutative ring over a finitely generated ideal, $A$, $B$ are given by their (finite) generating sets. I do not remember seeing this question in standard books on Groebner bases, but perhaps it is hidden there. - -REPLY [9 votes]: I think this problem can in fact be handled by Gröbner basis theory in the case $A$ is a polynomial ring. Since $I\cdot J \subseteq I\cap J$ for any two ideals, one can simply compute a Gröbner basis of $I\cap J$ (which is computed as the elimination ideal $( t\cdot I+(1-t)\cdot I ) \cap A$) and then checking whether each generator belongs to $I\cdot J$ (again using Gröbner basis algorithm). -EDIT: As Mark points out in his comment, this argument can be used to solve the problem for the general case $A=k[X]/R$ by considering the ideals $I+R$ and $J+R$ in $k[X]$.<|endoftext|> -TITLE: Iwasawa main conjectures vs Bloch-Kato conjectures -QUESTION [30 upvotes]: Let $p$ be a prime, $K$ be a number field, $S$ a finite set of finite places of $K$ containing the set $S_p$ of places above $p$ and the places at infinity, $G:=G_{K,S}$ the Galois group of the maximal extension of $K$ unramified outside $S$, $\rho: G_K \rightarrow Gl_d({\mathbb Q}_p)$ a geometric irreducible representation of $G_K$. For $n$ any integer, $\rho(n)$ is the Tate twist of $\rho$, that is $\rho$ tensor the cyclotomic character to the power $n$. -The Bloch-Kato Selmer group of $\rho$, denoted $H^1_f(G,\rho)$ is defined as an explicit subspace of $H^1(G,\rho)$ (continuous cohomology): -$$H^1_f(G,\rho) = \ker \left(H^1(G,\rho) \rightarrow \prod_{v \in S_K-S_p} H^1(I_v,\rho) \times \prod_{v \in S_p} H^1(D_v, \rho \otimes B_{crys})\right),$$ where $D_v$, $I_v$ are respectively a decomposition subgroup and an inertia subgroup at $v$ of $G$, -and the $\rightarrow$ is the product of the restriction maps. -The first statement of the Bloch-Kato conjecture is (for all $n \in \mathbb{Z}$): -CONJECTURE: $\dim H^1_f(G_K,\rho(n)) - \dim H^0(G_K,\rho(n)) = \text{ord}_{s=1-n} L(\rho^\ast,s).$ -Here $L(\rho,s)$ is the complex $L$-function (we assume it has a meromorphic continuation over $\mathbb{C}$) -There are other statements concerning the principal values of the L-function -at $1-n$, that I do not consider here. Note that this conjecture is obviously invariant by Tate twists. Also, the $H^0$ term is $0$ except if $\rho(n)$ is the trivial representation. -Now I come to my question: It is clear that the Iwasawa main conjectures (by which I mean -not only Iwasawa's original conjecture on the Kubota-Leopoldt $\zeta$-function, but its modern generalizations) belongs to the same circle of idea. But what exactly is the -relation? -To make my question more precise, let us consider to fix ideas -Greenberg's form of the main conjecture, as stated for examples in his paper in Motives. -A condition on $\rho$, called the Panchiskin condition, is needed to formulate the conjecture. Then a Selmer group is defined as a module over the Iwasawa algebra -$\Lambda$, and this module is conjectured to be co-finite and related to the $p$-adic $L$-function of $\rho$. Unfortunately, Iwasawa-theorist tend to use a different language -than Bloch-Kato-theorists: they work with modules like $\mathbb{Q}_p/\mathbb{Z}_p$ instead of $\mathbb{Z}_p$ or $\mathbb{Q}_p$ and properties like co-finite instead of finite (perhpaps -they are comathematicians). After one takes cohomology, families, etc, the translation between the two languages becomes far from transparent. Yet, I know that the Iwasawa main conjectures have consequences that can be stated in a way very similar to the Bloch-Kato's conjecture. - -Can you state such a consequence of Iwasawa's main conjecture in a language closer to Bloch-Kato, precisely : relating (probably in a weaker sense that in BK) the dimension of a suitablle Selmer groups defined as a subspace of $H^1(G,\rho(n))$ cut by local conditions with the order of vanishing of the p-adic L-function of $\rho^\ast$ (assuming it exists) at some points ($1-n$?). Or is such a thing written somewhere? - -I apologize that my question is at the same time technical and elementary. Yet an answer would help me a lot, and possibly may help other people who want to get a global picture -of this kind of conjectures, and of the progresses made so far. For example, my question contains as a special case: - -What does the Iwasawa main conjecture for ordinary elliptic curces implies for the BSD conjecture? - -REPLY [11 votes]: If I understand your question properly, then I think much is known. Let me sum up what I understand about this picture. -First a short answer to your question. Contrary to what you ask for, it is not expected that the dimension of a subspace of $H^{1}$ cut by local conditions should express the order of vanishing of the $p$-adic $L$-function. -Let us start with Bloch-Kato conjecture. This conjecture can be interpreted as a description of cohomological invariants of motives using special values of the $L$-function (many people think of it in the converse way, as description of special values of the $L$-function in terms of Galois invariants). The first question to ask is "which cohomological invariants are we trying to describe?" and the most reasonable answer is "the complex $C$ of motivic cohomology with compact support" (not known to exist in general). Then the order of vanishing of the $L$-function gives the Euler characteristic of $C\otimes_{\mathbb Q}\mathbb R$ whereas the $p$-adic valuation of the principal term of the $L$-function (divided by the period defined in Bloch-Kato) is a $\mathbb Z_{p}$-basis of the determinant of $C\otimes_{\mathbb Q}\mathbb Q_{p}$ (more precisely, of the inverse of the determinant). Even though you knew all this already, I found it necessary to recall it in order to state what forms the IMC takes in this context. -Assume now that our $p$-adic Galois representation $V$ comes from a pure motive and is crystalline at $p$ (I realize that you don't want to make such a strong assumption, but I think all I will say will continue to hold, at least conjecturally). As pointed out in comments already, and as you know, the IMC will say something about the interpolation of the Bloch-Kato conjecture in a $\mathbb Z_{p}$-extension (or more generally in a universal deformation space). I will discuss here only the case of the cyclotomic $\mathbb Z_{p}$-extension. Inside $D_{cris}(V)$ sits $D^{\phi=p^{-1}}$. Let $e$ denotes the dimension of this space over $\mathbb Q_{p}$. Then the cohomological object described by the special values of the (putative) $p$-adic $L$-function is the Selmer complex $S$ of $V$ with the unramified conditions at places $\ell≠p$ of ramifications of $V$ and with the Bloch-Kato condition at the level of complex at $p$. -Based on Bloch-Kato, we should thus expect the Euler characteristic of $S$ evaluated at a character (this is to say of $S\otimes_{\Lambda}\mathbb Z_{p}[\chi]$) to be the order of vanishing of the $p$-adic $L$-function and the $p$-adic $L$-function to give a basis of $\det_{\Lambda} S$. Alas, things are not so easy, because of the infamous trivial zeroes phenomena. So what you can show (possibly assuming plausible conjectures or restricting yourself to rank at most 2 along the way, I'll make an effort to state something really precise if you need to) is that, under Bloch-Kato, the Euler characteristic of $S\otimes_{\Lambda}\mathbb Z_{p}[\chi]$ is equal to the order of vanishing of the usual $L$-function twisted by $\chi$ (as expected) plus $e$ (this is the contribution of the trivial zeroes) $\textit{provided}$ the $\mathcal L$-invariant does not vanish (this is, or should be, equivalent to the semi-simplicity of the complex giving the local condition at $p$). -All this having been said, perhaps you want a concrete answer for a concrete representation. In that case, nothing is simpler than a brave old ordinary representation. For ordinary representation, the local condition at $p$ for the Selmer complex $S$ is simply $R\Gamma(G_{\mathbb Q_{p}},V)$. Hence, the order of vanishing of the $p$-adic $L$-function at a given $\chi$ should simply be the order of vanishing of the $L$-function plus the dimension of $H^{0}(G_{\mathbb Q_{p}},V^{*}(1)/F^{+}V^{*}(1))$ plus or minus simple terms (like the zeroes or poles of the Gamma factors). This reflects the fact that in the generic case, the order of vanishing of the $p$-adic $L$-function should be the dimension of the first cohomology of $S$ (which is not a subspace of $H^{1}$, hence my word of warning at the beginning). -Hope this helped somehow. -Now, let us move on to your second question. I think that if you knew only the IMC, then you couldn't say much about the order of vanishing part of Bloch-Kato. However, if you knew the IMC as well as non-degeneracy of the $p$-adic height pairing (required to formulate the Equivariant Tamagawa Number Conjecture) as well as the Equivariant Tamagawa Number Conjecture for each layer of the cyclotomic extension and/or the vanishing of the $\mu$-invariant, then the order part of Bloch-Kato would follow. Here is how I would try to prove this. First, I would define $S$ (no problem here,as we are in the ordinary case). Then I would construct a canonical trivialization of this complex at each finite layer using the non-degeneracy of the height pairing. Then I would use the ETNC (or I would deduce the ETNC from the IMC using the vanishing of the $\mu$-invariant) to show that the image of the determinant of $S$ at a finite layer under my canonical trivialization is really the value of the principal term of the analytic $L$-function (perhaps times the $\mathcal L$-invariant, but I would know this to be non-zero by semi-simplicity of my complexes). In this way, I would manufacture a complex $L$-function which would agree with the ordinary $L$-function at many (not necessarily classical) points (this would presumably require the IMC and ETNC not only for the cyclotomic extension but for the Hida family containing $E$) and would thus be equal to it. Now, I would know the order of vanishing of my algebraic complex $L$-function at a classical point, so I would know the order of vanishing of the complex $L$-function as well so (finally!) I could check Bloch-Kato. -So, yeah, if you knew the ETNC for the full Hida family and/or the vanishing of the $\mu$-invariant plus the non-degeneracy of the $p$-height pairing, you can, I think, collect the order part of Bloch-Kato as a bonus. Perhaps a moment of sober reflexion is in order now. -Again, hope this helped (but doubt it somehow).<|endoftext|> -TITLE: The algebraic fundamental group of a reductive algebraic group -QUESTION [22 upvotes]: For a connected reductive algebraic group $G$ over a field $k$, other than the \'etale fundamental group of $G$ (regarded just as a scheme), there seems to be another notion, usually called the algebraic fundamental group of $G.$ I am not sure of its definition, but I guess (at least when $G$ is split) it might be something like $I/\Gamma,$ where $I$ is the "topological fundamental group" of a maximal torus $T$ (if it makes sense), and $\Gamma$ is the subgroup generated by the inverse roots. In particular, for $GL_n$ this should produce $\mathbb Z,$ which agrees with the topological fundamental group when $k=\mathbb C.$ -Could anyone give some references on this? - -REPLY [7 votes]: In addition to Marty's reference, I would recommend to look into §6 "Le groupe fondamental algébrique des groupes algébriques linéaires connexes via les resolutions flasques" -of Colliot-Thélène's paper -MR2404747, Colliot-Thélène, Jean-Louis, Résolutions flasques des groupes linéaires connexes, J. Reine Angew. Math. 618 (2008), 77–133. -This reference is in any characteristic.<|endoftext|> -TITLE: Abelian varieties and Selberg class -QUESTION [7 upvotes]: Hello everyone, -I would like to know whether, assuming Selberg's orthonormality conjecture, it would be possible to establish a "natural" correspondence between abelian varieties and functions belonging to the Selberg class S in such a way: -1) One associates to a simple abelian variety a primitive function of S, -2) One associates to an abelian variety of dimension d a function of S of degree d, -3) If V is an abelian variety isogenous to a product of abelian varieties of lower dimensions V_1, V_2, ... V_n, then the function F of S related to V is the product of the F_i where F_i is the function of S related to V_i. -Thank you in advance. - -REPLY [2 votes]: (Assuming the abelian varieties are supposed to be defined over $\mathbf{Q}$). For every dimension $d\geq 1$, select a cuspidal automorphic representation $f_d$ of $GL(d)$ over the rationals. Then map any simple $A$ of dimension $d$ to $L(f_d,s)$, and extend by multiplicativity using the simple factors of a general $A$ up to isogeny. -(Of course this question is utterly artificial; it seems reasonable to think that the only natural $L$-function associated to an abelian variety is its Hasse-Weil $L$-function, which has degree $2\dim(A)$.)<|endoftext|> -TITLE: notation for finite sequence with one element is removed -QUESTION [6 upvotes]: Often you need a notation for a finite sequence with one element is removed; -i.e. $$(x_1,\dots,x_{i-1},x_{i+1}\dots, x_n).$$ -I know one notation -$$(x_1,\dots,\hat x_i,\dots, x_n)$$ -and I hate it. It is too long and it has no sense; i.e., unless you know the meaning you will never guess what is it. -Question: Did you see any other? - -REPLY [2 votes]: Borrowing from simplicial sets/complexes, $d_i: \mathbf{n-1} \to \mathbf{n}$ -is the map that skips $i$, so your sequences would be $x\circ d_i$.<|endoftext|> -TITLE: Mazur's unpublished manuscript on primes and knots? -QUESTION [35 upvotes]: The story of the analogy between knots and primes, which now has a literature, started with an unpublished note by Barry Mazur. -I'm not absolutely sure this is the one I mean, but in his paper, Analogies between group actions on 3-manifolds and number fields, Adam Sikora cites -B. Mazur, Remarks on the Alexander polynomial, unpublished notes. -He also cites the published paper -B. Mazur, Notes on étale topology of number fields, Ann. Sci. Ecole Norm. Sup. (4)6 (1973), 521-552. -I suppose an expert would recognize as relevance of this paper, but I don't see that even the word "knot" ever occurs there. -[My Question] Does anyone have a copy of Mazur's note that they would share, please? If not, has anyone at least actually seen it. -By the way, already years ago, I asked Mazur himself. I watched him kindly search his office, but he came up dry. -I realize that whatever insights the original note contains have doubtless been surpassed after c. 40 years by published results, but historical curiosity drives my desire to see the document that started the industry. - -REPLY [12 votes]: Thanks to this MO question, the paper Remarks on the Alexander Polynomial is now available on Barry Mazur's website. -Gratuitous addendum. For a recent overview, see -Morishita (Masanori), -Analogies between prime numbers and knots, -Sūgaku 58 (2006), no. 1, 40–63. -An English translation has appeared in -Sugaku Expositions 23 (2010), no. 1.<|endoftext|> -TITLE: Do strict pro-sets embed in locales? -QUESTION [13 upvotes]: It is well-known that the category of profinite groups (by which I mean Pro(FiniteGroups), i.e. the category of formal cofiltered limits of finite groups) is equivalent to a full subcategory of topological groups, namely those with "profinite topologies." In fact this is a specialization to groups of a more general statement: the category of pro-(finite sets) is equivalent to the category of topological spaces with profinite topologies, via the functor which says "take the limit in the category of topological spaces, where your finite sets all have the discrete topology." -It is proven in the paper "Prodiscrete groups and Galois toposes" by Moerdijk that more generally, the category of pro-groups with surjective transition maps (aka "strict" or "surjective" pro-groups) is equivalent to that of prodiscrete localic groups, i.e. groups in the category of locales which are a cofiltered limit of ordinary groups, regarded as localic groups with discrete topologies. Is there a non-group version of this? Can "strict pro-sets" be identified with "pro-discrete locales"? -Note that some hypothesis such as "surjective transition maps" is necessary. For instance, the pro-set $\cdots \xrightarrow{+1} \mathbb{N} \xrightarrow{+1} \mathbb{N} \xrightarrow{+1} \mathbb{N}$ is not isomorphic to the trivial pro-set ∅, but its limit in the category of locales is just as empty as its limit in the category of sets. - -REPLY [3 votes]: I think the answer is no. -First note that if $(S_i)_i$ is a pro-set, which we may WLOG assume to be indexed on a directed poset, then the corresponding prodiscrete locale $\lim S_i$ is presented by the following posite. Its underlying poset is the category of elements of the diagram $(S_i)_i$, i.e. its elements are pairs $(i,x)$ with $x\in S_i$ and we have $(i,x)\le (j,y)$ if $i\le j$ and $s_{i j}(x)=y$, where $s_{i j} \colon S_i \to S_j$ is the transition map. The covers are generated by $(i,x) \lhd s_{i j}^{-1}(x)$ for any $i,j,x$. -Thus, the open sets in $\lim S_i$ are the "ideals" for this coverage, i.e. sets $A$ of pairs $(i,x)$ which are down-closed and such that if $(j,y)\in A$ for some $j$ and all $y\in s_{i j}^{-1}(x)$, then $(i,x)\in A$. -Now consider morphisms $S\to 2$, where $2=\{\bot,\top\}$, regarded as a pro-set in the trivial way, giving rise to a discrete locale. A morphism of pro-sets $S\to 2$ is determined by a partition of some $S_i = S_i^\bot \sqcup S_i^\top$ (modulo a suitable equivalence relation as we change $i$). But a morphism of locales $\lim S_i \to 2$ consists of two ideals $A^\bot$ and $A^\top$ which are disjoint and whose union generates the improper ideal (which consists of all pairs $(i,x)$). A pro-set morphism $S\to 2$ induces a locale map $\lim S_i \to 2$ where $A^\bot$ and $A^\top$ are the ideals generated by $S_i^\bot$ and $S_i^\top$, but in general not every morphism $\lim S_i \to 2$ is induced by one $S\to 2$. -Specifically, consider the following pro-set, which is indexed on the natural numbers with the inverse ordering: -$$ \dots \to S_i \to \dots \to S_2 \to S_1 \to S_0 $$ -We define $S_i = (\mathbb{N} \times \{a,b\}) / \sim_i$, where $\sim_i$ is the equivalence relation generated by $(k,a)\sim_i (k,b)$ for $k\ge i$. The transition maps are the obvious projections, which are surjective. Define -$$ A^\bot = \{ (i,(k,a)) | k < i \} \quad\text{and}\quad A^\top = \{ (i,(k,b) | k < i \}.$$ -Then $A^\bot \cup A^\top$ generates the improper ideal, since for any $i$ we have $\{ (i+1, (i,a)), (i+1,(i,b)) \} \subset A^\bot \cup A^\top$, which covers $(i,(i,?))$, which covers $(i-1,(i,?))$, and so on down to $(0,(i,?))$. However, no $S_i$ can be partitioned as $S_i = S_i^\bot \sqcup S_i^\top$ in such a way that $S_i^\bot$ generates $A^\bot$ and $S_i^\top$ generates $A^\top$. Thus, this defines a locale map $\lim S_i \to 2$ which does not arise from a pro-set morphism $S\to 2$.<|endoftext|> -TITLE: What group is $\langle a,b \,| \, a^2=b^2 \rangle$? -QUESTION [28 upvotes]: In teaching my algebraic topology class, this group showed up as part of an easy fundamental group computation: $\langle a,b\mid a^2=b^2\rangle$. My first instinct was that this must be $\mathbb{Z}*\mathbb{Z}/2$ because clearly every element can be written as a product of $b$'s (only to the power 1) and powers of $a$. But this turns out to be far from clear (and likely wrong). I assume this must be a well-known group to group theorists, so I'm curious if it's isomorphic to something that can be described by other means (or what's known about it in general). -Thanks! - -REPLY [21 votes]: $a^2 = b^2$ says to me "two Mobius bands glued along their boundaries" which then says "the fundamental group of the Klein bottle." You can hear similar sounds from the presentations $\langle a, b \mid a^p = b^q \rangle$ which give rise to fundamental groups of torus knot complements. (Hmm. As long as $p, q$ are relatively prime.)<|endoftext|> -TITLE: What is the simplest, most elementary proof that a particular number is transcendental? -QUESTION [38 upvotes]: I teach, among many other things, a class of wonderful and inquisitive 7th graders. We've recently been studying and discussing various number systems (N, Z, Q, R, C, algebraic numbers, and even quaternions and surreals). One thing that's been hanging in the air is giving a proof that there really do exist transcendental numbers (and in particular, real ones). They're willing to take my word for it, but I'd really like to show them if I can. -I've brainstormed two possible approaches: -1) Use diagonalization on a list of algebraic numbers enumerated by their heights (in the usual way) to construct a transcendental number. This seems doable to me, and would let me share some cool facts about cardinality along the way. The asterisk by it is that, while the argument is constructive, we don't start with a number in hand and then prove that it's transcendental--a feature that I think would be nice. -2) More or less use Liouville's original proof, put as simply as I can manage. The upshots of this route are that we start with a number in hand, it's a nice bit of history, and there are some cool fraction things that we could talk about (we've been discussing repeating decimals and continued fractions). The downside is that I'm not sure if I can actually make it accessible to my students. -So here is where you come in. Is there a simple, elementary proof that some particular number is transcendental? Two kinds of responses that would be helpful would be: -a) to point out some different kind of argument that has a chance of being elementary enough, and -b) to suggest how to recouch or bring to its essence a Liouville-like argument. My model for this is the proof Conway popularized of the fact that $\sqrt{2}$ is irrational. You can find it as proof 8''' on this page. -I realize that transcendence is deep waters, and I certainly don't expect something easy to arise, but I thought I'd tap this community's expertise and ingenuity. Thanks for thinking on it. - -REPLY [2 votes]: Full details require a short paper rather than a long MO comment and Daniel Briggs and I have started discussing writing one jointly. But let me address some of what you ask right now. -A finite collection of polynomials has only a finite number of roots on the real line. So you can find an interval in the real that avoids all of them. Fixing the first digit of Liouville's number together with a minimum gap till the next digit does just that. Note that $1/10$ itself can't be the root of an irreducible polynomial of degree higher than $1$. Now you increase the degree and the height (maximum magnitude of the coefficients) you'll allow your polynomials and this gives you new roots to avoid, but you can still find a small "safe" interval inside the interval you had before. This is the essence of diagonalization. Each further specification of the number kills more polynomials. The trouble is that the set of roots of polynomials of bounded degree and height looks very complicated, so Liouville's trick takes extreme measures to avoid them all but without much computation (where Cantor would determined merely one more digit to avoid the fully calculated root of just one more polynomial). -Now of course you're right - changing a transcendental number by a rational leaves it transcendental. That just means that each polynomial gets killed many times (who says you can't beat a dead horse in mathematics).<|endoftext|> -TITLE: Simple(st) example of an infinite $p$-group with trivial center -QUESTION [10 upvotes]: The only examples I have encountered of infinite $p$-groups with trivial center employ non-elementary methods in their construction. For instance, Example 9.2.5 of Scott's Group Theory is a perfectly satisfactory example, but it requires the wreath product (which, though an invaluable group-theoretic tool, is not what I consider an "elementary method"). -Does anyone know of an example (of an infinite $p$-group with trivial center) that can be constructed and proven to have the claimed properties in a way that is friendly to, say, students of a first course in group theory? Perhaps a large product of finite groups or an easy-to-describe matrix group? -(I also welcome arguments for the nonexistence of such an example!) - -REPLY [18 votes]: Here is a matrix example, hopefully correct and also hopefully sufficiently simple: -Consider the group $U$ of $\infty\times \infty$ upper unipotent matrices with entries in -$\mathbb F_p$, with all but finitely many entries equal to zero; -so they have the form -$$\begin{pmatrix} 1 & a_{1 2} & a_{1 3} & \cdots \\\ 0 & 1 & a_{2 3} & \cdots \\\ \vdots & \vdots & \vdots \end{pmatrix},$$ -with $a_{i,j} \in \mathbb F_p$ and all but finitely many $a_{i j } = 0$. -Note that $U$ admits a homomorphism onto the upper unipotent matrices $U_n$ in $GL_n(\mathbb F_p)$ -for any $n$, given by forgetting the $a_{i,j}$ for $i$ or $j$ greater than $n$. -If $z$ is in the centre of $U$, then its image lies in the centre of $U_n$ for each $n$. -But the centre of $U_n$ has trivial image in $U_{n-1}$, and so the centre of $U$ actually -has trivial image in each $U_n$. Since $U$ evidently embeds into the projective limit of -the $U_n$, we see that $U$ has trivial centre. (Intuitively, we've made the nilpotency class infinite, and so pushed the centre away to infinity.)<|endoftext|> -TITLE: Does the fact that this vector space is not isomorphic to its double-dual require choice? -QUESTION [44 upvotes]: Let $V$ denote the vector space of sequences of real numbers that are eventually 0, and let $W$ denote the vector space of sequences of real numbers. Given $w \in W$ and $v \in V$, we can take their "dot product" $w \cdot v$, and for any give $w \in W$, this defines a linear functional $V \to \mathbb{R}$. In fact, under this association, we see that $W \cong V^{\ast}$. Assuming choice, $W$ has a basis and has uncountable dimension, whereas $V$ has countable dimension. So $\mathrm{dim} (V) < \mathrm{dim} (V^{\ast}) \leq \dim(V^{\ast \ast})$ so $V$ is not isomorphic to its double-dual. In particular, the canonical map $\widehat{\cdot} : V \to V^{\ast \ast}$ defined by $\widehat{x} (f) = f(x)$ is not an isomorphism. -Question 1: Can you explicitly write down an element of $V^{\ast \ast}$ that isn't of the form $\widehat{x}$? -Question 2: What's the situation if we don't assume choice? (i.e. might the canonical map be an isomorphism, might $W$ not even have a basis, might $V$ be isomorphic to its double-dual but not via the canonical map, etc?) - -In light of Daniel's and Yemon's comments, and Konrad's related question here, I'd like to reorganize my question(s). So consider the following statements: - -The canonical map $\widehat{\cdot} : V \to V^{\ast \ast}$ is injective but not surjective. -The canonical map is not an isomorphism. -There is no isomorphism from $V$ to its double-dual. -The double-dual is non-trivial. -$V^{\ast}$ has a basis. - -This is the question I'm really interested in, and which clarifies and makes more precise my original Question 1: -Revised Question 1: Are there models without choice where (1) holds? If so, can we find some $x \in V^{\ast \ast}$ witnessing non-surjectivity to describe a witness to non-surjectivity in choice models? Are there models of choice where (1) fails? -The following question is a more precise version of Question 2. -Revised Question 2: Under AC, all 5 statements above are true. There are $2^5 = 32$ ways to assign true-false values to the 5 sentences above that might hold in some model of ZF where choice fails. Not all 32 are legitimate possibilites, some of them are incompatible with ZF since, for instance, (1) implies (2), (5) implies (4), and the negation of (4) implies (1) through (3), etc. Which of the legitimate possibilites actually obtains in some model of ZF where choice fails? -This second question is rather broad, and breaks down into something on the order of 32 cases, so seeing an answer to any one of the legitimate cases would be cool. - -REPLY [6 votes]: I have recently noticed$^1$ that in models of ZF+DC+"All sets of real numbers have the Baire property" (e.g. Solovay's model or Shelah's model mentioned by Andreas) there is a very interesting property for Banach spaces: - -If $V$ is a Banach space, $W$ is a normed space and $T\colon V\to W$ is linear then $T$ is continuous. - -In particular this means that if $V$ is a Banach space over $\mathbb R$ then every functional is continuous. This means that the algebraic dual of a Banach space is the same as the continuous dual of the space. -For example, $\ell_p$ for $p\in(1,\infty)$ have this property, using DC we can develop the basic tools of functional analysis just fine (except Hahn-Banach, though). This implies that $\ell_p$ is reflexive in the algebraic sense, not only in the topological sense. -We further remark$^2$ that the assertion $\ell_1\subsetneq\ell_\infty^\prime$ (where the $\prime$ denotes all the continuous functionals, but in our model these are all the functionals) implies the negation of "All sets of real numbers have the Baire property". The result is that $\ell_1$ is also reflexive in the algebraic sense and in the topological sense. -So with this in mind we have $V_p=\ell_p\oplus\ell_q$ (where $\frac1p+\frac1q=1$) is a self-dual as well algebraically reflexive space, and for distinct $p_1,p_2\in[1,2]$ we have $V_{p_1}\ncong V_{p_2}$ as well, so there are many non-isomorphic examples for this. - -Notes: - -After sitting to write all the details of the above I found out it is mentioned in Eric Schechter's book, Handbook of Analysis and its Foundations in the last section of chapter 27. -The above reference does not discuss the case of $p=1$, which is discussed later in the end of chapter 29.<|endoftext|> -TITLE: $G_2$ and Geometry -QUESTION [34 upvotes]: In a recent question Deane Yang mentioned the beautiful Riemannian geometry that comes up when looking at $G_2$. I am wondering if people could expand on the geometry related to the exceptional Lie Groups. I am not precisely sure what I am looking for, but ostensibly there should be answers forth coming from other who have promised such answers. I understand a bit about how the exceptional Lie groups come up historically, and please correct the following if it is incorrect, but when looking at the possible dynkin diagrams you see that there is no reason for $E_6$,$E_7$,$E_8$,$G_2$, and $F_4$ to not occur as root systems. While root systems are geometric, this is not what I am asking about. -Thanks - -REPLY [2 votes]: A review on the history of $G_2$ and its relation with the 7-dimensional geometry, is given in the following article of AMS Notices: -link text<|endoftext|> -TITLE: Is there a good notion of "induction" for representations of 2-categories? -QUESTION [15 upvotes]: One of the most important observations in the representation theory of algebras is that if one has a subalgebra $A\subset B$, then these is a functor $B\otimes_A -\colon A\operatorname{-pmod}\to B\operatorname{-pmod}$ called induction. -This is a special case of a notion for representations of a category. (The representations of a category are the functors from that category into vector spaces.) In this case, one replaces the category with its idempotent completion (which doesn't change representations, since vector spaces are idempotent complete), which we'll assume for now is compactly generated. In this case, the induction of a representation $V$ is $$\operatorname{Ind} V(Y)=V(X)\otimes_{\operatorname{End}_A(X)}\operatorname{Hom}_B(X,Y).$$ - -Now, I can just as well talk about representations of 2-categories, which are 2-functors to the 2-category of categories (with whatever additions you like; additive and enriched over a field would be good choices). - -Has anyone written up the basics of induction in this situation? - -I don't strictly need to state things this way for what I want to do, but it would be better to give credit if someone else has done it and to have things "work out of the box." - -REPLY [6 votes]: What's going on here formally in the category case is that the inclusion $A \subset B$ extends to a map $f: A \to [B^{\operatorname{op}}, \mathcal{V}]$ (where $\mathcal{V}$ is the category you're enriching over) via the Yoneda embedding. The Yoneda embedding is the free cocompletion of a category (i.e., the completion under weighted colimits), so this map extends to a cocontinuous functor $\widehat{f}: [A^{\operatorname{op}}, \mathcal{V}] \to [B^{\operatorname{op}}, \mathcal{V}]$, given by the weighted colimit formula you indicated. This functor is right adjoint to $\operatorname{Hom}(f-, -): [B^{\operatorname{op}}, \mathcal{V}] \to [A^{\operatorname{op}}, \mathcal{V}]$, which is just the restriction functor in this case. (What I've said works more generally for any such $f$, not just one induced by an inclusion of categories.) -Everything I have said should extend more or less identically to the setting of bicategories. A pseudofunctor between bicategories gives you an induction pseudofunctor between the corresponding presheaf categories by a weighted bicolimit formula analogous to the weighted colimit formula in the 1-categorical setting. This should be right biadjoint to the obvious restriction pseudofunctor, although I haven't checked the details (and I don't know if this is written up anywhere). -The situation you are interested in seems a bit more delicate, because you want to work in the setting where your hom-categories are enriched. To formulate all of this properly, one would need a notion of "weak enrichment," which generalizes the theory of bicategories. I've heard that there are people working on this, but I don't know of any available sources yet. -Note that while in principle every "weakly $\mathcal{V}$-$\operatorname{Cat}$-enriched category" may be equivalent to a category enriched in the underlying 1-category $\mathcal{V}$-$\operatorname{Cat}$ (and indeed, you may only be interested in the latter kinds of categories), the $\mathcal{V}$-$\operatorname{Cat}$-enriched setting probably isn't flexible to properly handle the kind of "representations" you would like to handle. -Edit: I really just wanted to fix the spelling of "principle" above, but as long as I'm bumping a two-year-old question, I might as well link to a recent paper by Garner and Shulman that develops the theory of bicategories enriched in a monoidal bicategory. In particular, they show that the bicategory of (certain) modules over an enriched bicategory is the free cocompletion under (certain) weighted colimits, which allows you to describe induction as a weighted colimit as in the 1-categorical setting.<|endoftext|> -TITLE: Does every symmetric group S_n have a single element of maximal word norm? -QUESTION [7 upvotes]: Generate $S_n$ by transpositions $s_i$ of (i) and (i+1). Both $S_3$ and $S_4$ have single elements of maximal word norm associated with this presentation. In fact, the Cayley graph of $S_3$ can be seen as a tiling of $S^1$, and the Cayley graph of $S_4$ a tiling of $S^2$. The element of maximal length is then antipodal to e. -Does every symmetric group $S_n$ have a single element of maximal word norm? If so, is there a formula for its length l(n)? - -REPLY [33 votes]: It is amazing how a fact that I was taught in a middle school can be proved using big theories where I don't understand half of the words. Let me add a straightforward proof (for $S_n$ and only $S_n$). -For a permutation $\sigma:\{1,\dots,n\}\to\{1,\dots,n\}$, let $\lambda(\sigma)$ denote the number of inversions in $\sigma$, that is the number of pairs $(i,j)$ such that $i\sigma(j)$. Then $\lambda(\sigma)$ equals the length of $\sigma$ with respect to the generating set $\{s_i\}$. -Indeed, left-multiplying $\sigma$ by $s_i$ only interchanges $\sigma(i)$ and $\sigma(i+1)$, and hence changes $\lambda(\sigma)$ by at most 1. Therefore the length is bounded below by $\lambda$. On the other hand, if $\sigma$ is not the identity, there exists $i$ such that $\sigma(i+1)<\sigma(i)$, then left-multiplying by $s_i$ decreases $\lambda(\sigma)$ by 1. Repeating this procedure, one reaches the identity from $\sigma$ by exactly $\lambda(\sigma)$ multiplications by generators. -Now it is clear that the maximum length equals $n(n-1)/2$ and is attained only at the order reversing permutation (the one given by $\sigma(i)=n+1-i$ for all $i$).<|endoftext|> -TITLE: who can tell me how to understand and compute Chern roots -QUESTION [6 upvotes]: As i learn the local index theory , Chern roots appears and i cannot understand what it is and i can't find any references about it .Can anyone tell me something about it and give me some references,also tell me how to compute it ? - -REPLY [4 votes]: Chern root algorithm and local index theorems: -http://books.google.com/books?id=sqwa47R6Ds4C&pg=PA324&lpg=PA324<|endoftext|> -TITLE: Is the non-triviality of the algebraic dual of an infinite-dimensional vector space equivalent to the axiom of choice? -QUESTION [56 upvotes]: If $V$ is given to be a vector space that is not finite-dimensional, it doesn't seem to be possible to exhibit an explicit non-zero linear functional on $V$ without further information about $V$. The existence of a non-zero linear functional can be shown by taking a basis of $V$ and specifying the values of the functional on the basis. -To find a basis of $V$, the axiom of choice (AC) is needed, and indeed, it was shown by Blass in 1984 that in Zermelo-Fraenkel set theory (ZF) it is equivalent to the axiom of choice that any vector space has a basis. However, it's not clear to me that the existence of a non-zero element of $V^*$ really needs the full strength of AC. I couldn't find a reference anywhere, so here is my question: - -Consider the following statement: -(D) For any vector space $V$ that is not finite-dimensional, $V^*\neq \{0\}$. -Is (D) equivalent to AC in ZF? If not, is there some known axiom that is equivalent to (D) in ZF? - -Note that this question is about the algebraic dual $V^*$. There are examples of Banach spaces, for example $\ell^\infty/c_0$, where it is possible (in the absence of the Hahn-Banach theorem, itself weaker than AC) for their topological dual to be $\{0\}$; see this answer on MO. I'm not aware of any result for the algebraic dual. -This question was inspired by, and is related to this question on MO. - -Edit: Summary of the five answers so far: - -Todd's answer + comments by François and Asaf: -in Läuchli's models of ZF there is an infinite dimensional vector space $V$ such that all proper subspaces are finite dimensional. In particular, $V$ does not have a basis and $V^*=\{0\}$. Also, according to Asaf, in these models Dependent Choice can still hold up to an arbitrarily large cardinal. -Amit's answer + comment by François: -in Shelah's model of ZF + DC + PB (every set of real numbers is Baire), $\Bbb R$ considered as a vector space over $\Bbb Q$ has a trivial dual. -François's answer (see also godelian's answer) + Andreas' answer -in ZF the following is equivalent to BPIT: all vector spaces over finite fields have duals large enough to separate points. - -So DC is too weak, and BPT is strong enough for finite fields (and in fact equivalent to a slightly stronger statement). How far does Choice fail in Blass' model? Update: according to Asaf Karagila, $DC_{\kappa}$ can hold for arbitrarily large $\kappa$. - -REPLY [18 votes]: To add the proof for my claim in Todd's answer, which essentially repeats Läuchli's original [1] arguments with minor modifications (and the addition that the resulted model satisfies $DC_\kappa$). -We will show that it is consistent to have a model in which $DC_\kappa$ holds, and there is a vector space over $\mathbb F_2$ which has no linear functionals. - -Assume that $M$ is a model of $ZFA+AC$ and that $A$, the set of atoms has $\lambda>\kappa$ many atoms, where $\lambda$ is a regular cardinal. Endow $A$ with a structure of a vector space over $\mathbb F=\mathbb F_2$. Now consider the permutation model $\frak M$ defined by the group of linear permutations of $A$, and by ideal of supports generated by subsets of dimension $\le\kappa$. -Denote by $\operatorname{fix}(X)$ the permutations which fix every element of $X$, by $\operatorname{sym}(X)$ the permutations that fix $X$ as a set, and by $[E]$ the span of $E$ as a subset of $A$. We say that $E\subseteq A$ is a support of $X$ if $\pi\in\operatorname{fix}(E)\Rightarrow\pi\in\operatorname{sym}(X)$. -Final word of terminology, since $A$ will play both the role of set of atoms as well the vector space, given $U\subseteq A$ the complement will always denote a set complement, whereas the direct complement will be used to refer to a linear subspace which acts as a direct summand with $U$ in a decomposition of $A$. -Claim 1: If $E$ is a subset of $A$ then $\operatorname{fix}(E)$ is the same as $\operatorname{fix}([E])$. -Proof: This is obvious since all the permutations considered are linear. $\square$ -From this we can identify $E$ with its span, and since (in $M$) the $[E]$ has the same cardinality of $E$ we can conclude that without loss of generality supports are subspaces. -Claim 2: $\frak M$$\models DC_\kappa$. -Proof: Let $X$ be some nonempty set, and $\lt$ a binary relation on $X$, both in $\frak M$. In $M$ we can find a function $f\colon\kappa\to X$ which witness $DC_\kappa$ in $V$. -Since $\frak M$ is transitive, we have that $\alpha,f(\alpha)\in\frak M$ and thus $\langle\alpha,f(\alpha)\rangle\in\frak M$. Let $E_\alpha$ be a support for $\lbrace\langle\alpha,f(\alpha)\rangle\rbrace$ then $\bigcup_{\alpha<\kappa} E_\alpha$ is a set of cardinality $<\kappa^+$ and thus in our ideal of suports. It is simple to verify that this is a support of $f$, therefore $f\in\frak M$ as wanted. $\square$ -Claim 3: If $x,y\in A$ are nonzero (with respect to the vector space) then in $M$ there is a linear permutation $\pi$ such that $\pi x=y$ and $\pi y=x$. -Proof: Since $x\neq y$ we have that they are linearly independent over $\mathbb F$. Since we have choice in $M$ we can extend this to a basis of $A$, and take a permutation of this basis which only switches $x$ and $y$. This permutation extends uniquely to our $\pi$. -Claim 4: If $U\subseteq A$ and $U\in\frak M$ then either $U$ is a subset of a linear subspace of dimension at most $\kappa$, or a subset of the complement of such space. -Proof: Let $E$ be a support of $U$, then every linear automorphism of $A$ which fixes $E$ preserves $U$. If $U\subseteq [E]$ then we are done, otherwise let $u\in U\setminus [E]$ and $v\in A\setminus [E]$, we can define (in $M$ where choice exists) a linear permutation $\pi$ which fixes $E$ and switches $u$ with $v$. By that we have that $\pi(U)=U$ therefore $v\in U$, and so $U=A\setminus[E]$ as wanted. $\square$ -Claim 5: If $U\subseteq A$ is a linear proper subspace and $U\in\frak M$ then its dimension is at most $\kappa$. -Proof: Suppose that $U$ is a subspace of $A$ and every linearly independent subset of $U$ of cardinality $\le\kappa$ does not span $U$, we will show $A=U$. By the previous claim we have that $U$ is the complement of some "small" $[E]$. -Now let $v\in A$ and $u\in U$ both nonzero vectors. If $u+v\in U$ then $v\in U$. If $u+v\in [E]$ then $v\in U$ since otherwise $u=u+v+v\in[E]$. Therefore $v\in U$ and so $A\subseteq U$, and thus $A=U$ as wanted.$\square$ -Claim 6: If $\varphi\colon A\to\mathbb F$ a linear functional then $\varphi = 0$. -Proof: Suppose not, for some $u\in A$ we have $\varphi(u)=1$, then $\varphi$ has a kernel which is of co-dimension $1$, that is a proper linear subspace and $A=\ker\varphi\oplus\lbrace 0,u\rbrace$. However by the previous claim we have that $\ker\varphi$ has dimension $\kappa$ at most, and without the axiom of choice $\kappa+1=\kappa$, thus deriving contradiction to the fact that $A$ is not spanned by $\kappa$ many vectors. - -Aftermath: There was indeed some trouble in my original proof, after some extensive work in the past two days I came to a very similar idea. However with the very generous help of Theo Buehler which helped me find the original paper and translate parts, I studied Läuchli's original proof and concluded his arguments are sleek and nicer than mine. -While this cannot be transferred to $ZF$ using the Jech-Sochor embedding theorem (since $DC_\kappa$ is not a bounded statement), I am not sure that Pincus' transfer theorem won't work, or how difficult a straightforward forcing argument would be. -Lastly, the original Läuchli model is where $\lambda=\aleph_0$ and he goes on to prove that there are no non-scalar endomorphisms. In the case where we use $\mathbb F=\mathbb F_2$ and $\lambda=\aleph_0$ we have that this vector space is indeed amorphous which in turn implies that very little choice is in such universe. -Bibliography: - -Läuchli, H. Auswahlaxiom in der Algebra. Commentarii Mathematici Helvetici, vol 37, pp. 1-19.<|endoftext|> -TITLE: Classification of real forms up to inner automorphisms -QUESTION [6 upvotes]: I hope to know the classification of real forms of complex simple Lie algebras of types $A$, $D$, $E$ up to inner automorphisms. -Let $\mathfrak{g}_1$ and $\mathfrak{g}_2$ be real forms of a complex simple Lie algebra $\mathfrak{g}$. -We say that they are equivalent if there is an isomorphism $\mathfrak{g}_1 \to \mathfrak{g}_2$ -which extends to an inner automorphism of $\mathfrak{g}$. -However, there may be real forms which are isomorphic but not equivalent. - -Where can we find the classification of such equivalence classes of real forms? - -REPLY [3 votes]: The one case (for $\frak g$ simple) where the two definitions of real forms (up to $Aut(\frak g)$ versus $Int(\frak g)$) don't agree is the following. -There is a real form of $\frak g=\frak s\mathfrak o(2n,\mathbb C)$ denoted -$\frak s\frak o^*(2n)$. -For $n\ge4$ even $\frak g$ has two subalgebras -isomorphic to $\frak s\frak o^*(2n)$ -which are related by $Aut(\frak g)$, but not by $Int(\frak g)$. -If $n\ge 3$ is odd there is only one such algebra up to $Int(\frak g)$. -The difference is because the center of the simply connected group is $\mathbb Z/2\times\mathbb Z/2$ if $n$ is even, and $\mathbb Z/4$ if $n$ is odd. -$n=4$ is particularly interesting: the triality automorphism of $D_4$ interchanges these -two copies of $\mathfrak s\mathfrak o*(2n)$, as well as $\frak s\frak o(6,2)\simeq \frak s\frak o^*(2n)$.<|endoftext|> -TITLE: Behaviour of power series on their circle of convergence -QUESTION [66 upvotes]: I asked myself the following question while preparing a course on power series for 2nd year students. Let $F$ be the set of power series with convergence radius equal to $1$. What subsets $S$ of the unit circle $C$ can be realised as -$$ -S:=\{x \in C: f\text{ diverges in }x\} -$$ -for $f \in F$? Any finite subset (and possibly any countable subset) of $C$ can be realised that way. Who knows more on this? - -REPLY [67 votes]: [Edit (Jan.12/12): I've recently been made aware of additional results. I have added an update at the end.] -Hi Piotr, -As far as I know, the question is still open, and not much seems known beyond the classic results of George Piranian and Fritz Herzog. They are contained in two joint papers, available at the publications page of Piranian's website (although the quality of the scans is not optimal), - -"Sets of convergence of Taylor series. I." Duke Math. J. 16, (1949), 529-534, and -"Sets of convergence of Taylor series. II." Duke Math. J. 20, (1953), 41-54. - -Suppose $f(z)=\sum_na_nz^n$, and let $f_n(z)=\sum_{k\le n}a_kz^k$. The Cauchy convergence criterion says that $f(z)$ converges iff $$ \forall n\exists N\forall m,k\ge N |f_m(z)-f_k(z)|\le 1/n. $$ -This shows clearly that the set of $z$ on the unit circle where $f$ converges is a countable intersection (indexed by $n$) of a countable union (indexed by $N$) of closed sets (namely, the intersection over $m,k\ge N$ of the sets $\{z\in C\mid|f_m(z)-f_k(z)|\le 1/n\}$). -The conjecture is that any subset of $C$ that is a countable intersection of a countable union of closed sets is the set of convergence of some $f$. [Edit (Jan.12/12): This is not the case. See the update at the end.] These things are called $F_{\sigma\delta}$, though I've never quite liked the name. -I have been interested in the problem (on and off) for a bit, so I have asked around what is known. Both Donald Sarason and Steve Krantz agreed that the state of the art seems to be whatever you find in the two papers listed above (Herzog and Piranian have a few papers on closely related topics, some may also be relevant). Alekos Kechris also pointed out that Ted Kaczynski, who was Piranian's student, had worked on related problems. There is a family of questions in the neighborhood of this one, that descriptive set theorists find interesting, so Alekos and other descriptive set theorists had followed some of this work for a little while. -In his thesis, Kaczynski proved that if we look at analytic functions $f$ the upper half plane, then the $F_{\sigma\delta}$ subsets of the real axis are precisely the sets $A$ for which there is such an $f$ so that $A$ is the set of points $p$ on the real axis such that there is an arc $\gamma$ ending at $p$ such that the limit of $f(z)$ as $z$ approaches $p$ along $\gamma$ exists. In his papers he finds several extensions of this result. It seems to me that his techniques are highly relevant to the problem you are asking. -I emailed Piranian a few years ago, but sadly he never replied to my email, it would have been nice to meet him. -The results in the two papers above are as follows: In the first one, Herzog and Piranian trace the history of the problem (for example, Mazurkiewicz had shown that any closed subset of $C$ is the set of convergence of some $f$). They prove that every $F_\sigma$ set is a set of convergence. Their technique is nice (it has a recursion theoretic flavor to it, at least for me, ensuring convergence in some regions and divergence in some others; it is akin to a priority argument), and it looks to me like the "right" kind of approach to this problem. A very simple version of their argument can be found in Rudin's analysis book, where it is shown that $$ \sum_n \frac{z^n}n $$ converges for all $z\in C$ other than $z=1$. The general argument requires some easy trigonometric estimates (that they establish in sections 2,3). -In the second paper, they build some examples of $F_{\sigma\delta}$ sets that are not $F_\sigma$ and yet are sets of convergence. These constructions are more ad hoc, at least to my untrained eyes, they are certainly much more involved than the results in the first paper. There are two proofs given there: The first elaborates the techniques of the paper by Erdös-Herzog-Piranian mentioned by Gideon in a comment to the question, and so the functions that are obtained are injective in the unit disk. Granting injectivity is one of the main reasons for the difficulty of the approach. The second proof elaborates the approach of the first paper (so injectivity is not ensured here), and uses ideas of Fejér (as mentioned by Theo in a comment below). The first argument allows one to deal with questions of uniform convergence; the second seems more general but does not grant this control. As the authors point out (and thanks to Theo for mentioning this in his comment) Fejér's techniques suffice to show that there are non-$F_\sigma$ sets of convergence. Unfortunately, I am not convinced these techniques alone suffice to establish the full characterization. -Hope this helps! I am very curious to see if there are additional results I've missed. - -Update (Jan.12/12): Essentially the same question was asked on the sister site. Dave Renfro mentioned a reference I wasn't aware of: Thomas W. Körner, "The behavior of power series on their circle of convergence", in Banach Spaces, Harmonic Analysis, and Probability Theory, Springer Lecture Notes in Mathematics #995, Springer-Verlag, 1983, 56-94. -This is a nice survey, its only problem is that it lists no references. Körner proves a result there that goes beyond what I discuss above and shows that the question of what sets are sets of convergence is not purely topological. - -Theorem (Körner). There is a $G_\delta$ subset of $C$ that is not a set of convergence. - -(This is theorem 6.2 in the paper.) The theorem builds on a result of Marcinkiewicz, who showed that if $a_n\to0$ as $|n|\to\infty$, then $\sum_{-N}^N a_n \exp(int)$ cannot diverge boundedly everywhere as $N\to\infty$. This can be refined to show that, under the same assumptions, if $\sum_{-N}^N a_n \exp(int)$ is bounded for all $t$ in some interval $I$, then there is a subinterval $J\subseteq I$ where $\lim_{N\to\infty}\sum_{-N}^N a_n \exp(int)$ converges almost everywhere. -The construction makes use of (Cantor-like) sets of positive measure, and in particular leaves the following open: - -Is every $G_{\delta\sigma}$ set of Lebesgue measure zero the complement of a set of convergence? - -(Recall that Katznelson and Kahane have shown that every set of Lebesgue measure zero has a Fourier series diverging on it, and possibly elsewhere, see also this question.) -I contacted Körner, who indicated he didn't know of additional results. In particular, he mentioned the question above. He also suggested looking into some recent results of Matheron (in particular, the Matheron-Zelený 2005 paper in Fundamenta), which may suggest descriptive set theoretic approaches to the characterization problem.<|endoftext|> -TITLE: question in prime numbers -QUESTION [8 upvotes]: Is it true that in any successive (natural) $2p_n$ numbers there are at least three numbers that are not divisible by any prime less (not equal) than $p_n$? Here, $p_n$ denotes the $n$-th prime number. -For -example in any six successive numbers there are at least 3 numbers that are not divisible by 2,in any 10 successive numbers there are 3 numbers that are not divisible by 2 or 3, in any 14 successive numbers there are at least 3 that are not divisible by 2, 3, or 5. - -REPLY [11 votes]: Late update A better question might be; "What is the Largest number of consecutive integers such that each is divisible by a prime $\le p_n$ ? Computing a few values and checking the handy OEIS yields the link above. Matching the values with the appropriate primes shows that one can get over twice (at p=67) or even thrice the next prime with no gaps. -[2, 1], [3, 3], [5, 5], [7, 9], [11, 13], [13, 21], [17, 25], [19, 33], [23, 39], [29, 45], [31, 57], [37, 65], [41, 73], [43, 89], [47, 99], [53, 105], [59, 117], [61, 131], [67, 151], [71, 173], [73, 189], [79, 199], [83, 215], [89, 233], [97, 257], [101, 263], [103, 281], [107, 299], [109, 311], [113, 329], [127, 353], [131, 377], [137, 387], [139, 413], [149, 431], [151, 449], [157, 475], [163, 491], [167, 509], [173, 537], [179, 549], [181, 573], [191, 599], [193, 615], [197, 641], [199, 657], [211, 685], [223, 717], [227, 741] - -older answer -The first counter-example is for $p=17$. The interval of length $40$ starting at $87890$ only yields three integers with least prime factor greater than $17$: the primes $87911,87917$ and also $87929=23\cdot 3823$. So we can use the inteval of length $38$ starting at $87890$ or at $87891$. -I may be mistaken, but I think that there are not any examples for $p=19$ and $p=23$. -On the other hand, of the 89 consecutive integers from $43559563512434$ to $43559563512522$ inclusive, all but one have a prime factor $41$ or less. The exception is $43559563512481=9393910613\cdot 4637$ -There are essentially 6 ways to get a run of length $39$ (or 3 up to reflection): One can start at $87890,177980,182342,328130,332492$ or $422582 \mod 510510$. -Discussion (Since somebody asked) For a prime $p$ call an integer $p$-good if it has a prime factor $p$ or less and $p$-bad otherwise. Let $M(p,b)$ be the longest possible run of consecutive integers which are all $p$-good except for $b$ exceptions (called holes). Note that $M(p,b)$ is odd. To find the values of $M(p,b)$ and all intervals attaining it, it suffices to look for runs in the integers from $0$ to $p\sharp-1$ where $p\sharp$ is the product of the primes up to $p$. That is true if we look cyclically or go a reasonable distance past $p\sharp$. I wondered at the particular question asked by the OP (" Is it the case that $M(p_{n-1},2)<2p_n$ for all $n$?" But now I realize that must be inspired by the fact that the $M(p_{n-1},2)\ge 2p_n-1$ is witnessed by the run from $p_{n-1}\sharp-p_n+1$ to $p_{n-1}\sharp+p_n-1$ (with the two bad numbers being $p_{n-1}\sharp \pm 1$). -For my computations I generated a list of the $p-bad$ numbers up to $p\sharp$. This can be done iteratively. I did this up to $p=23$ using a fairly uninspired Maple program on a unimpressive computer. $p=19$ took 9 seconds and $p=23$ took 250. The lists are centrally symmetric but I did not exploit that. A faster language, better program, faster machine, and storage in a file might allow one to go somewhat further. I then made the list of jumps from one $p$-bad number to the next. The jumps for $p=17$ with their multiplicities were -$[[2, 22274], [4, 22275], [6, 26630], [8, 6812], [10, 7734], [12, 4096]$ -$[14, 1406], [16, 432], [18, 376], [20, 24], [22, 78], [24, 20], [26, 2]]$. -I then had Maple look for 3 successive gaps adding to at least $40$. There were 3 up to the middle. -I did not care to try $p=29$ by the same method! I looked for $M(23,3)$ using the same list of gaps in hopes of finding something good enough that I could get a good bound for $M(29,2)$ use the Chinese remainder theorem to shift everything preserving the 23-good members and plug a hole by making a 23-bad member 29-good. I didn't find anything by eyeballing. Similarly for $M(23,4)$ using $p=29,31$ and $M(23,5)$ using $p=29,31,37$. I did not think to look if there was something attaining $M(23,5)$ so that two of the holes were separated by 58 so that they could be plugged using up just one prime. But something with 2 of 6 holes separated by 62 did get noticed. -Note: I might have overlooked things. I did not really look at the two largish gaps around $p\sharp$. Also, The two holes with one prime trick might make a good example out of a run which is 2 or 4 less than $M(23,b)$.<|endoftext|> -TITLE: First correct proof of FLT for exponent 3? -QUESTION [40 upvotes]: It is well known that Euler gave the first proof of FLT ($x^n + y^n = z^n$ has -no nontrivial integral solutions for $n > 2$) for exponent $n=3$, but that his proof -had gaps (which are not as easily closed as Weil seems to suggest in his excellent -Number Theory - An Approach through History). Later proofs by Legendre and Kausler -had the same gap, and in fact I do not know any correct proof published before Kummer's proof for all regular primes. Gauss had a beautiful proof, with the 3-isogeny clearly visible, which was published posthumously by Dedekind, and of course Dirichlet could have given a correct proof (he gave one for $n = 5$ in his very first article but apparently did not dare to provoke Legendre by suggesting his proof in Theorie des Nombres was incomplete) but did not. -The problem in the early proofs is this: if $p^2 + 3q^2 = z^3$, one has to show that -$p$ and $q$ can be read off from $p + q \sqrt{-3} = (a + b\sqrt{-3})^3$. The standard proofs use unique factorization in ${\mathbb Z}[\zeta_3]$ or the equivalent fact that there is one class of binary quadratic forms with discriminant $-3$; Weil uses a (sophisticated, but elementary) counting argument. -I wonder whether there is any correct proof for the cubic Fermat equation before Kummer's -proof for all regular prime exponents (1847-1850)? - -REPLY [6 votes]: I had a look at Paulo Ribenboim's "13 lectures on Fermat's last Theorem" (Springer Verlag, 1979). In section 3 of Chapter III, he discusses (with full bibliographical details) the controversy around Euler's proof, and then provides a proof, using purely elementary number theory, which he attributes to Euler.<|endoftext|> -TITLE: Nonexistence of boundary between convergent and divergent series? -QUESTION [77 upvotes]: The following is a FAQ that I sometimes get asked, and it occurred to me that I do not have an answer that I am completely satisfied with. In Rudin's Principles of Mathematical Analysis, following Theorem 3.29, he writes: - -One might thus be led to conjecture that there is a limiting situation of some sort, a “boundary” with all convergent series on one side, all divergent series on the other side—at least as far as series with monotonic coefficients are concerned. This notion of “boundary” is of course quite vague. The point we wish to make is this: No matter how we make this notion precise, the conjecture is false. Exercises 11(b) and 12(b) may serve as illustrations. - -Exercise 11(b) states that if $\sum_n a_n$ is a divergent series of positive reals, then $\sum_n a_n/s_n$ also diverges, where $s_n = \sum_{i=1}^n a_n$. Exercise 12(b) states that if $\sum_n a_n$ is a convergent series of positive reals, then $\sum_n a_n/\sqrt{r_n}$ converges, where $r_n = \sum_{i\ge n} a_i$. -Although these two exercises are suggestive, they are not enough to convince me of Rudin’s strong claim that no matter how we make this notion precise, the conjecture is false. Are there any stronger theorems in this direction? -Edit. For example, are there any theorems about the topology/geometry of the spaces of all convergent/divergent series, where a series is viewed as a point in $\mathbb{R}^\infty$ or $(\mathbb{R}^+)^\infty$ in the obvious way? - -REPLY [8 votes]: As Dylan Wilson pointed out, the following question appears in Folland's real analysis book: -(second edition, pg. 164) (33) There is no slowest rate of decay of the terms of an absolutely convergent sequence; that is, there is no sequence $\{ a_n \}$ of positive numbers such that $\sum a_n|c_n| < \infty$ iff $\{ c_n \}$ is bounded.<|endoftext|> -TITLE: Nondifferentiability set of an arbitrary real function -QUESTION [10 upvotes]: A theorem by Zygmunt Zahorski states that a necessary and sufficient condition for a subset of $\mathbb{R}$ to be the nondifferentiability set of a continuous real function is that it is the union of a $G_\delta$ set and a $G_{\delta \sigma}$ set of zero measure. -On the other hand it is not hard to see that the nondifferentiability set of an arbitrary real function is always a $G_{\delta \sigma}$ set. -My question is: why can't we leave the word 'continuous' out in Zahorski's theorem? In other words, what would be an example of a (necessarily discontinuous) real function whose nondifferentiability set is not a union of a $G_\delta$ set and a $G_{\delta \sigma}$ set of zero measure? - -REPLY [6 votes]: Apparently, continuity is not essential. -According to A. Brudno, Continuity and differentiability (Russian), Rec. Math [Mat. Sbornik], N.S. 13 (55), (1943), 119–134 (MathSciNet review here, online article here), the set of non-differentiability of any real function is the union of a $G_{\delta}$-set with a $G_{\delta\sigma}$-set of measure zero. -I quote from Brudno's English summary: - -In the present paper we investigate the structure of the set of points, in which the function of one real variable is not differentiable. All the functions are only supposed to be finite in every point. -(...) -Theorem IV. In order that the set $Q$ should be the totality of points, in which a function f(x) does not possess a derivative, it is necessary and sufficient that - $Q = G_{\delta} + G_{\delta\sigma}$ $(\operatorname{mes}G_{\delta\sigma} = 0)$. - -Since I do not read Russian, I cannot tell you anything about the methods of proof apart from the fact that it involves an investigation of the sets where the Dini derivatives are infinite or distinct.<|endoftext|> -TITLE: Is there a category structure one can place on measure spaces so that category-theoretic products exist? -QUESTION [64 upvotes]: The usual category of measure spaces consists of objects $(X, \mathcal{B}_X, \mu_X)$, where $X$ is a space, $\mathcal{B}_X$ is a $\sigma$-algebra on $X$, and $\mu_X$ is a measure on $X$, and measure preserving morphisms $\phi \colon (X, \mathcal{B}_X, \mu_X) \to (Y, \mathcal{B}_Y, \mu_Y)$ such that $\phi_\ast \mu_X(E) = \mu_X(\phi^{-1}(E)) = \mu_Y(E)$ for all $E \in \mathcal{B}_Y$. -The category of measurable spaces consists of objects $(X, \mathcal{B}_X)$ and measurable morphisms $\phi \colon (X, \mathcal{B}_X) \to (Y, \mathcal{B}_Y)$. -Products exist in the category of measurable spaces. They coincide with the standard product $(X \times Y, \mathcal{B}_X \times \mathcal{B}_Y)$, where $X \times Y$ is the Cartesian product of $X$ and $Y$ and $\mathcal{B}_X \times \mathcal{B}_Y$ is the coarsest $\sigma$-algebra on $X\times Y$ such that the canonical projections $\pi_X \colon X \times Y \to X$ and $\pi_Y \colon X \times Y \to Y$ are measurable. Equivalently, $\mathcal{B}_X \times \mathcal{B}_Y$ is the $\sigma$-algebra generated by the sets $E \times F$ where $E \in \mathcal{B}_X$ and $F \in \mathcal{B}_Y$. -However, in the category of measure spaces, products do not exist. The first obstacle is that the canonical projection $\pi_X \colon X \times Y \to X$ may not be measure preserving. A simple example is the product of $(\mathbf{R}, \mathcal{B}[\mathbf{R}], \mu)$ with itself, where $\mathcal{B}[\mathbf{R}]$ is the Borel $\sigma$-algebra on $\mathbf{R}$. In this case, -$(\pi_{\mathbf{R}})_\ast\mu\times \mu([0,1]) = \mu\times \mu(\pi_\mathbf{R}^{-1}([0,1])) = -\mu \times \mu([0,1]\times \mathbf{R}) = \infty \neq 1 = \mu([0,1])$. -In addition, there may be multiple measures on $X\times Y$ whose pushforwards on $X$ and $Y$ are $\mu_X $ and $\mu_Y$. Terry Tao mentions that from the perspective of probability, this reflects that the distribution of random variables $X$ and $Y$ is not enough to determine the distribution of $(X, Y)$ because $X$ and $Y$ are not necessarily independent. -Given that the products in the usual category fail to exist, is it possible to define a new categorical structure on the class of measure spaces such that products do exist? - -REPLY [73 votes]: To clarify Chris Heunen's answer, let me point out that most notions of measure theory -have analogs in the category of smooth manifolds. -For example, the analog of a measure space (X,M,μ), where X is a set, M is a σ-algebra of measurable subsets of X, and μ is a measure on (X,M), -is a smooth manifold X equipped with a density μ. -Likewise, the analog of a measure-preserving morphism is a volume-preserving smooth map. -The category of smooth manifolds equipped with a density together -with volume preserving maps as morphisms does not have good categorical properties. -The problem comes from the fact that preservation of volume is too strong a condition to allow for good categorical properties. -If one drops the data of a density and the property of volume preservation, -then the resulting category of smooth manifolds has relatively good categorical properties, -such as existence of finite products and, more generally, existence of all finite transversal limits. -The same is true for the category of measure spaces. -However, in this case we cannot simply drop the data of a measure from the definition -and expect to get a category with good properties. -The reason for this is that the data of a measure in fact combines two independent pieces of data. -The first piece of data tells us which sets have measure zero and which ones don't. -The second piece of data tells us the actual values of measure on sets of non-zero measure. -The analog of dropping the data of a density for measure theory is dropping the second piece -of data described above, but not the first one. -This can already be seen for smooth manifolds: If we have a smooth manifold, -we don't need a density on it to say which sets have measure 0. -Thus one is naturally led to the notion of a measurable space that knows which measurable -sets have measure 0. -I describe it in more detail in [1], -therefore here I offer only a brief summary of the main definitions. -A measurable space is a triple (X,M,N), -where X is a set, M is a σ-algebra of measurable subsets of X, and N⊂M is a σ-ideal -of measure 0 sets. -A morphism of measurable spaces f: (X,M,N)→(Y,P,Q) is an equivalence class of maps of sets g: X→Y -such that the preimage of every element of P is an element of M and the preimage of every -element of Q is an element of N. Two maps g and h are equivalent if they differ on a set of measure 0. -Here for the sake of simplicity I assume that both measurable spaces are complete -(any subset of a measure 0 subset is again a measure 0 subset). -Every measurable space is equivalent to its completion [2], -hence we do not lose anything by restricting ourselves to complete measurable spaces. -In general, one has to modify the above definition to account for incompleteness, as explained -in the link above. -Finally, one has to require that measurable spaces are localizable. -One way to express this property is to say that the Boolean algebra M/N of equivalence -classes of measurable sets is complete (i.e., it has arbitrary suprema and infima). -Many basic theorems of measure theory fail without this property. -In fact, as explained in the link above, theorems such as Radon-Nikodym theorem and Riesz -representation theorem are equivalent to the property of localizability. -Remarkably, the category of localizable measurable spaces is equivalent to the opposite -category of the category of commutative von Neumann algebras [7]. -This statement can be seen as another justification for the property of localizability. -Henceforth I assume that all measurable spaces are localizable. -Being in possession of a good category of measurable spaces we can prove that it -admits finite products, and, more generally, arbitrary finite limits. -Let me point out that at this point measure theory diverges from smooth manifolds: -The functor that sends a smooth manifold to its underlying measurable space -is colax monoidal but not strong monoidal -with respect to the monoidal structures given by the categorical products. -More precisely, the category of measurable spaces admits two natural product-like monoidal -structures. One is given by the categorical product mentioned above, -and the other one is the spatial product, which is often simply called the product in many textbooks on measure theory. By the universal property of product there is a canonical map from the spatial product -to the categorical product, which is a monomorphism -but not an isomorphism unless one of the spaces is atomic, i.e., a disjoint union of points. -The forgetful functor F from the category of smooth manifolds to the category of measurable -spaces is strong monoidal with respect to the categorical product on the category -of smooth manifolds and the spatial product on the category of measurable spaces. -Therefore it is also colax monoidal with respect to the categorical product on measurable spaces, -but not strong monoidal because F is essentially surjective on objects and the map -from the spatial product to the categorical product is not always an isomorphism. -Here is an instructive example for the last statement. -Let Z=F(R), where R is the real line considered as a smooth manifold. -Consider the spatial product of Z and Z, which is canonically isomorphic to F(R×R). -The spatial product maps monomorphically to the categorical product Y=Z×Z. -However, Y is much bigger than F(R×R). -For example, the diagonal map Z→Y=Z×Z is disjoint from F(R×R) in Y. -The space Y also has a lot of other subspaces -whose existence is guaranteed by the universal property. -Note that set-theoretically F(R×R) also has a diagonal subset. -However, this subset has measure 0 and therefore is invisible in this formalism. -Let me finish by mentioning that locales arguably provide much better formalism for measure theory, -which in particular does not suffer from problems with sets of measure 0, -e.g., we don't need to pass to equivalence classes or even mention the words “almost everywhere”. -The relevant functor sends a measurable space (X,M,N) to the locale M/N (as explained -above M/N is a complete Boolean algebra, hence a locale). -We obtain a faithful functor from the category of measurable spaces to the category of locales. -Let's call its image the category of measurable locales. -Then measurable morphisms of measurable locales correspond bijectively -to equivalence classes of measurable maps. -Note that we don't need to pass to equivalence -classes to define a measurable morphism of measurable locales. -Every measurable space (or locale) can be uniquely decomposed into its atomic and diffuse part. -The atomic part is a disjoint union of points and the diffuse part does not have any isolated points. -An example of a diffuse space is given by F(M) where M is a smooth manifold of non-zero dimension. -(In fact all these spaces are isomorphic as measurable spaces if the number of connected -components of M is countable.) -Here is the punchline: If Z is a diffuse measurable locale, then it does not have any points, -in particular it is non-spatial. -This can serve as an explanation of why we cannot construct a reasonable concrete category -of measurable spaces and why we always have to use equivalence classes if we want to stay -in the point-set measure theory. -References that advocate (but do not evangelize) the viewpoint described in this answer: - -[1] Is there an introduction to probability theory from a structuralist/categorical perspective? -[2] What's the use of a complete measure? -[3] Monoidal structures on von Neumann algebras -[4] Why do probabilists take random variables to be Borel (and not Lebesgue) measurable? -[5] Is there a measure zero set which isn't meagre? -[6] When is $L^2(X)$ separable? -[7] Reference for the Gelfand-Neumark theorem for commutative von Neumann algebras -[8] Decomposition of an abelian von Neumann algebra -[9] Subfactor theory and Hilbert von Neumann Algebras -[10] Problems where we can't make a canonical choice, solved by looking at all choices at once -[11] Integration of differential forms using measure theory? -[12] Can we characterize the spatial tensor product of von Neumann algebras categorically? -[13] Which complete Boolean algebras arise as the algebras of projections of commutative von Neumann algebras? -[14] Conditional Expectation for $\sigma$-finite measures<|endoftext|> -TITLE: Killing form vs its counterpart in a given represenation -QUESTION [8 upvotes]: Let $\mathfrak{g}$ be a semi-simple Lie algebra and let $\phi:\mathfrak{g}\rightarrow\mathfrak{gl}(V)$ be its finite-dimensional complex irreducible representation. You can define two non-degenerate symmetric forms on $\mathfrak{g}$: - -Standard Killing form: $K(X,Y)=tr(ad_X\circ ad_Y)$ -"Killing-like" form associated with $\phi$: $K_{\phi}(X,Y)= tr(\phi(X)\phi(Y))$ - -In general, is there any connection between $K_\phi$ and $K$? I know for instance that for defining representation of $\mathfrak{gl}(N,\mathbb{C})$ both forms are proportional. Is it true for general semi-simple Lie algebra? If not, is there a separate name for $K_{\phi}$? -I'm asking this question because in math-physical literature connected to research I'm doing people tend to confuse these two forms: $K_\phi$ is used to define second order Casimir invariant of a given representation. Yet, in some articles there is simply $K$ instead of $K_{\phi}$. - -REPLY [5 votes]: They are proportional if $g$ is simple. The form $K_\phi$ defines a homomorphism from the adjoint to the coadjoint representation. If the adjoint representation is irreducible, i.e. $g$ is simple, you know all such homomorphisms are proportional by Schur's lemma.<|endoftext|> -TITLE: Words in two infinitismal rotations -QUESTION [6 upvotes]: I asked this as subquestion in a comment pursuant to my Banach-Tarski -question. I think it is worth promoting here to a question in its own right. -Consider these two matrices over ${\Bbb R}[[\epsilon]]$: -$$A = \left[ \begin{array}{ccc} -\cos(\epsilon) & \sin(\epsilon) & 0 \\ --\sin(\epsilon) & \cos(\epsilon) & 0 \\ - 0 & 0 & 1 \end{array} \right] -\ {\rm and}\ -B = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ -0 & \cos(\epsilon) & \sin(\epsilon) \\ -0 & -\sin(\epsilon) & \cos(\epsilon) \end{array} \right].$$ -(By $\sin$ and $\cos$ I mean the formal Taylor series.) -If a word $w$ in $A$ and $B$ equals the identity modulo $\epsilon^k$, -must $w$ belong to the $k$th term of the derived series of the free group -on the symbols $A$ and $B$? - -REPLY [7 votes]: This question is related (loosely) to the theory of discrete groups generated by small elements, which was developed and used for various purposes by Margulis and others. If $A$ and $B$ are elements of a Lie group, then in exponential coordinates around the identity, multiplication looks like addition to first order, and the commutator $A*B*A^{-1}*B^{-1}$ vanishes to first order, and looks like Lie bracket to second order: the errors are $o(|A||B|)$. -The Lie algebra of $SO(2)$ has a vector space basis $A, B, C$ where the Lie brackets are cross product, $[A,B]=C, [B,C]=A$ and $[C,A]=B$. In this Lie algebra, there is a commutator relation $[A,[B,[A,B]]] = 0$. This implies that the corresponding group word applied to the two matrices $A$ and $B$ above is $o(\epsilon^4)$, but it is only in the third term of the derived series for the free group. The mismatch will grow with more complicated commutators: $so(3)$ is just not big enough to accomodate the larger and larger dimension of the free nilpotent Lie algebras corresponding to the quotients of the free group by the terms in its lower central series. No finite dimensional Lie algebra is big enough.<|endoftext|> -TITLE: Regular simplex in projective space -QUESTION [7 upvotes]: Is there a reference or a very short argument proving the following statement? -Let $C$ be a set consisting of $r$ points in the real projective space $\mathbb RP ^k$ -with its usual round metric. Assume that the distances between all pairs of points in $C$ are the same. Assume further that $r>k+1$. Then $r$ must be equal to $k+2$ and $C$ must be the image of the set of vertices of a regular simplex inscribed in the unit sphere. - -REPLY [17 votes]: Equidistant points in projective space correspond to equiangular lines in Euclidean space. The maximum number of equiangular lines in $\mathbb R^n$, $f(n)$ is a function which might not be as nice as you think, some computed values are in OEIS. For example, your claim fails even in the case $k=2$ where you can take the six large diagonals of the icosahedron and get 6 equidistant points in $\mathbb RP^2$. An absolute upper bound is given by $f(n)\le \binom{n+1}{2}$ (attributed to Gerzon in "Equiangular lines" by Lemmens and Siedel, proved again later by Koornwinder in "A note on the absolute bound for systems of lines" in both real and complex case).<|endoftext|> -TITLE: Technology for various models of spectra -QUESTION [8 upvotes]: There are a couple different models for spectra, or constructions of the categories of spectra that have the desired properties (homotopically and otherwise). The construction of the Categories of $S$-algebras in EKMM (Rings, Modules, and Algebras in stable homotopy theory by Elmendorf-Mandell-May-Kriz) is one such model. In EKMM, the authors develop a lot of "technology" for their category of spectra. In particular, they construct several spectral sequences and talk about Algebraic K-theory of S-algebras. I am curious to know if such developments have been pursued in the various other models, such as Symmetric Spectra. -All of these models yield the same homotopy category (I believe the paper of Mandell-May-Schwede-Shipley does this, or does a lot of it), but the impression I have gotten is that $S$-algebras have a lot of the technology already developed that one might want in stable homotopy theory. I believe Schwede has worked on computing homotopy groups of Symmetric Spectra, is this a sign that more technology is on its way? -I suppose the above is all background in some sense, so let me clearly state my question(s): -Are there developed notions of spectral sequences for other models of spectra? -Are there simple reasons why one would or would not expect such developments to occur (other than we already have them for $S$-algebras)? -Also, I don't mean for this to spark a debate about different models. I am just curious. -Thanks -EDIT: Clearly, I was unclear. I am wondering if people have used, written down, proved results regarding convergence etc, spectral sequences based on other models of spectra. Is it as Tyler points out that all constructions of spectral sequences essentially occur in the homotopy category? That would certainly answer my question. I thought that since computing homotopy groups of Symmetric Spectra (non-naively) was subtle that such constructions might be subtle as well. - -REPLY [5 votes]: I don't think I noticed this question before. One point is that now that we -have multiplicatively well-behaved Quillen equivalences between all reasonable -models for the stable category, hence between reasonable models for categories -of ring and module spectra, it is formal to transport constructions like spectral -sequences from one model to any other. Another point is just historical: -EKMM got there first and skimmed off the easy applications. In view of the -first point, there is no reason for anyone to want to reinvent the wheel. -A technical point is that, for spectral sequences especially, it is convenient -to work with CW spectra (or CW R-modules for a ring spectrum R), and these are -nowhere written down in any context other than EKMM (harking back to LMS). -They are only natural objects when all spectra are fibrant; more precisely, -unless all spectra are fibrant, CW spectra will not be nicely related to the -cell spectra relevant to the model structures of interest. (Section 24.2 of -Parametrized Homotopy Theory http://www.math.uchicago.edu/~may/EXTHEORY/MaySig.pdf -has some discussion of this.) This advantage of EKMM has directly related -offsetting disadvantages; in particular, the sphere spectrum is not cofibrant. -It is an old theorem of Gaunce Lewis that you can't have everything. -Repeating myself from other answers, it is best to be eclectic.<|endoftext|> -TITLE: Is this method of "fractional sums" using a Fourier series viable? -QUESTION [5 upvotes]: Hi. -I have this idea about developing what I call a "continuum sum", that is, a method to "add up a non-integer number of terms", i.e. to see if there is a "natural" way to assign a meaning to the expression -$\sum_{n=a}^b f(n)$ -where $a$ and $b$ are non-integer fractional, real, or even complex numbers, for a target function $f$ defined on one of those domains. -What does that mean, exactly? Well, I consider this the problem of constructing a "summation operator" $\Sigma$ which is an "inverse" of the unit forward difference operator $\Delta$, but with both acting on functions on the real or complex continuum instead of just on the integers. This relationship is the same as how the integral is the inverse of the derivative. That is, applying $\Sigma$ to a function $f$ is equivalent to solving the functional equation $F(z + 1) - F(z) = f(z)$ for $F$. With such an $F$ in hand, we can then say $\sum_{n=a}^b f(n) = F(b+1) - F(a)$. -Solutions to this equation, however, are not unique. If we fill any unit real interval with some function, then the equation entails the function at the whole real line (provided $f$ is defined there.). So there are as many solutions as there are functions on the real numbers -- an uncountable $\beth_2$ possible solutions (which is even bigger than the continuum itself at $\beth_1$). In general, given any solution $F$, we can express any other $G$ as $G(x) = F(x) + \theta(x)$, where $\theta(x)$ is a 1-periodic "wobble" function. -And thus our problem is: is there some solution of this equation for a given function $f$, which is more "good" or "natural" than others? To attempt to ease the problem, it seems best to impose some restrictions on the candidate pool of input functions $f$ to consider, as in general, the tamer the function, the easier the analysis of the equation and the more techniques are available. But we don't want too much up-front restriction, or the methods available may then become relatively useless (e.g. a method that only summed linear functions and nothing else would have little to no use.). -There was a method proposed by Markus Mueller in an article called "Fractional sums and Euler-like identities" that attempts to do something like this, but its restrictions seem a little too stiff. For example, it does not appear the method is of any use in summing, e.g. -$\sum_{n=a}^b n! = \sum_{n=a}^b \Gamma(n+1)$ -though a solution does exist, namely -$\sum_{k=1}^{n} k! = \frac{-e + \mathrm{Ei}(1) + \pi i + E_{n+2}(-1) \Gamma(n+2)}{e}$ -using the exponential integral and En-function. -What is wanted here is to come up with some kind of "grand unified theory" of these kinds of sums: a solution for the functional equation that would be able to recover most if not all of these kind of formulae, and also enable the summation of many other functions. -For my approach, I suppose that $f$ be at least a holomorphic function of a complex variable, and that $F$ is to be as well. This is still pretty broad and does not uniquely determine a solution, but is tight enough to maximize the available tools for analysis. At this point, though, I'll suppose even further that $f$ and $F$ be entire. Then we can see if the techniques generalize to non-entire cases. -The approach I settled on (how I got to this is omitted here for the sake of brevity), is using Fourier series. If $f(z)$ is a periodic function with period $P$, then -$f(z) = \sum_{n=-\infty}^{\infty} a_n e^{\frac{2\pi i}{P} nz}$. -Now we have a simple formula for the sum of an exponential: $\sum_{n=0}^{z-1} e^{un} = \frac{e^{un} - 1}{e^u - 1}$. We can apply this to the above to get -$\sum_{n=0}^{z-1} f(n) = \sum_{n=-\infty}^{\infty} \frac{a_n}{e^{\frac{2\pi i}{P} n} - 1} \left(e^{\frac{2\pi i}{P} nz} - 1\right)$ -with the term at $n = 0$ on the right (for which the given expression fails directly with a division by 0) interpreted as $a_0 z$. This series can converge even when the given function fails some of the mentioned requirements, however it does not work for functions with harmonics of period 1. -My idea, then, was to consider a sequence of periodic entire functions $f_i$ that converge to a given function $f$ that's entire and aperiodic, but not necessarily of exponential type less than $2\pi$. Then take their continuum sums by the above formula and take the limit. If this limit exists, call it the continuum sum of $f$ itself. The questions I have, then, are, what conditions are needed on $f$ for this to work, and also, more importantly, is the limit independent of the chosen sequence of functions, and if so, what is the proof, and if not, what is a counterexample? This is why I mean by it being "viable" or not. If the limit doesn't work, this isn't of much use. I'm not necessarily interested in a complete proof but more on advice about how one would go about approaching a proof of this, useful reference material, etc. as I'd like to do some of it myself. However, if the hypothesis is false, I'd like a full counterexample. - -Add: The justification for considering this approach as "natural" is based on two approaches. One is Faulhaber's formula, which gives a sum of powers by the Bernoulli polynomials, and this sum has a simple uniqueness criterion: it sends polynomials to polynomials. One can then apply this to Taylor series. The trouble is, that such a method looks only to work on an extremely limited set of analytic functions: entire functions of exponential type less than $2\pi$. This limit seems a bit too onerous. This is one of the "some methods" I mentioned as having experimented with for defining the continuum sum. It is somewhat long to give the whole derivation, but for $e^{uz}$, summed from $0$ to $z-1$, and $|u| < 2\pi$, it yields $\frac{e^{uz} - 1}{e^u - 1}$. Another justification is much simpler. We know that $\Delta e^{uz} = \left(e^u - 1\right) e^{uz}$. Thus, it'd seem sensible to presume $\Sigma \left(e^u - 1\right) e^{uz} = e^{uz}$. This leads to (assuming $\Sigma$ is linear), $\Sigma e^{uz} = \frac{e^{uz}}{e^u - 1}$, and then the sum from $0$ to $z-1$ is $\frac{e^{uz} - 1}{e^u - 1}$. I suppose one could get a third justification in that both of these methods give the same result. Finally, because $\Sigma$ is linear, it is no big step to obtain the result for periodic functions. - -REPLY [2 votes]: Let me rephrase your question in the special case $a=0$ and $b=t$. Your sum is given for $t$ a nonnegative integer, and you want a general and natural procedure to define a function $F(t)$ on $t\ge0$ such that $F(n)$ is equal to the sum when $n=0,1,2,\dots$. -Now, there is a 'natural' and I would say standard wey to give a meaning to a sum $\sum a_n$ up to a real value, just by smoothing (if you prefer, convolution): fix a smooth function $\chi(t)$ equal to 1 for $t\le1/4$ and vanishing for $t>3/4$, and define -$$F(t)=\sum a_n\chi(n-t).$$ -In your original problem, the values $a_n$ arise as special values of a function $f(n)$, but I think this point of view is misleading; there are $\infty$ functions with the same values at integers, sometimes even nicer than the one of your choice, so what does it mean to deduce the form of $F$ in a natural way from $f$? when your data does not keep into account the precise form of $f$? I'm only trying to say that your original question is ill-defined.<|endoftext|> -TITLE: Why are profinite topologies important? -QUESTION [34 upvotes]: I hope this is not too vague of a question. Stone duality implies that the category Pro(FinSet) is equivalent to the category of Stone spaces (compact, Hausdorff, totally disconnected, topological spaces). This equivalence carries over to profinite vs Stone-topological algebras for a number of algebraic theories, such as groups, monoids, semigroups, and rings. The case of profinite groups is especially well-known. -My question is: why are such equivalences important? Where in mathematics do we gain something by identifying a pro-(finite group) with a Stone topological group? I mean something other than "concreteness" or "familiarity"—certainly it may be easier (for some people) to think about a Stone topological group than about a cofiltered diagram of finite groups, but are there important things that we couldn't prove about cofiltered diagrams of finite groups without knowing that they are equivalent to Stone topological groups? -I am especially interested because this manifestation of Stone duality seems to be "fragile" for generalizations in several directions. For instance, Theo JF commented on this question that Stone-topological groupoids are not equivalent to pro-(finite groupoids). The equivalence is also false if we generalize from finite sets/groups to ones of larger cardinality. It is true that pro-groups with surjective transition maps can be identified with pro-discrete locales, but I don't know anything about whether this is true for pro-sets (cf. question linked above), or any type of algebras other than groups. So in all the cases where the generalization fails, what is lost if we just work with pro-objects and ignore the missing topological aspect? - -REPLY [5 votes]: Compact semigroups have nice properties (existence of idempotents, minimal ideals, etc). If you view a profinite semigroup as a compact semigroup you get all this for free. I am not even sure how to think of these things for a pro-object in the category of finite semigroups. -Also one wants to think about free profinite groups and semigroups as missing free objects from the categories of finite groups and semigroups. That is the categories of profinite groups and semigroups have natural forgetful functors to sets which have left adjoints. It is not so natural from the pro-object viewpoint.<|endoftext|> -TITLE: The single-plus construction is not the left adjoint of the inclusion of separated presheaves? -QUESTION [8 upvotes]: Convention: -Before I start, please note that it is not going to be sufficient to assume that the topology is subcanonical or that the site $C$ has finite limits, since the application I have in mind does not satisfy either of these two criteria. Therefore, we take our definition of a topology to be the one in terms of covering sieves (a function $J$ assigning to each object $U$ a family of subfunctors $S\subseteq h_U$ satisfying conditions). Each sieve has a canonical "cover" attached to it, $\{U_i\to U\}_{i\in I}$ (although because our case is somewhat pathological, the resulting system of covers does not meet the requirements for being a Grothendieck pretopology (the requirements are not meaningful in this setting because our underlying category does not have pullbacks)). We will make use of these covers in a construction, but it is to be understood that these arise from an actual Grothendieck topology. -Let $(C,\tau)$ be a Grothendieck site. It's well-known that the double-plus construction $(-)^{++}$, which is actually the composite $((-)^+)^+$ is the left adjoint to the inclusion functor $Sh(C,\tau)\hookrightarrow Psh(C)$. Recall that $(-)^+$ replaces the functor $F(-)$ by the functor $H_\tau^0(-,F)$, or the $0^{th}$ sheaf of Čech cohomology, which can be constructed as the colimit $H^0_{\tau}(U,F)=L_\tau F(U):=Colim_{S\in J(U)^{op}} Hom(S,F)$. -Up until yesterday or so, I believed that the functor $(-)^+:Psh(C)\to SepPsh(C)$ is left adjoint to the inclusion $\iota:SepPsh(C,\tau)\hookrightarrow Psh(C)$, but I feel very inclined to doubt it now, since by a computation I did with a toy example, it doesn't seem to have the right universal property (I'm unsure, because the one place that I've found that addresses the issue says that it is). -However, when I was looking back through Vistoli's notes in the section on sheafification, I noticed that his construction of the sheafification is actually as $((-)^{sep})^+$, where $(-)^{sep}$ is defined by quotienting out by the following equivalence relation: $R(U)\subset F(U)\times F(U)$ where a pair $(a,b)\in F(U)\times F(U)$ is in $R(U)$ (that is to say, $a\sim b$) if there exists a cover $\{U_i\to U\}_{i\in I}$ such that $a|_{U_i}=b|_{U_i}$ for each $i\in I$. -First thing: It's clear that the relation is reflexive and symmetric, but is it also clear that it's transitive? If it's not, does Vistoli's construction work if we saturate the relation to an equivalence relation? -Second thing: Is the $(-)^+$ functor actually the left-adjoint of the inclusion $\iota$? If not, is $(-)^{sep}$? If the left adjoint is indeed $(-)^{sep}$, does it preserve monomorphisms (this is not to show that it's left-exact, since that will definitely fail (since separated presheaves do not form a topos). I have something totally different in mind unrelated to the rest of this question). -The toy example I mentioned above is the one to keep in mind, so I'll mention it here: -Consider the category $[1]$, which is the category with two objects $0$ and $1$ with one nonidentity morphism $0\to 1$. Let $S=h_0\subseteq h_1$ be the only nonidentity covering sieve of $1$. Then separated presheaves on this site are exactly the functors $F:[1]^{op}\to Set$ such that the restriction map $res_{0}:F(1)\to F(0)$ is injective. Given a presheaf $F$, $F^{sep}$ is given by the quotient where $F^{sep}(0)=F(0)$ and $F^{sep}(1)$ is the quotient of $F(1)$ by the equivalence relation noted above. -Notice also that sheaves for this topology are the presheaves such that $F(1)\to F(0)$ is a bijection. -We can easily verify the universal property of Vistoli's construction, since given a morphism $F\to G$ where $G$ is separated, we see that it obviously factors uniquely through the quotient. The $(-)^+$ construction does not seem to have this property in any obvious way. - -REPLY [11 votes]: About your first question: the relation is transitive. If $a$ is equivalent to $b$ using a covering, and $b$ is equivalent to $c$ using another covering, it is easy to see that $a$ is equivalent to $c$ using a common refinement. And yes, $(-)^{\rm sep}$ is left adjoint to the inclusion of separated presheaves into presheaves, that's easy to check.<|endoftext|> -TITLE: Is the generalized Erdős–Heilbronn problem true for finite cyclic groups? -QUESTION [5 upvotes]: The generalized Erdős–Heilbronn (GEH) theorem, which is proved by da Silva and Hamidoune in 1994, states that: -Theorem. If p is a prime and $X$ is a subset of $\mathbb{Z}_p$, then $|\hat{k}X| \geq \min \lbrace k|X|-k^2+1 , p \rbrace$ for $\hat{k}X = \lbrace x_1+\ldots+x_k \mid x_i \in X , x_i \neq x_j \rbrace$. -Also, for the case that $k=2$ (which is called the Erdős–Heilbronn problem), the above statement holds for $X$ as a subset of any finite group $G$; the result is proved by Balister and Wheeler in 2009. - -Problem 1. Is the generalized Erdős–Heilbronn problem also true for any finite groups? In particular, is it true for finite cyclic groups? - -This question is inspired by the construction of a counter-example to some variants of Ramsey theorem. In the construction we may not need the full strength of the GEH, so a related question is: - -Problem 2. Is there any weaker results to the GEH, which have already been proved? - -REPLY [5 votes]: Arithmetic progressions usually give small sumsets so...How about $X=\lbrace 0,3,6,9,12 \rbrace \subset \mathbb{Z}_{15}$? Then for $k=2,3,4$ one has $\hat{k}X=X$. That is ok for $k=4$ but not for $k=2,3$. (Doesn't that contradict what you said about $k=2$?) -In fact $X=\lbrace 0,d,2d,\cdots,nd\rbrace \subset \mathbb{Z}_{nd+d}$ for $n \ge 5$ is always a counter-example for $k=3$. -The same type of examples should work as well for larger $k$ and should also work with a few elements removed. -later One can't hope for better than $|\hat{k}X| \geq \min \lbrace k|X|-k^2+1 , |H| \rbrace$ where $H$ is the subgroup generated by $X$. After all, $\hat{k}X \subset H$. I don't know if that is true though. A weaker claim, which seems somewhat more likely, is to replace $|H|$ by the size of the smallest subgroup of size at least $|X|$. I have not read the Balister and Wheeler paper you mention but it seems (from the review posted) to replace $|H|$ with the size of the smallest non-trivial subgroup, which would be prime of cyclic order.<|endoftext|> -TITLE: Outer automorphisms of free groups into bigger free groups -QUESTION [9 upvotes]: This may be very either very simple or very unknown, but here goes: Let $F_n$ be the free group on $n$ generators and $Out(F_n)$ its outer automorphism group. -Can embeddings $Out(F_n) \hookrightarrow Out(F_m)$ be characterized for $m > n$? - -REPLY [13 votes]: There is a bunch of negative results here: -1) Khramtsov has shown that there are no embeddings for $m = n+1$ (when $n>1$) -2) Bridson and Vogtmann have shown that any homomorphism $Out(F_n) \to Out(F_m)$ factors through $\mathbb{Z} / 2 \mathbb{Z}$ if -a) $ m < n, n>2$ -b) $n > 8, n< m < 2n-2$ for $n$ odd, and $n < m< 2n$ for $n $ even. This is a slightly stronger version of a result of Potapchik--Rapinchuk. -3) I have a result showing that in fact all such homomorphisms factor through $\mathbb{Z} / 2 \mathbb{Z}$ if $n > 5$ and $ n < m < {n \choose 2}$. -There are two positive results as well, one by Aramayona--Leininger--Souto and one by Bogopol'skii--Puga and Bridson--Vogtmann; in both cases $m$ grows at least exponentially with $n$. -As far as I am aware, this is all that is known. I'd be happy to give you further details if you are interested.<|endoftext|> -TITLE: Dynamical systems, minimal sets and the Axiom of Choice -QUESTION [12 upvotes]: Perhaps the most important application of the Axiom of Choice within the theory of dynamical systems (meaning here, compact Hausdorff spaces with a self-map) yields, within every dynamical system, the existence of at least one non-empty minimal set (meaning a closed, invariant subset itself containing no proper closed, invariant subset). Since every point in a minimal set is almost periodic, this gives the existence of almost periodic points in every dynamical system. -Can anyone tell me please, how much of AC one gets back, over ZF, by taking the existence of minimal sets in dynamical systems as an axiom? -Are there interesting classes uncountable compact Hausdorff spaces where one has the existence of minimal sets already from ZF? - -REPLY [2 votes]: This is somewhat of a non-answer to the second question (I do need a little bit of AC) so I apologize. But I thought it might be worthwhile for its generality. -The uncountable example I had in mind is $\beta \mathbb{N}$, the set of ultrafilters on the natural numbers, i.e., the Stone–Čech compactification of the discrete space $\mathbb{N}$. - -$(\beta \mathbb{N},+)$ is a semigroup (with + extending the usual addition on $\mathbb{N}$ to $\beta \mathbb{N}$ in such a way that addition with a fixed right hand side is continuous) -The shift $s(p)= 1+ p$ makes $(\beta \mathbb{N},s)$ a dynamical system (because addition with natural numbers is left-continuous, too) -Its minimal systems are exactly the minimal left ideals of the semigroup -Its cardinality is $2^{2^{\aleph_0}}$ (i.e., the size of the power set of the reals) -The minimal left ideals are universal minimal systems for discrete time, so one minimal left ideal induces minimal systems everywhere - -To prove all this, if I'm not mistaken, you require "only" - -The ultrafilter lemma for (the power set of) $\mathbb{N}$ so that you actually get the space (the ultrafilter lemma is strictly weaker than AC) -An application Zorn's Lemma to find a minimal left ideal - -So that's not "a lot" of AC to get one minimal system everywhere (which is why I thought it'd be worthwhile). -On the other hand, if you assume, e.g., ZF+AD you do not find any free ultrafilters on $\mathbb{N}$, so no $\beta \mathbb{N}$, and I have no idea what the dynamics look like then...<|endoftext|> -TITLE: Can ⨁_I A be isomorphic to ∏_I A for infinite I? -QUESTION [11 upvotes]: Suppose $A$ is a non-zero ring (say commutative unital) and $I$ is an infinite set. Can it happen that there is an isomorphism of $A$-modules $\bigoplus_{i\in I}A\cong \prod_{i\in I}A$? - -The obvious morphism $\bigoplus_{i\in I}A\to \prod_{i\in I}A$ is obviously not an isomorphism, but could there be another one? - -If $A$ is a field, then no. -In the case when $A$ is a field, I think the following argument does the trick. I'd be happy to know if there's a simpler one. If you're like me and my friends, you'll think this is obvious, but you won't be able to prove it. -Case 1: $A$ is finite or countable. Then $\bigoplus_{i\in I}A$ has cardinality $|I|$ and $\prod_{i\in I}A$ has cardinality $|A|^{|I|} = 2^{|I|}>|I|$. There is no bijection between the two sets, so there's no $A$-module isomorphism. -Case 2: $A$ has arbitrary size. Let $k\subseteq A$ be the prime field of $A$ (either $\mathbb F_p$ or $\mathbb Q$). There is a natural inclusion $\prod_{i\in I}k\hookrightarrow \prod_{i\in I}A$. Suppose $S\subseteq \prod_{i\in I}k$ is linearly independent with $|S|>|I|$. Then I claim the image of $S$ under this inclusion is $A$-linearly independent. This would show that $\dim_A (\prod_{i\in I} A)>|I|=\dim_A(\bigoplus_{i\in I}A)$, so the two cannot be isomorphic. -To show the claim, suppose there is a non-trivial $A$-linear relation on $S$. This can be used to construct a $k$-linear map (the "coefficients map" of the relation) $\phi:\prod_{i\in I}k\to A$ so that $\phi(s_1+\cdots +s_n)=0$, but $\phi(s_r)$ is not zero for some $r\in \{1,\dots, n\}$. Expressing $A$ as $k^{\bigoplus J}$ for some $J$, we get $J$ many $k$-linear compositions -$$ -\phi_j:\prod_{i\in I}k \xrightarrow{\phi} A\cong k^{\bigoplus J} \xrightarrow{p_j}k -$$ -Since $\phi(s_r)$ is not zero for some $r\in \{1,\dots, n\}$, there is some $j$ so that $\phi_j(s_r)$ is not zero. But $\phi_j(s_1+\cdots +s_n)=0$, so this gives a non-trivial $k$-linear relation on $S$, contradicting the assumption that $S$ is $k$-linearly independent. -If I'm not mistaken, better bookkeeping in the above argument actually shows that $\dim_A (\prod_{i\in I}A)=2^{|I|}$. - -The obvious way to try to settle the question for a general (commutative unital) ring $A$ is to tensor with $A/\mathfrak m$, where $\mathfrak m$ is a maximal ideal of $A$. So I may as well pose the additional question - -If $A$ is a (commutative unital) ring and $I$ is an infinite set, is it necessarily true that $(A/\mathfrak m) \otimes_A (\prod_{i\in I} A) \cong \prod_{i\in I} (A/\mathfrak m)$ as $(A/\mathfrak m)$-modules? - -REPLY [9 votes]: The answer to the first question is no. Here is a proof. -Let $A$ be a commutative ring and $\mathfrak{m}$ be a maximal ideal of $A$. Let $k=A/\mathfrak{m}$. Suppose we have an isomorphism of $A$-modules $A^I \cong A^{(I)}$, where $A^{(I)} := \bigoplus_{i \in I} A$. Tensoring with $k$ over $A$ yields an isomorphism of $k$-vector spaces -\begin{equation} -\frac{A^I}{\mathfrak{m} \cdot A^I} \cong k^{(I)} -\end{equation} -where $\mathfrak{m} \cdot A^I$ denotes the $A$-submodule generated by the $m\cdot x$ with $m \in \mathfrak{m}$ and $x \in A^I$. Since $\mathfrak{m} \cdot A^I \subset \mathfrak{m}^I$, we have a surjective $k$-linear map -\begin{equation} -\frac{A^I}{\mathfrak{m} \cdot A^I} \to k^I -\end{equation} -So we get a contradiction by considering the dimension over $k$ (and using the case you first settled). -EDIT : The answer to the second question is also negative in general. Let me prove that for $I=\mathbf{N}$, there exist a couple $(A,\mathfrak{m})$ such that the statement is false. The idea is to take $A=k[X_j, j \in J]$ and $\mathfrak{m}=\langle X_j, j \in J \rangle$ with a sufficiently big set $J$ (the argument should also work if $\mathfrak{m}/\mathfrak{m}^2$ is a sufficiently big $k$-vector space). We want to prove that -\begin{equation} -\dim \frac{A^{\mathbf{N}}}{\mathfrak{m} \cdot A^{\mathbf{N}}} > \dim k^{\mathbf{N}}. -\end{equation} -It is sufficient to show that $\dim \frac{\mathfrak{m}^{\mathbf{N}}}{\mathfrak{m} \cdot A^{\mathbf{N}}} > \dim k^{\mathbf{N}}$. We have -\begin{equation} -\dim \frac{\mathfrak{m}^{\mathbf{N}}}{\mathfrak{m} \cdot A^{\mathbf{N}}} \geq \dim \frac{\mathfrak{m}^{\mathbf{N}}}{\mathfrak{m} \cdot A^{\mathbf{N}} + (\mathfrak{m}^2)^{\mathbf{N}}} -\end{equation} -The latter space can be identified with the cokernel of the natural (injective) map $k^{\mathbf{N}} \otimes V \to V^{\mathbf{N}}$, where $V=\mathfrak{m}/\mathfrak{m}^2$. -Lemma The cokernel of $f : k^{\mathbf{N}} \otimes V \to V^{\mathbf{N}}$ has dimension $\geq \dim V$. -Proof. A basis of $V$ is given by the family $e_j = \overline{X_j}$ indexed by $j \in J$. Write $J$ as a disjoint union $J = \sqcup_{s \in S} J_s$ where each $J_s$ is countable. We have $\operatorname{card} S = \operatorname{card J}$. Fix isomorphisms $\phi_s : \mathbf{N} \to J_s$. For any $s \in S$, let -\begin{equation} -u^{(s)} = (e_{\phi_s(n)})_{n \in \mathbf{N}} \in V^{\mathbf{N}}. -\end{equation} -Let us say that a sequence $u \in V^{\mathbf{N}}$ has finite rank if the vector space generated by the $u_n$ is finite dimensional. Then the image of $f$ is the subspace of finite rank sequences in $V^{\mathbf{N}}$. It is not difficult to see that a finite linear combination $\sum_s \lambda_s u^{(s)}$ with all $\lambda_s \neq 0$ cannot have finite rank, whence the lemma. -It remains to take $J$ sufficiently big, for example we can take $J=2^{k^{\mathbf{N}}}$.<|endoftext|> -TITLE: n-dimensional voronoi diagram -QUESTION [13 upvotes]: Hi, I need to compute the voronoi diagram of a set of points in $R^n$. -I'm quite unschooled on the topic, could someone point me to the right references so that I can -a) understand the theory behind it; -b) implement an actual algorithm to compute it. - -P.S. Googling I found out I need to know about euclidean graphs, but I - couldn't find any decent introduction to it. Pointers are appreciated! - -REPLY [3 votes]: R (a favorite open source, object oriented interactive statistical package) has what appears to be a suitably fast algorithm, but I've only used it in a problem with 20 some odd points or so. -As the others have mentioned above, the Voronoi tessellation or mosaic is computed using the (dual graph which is the) Delaunay triangulation. The R function 'veronoi.mosaic' in the package 'tripack' calls a FORTRAN routine. If its convenient enough to do it in R then just call the veronoi.mosaic function (details on getting R and getting the tripack package follow). Otherwise, if you'd rather just use the internal FORTRAN subroutine, you can figure out the meaning of the arguments by looking at the veronoi.mosaic function in R (at the point at which the FORTRAN subroutine 'veronoi' is called) and work backwards. -In any case, all of its open source and works like a charm. -getting R: -http://cran.r-project.org/ -the tripack package -http://cran.r-project.org/web/packages/tripack/index.html -Enjoy! -Best Regards, -Grant Izmirlian<|endoftext|> -TITLE: An algebraic vector bundle is trivialized by open sets. How many does one need? -QUESTION [34 upvotes]: Consider an algebraic vector bundle $E$ on a scheme $X$. By definition there is an open cover of $X$ consisting of open subsets on which $E$ is trivial and if $X$ is quasi-compact, a finite cover suffices. The question then is simply: what is the minimum number of open subsets for a cover which trivializes $E$ ? Now this is silly because the answer obviously depends on $E$ ! If $E$ is trivial to begin with, the cover consisting of just $X$ will do, of course, but if you take $\mathcal O(1)$ on $\mathbb P^n_k$ you won't get away with less than $n+1$ trivializing open subsets . Here is why. -Suppose you have $n$ open subsets $U_i\subset \mathbb P^n_k$ over which $\mathcal O(1)$ is trivial. Take regular nonzero sections $s_i\in \Gamma(U_i,\mathcal O(1) )$ and extend them rationally to $\mathbb P^n_k$. Each such extended rational section $\tilde {s_i}$ will have a divisor $D_i$ and the complements $\tilde U_i= X\setminus |D_i|$, $(U_i\subset \tilde U_i)$, of the supports of those divisors will give you a cover of $\mathbb P^n_k$ by $n$ affine open subsets trivializing $\mathcal O(1)$. But this is impossible , because $n$ hypersurfaces in $\mathbb P^n_k$ cannot have empty intersection. -This, conversations with colleagues and some vague considerations/analogies have led me to guess ( I am certainly not calling my rather uninformed musings a conjecture) that the following question might have a positive answer: - -Is it true that on a (complete) algebraic variety of dimension $n$ every vector bundle is trivialized by some cover consisting of at most $n+1$ open sets? - -For example, the answer is indeed yes for a line bundle on a (not necessarily complete) smooth curve $X$: every line bundle $L$ on $X$ can be trivialized by two open subsets . -Edit Needless to say I'm overjoyed at Angelo's concise and brilliant positive answer. In the other direction ( trivialization with too few opens to be shown impossible) I would like to generalize my observation about projective space. So my second question is: - -Consider a (very) ample line bundle $L$ on a complete variety $X$ and a rational section - $s \in \Gamma _{rat} (X, L) $. Is it true that its divisor $D= div (s)$ has a support $|D|$ whose complement $X\setminus |D|$ is affine ? Let me emphasize that the divisor $D$ is not assumed to be effective, and that is where I see a difficulty. - -REPLY [10 votes]: This is an answer to Georges' updated question at the end of his post. - -An equivalent formulation of the question is the following: - -Question - Let $L$ be an ample Cartier divisor on a projective scheme $X$ and suppose there exist effective divisors $D_1, D_2$ such that $L\sim D_1-D_2$. Then is it true that $X\setminus \left({\rm supp}\,D_1 \cup {\rm supp}\,D_2\right)$ is affine? - -I think this is true in some cases, but not in general. - -Claim 1 The answer to the question is YES if $X$ is a projective curve. - -Proof -Both $D_1$ and $D_2$ are effective and hence ample and similarly so is $A=D_1+D_2$. Clearly - $X\setminus \left({\rm supp}\,D_1 \cup {\rm supp}\,D_2\right)=X\setminus {\rm supp}\, A$, which is affine. $\square$ - -Claim 2 There are many examples for smooth projective varieties for which there exists $L, D_1, D_2$ as above such that $X\setminus \left({\rm supp}\,D_1 \cup {\rm supp}\,D_2\right)$ is not affine. In fact, this happens on any smooth projective surface containing a $(-1)$-curve. - -Remark I am pretty sure one does not need smoothness and there are also singular examples. (Actually the example below only needs one smooth point.) -Proof -Let $Y$ be an arbitrary projective variety (reduced) of dimension at least $2$ and $H$ an effective (very) ample Cartier divisor on $Y$. Let $\sigma : X\to Y$ be the blow up of a smooth point $p\in Y$ that is not contained in $H$ and let the exceptional divisor of $\sigma$ be $E\subset X$. -Then for some $m>0$ positive integer, $L=m\sigma^*H-E$ is ample. (I suspect that most people know this, but if you need a hint for this statement, an explicit estimate on $m$ can be found in Lemma 2 of this answer to another MO question.) -Now let $D_1=m\sigma^*H$ and $D_2=E$. Notice that by the choice of the point that was blown up, $D_1$ and $D_2$ are disjoint. It follows that $X\setminus \left({\rm supp}\,D_1 \cup {\rm supp}\,D_2\right)\simeq (Y\setminus {\rm supp}\, H)\setminus \{p\}$. Furthermore, since $H$ is ample on $Y$, it follows that $Y\setminus {\rm supp}\, H$ is affine, and hence -$(Y\setminus {\rm supp}\, H)\setminus \{p\}$ is not. $\square$ -It is actually true, that for any line bundle there always exists a rational section for which the complement of its divisor is affine. - -Claim 3 - Let $L$ be an arbitrary Cartier divisor on a projective scheme $X$. Then there exist effective very ample divisors $D_1, D_2$ such that $L\sim D_1-D_2$. - -Proof -Choose an arbitrary ample Cartier divisor $A$ on $X$. For large enough $r_1\gg 0$ -$L+r_1A$ is basepoint-free by the definition (or one of the basic properties depending on what you choose as definition) of ampleness. Then for an even larger $r\gg r_1$ we may assume that $L+rA$ is both basepoint-free and ample and hence very ample and also that $rA$ is very ample as well. Now choose $D_1=L+rA$ and $D_2=rA$. $\square$ -And we get as an easy consequence: - -Corollary - With the notation of Claim 3, we may choose $D_1$ and $D_2$ such that - $X\setminus \left({\rm supp}\,D_1 \cup {\rm supp}\,D_2\right)$ is affine. - -Proof -Replace $D_1$ and $D_2$ with general members of their complete linear systems. Then we may assume that they do not have a common component and hence ${\rm supp}\,(D_1+D_2)={\rm supp}\,D_1 \cup {\rm supp}\,D_2$. Since $D_1+D_2$ is also ample, this proves that claim. $\square$<|endoftext|> -TITLE: Dimension of infinite product of vector spaces -QUESTION [31 upvotes]: This question is motivated by the question link text, which compares the infinite direct sum and the infinite direct product of a ring. -It is well-known that an infinite dimensional vector space is never isomorphic to its dual. More precisely, let $k$ be a field and $I$ be an infinite set. Let $E=k^{(I)}=\oplus_{i \in I} k$ be the $k$-vector space with basis $I$, so that $E^{*}$ can be identified with $k^I = \prod_{i \in I} k$. Then a stronger result asserts that the dimension of $E^{*}$ over $k$ is equal to the cardinality of $k^I$. This is proved in Jacobson, Lectures in Abstract Algebra, Vol. 2, Chap. 9, $\S$ 5 (Jacobson deduces it from a lemma which he attributes to Erdös and Kaplansky). Summarizing, we have -\begin{equation} -\operatorname{dim}_k (k^I) = \operatorname{card} k^I. -\end{equation} -Now, if $V$ is any $k$-vector space, we can ask for the dimension of $V^I$. Does the Erdös-Kaplansky theorem extend to this setting ? - -Is it true that for any vector space $V$ and any infinite set $I$, we have $\operatorname{dim} V^I = \operatorname{card} V^I$ ? More generally, given a family of nonzero vector spaces $(V_i)$ indexed by $I$, is it true that $\operatorname{dim} \prod_{i \in I} V_i = \prod_{i \in I} \operatorname{card} V_i$ ? - -If $V$ is isomorphic to $k^J$ for some set $J$, then the result holds as a consequence of Erdös-Kaplansky. In the general case, we have $V \cong k^{(J)}$, and we can assume that $J$ is infinite. In this case I run into difficulties in computing the dimension of $V^I$. I can only prove that $\operatorname{dim} V^I \geq \operatorname{card} k^I \cdot \operatorname{card} J$. - -REPLY [18 votes]: Here is a self-contained version of Todd Trimble's wonderful answer. -Let $K$ be a field. "Vector space" shall mean "$K$-vector space", "linear" shall mean "$K$-linear", $\dim$ shall mean $\dim_K$, $\operatorname{Hom}$ shall mean $\operatorname{Hom}_K$, and $|X|$ shall denote the cardinal of $X$ for any set $X$. -Let $V$ be the product of a family of nonzero vector spaces $(V_i)_{i\in I}$: -$$ -V=\prod_{i\in I}V_i. -$$ -As we have -$$ -\dim V=\sum_{i\in I}\dim V_i -$$ -if $I$ is finite, we can (and will) assume from now on that $I$ is infinite. - -Main Theorem. We have, in the above notation, $\dim V=|V|$. In words: the dimension of the product of an infinite family of nonzero vector spaces is equal to its cardinal. - -As a corollary, let us express explicitly the dimension of the product $V$ of the $V_i$ in terms of the $d_i:=\dim V_i$. Setting -$$ -\mu:=\max(\aleph_0,|K|),\quad\alpha:=|\{i\in I\ |\ d_i < \mu\}|, -$$ -we get -$$ -\dim\prod_{i\in I}V_i=|K|^\alpha\prod_{d_i\ge\mu}d_i. -$$ -Let us prove the Main Theorem. - -Lemma. If $V$ is a vector space which is infinite as a set, then we have - $$ -|V|=|K|\cdot\dim V. -$$ - -Proof. It is easy to see that, if $S$ is an infinite generating subset of a group $G$, then $S$ and $G$ are equipotent. Putting -$$ -G:=V,\qquad S:=\{\lambda b\ |\ (\lambda,b)\in K\times B\}, -$$ -where $B$ is a basis of $V$, we get the conclusion. QED -Let $V$ be an infinite dimensional vector space. -Say that $V$ is large if $\dim V\ge \max (|K|, \aleph_0)$. -By the lemma, $V$ is large if and only if $\dim V=|V|$. - -Erdős-Kaplansky Theorem. The vector space $K^{\mathbb N}$ is large. - -It is clear that the Erdős-Kaplansky Theorem implies the Main Theorem. So we are left with proving the Erdős-Kaplansky Theorem. -Proof of the Erdős-Kaplansky Theorem. Let $B$ be a $K$-basis of $K^{\mathbb N}$, and suppose by contradiction $|B|<|K|$. Let $K_0$ be the prime subfield of $K$, and put -$$ -K_1:=K_0(\{b_j\ |\ b\in B,\ j\in\mathbb N\}). -$$ -As $|K_1|<|K|$ and $K$ is infinite, there is an $x$ in $K^{\mathbb N}$ whose coordinates are $K_1$-linearly independent. There are $c_1,\dots,c_n$ in $B$ such that $x$ is a $K$-linear combination of the $c_j$. Since $c_{ij}$ is in $K_1$ for all $i,j$, there is a nonzero $\lambda$ in $K_1^{n+1}$ such that -$$ -\sum_{j=0}^n\lambda_j\,c_{ij}=0 -$$ -for $1\le i\le n$, and we have -$$ -\sum_{j=0}^n\lambda_j\,x_j=0, -$$ -in contradiction with the choice of $x$. QED<|endoftext|> -TITLE: Is the following graph well known? -QUESTION [7 upvotes]: The vertices are all (ordered) $k$-tuples with distinct components from a universe of size $n$. -Two vertices are adjacent if their Hamming distance is 1 (i.e. if they differ in exactly one component). - -Notice that this graph has $\frac{n!}{(n-k)!}$ vertices. It is an $k(n-k)$-regular graph. -Does the graph have a name? -A closely related graph is the Hamming graph (denoted by $H(n,k)$), which does not exclude the tuples with duplicate components. It has $k^n$ vertices. It is closely related because the graph I am looking for is the induced subgraph of $H(n,k)$ after removing the vertices corresponding to tuples with repeat components. -Another similar graph is the Johnson graph, whose vertices are all subsets of size $k$ from a universe of size n with an edge between two vertices if their set intersection is $n-1$. It has $\binom{n}{k}$ vertices. -The Kneser graph (and its generalization) have similar definitions as well, but like the Johnson graph, they also have $\binom{n}{k}$ vertices. - -REPLY [5 votes]: They might have a name, I don't know. For the next few lines let us call each a Partial Permutation graph $PP(n,k)$ (assume $k4$ and $k=2$. In all 4 cases the maximum distance is 2. In $J(n,2)$ any two non-adjacent vertices such as $\lbrace a,b \rbrace$ & $\lbrace c,d \rbrace$ have 4 mutual neighbors. In $H(n,2)$, $[a,b]$ & $[c,d]$ have two mutual neighbors ($[a,d]$ and $[c,b]$) as do $[a,a]$ & $[c,d]$ or $[a,a]$ & $ [c,c]$ and more importantly $[a,b]$ & $[c,a]$ as well as $[a,b]$ & $[b,a]$. I'll leave Knesser graphs for the reader to consider. In $PP(n,2)$ $[a,b]$ & $[c,d]$ still have two mutual neighbors but $[a,b]$ & $[c,a]$ only have one while $[a,b]$ & $[b,a]$ have none. So there are three ways to be at distance $2$. It is a 5-class association scheme of diameter 2 however. For $k=3$ or $k=4$ there begin to be a great number of associate classes. If I calculate correctly, for $n>2k$ $PP(n,k)$ has diameter $k$ but $\binom{k+1}{2}-1$ associate classes (or orbits on pairs of vertices under the action of the automorphism group if you prefer) -In the special case that $k=n$ one would have a graph with $n!$ vertices each of degree $\binom n2$ (one would have to let adjacency be differing in only two positions). Since that is a Cayley graph for $S_n$, it is distance transitive. In the more restricted case that adjacency is that two permutations differ only by the swap of two adjacent positions it is the skeleton of the permutohedron. -later I certainly looked at those graphs at some point. At the time I did not realize that they (seem) to always have all eigenvalues integral (based on a criminally small number of test cases). I wonder if there is an easy way to see that happens (if it does..) -much later The graph PP(7,3) actually has 10 associate classes. I had fun so here are the details. A disclaimer, I'm sure that group representation methods are much more efficient if one knows how to use them.I found the classes by making the 210 by 210 adjacency matrix then raising it to the 4th power. Of the 44100 entries, 10 distinct values occur and they reveal what the classes are (although once you know it is obvious!). Then the adjacency matrix (or simple combinatorics) reveals the numbers. The relation depends on the number of entries equal in value and in the same place and also the number of entries equal but not in the same place. And then more. Note that with respect to 123, both 214 and 241 have one new symbol, but 123 and 214 are at distance 4 (say 123 124 154 254 214) while 123 143 243 241 is a distance 3 path. -distance 0: 123 (itself) -distance 1: 124 -distance 2: 134 145 -distance 3: 234 245 456 132 -distance 4: 214 231 -distance 0: 123 (itself) -distance 1: 124 -distance 2: 145 134 -distance 3: 132 456 451 432 -distance 4: 214 231 -Each vertex (such as u=123) is adjacent to 12 other vertices. If v is a vertex in class i with respect to u then row i in this matrix shows in column j how many of the 12 neighbors of u are in class j with respect to v. Then we find that the eigenvalues of this small matrix are [12,8,5,4,3,1,0,-3,-3,-3] The repeated eigenvalue -3 was what made me suspect that there were 8 classes. ` -$$ \left[ \begin {array}{cccccccccc} -0&12&0&0&0&0&0&0&0&0 -\\ 1&3&6&2&0&0&0&0&0&0\\ 0&2&4&2&0 -&2&2&0&0&0\\ 0&1&3&3&1&0&3&1&0&0 -\\ 0&0&0&8&0&0&0&0&4&0\\ 0&0&3&0&0 -&3&6&0&0&0\\ 0&0&1&1&0&2&5&2&1&0 -\\ 0&0&0&1&0&0&6&3&1&1\\ 0&0&0&0&1 -&0&6&2&3&0\\ 0&0&0&0&0&0&0&12&0&0\end {array}\right] $$ -Further analysis is possible, such as finding nice eigenvectors and substructures, but I still suspect that someone knows all this.<|endoftext|> -TITLE: Quotient space of $\mathbb{C}^5$ under the action of $SL(2,\mathbb{C})$ -QUESTION [10 upvotes]: One sees that given the $SL(2,\mathbb{C})$ action on $\mathbb{C}^5$, thought of as the space of polynomials of the form, -$$a_0 x^4 + 4a_1 x^3 y + 6a_2x^2y^2 + 4a_3xy^3 + a_4 y^4$$ -the ring of invariants is generated by the following functions, -$$g_2(a) = a_0a_4 - 4a_1 a_3 + 3a_2^2$$ -and -$$g_3(a) = a_0a_2a_4 - a_0a_3^2 - a_1^2a_4 + 2a_1a_2a_3 - a_2^3$$ -But if these $g_2$ and $g_3$ satisfy the discriminant $=0$ condition then there are inequivalent $SL(2,\mathbb{C})$ polynomials which map to the same $(g_2,g_3)$ point. -But if I look at say Theorem 5.9 in the book by Mukai then I get to see that the closure equivalent classes of orbits of the action of a linearly reductive group like $SL(2,\mathbb{C})$ on $\mathbb{C}^5$ are in bijective correspondence to the the points of $\mathbb{C}^5//SL(2,\mathbb{C})$ (which is defined as the spectrum of the invariant polynomials under $SL(2,\mathbb{C})$) -Also look at the theorem at the end of page 11 of this paper. -In the above paper "//" is defined as identifying points in the affine variety if one lies in the closure of the orbit through the other. - -Are these two notions of "//" equivalent? If yes, how? -In the light of the above two theorems, can one say that the $SU(2)$ invariant polynomials among binary homogeneous quartics are in bijection with those closure equivalent classes of orbits of $SU(2)^{\mathbb{C}} = SL(2,\mathbb{C})$ which are labeled by the pairs of invariants $(g_2,g_3)$ such that $g_2^3 - 27 g_3^2 \neq 0$ ? -Hence if I am interested in only closure equivalent orbits can I just forget those pairs of values of the invariants which lie on the discriminant $0$ curve? -Conversely given a $(g_2,g_3)$ for which the above discriminant condition is satisfied can one write down the family of $SU(2)$ invariant polynomials explicitly? -I would anyway like to know how to distinguish the orbits corresponding to the discriminant $0$ condition. - - -In light of the various extremely helpful and references that have come up, I realize that there there is a notion of a "discriminant" for homogeneous polynomials of degree $d$ in $n$ variables. (call this space of polynomials as $P(n,d)$) This discriminant is in some sense a "homogeneous invariant" and for the $n=2$ case in which I am interested in, it generates the subalgebra of the coordinate ring over of these polynomials which is invariant under this group action (call that $\mathbb{C}[P(n=2,d=4)]^{SL(2,\mathbb{C})}$). (this is lucky!) -I guess the discriminant in this case has to be a polynomial in polynomials of homogeneous degree $4$ in $2$ variables. The above I guess implies that $\mathbb{C}[P(n=2,d=4)]^{SL(2,\mathbb{C})}$ is generated by just one such polynomial in polynomials. -I would like to know how is a ``discriminant" defined for such polynomials. (searching and asking around I am only getting definitions for the single variable case) -I want to know if knowing the generator of $\mathbb{C}[P(n=2,d=4)]^{SL(2,\mathbb{C})}$ tells me about the initial objective of knowing $P(n=2,d=4)\text{ }mod\text{ }SL(2,\mathbb{C})$ -I wonder if this unique generator of the invariant subalgebra is related to the null-cone that was pointed out by Bart in his comment. -Also I would like to be pointed out if there is any mistake in what I said above! - -I had recently tried asking a similar question here. But I think I could not precisely convey what I was looking for. Let me here try to give a specific situation that I need to understand coming from certain other considerations in Superconformal Quantum Field Theories. -I can think of $\mathbb{C}^5$ as being the space of all homogeneous degree $4$ polynomials in $2$ variables. On this space $SL(2,\mathbb{C})$ has the standard action. -I want to know what is the most explicit (or the best!) way to describe the quotient space thus obtained. I want to understand how do the orbits look like. -[EDIT: I was initially asking if there exists fixed subspaces etc but then from the comments I realized that I was missing the elementary fact that it is an irreducible representation! Hence nothing like this can exist.] -I tried something naive. I wrote down the most general element of $SL(2,\mathbb{C})$ using its canonical polar decomposition and then acted it on the most general homogeneous degree $4$ polynomial in $2$ variables and tried to see how the coefficients change. Unfortunately the equations are very complicated and I didn't see any hope of me being able to solve them to find the fixed points. -Apart from this specific example I would also like to know of references to simpler examples than this where a similar question is asked and answered. - -REPLY [15 votes]: The description of orbits in the $d=4$ and $n=2$ is not difficult. -It is better to work projectively i.e. look at the orbits in the -projective space $P(\mathbb{C}^5)$ -of nonzero binary quartics up to a constant. -A nonzero binary quartic $F$ can be written as a product of linear forms and therefore -corresponds to a collection of four points on the projective line with possible repetitions -but no ordering. If all points are distinct one can transform them by an $SL_2$ element -into 0, 1, $\infty$ and some other guy $\lambda$. Orbits in this case are in one-to-one -correspondence with the $j$-invariant -$$j(F)=\frac{4}{27}\times\frac{(\lambda^2-\lambda+1)^3}{\lambda^2(\lambda-1)^2}$$ -Essentially $\lambda$ is the cross ratio of the four points but this depends on the chosen -ordering. The above rational fraction is what is needed in order to kill the dependence on the ordering. The same $j$-invariant can be expressed in terms of invariants of $F$: -$$ -j(F)=\frac{S^3}{S^3-27 T^2} -$$ -where $S$ is the invariant of degree 2 and $T$ is the invariant of degree 3 (both defined up to normalization by a constant). They generate the ring of invariants you are talking about. -What you said above is false: this ring is not generated by the discriminant which is not -a fundamental invariant. It is given by the denominator $S^3-27 T^2$. -Finally the other orbits correspond to three points or less. These points can be placed anywhere we want on the projective line by an $SL_2$ element. -So these orbits are characterized by the multiplicities $(2,1,1)$, $(2,2)$, -$(3,1)$ and $(4)$. The last two form the null cone of binary forms with a root -of multiplicity $>$ half the degree of the form. All invariants vanish on these ones. -What one needs to distinguish them are covariants: joint invariants of the form and -an extra auxiliary point. -For $n>2$ one needs mixed concomitants, i.e., joint invariants of the form and an extra auxiliary complete flag. - -What I said above is a description of the quotient $Q=P(\mathbb{C}^5)/G$ where $G=SL_2$. -Now if you want the affine version $A=\mathbb{C}^5/G$ it can be described as follows. -If $F$ is a nonzero binary quartic denote by $[F]$ the corresponding point in projective -space, i.e., its class up to multiplication by a nonzero constant. -On $\mathbb{C}^5$ one has the composition of maps $F\rightarrow [F] \rightarrow G[F]$ -which ends in $Q$. Clearly the preimage of each point in $Q$ is a union of orbits -in $\mathbb{C}^5$. Figuring out $A$ from the previous description of $Q$ can be done by looking -at each $G[F]$ and asking if different multiples of that $F$ are related by $G$. -The resulting list of affine orbits is: - -the orbit of $0$, with of course $S=T=0$. -the orbit of $x^4$ (all multiples are related by $G$), with $S=T=0$. -the orbit of $x^3 y$ (all multiples are related by $G$), with $S=T=0$. -orbits of $6\alpha x^2 y^2$, $\alpha\neq 0$, with $(S,T)=(3\alpha^2, -\alpha^3)$. Here different $\alpha$'s give different orbits. -orbits of $12\beta x^2 y (x+y)$, $\beta\neq 0$, with $(S,T)=(12\beta^2, -8\beta^3)$. Here different $\beta$'s give different orbits. -orbits of $\gamma xy(x-y)(x-\lambda y)$ with $\gamma\neq 0$, and $\lambda$ in a fundamental domain of the complex plane minus the points $0,1,-1,2,\frac{1}{2}, -\omega, --\omega^2$, where $\omega=e^{\frac{2i\pi}{3}}$, with respect to the group ($\simeq S_3$) of six transformations -generated by $\lambda\rightarrow 1-\lambda$ and $\lambda\rightarrow \frac{1}{\lambda}$. -One then has -$$ -(S,T)=\left(\frac{\gamma^2}{12} (\lambda^2-\lambda+1) -,\frac{\gamma^3}{2^4\times 3^3}(\lambda+1)(2\lambda^2-5\lambda+2)\right) -$$ -Again different pairs $(\gamma,\lambda)$ give different orbits. -orbits of $\mu xy(x-y)(x+\omega y)$, with $\mu$ in a fundamental domain of $\mathbb{C}\backslash\{0\}$ with respect to multiplication by a cube root of unity. -One then has -$$ -(S,T)=(0,\frac{i\sqrt{3}}{2^4\times 3^2} \mu^3) -$$ -One needs this fundamental domain to avoid repetition because multiplying such a form -by $\omega$ does give an $SL_2$ equivalent form, and this only happens when one multiplies -by a cube root of unity. -orbits of $\nu xy(x-y)(x+y)$, with $\nu$ in a fundamental domain of $\mathbb{C}\backslash\{0\}$ with respect to multiplication by $-1$. -One has $(S,T)=(\frac{\nu^2}{4},0)$. -Same remark that minus such a form is $SL_2$ equivalent to the original one. - -Now if you look at the map from $A$ to $\mathbb{C}^2$ given by the pair $(S,T)$ -you see the following. -Each pair away from the $S^3-27 T^2=0$ discriminant curve has exactly one preimage. -This corresponds to the cases $6,7,8$ above. The last two are called equianharmonic and harmonic configurations respectively. -Then if $(S,T)\neq (0,0)$ is on the discriminant curve, one has exactly two preimages, -cases $4,5$ with $\alpha=2\beta$. Finally the point $(0,0)$ has three preimages, the cases -$1,2,3$. -If you want to distinguish between the first five cases invariants will not do, you need covariants. These cases are examples of so called coincident root loci, you can lookup the papers by my collaborator J. Chipalkatti for the equations for such things. -See also Polynomial with two repeated roots -Case $2$ is characterized by the vanishing of the Hessian of $F$, and the inequality $F\neq 0$. -You can find equations for case $3$ in -my article with Chipalkatti: -"The bipartite Brill-Gordan locus and angular momentum". Transform. Groups 11 (2006), no. 3, 341--370. A preprint version is http://arxiv.org/abs/math.AG/0502542 -(you have to add the inequality that the Hessian is not zero). -For case $4$ you can find defining equations in -this other article with Chipalkatti: -"Brill-Gordan Loci, transvectants and an analogue of the Foulkes conjecture". Adv. Math. 208 (2007), no. 2, 491--520. A preprint version is http://arxiv.org/abs/math.AG/0411110<|endoftext|> -TITLE: What is the virtue of profinite groups as mathematical objects? -QUESTION [20 upvotes]: In my own research I use profinite groups quite frequently (for Galois groups and etale fundamental groups). However my use of them amounts to book-keeping: I only care about finite levels (finite Galois extensions; finite covers) and so I take their inverse limit. Then various "topological" arguments just mean: look at the finite levels (this occurs because the quotients by the open subgroups exactly correspond to the finite levels I care about.) -There are people who use profinite groups more intently than I do, and I suspect that they see some value in them as mathematical objects. So my question is this: -Question -What theorems/properties are good about profinite groups that don't arise trivially from its book-keeping nature (so for example the profinite Sylow theorems are disqualified, because they arise trivially from the finite-level group theory). What, if anything, rewards us for dealing with this new type of object rather than with the finite levels? (again, except that it makes it easier to write down notes.) - -REPLY [8 votes]: Gildenhuys, Ribes and Zakesskii and others have developed a Bass-Serre theory for profinite groups acting on profinite trees. Using this theory, Ribes and Zalesskii showed that if $H_1,\ldots,H_n$ are finitely generated subgroups of a free group then the subset $H_1\cdots H_n$ is closed in the profinite topology. This was a conjecture by people working in semigroup theory and automata theory that is essentially equivalent to a conjecture of Rhodes in semigroup theory. The Ribes and Zalesskii proof does not use approximation by finite groups, but rather the geometry of these profinite trees. Other proofs using geometric group theory and using model theory now exist. -Recently Almeida assigned profinite groups to irreducible symbolic systems in a way that is functorial up to inner automorphism. Again finite groups do not explicitly appear in the discussion. -UPDATE: Another aspect of the theory of profinite groups as objects in their own right is the study of just infinite profinite groups. An infinite profinite group is just infinite if all its non-trivial closed normal subgroups are open. For example, the p-adic integers are just infinite. Just infinite is the analogue of simple for infinite profinite groups. Every finitely generated infinite profinite group has a just infinite quotient. There is a trichotomy due to Wilson (and refined by Grigorchuk) describing what they can look like. The study of just infinite profinite groups is connected to the theory of profinite branch groups and actions on rooted trees. See the handbook chapter by Bartholdi, Grigorchuk and Sunik.<|endoftext|> -TITLE: Direct construction of the Stone-Čech compactification using ultrafilters? -QUESTION [13 upvotes]: If $X$ is a set (regarded as a discrete space), its Stone-Čech compactification can be identified with the set of ultrafilters on $X$ with its natural (Stone) topology. If $X$ is a general topological space, given any continuous function $f : X \to C$ from $X$ to a compact Hausdorff space $C$, we can push forward ultrafilters on $X$ to ultrafilters on $C$, which must have unique limits. So define an equivalence relation on ultrafilters as follows: $F \sim F'$ if their pushforwards under all continuous functions from $X$ to some compact Hausdorff space $C$ have the same limits. Then it seems to me that the set of equivalence classes of ultrafilters on $X$ under $\sim$, with an appropriate topology (perhaps the quotient topology from the space of ultrafilters on $X$?), ought to be the Stone-Čech compactification of $X$ in general. -So does this construction actually work, and if so, is there a simpler definition of the equivalence relation $\sim$? -Edit: also, if this construction works, is it explicitly written down anywhere in the literature? The references I found seem to work with a different notion of ultrafilter (ultrafilters of zero sets or something like that), and I'm wondering whether (or why) this is necessary. -Edit #2: Okay, so as it stands, the problem that I see with the definition of $\sim$ is that I am attempting to quantify over the class of compact Hausdorff spaces. How do I fix this? - -REPLY [10 votes]: This is actually somewhat complicated in general, and requires some hypotheses on the space $X$. Namely, $X$ must be completely regular, so if your construction does not use this hypothesis it will not in general give the Stone-Cech compactification. See section 1.34 onward of Russel C. Walker's book "The Stone-Cech Compactification." I believe the ultrafilter construction is initially due to Waldhausen, but Walker's exposition is very readable and I can't find the Waldhausen paper. -The upshot is that in general one wants to consider ultrafilters of zero sets of $X$, rather than ultrafilters of closed sets in general. This is necessary because the closed sets of a non-normal space don't behave nicely--in particular, they aren't separated from one another by continuous real-valued functions. But if you look at the universal property of the Stone-Cech compactification, this is exactly the relevant type of separation, so one needs to look at zero sets instead. -EDIT: In response to Qiaochu's comment, I'm going to make some fuzzy remarks on how much of the topology of a topological space $X$ the category Top sees, as opposed to how much of the topology of $X$ the category of compact Hausdorff spaces sees (call this KHaus). I won't be as formal as possible because I think doing so would risk obfuscating some subtle points--but to point in the direction of a formalization, note that there is a functor $F: Top\to [Top, Sets]$ (where $[-,-]$ denotes the functor category), and a functor $G: Top\to [KHaus, Sets]$ where both functors are given by $X\mapsto \operatorname{Hom}(X, -)$. The question I want to address is---how much of the topology on $X$ can we recover from the functors $F(X), G(X)$? -Now for the first question, Top sees everything about a topological space $X$. This is obvious from the Yoneda lemma, but more explicitly, one can consider the two-point space $A=\{x, y\}$, where the open sets are $\emptyset, \{x\}, \{x, y\}$. Then $\operatorname{Hom}(X, A)$ gives exactly the open/closed sets of $X$, and one can even extract the (closed) sets themselves by taking the pullback of diagrams of the form $X\to A\\leftarrow \{y\}$. -But note that the space $A$ is not contained in KHaus, since $x$ is not a closed point. One should think of KHaus as only seeing "zero sets," or more generally level sets, of functions on $X$. Indeed, given any function $f: X\to Y$ in KHaus, one may extract the level set $f^{-1}(y)$ by taking the pullback of $X\to Y\leftarrow y$ (which may not exist in KHaus, but one can see the functor it represents). On the other hand, if $S$ is a closed subset of $X$ which is not the level set of any function, how do we see it? For example, $\operatorname{Hom}(A, Y)=\operatorname{Hom}(pt, Y)$ for any compact Hausdorff space $Y$---KHaus does not distinguish these two spaces. -The upshot is that KHaus only "sees" zero sets (or more generally, level sets of functions), so any natural functor defined out of KHaus will be built from them, rather than arbitrary closed sets. -EDIT 2: This is an attempt to address Qiaochu's Edit #2. I claim that it suffices to quantify over spaces of cardinality at most that of $X^{\mathbb{N}}$. Indeed, let $C$ be any compact Hausdorff space, and $f: X\to C$ a function. Then it factors through $\overline{f(X)}$ which is closed in $C$ and thus compact Hausdorff, and has cardinality at most $X^\mathbb{N}$. So it suffices to quantify over compact Hausdorff spaces of bounded cardinality. -To see the claim about the cardinality of $\overline{f(X)}$, note that $f(X)$ has cardinality at most that of $X$, and there is a surjective map from convergent sequences in $f(X)$ to $\overline{f(X)}$ (namely taking the limit). But convergent sequences in $f(X)$ are a subset of $f(X)^{\mathbb{N}}$. -Now the class of isomorphism classes of compact Hausdorff spaces of bounded cardinality (or indeed, of topological spaces of bounded cardinality) is clearly a set, which lets the quantification go through. -EDIT 3: Qiaochu points out that I assumed the existence of a locally countable base in the last edit---we can't say that every point is the limit of a sequence; instead one needs to say it's the limit of an ultrafilter. This gives a larger bound on cardinality, but still a bound.<|endoftext|> -TITLE: Omega_{1} unions of null sets: Martin's Axiom -QUESTION [6 upvotes]: Hello, -let $U$ be the assertion "The union of $\aleph_{1}$ null sets of reals is null", i.e. -$U$ = Given any $\omega_{1}$-sequence of null sets $X_{\alpha}$, for $\alpha<\omega_{1}$, then $\bigcup_{\alpha <\omega_{1}} X_{\alpha}$ is null. -$U$ is known to be independent of ZFC. -It clearly does not hold if $CH$ holds. However it follows, for example, from -$MA_{\aleph_{1}}$ (Martin's Axiom at $\aleph_{1}$). -Three useful consequences of $U$ (and therefore of $MA_{\aleph_{1}}$ as well) are listed below. Let $\Omega$ be the set of Lebesgue measurable sets of reals in [0,1] and $\mu$ the uniform measure on [0,1]. Then under $U$: -1) $\omega_{1}$-completeness of $\Omega$: for any $\omega_{1}$-sequence of disjoint measurable sets $X_{\alpha}\in \Omega$, $\bigcup_{\alpha < \omega_{1}} X_{\alpha} \in \Omega$. -2) $\omega_{1}$-continuity of $\mu$: $\bigsqcup_{\alpha <\omega_{1}} \mu (X_{\alpha})= \mu ( \bigcup_{\alpha<\omega_{1}} X_{\alpha} ) $, where $\bigsqcup$ is the join operation. -3) Every $\Delta^{1}_{2}$ set is in $\Sigma$. -Now i recently wrote a paper (in computer science) that uses $U$ as hypothesis for the main theorem. I specified that the proof is valid in $ZFC+MA_{\aleph_{1}}$. However I would like to know if: -A) Has the assertion $U$ been studied independently from $MA$? Does it have a distinguished name? -B) is U strictly weaker than $MA_{\aleph_{1}}$? I guess it is, but don't know. -C) Do you have any reference to papers/book chapters/etc where the problem $U$ is considered and analyzed? -Than you in advance, -Matteo Mio - -REPLY [9 votes]: These questions and many other similar questions are -intensely studied in the field known as cardinal -characteristics of the continuum. Some ideas are mentioned -in this MO -answer (and -also in this MO answer and this one) and -you may consult Andreas Blass's excellent survey -article. -In particular, the answer to all your questions is Yes. -Your statement $U$ is exactly expressing that the -additivity of the null ideal is at least $\omega_2$. -The additivity of an ideal is the smallest cardinal $\kappa$ for which the union of any fewer than -$\kappa$ many sets in the ideal is still in the ideal. Since the union of countably many measure $0$ sets has measure $0$, the additivity of the Lebesgue null ideal is always at least $\omega_1$, but as you have observed, it is consistent with ZFC that it is larger. -Other similar cardinal characteristics would include the -the additivity of the meager ideal, the covering number of the null ideal (the smallest number of null sets covering -$\mathbb{R}$), the covering number of the meager ideal, and many others. -These cardinal characteristics and others appear in -Cichon's -diagram, -which exhibits and explores their relationships. -Under the Continuum Hypothesis, all these cardinal -characteristics agree, and have value continuum, but when -CH fails, it is possible to separate them consistently with -ZFC. Martin's Axiom pushes all these cardinal characteristics to the top, with value continuum. (In particular, under MA with large continuum one can strengthen your statement $U$ to allow that the union of $\aleph_{17}$ many null sets is null, or $\aleph_{\omega^2+5}$ many, etc.) To separate the cardinals from each other, therefore, rather than merely just from $\omega_1$, one uses forcing, and the most powerful results are obtained by forcing constructions carefully designed specifically for the purpose. Indeed, the study of fundamental -questions in the field of cardinal characteristics of the continuum, concerning the problems of separating the cardinal characteristics, led to important advances in the theory of forcing, such as a -much deeper understanding of forcing iterations and perhaps -even to the rise of proper forcing.<|endoftext|> -TITLE: Is this a (well known) open problem?(infinitness and more on $anm \pm n\pm m$ ) -QUESTION [19 upvotes]: Consider the following question: - -1) For a given natural number $a$, are there finitely or infinitely many natural numbers that are not of the form $anm \pm n\pm m$, where $m$ and $n$ range over positive integers? (For $a=1$ or $a=2$ you have all the natural numbers.) - -Does this problem appear in the literature? -As one can see at MO Scribe's question Chen's Theorem with congruence conditions. it is the same like asking if there are infinitely many $k$ such that both $ak+1$ and $ak-1$ do not have any non-trivial factors of the form $\pm 1 \mod a$. -I give a proof that for $a=6$ the question is equivalent to the twin prime conjecture so it is known that we don't have any proof. But what about other values of $a$? Is the problem for $a=100$ or more of the same difficulty? - -2) Is the density (Szemeredi's or Schnirelman's), of the numbers that are not of the form, zero for any value of $a$? -3) From Viggo Brun's theorem we have that the sum of the reciprocals of the twin primes converges. Does the sum of the reciprocals for any value of $a$ of the numbers that are not of this form converge? -4)For a given natural number $a$ are there infinitely many $k$ such that both $ak+1$ , $ak-1$ do not have any $prime$ factors of the form $\pm 1 \mod a$? (the same questions for these $k$ as in 2) and 3) ) -5) And the most easy : for which $a$ do we have a proof that there are infinetely many $k$ such that such that both $ak+1$ , $ak-1$ are either prime or can be written as a product of two numbers both not of the form $\pm 1 \mod a$? I guess that if $φ(a)$ is big enough we can have such a proof (the same questions for these $k$ as in 2) and 3) ) - -NOTE: In 1) and 4) both both $ak+1$ and $ak-1$ can be either primes or the product of primes not of the form $\pm 1 \mod a$ ,but in the 1) no subproduct of them can be of this form. -As MO Scribe noticed the conjectural answer is that there should be infinetely many such pairs because we are aspecting to have infinitely many prime pairs of any reasonable congruence condition. -https://math.stackexchange.com/questions/15075/do-we-have-a-proof-of-the-infiniteness - -There are infinitely many twin primes if and only if there are infinitely many natural numbers that are not of the form $6nm \pm n \pm m$. -Proof: Every number that is not a multiple of $2$ or $3$ is of the form $6N\pm 1$. So the only pairs that are not divisible by $2$ or $3$ are $(6N-1,6N+1)$ for any $N$. Now are there infinitely many such prime pairs (twin primes)? -If the number $6N-1$ is prime it should not be written as a product of some numbers $6n+1,6m-1$ for any $n,m > 0$. So $(6n+1)(6m-1)=6(6nm-n+m)-1$, which means that $N$ should not be of the form $6nm-n+m$ for any $n,m>0$. -Similarly, if $6N+1$ is a prime it should not be a product of some numbers $(6n-1)(6m-1) =6(6nm-n-m)+1$, or $(6n+1)(6m+1) =6(6nm+n+m)+1$. Which means that we have a prime couple of the form $(6N-1,6N+1)$ if and only if $N$ is not of the form $6nm \pm n \pm m$ for any $n,m$. -NOTE: After i have edited this observasion-question I realised that it is well known that for $a=6$ it is equivalent to twin prime conjecture (as it is written on the answer by Luis below too), a notice that S.Golomb seems to have done first, but my question is focused to the other values of $a$. -If someone want to add something http://mathoverflow.tqft.net/discussion/921/reedited/#Item_1 - -REPLY [22 votes]: $\newcommand\Z{\mathbf{Z}}$ -$\newcommand\Q{\mathbf{Q}}$ -(Caveat: normally I wouldn't answer a question with such a limited knowledge of the general theory, but classical analytic number theory seems not so well represented by active MO members.) -Suppose that $A$ is a finite abelian group. Then I claim that given any set of at least $|A| + 1$ (not necessarily distinct) elements of $A$ one can find a proper subset whose sum is the identity. Proof: Denote the elements $a_i$ for $i = 1$ to $|A|+1$. By the pigeonhole principle, either one of the $|A|$ sums $\sum_{i=1}^{r} a_i$ for $r = 1$ to $|A|$ is the identity, or two of the sums are the same element of $|A|$, in which case consider the difference. We deduce from this the following: Let $n$ be any integer coprime to $a$ with more than $r:=|(\Z/a \Z)^{\times}|$ prime factors. Then $n$ has a proper divisor of the form $1 \mod a$. -Suppose that $k$ cannot be represented by the form $amn \pm m \pm n$, and suppose that $a > 2$. It is simple to deduce that this is equivalent to asking that $ak+1$ and $ak-1$ have no proper divisors of the form $\pm 1 \mod a$. It follows that $ak+1$ and $ak-1$ each have at most $r=|(\Z/a \Z)^{\times}|$ prime -factors. The integers with at most $r$ prime factors -are sometimes called $r$-almost primes. If $\pi_r(x)$ counts the number of $r$-almost -primes $\le x$ then -$$\pi_r(x) \sim \frac{x (\log \log x)^{r-1}}{\log(x)}.$$ -(Compare this to the prime number theorem when $r = 1$.) In particular, we see that the - $r$-almost primes have zero density (in any sense), and thus: - -2) The density of integers that can not be represented in the form $amn + m + n$ is zero. - Similarly, the density of integers that cannot be represented in the form $amn + m -n$ is zero. In particular, the density of the $a$-asterios numbers, the integers that can neither - be represented in the form $amn+m+n$ nor $amn+m-n$, is zero. - -Let $\pi_{r,2}(x)$ denote the number of twin $r$-almost primes less than or equal to $x$, that is, the number of integers $n \le x$ such that $n$ and $n+2$ -are both $r$-almost primes. -(For example, $\pi_{1,2}(x)$ counts the number of twin primes less than $x$.) -What do we know about this function? Brun was the first to give an upper bound -for $\pi_{r,2}(x)$ using sieving techniques. Refinements by others (in particular -Selberg) allowed one to obtain the estimate -$$\pi_{1,2}(x) \ll \frac{x}{(\log x)^2},$$ -which gives the correct (conjectural) order of magnitude. -Without going into the Selberg sieve, let me say that what these arguments -really -give is decent upper and lower bounds of the following kind (for large $x$): -$$\frac{A x}{(\log x)^2} < -\left\{n < x, \ p \nmid n(n+2) \ \text{if} \ p < x^{\alpha}\right\} -< \frac{B x}{(\log x)^2}$$ -for non-zero constants $A$ and $B$, -where $0 < \alpha < 1$ is some fixed small constant, which we might imagine for the -sake of argument is $1/10$. Since every twin prime $>x^{\alpha}$ contributes to -this sum, this gives the correct (up to a constant) upper bound for -$\pi_{1,2}(x)$. It also gives a lower bound for $\pi_{10,2}(x)$, since if every -factor of $n < x$ is at least $x^{1/10}$, then $n$ has at most $10$ prime factors. -The motto I learnt about sieving was the following: upper bounds are easy, lower bounds are hard. Thus, since we are interested in bounding $\pi_{r,2}(x)$, it seems that we are in good shape. -However, there is a subtlety here. Let $\pi(x,z)$ denote the number of integers $n$ less -than $x$ such that every prime factor of $n$ is at least $z$. It's clear by the -argument of the last paragraph that -$\pi_r(x) \ge \pi(x,x^{1/r})$. One might imagine that these numbers are roughly of the -same magnitute. However, it turns out that $\pi_r(x)$ is much bigger than $\pi(x,x^{1/r})$. -The latter is comparable to the number of primes less than $x$, where the former has -an extra factor of $(\log \log x)^{r-1}$. The reason is that $\pi_r(x)$ is dominated -by numbers with (a few) small prime factors. In fact, -as Kowalski pointed out to me, it is not even obvious that -one can easily obtain the -correct upper bound for $\pi_r(x)$ simply by sieving over primes. -From the asymptotic for $\pi_{r}(x)$, one expects that -$$(*): \qquad \pi_{r,2}(x) =^{?} \ O\left(\frac{x (\log \log x)^{2r-2}}{(\log x)^2}\right).$$ -(EDIT: My resident expert reports that this is known. Here is a sketch of the idea in the simpler case where we want to count pairs $n$ and $n+2$ where $n$ is a $2$-almost prime and $n+2$ is prime. -First, for a small prime $p$, we want to find an upper bound for the number of $n < x$ such that $n$ is divisible by $p$ and both $n/p$ and $n+2$ are prime. This is a similar problem to counting twin primes, and in a similar way one obtains a bound of the form $O(x/\log x)$ (key point: the implied constant does not depend on $p$). If we wish to bound the number of pairs $(n,n+2)$ such that $n+2$ is prime and $n$ is a $2$-almost prime, we may instead count the triples $(p,n,n+2)$ where -$p < x$ is prime, $n < x$ is divisible by $p$, $n/p$ is prime, and $n+2$ is prime. If for each $p < x$ we have a upper bound of $Ax/\log x$ (for the same $A$), in total we obtain the upper bound: -$$ \frac{Ax}{\log x} \cdot \sum_{p < x} \frac{1}{p} \sim \frac{Ax \log \log x}{\log x}.$$ -Of course, the devil is in the details! END EDIT) -All one needs to answer 3) is that - the exponent of $\log(x)$ in the denominator is $> 1$. - -3). Assuming the expected result (*), the inverse sum of the $a$-asterios primes converges. - -Consider the set of integers $S_a$ -which do not have any prime factors of the form -$\pm 1 \mod a$. This is a reasonable thing to do whenever -$|(\Z/a\Z)^{\times}| > 2$. This is a weaker condition, so there are more of these numbers -and consequently obtaining upper bounds is harder. -We may form the Dirichlet series -$$L(s) = \sum_{S_a} \frac{1}{n^s}$$ -which has an Euler product: -$$L(s) = \prod_{p \not\equiv \pm 1} \left(1 - \frac{1}{p^s}\right)^{-1}$$ -Now let $K = \Q(\zeta_a)^{+}$ be the totally real subfield of $\Q(\zeta_a)$. -It has degree $r/2 = \phi(a)/2$, where $r > 2$ unless $a = 1,2,3,4$ or $6$. -(What we say now only makes sense for $r \ge 2$, in which case $r/2 \in \Z$.) -A prime splits completely in $K$ if and only if it is of the form -$\pm 1 \mod a$. Looking at the Euler product of $\zeta_K(s)$, we see that, -up to a constant - which can be explicitly written as some product over primes, -$$\zeta_K(s) L(s)^{r/2} \sim \zeta_{\Q}(s)^{r/2}$$ -as $s \rightarrow 1^{+}$, and hence $L(s) \sim (s-1)^{(2-r)/r}$ -(up to some constant) - as $s \rightarrow 1^{+}$. We deduce (Perron's formula) that the number -of integers $\le x$ -all of whose prime factors are not of the form $\pm 1 \mod a$ is -asymptotic to -$$ \kappa \cdot \frac{x}{(\log x)^{2/r}},$$ -for some non-zero -constant $\kappa$. This is the same analysis that gives the asymptotic -formula for the number of integers $\le x$ which can be written as a sum -of two squares (a result of Landau). -We immediately deduce: - -4a) The number of integers $a$ such that $ak+1$ (or $ak-1$) has no - prime factors of the form $\pm 1 \mod a$ has zero density. - -If $r > 2$ (so $a \ge 3$ and $a \ne 3,4,6$) then the power $2/r$ of - $(\log x)$ is at most $1/2$. Thus, we actually -are led to the following guess: - -4b) If $r = \phi(a) > 2$, then one would heuristically expect the inverse - sum of integers $k$ such that none of the prime factors of $ak-1$ and $ak+1$ - are $\pm 1 \mod a$ diverges. If $r = 2$, so $a = 3$, $4$, or $6$, then (Brun) the series converges. -Here is a related problem of the very same kind: can one count the number of - integers $n \le x$ such that both $n$ and $n+1$ can be expressed as the sum of - two squares, and prove that there are $\sim x/\log(x)$ such integers - (perhaps up to non-zero constant factors)? - -Any integer $n$ can be written as the product of two numbers not -of the form $\pm 1$ unless every prime factor of $n$ is of the -form $\pm 1 \mod a$. The integers all of whose prime factors are -$\pm 1 \mod a$ can be analyzed exactly as in the last paragraph. -In this case, the number of integers all of whose prime factors are of the -form $\pm 1 \mod a$ is asymptotic to -$$ \kappa \cdot \frac{x}{(\log x)^{(1-2/r)}}.$$ -Suppose that $r > 2$. Then the set of such integers has density zero, and thus the -set of integers which have a factor not of the form $\pm 1 \mod a$ -has density one. Any set of density one has infinitely many -"twins" satisfying any fixed congruence condition. -Hence: - -5) If $r = \phi(a) > 2$, then there are infinitely many $k$ such that both - $ak+1$ and $ak-1$ have a factor not of the form $\pm 1 \mod a$. Indeed, such numbers have density one. - -Finally, I have nothing to say about problem 1) besides the remarks I made in my rephrasing of the original question here: -Chen's Theorem with congruence conditions.<|endoftext|> -TITLE: What kind of colimits are preserved by a certain Yoneda embedding? -QUESTION [6 upvotes]: (This question is related to this one) -Let $k$ be a field and consider the category $Sch/k$ of schemes over $k$, say also separable and of finite type. The Yoneda embedding -$$ -Y:Sch/k \to Pre(Sch/k) -$$ -does not respect colimits but if you factorize $Y$ through -$$ -Y:Sch/k \to Shv(Sch/k) -$$ -with respect to the Zariski Grothendieck topology (given by the Zariski open immersions), it does respect some colimits (if you want you may replace $Sch/k$ by the category of commutative algebras over $k$ of finite type). -In particular a pushout of the form -$$ - U~\xleftarrow{f}~ U\cap V ~\xrightarrow{g}~ V -$$ -in $Sch/k$ where $f$ and $g$ are open immersions is also a pushout in $Shv(Sch/k)$. -The pushout of two closed immersions -$$ - B\leftarrow A \rightarrow C -$$ -in $Sch/k$ exists in general but let's consider the situation where $A,B,C$ are affine because in this case the existence is 'immediate' (whatever that is) by the antiequivalence to $k$-algebras. For example the coordinate cross $Spec k[X,Y]/(XY)$ is the pushout of $ \mathbb{A^1}\leftarrow Spec k\to \mathbb{A^1}$. -My question is - -What happens to these 'closed' pushouts under the Yoneda embedding into Shv(Sch/k)? Are they preserved or is there an affine counterexample? - -Edit: What happens if one takes sheaves with respect to the etale topology? - -REPLY [4 votes]: No, the embedding of schemes in the big Zariski site does not preserve colimits. It is possible to see this in the example you suggest by computing the "tangent space" at the origin. Let $X$ be the colimit in the category of schemes; the tangent space of $X$ at the origin is a $2$-dimensional vector space. Let $Y$ be the colimit in the category of sheaves; I claim that the "tangent space" of $Y$ at the origin is the union of the coordinate axes inside of the tangent space of $X$. -The tangent space of $Y$ is the space of sections of $Y$ over $Z := \mathrm{Spec}\: k[\epsilon] / \epsilon^2$. The Zariski topology is trivial on $Z$, so sections of $Y$ over $Z$ are the same as sections of the presheaf colimit over $Z$. The "tangent space" of the presheaf colimit is the union of the tangent spaces of the two copies of $\mathbf{A}^1$.<|endoftext|> -TITLE: History of Irrationality results -QUESTION [18 upvotes]: The Greeks knew that numbers of the form $\sqrt{n}$ for nonsquare -integers $n$ are not rational. Much later, Lambert (1768) proved that -the values of $e^x$ and $\tan x$ are irrational for nonzero rational numbers $x$ (and conjectured that these values are transcendental, hence cannot be constructed using ruler and compass). My question is; what happened in between? -Here's what little I have found: - -Fibonacci (Flos) showed that the real root of $x^3 + 2x^2 + 10x = 20$ is neither -rational, nor the square root of a rational, nor equal to one of the other irrational -numbers occurring in Euclid X. -M. Stifel (Arithmetica integra, 1544) at least claimed that e.g. cube roots of noncube -integers are not rational. -Fermat claimed to have a proof that if $a$ and $b$ are positive rational numbers such -that $a^2 + b^2 = 2(a+b)x + x^2$, then neither $x$ nor $x^2$ are rational. - -Apart from occasional claims that Euler proved the irrationality of $e$ there seem to be no results in this direction between Euclid and Lambert. -Are there any irrationality proofs going beyond the square roots of integers and known before Euler and Lambert? -Edit Following up on Michael Hardy's suggestion, I haven't found anything predating Euler. On the other hand, Euler, in his Introductio in analysin infinitorum, claims that logarithms $\log_a b$ are "neither rational nor irrational" for integers $a, b > 1$. He does not prove that the logarithms are irrational (probably because he regarded it as trivial), and claims that they are not irrational (meaning it is not the square root of a nonsquare rational) since otherwise we would have $a^{\sqrt{m}} = b$, "which is impossible" (again, no proof, but this time it is not at all obvious but a very special case of Gelfond-Schneider). - -REPLY [7 votes]: One of the Pythogoreans, probably Hippasos of Metapont, showed the irrationality of $\sqrt 2$ about 500 BC. According to Platon, Theodorus of Cyrene knew that the square roots of all integers up to 17 are integers or irrationals, and Theaetetus proved this for all integers. Much of what Euclid writes about incommensurable magnitudes, distinguishing different kinds of irrationalities, stems from these scholars and Eudoxus of Knidos who also invented the theory of exhaustion. (Note that Euclid's understanding of irrational numbers is not the same as today, but this would deviate too much from answering this question.). Appolonius of Perga extended this theory, as can be obtained from an Arabic translation of a commentary of Pappus on Euclid's book X. No details, however, are known. -The Indian mathematician Bhāskarāchārya, a contemporary of Fibonacci, calculated square roots of sums of rational and irrational numbers and, like Fibonacci, treated polynomial equations of higher than second degree. Also in the 13th century Johannes de Sacrobosco (Johannes Campanus) proved the irrationality of the golden ratio (this result had been known to the old Greek but lost in the pass of times) by the method of descente infinie. -Simon Stevin, the inventor of the decimal system, knows that there are two cases which do not allow an exact decimal representation, fractions like 5/6 and irrationals. Marin Mersenne in a critique of a paper of Alfons Anton de Sarasa, mentions the difficulty to obtain by geometrical means the logarithm of a certain quantity, if two other logarithms are given, may they be rational or irrational. The correspondence between Leibniz and Newton is abundant with irrationalities and the praise of the own method to handle them better. But proofs of irrationalities are not contained. May it be that enough irrational numbers were already available, or that the proof of irrationality in case of logarithms is so easy (Euler mentiones it en passant). -The French algebraist de Lagny showed that a certain kind of polynomial equations have irrational roots (Histoire de l'Académie de Sciences, 1705, p. 294). -It is controversial whether Euler has implicitly proved the irrationality of $e$ and $\pi$ by means of continued fractions. Anyhow, his introductio in analysin infinitorum is full of irrationalities. Vol. 1, contains, in chapter 6, the assertion that with exception of the powers of the base, the logarithm of a number $h$ is not rational (and cannot be irrational, hence must be transcendental). In § 508 of vol. 2, he explains: The algebraic equations are either rational and do not contain other than integer exponents or irrational with broken exponents. But in the latter case, they can be made rational. If an equation of a graph neither is rational nor can be made rational, it is transcendental. If an equation contains powers, the exponents of which are neither integers nor fractions, it cannot be made rational. The graphs of these equations are the first and, so to say, simplest kind of transcendental graphs, namely such resulting from equations with irrational exponents. § 509 starts with the example $y = x^{\sqrt 2}$. -And then came Lambert. -Etymology -In Earliest Known Uses of Some of the Words of Mathematics we read: - Cajori (1919, page 68) writes, "It is worthy of note that Cassiodorius was the first writer to use the terms 'rational' and 'irrational' in the sense now current in arithmetic and algebra." The first citation of rational in the OED2 is by John Wallis in 1685. Irrational is used in English by Robert Recorde in 1551 in The Pathwaie to Knowledge: "Numbres and quantitees surde or irrationall." -M. Cantor in his Vorlesungen über die Geschichte der Mathematik, vol 2, credits the Italien (living is Spain) Gerard of Cremona (c. 1114 - 1187), the English mathematician and bishop Thomas Bradwardine (c. 1290 - 1349), the French mathematician and bishop Nicole d'Oresme (c. 1320 – 1382), and the German mathematician and bishop Albert of Saxony (c. 1320 - 1390) with early use of the word "irrational" in mathematical context, arguing how fast this notion spread in the world of mathematics in the 14th century. Nevertheless Vieta, Fermat, Newton and others used the word "asymetriae" or "quantitates surdae".<|endoftext|> -TITLE: How many cubes cover a bigger cube? -QUESTION [65 upvotes]: How many $n$-dimensional unit cubes -are needed to cover a cube with side -lengths $1+\epsilon$ for some -$\epsilon>0$? - -For n=1, the answer is obviously two. For n=2, the drawing below shows that three unit cubes suffice, but it is impossible using only two cubes. In general, a total of $n+1$ cubes is enough, as is shown in Agol's answer, but is this the smallest number possible? - -(source: folk.uio.no) - -REPLY [4 votes]: I had this very same question posted on my web page http://webhome.auburn.edu/~kuperwl/ for more than 20 years. To the best of my knowledge, the complete answer is still unknown.<|endoftext|> -TITLE: Reference for two facts about perverse sheaves on G/B -QUESTION [5 upvotes]: I wonder whether there is a reference for the following two things: -The Grothendieck group of B-equivariant semisimple? perverse sheaves on $G/B$ is the Hecke-algebra. -The category of B-equivariant perverse sheaves on $G/B$ is equivalent to those modules of category $\mathcal O$, where the center acts trivial. - -REPLY [2 votes]: A couple of things (I meant to just have this as a comment but it got too long): I think with regard to the first fact, a lot of people realised it all around the same time -- the article by Springer which Ben mentions does all the calculations needed, as he says, but doesn't state the result, while an article not long after by Lusztig and Vogan in 1983 states the result at the end, and it's mentioned elsewhere in the literature that Brylinski also proved it (and I can't believe Beilinson and Bernstein didn't know also). It's now, however, textbook stuff thanks to the book by Hotta Tanisaki and Takeuchi. -As to the second, I think Ben is overcomplicating things a little. If I say B-equivariant perverse sheaves, that will force the $\mathfrak h$-semisimplicity, which is what you need e beyond say the equivalence which is proved in the paper of Kashiwara-Brylinski. Of course Ben is right to point out that more care needs to be taken with the derived categories, but there are other references for that -- e.g. Bernstein and Lunts paper. The other classic reference that should be mentioned here is the paper of Beilinson and Bernstein on the Jantzen filtration, which discusses a lot of the issues related to central characters (e.g. considering the centre acting trivially or nilpotently etc.)<|endoftext|> -TITLE: Hearing the 17 planar symmetry groups -QUESTION [16 upvotes]: Though I'm sure it's not really hard to work out for myself, does anyone know a reference for the spectra of the Laplacian on the 17 flat compact orbifolds that underlie the 17 planar symmetry groups. -I'm thinking Neumann boundary conditions to model reflection lines. -And I do realize that these spectra may vary according to a certain number of moduli depending on the group. -(Feel free to add more tags if appropriate.) - -REPLY [5 votes]: I don't think that it's so hard to work out the solution non-rigorously, nor all that onerous to do it rigorously following Ian Agol's suggestion in the comments. A reference could be nice, but I don't think that it's quite necessary, because the answer itself is not all that different from its derivation. -First, in the Euclidean plane $\mathbb{R}^2$, the plane wave -$$f(\vec{x}) = \exp(i\vec{k} \cdot \vec{x})$$ -is an eigenfunction of the Laplacian $-\nabla$ with eigenvalue $\vec{k} \cdot \vec{k}$. Suppose that the orbifold is expressed as $X = \mathbb{R}^2/\Gamma$, where $\Gamma$ is a discrete, cocompact group. Then if you average $f$ with respect to $\Gamma$, you will get a Laplace eigenvector on $X$ with the same eigenvalue. Since these plane waves are complete (in an analytic sense) upstairs, their averages are at least complete downstairs. -Let $A \subseteq \Gamma$ be the subgroup of translations of $\Gamma$. Then the $A$-average of $f$ is either $f$ again or it vanishes. It's $f$ precisely when $\vec{k} \in 2\pi A^*$, where $A^*$ is the dual lattice of $A$. -Then there is a rotation group $R = \Gamma/A$ which is some finite group. You can now average $f$ over any lift of $R$ and see what you get. If $R$ lifts to a finite subgroup of $\Gamma$, then that subgroup, call it $R$ again, fixes a point, say the origin. In this case you get a basis of eigenfunctions using the $R$-orbits of $f$ and $\vec{k}$. The answer is all $\vec{k} \cdot \vec{k}$, where $\vec{k}$ represents each $R$-equivalence class in $2\pi A^*$. -For example, suppose that $\mathbb{R}^2/A$ is the standard square torus and $A$ is the standard square lattice, so that $A^* = A$. Suppose that $R$ is generated by a rotation by 90 degrees at the origin. Then $\vec{k} = 2\pi(n,m)$ and the eigenvalue is $4\pi^2(n^2 + m^2)$, where you should be careful to choose the pair of integers $(n,m)$ with $n \ge 0$ and $m > 0$, or $n = m = 0$. -If $\Gamma$ is a non-split extension of $R$ by $A$, then the answer is a little more complicated, but it's not that much more complicated. In fact the basic analysis is the same for compact Euclidean orbifolds in any dimension.<|endoftext|> -TITLE: "Explicit" embedding of $\ell^1$ as a closed subalgebra of a direct sum of matrix algebras -QUESTION [14 upvotes]: For sake of brevity let $A$ denote the Banach algebra formed by equipping $\ell^1({\mathbb N})$ with pointwise multiplication. This algebra is clearly not isomorphic as a Banach algebra to any uniform algebra (since it has idempotents of arbitrarily large norm). -It has been known since the 1970s that there exists a Hilbert space $H$ and a continuous injective algebra homomorphism $\theta:A\to B(H)$ which has (norm-)closed range. That is, $A$ is isomorphic to a non-self-adjoint operator algebra. However, the proof is non-explicit; the only argument I have seen uses the fact that multiplication $A \otimes A \to A$ extends continuously to the injective tensor product of $A$ with itself, and then uses abstract GNS-flavoured techniques to build the required space $H$. See - -Varopoulos, N. Th. Some remarks on $Q$-algebras. Ann. Inst. Fourier (Grenoble) 22 (1972), no. 4, 1–11. -Davie, A. M. Quotient algebras of uniform algebras. J. London Math. Soc. (2) 7 (1973), 31–40. - -For each $n$ let $A_n$ be the subalgebra of $A$ generated by the first $n$ minimal idempotents (so $A_n$ is just ${\mathbb C}^n$ with the $\ell^1$-norm and pointwise product). -UPDATE I originally mis-stated part of the question, as Andreas, Bill and Gideon were all quick to point out; my apologies for the confusion. -UPDATE, ENCORE My apologies for having omitted yet another condition in my haste to get the original edits done - thanks to fedja for catching this -Taking the existence of $\theta$ as read, it is straightforward to show that given $a\in A_n$ there exists a subspace $H_n\subset H$ of dimension $n+1$ and a homomorphism $\theta_n A_n \to B(H_n)$ such that -$$ c\Vert a\Vert_1 < \Vert\theta_n(a)\Vert \leq C \Vert a\Vert_1 $$ -where the constants $c$ and $C$ are strictly positive and independent of $n$ and $a$. -By a crude compactness argument, this implies the following: - -There exist constants $0 < c < C$ such that, for each $n\geq 1$, we can find $m(n)$ and idempotents $E_1,\dots, E_n$ in $M_{m(n)}({\mathbb C})$, which satisfy -$$ E_jE_k=0 \quad\hbox{for $j\neq k$} $$ -and -$$ c\sum_{j=1}^n |a_j| \leq \left\Vert \sum_{j=1}^n a_j E_j \right\Vert \leq C \sum_{j=1}^n |a_j| $$ -for all $a_1,\dots, a_n\in {\mathbb C}$. - -My question is this: does anybody know of an explicit construction of such $E_1,\dots, E_n$? I would be interested just to know of a construction which works for a sequence $n_1 < n_2< \dots$ Also, how small can we make the containing matrix algebra? (The compactness argument referred to above produces an embedding into the block-diagonal matrices with block size $n+1$ -- I haven't tried to work out just how `long' the diagonal has to be.) -These seem to me like questions that must have been looked at before, so perhaps the answer would be as simple as supplying an appropriate reference. -(The choice to work over complex scalars here is not important; a solution just for real scalars would be just as welcome.) -Remark: perhaps it might be useful to use the fairly well-known equivalence between the $\ell^1$-norm and the norm -$$ \Vert x \Vert_S := \sup_{F \subseteq {\mathbb N}} \left\Vert \sum_{i\in F} x_i \right\Vert $$ - -REPLY [11 votes]: You can manage to get $m(n)=(n+2)2^{n}$ (this can certainly be made a bit better). By your remark, it is enough to construct projections $E_1,\dots,E_n$ in $M_{n+2}(\mathbb C)$ such that: - -$E_i E_j=0$ for $i \ne j$ -$\|E_i\|\leq 3$ -$\| \sum s_i E_i\| \geq |\sum s_i|$ for any complex numbers $s_1,\dots s_n$. - -(Then consider $\oplus_{F \subset (1,\dots,n)} M_{|F|+2}$ inside $M_{m(n)}(\mathbb C)$). -We use the following elementary fact: there exists $\Omega$ a unit vector in the euclidean space $\mathbb C^{n+2}$ and $x_1,\dots,x_n$ and $y_1,\dots,y_n$ of norm $\sqrt 3$ in $\mathbb C^{n+2}$ such that $\langle x_i,y_j\rangle=1$ if $i=j$ and $0$ if $i\ne j$, and such that $\langle x_i,\Omega\rangle=\langle y_i,\Omega\rangle=1$. -(if $\Omega,e_0,\dots,e_n$ is an orthonormal basis, take $x_i=\Omega+e_0+e_i$ and $y_i=\Omega-e_0+e_i$). -Then the matrices $E_i=x_i^* \otimes y_i$ (the rank one operator mapping $x$ to $\langle x_i,x\rangle y_i$ satisfy the conditions above. The only point perhaps not completely evident is the third point, but this is because $\langle \Omega,\sum_i s_i E_i \Omega \rangle = \sum_i s_i$.<|endoftext|> -TITLE: What do cohomology operations have to do with the non-existence of commutative cochains over $\mathbb{Z}$? -QUESTION [24 upvotes]: Let $X$ be a topological space. In elementary algebraic topology, the cup product $\phi \cup \psi$ of cochains $\phi \in H^p(X), \psi \in H^q(X)$ is defined on a chain $\sigma \in C_{p+q}(X)$ by $(\phi \circ \psi)(\sigma) = \phi(_p\sigma)\psi(\sigma_q)$ where $p_\sigma$ and $\sigma_q$ denote the restriction of $\sigma$ to the front $p$-face and the back $q$-face, respectively. (More generally, any diagonal approximation $C_{\ast}(X) \to C_{\ast}(X) \otimes C_{\ast}(X)$ could be used; this is the Alexander-Whitney one.) The cup product defined by the Alexander-Whitney diagonal approximation as above is associative for cochains but skew-commutative only up to homotopy (this results from the fact that the two diagonal approximations $C_{\ast}(X) \to C_{\ast}(X) \otimes C_{\ast}(X)$ given by Alexander-Whitney and its "flip" (with the signs introduced to make it a chain map) agree only up to chain homotopy. -The commutative cochain problem attempts to fix this: that is, to find a graded-commutative differential graded algebra $C_1^*(X)$ associated functorially to $X$ (which may be restricted to be a simplicial complex) which is chain-equivalent to the usual (noncommutative) algebra $C^{\ast}(X)$ of singular cochains. -In Rational homotopy theory and differential forms, Griffiths and Morgan mention briefly that there is no way to make the cup-product skew-commutative on cochains (that is, to solve the commutative cochain problem) with $\mathbb{Z}$-coefficients, and that this is because of the existence of cohomology operations. It is also asserted that these cohomology operations don't exist over $\mathbb{Q}$ (presumably excluding the power operations). Could someone explain what this means? - -REPLY [3 votes]: A nice example is the following. Let $G=\mathbb{Z}/p$ and look at the DGA $C^*$ of singular cochains on $BG$ with coefficients in $\mathbb{F}_p$. Let us assume $p>2$. -Its cohomology is an exterior algebra in a generator $x$ in degree $1$ tensored with a polynomial algebra of a generator $y$ in degree $2$. -It cannot be formal, since $y$ is a Massey power of $x$. Thus it also cannot be quasiisomorphic to a graded commutative DGA. If there was such a DGA $D^*$, then we can define a quasiisomorphism $H^*(D^*)\rightarrow D^*$ by sending the cohomology classes $x$ and $y$ to any representing cocycles. Thus $C^*$ would be formal.<|endoftext|> -TITLE: Did Erdős publish his proof of the multiplicative version of the Erdős-Turán conjecture? -QUESTION [12 upvotes]: I read in an article of Erdős ("Extremal problems in number theory") that he had a proof of the multiplicative version of the Erdős-Turán conjecture. The statement of this theorem is - -Let $a_1 < a_2 < \cdots$ and denote by $g(n)$ the number of solutions to $n=a_ia_j$. Then $g(n)>0$ for all $n>n_0$ implies - $$\limsup_{n\to \infty} \ g(n)=\infty.$$ - -He also claims that this follows under the weaker assumption: let $A(x)=\sum_{a_i < x} 1$. Assume that for every $k$ we have $$\limsup_{x\to \infty} \ A(x)\left(x\left(\frac{\log\log x}{\log x}\right) ^k \right)^{-1}=\infty$$ than the same conclusion follows. However he also says that the proofs are difficult and haven't been published yet. -Since that article is from 1965 I am assuming he must have published something about this theorem afterward, but I don't have a reference. I have seen a proof of the first statement before (not by Erdős), but not the second one. Does anyone know if these proofs were published, simplified or generalized? - -REPLY [19 votes]: There is a generalization of Erdos' theorem in my paper "Multiplicative representations of integers," Israel Journal of Mathematics 57 (1987), 129--136.<|endoftext|> -TITLE: Inverse Length 3 Arithmetic Progression Problem for sets with positive upper density -QUESTION [6 upvotes]: It is a famous theorem of Roth, which Szemerédi famously generalized, that if a set of natural numbers has positive upper density then it contains arithmetic progressions of length $k$. The famous Green-Tao Theorem generalized this property to the primes. My question is, is there any progress on the 'inverse' problem? -First Question: Suppose $A \subset \mathbb{N}$ has positive upper density. Does it follow that with at most finitely many exceptions, all elements $a \in A$ is in an arithmetic progression of length at least $3$ in elements of $a$? That is, with at most finitely many exceptions, is it true that for each $a \in A$ there exists $k > 0$ such that $a, a+k, a+2k \in A$? -Edit: This question has been answered; see below by two constructions. However a second question may be asked. -A related problem (which I believe to be harder) is the same question for the primes, which has been asked here before: -Are all primes in a PAP-3? -Another related problem is found here (the constructions given all have density less than 1/2. Is it possible to find counterexamples with large density?) -Do there exist sets of integers with arbitrarily large upper density which contains infinitely many elements that are not in an arithmetic progression of length 3? - -REPLY [11 votes]: Or just take all powers of $3$ and add to them all numbers that are congruent to $1$ modulo $3$.<|endoftext|> -TITLE: Is a subset that contains no positive measurable subsets contained in a null measurable set? -QUESTION [6 upvotes]: Let $(X, \mathcal B, \mu)$ be a "good" measure space, e.g. $\mu$ is a positive Radon measure on a locally compact topological space $X$ with Borel $\sigma$-algebra $\mathcal B$. Let $A\subset X$ such that every measurable subset of $A$ has zero measure. Is it true that there is a zero measure set $B$ such that $A\subset B$? - -REPLY [16 votes]: You are asking whether every set with inner measure $0$ has measure $0$ with respect to the completion measure. The Lebesgue measure, for example, does not have this property, since the usual Vitali set is non-measurable, but has inner measure $0$. -I claim that a ($\sigma$-finite) measure space has your property if and only if every set -is measurable with respect to its completion measure. -For the forward direction, suppose that $A$ is any subset -of the space $X$. Let $a$ be the inner measure of $A$, the -supremum of $\mu(A_0)$ among all measurable $A_0\subset A$ (and we may assume wlog this is finite). -By taking a union, it follows that the inner measure is -realized, so that there is some measurable $A_0\subset A$ -with $\mu(A_0)=a$. It follows that $A-A_0$ has inner -measure $0$. By the hypothesis, it follows that -$A-A_0\subset B$ for some measurable set $B$ with -$\mu(B)=0$, and so $A-A_0$ has measure $0$ with respect to -the completion. Consequently, $A$ differs from the -measurable set $A_0$ on a completion-measure zero set -$A-A_0$, and hence is measurable with respect to the -completion measure. -Conversely, suppose that every set is measurable with -respect to the completion of $\mu$. Suppose that $A$ has -inner measure $0$. By assumption, there is some measurable -set $A_0$ such that the symmetric difference $A\triangle -A_0\subset A_1$ for some measurable $A_1$ with -$\mu(A_1)=0$. It follows that $A_0-A_1$ is a measurable -subset of $A$, and hence measure $0$, and so $A\subset -B=A_0\cup A_1=(A_0-A_1)\cup A_1$ shows that $A$ is contained -in a measure $0$ set $B$, as desired. -In particular, a complete measure has the property if and -only if it measures every set.<|endoftext|> -TITLE: Boolean Prime Ideal Theorem versus the Axiom of Determinacy -QUESTION [7 upvotes]: I'm assuming someone must have scooped me on this simple argument. Where does it (first) appear in the literature? -Fix an ultrafilter $\mu$ on $\omega$, the natural numbers. -Alice and Bob play a nim-like game. At the start each player "holds" the empty set and the -the starting "position" consists of $\omega$. Beginning with Alice, each player in turn will remove a non-empty finite initial segment from the current position (leaving some final segment of $\omega$) and deposit the removed segment into his or her holdings. Play proceeds for $\omega$ rounds. -Now the object of the game: to finish with holdings that belong to the ultrafilter $\mu$. -Strategy stealing obviates the possibility of either player possessing a winning strategy; the existence of $\mu$ thus contradicts AD. In more detail, if either player has a winning strategy, the game must admit infinitely many winning positions, but that would allow the other player to possibility of moving to a winning position on his or her very first move. - -Many sources work much harder than this to prove the weaker results that AC contradicts -AD. (Afterthought: I'd love to see a big-list question collecting theorems where unnecessarily complicated proofs permeate the literature despite the availability of simpler treatments...MO appropriate?) - -REPLY [2 votes]: I "discovered" this argument while I was a grad student (probably in 1969), but, alas, it was already known to Gene Kleinberg (if I remember correctly). My first guess for the original discoverer would be Fred Galvin; my second guess would be folklore.<|endoftext|> -TITLE: Axiom of Symmetry, aka Freiling's argument against CH -QUESTION [31 upvotes]: Hello! I would like to open a discussion about the Axiom of Symmetry of Freiling, since I didn't find in MO a dedicated question. I'll first try to summarize it, and the ask a couple of questions. -DESCRIPTION -The Axiom of Symmetry, was proposed in 1986 by Freiling and it states that -$AS$: for all $f:I\rightarrow I_{\omega}$ the following holds: $\exists x \exists y. ( x \not\in f(y) \wedge y\not\in f(x) )$ -where $I$ is the real interval $[0,1]$, and $I_{\omega}$ is the set of countable subsets of $I$. -It is known that $AS = \neg CH$. What makes this axiom interesting is that it is explained and justified using an apparently clear probabilistic argument, which I'll try to formulate as follow: -Let us fix $f\in I\rightarrow I_{\omega}$. We throw two darts at the real interval $I=[0,1]$ which will reach some points $x$ and $y$ randomly. Suppose that when the first dart hits $I$, in some point $x$, the second dart is still flying. Now since $x$ is fixed, and $f(x)$ is countable (and therefore null) the probability that the second dart will hit a point $y\in f(x)$ is $0$. -Now Freiling says (quote), - -Now, by the symmetry of the situation (the real number line does not really know which dart was thrown first or second), we could also say that the first dart will not be in the set $f(y)$ assigned to the second one. - -This is deliberately an informal statement which you might find intuitive or not. -However, Freiling concludes basically saying that, since picking two reals $x$ and $y$ at random, we have almost surely a pair $(x,y)$ such that, $x \not\in f(y) \wedge y\not\in f(x) )$, then, at the very least, there exists such a pair, and so $AS$ holds. -DISCUSSION -If you try to formalize the scenario, you'd probably model the "throwing two darts" as choosing a point $(x,y) \in [0,1]^{2}$. Fixed an arbitrary $f\in I\rightarrow I_{\omega}$, Freiling's argument would be good, if the set -$BAD = ${$(x,y) | x\in f(y) \vee y \in f(x) $} -has probability $0$. $BAD$ is the set of points which do not satisfy the constraints of $AS$. If $BAD$ had measure zero, than finding a good pair would be simple, just randomly choose one! -In my opinion the argument would be equally good, if $BAD$ had "measure" strictly less than $1$. In this case we might need a lot of attempts, but almost surely we would find a good pair after a while. -However $BAD$ needs not to be measurable. We might hope that $BAD$ had outermeasure $<1$, this would still be good enough, I believe. -However, if $CH$ holds there exists a function $f_{CH}:I\rightarrow I_{\omega}$ such that $BAD$ is actually the whole set $[0,1]^{2}$!! This $f_{CH}$ is defined using a well-order of $[0,1]$ and defining $f_{CH}(x) = ${$y | y \leq x $}. Under $CH$ the set $f(x)$ is countable for every $x\in[0,1]$. Therefore -$BAD = ${$ (x,y) | x\in f_{CH}(y) \vee y \in f_{CH}(x) $}$ = ${$ (x,y) | x\leq y \vee y \leq x $}$ =[0,1]^{2}$ -So it looks like that under this formulation of the problem, if $CH$ then $\neg AS$, which is not surprisingly at all since $ZFC\vdash AS = \neg CH$. -Also I don't see any problem related with the "measurability" of $BAD$. -QUESTIONS -Clearly it is not possible to formalize and prove $AS$. However the discussion above seems very clear to me, and it just follows that if $CH$ than $BAD$ is the whole set $[0,1]^{2}$. without the need of any non-measurable sets or strange things. And since picking at random a point in $[0,1]^{2}$ is like throwing two darts, I don't really think $AS$ should be true, or at least I don't find the probabilistic explanation very convincing. -On the other hand there is something intuitively true on Freiling's argument. -My questions, (quite vague though, I would just like to know what you think about $AS$), -are the following. -A) Clearly Freiling's makes his point, on the basis that the axioms of probability theory are too restrictive, and do not capture all our intuitions. -This might be true if the problem was with some weird non-measurable sets, but in the discussion above, non of these weird things are used. -Did I miss something? -B) After $AS$ was introduced, somebody tried to tailor some "probability-theory" to capture Freling's intuitions? More in general, is there any follow up, you are aware of? -C) Where do you see that Freiling's argument deviates (even philosophically) from my discussion using $[0,1]^{2}$. -I suspect the crucial, conceptual difference, is in seeing the choice of two random reals as, necessarily, a random choice of one after the other, but with the property that this arbitrary non-deterministic choice, has no consequences at all. -Thank you in advance, -Matteo Mio - -REPLY [3 votes]: Actually, in regards to your question B), there is a large cardinal axiom that implies $AS$. Your link to the wikipedia article regarding Freiling's axiom of symmetry states the following, in the section "Objections to Freiling's Argument": -"The naive probabilistic intuition used by Freiling tacitly assumes that there is a well-behaved way to associate a probability to any subset of the reals." -This is important, because the truth or falsity of $CH$ is intimately connected with the ability to assign to each subset of the reals, a probability measure. -Consider the following definition, from the Wikipedia article on measurable cardinals: -"A cardinal $\kappa$ is real-valued measurable iff there is a $\kappa$-additive probability measure on the power set of $\kappa$ which vanishes on singletons" (i.e. singletons have probabiity measure zero). -Axiom. Let $\mathfrak c$ be the cardinality of the continuum. $\mathfrak c$ is real-valued measurable. -Consider also the following equivalences from the same wikipedia article: -"A real-valued measurable cardinal ='$\mathfrak c$' exists iff there is a countably additive extension of the Lebesgue measure to all sets of reals iff there is an atomless probability measure on $\mathscr P$($\mathfrak c$). -Note also that the wikipedia article on Freiling's axiom of symmetry linked to your question states that $AS$ is equivalent to $\lnot$$CH$ by a theorem of Sierpinski, and also states that back in 1929, Banach and Kuratowski proved that $CH$ implies that $\mathfrak c$ is not real-valued measurable. -So consider the contrapositive of that statement, that if $\mathfrak c$ is real-valued measurable, then $\lnot$$CH$. Since $AS$ is equivalent to $\lnot$$CH$, then "$\mathfrak c$ is real-valued measurable" immediately implies $AS$. By the definition of real-valued measurable and the aforementioned equivalences found in the wikipedia article on measurable cardinals, Freiling's prereflective probabilistic argument seems to be essentially correct. -This is further confirmed by Noa Goldring in his paper "Measures, Back and Forth Between point Sets and Large Sets", (The Bulletin of Symbolic Logic, Vol. 1, Number 2, June 1995, pp. 182-183 footnotes 17 and 18--also ibid, pp. 171-188.<|endoftext|> -TITLE: The functional equation $f(f(x))=x+f(x)^2$ -QUESTION [45 upvotes]: I'd like to gather information and references on the following functional equation for power series $$f(f(x))=x+f(x)^2,$$$$f(x)=\sum_{k=1}^\infty c_k x^k$$ -(so $c_0=0$ is imposed). -First things that can be established quickly: - -it has a unique solution in $\mathbb{R}[[x]]$, as the coefficients are recursively determined; -its formal inverse is $f^{-1}(x)=-f(-x)$ , as both solve uniquely the same functional equation; -since the equation may be rewritten $f(x)=f^{-1}(x)+x^2$, it also follows that $f(x)+f(-x)=x^2$, the even part of $f$ is just $x^2/2$, and $c_2$ is the only non-zero coefficient of even degree; -from the recursive formula for the coefficients, they appear to be integer multiples of negative powers of $2$ (see below the recursive formmula). Rmk. It seems (but I did not try to prove it) that $2^{k-1}c_k$ is an integer for all $k$, and that $(-1)^k c_{2k-1} > 0$ for all $k\geq 2$. - - -Question : how to see in a quick way - that this series has a positive radius - of convergence, and possibly to - compute or to evaluate it? -[updated] - A more reasonable question, after the numeric results and various comments, seems to be, rather: - how to prove that this series does - not converge. - -Note that the radius of convergence has to be finite, otherwise $f$ would be an automorphism of $\mathbb{C}$. Yes, of course I did evaluate the first coefficients and put them in OEIS, getting the sequence of numerators A107700; unfortunately, it has no further information. -Motivation. I want to understand a simple discrete dynamical system on $\mathbb{R}^2$, namely the diffeomorphism $\phi: (x,y)\mapsto (y, x+y^2)$, which has a unique fixed point at the origin. It is not hard to show that the stable manifold and the unstable manifold of $\phi$ are -$$W^s(\phi)=\mathrm{graph}\big( g_{|(-\infty,\ 0]}\big)$$ -$$W^u(\phi)=\mathrm{graph}\big( g_{|[0, \ +\infty)}\big)$$ -for a certain continuous, strictly increasing function $g:\mathbb{R}\to\mathbb{R}$, that solves the above functional equation. Therefore, knowing that the power series solution has a positive radius of convergence immediately implies that it coincides locally with $g$ (indeed, if $f$ converges we have $f(x)=x+x^2/2+o(x^2)$ at $x=0$ so its graph on $x\le0$ is included in $W^s$, and its graph on $x\ge0$ is included in $W^u$: therefore the whole graph of $f$ would be included in the graph of $g$,implying that $f$ coincides locally with $g$). If this is the case, $g$ is then analytic everywhere, for suitable iterates of $\phi$ give analytic diffeomorphism of any large portion of the graph of $g$ with a small portion close to the origin. -One may also argue the other way, showing directly that $g$ is analytic, which would imply the convergence of $f$. Although it seems feasible, the latter argument would look a bit indirect way, and in that case I'd like to make sure there is no easy direct way of working on the coefficients (of course, it may happen that $g$ is not analytic and $f$ is not convergent). -Details: equating the coefficients in both sided of the equation for $f$ one has, for the 2-Jet -$$c_1^2x+(c_1c_2+c_2c_1^2)x^2 =x + c_1^2x^2,$$ -whence $c_1=1$ and $c_2=\frac 1 2;$ and for $n>2$ -$$2c_n=\sum_{1\le j\le n-1}c_jc_{n-j}\,-\sum_{1 < r < n \atop \|k\|_1=n}c_rc_{k_1}\dots c_{k_r}.$$ -More details: since it may be of interest, let me add the argument to see $W^s(\phi)$ and -$W^u(\phi)$ as graphs. -Since $\phi$ is conjugate to $\phi^{-1}=J\phi J $ by the -linear involution $J:(x,y)\mapsto (-y,-x)$, we have $W^u(\phi):=W^s(\phi^{-1})=J\ -W^s(\phi)$, and it suffices to study $\Gamma:=W^s(\phi)$. For any -$(a,b)\in\mathbb{R}^2$ we have $\phi^n(a,b)=(x_n,x_{n+1})$, with $x_0=a$, $x_1=b$, -and $x_{n+1}=x_n^2+x_{n-1}$ for all $n\in\mathbb{N}$. From this it is easy to see -that $x_{2n}$ and $x_{2n+1}$ are both increasing; moreover, $x_{2n}$ is bounded -above iff $x_{2n+1}$ is bounded above, iff $x_{2n}$ converges, iff $x_n\to 0$, iff -$x_n\le 0 $ for all $n\in\mathbb{N}$. -As a consequence $(a,b)\in \Gamma$ iff $\phi^n(a,b)\in -Q:=(-\infty,0]\times(-\infty,0]$, whence $ \Gamma=\cap_{ n\in\mathbb{N}} -\phi^{-n}(Q)$. The latter is a nested intersection of connected unbounded closed -sets containing the origin, therefore such is $\Gamma$ too. -In particular, for any $a\leq 0$ there exists at least a $b\leq 0$ such that -$(a,b)\in \Gamma$: to prove that $b$ is unique, that is, that $\Gamma$ is a graph -over $(\infty,0]$, the argument is as follows. Consider the function -$V:\Gamma\times\Gamma\to\mathbb{R}$ such that $V(p,p'):=(a-a')(b-b')$ for all -$p:=(a,b)$ and $p':=(a',b')$ in $\Gamma$. -Showing that $\Gamma$ is the graph of a -strictly increasing function is equivalent to show that $V(p,p')>0$ for all pair of -distinct points $p\neq p'$ in $\Gamma$. -By direct computation we have -$V\big(\phi(p),\phi(p')\big)\leq V(p,p')$ and $\big(\phi(p)-\phi(p')\big)^2\geq -\|p-p'\|^2+2V(p,p')(b+b')$. Now, if a pair $(p,p')\in\Gamma\times\Gamma$ has -$V(p,p')\le0$, then also by induction $V\big(\phi^n(p),\phi^n(p')\big)\leq 0$ and -$\big(\phi^n(p)-\phi^n(p')\big)^2\geq \|p-p'\|^2$ for all $n$, so $p=p'$ since both -$\phi^n(p)$ and $\phi^n(p')$ tend to $0$. This proves that $\Gamma$ is a graph of a -strictly increasing function $g:\mathbb{R}\to\mathbb{R}$: since it is connected, $g$ -is also continuous. Of course the fact that $\Gamma$ is $\phi$-invariant implies that $g$ solves the functional equation. - -REPLY [2 votes]: It seems interesting to complexify the map you are studying. Thus we can study now $$F:\binom{x}{y} \mapsto \binom{y}{x+y^2}$$ -as a map from $\mathbb{C}^2$ to itself ( the so-called "complex Henon map"). Now near the origin, the second iterate $F \circ F$ is tangent to the identity. There is now a whole body of results concerning germs of maps of $\mathbb{C}^N$ that are tangent to the identity, that can be applied here. -In particular, such germs can have "parabolic curves" (analytic images of a disk $\Delta$ in $\mathbb{C}^2$ such that $(0,0) \in \partial \Delta$), included in the stable manifold of the origin. Such curves should be seen as generalizations of the "Leau-Fatou flowers" appearing in the study of parabolic fixed points in $\mathbb{C}$. -A very nice survey about local dynamics in $\mathbb{C}^N$ has been written by M. Abate (in LNM 1998, Springer "Holomorphic dynamical systems"). -When studying such Henon maps in $\mathbb{C}^2$ it is also usually helpful to plot slices of the set $K^+$ (points with bounded forward orbits): here is a horizontal slice y=0 of $K^+$. In blue, you see the set of points with unbounded forward orbit. -(source)<|endoftext|> -TITLE: The influence of string theory on mathematics for philosophers. -QUESTION [29 upvotes]: I've agreed, perhaps unwisely, to give a talk to Philosophers about string theory. -I'd like to give the philosophers an overview of the status and influence of string theory in physics, which I feel competent to do, but I would also like to say something about the influence it has had in mathematics where I am on less -familiar ground. I've read the Jaffe-Quinn manifesto and the responses in -http://arxiv.org/abs/math.HO/9404229. What I would like from MO are pointers to more recent discussions of this issue in the mathematical community so that I can get a sense of where things stand 16 years later. - -REPLY [22 votes]: The Jaffe-Quinn manifesto really had little to do with string theory, but a lot to do with topological quantum field theory, especially 3d tqft. I remember Frank Quinn talking about this at length during a hike at the 1991 Park City summer school. He was lecturing there on topological qft, see -"Lectures on Axiomatic Topological Quantum Field Theory" in "Geometry and Quantum Field Theory, IAS/Park City Mathematics Series, Volume 1", edited by Daniel Freed and Karen Uhlenbeck. -The sort of thing that was worrying Quinn was: - -Witten's great paper on "Supersymmetry and Morse Theory", which was published in a mathematics journal, the Journal of Differential Geometry. -Witten's Fields medal winning work on the Jones polynomial and Chern-Simons theory. - -Quinn explained that at the beginning of his career he had been heavily influenced by the work of Thurston and Sullivan, but found that trying to emulate them had led him to lose track of what he precisely understood and what he didn't, requiring a painful period of getting back to a more rigorous way of working. He was worried that losing the distinction between works like Witten's and truly rigorous work would lead others to the problematic situation he had found himself in as a young mathematician. In the end, I think Atiyah's response won the day: he argued that mathematicians were fully capable of protecting their virtue while interacting with physicists. Shortly after this exchange, those topologists in the math community who were skeptical about the importance of what Witten was bringing to mathematics were conclusively won over by the Seiberg-Witten equations. -But the example set by Quinn of how to do TQFT in the end has largely won out. There was an attempt to teach mathematicians the actual QFT behind Seiberg-Witten at the IAS in 96/97, but I don't think it was very successful. These days both TQFT and the Seiberg-Witten equations remain very important ideas in topology, but they're pursued with conventional standards of rigor. Mathematicians have gotten used to taking physicist's QFT arguments and extracting and generalizing those parts that can be made rigorous and fit into the evolving mathematical tradition. -As others have mentioned, for the case of string theory, mirror symmetry is probably the best example of an idea coming out of it that has had a huge influence in mathematics. Yau's recent popular book "The Shape of Inner Space" contains lots of other examples of the interaction of math and physics surrounding Calabi-Yau manifolds.<|endoftext|> -TITLE: Free symmetric monoidal $(\infty,n)$-categories with duals -QUESTION [17 upvotes]: The reading of (Hopkins-)Lurie's On the Classification of Topological Field Theories (arXiv:0905.0465) suggests that a stronger version of the cobordism hypothesis should hold; namely, that (under eventually suitable technical assumptions), the inclusion of symmetric monoidal $(\infty,n)$-categories with duals into $(\infty,n)$-categories should have a left adjoint, the ``free symmetric monoidal $(\infty,n)$-category with duals on a given category $\mathcal{C}$ '', and that this free object should be given by a suitably $\mathcal{C}$-decorated $(\infty,n)$-cobordism $Bord_n(\mathcal{C})$. This would be an higher dimensional generalization of Joyal-Street-Reshetikhin-Turaev decorated tangles. -Such an adjunction would in particular give a canonical symmetric monoidal duality -preserving functor $Z: Bord_n(\mathcal{C})\to \mathcal{C}$ which seems to appear underneath the constructions in Freed-Hopkins-Lurie-Teleman's Topological Quantum Field Theories from Compact Lie Groups (arXiv:0905.0731). -Yet, I've been unable to find an explicit statement of this conjectured adjointness in the above mentioned papers, and my google searches for "free symmetric monoidal n-category" only produce documents in which this continues with "generated by a single fully dualizable object", as in the original form of the cobordism hypothesis. Is anyone aware of a formal statement or treatment of the cobordism hypothesis from the left adjoint point of view hinted to above? - -REPLY [14 votes]: The existence of a left adjoint follows by formal nonsense. If you have a symmetric monoidal $(\infty,n)$-category which is can be built by first freely adjoining some objects, -then some $1$-morphisms, then some $2$-morphisms, and so forth, up through $n$-morphisms -and then stop, then there is an explicit geometric description of the $(\infty,n)$-category you get by "enforcing duality" in terms of manifolds with singularities. This is sketched in one of the sections of the paper you reference. I don't know of a geometric description for what you get if you start with an arbitrary symmetric monoidal $(\infty,n)$-category and then "enforce duality".<|endoftext|> -TITLE: 4-coloring maps of pentagons -QUESTION [6 upvotes]: Is there a simple proof for the 4-color theorem when restricted to (finite) maps all of whose -(internal) regions are pentagons? I am in fact most interested in convex pentagons, if that additional -structure helps. These maps lead to graphs of maximum degree $\Delta=5$, so a result for -such graphs will answer my question. -If all regions are quadrilaterals, then four colors are sometimes necessary, -even if all quadrilaterals are convex, e.g., - - -But here there is a simple proof of 4-colorability: -Identify a quadrilateral with an exposed edge, remove it, 4-color the remainder -by induction, and replace the quad, coloring it with a color different from its -at most three neighbors. -Obviously this simple proof fails for pentagons. -Especially the $\Delta=5$ case is likely known to the experts. So a reference would suffice. -Thanks! - -REPLY [10 votes]: If $v$ is a vertex of degree at most 4 in a planar graph $G$ then one can extend a proper 4-coloring of $V(G) \setminus \{v\}$ to $v$ after possibly modifying it using the classical Kempe chain argument. See for example paragraph 5 of the "Summary of proof ideas" section of the Wikipedia entry on the 4-color theorem: http://en.wikipedia.org/wiki/Four_color_theorem -As pentagons with exposed edges correspond to vertices of degree 4 in the dual graph, one can color the map by induction using this trick.<|endoftext|> -TITLE: Definition of fiber bundle in algebraic geometry -QUESTION [23 upvotes]: If we have a map p: X --> Y of topological spaces, we can make a definition expressing that the topological type of the fibers of p varies continuously (edit: better to say "locally constantly", thanks Dave) with the base: we can say that p is a fiber bundle. -My question is, can we capture this notion algebro-geometrically, in the case where X and Y are varieties over a field of characteristic zero and p is a map of varieties? I'm looking for a definition hopefully having the following properties (side question: do these seem reasonable?): -1) If X and Y are over the complex numbers, then p is an algebro-geometric fiber bundle if and only if it is a topological fiber bundle on complex points; -2) If f: X-->Y is arbitrary then there is an algebraic stratification of Y such that over each stratum f is a fiber bundle. -Examples should include smooth maps having smooth proper compactifications for which the boundary divisors are in strict normal crossings position, but I would rather the definition not be along these lines because, for instance, I don't want to need resolution of singularities to check that the structure map to the ground field is a fiber bundle. -Edit: In response to several comments, yes, another example would be the normalization map for a cuspidal singularity. In fact I would like the definition to be "topological", in the sense that it factors through h-sheafification. -Edit 2: Whoops, it looks like I used some bad terminology, which probably led to misinterpretations. Sorry folks! To fix things I've replaced all instances of "fibration" with "fiber bundle". -Any thoughts are appreciated! - -REPLY [11 votes]: OK, let me venture to give a definition. Say that a morphism $f:X\to Y$, of varieties over -a field, is an algebraic fibration if there exists a factorization -$X\to \overline{X}\to Y$, such that that the first map is an open immersion, and the -second map is proper and there exists a partition into Zariski locally closed strata $\overline{X}=\coprod\overline{X}_i$, such that restrictions -$\overline{X}_i\to Y$ are smooth and proper. $X$ should be a union of strata. -Perhaps, one should also insist that this is Whitney stratification. - - -Shenghao's comment made me realize that my original attempt at an answer was problematic. -Rather than trying to fix it, let me make a fresh start. -Let us say that $f:X\to Y$ is an algebraic fibration if there exist a simplicial scheme $\bar X_\bullet$ with a divisor $D_\bullet\subset \bar X_\bullet$ such that - -There is map $\bar X_\bullet -D_\bullet\to X$ satisfying cohomological descent, in the -sense of Hodge III, for the classical topology (over $\mathbb{C}$) or etale topology (in general). -The composite $\bar X_n\to Y$ is smooth and proper, and $D_n$ has relative normal crossings for each $n$. - -These conditions will ensure that $R^if_*\mathbb{Z}$ (resp. $R^if_*\mathbb{Z}/\ell \mathbb{Z}$) -are locally constant etc. -I think that this would also apply Shenghao's question -What would be a characteristic-$p$ analogue for $C^{\infty}$-fiber bundles? -Although I won't claim that this is in any sense optimal. -Oh, and I forgot to say that when $Y=Spec k$ is a point, every $X$ can be seen to be fibration (as it should) by De Jong<|endoftext|> -TITLE: Boundary links and ribbon links. -QUESTION [6 upvotes]: This question is about the relation between the notions of boundary link and ribbon link. -For the definition of ribbon link see: ribbon links - counterexamples. -An n-component link $L=L_1\cup\dots\cup L_n$ is said to be a boundary link if there exists an orientable surface consisting of n disjoint components $S=S_1\cup\dots\cup S_n$ such that -for each $i\leq n$ we have $\partial S_i=L_i$. -There exists boundary links which are not ribbon and ribbon links which are not boundary (see Rolfsen book chapter 5 pag. 140) -It can be shown that every pure ribbon link is a boundary link. (A ribbon link is said to be pure if every ribbon singularity involves only one disc.) -Now my problem is: when is a boundary link ribbon? Clearly every component of such a link must be a ribbon knot so my question is: - -$\textbf{Is every boundary link with ribbon components a ribbon link?}$ - -Assume the answer to the previous question is yes. Then the following question -makes sense: - -$\textbf{Is it true that a link is purely ribbon iff it is a boundary link}$ -$\textbf{ with ribbon components?}$ - -REPLY [6 votes]: The answer to your first question is no. There are non-ribbon boundary links whose components are unknotted! Indeed the Bing double of a knot is a boundary link with unknotted components, but it has recently been proven by several authors that the Bing double of the figure-8 knot is not slice (hence not ribbon.) See for example this paper. -Edit: The Bing double of the trefoil is also not slice, which is a less subtle calculation using the signature.<|endoftext|> -TITLE: Radix notation and toposes -QUESTION [9 upvotes]: In classical logic plus ZF, the field of real numbers admits infinitely many isomorphic realizations as a numeral system --- as the radix varies. The intuitionistic status of these systems seems less clear however. First of all, the notion of "field" has several -distinct intuitionistic interpretations (e.g. non-zero implies invertible; non-invertible implies zero; invertible or zero). And even with a way to set up things to make these numeral systems fields in some appropriate sense, one still may lack for the maps between them that make them isomorphic. -If a topos $T$ has a natural number object, surely one can make sense of "the numeral system with radix $n$" at least with $n$ any fixed (real world) natural number > 1. So fixing $T$ one gets an equivalence relation on the real world natural numbers according to the isomorphism of numeral systems of various radix. (Well, depending upon the topos and the interpretation of numeral system, perhaps one might have to discard some radices altogether, where one doesn't have total operations.) -My question: Can one realize any equivalence relation on the natural numbers by choosing some appropriate topos? Failing a full answer, what example topos distinguish some systems but not others? - -REPLY [6 votes]: No, and actually, you cannot realise any non-trivial equivalence relation this way. If any of the (non-trivial) pairs of radix systems are isomorphic, they all are. It is "well-known" that not every real number has a decimal expansion. This extends to this situation. -Specifically, for any pair n and m (unless one divides the other), the assumption that every radix-n number has a radix-m expansion impies a weak form of the law of the excluded middle, called the limited principle of omniscience (LPO): this says that for every sequence of integers, either it is constant zero, or it has a non-zero element. Given a sequence of integers, you can construct a radix-n number such that the first digit of the corresponding radix-m number decides the answer. (E.g. given base-10 0.3333333... convert it to base-3, it's 0.100000... or 0.022222...) -Conversely if LPO holds, you can show all radix-n systems are equivalent. -It's also been pointed out that these number systems aren't necessarily even rings. The same thing happens - if any of them are rings, then LPO holds. And if LPO holds, then they are all fields (in any sense). -This reasoning goes through in any topos (even Pi-pretopos) with a natural numbers object.<|endoftext|> -TITLE: Are the numbers of elements of two distinct prime orders not equal in finite groups? -QUESTION [5 upvotes]: Are the numbers of elements of two distinct prime orders not equal in finite groups? - -REPLY [15 votes]: The answer is "no": -One can construct a counterexample of the form $G\times H$. Say $p$ and $q$ are distinct primes, and for some integer $m$ one can find two groups $G, H$ with -with $q\nmid|G|$ and $m$ elements of order $p$ in $G$, and $p\nmid|H|$ and $m$ elements of order $q$ in $H$. Then $G\times H$ has $m$ elements of order $p$, and $m$ of order $q$. -Now take $p=5$ and $q=3$. Then $G=C_{11}\rtimes C_5$ has $44$ elements of order $5$ (every element not in $C_{11}$), and $H=C_{21}\rtimes C_3$ has $44$ elements of order $3$ ($2$ in $C_{21}$ plus all $42$ outside it). So $G\times H$ has 44 elements of order $3$ and of order $5$. -To check this in Magma: -P:=DirectProduct(SmallGroup(55,1),SmallGroup(63,3)); -assert #[g: g in P | Order(g) eq 5] eq 44; -assert #[g: g in P | Order(g) eq 3] eq 44;<|endoftext|> -TITLE: "The Z/2-cohomology functor from Top to GrVecSpaces factors through Unstable-A-Mod, and this is the largest algebraic category through which it factors." -QUESTION [9 upvotes]: (Here A is the Steenrod algebra.) -I have it on good authority (p. 23) that this is true, but I can't quite make sense of it. The crux of the matter is something I've been wondering for a while: - -Why exactly is it that cohomology - operations are "the best we can do" - (i.e./more generally, why is it that - the finest structure we can impose on - $E$-cohomology is that it be a module - over $E^*E$)? - -There might be something tied up in the qualification that we're talking about algebraic categories, which are defined here. I've never heard of these before. Presumably their properties show up in GrVecSpaces as manifestations of actual things on the topological side, which makes it seem like we're only restricting ourselves to categories that are going to preserve some information that we obviously (?) want to preserve, but probably there's more to it than that. And in any case I don't understand why those should be exactly the properties of GrVecSpaces that we care about. -(I'd love to hear about category theory stuff of course, but my main goal is to understand the boxed question.) - -REPLY [4 votes]: A little more information is at the page on infinitary Lawvere theories on the nLab. Basically, Charles is right: as cohomology is defined by the Eilenberg-Mac Lane spaces, everything about the resulting algebraic category is contained in the Eilenberg-Mac Lane spaces and their morphisms. In particular, if that category had more structure then that would be visible in the cohomology of the Eilenberg-Mac Lane spaces and would then be already in the algebraic category we were already in. In short it's because (as I proved on that page, in generality) the cohomology of the Eilenberg-Mac Lane spaces are the free algebras in this algebraic category. -Furthermore, to extend a point that Sam Isaacson makes, it's not true to say that $E^\star(X)$ is a module over $E^\star(E)$ with the usual interpretation of module as an abelian group with a (suitable) action of a ring. (Charles' answer is correct in his language, but you may not have noticed his choice of language.) What is correct is to say that $E^\star(X)$ is a module for the Tall-Wraith monoid $E^\star(\underline{E}_\star)$. This use of module generalises the idea of a module over an algebra. In particular, the fact that $E^\star(X)$ is an $E^\star$-algebra is included in this description. More on the nLab page Tall-Wraith monoids and the papers linked therein. -(All of this holds for any generalised cohomology theory.)<|endoftext|> -TITLE: Induced representations and $(\varphi, \Gamma)-modules -QUESTION [10 upvotes]: Let $K$ be a finite extension of $\mathbb{Q}_p$ and $L$ be a finite Galois extension of $K$. Let $V$ be a $p$-adic representation of $G_L$ Let $D$ be the $(\varphi, \Gamma)$-module associated to $V$ by Fontaine's functor. Is there an easy way to describe the $(\varphi, \Gamma)$-module associated to $Ind_{G_L}^{G_K} V$ in terms of $D$ ? - -REPLY [11 votes]: If $L/K$ is unramified, then the answer is exactly what Matt said. In the general case, you have to take into account the fact that $\Gamma_K$ is larger than $\Gamma_L$ and the construction is given in 2.2 of Ruochuan Liu's "Cohomology and Duality for (phi,Gamma)-modules over the Robba ring", see http://arxiv.org/abs/0711.4346<|endoftext|> -TITLE: A graph with few edges everywhere -QUESTION [21 upvotes]: Given a graph $G(V,E)$ whose edges are colored in two colors: red and blue. -Suppose the following two conditions hold: - -for any $S\subseteq V$, there are at most $O(|S|)$ red edges in $G[S]$ -for any $S\subseteq V$, if $G[S]$ contains no red edges, then it contains $O(|S|)$ blue edges - -My question is: can we conclude from this that the total number of blue edges is linear? -I have no strong intuition for this, but it seems that it might be possible (some averaging/probabilistic argument?). To try to give an intuition, we can rephrase it as follows. The red graph is very sparse, even locally. The blue graph is also sparse in all regions that are free of red edges. Due to the sparseness of the red graph those 'regions' are numerous, so we hope this might imply that the blue graph is also sparse. -One can maybe consider first an easier version, if we assume that the red degree of every vertex is $O(1)$. In this case I also don't know the answer. -Note that it's already too weak if we replace the first condition with just: the total number of red edges is linear. Look at the example: a blue $K_{\sqrt n,n-\sqrt n}$ with a red $\sqrt n$-clique added in the corresponding part. This graph has $\Omega(n^{3/2})$ blue edges (example by D. Palvolgyi). We can still ask in this version whether one can do better than $n^{3/2}$. - -REPLY [12 votes]: I think one can push through the probabilistic arguments of Tim Gowers and Fedor Petrov in the general case, as follows. -Let $c$ be a constant such that the number of red edges in $G[S]$ -is at most $c|S|$ for every $S \subseteq V(G)$. One can order the vertices of $G$: $v_1, v_2, \ldots, v_n$, so that every vertex has at most $2c$ neighbors with lower indices. (Define the ordering starting with the highest index. If $v_n, \ldots,v_{i+1}$ are defined, set $v_i$ to be the vertex with the smallest degree in the subgraph induced by the vertices which are not yet indexed. This is a standard trick.) -Now we define a random subset $S$ of $V(G)$ recursively: if $S \cap$ {$v_1, \ldots, v_i$} is chosen put $v_{i+1}$ in $S$ with probability $1/2$ if it is not joined by a red edge to any of the vertices already in $S$, otherwise don't put it in $S$. Then $S$ is red-free -and, just as in Fedor's answer, we can see that the probability that a pair of vertices $u$ and $v$ joined by a blue edge both lie in $S$ is at least $2^{-4c-2}$. Therefore the number of blue edges is at most -$2^{4c+2}c' \mathbf{E}[|S|] \leq 2^{4c+1}c'|V(G)|,$ -where $c'$ is the constant implicitly present in the condition on the density of the blue edges.<|endoftext|> -TITLE: Are there graph models for other moduli spaces? -QUESTION [15 upvotes]: Recall that a ribbon graph is a graph with a cyclic ordering at each vertex and such that each vertex has valence greater than or equal to 3. This cyclic ordering exactly gives one the information to thicken the graph to a surface with boundary and it therefore makes sense to talk about the boundary components of a ribbon graph. -We can make these ribbon graphs into a category $\mathsf{RibbonGr}$ by demanding that morphisms are those surjective morphisms of the underlying graphs which collapse trees while preserving the boundary cycles. This is a theorem of Strebel-Penner-Kontsevich-Igusa (bonus sub-question: is this the correct attribution?) that the geometric realisation of $\mathsf{RibbonGr}$ is homotopy equivalent to the disjoint union of moduli spaces of compact Riemann surfaces with boundary over the set of isomorphism classes of compact Riemann surfaces of genus $g$ and $n$ boundary components except $(g,n) = (0,0), (0,1), (0,2)$. -I like to call this the graph model of the moduli spaces. It can be extended to include labels on the ribbon graphs and Riemann surfaces, e.g. in Godin's work on higher string operations. However, this theorem made me wonder: are there any other generalizations of this theorem? Are there other categories of graphs or some (higher-dimensional) generalization of graphs which geometrically realize to spaces homotopy equivalent to moduli spaces? - -REPLY [12 votes]: Yes, graphs come up in various moduli contexts quite a bit! As the previous two answers indicated, $Out(F_n)$ is closely related to graphs. This goes back to the original paper of Vogtmann and Culler where outer space was first introduced. $BOut(F_n)$ can be modelled as the moduli space of metric graphs with first Betti number equal to $n$. If you consider the moduli space of graphs as an orbifold then this has the right integral homotopy type (but if you take the coarse quotient space then it is just a rational classifying space because $Out(F_n)$ acts on outer space with finite stabilisers). -Given a cyclic operad $P$ in the category of topological spaces one can talk about the space of graphs with vertices labelled by $P$. An ordinary graph is the same as a $Comm$-labelled graph since the commutative cyclic operad is just a point in each arity. A ribbon graph is the same as an $Assoc$-labelled graph since in cyclic arity $n$ the associative cyclic operad is the set of cyclic oderings on $n$ letters. -So the Culler-Vogtmann work can be interpreted as saying that $BOut(F_n)$ is the moduli space of rank $n$ $Comm$-graphs. Moduli spaces of Riemann surfaces with marked points are are homotopy equivalent to spaces of $Assoc$-graphs. There are some other results of this type. A Mobius graph is like a ribbon graph but with edges possibly given a half-twist; these are the same as graphs labelled by the cyclic operad $InvAss$ for associative algebras with an involution (perhaps could be called the hermitian associative operad). The spaces of Mobius graphs are homotopy equivalent to the moduli spaces of surfaces with a Klein structure (an unoriented version of complex structure), or equivalently, the classifying spaces of the mapping class groups of unorientable surfaces. Another result of this type is that the space of graphs labelled by the framed little 2-discs cyclic operad is homotopy equivalent to the moduli space of 3-dimensional oriented handlebodies. -To expand on some of Jim's comments a bit further: The connection to graph homology is as follows. First you need to know that there is a duality functor for cyclic operads in chain complexes. Sometimes it is called dg-duality, and sometimes it is called the Bar construction or Koszul duality. (Strictly speaking, the duality is given by a bar construction that produces a cyclic cooperad, followed by applying linear duality to turn in back into a cyclic operad; this construction agrees up to quasi-isomorphism with the koszul duality construction for cyclic operads that are Koszul.) The associative cyclic operad is self-dual. The commutative cyclic operad is dual to the Lie cyclic operad. -The general theorem is that if $P$ is a cyclic operad in topological spaces, then $C_*P$ is a cyclic operad in chain complexes with dual $D(C_*P)$, and $D(C_*P)$-graph homology computes the cohomology of the space of $P$-labelled graphs. This is why $Lie$ graph homology computes the cohomology of $Out(F_n)$ and $Ass$ graph homology computes the cohomology of moduli spaces of Riemann surfaces.<|endoftext|> -TITLE: Chern classes of pushforwards -QUESTION [5 upvotes]: Let $f:X\to Y$ be a proper morphism of normal varieties (smooth as DM stacks, but I only care about the coarse spaces). The map $f$ is generically finite, but not flat (so no hope of smoothness and Grothendieck-Riemann-Roch applying) and I have a fairly detailed understanding of the fiber over any point, images of $f$ restricted to divisors and so forth. -Now, take a divisor $D$, and identify it with an invertible sheaf in the standard way. I'm looking for a way to compute the first Chern class of $f_*D$ on $Y$. - -REPLY [6 votes]: Do you know the (generic) degree of your map $f$? As you probably know, standard intersection theory says $f_*[D] = n[f(D)]$ as classes in $A_{d-1}Y$, where $n$ is the degree of $f$ (restricted to $D$) and $d=\dim X = \dim Y$. No flatness or smoothness hypotheses on $f$ are needed for this; the sticky point is in identifying these divisors with line bundles. But since you're dealing with smooth DM stacks, that should be ok (over ${\Bbb Q}$ at least). -EDIT (incorporating the comments): For a proper map $f$, there is a map defined at the cycle level by $$f_*[D] = n\cdot [f(D)],$$ where $n$ is the degree of $D$ over $f(D)$ (i.e., degree of the induced field extension) when these have equal dimension, and $n=0$ when $\dim f(D)< \dim D$. This passes to rational equivalence, so defines a map $A_{d-1}X \to A_{d-1}Y$. In particular, if $f$ collapses a divisor $D$, then $f_*[D]=0$. -All this is in Fulton's Intersection Theory, Section 1.4.<|endoftext|> -TITLE: ZF and and the number of non-principal ultrafilters -QUESTION [12 upvotes]: ZF (if consistent) has models where $\omega$ supports no non-principal ultrafilters and others where it supports $2^c$ such. Can other cardinals occur? Or does the existence of just one somehow entail the existence of $2^c$. - -REPLY [10 votes]: Much of the following is David Feldman's argument in one of the comments, but with the topology replaced by combinatorics. There is, however, one slightly tricky point, arising from the fact that, in the absence of choice, the existence of a surjection from one set to another does not imply that the former set has at least the same cardinality as the latter. -Without any form of choice, we can get a family $X$ of $c$ independent subsets of $\omega$ (where "independent" means that the intersection of any finitely many of these sets and the complements of any finitely many others is infinite). Then any subfamily $Y$ of $X$ gives rise to the filter $F(Y)$ generated by the sets in $Y$ and the complements of the sets in $X-Y$. These filters are pairwise incompatible (i.e., the union of any two contains two complementary subsets of $\omega$), so no ultrafilter extends more than one of the $F(Y)$'s. Thus, we have a function $f$ assigning to each ultrafilter $U$ on $\omega$ the unique $F(Y)$ included in $U$ (or some fixed $F(Y)$, say $F(X)$, if $U$ includes no $F(Y)$). -Now if we assume the Boolean Prime Ideal Theorem (without which we're not guaranteed any nontrivial ultrafilters), we can extend each $F(Y)$ to an ultrafilter and thereby conclude that $f$ maps onto the whole set of $F(Y)$'s, a set of size $2^c$. By itself, that doesn't yet ensure that there are at least $2^c$ ultrafilters on $\omega$; existence of a surjection from one set to another does not in general (without choice) imply existence of an injection in the other direction. -In this case, however, it seems that BPIT enables us to select, for each $Y\subseteq X$, one ultrafilter extending $F(Y)$. That will give a one-to-one map from the set of $2^c$ $Y$'s into the set of ultrafilters. To prove that BPIT gives the desired selection, proceed as follows. Let $P$ be the product of $2^c$ copies of $\omega$, indexed by the subsets $Y$ of $X$; let $p_Y$ be the projection from $P$ to the $Y$'th factor $\omega$. For each $Y$, the filter $F(Y)$ can be "cylindrified" to produce the filter $F'(Y)$ on $P$ generated by the sets ${p_Y}^{-1}(A)$ for $A\in F(Y)$. These filters $F'(Y)$, for all $Y\subseteq X$, are coherent; their union generates a filter $F''$ on $P$. By BPIT, there is an ultrafilter $U$ on $P$ extending $F''$. The projections $p_Y(U)=\{A\subseteq\omega:{p_Y}^{-1}(A)\in U\}$ are ultrafilters on $\omega$ extending the corresponding $F(Y)$. -Conclusion: BPIT implies the existence of $2^c$ ultrafilters on $\omega$.<|endoftext|> -TITLE: Hodge structures on algebraic spaces -QUESTION [12 upvotes]: Let $X$ be a proper smooth algebraic space over $\mathbb C$ (which amounts, due to Artin, to giving a Moishezon space: a compact complex manifold whose dimension equals the transcendence degree of its field of meromorphic functions). I'd like to know if the classical Hodge theory holds on the cohomology of $X$ (e.g. degeneration of the Hodge vs de Rham spectral sequence and Hodge symmetry) and, if it does, could someone give a reference. -I will sketch a "proof" in the following. There are details to check. The case for proper smooth algebraic varieties over $\mathbb C$ is proved by Deligne, Théorème de Lefschetz et critères de dégénérescence de suites spectrales (Pub. math. de IHES, tome 35, 1968). Some key ingredients in Deligne's proof are: Chow's lemma, Hironaka's resolution of singularities and Serre duality. One has Chow's lemma for algebraic spaces (Knutson, Algebraic spaces, chp IV, 3.1), so there exists a projective birational morphism from a projective smooth algebraic variety to $X.$ -In the beginning we used two terminologies (algebraic space and Moishezon space) to remind myself that there is a GAGA issue when defining $H^q(X,\Omega^p)$ (of course we assume $X$ is not Kähler, so there's no *-operator and we don't use $(p,q)$-forms in defining $H^{q,p};$ we use $H^p(X,\Omega^q)$ as its definition). In his thesis, B. Toen proved (among many other things) GAGA for proper DM-stacks; he refered (SGA1, XII) on p.176 for GAGA for proper algebraic spaces, but I couldn't find the statement for algebraic spaces there. The proof in SGA1 uses Chow's lemma to reduce to the projective case which then follows from Serre's result, and one can probably use Chow's lemma for algebraic spaces to generalize the proof to algebraic spaces. I'm not sure if this is what Toen was thinking, since Knutson's book is not in the references in Toen's thesis. (I'll be grateful if someone can provide a reference where this is written down.) -For Serre duality, I learned from wiki that it works for holomorphic vector bundles on compact complex manifolds (not necessarily Moishezon), which suffices for our argument. Again, can someone give a reference for this version of Serre duality? -This is the sketched proof for the degeneration of Hodge vers de Rham and the Hodge symmetry, on proper smooth $\mathbb C$-algebraic spaces. The degeneration part (but not the Hodge symmetry part) can also be proved using reduction mod p and Deligne-Illusie. -I'll appreciate if someone can point out a mistake or give a reference. -Edit: This seems to be well-known to the experts. See Deformations of Kähler manifolds where Hodge decomposition fails? - -REPLY [7 votes]: If the Hodge to de Rham sequence degenerates for a smooth compact complex manifold $M$, it degenerates for any smooth compact manifold bimeromorphic to $M$, see theorem 5.22 of Deligne, Griffiths, Morgan, Sullivan, Real homotopy theory of K\"ahler manifolds (theorem 5.22 is about the $dd^c$ lemma, but this is equivalent as shown earlier in the same paper). Moishezon spaces are bimeromorphically projective, hence bimeromorphically K\"ahler. -Re the Serre duality for arbitrary smooth compact complex manifolds: I don't know a reference off hand, but if memory serves it can be found somewhere in Chapter 1 of Peters-Steenbrink, Mixed Hodge structures. -Again, if memory serves, in Chapter 2 of the same book the authors sketch a proof that the Hodge symmetry holds for the cohomology of any bimeromorphically K\"ahler manifold.<|endoftext|> -TITLE: Are completions stalks under some Grothendieck topology? -QUESTION [20 upvotes]: Let $R$ be a ring, and $\mathfrak{p}$ be a prime ideal. The stalk at $\mathfrak{p}$ with respect to the etale topology is $(R_{\mathfrak{p}})^{sh}$ (the strict henselization of $R_{\mathfrak{p}}$). The stalk at $\mathfrak{p}$ with respect to the Nisnevich topology is $(R_{\mathfrak{p}})^h$ (the henselization of $R_{\mathfrak{p}}$). -Grothendieck also spoke of formal neighborhoods, and I wonder if this fits into the pattern above. To be precise: is there some Grothendieck topology for which the stalk at $\mathfrak{p}$ would be the completion of $R_{\mathfrak{p}}$ with respect to $\mathfrak{p}$? If so, what is it? - -REPLY [20 votes]: You may want to look at the Artin approximation theorem. Roughly, it says (in the context of varieties, say) that any phenomenon you observe at the level of completions is already achieved etale locally. -E.g. suppose that $X$ is an algebraic curve over $k$ (an algebraically closed field) and $x$ is a closed point of $X$ such that completed local ring $\hat{\mathcal O}_{X,x}$ has two irreducible components in its spectrum. Then Artin's theorem says that there is an etale n.h. of $x$ which is the union of two branches passing through $x$. -This has the practical consequence that notions such as $x$ is a node'' can be defined either by a condition on the completion $\hat{\mathcal O}_{X,x}$ or by an etale local condition. The former is more elementary, and usually easier to check; the latter is often more powerful in proofs, because it is more tightly connected to the geometry of the whole curve $X$.<|endoftext|> -TITLE: Set theoretic question about real valued functions -QUESTION [10 upvotes]: It's a question I've been thinking about but I can't find an easy answer. I think it will be interesting. Can there be a countable collection of real valued functions $f_1, f_2 , ... $ such that for any subset $K$ of $\mathbb R$ of cardinality continuum, the set of those $n$ such that $f_n(K)$ is not the whole of $\mathbb R$ is finite? - -REPLY [6 votes]: I have three observations. -First, I have noticed that there can be no such sequence of -functions $f_n$, if one insists that every $f_n$ is a -measurable function. In particular, there can be no sequence of Borel functions with your property. To see this, suppose that -$f_n:\mathbb{R}\to\mathbb{R}$ is a countable sequence of -measurable functions. If infinitely many $f_n$ are not -surjective, then clearly the property will fail and we are -done. So we may assume that all but finitely many of the -functions are surjective, and by throwing away finitely -many functions, we reduce to the case where all $f_n$ are surjective. -For each natural number $n$ and real $y$, let -$A^y_n=f_n^{-1}(y)=\{\ x\ \mid\ f_n(x)=y\ \}$. For any fixed -$n$, the sets $A^y_n$ for various $y$ partition -$\mathbb{R}$ into continuum many disjoint measurable sets. -Since one cannot have uncountably many disjoint positive -measure sets, it follows that there must be some $y_n$ (and -in fact many such $y_n$) with $\lambda(A^{y_n}_n)=0$, where -$\lambda$ is the Lebesgue measure. Let -$K=\mathbb{R}-\bigcup_n A^{y_n}_n$, which is the complement -of a measure $0$ set, and so in particular, $K$ has size -continuum. Observe that $f_n[K]$ omits $y_n$ by -construction, and so there is no $n$ for which -$f_n[K]=\mathbb{R}$, showing that the desired property -fails. I guess the argument actually doesn't need that -every $f_n$ is measurable, but only that $f_n^{-1}(y)$ is a -measurable set for every $y$. -Second, this idea combines with the idea that Gowers had briefly posted yesterday, namely, the idea of making a large cardinal assumption, and allows us now to prove the following theorem. -Theorem. If the continuum is a real-valued -measurable cardinal (a hypothesis that is equiconsistent -over ZFC with the existence of a measurable cardinal), then -there is no such sequence of functions -$f_n:\mathbb{R}\to\mathbb{R}$. -Proof. Suppose that the continuum is a real-valued -measurable cardinal. This means that there is a -countably-additive real-valued measure $\mu$, measuring -every subset of $\mathbb{R}$ and giving points measure $0$. -We may assume that $\mu$ extends the Lebesgue measure. -Using this measure, every function $f_n$ is measurable, of -course, and so the idea of the paragraph above goes -through, using the new measure in place of the Lebesgue measure. QED -In particular, this shows that if large cardinals are consistent with ZFC, then a negative answer to your question is also consistent with ZFC. So one shouldn't expect to be able to build a sequence of functions provably exhibiting the property. -Finally, third, I noticed that if the Continuum Hypothesis holds, -then there is a sort-of-near-positive solution in the sense that there is a sequence of functions -$f_n:\mathbb{R}\to\mathbb{R}$ such that for every -uncountable set $K\subset\mathbb{R}$, the combined range is -onto, meaning $\bigcup_n f_n[K]=\mathbb{R}$. (And in fact, -the existence of such $f_n$ is equivalent to CH.) To see -this, let $\lt$ well-order $\mathbb{R}$ in order type -$\omega_1$. In particular, every real $x$ has only -countable many predecessors in $\lt$, so enumerate them -as $f_n(x)$ for $n\in\omega$. If $K\subset\mathbb{R}$ is -any uncountable set, then $K$ is unbounded in the -well-order, and so every real $y$ is a predecessor of some -element of $K$, and so $y=f_n(x)$ for some $x\in K$. Thus, -$\bigcup_n f_n[K]=\mathbb{R}$, as desired. -Your question is very interesting!<|endoftext|> -TITLE: The Invariant Subspace Problem: examples -QUESTION [9 upvotes]: Question. Is there a concrete example of a bounded linear operator on a Hilbert space for which it is not known if it has a non-trivial closed invariant subspace? -[Added 24.01.2011: According to Bernard Beauzamy (Introduction to Operator Theory and Invariant Subspaces, Elsevier (1988), p. 345), - -the operator which is "closest" to a counter-example is the one built by the present author: it has one hypercyclic point $x_0$, and for every polynomial $p$ with complex coefficients, $p(T)x_0$ is also hypercyclic. Therefore, the operator has a vector space of hypercyclic points (thus solving a question raised by P. Halmos), but it may still have points which are not cyclic at all, thus having Invariant Subspaces. - -Beauzamy refers to his manuscript "The orbits of a linear operator". I have not been able to find an electronic version of this manuscript (or paper) online. Does anyone know where one may find a description of the example? Is it presently known whether the operator in Beauzamy's example has an invariant subspace?] - -REPLY [5 votes]: It seems likely that the author of the question has found the reference in the meantime. I will provide it here for the sake of completeness. -The article containing the construction of the operator described in - Bernard Beauzamy (Introduction to Operator Theory and Invariant Subspaces, Elsevier (1988), p. 345 -can be found here: - -Bernard Beauzamy, An operator on a - separable Hilbert space with all - polynomials hypercyclic. Studia Math. - 96 (1990), no. 1, 81–90. MR1055079 - -Direct link to the document: -http://matwbn.icm.edu.pl/ksiazki/sm/sm96/sm9618.pdf -It seems still not to be known whether this bounded operator on a separable Hilbert space admits a non-trivial closed invariant subspace. Note that there are quite a number of articles containing a reference to the above article by Beauzamy (cf. subscription-only databases MathSciNet and ISI Web of Knowledge).<|endoftext|> -TITLE: Factoring a field extension into one which adds no roots of unity, followed by one which adds only roots of unity -QUESTION [12 upvotes]: I am asking my question here, since it's been voted up a fair bit on math.SE, but without answers, so it may be harder than I assumed it was. -Can we always break an arbitrary field extension $L/K$ into an extension $F/K$ in which the only roots of unity of $F$ are those in $K$, i.e. $\mu_F=\mu_K$, followed by an extension $L/F$ which is of the form $L=F(\{\omega_i\})$, where the $\omega_i$ are roots of unity? If not, are there reasonable hypotheses (e.g. separable, finite) on $L/K$ that would make this true? -My motivation was simply that the other order, i.e. breaking an arbitrary extension $L/K$ into one where $F=K(\{\omega_i\})$ for some roots of unity $\omega_i$, followed by $L/F$ where $\mu_L=\mu_K$, is obvious - specifically, set $F=K(\mu_L)$. -Now, my first (naive) approach was to try to construct the "maximum" intermediate field which does not add roots of unity by taking the compositum of all such intermediate extensions. However, this doesn't exist even for number fields, e.g. setting $K=\mathbb{Q}$, $L=\mathbb{Q}(\zeta_3,\sqrt[3]{2})$, $E_1=\mathbb{Q}(\sqrt[3]{2})$, and $E_2=\mathbb{Q}(\zeta_3\sqrt[3]{2})$, we have $\mu_{E_1}=\mu_{E_2}=\mu_K$, but $\mu_{E_1E_2}=\mu_{L}\supsetneq\mu_K$. -Note that $K=\mathbb{Q}$ and $L=\mathbb{Q}(\zeta_3,\sqrt[3]{2})$ isn't a counterexample to the actual problem - for example, $F=E_1$ works, because $\mu_{E_1}=\mu_{\mathbb{Q}}$, and $L=E_1(\zeta_3)$. -So, to prove the claim / construct a counterexample, it seems to me that we want to look at intermediate fields $E$ which are maximal among those such that $\mu_E=\mu_K$, and determine whether or not there always exists at least one such $E$ such that $L=E(\text{some roots of unity})$. - -Here is Arturo Magidin's comment on the original question: - -not a proof/counterexample, but an observation: suppose $L$ is Galois over $K$; we can let $M$ be the extension of $K$ obtained by adding all roots of unity in $L$; this is Galois over $K$, so corresponds to a normal subgroup $H$ of $G=\text{Gal}(L/K)$. If we can break up the extension as you mention, then $L$ is Galois over $F$, and $\text{Gal}(L/F)=\text{Gal}(M/K)=G/H$. So $G$ would necessarily have normal subgroup $H$ and a subgroup isomorphic to $G/H.$ - -REPLY [6 votes]: $\newcommand{\Q}{\mathbf Q} \newcommand{\F}{\mathbf F} -$Here is an example which is essentially in the same spirit as Peter McNamara's but perhaps somewhat simpler to understand. -Let $K$ be a finite extension of $\Q_p$, let $l$ be a prime $\neq p$, and suppose that $K$ does not contain a primitive $l$-th root $\zeta$ of $1$, so that the extension $K'=K(\zeta)$ of $K$ has degree $>1$. Note that $K'|K$ is unramified, of group $\Gamma$ (say) which is naturally a subgroup of $\F_l^\times$. -Cyclic extensions $L|K'$ of degree $l$ correspond to $\F_l$-lines $D$ in $K'^\times/K'^{\times l}$ under $L=K'(\root l\of D)$. Such an $L$ is of the form $L=F(\zeta)$ for some degree-$l$ extension $F$ of $K$ if and only if the line $D$ is $\Gamma$-stable. -Now, as a $\Gamma$-module, the $\F_l$-space $K'^\times/K'^{\times l}$ is isomorphic to $\mu_l\oplus\mathbf{F}_l$, so it certainly contains lines which are not $\Gamma$-stable. -In other words, there are degree-$l$ cyclic extensions $L$ of $K'=K(\zeta)$ which are not of the form $L=F(\zeta)$ for a degree-$l$ extension $F$ of $K$. -The simplest instance of this phenomenon occurs for $K=\Q_2$, $l=3$: the extension $L=\Q_2(\zeta,\root3\of{2\zeta})$ of $\Q_2$ is not of the form $F(\zeta)$ for any cubic extension $F$ of $\Q_2$ basically because otherwise $L|\Q_2$ would be galoisian, which it is not. -Note finally that in this entire discussion the finite extension $K$ of $\Q_p$ can be replaced by $k((T))$ for any finite extension $k$ of $\F_p$.<|endoftext|> -TITLE: Does every polyomino tile R^n for some n? -QUESTION [78 upvotes]: This is a question posed by Adam Chalcraft. I am posting it here because I think it deserves wider circulation, and because maybe someone already knows the answer. -A polyomino is usually defined to be a finite set of unit squares, glued together edge-to-edge. Here I generalize it to mean a finite set of unit hypercubes, glued together facet-to-facet. -Given a polyomino $P$ in $\mathbb{R}^m$, I can lift it to a polyomino in a higher-dimensional Euclidean space $\mathbb{R}^{m+n}$ by crossing it with a unit $n$-cube: the lifted polyomino is just $P\times [0,1]^n$. -Obviously, not all polyominos tile space. - -Is it true that given any polyomino $P$ in $\mathbb{R}^m$, there exists some $n$ such that the lifted polyomino $P\times [0,1]^n$ tiles $\mathbb{R}^{m+n}$? - -Many people's first instinct is that multiply-connected polyominos (those with "holes" in them) can't possibly tile, but you can get inside holes if you lift to a high enough dimension. - -REPLY [37 votes]: A positive answer to this question has just appeared in the arXiv: -Tiling with arbitrary tiles; -Vytautas Gruslys, Imre Leader, Ta Sheng Tan; -http://arxiv.org/abs/1505.03697<|endoftext|> -TITLE: What properties should a good definition of (weak) $n$-category satisfy? -QUESTION [11 upvotes]: My (perhaps inaccurate) impression is there are many competing definitions of the notion of a (weak) $n$-category, none of which are generally accepted. I've run across several properties such definitions should satisfy, e.g. that weak $n$-groupoids should model homotopy $n$-types, or that the collection of $n$-categories should naturally be an $n+1$-category. But I haven't found anything like a comprehensive list of conjectural properties a "good" theory of weak $n$-categories should satisfy. - -Is there such a list somewhere in the literature? - -If not, I'd of course be thrilled with an answer containing such a list. - -REPLY [5 votes]: Also read some of the original papers by Baez and Dolan, for example Categorification (in Higher Category Theory, Contemporary Mathematics 230, AMS 1998), where the Tangle Hypothesis is explained among other things. -It is hoped that a satisfactory definition on n-category would provide a framework for proving the homotopy hypothesis, the tangle hypothesis, and the stabilization conjecture, as discussed at the n-Category Café here. -I wouldn't say that none of the proposed definitions is generally accepted, but rather that none has emerged as clearly the best definition. In some sense all are accepted as capturing some useful intuition, although some seem easier to work with than others for some specific purpose. Perhaps the most developed notions are the more geometric ones which take advantage of decades of work in homotopy theory.<|endoftext|> -TITLE: Who uses radicial morphisms? -QUESTION [24 upvotes]: In editing the algebraic geometry notes posted here, prompted by Brian Conrad, I am introducing the notion of radicial morphism. This seems to me to not be a notion that absolutely everyone should see in a first serious schemes course, given the volume of definitions a student must digest (even if this idea is wafer-thin). So I would like to make clear to the learner who this notion is for. But the problem is: I don't know. I have two bits of data: I've never used it professionally, and I know some arithmetic geometers (e.g. in my department) have used it. Thus I ask: - -Who uses radicial morphisms? - -In the answers, I expect to hear multiple interesting answers to the implicit question: - -What are radicial morphisms good for? - -REPLY [3 votes]: Here is a fun example of a place where radicial morphisms get used: in the valuative criterion for when a map of schemes is a locally closed embedding (i.e. the composition of a closed embedding and an open embedding), due to Mochizuki. -Roughly speaking, this valuative criterion * says that if we are given a map $X \to Y$ and a map from the spectrum of a DVR $T$ to $Y$ such that both points of the DVR map to $X$ with the natural diagrams commuting, then we have a map from $T$ to $X$ making the diagrams commute. -Finite type maps of noetherian schemes satisfy * if and only if they can be expressed as the composition of a radicial map and an open immersion. From this, one deduces that finite type maps of noetherian schemes are locally closed immersions if and only if they are monomorphisms and satisfy *. -The only proof I know for this valuative criterion for locally closed immersions, which seems like a natural thing to want, goes by first proving the statement about radicial maps. -Of course, none of this should go in a first course on schemes, but I just thought it was a fun application of radicial maps.<|endoftext|> -TITLE: Nonstandard models of PA of large cardinal size -QUESTION [6 upvotes]: It is easy to overlook the fact that the existence of a given large cardinal provides us with a true arithmetical statement that would otherwise be false if the large cardinal notion were not consistent with ZFC (See On statements independent of ZFC + V=L ). The arithmetical statement that I'm referring to here is CON(ZFC + large cardinal notion). This is an example of where we use a fact that is external to the set of Natural numbers, the existence of the large cardinal, to prove a result that's internal to $\mathbb{N}$. If we want a less esoteric statement of looking externally (not using a large cardinal notion) to prove an internal result about the Natural numbers, we can consider Goodstein's Theorem. Goodstein's Theorem states that a certain infinite collection of sequences, almost all of which grow very quickly on some initial segment, eventually descend to 0. The amazing result is that while this fact is not provable in Peano arithmetic, we can give a very simple proof of it in ZFC. In this case, we consider sequences of infinite ordinals to prove a true arithmetical statement representable in PA. -Let me now switch gears a little to say something about large cardinals. Assuming increasingly strong large cardinal axioms opens up the possibility for increasingly complex transitive set models of ZFC and then increasingly complex definable inner models of ZFC. Nevertheless, even if we assume a sufficiently strong large cardinal hypothesis, say the existence of a weakly compact cardinal $\kappa$, the elementary embeddings that arise fix all "small" elements (hereditary size less than $\kappa$ in the domain). I would therefore like to consider nonstandard models of PA that will not be fixed by such embeddings in an effort to make use of the large cardinal assumptions in a meaningful way for revealing internal truth by external examination. In order to avoid asking a rather vague question, let me pose it as follows: - -Has anyone considered models of PA of large cardinal size? - -As an example of what I have in mind, assume that we have a nonstandard model $M$ of PA of size $\kappa$ containing an unbounded well-order ( See Uncountable nonstandard models of PA ) for $\kappa$ supercompact. Can we use the fact that $\kappa$ is supercompact to reveal any interesting number-theoretic properties true in $M$? - -REPLY [5 votes]: See Section 4.2 of Harvey Friedman's book draft for the details of a combinatorial, non-metamathematical application of n-Mahlo-sized extensions of structures of the form $(\omega,<,0,1,+,f,g)$. -In this example, $f$ and $g$ are not allowed to grow fast enough for either to be chosen to be the multiplication function. More generally, I am not aware of any "interesting" number-theoretic applications of large cardinals. That said, perhaps you should ask Friedman about large-cardinal-sized models of PA.<|endoftext|> -TITLE: Applications of periodic continued fractions -QUESTION [7 upvotes]: Some answers from Applications of finite continued fractions in fact are Applications of periodic continued fractions. I think that it should be separate question. -What can you add to the following list of applications? -1) Calculation and approximation of quadratic irrational numbers. calculation of corresponding covex hulls. -2) Pell equation and calculation of fundamental units in quadratic fields. -3) Reduction of quadratic forms. Calculation of class numbers of imaginary quadratic field. -4) Legendre's factorization method. - -REPLY [3 votes]: The conjugacy problem in $SL(2,Z)$. For matrices $M \in GL(2,Z)$ having trace of absolute value $>2$, the slope of its expanding eigenvector has an eventually periodic continued fraction expansion (it is a quadratic irrational), and the primitive period loop is a conjugacy invariant in $SL(2,Z)$. Throw in the absolute value of the trace itself and you have a complete conjugacy invariant in $SL(2,Z)$.<|endoftext|> -TITLE: Is there a "deep" reason that the first Perrin pseudoprime is large? -QUESTION [13 upvotes]: Let $f(x) \in \mathbb{Z}[x]$ be a monic irreducible polynomial with roots $\alpha_1, ... \alpha_k$, and let $\Delta$ be the discriminant of $f$. For any prime $p \nmid \Delta$, the Frobenius morphism permutes the roots of $f$ in $\mathbb{F}_p$, hence in particular -$$\alpha_1^p + ... + \alpha_k^p \equiv \alpha_1 + ... + \alpha_k \bmod p.$$ -(This result is also true for primes dividing the discriminant, but I don't know an algebraic argument, just a combinatorial one.) A positive integer $p \ge 2$ with this property is a kind of pseudoprime with respect to $f$, which I'll call an $f$-pseudoprime. (This is related to but weaker than the notion of a Szekeres pseudoprime or a Frobenius pseudoprime with respect to $f$.) In particular, a Fermat pseudoprime with base $a$ is an $(x-a)$-pseudoprime. If $f$ is quadratic one gets a notion of pseudoprime related to (equivalent to?) the Lucas / Fibonacci pseudoprimes. -A Perrin pseudoprime is an $(x^3 - x - 1)$-pseudoprime, and the smallest Perrin pseudoprime which is not prime is $271441 = 521^2$. In another MO thread Kevin O'Bryant mentioned that Freeman Dyson and others consider the size of this pseudoprime surprising and suspect there might be a good explanation of why it is larger than one might naively expect (akin to the explanation of why $e^{\pi \sqrt{163} }$ is close to an integer). -I'm not convinced this is a phenomenon requiring a deep explanation. Hence what I would like to see is - -A heuristic relating the size of the smallest $f$-pseudoprime which is not prime to the "complexity" of $f$ (to be defined freely) - -and either - -A computation showing that the result for $f = x^3 - x - 1$ is consistent with the heuristic, or -A computation showing that the result for $f = x^3 - x - 1$ is not consistent with the heuristic, and some speculation about why this should be the case. - -REPLY [11 votes]: A Carmichael number all of whose factors have $f$ splitting completely is an $f$-pseudoprime (sufficient, but not necessary). So if you take any heuristic that there are infinitely many Carmichael numbers (in particular with a fixed number of factors) and combine it with the idea that the probability of getting all factors splitting completely for a polynomial of degree $d$ is $(1/d!)^k$ (where $k$ is the number of factors), you can get a more elementary heuristic. My paper essentially followed that, but had to deal with the fact that we don't have infinitely many Carmichael numbers proven for any particular $k$. -As for the original question, has anyone computed the smallest $f$-pseudoprime with respect to a whole bunch of degree 3 irreducible polynomials and seen if the Perrin case is an outlier? That would be my first step.<|endoftext|> -TITLE: Algebraicity of Eigenvectors in a Hilbert space -QUESTION [11 upvotes]: Let $(e_j)_{j\in\mathbb N}$ be an orthonormal basis of a Hilbert space $V$. Let $T:V\to V$ be continuous, selfadjoint linear operator. -Assume that for all $i,j\in\mathbb N$ the number $\langle Te_i,e_j\rangle$ is rational. -Let $\lambda\in\mathbb C$. Suppose there exists an eigenvector $v\in V$ with $Tv=\lambda v$. -The question is this: Does there exist an eigenvector $w$ for the eigenvalue $\lambda$ such that $\langle w,e_j\rangle$ lies in the field ${\mathbb Q}(\lambda)$ for every $j\in\mathbb N$? -This is true for finite dimensional spaces, even without the self-adjointness, since $\lambda$ is a zero of the characteristic polynomial. -If it turns out to be false as stated, you may add further conditions on $T$, like the one that it preserves the linear span of the basis vectors. - -REPLY [15 votes]: The answer is no. -Update. Here is an explicit example (which also preserves the $\mathbb Q$-span of the basis). Let $V=\ell_2$ over $\mathbb R$ and $(e_j)$ the standard basis. Define $T:V\to V$ by -$$ -\begin{aligned} - (Tx)_1 &= x_1+3x_2+x_3 \\ - (Tx)_2 &= 3x_1+10x_2+6x_3+x_4 \\ - (Tx)_n &= x_{n-2}+6x_{n-1}+11x_n+6x_{n+1}+x_{n+2}, \qquad n>2 . -\end{aligned} -$$ -Let $\alpha$ be the root of $p(x)=x^2+3x+1$ such that $|\alpha|<1$. Then the vector $v\in\ell_2$ defined by $v_n=\alpha^n$ satisfies $Tv=0$, so it is an eigenvector for $\lambda=0$. On the other hand, no nonzero rational vector $w$ satisfies $Tw=0$ because the relation $Tw=0$ implies the recurrence relation -$$ - w_{n+4}=-6w_{n+3}-11w_{n+2}-6w_{n+1}-w_n , -$$ -which does not have rational solutions tending to 0. -Below is the original answer. -Let $V=\ell_2$ over $\mathbb R$ and $(e_j)$ the standard basis, so that $\langle x,e_j\rangle=x_j$ for every $x\in\ell_2$. Let $v\in V$ be a unit vector with irrational ratios of coordinates (for definiteness, take a geometric progression like $(c\pi^{-n})_{n\in\mathbb N}$). Then there exist a continuous self-adjoint $T:V\to V$ with rational coordinates such that $\ker T=\langle v\rangle$. So we have $\lambda=0$ but no $\lambda$-eigenvector is rational. -I represent $T$ by its (infinite) matrix $(t_{ij})_{i,j\in\mathbb N}$, $t_{ij}=\langle Te_i,e_j\rangle$. This matrix should be symmetric, and continuity of the resulting operator should be taken care of. -Begin with a self-adjoint $A$ such that $\ker A=\langle v\rangle$ with irrational components $a_{ij}$. For example, let $A$ be the projection to the orthogonal complement of $v$, so $a_{ij}=\delta_{ij}-v_iv_j$. I am going to approximate the matrix $(a_{ij})$ by a rational matrix $(t_{ij})$ such that the kernel stays the same. -Since $Av=0$, each row of our matrix is orthogonal to $v$. Approximate the first row $a_i=(a_{1i})$ by a rational vector $t_1=(t_{1i})$ such that $\langle t_1,v\rangle=0$ and $|t_{1i}-a_{1i}|$ is bounded by a rapidly decaying geometric progression (see below for details). Replace the first row and the first column of the matrix by this approximation. Then adjust the diagonal elements $a_{ii}$ so that the rows remain orthogonal to $v$, namely change $a_{ii}$ to $a_{ii}-(a_{1i}-t_{1i})v_1v_i^{-1}$. Recall that $v_i$ is a not-so-fast decaying geometric progression, so this adjustment is a small change. -Now remove the first row and the first column from the matrix, and the first component from $v$. Apply the same procedure to the truncated data: namely approximate the first remaining row and the first remaining column by a rational vector whose scalar product with $v$ is the same as before, then adjust the diagonal elements so that the scalar products of all rows with $v$ are preserved. Repeat ad infinitum. Note that every element of the original matrix is changed only finitely many times, and $v$ belongs to the kernel of the matrix after each step. -The approximations are controlled and the adjustments are bounded in terms of approximations, so the sum $\sum (a_{ij}-t_{ij})^2$ can be made arbitrarily small. This implies that $T$ is continuous and $\|T-A\|$ is small (say, less than 1/2), so we have $Tv=0$ and $\|Tw\|\ge\frac12\|w\|$ for all $w$ orthogonal to $w$. Therefore $\ker T=\langle v\rangle$. However $v$ cannot be rescaled to a rational vector. -It remains to show that every vector $w\in\ell_2$ can be approximated by a rational vector $w'$ such that $\langle w',v\rangle=\langle w,v\rangle$ and $w'_i-w_i$ is bounded by a small, fast decaying progression. Let $w'_1=w_1+\varepsilon_1$ be a rational approximation of $w_1$, then let $w'_2=w_2-\varepsilon_1v_1v_2^{-1}+\varepsilon_2$ be a rational approximation of $w_2-\varepsilon_1v_1v_2^{-1}$, then let $w'_3=w_3-\varepsilon_2v_2v_3^{-1}+\varepsilon_3$ be a rational approximation of $w_3-\varepsilon_2v_2v_3^{-1}$, and so on. The $\varepsilon_i$ at each step can be chosen arbitrarily small, and the resulting vector $w'$ satisfies $\langle w'-w,v\rangle=0$.<|endoftext|> -TITLE: Are class numbers encoded in the absolute Galois group of ${\mathbb Q}$? -QUESTION [22 upvotes]: The absolute Galois group $G_{\mathbb Q}=\text{Gal}(\bar{\mathbb Q}/\mathbb Q)$, as a profinite group, encodes a lot of things: the whole lattice of number fields (closed subgroups of finite index), Galois extensions (normal subgroups), abelian extensions etc. Is it possible to recognize unramified abelian extensions from the group-theoretic data? So, - -Question: Given a closed subgroup $H\subset G_{\mathbb Q}$ of finite index, is there an explicit group-theoretic way to recover the class group or the class number of the fixed field $K=\overline{\mathbb Q}^H$? - -(Because all continuous automorphisms of $G_{\mathbb Q}$ are inner, in theory $K$ itself can be recovered from $H$ and its class group computed, but this is not what I'd like to call "group-theoretic" or "explicit".) - -REPLY [17 votes]: Dear Tim, -As you're probably aware, this is part of the 'anabelian' etcetera. -It suffices to recover all intertia subgroups $I_v\subset H$, because their union will then be a normal subgroup $N$ such that $H/N$ is the Galois group of the maximal extension of $K$ unramified everywhere. We can get the ideal class group then by (topological) abelianization. The fact that we can get all the decomposition groups $D_v\subset G$ is Neukirch's theorem (together with Artin-Schreier at infinity). This says the maximal subgroups isomorphic to a local Galois group are exactly the decomposition groups. If you want to make this purely group-theoretic for the finite places, you invoke the theorem of Jannsen-Wingberg that lays out a presentation for all local Galois groups and consider maximal elements in the lattice of subgroups isomorphic to such an explicit presentation. Once you have the $D_v$, there is a standard group-theoretic recipe for $I_v$, which escapes me for the moment. But I'll get back to you with it, if you don't figure it out in the meanwhile. -Added: -OK, so here is the easy part. Now let $F$ be a finite extension of $\mathbb{Q}_p$ and -$D=Gal(\bar{F}/F)$. We know that $D^{ab}$ fits into an exact sequence -$$0\rightarrow U_F\rightarrow D^{ab}\rightarrow \hat{\mathbb{Z}}\rightarrow 0,$$ -so we recover $p$ as the unique prime such that the topological $\mathbb{Z}_p$-rank of -$D^{ab}$ is $r_D\geq 2$. The order $q_D$ of the residue field is 1 greater than the order of -the prime-to-$p$ torsion subgroup of $D^{ab}$. Also, we know -$r_D=1+[F:\mathbb{Q}_p]$. Now we apply the same reasoning to the subgroups of finite index in -$D$ to figure out those corresponding to unramified extensions. That is, consider -the subgroups $E$ of finite index such that $q^{r_D-1}_E=q_D^{r_E-1}$. Then the inertia subgroup -of $D$ is the intersection of all of these.<|endoftext|> -TITLE: $2$-torsion line bundles on abelian varieties -QUESTION [7 upvotes]: Let $\mathcal{A}_{g,D}$ be the moduli space of abelian varieties of dimension $g$ and polarization $D$ of type $(d_1, \ldots, d_g)$. -Let $\mathcal{M}$ be the moduli space parametrizing pairs $(A, \mathcal{L})$, where $A \in \mathcal{A}_{g,D}$ and $\mathcal{L}$ is a non-trivial $2$-torsion line bundle on $A$, i.e. a non-zero element of $\textrm{Pic}^0(A)$ such that $\mathcal{L}^{\otimes 2}=\mathcal{O}_A$. -Then there is a covering -$\pi \colon \mathcal{M} \longrightarrow \mathcal{A}_{g,D}$ -of degree $2^{2g}-1$, given by $[A, \mathcal{L}] \to A$. -Question Is the monodromy group of $\pi$ transitive? Or, equivalently, is $\mathcal{M}$ connected? -The answer is yes when $g=1$, i.e. for elliptic curves. In fact in this case $\mathcal{M}$ is a particular case of a more general construction called the moduli space of spin curves, which was studied by several authors (Cornalba, Verra, Farkas, etc). -What about the case $g \geq 2$? Is there any reference? I'm particularly interested to the case where $g=2$ and $D=(1,2)$. -EDIT Let me explain better the case I'm interested in, hoping that this can be helpful. Let $(A, D)$ be an abelian surface with polarization of type $(1,2)$, which I assume to be not of product type. The linear system $|D|$ is a pencil, that is $h^0(D)=2$, its general element is irreducible and up to a translation we can take $\mathcal{O}_A(D)$ symmetric, i.e. $(-1)_A^* \mathcal{O}_A(D)= \mathcal{O}_A(D)$. -Therefore the base locus of $|D|$ is given by the zero element $o$ of $A$ and by three $2$-division points $e_1$, $e_2$, $e_3$ such that $e_1+e_2=e_3$. -There are exactly three $2$-torsion line bundles $\mathcal{L}_1$, $\mathcal{L}_2$, $\mathcal{L}_3$ on $A$ such that there exists an element in the "translated pencil" $|D \otimes \mathcal{L}_i|$ having a double point at $o$ (which is easily proven to be a node). The set -$\{\mathcal{O}_A, \mathcal{L}_1, \mathcal{L}_2, \mathcal{L}_3\}$ -form a subgroup of $\textrm{Pic}^0(A)$ isomorphic to $\mathbb{Z}/(2) \times \mathbb{Z}/(2)$, which is exactly the image of the map -$\phi \colon A[2] \longrightarrow \textrm{Pic}^0(A)[2]$, -see Laurent Moret-Bailly's answer. - -REPLY [7 votes]: Let's stick to $D=(1,2)$. Associated to any point of $\mathcal{M}$ (say, a complex point) we have an abelian surface $A$ and an isogeny $\phi:A\to A'$ where $A'=\mathrm{Pic}^0(A)$ is the dual. The kernel of $\phi$, in our case, is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^2$. So, the image of $A[2]$ in $A'$ is a canonically defined subgroup $H$ of $A'[2]$, isomorphic to $(\mathbb{Z}/2\mathbb{Z})^2$. On the other hand, the line bundle $\mathcal{L}$ is a nontrivial point in $A'[2]$. Clearly, those $(A,\phi, \mathcal{L})$ such that $\mathcal{L}\in H$ form a nontrivial open and closed subset of $\mathcal{M}$, so $\mathcal{M}$ is not connected.<|endoftext|> -TITLE: Gromov-Witten invariants counting curves passing through two points -QUESTION [11 upvotes]: Let us say that a closed symplectic manifold $X$ is $GW_g$-connected if there is a nonvanishing Gromov-Witten invariant of the form -$GW_{g,n}^{X,A}(\beta,point, point,\alpha_3,\ldots,\alpha_n)$ --in other words a nonvanishing invariant that formally counts (pseudo-)holomorphic curves of genus g passing through two generic points and satisfying some other constraints $\beta$ coming from $\bar{M}_{g,n}$ -and $\alpha_i$ coming from other incidence conditions in $X$. -When $g=0$ this is something like saying that $X$ is rationally connected in the algebro-geometric sense, and there's been recent work (such as arXiv:1006.2486) relating to the question of whether the notions are the same. But in higher genus an analogous statement should fail--for instance in the product X of two elliptic curves there's a (reducible) genus two curve passing through any two points, but -X is certainly not $GW_2$-connected. - -Question: For which symplectic four-manifolds (or Kahler surfaces) $X$ does there exist g such that $X$ is $GW_g$-connected? - -My personal motivation for this question comes from the fact that if $X$ is $GW_g$-connected for some g then by a result of Lu $X$ has finite Hofer-Zehnder -capacity; however the question seems reasonably interesting aside from that. I restrict to dimension four here only because I expect doing so to make the question more tractable; insights into higher-dimensional cases would also be welcome. - -Here are some preliminary observations in the direction of an answer: -It's an easy consequence of a result of McDuff that the only symplectic four-manifolds that are $GW_0$-connected are the rational ones (i.e. those related to -$\mathbb{C}P^2$ by blowups and blowdowns). -For larger g, I've convinced myself that it's likely that any ruled surface over a curve of genus g ought to be $GW_g$-connected, though I haven't written -down a careful proof--if someone knows where one can be found or knows that I'm wrong about this I'd be glad to hear about it. -I'd expect that symplectic four-manifolds with $b^+>1$ (for complex surfaces this means $p_g>0$) should rarely if ever have this property, since they -typically don't have GW invariants counting curves with nontrivial incidence constraints. In fact for Kahler surfaces with $p_g>0$ this follows from a result of Lee and Parker. -For symplectic manifolds with $b^+=1$ which are not rational or ruled I'm not really sure what to expect. These usually have a decent supply of nontrivial -Gromov-Witten invariants (as can be seen from Taubes-Seiberg-Witten theory), but it's not clear to me in general whether one should expect a nonvanishing -invariant with two point constraints. -EDIT: Since originally posting this question I looked a little more carefully at the literature on four-manifolds with $b^+=1$, and found that work of Li, Liu, and others based on Taubes-Seiberg-Witten theory is enough to show that any closed symplectic four-manifold with $b^+=1$ is $GW_g$-connected for some $g$. I've provided details of the argument in the appendix of this preprint. -So it seems likely that the answer to the original question is that a closed symplectic four-manifold is $GW_g$-connected iff it has $b^+=1$: the backward implication is always true (by Li-Liu), and the forward implication is definitely true if one restricts to Kahler surfaces (by Lee-Parker), and there are also many non-Kahler examples for which it can be checked. It seems a good deal harder to say anything about higher-dimensional cases. - -REPLY [6 votes]: I am not familiar with symplectic geometry so let's assume everything here is at least K\'ahler. -If $g=1$, then the condition $\langle [pt], [pt], \ldots \rangle^X_{1, [C]}\neq 0$ implies that the variety is uniruled, which is equivalent to $\langle [pt], \ldots \rangle^X_{0, {C}}$. -I hope it is true that for a rationally connected fibration over a curve of any genus, your condition $\langle \beta, [pt], [pt], \ldots \rangle^X_{g, [C]}\neq 0$ is always true. And it is true when the fiber dimension is at most $2$. Basically as long as you know that there is a section which gives non-zero GW invariant, you can glue this section with curves in a general fiber which is minimal among all curves with non-vanishing GW invariant $\langle [pt], [pt], \ldots \rangle$. -For ruled surface, what you said is true. The methods used in the paper here certainly work.<|endoftext|> -TITLE: Algorithm for k-medians in a convex polygon -QUESTION [6 upvotes]: Are there any known approximation algorithms or exact solution schemes for the k-medians problem in a convex polygon? That is, placing a collection of points $p_1,\dots,p_k \subset \mathbb{R}^2$ in a convex polygon $C$ so as to minimize $$\iint_C \min_i \|x-p_i\| dx$$ - -REPLY [4 votes]: A while ago I wrote, but never published, an approximation algorithm for this problem. Using some new results and updating the citations, it looks like I can get the approximation constant down to 9.026 (assuming I didn't make any mistakes). It's not clear to me if that's publication-worthy, but I uploaded a draft to -http://www.tc.umn.edu/~jcarlsso/fermat-weber.pdf -if anyone is interested.<|endoftext|> -TITLE: The Frobenius morphism -QUESTION [30 upvotes]: I found the following list on the "Frobenius Page" by David Ben-Zvi, described by the author as "an outdated collection of intuitive ways to think about raising to the p-th power". - - -Generates a copy of the integers (or rather their profinite completion) in local Galois groups. Geometrically, this gives us an "infinitesimal circle" above every point, for which Frobenius is the monodromy. E.g. an algebraic curve over a finite field can be thought of as a 3-manifold fibering over the circle, with Frobenius as the circle direction. -Provides natural grading on "mixed" objects (e.g. cohomologies of singular schemes, perverse sheaves etc.) by its eigenvalues. Purity: only one graded piece. Key for results such as the Decomposition Theorem. -Pullback of deRham complex becomes $\mathcal{O}$-linear. (Derivative of pth power is zero - simplifies calculus!) Leads to the Cartier operator on cohomology. Linear structure enables e.g. Deligne-Illusie to prove degeneration of Hodge-to-deRham. -On same note, sheaves acquire a canonical connection after Frobenius pullback - when Frobenius is an isomorphism. Counterexample: $X$ abelian variety, $H^0(X,\Omega)$ in $H^1(X)$ (first piece of Hodge filtration) is killed by Frobenius - since every one-form is locally exact. -Its failure to commute with a connection gives rise to p-curvature, a canonical $\mathcal{O}^p$ - linear form. This measures the obstruction to exponentiating a connection through the p-th order (p!=0 => need for divided powers..) -p-curvature of the Gauss-Manin connection as Kodaira-Spencer class (Katz). -Provides an automorphism of the additive group - generates ring of additive polynomials. Analog of ring of differential operators. Embeddings of coordinate rings into this ring (Drinfeld modules) provide important moduli problems, key to Langlands program for function fields. -Gives rise to automorphisms of flag varieties (for example) => ask for a flag and its Frobenius translates to have prescribed relative position: Deligne-Lusztig varieties (Borel-independent analogs of Schubert cells). -The kernels of Frobenius and its iterates give natural infinitesimal neighborhoods of the identity on group schemes. -F-crystals: both "flat connection" and Frobenius structure. F is horizontal, but nonetheless eigenvalues not constant -> Newton polygons, and related stratifications of moduli schemes. -As a contracting map - can apply fixed point formulas after applying Frobenius. -Dwork's principle - use Frobenius to fix constant terms to do p-adic integration (use to patch together local pieces). -Frobenius liftings to characteristic zero - p-adic analogs of Kahler metrics (Mochizuki). -Together with Verschibung and homothety generates the ring of Cartier operators. Appropriate modules over this (Dieudonne modules) are essential to classification of group schemes. - - -I think it would be very interesting if some MO users (among whom maybe even the author himself) would - -Expand the description of each point 1.-10. also explaining his/her own intuitive understanding of the phenomenon, and giving some examples and reference to interested readers. -Add new items to the list. - -REPLY [3 votes]: Connes, Consani and Marcolli developed a theory of a "Frobenius in char. 0", related there to inherent time evolutions of von Neumann-algebras and providing an unusual way to view the char p cases. (link)<|endoftext|> -TITLE: L-functions and higher-dimensional Eichler-Shimura relation -QUESTION [16 upvotes]: From what I have been reading I understand that it is a part of the Langlands program to express Zeta-function of a Shimura variety associated to the group G in terms of L-functions attached to G. I am not sure to what extent this already has been proven. -In the cases I "know", which are the modular curves and Shimura curves - the essential step is to use the Eichler-Shimura congruence relation, which gives a connection between the Frobenius action and the Hecke correspondences. -Now, in the higher-dimensional case, say that of Shimura varieties of PEL-type, is there a generalization of the Eichler-Shimura relation, and does this generalization play the same role in the proof of the statement? -Maybe someone could give a rough overview of what is used in the proof in the known cases (as far as I understand, PEL case is known)? -Thank you. - -REPLY [24 votes]: Surprisingly, the case of modular curves is misleading! General theory of correspondences, plus the theory of the mod $p$ reduction of curves like $X_0(Np)$ ($p$ doesn't divide $N$) give a relationship between the Hecke operator $T_p$ and the Frobenius endomorphism $Frob_p$ acting on, say, the Jacobian of a modular curve. But now there's a coincidence: from this data we can figure out a degree two polynomial satisfied by $Frob_p$ on the chunk of the Tate module of the Jacobian which corresponds to a given weight 2 eigenform, and we know for other reasons that this chunk is 2-dimensional, and it's not hard to tease out from this a proof that the degree two poly we know must be the char poly of Frobenius. I would recommend Brian Conrad's appendix to the Ribet-Stein paper for a modern treatment of this story. -But pretty much the moment one leaves the relative safety of curves, this trick doesn't work, because the arithmetic geometry gives you a polynomial which kills Frobenius, but you don't have enough information to check that this poly is the char poly. So you have to do something vastly more complicated! This vastly more complicated thing was figured out by Langlands in the case of modular curves (the details are mostly buried in his Antwerp paper). Another place to look for the modular curve case is Clozel's Bourbaki article from about 1993, although he skips many of the details when it comes to cusps and supersingular points. -OK so Langlands' proof looks like this. We need to check that the $L$-function of a modular curve is a product of $L$-functions coming from automorphic forms. The $L$-function of a modular curve is a product of local terms, and the local term at $p$ can be computed if one knows the number of $k$-points on the curve for $k$ running through all finite fields of char $p$. The goal is to relate these numbers to numbers coming from the theory of automorphic forms---basically numbers related to the Hecke operators at $p$. Langlands observes however that he can count the number of $k$-points on a modular curve: there are cusps, which aren't too hard, and then there are supersingular points, which come out as some explicit number, and then there are ordinary points, which he counts isogeny class by isogeny class. Langlands can compute isogeny classes using Honda-Tate theory and then he writes a formula for the size of each isogeny class, via some manipulations with lattices, as an integral over some adele group (the point is that you use the Tate module of the curve at $\ell$ to understand $\ell$-power isogenies, and then rewrite such things as integrals over $\ell$-adic groups, and then take the product over all $\ell$). -The resulting equation is a horrific mess, which can basically be almost completely written in terms of twisted orbital integrals on $GL(2)$ and various subgroups. The reason the orbital integrals are twisted is that the arguments for isogenies at $p$ use the Dieudonne module of the curve, so one has to do semilinear algebra here to count subgroups of $p$-power order. -Next Langlands uses the Fundamental Lemma for $GL(2)$, which he can prove with his bare hands, to turn his twisted orbital integrals into orbital integrals. -Finally, Langlands looks at the resulting mess and says "oh look, this is just precisely the geometric side of the trace formula! So it equals the spectral side.". And then he looks to see what he has proved, and he's proved that the size of $X_0(N)(k)$ equals the trace -of some Hecke operator (depending on $k$) and he figures out how this trace relates to $T_p$ and then he's proved Eichler-Shimura. -I know of no complete reference for the above argument. In particular, checking that the "nasty" terms in the trace formula coming from the fact that we're in the non-compact case, match up precisely with the number of cusps, is something so deeply embedded in Langlands Antwerp that I have never really managed to get it out. -This method is much more complicated than Eichler-Shimura, but it generalises much better. By Corvallis they basically had got it working for forms of $GL(2)$ over a totally real field, where already the arithmetic geometry methods were struggling. See the articles of Kottwitz, Milne, Casselman etc. -Later on Kottwitz pushed the method through for a wide class of PEL Shimura varieties, but I'm not sure he did it for all of them. This is what prompted Clozel's Sem Bourbaki talk in 1993 or so---I mentioned this earlier. I think Milne has written a lot about this area recently though, in particular I believe he's trying to debunk the myth that the result is proved for all PEL Shimura varieties. See in particular Milne's paper "Points on Shimura varieties over finite fields: the conjecture of Langlands and Rapoport" (2008) available -here for example.<|endoftext|> -TITLE: How does the Frobenius act on the prime-to-$p$ $\pi_1(\mathbb{P}^1_{\overline{\mathbb{F_p}}}\setminus \{a_1,...,a_r\})$? -QUESTION [16 upvotes]: From Grothendieck's work we know that the prime-to-p fundamental group $\pi_1(\mathbb{P}^1_{\overline{\mathbb{F}_p}}\setminus\{a_1,...,a_r\})$ where $a_1,...,a_r \in \mathbb{F}_p$ is isomorphic to the prime-to-p part of the profinite completion of $\langle \alpha_1,...,\alpha_r|\alpha_1...\alpha_r=1\rangle$. -The question is: how does the Frobenius automorphism of $\overline{\mathbb{F}_p}$ act on the prime-to-p $\pi_1(\mathbb{P}^1_{\overline{\mathbb{F}_p}}\setminus\{a_1,...,a_r\})$ where $a_1,...,a_r \in \mathbb{F}_p$? -I don't actually expect an answer. I gather that this is not well understood. -My question is: what is known about it? Where can I read more? And in general any insight about this question is very welcome. -I put a community wiki stamp on this because there's no one right answer. - -REPLY [22 votes]: Thanks to Masoud Kamgarpour for bringing this question to my attention; to make up for being 8 years late, I've written way too much. I'll try to (1) clarify the question a bit, (2) list what little explicit knowledge we have about it, and (3) remark on what information we get from the Langlands program. -Let $k$ be a finite field of characteristic $p$ and let $X=\mathbb{P}^1_{k}\setminus\{a_1, \cdots, a_r\},$ where the $a_i\in \mathbb{P}^1_k(k)$. -First, let me try to clarify the question a little bit, as it is not quite well-defined, since the object $\pi_1^{\text{et}}(X_{\bar k})$ is only well-defined up to inner automorphism. To get a well-defined object, one must choose a geometric basepoint $\bar x\in \mathbb{P}^1_k$. Then there is an exact sequence $$1\to \pi_1^{\text{et}}(X_{\bar k}, \bar x)\to \pi_1^{\text{et}}(X, \bar x)\to \text{Gal}(\bar k/k)=\widehat{\mathbb{Z}}\cdot \text{Frob}\to 1.$$ -Thus $\text{Gal}(\bar k/k)$ has an outer action on $\pi_1^{\text{et}}(X_{\bar k}, \bar x)$, via conjugation, but to get a well-defined honest action, one needs to choose a splitting of this exact sequence. The action really depends on the choice of splitting. -How does one "choose a splitting"? Natural choices come from choosing the basepoint $\bar x$ to be a rational (or rational tangential) basepoint. That is, let $x\in X$ be a rational point, and let $\bar x$ be the geometric point arising from a choice of algebraic closure of $k$. The map $x\to X$ induces a map $\text{Gal}(\bar k/k)\to \pi_1^{\text{et}}(X, \bar x)$ splitting the exact sequence above, by the functoriality of the etale fundamental group. So we now have an honest action of $\text{Gal}(\bar k/k)$ on $\pi_1^{\text{et}}(X_{\bar k}, \bar x)$, which I claim depends on $x$. -This action is, broadly speaking, quite mysterious. The rest of this answer will aim to explain what is known about it. -First, as the OP explains, it is a result of Grothendieck that (by comparison to the situation over $\mathbb{C}$), the maximal prime-to-$p$ quotient of $\pi_1^{\text{et}}(X_{\bar k}, \bar x)$ is isomorphic to the maximal prime-to-$p$ quotient of $$\langle \gamma_1, \cdots, \gamma_r \mid \prod \gamma_i=1\rangle,$$ that is, the free profinite prime-to-$p$ group on $r-1$ letters. Let $\ell$ be a prime different from $p$; at the cost of losing some information, we will work with the maximal pro-$\ell$ quotient of $\pi_1^{\text{et}}(X_{\bar k}, \bar x)$, which we denote by $\pi_1^\ell$. This is the free pro-$\ell$ group on $r-1$ letters; again the Galois action on this group will depend on the choice of basepoint, though I will suppress it from the notation in the rest of this answer. -Let $$R=\mathbb{Z}_\ell[[\pi_1^\ell]]:=\varprojlim_H \mathbb{Z}_\ell[H],$$ where $H$ runs over the finite quotients of $\pi_1^\ell$ be the group ring of $R$. Instead of describing (what we know about) the action of Frobenius on $\pi_1^\ell$, we will describe its action on $R$; this doesn't lose any information as one can recover $\pi_1^\ell$ as the set of group-like elements in $R$, given the usual Hopf-algebra structure of a group ring. Let $\mathscr{I}\subset R$ be the augmentation ideal. That is, $\mathscr{I}$ is the kernel of the map $R\to \mathbb{Z}_\ell$ sending every group element to $1$. Abstractly, $R$ is isomorphic to a non-commutative power series ring on $r-1$ variables $\mathbb{Z}_\ell\langle\langle X_1, \cdots, X_{r-1}\rangle\rangle$, via the map sending generators $\gamma_i$ of $\pi_1^\ell$ to $(X_i-1)$; under this isomorphism, $\mathscr{I}$ is the two-sided ideal generated by the $X_i$. -First one describes the Galois action on $$\text{gr}_{\mathscr{I}^\bullet} R=\bigoplus \mathscr{I}^n/\mathscr{I}^{n+1}.$$ It is a general fact from the theory of pro-$\ell$ groups that $$\mathscr{I}/\mathscr{I}^2\simeq \pi_1^{\ell, \text{ab}}$$ canonically; hence $$\text{Hom}(\mathscr{I}/\mathscr{I}^2, \mathbb{Z}_\ell)\simeq \text{Hom}(\pi_1^{\ell, \text{ab}}, \mathbb{Z}_\ell)\simeq H^1_{\text{et}}(X_{\bar k}, \mathbb{Z}_\ell)=\mathbb{Z}_\ell(-1)^{\oplus (r-1)}.$$ So Galois acts on $\mathscr{I}/\mathscr{I}^2$ via the cyclotomic character; in other words, Frobenius acts via multiplication by $q$. -But $\mathscr{I}^n/\mathscr{I}^{n+1}\simeq (\mathscr{I}/\mathscr{I}^2)^{\otimes n}$ via the multiplication map, so Frobenius acts on $\mathscr{I}^n/\mathscr{I}^{n+1}$ via multiplication by $q^n$. -So we're done describing the Frobenius action on $\text{gr}_{\mathscr{I}^\bullet} R$; what this tells us is that the interesting data is contained in the extensions between the $\mathscr{I}^n/\mathscr{I}^{n+1}$. More prosaically, we've found that $$\text{Frob}(X_i^n)=q^nX_i^{n}+(\text{terms of degree $>n$}).$$ Giving a good description of these higher-order terms is an open problem, but a lot of work has been done, essentially after replacing $\pi_1^\ell$ with a metabelian quotient thereof. Of course one cannot in practice ``choose generators of $\pi_1^\ell$," so actually computing the coefficients of the terms in the expression above is not a well-stated goal --- but some of their invariants (e.g. the $\ell$-adic valuations of the coefficients) are well-defined enough to make reasonable questions. -Let me briefly summarize some of the literature. -Section 19 of Deligne's classic paper "Le groupe fondamental de la droite projective moins trois points" essentially boils down to computing the action of $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ on a quotient of $$\mathbb{Z}_\ell[[\pi_1^\ell(\mathbb{P}^1_{\overline{\mathbb{Q}}}\setminus \{0,1\infty\}, \text{tangential basepoint at $0$)}]]\simeq \mathbb{Z}_\ell\langle\langle X, Y\rangle\rangle,$$ where $1+X$ is a loop around $0$ and $1+Y$ is a loop around $1$. Here we quotient by the left ideal generated by $Y$ and all monomials whose $X$-degree is at least 2. If one thinks a bit, one can extract the actions of individual Frobenii on this quotient from his paper. (Deligne doesn't quite phrase things this way, but I find this to be the easiest way to think about what he does.) -Ihara, Nakamura, Wojtkowiak, Anderson and many other authors have performed variants of this computation (in fact, some version of it is originally due to Wojtkowiak, if I remember correctly); Ihara, Nakamura, and Anderson give beautiful symmetric formulae for the Galois action on the quotient of $$G:=\pi_1^\ell(\mathbb{P}^1_{\overline{\mathbb{Q}}}\setminus \{0,1\infty\}, \text{tangential basepoint at $0$})$$ by the normal subgroup generated by $[[G, G],[G,G]]$. Ihara's ICM notes and the references therein are a good summary; one can think of this work as (equivalently) computing the Galois action on the Tate modules of the Jacobians of Fermat curves (which are cofinal among the abelian covers of $\mathbb{P}^1\setminus \{0,1\infty\}$). -The only work I know of that goes substantially beyond metabelian quotients (in approaching these kind of explicit questions), is Anderson-Ihara's paper "Pro-$\ell$ branched coverings of $\mathbb{P}^1$ and higher circular $\ell$-units II." In principle this paper gives an algorithm for computing the power series $$\text{Frob}(X_i^n)=q^nX_i^{n}+(\text{terms of degree $>n$})$$ above, to arbitrary precision (i.e. modulo the ideal $(X_1^N, \cdots, X_{r-1}^N, \ell^N)$, for arbitrary finite $N$). Unfortunately these results are not (at least in my experience) useful for answering theoretical questions about the Galois action. The paper doesn't appear to be available online, but you can find it in the library (or if you can't, I email me and I'll send you a scan). -As far as I know, the state of our knowledge on these precise questions has not progressed dramatically since 1990, when Ihara's ICM notes were written. Of course there has been lots of very interesting incremental progress, e.g. in the beautiful papers of Nakamura and Wojtkowiak; many people (myself included) are still actively thinking about these questions. - -That's the summary of explicit work on this issue, at least to my knowledge. After 1990, work on this question has been largely inexplicit, and has focused on exploiting the work of L. Lafforgue on the Langlands program for function fields over $\mathbb{F}_q$ (as well as Deligne's Weil II results). The basic idea is that one can study the outer action of $\text{Gal}(\bar k/k)$ on $\pi_1(X_{\bar k}, \bar x)$ via its action on the set of continuous $\overline{\mathbb{Q}_\ell}$-representations of $\pi_1(X_{\bar k}, \bar x)$; inner automorphisms act trivially on the set of (isomorphism classes of) representations, so this action is actually independent of the basepoint $\bar x$. -Unfortunately it is hard to translate the information that can be extracted from Lafforgue into answers to the sort of explicit questions we asked above. We know: - -The set of rank $n$ semisimple continuous representations of $\pi_1(X_{\bar k}, \bar x)$ with "bounded wild ramification" (that is, bounded Swan conductor) and fixed by $\text{Frob}^m$ is finite. In particular, any representation of the prime-to-$p$ quotient of $\pi_1$ is tame; hence there are only finitely many of any given rank fixed by $\text{Frob}^m$ (for given $m$). -Any irreducible continuous representation of $\pi_1(X_{\bar k}, \bar x)$ fixed by $\text{Frob}$ extends to a representation of $\pi_1(X_{k}, \bar x)$ which is pure of weight zero. - -(In fact, (1) follows from (2) via a sphere-packing argument, which is alluded to in e.g. this paper of Esnault and Kerz, but which as far as I know does not appear in the literature; I can sketch it if you'd like. One can also deduce (1) directly via automorphic methods; Esnault and Kerz sketch this argument.) - -The number of representations of rank $n$ fixed by $\text{Frob}^m$ with fixed monodromy at infinity has a well-behaved generating function in many cases; namely it looks like the set of points of a variety over a finite field (see e.g. this paper of Deligne and Flicker and this paper of Hongjie Yu, for example). As far as I know, the complete story isn't really known here yet. -Let $$\rho: \pi_1(X_{\bar k}, \bar x)\to GL_n(\overline{\mathbb{F}_\ell((T))})$$ be a continuous representation, fixed by Frobenius. Then $\rho$ has finite image. This was a conjecture of de Jong, proven by Gaitsgory, using a nontrivial variant of Lafforgue's work. - -This is of course not an exhaustive list; the Langlands program gives much finer information, though it is not necessarily easy to compute with. Much ongoing work aims at bootstrapping these results. For a recent exciting example, see the work of Deligne, Drinfeld, Esnault, Kedlaya, and many others on Deligne's ``companion conjectures" (finally completely proven as of a few days ago) --- see here for a fairly complete reading list on this topic. Of course this is a very active field, and there are other recent papers (my own included) which aim to understand something concrete about $\pi_1$ via the Langlands program. - -A final remark: note that properties 1-4 above are purely group-theoretic properties of the outer action of Frobenius on $\pi_1(X_{\bar k})$. In my view, the biggest (if not well-stated) open question about this action is the following: - -Can one give a purely group-theoretic proof of 1-4 above? That is, what (explicit) property of the outer action of Frobenius on $\pi_1(X_{\bar k})$ makes 1-4 above true? - -For me, at least, this is what it would mean to actually understand the action of Frobenius on the fundamental group.<|endoftext|> -TITLE: Problems where we can't make a canonical choice, solved by looking at all choices at once -QUESTION [45 upvotes]: It's a common theme in mathematics that, if there's no canonical choice (of basis, for example), then we shouldn't make a choice at all. This helps us focus on the heart of the matter without giving ourselves arbitrary stuff to drag around. -However, in this question, I'm looking for examples of problems solved by a specific type of "not making a choice" - namely, making all available choices, and looking at all the end results together as a whole. We can't necessarily discern any individual piece, but the average behavior, or some other information about the big picture, provides (or at least points towards) a solution. -I really wish I had an example of this phenomenon to provide, but even one escapes me at the moment, which is what spurred me to ask this question. I imagine combinatorics is full of examples; unfortunately I haven't really studied that field in any depth yet. -Something close to what I'm after is Burnside's (a.k.a. not-Burnside's) Lemma. There's no good way of directly counting orbits, i.e. choosing to look at a particular orbit one at a time, so we just look at the average number of fixed points of elements of $G$ (I'm reluctant to call this an example of the kind of result I'm looking for because I'm not entirely clear on why the fixed points of an element should be thought of as substitutes for orbits. Perhaps that's a separate question). -This question is in a similar vein. - -REPLY [10 votes]: General topology as found in textbooks seems to be chock-full of examples where the axiom of choice seems to be (unconsciously?) invoked, and unnecessarily if one follows Eilenberg's advice to avoid subscripts! Andrej Bauer in his BAMS expository article on constructive mathematics, page 484, cites an example from Engelking's topology book. Here is an example from the famous textbook by Munkres (page 166 in the 1975 edition). - -Every compact subset of a Hausdorff space is closed. -Proof. Let $Y$ be a compact subset of the Hausdorff space $X$. We shall prove that $X - Y$ is open, so that $Y$ is closed. -Let $x_0$ be a point of $X - Y$. For each point $y$ of $Y$, let us choose disjoint neighborhoods $U_y$ and $V_y$ of the points $x_0$ and $y$, respectively (using the Hausdorff condition). The collection $\{V_y|y \in Y\}$ is a covering of $Y$ (...) - -Okay, stop right there. We can see where he's going with this, but do we have to make infinitely many choices like this? No, of course not. You could say instead - -Let $x$ be a point of $X - Y$. The collection $\{V \subset X \;\text{open}| \exists U\; \text{open}: x \in U\; \text{and}\; U \cap V = \emptyset\}$ covers $Y$, because for each $y \in Y$ there are disjoint open neighborhoods $V$ of $y$ and $U$ of $x$, using the Hausdorff condition. (...) - -If you are sensitive to the axiom of choice and look around, you find the same choice-y subscript-y device used all over the place, esp. in proofs of lemmas involving compactness arguments (such as the tube lemma). -What seems to be going on in such cases is that one (knowingly or not) has an entire or total relation $R$ from a set $A$ to a set $B$ (meaning that for all $a \in A$ there exists a $b \in B$ with $R(a, b)$), but one has resorted to the axiom of choice which guarantees the existence of a function $f \subseteq R$, typically denoted with a subscript: $a \mapsto b_a$. The advice is to look at all choices instead, i.e., work with the entire relation $R$ directly, and dispense with choice functions altogether. -Or, don't. It's your choice. But if some astute student asks whether you secretly used AC there and is that actually necessary? then you'll be glad if you had the other argument ready.<|endoftext|> -TITLE: Intuitive explanation of Burnside's Lemma -QUESTION [29 upvotes]: Burnside's Lemma states that, given a set $X$ acted on by a group $G$, -$$|X/G|=\frac{1}{|G|}\sum_{g\in G}|X^g|$$ -where $|X/G|$ is the number of orbits of the action, and $|X^g|$ is the number of fixed points of $g$. In other words, the number of orbits is equal to the average number of fixed points of an element of $G$. -Is there any way in which the fixed points of an element $g$ can be thought of as orbits? I had wondered aloud on my recent question here how (or if) Burnside's Lemma can be interpreted as having the same kind of object on both sides, so as to be a "true" average theorem, e.g. -"number of orbits = average over $g\in G$ of (number of orbits satisfying (something to do with $g$))" -or -"number of orbits = average over $g\in G$ of (number of orbits of some new action which depends on $g$)" -Since Qiaochu stated the comments to my question that he suspects Burnside's Lemma can be categorified, and that this may be related, I have also added that tag. - -REPLY [37 votes]: One can view Burnside's lemma as a special case of the mean ergodic theorem, which links time averages to spatial averages, which may qualify as "equating two objects of the same type". On the other hand, the mean ergodic theorem is more complicated than Burnside's lemma, so this may not qualify as an intuitive explanation. -Nevertheless: given a measure-preserving action of an amenable group $G$ on a space $X$, the mean ergodic theorem tells us that -$$ {\bf E}_{g \in G} \langle T_g f, f \rangle_{L^2(X)} = \| \pi(f) \|_{L^2(X)}^2,$$ -where $\pi(f)$ is the orthogonal projection of $f$ to the $G$-invariant functions, and $T_g f(x) := f(g^{-1} x)$, and ${\bf E}_{g \in G}$ is a mean on $G$. -If one applies this to the one-sided action $g: (x,y) \to (gx,y)$ on the product space $X \times X$ equipped with counting measure, with $f$ equal to the Kronecker delta function $f(x,y) = \delta_{x,y}$, $\pi(f)$ is equal to $1/|O|$ on the square $O \times O$ of each orbit $O$, and so one obtains -$$ {\bf E}_{g \in G} |X^g| = |X/G|$$ -which is Burnside's lemma. - -REPLY [6 votes]: I saw a discussion of this lemma in a combinatorics article just very recently. It goes along lines like this: We want to count orbits of X under the action of G. We don't know how to pick representatives of each orbit, so let's just count all of them (thus relating this to your other question?), appropriately discounted. So, -$$|X/G| = \sum_{x \in X} \frac{1}{|Gx|}$$ -where $Gx$ is the orbit of $x \in X$ under the action of G. An application of the orbit-stabiliser theorem yields -$$|X/G| = \frac{1}{|G|} \sum_{x \in X} |G_x|$$ -and then the usual formula follows by recognising that the sum is nothing more than the cardinality of the set $\{ (g, x) \in G \times X : g \cdot x = x \}$ (Z in the notation of José's post). -I suppose the natural (pun intended) question to ask is whether there's a natural bijection between the $G \times X/G$ and the disjoint union $Z = \displaystyle \coprod_{g \in G} X^g$. If there is one, it's not immediately obvious to me... I suspect there isn't, simply because $G \times X/G$ is a disjoint union over G of sets of the same size, whereas $Z$ is a disjoint union over G of sets of different sizes.<|endoftext|> -TITLE: When can Witten-esque moduli spaces be used to define invariants of geometric structures? -QUESTION [14 upvotes]: I am trying to understand the big picture around Seiberg-Witten invariants of 4-manifolds. Of course, this points to Taubes work on Gromov-Witten invariants of symplectic manifolds. -It is striking to me that both developments arise, essentially, by studying spaces of functions that satisfy non-linear elliptic equations. For Seiberg-Witten, one studies moduli spaces of sections of bundles over a 4-manifold modulo a gauge group action. In the case of Gromov-Witten, one analyzes spaces of special maps from surfaces to symplectic manifolds. -Perhaps someone has cataloged or systematically studied which elliptic moduli spaces produce "useful" invariants of the underlying geometric structures. Is it a work of genius ala Gromov and Witten to suggest these equations or do we have some methodical process whereby we can define (or guess) such a useful equation and resulting moduli space? For example, might there be a Witten-like equation which helps elucidate contact structures on 7-manifolds, etc? -So my question is, what other Witten-like moduli spaces can be used to define invariants of geometric structures? - -REPLY [20 votes]: One general lesson we can learn from the work of Witten and others is that supersymmetric field theories are a systematic source of deep invariants (in the form of moduli spaces from the classical theories and meaningful linearizations of them from the quantum theories, to oversimplify egregiously). In other words one general principle is to look for representations of super-Lie algebras extending the symmetries of space(time) on spaces of fields (functions, spinors, maps, connections, and other local expressions). -While this might initially seem as 1. an ad hoc principle, and 2. as hopelessly general, it turns out to be neither. It is a unifying theme to many (most?) of the applications of stringy physics to math; and there are remarkable classification results (due to Nahm and others) saying that there are far far fewer such theories than you might originally think. Namely there are strict bounds in terms of the dimension of space to how big a supersymmetry algebra you can find, if you bound the spin of the fields you consider (ie if you want to consider equations involving only scalars, or only scalar and one-forms, or only up to two-forms [gravity]-- physicists tell us to look no further than that, and therefore at no higher than 11 dimensions..) -Since there turn out to be very very few "maximally supersymmetric" equations you can write down, kind of like exceptional Lie groups or other exceptional objects in math, each turns out to be amazingly meaningful and applicable and provides a powerful guiding principle for mathematicians asking the kind of question you're asking. -For example you can ask "hmm.. how can I interpret Khovanov homology in terms of elliptic equations or field theory somehow?" Well if you're Witten this summer, you answer: well -I need a supersymmetric field theory, I realize it needs to have more than four dimensions, -there's basically one super special theory I know that lives in 6 dimensions, so I should look for which geometry to stick in my two extra dimensions to get what I want, -and there aren't so many options, I find that taking the plane and modding out by rotations is the most reasonable choice, and bam! I have a new construction of Khovanov and in fact an extension of it to a full-fledged 4d TFT.. -I recommend the expository articles of Dan Freed (with Deligne in the IAS QFT volumes, and his Five Lectures of Supersymmetry, and a PCMI proceedings) for intros for mathematicians to supersymmetry, classical and quantum.<|endoftext|> -TITLE: Secret Santa (expected no of cycles in a random permutation) -QUESTION [11 upvotes]: In a Secret Santa game, each of $n$ players puts their name into a hat and then each player picks a name from the hat, who they buy a Christmas present for. Obviously, if someone picks their own name then they put it back and draw again (if they're the only person remaining in the hat then everyone heaves an exasperated sigh, and you start again). -My question: how many loops of present-giving do you expect to see? (e.g. A -> B -> C -> A) -More mathematically: what is the expected number of cycles in a random $n$-permutation which contains no fixed points? -The twelve-fold way doesn't appear to answer this question. Some brief Monte Carlo suggests that the limiting answer as $n\to\infty$ is $c\log n$ for some constant $c\approx 0.95$ (see attached graph, where circles are Monte Carlo approximation, straight line is $\log n$) but I'd like to see an analytical argument for this. -alt text http://www.freeimagehosting.net/uploads/330281fbea.png - -REPLY [5 votes]: I want to elaborate on and correct a comment that I made. -The two questions are not the same. The Secret-Santa-game method used to choose a derangement is not uniform on derangements. For $n=4$, you get the following: -$P[\text{Restart}] = 5/36$ -$P[(12)(34)] = 1/9$ -$P[(13)(24)] =1/12$ -$P[(14)(23)]=1/12$ -$P[(1234)]=1/18$ -$P[(1243)]=1/9$ -$P[(1324)]=1/12$ -$P[(1342)]=1/12$ -$P[(1423)]=1/12$ -$P[(1432)]=1/6$ -If you simulate this, the chance of two transpositions is $36/31(1/9+1/12+1/12)= 10/31$ rather than $1/3$, and the average number of cycles should be about $41/31$ instead of $4/3$. I suspect that this makes no difference to the asymptotics, and that no one will care about the nonuniform method's exact answer.<|endoftext|> -TITLE: Prepending strings to primes. -QUESTION [9 upvotes]: Hello, -we all know that 31,331,3331,33331,333331,3333331,33333331 all are primes, and that 333333331 is not. Here we prepend the digit 3 to 31, to get a list of 7 primes.This gives me the following thought: -Let - $$D = \{\text{all possible nonnull finite digit strings}\},$$ - $$D' = \{\text{all things in D that do not start with 0}\}.$$ -Define a function $m: D' \times D \to N \cup {\infty}$ by: - $$m(A,B)= \min \{ k\geq 1 \colon A^kB \text{ is composite} \} - 1,$$ -i.e., the number of consecutive primes at the beginning of the list $AB, AAB, AAAB, \dots$. For example, $m(3,1)=7$. - -Which values does $m$ take? Is it unbounded? Is it ever $\infty$? - -This question has been posted to math.stackexchange too, and I got one comment talking about that it might involve Tao-Ziegler extension to the Green-Tao theorem, and so I thought it might be more appropriate here. So please, excuse me if it's posted wrongly, or if one shouldn't post to both channels. - -REPLY [2 votes]: To supplement David's answer, the 2nd option, I can add -the following heuristics. -If we fix a positive integer $n$ and take irreducible polynomials -$$ -f_k(x) -=A\frac{x^{k+1}-1}{x-1}+B -=A(x^k+x^{k-1}+\dots+x)+B, -\qquad k=1,\dots,n, -$$ -subject to the condition that $f_1(x)\dots f_n(x)/m$ is not -an integer-valued polynomial for all $m>1$ (take, for -example, $A=3$ and $B=1$), then -Schinzel's Hypothesis H -implies that the numbers $f_1(R),\dots,f_n(R)$ are simultaneously prime for infinitely many -positive $R\in\mathbb Z$. There is no guarantee however that $R$ assumes the form $10^k$. -I owe this idea to Frictionless Jellyfish -who removed his/her related comment to -this post -(which I would definitely accept as answer).<|endoftext|> -TITLE: Is the tensor product of regular rings still regular -QUESTION [9 upvotes]: An imprecise version of the question is that when $A$ and $B$ are regular rings, is $A \otimes B$ also regular? Please allow me to put more restrictions, here I am only interested in the case when $A$ and $B$ are finitely generated $k$-algebras. When $k$ is perfect, the answer is yes, see http://arxiv.org/abs/math/0210359. In fact, we can view $A$ and $B$ as coordinate rings of some affine $k$-varieties say $X$ and $Y$ respectively. Since $k$ is perfect, regularity is equivalent to being smooth and it can be shown easily $X \times_k Y$ is smooth. Hence the conclusion. Now when $k$ is not perfect, and assume in addition $X$ and $Y$ are both geometrically (absolutely) integral, moreover they contain some (regular but) not smooth point (so the above method doesn't apply), then is $X \times_k Y$ still regular? -Example, $k=\mathbb{F}_p(t)$, $p>2$, $A=k[x,y]/(x^p-x^{p-1}y-t)$, $X=\operatorname{Spec} A$. Then it is easy to show $X$ is geometrically integral, the maximal ideal generated by $(Y)$ in $A$ is regular but not smooth, and that $A$ is regular. The question is that is $X\times_k X$ regular? -Note, it is easy to produce a counter example when $X$ is not assumed to be geometrically integral, e.g. $A=k[x]/(x^p-t)$, then it is easy to show $X\times_k X$ is a $0$-dimension local ring but not a domain, therefore it can not be regular. - -REPLY [10 votes]: This is inspired by Tom Goodwillie's answer. - -Let $X$ be an algebraic variety over a field $k$ (i.e. a scheme of finite type over $k$). If $X\times_k X$ is regular, then $X$ is smooth over $k$. - -Proof. The first projection $X\times_k X\to X$ is faithfully flat, so the regularity of $X\times_k X$ implies that of $X$. To prove the smoothness of $X$, we can suppose $X$ is connected and affine. The diagonal morphism $\Delta: X\to X\times_k X$ is then a closed immersion from a regular scheme to a regular scheme. Therefore $\Delta$ is locally complete intersection. If $J$ is the ideal sheaf on $X\times_k X$ defining $\Delta(X)$, then $\Delta^*(J/J^2)$ is locally free of rank the codimension of $X$ in $X\times_k X$ which is -equal to $\dim X$. -Now $\Delta^*(J/J^2)$ is isomorphic to the sheaf of differential forms $\Omega^1_{X/k}$ on $X$ (see Hartshorne), so the latter is locally free of rank $\dim X$. This implies that $X$ is smooth.<|endoftext|> -TITLE: Surgery in complex geometry -QUESTION [11 upvotes]: I've been thinking about surgery on complex manifolds. Not very seriously, but just to the point that I think it's odd how there's almost no mention of it in the literature. I figure there's something I'm missing. -In "Complex manifolds and deformations complex structures" Kodaira defines a surgery operation which takes two complex manifolds, makes certain choices, and returns a third complex manifold obtained by joining the two along a submanifold. The details are on page 52 of Kodaira's book, which may be found here. -After a quick look it seems to me that Kodaira uses this surgery to construct the Hirzebruch surfaces, the Hopf surface, and the blow-up of a point in $\mathbb C^n$. Elsewhere, the only similar examples I can find are the ones of a blow-up of a point or subvariety on a complex manifold (see, for example, page 93 of Zheng's "Complex differential geometry"). -So my question is, why don't we hear more about surgery of complex manifolds? Heuristically I'd expect that it doesn't work all that well, because else I'd already seen it used to construct (counter-)examples, but I also can't figure out why it doesn't work. - -REPLY [9 votes]: There is an operation in algebraic geometry, called a flip, which is a (kind of a special) surgery, so one could say that you hear about it, but under a different name. You can see the definition of a flip on page 41 of Birational geometry of algebraic varieties by János Kollár and Shigefumi Mori (unfortunately that exact page is not available on google books). See also this and this MO answers. -One possible reason for the general lack of mentioning surgery in general may be that it is extremely hard to prove that it exists (at least for flips, but I suppose if one came up with a more general definition it would be also hard to prove existence). Shigefumi Mori was awarded the Fields Medal for proving the existence of flips in dimension 3 and it was only proved very recently by Hacon and McKernan that flips exist in any dimension. (Of course, here one would have to mention the works of Shokurov and Siu that influenced them and give references and try to give credit to everyone, but I don't want to write a book here, so let me just leave researching this in detail for the reader for now.)<|endoftext|> -TITLE: Do singular values of a point set determine its shape? -QUESTION [8 upvotes]: Suppose I have $k$ points in $d$ dimensions. Let A be a $k\times d$ matrix with $i$th row giving the coordinates of $i$th point. Do singular values of this matrix have an interpretation as some kind of geometric invariant? -I tried a set of convex point sets formed by centroids of vertices of 8 dimensional 7-simplex, and there were 49 distinct singular value sets, which is the same as the number of such point sets not equivalent under coordinate permutation, obtained by Peter Shor in earlier post, I'm wondering if it's a coincidence - -REPLY [7 votes]: The singular values of this matrix are the square roots of the eigenvalues of the Gram matrix of the vectors, and can be viewed as the dimensions of the best approximating ellipsoid to the vectors (as per principal component analysis).<|endoftext|> -TITLE: What's the best way to test if a sphere is a polytope? (algorithms for the Simplicial Steinitz Problem) -QUESTION [8 upvotes]: The problem of recognizing whether a simplicial face lattice is polytopal is sometimes called the Steinitz problem. -Sturmfels and Bokowski advanced a set of methods in the late 80s to test whether the face lattice of a simplicial sphere was also realizable as a polytope. -The method uses oriented matroids. The problem is NP-hard, so their algorithm requires exponential time in the worst case, but they reported that the algorithm often converged quickly. -In the intervening two decades, I'm sure that newer approaches have been developed. Is there a better method known today? More interestingly, are there any software implementations available that solve this problem -- even using the older approach? - -REPLY [5 votes]: The state-of-the-art methods for proving non-polytopality differ from those of proving polytopality of a simplicial sphere $S$. Let me give you an overview: - -If $S$ is non-polytopal, then one way of proving non-polytopality is to generate all compatible oriented matroids and then show that none of these oriented matroids is realizable. For this, the best methods seems to be the biquadratic final polynomial method by Richter-Gebert and Bokowski. (This mehtod is explained in the book you cite.) For some simplicial spheres with relatively few vertices in low dimension already, there are very many compatible oriented matroid (let's say a million). In these cases it is faster to run the biquadratic final polynomial already on a partial chirotope, consisting of those signs that are determined by the simplicial sphere directly. -If $S$ is polytopal, the best way of proving polytopality is giving an explicit realization. To this end we consider the semi-algebraic set that corresponds to the realization space of a polytope with $S$ as a boundary. We set up the corresponding system of non-linear equations and use non-linear solvers to find a solution. The solvers we use employ an branch-and-bound approach together with linear underestimators. Again In a next step we convert the numerical solution into an exact solution, such that it can be checked in exact arithmetic that we found indeed a realization of $S$. - -More details can be found in my preprint: -Realizability and inscribability for simplicial polytopes via nonlinear optimization -As one application of these techniques we consider simplicial $3$-spheres with $10$ vertices and decide whether they are polytopal for all of them. There are precisely $247,882$ such spheres; $162,004$ are polytopal and the $85,878$ are non-polytopal. Therefore we obtain a complete classification of all combinatorial types of $4$-polytopes with $10$ vertices.<|endoftext|> -TITLE: Reference request: discrete harmonic functions and ends of graphs -QUESTION [6 upvotes]: Let $G$ be an infinite locally finite connected graph with finitely many ends. A real-valued function $f : G \to \mathbb{R}$ is harmonic if -$$f(v) = \frac{1}{d_v} \sum_{v \sim w} f(w)$$ -where $v \sim w$ means that $v, w$ are connected by an edge. Playing around with a few examples leads me to suspect that the dimension of the space of harmonic functions on $G$ is the number of ends. (Heuristic: given a harmonic function, start with a vertex $v$ and move to a neighbor $w$ of $v$ such that $f(w) \ge f(v)$. If $f$ is nonconstant this should give a path converging to an end, and this should be possible for any end. Moreover a harmonic function should be determined by its "values at the ends.") Does anyone know if this is true and, if so, does anyone know of a reference for this fact? -(Tags are because a major application is to Cayley graphs of finitely generated groups and I would be interested in seeing how far one can push this method to prove basic facts about ends of such graphs.) - -REPLY [12 votes]: Life is much more complicated than that. In nice situations (for instance, if your graph is $\delta$-hyperbolic), then you can attach a more refined boundary than just the ends and (if you are lucky) solve the Dirchlet problem. A lot depends on what kinds of regularity conditions you assign to functions on the boundary at infinity. -This is by now a well-established part of geometric group theory. For instance, it plays a key role in Kleiner's recent new proof of Gromov's theorem on groups with polynomial growth. See here. -One textbook reference that covers some of this information is -Woess, W., Random Walks on Infinite Graphs and Groups, Cam- bridge Tracts in Math. 138, Cambridge Univ. Press, 2000 -EDIT : By the way, since you are in Cambridge, Curt McMullen at Harvard is a good person to talk to about this kind of stuff. I learned most of what I know about the subject from a course he taught last year.<|endoftext|> -TITLE: If the n-th root of unity exists locally, does it exist globally? -QUESTION [8 upvotes]: My question is: is it true that $\zeta_n$ is in $K$ (a number field) iff $\zeta_n$ is in all but finitely many of the $K_{\mathfrak{p}}$? - -REPLY [6 votes]: The answer is yes, and this follows (as BCnrd points out in the comments above) immediately from the Chebotarev Density Theorem. See e.g. Corollary 9 here. -More generally, let $K$ be any global field and $f \in K[t]$ be any irreducible, separable polynomial of degree $d > 1$. Then there are infinitely many places $v$ of $K$ such that $f$ does not split completely over $K_v$. Indeed, let $\alpha$ be a root of $f$ in a separable closure of $K$, let $L = K[\alpha]$ and let $M$ be the Galois closure of $L/K$. A finite place $v$ of $K$ splits completely in $L$ iff it splits completely in $M$ (Exercise 5.1.3), so by Chebotarev's Theorem, the set of primes which do not split completely in $L$ has density -$1 - \frac{1}{[M:K]} > 0$. -If $L/K$ is already Galois, the argument is simpler and the conclusion is stronger: there are infinitely many places $v$ of $K$ such that $f$ does not have any $K_v$-rational roots. This is the case in the OP's question, which we recover by taking $f$ to be the minimal polynomial over $K$ of any one primitive $n$th root of unity. -On the other hand for every composite number $d$, there is a degree $d$ irreducible polynomial $f \in K[t]$ which is reducible in every completion $K_v$: this is a 2005 theorem of Guralnick, Schacher and Sonn.<|endoftext|> -TITLE: Manifolds with rectifiable curves -QUESTION [9 upvotes]: To begin with, observe that the notion of a rectifiable curve makes sense in, say, a smooth or a PL-manifold but not in merely a topological manifold. Indeed if $f:[0,1]\rightarrow U\subset{\Bbb R}^n$, $U$ open, represents a rectifiable curve in $U$, and $g:U\rightarrow V$ a mere homeomorphism, we would generally have no reason to expect a rectifiable curve from the composition $g\circ f$. -Certain homomorphisms (e.g. smooth, PL) do preserve rectifiable curves; the totality of these forms a pseudogroup and we may use this pseudogroup to equip manifolds with just enough geometric structure to distinguish a class of curves as rectifiable. -Do manifolds with just this much structure (I'll call it a rectifiability structure till I know better) occur in the literature? And the usual questions: do all manifolds support a rectifiability structure? if one exists, is it unique up to homeomorphism? -I'd be interested in similar questions relative to higher dimensional objects - manifolds where embedded $n$-disk "have" an $n$-volume ("have" in scare quotes because I only mean relative to each patch of any atlas). -Just a final remark: biLipschitz homeomorphisms preserve rectifiability (right?), but not all rectifiability preserving homeomorphisms are biLipschitz (e.g. $x \mapsto x^{1/3}$), so if I've reinvented the wheel, I haven't reinvented that wheel. - -REPLY [7 votes]: This is just Lipschitz structure: only locally Lipschitz maps preserve rectifiability of all curves. -What is wrong with the $x\mapsto x^{1/3}$ map is explained in Tapio Rajala's answer. (A more explicit example is the path $t\mapsto t^2\sin(1/t)$, $t\in[0,1]$, and you can make a non self-overlapping example if you go to dimension 2). Here are the missing details of Tapio's answer in the general case. -If a homeomorphism $f:U\to U$ is not Lipschitz on some compact set, then there exist sequences $x_n$ and $y_n$ converging to some $p\in U$ rapidly (e.g. such that $|x_n-p|<2^{-n}$ and $|y_n-p|<2^{-n}$) and such that $|f(x_n)-f(y_n)|>n^2|x_n-y_n|$. Consider a curve that oscillates between $x_n$ and $y_n$ approximately $(n^2|x_n-y_n|)^{-1}$ times, then goes to $x_{n+1}$ and oscillates between $x_{n+1}$ and $a_{n+1}$, and so on. It is rectifiable (its length is bounded above by something like $2\sum_n (n^{-2}+2^{-n})$) but its $f$-image is not: each oscillating part adds at least 1 to the length.<|endoftext|> -TITLE: Cardinal numbers vs collections of cardinal numbers -QUESTION [8 upvotes]: Is there any literature, especially when doing mathematics without the axiom of choice, that discusses using collections of cardinal numbers in place of individual cardinal numbers, when discussing cardinal numbers with certain properties? -That's a little vague, and I will presently give a few motivational examples that should help to clarify what I'm thinking of. But first, a quick definition of the translation between cardinal numbers and collections of cardinal numbers assuming the axiom of choice: If K is a cardinal number, then {L | L < K} is a collection of cardinals which is small and downward-closed. Conversely, if C is a collection of cardinals which is small and downward-closed, then there exists a unique cardinal K (to wit, the smallest cardinal that does not belong to C) such that C = {L | L < K}. So the correspondence between these two perspectives is straightforward —if we assume choice. -Now I'll consider some types of cardinal numbers. A weak limit cardinal is an infinite cardinal K such that L+ < K whenever L < K. A strong limit cardinal is an infinite cardinal K such that 2L < K whenever L < K. A regular cardinal is an infinite cardinal K such that Σi ∈ I Li < K whenever |I| < K and each Li < K. An inaccessible cardinal is an uncountable regular limit cardinal. A weakly compact cardinal is an inaccessible cardinal K such that the height of a tree is less than K whenever every level has width less than K and every branch has length less than K (the tree property). Etc. -Although these are all properties of an individual cardinal number K, they all refer to that cardinal only through the collection of smaller cardinal numbers. (The exceptions are the adjectives ‘infinite’ and ‘uncountable’, which are really only there to rule out trivial cases.) -If you don't assume the axiom of choice, then it's easy to still consider these conditions on a small, downward-closed collection of cardinals, but now this is more general than a collection of the form {L | L < K}. If you go beyond doubting the axiom of choice and drop the law of excluded middle (so doing constructive mathematics, in a moderate sense), you can get more flexibility by messing with the interpretation of ‘downward-closed’; for example, the collection {0, 1, 2, …} of all finite (in the strictest sense) cardinal numbers is closed under taking decidable sub-cardinals but not arbitrary sub-cardinals and so gives a ‘regular’ collection of cardinals which is different from {L | L < ℵ0}. -I'm interested in understanding what regular cardinals and inaccessible cardinals should be in constructive mathematics. (Bigger than that, I don't really even understand them in classical mathematics.) It seems to me that they have to be collections of cardinals rather than individual cardinal numbers. But this perspective already makes sense in classical mathematics, especially without the axiom of choice. Has anybody studied this? - -REPLY [3 votes]: As mentioned, in constructive set theory we generally talk about large sets rather than large cardinals. -The rough idea is: a regular set is basically a transitive model of Replacement. An inaccessible set is basically a regular model of Exponentiation: if $a \in X$ and $b \in X$ then ${}^ab \in X$ (absent the law of the Excluded Middle this does not imply $Pow(a) \in X$). For the details I am glossing over, see e.g. Aczel and Rathjen's Notes on constructive set theory section 10. -It is provable in ZFC that the inaccessible sets in the constructive sense are exactly the $V_\kappa$ with $\kappa$ a strong inaccessible cardinal in the classical sense. -Of course we may ask what, in the name of the good lord that made the integers, is a "regular set", concretely, and I just happen to have a paper on arXiv on that (section 10). I have some private scribblings on extending this to inaccessibles. (It is entirely sane to believe in inaccessible sets but not $Pow(\mathbb{N})$...) -Aczel and Rathjen's survey also includes a definition for Mahlo sets - basically an inaccessible $X$ such that for any function $f\in {}^XX$ there is an inaccessible $Y \in X$ which is closed under $f$ (not quite accurate - see the paper for details). Over ZFC these are again the $V_\mu$ for $\mu$ a strongly Mahlo cardinal. -The paper Mike Shulman cited gives even larger set notions. These are in the context of impredicative set theory (IZF rather than CZF).<|endoftext|> -TITLE: Light reflecting off Christmas-tree balls -QUESTION [85 upvotes]: 'Twas the night before Christmas and under the tree -Was a heap of new balls, stacked tight as can be. -The balls so gleaming, they reflect all light rays, -Which bounce in the stack every which way. -When, what to my wondering mind does occur: -A question of interest; I hope you concur! -From each point outside, I wondered if light -Could reach deep inside through gaps so tight? - - -           - -More precisely, -let $\cal{B}$ be a finite collection of congruent, perfect mirror balls -arranged, say, in a cubic close-packing cannonball stack. -Let $H$ be the set of points inside the closed convex hull of $\cal{B}$, -$H^+ = \mathbb{R}^3 \setminus H$ the points outside, -and $H^- = H \setminus \cal{B}$ the points in the crevasses inside. - -Q1. - Is it true that every point $a \in H^+$ can illuminate - every point $b \in H^-$ - in the sense that there is a light ray from $a$ that - reaches $b$ after a finite number of reflections? - -I believe the answer to Q1 is 'No': If $a$ is sufficiently -close to a point of contact between a ball of $\cal{B}$ and $H$, -then all rays from $a$ deflect into $H^+$. -If this is correct, the question becomes: which pair of points -$(a,b)$ can illuminate one another, for a given collection $\cal{B}$? -Specifically: - -Q2. - Is there some finite radius $R$ of a sphere $S$ enclosing - a collection $\cal{B}$ - such that every point $a$ outside $S$ - can illuminate every point $b \in H^-$? - More precisely, are there conditions on $\cal{B}$ - that ensure such a claim holds? - -If the centers of the balls in $\cal{B}$ are collinear, -then points in the bounding cylinder do not fully illuminate. -If the centers of the balls are coplanar, then points -on that plane do not fully illuminate. So some configurations -must be excluded. -Perhaps a precondition analogous to this might suffice: -If the hull $H$ of $\cal{B}$ encloses a sphere of -more than twice the common radius of the balls, then ... ? -Failing a general result, -can it be established for stackings as illustrated above? -The answers (especially Bill Thurston's) -in response to the earlier MO question -on lightrays bouncing between convex bodies may be relevant. -Even speculative 'answers' are welcomed! -Edit (23Dec). -Although I remain optimistic that there is a nice theorem -lurking here, fedja's observation that points near the boundary -of the hull remain dark makes it a challenge to formulate -a precise statement of a possible theorem. -Something like this: - -If $\cal{B}$ is sufficiently "fat," - then every point $a$ sufficiently far from $\cal{B}$ - illuminates every point $b$ in $H^-$ that is not too close - to the boundary of $H$. - -Edit (24Dec). There is an associated computational question, interesting -even in two dimensions: - -Given $a$ and $b$, what is the complexity of deciding if $a$ can illuminate $b$? - -Is it even decidable? - -REPLY [8 votes]: Can't add to the comments (insufficient reputation yet) - I do like the problem and the poetic insight a lot. But here is another less poetic insight. -Q2 can be reversed to look at whether a light at any point $b \in H^-$ will illuminate every point sufficiently distant from it. This seems intuitively more tractable.<|endoftext|> -TITLE: The exceptional Lie algebra $\mathfrak{g}_2$ and binary cubics -QUESTION [8 upvotes]: How is the exceptional 14-dimensional Lie algebra $\mathfrak{g}_2$ related to the covariant algebra for the binary cubic? -Here are some details on this question. This algebra is generated by 4 forms, these being the form itself $Q$, the discriminant $\Delta$, the Hessian $H$, and a transvectant $T$ of $Q$ and $H$. These four forms ${Q,H,T,\Delta}$ obey a syzygy because $Q^2\Delta$ is a linear combination of $H^3$ and $T^2$. (This is well-known. One may find an exposition in Peter Olver's book on classical invariant theory, for example.) -Let $(d,w)$ denote the degree and weight of a form. Then the degrees and weights of ${1,Q,H,T,\Delta}$ are respectively ${(0,0),(1,3),(2,2),(3,3),(4,0)}$. This sequence is peculiar because these nearly give the branching weights of the adjoint representation of $\mathfrak{g}_2$ with respect to a Lie subalgebra generated by a pair of opposite short root vectors. The only one "missing" is a pair of the form $(2,0)$, which cannot occur. -Is this just a happy coincidence, or is there some "deep" reason reason for this? - -REPLY [5 votes]: Well, I don't know about "deep", but here is another perspective. My point of view is much different than the classical invariant point of view -- I use the word "syzygy" about as often as I use the word "zeugma". -A good reference for everything I'm about to write is the Duke paper of Wee Teck Gan, Benedict Gross, and Gordan Savin, "Fourier coefficients of modular forms on $G_2$". -In the Lie algebra $\mathfrak{g}_2$, there is a maximal parabolic subalgebra $\mathfrak{p} = \mathfrak{l} \oplus \mathfrak{u}$, where $\mathfrak{u}$ is a two-step nilpotent Lie algebra with one-dimensional center $\mathfrak{z}$. Correspondingly, there is a parabolic subgroup $P = LU$ of the simple complex Lie group $G_2$. The Levi subalgebra $\mathfrak{l}$ contains root spaces for a short root and its negative. The derived subalgebra $[\mathfrak{l}, \mathfrak{l}]$ is generated by a pair of opposite short root vectors. -The associated Levi subgroup $L$ is isomorphic to $GL_2$, and $L$ acts by the adjoint action on $\mathfrak{g}_2$. Since $L$ normalizes $U$, we get an action of $L$ on the 5-dimensional vector space $\mathfrak{u}$, and this action stabilizes the one-dimensional $\mathfrak{z} \subset \mathfrak{u}$ (note also $\mathfrak{z} = [\mathfrak{u}, \mathfrak{u}]$). The resulting action of $L$ on $\mathfrak{u} / \mathfrak{z}$ is a 4-dimensional representation of $GL_2$. -This representation of $GL_2$ is isomorphic to the representation of $GL_2$ on the space of binary cubic forms $C(x,y) = ax^3 + bx^2 y + c x y^2 + d y^3$, in which $g \in GL_2$ acts on a cubic form via: -$$ \[ gC \](x,y) = det(g)^{-1} \cdot C( (x,y) \cdot g).$$ -See Proposition 3.1 of Gan-Gross-Savin for more. -One could go the other way as well, and construct the Lie algebra $\mathfrak{g}_2$ beginning with the space of cubic forms (which has some additional structure). I think that the machinery of "structurable algebras" developed by Bruce Allison would work for that (and general construction of Lie algebras with 5-term grading).<|endoftext|> -TITLE: Fermat's Last Theorem and Computability Theory -QUESTION [14 upvotes]: This question stems from the paper "Computably categorical fields via Fermat's Last Theorem," by Russell Miller and Hans Schoutens (available online at http://qcpages.qc.cuny.edu/~rmiller/Fermat.pdf). In this paper, they construct a field $F$ by starting with the field $\mathbb{Q}(x_0, x_1, x_2, . . .)$ of infinite transcendence degree over $\mathbb{Q}$, and then adjoining elements $y_i$ such that $(x_i, y_i)$ is a solution to the polynomial $X_i^{p_i}+Y_i^{p_i}=1$ for some odd prime $p_i$ (see paragraph 2, page 3); they then show that the resulting field has interesting computability-theoretic properties. In particular, they show that this field is computably categorical (i.e., any two computable presentations are computably isomorphic). I have only started reading this paper, but I have two questions, a simple one and a probably not-so-simple one: -Question 1: It is unclear to me exactly how much of Fermat's Last Theorem (FLT) is required for this paper, but certainly we need at least the existence of infinitely many primes $p$ such that $X^p+Y^p=1$ has no nontrivial rational solutions. How difficult is this fact to prove? (And, historically, when was it first known?) -Question 2: How much of FLT is actually required for the paper? I would be very interested if full FLT was required; although, as the authors point out, there has been at least one previous attempt made to prove the same result that apparently did not rely on FLT. -Thank you very much in advance, -Noah S - -REPLY [16 votes]: The Fermat curves (and FLT) are used in the quoted paper to prove the following result: -Theorem 3.1: There exists a (recursive) infinite sequence $\mathcal{C} = \{C_i\}$ of curves $C_i/\mathbb{Q}$ such that: -(i) for all $_i$, one can effectively determine the set $C_i(\mathbb{Q})$ and -(ii) $\mathcal{C}$ is non-covering: for all $i \neq j$, there does not exist a finite $k$-morphism $C_i \rightarrow C_j$. -As the authors remark, any family of nonisomorphic curves of common genus $g \geq 2$ is noncovering, so it would suffice to exhibit such an infinite family of curves such that each one provably has no $\mathbb{Q}$-rational points. -This is easy, and we do so below for each $g \geq 2$: -Start with the (smooth projective model of the) curve given by the following equation. -$C: y^2 = - (x^{2g+2} + 1)$. -This is a hyperelliptic curve of genus $g$ which is visibly without $\mathbb{R}$-points, so certainly $C(\mathbb{Q}) = \varnothing$. -Now, among many other constructions, one can perturb the coefficients of $C$ slightly to get infinitely other genus $g$ curves without $\mathbb{Q}$-rational points. Indeed, write $P(x) = -x^{2g+2} - 1$ and for $a = (a_1,\ldots,a_{2g+2}) \in \mathbb{Q}^{2g+2}$, consider the polynomial -$P_a(x) = P(x) + \sum_{i=0}^{2g+2} a_i x^i$. -There exists an effectively computable $\epsilon > 0$ such that if for all $i$, $|a_i|_{\infty} < \epsilon$, then $P_a(x)$ is negative for all real $x$, so the curve $C_a: y^2 = P_a(x)$ still has no $\mathbb{R}$-points and hence no $\mathbb{Q}$-points. Here $|x|_{\infty}$ is the standard Archimedean norm. -It is no problem to find infinitely many choices of $a$ leading to nonisomorphic curves $C_a$. For instance, one can find choices of $a$ which give rise to curves with pairwise distinct moduli. Alternately, by a standard weak approximation / Krasner's Lemma argument one can choose the coefficients so as to make the curve $C_a$ have / not have $\mathbb{Q}_p$-points for each prime $p$ in any finite set $S$ of rational primes. More details for this can be supplied on request. -There are many other possible constructions. For instance, in this paper I produced for each number field $K$ and each $g \geq 2$ a bielliptic curve $C_{/\mathbb{Q}}$ of genus $g$ which violates the Hasse Principle. In fact by varying the elliptic curve that $C$ maps $2:1$ onto, one gets infinitely many such examples. Note that here the curves have points over every completion but still no $\mathbb{Q}$-points. In general, producing curves which do not have $\mathbb{Q}$-points "for local reasons" is much easier. -In my opinion emphasizing the connection to Fermat's Last Theorem here is a bit misleading.<|endoftext|> -TITLE: If the Riemann Hypothesis fails, must it fail infinitely often? -QUESTION [45 upvotes]: That is must there either be no non-trivial zeros off the critical line or -infinitely many? -I'm sure that no one believes otherwise, but I've never seen a theorem in the -literature addressing this. Folklore perhaps? - -REPLY [12 votes]: On a somewhat related note I would like to call attention to the following papers: -A. Booker, Poles of Artin $L$-functions and the strong Artin conjecture, Annals of Math. 158 (2003), 1089-1098. -Here the author proves that for 2-dimensional Galois representations $\rho$ the Artin conjecture implies the strong Artin conjecture by showing that if some character twist $L(s,\rho\otimes\chi)$ has a pole then $L(s,\rho)$ has infinitely many poles (hence in fact all twists have infinitely many poles). -P. Sarnak, A. Zaharescu, Some remarks on Landau-Siegel zeros, Duke Math. J. 111 (2002), 495-507. -Here the authors prove that if all zeros of all quadratic Dirichlet $L$-functions are on the critical line or on the real axis, then the possible real zeros are much farther from $s=1$ than we can prove at present without any hypothesis.<|endoftext|> -TITLE: Why Are Weber Polynomial Coefficients Smaller than Hilbert Polynomial Coefficients? -QUESTION [5 upvotes]: The title says it all. Singular moduli of the j-function satisfy polynomials, but as the class number grows, these polynomial coefficients become very large. Weber functions are modular (not over the full modular group), and their values also satisfy polynomials. But the Weber polynomials tend to have much smaller coefficents. Why? - -REPLY [8 votes]: Simply because they satisfy an equation of the form $P(f)-fj$ for some polynomial $P$. This immediately implies that the height of $f(z)$ will be around $1/deg(P)$ of the height of $j(z)$, or more precisely, asymptotic to it as the discriminant of $z$ goes to infinity. -See A. Enge and F. Morain's "Comparing invariants for class fields of imaginary quadratic -fields", ANTS-V 2002.<|endoftext|> -TITLE: A bounded homogeneous space which fails to be symmetric? -QUESTION [6 upvotes]: Do we have examples of a contractible bounded open set $D\subseteq\mathbf{C}^n$ such that $Hol(D)$ (the group of biholomorphisms $f:D\rightarrow D$) acts transitively on $D$ but such that there exists no symmetry at a given point $x\in D$ (so at all points by homogeneity). By a symmetry at $x$ I mean an element $s\in Hol(D)$ such that in a small neighborhood of $x$ only $x$ is fixed and $s^2=1$. - -REPLY [14 votes]: E.Cartan proved in 1936 that for dimension 1 and 2 bounded homogeneous spaces are symmetric. For dimension 3 he did not publish the proof considering it loo long in comparison to the interest of the result. This has now changed with P-Sapiro's example for dimension 4. So the proof for dimension 3 is presumably somewhere in E.Cartan's Nachlass, unpublished.<|endoftext|> -TITLE: Solutions to some equations in a free group -QUESTION [11 upvotes]: Let $F$ be the free group on (say) two generators, $a$ and $b$. Let $A$ and $B$ be (freely reduced) elements of $F$. Let $W(X, Y)$ denote a word on the words $X, Y$. --Is it ever true that the equation $W(a, b) = W(A, B)$, has finitely many non-conjugate solutions (by conjugate solution I mean there exists a word $V$ such that $V^{-1}AV = A^{\prime}$ and $V^{-1}BV=B^{\prime}$)? -For example, take $W(a, b) = a^{-1}b^nab^m$. We therefore want to find $A$, $B$ such that $a^{-1}b^nab^m = A^{-1}B^nAB^m$. We can take $A=b^ia$ and $B=b$ for all $i$, and so this equation has infinitely many (non-conjugate) solutions. -In fact, $a\mapsto b^ia$, $b \mapsto b$ defines an automorphism of $F$ (as free groups are Hopfian). Further, different $i$'s give different coset representatives of Out(F), and so a related question would be, --Does there exist a word $W \in F$ such that there are only finitely many outer automorphisms $\phi$ such that $W\phi = W$? -I cannot seem to get anywhere with this. The only examples I can find are, essentially, trivial. For example, $W(a, b) = a$. However, this doesn't quite work, as then $b$ can be whatever we want (essentially, exclude this because its boring). -Any help/ideas of papers to look at would be greatly appreciated. - -REPLY [10 votes]: These equations have been considered here: Hmelevskiĭ, Ju. I. -Systems of equations in a free group. I, II. (Russian) -Izv. Akad. Nauk SSSR Ser. Mat. 35 (1971), 1237–1268; ibid. 36 (1972), 110–179. All solutions have been described there (see Theorem 3). - Update. See also more recent Touikan, Nicholas W. M.(3-MGL) -The equation $w(x,y)=u$ over free groups: an algebraic approach. -J. Group Theory 12 (2009), no. 4, 611–634.<|endoftext|> -TITLE: Poincare duality and the $A_\infty$ structure on cohomology -QUESTION [28 upvotes]: If $X$ is a topological space then the rational cohomology of $X$ carries a canonical $A_\infty$ structure (in fact $C_\infty$) with differential $m_1: H^\ast(X) \to H^{\ast+1}(X)$ vanishing and product $m_2: H^\ast(X)\otimes H^\ast(X) \to H^\ast(X)$ coinciding with the cup product. -If $X$ is a closed manifold then its cohomology algebra satisfies Poincare duality. This is a condition that refers only to the $m_2$ part of the $A_\infty$ structure. There are things more general than manifolds that satisfy rational Poincare duality, such as rational homology manifolds. In general, a space that satisfies rational Poincare duality is called \emph{rational Poincare duality space}. -Obviously, the statement that $X$ is a rational Poincare duality space places no additional restrictions on the $A_\infty$ structure beyond the condition on the product. -Question 1 Can one construct from the higher multiplications obstructions for a rational Poincare duality space to be rationally equivalent to a rational homology manifold, or a topological or smooth manifold? -Here is a second and somewhat related question. Poincare duality says that $H^\ast(X)$ is self-dual (with an appropriate degree shift). Thus the adjoints of the higher multiplication maps make the cohomology into an $A_\infty$ coalgebra (with appropriate adjustment of the grading). -Question 2 How does this $A_\infty$ coalgebra structure interact with the $A_\infty$ algebra structure on the cohomology? - -REPLY [16 votes]: It was shown in the IHES paper in the 70's "Infinitesimal computations in Topology" that over Q homological Poincare duality is sufficient for a simply connected Q homotopy type to be realized by a smooth closed simply connected manifold in dimensions 5,6,7,9.10,11,13,... etc. - In dim 8 ,12, 16, etc necessary and sufficient conditions -to realize are given. These are tantamount to the Thom Hirzebruch signature formula holding for some choice of fundamental class where the quadratic form over Q is a sum of positive and negative squares. -These statements imply the answer to question 1 is: There are no implications -on the higher infinity structures implied by a closed manifold representative -of a rational homotopy type.<|endoftext|> -TITLE: The Hardy Z-function and failure of the Riemann hypothesis -QUESTION [46 upvotes]: David Feldman asked whether it would be reasonable for the Riemann hypothesis to be false, but for the Riemann zeta function to only have finitely many zeros off the critical line. I very rashly predicted that this question would be essentially as hard as the Riemann hypothesis itself. However, on further reflection, I stumbled upon a natural and reasonable conjecture which has a serious bearing on whether this dichotomy holds, which I have never seen in print. -So, let $f:\mathbf{R}\to \mathbf{R}$ be continuous. There are various notions of quasi-periodic and almost periodic function in the literature. The following (quite weak) one is more than enough for my purposes: - -Definition. A function $f: \mathbf{R} \to \mathbf{R}$ is locally quasiperiodic if, for every bounded interval $I \subset \mathbf{R}$ and every $\delta>0$, there exists an unbounded sequence $t_n \in \mathbf{R}$ such that $\sup_{t\in I} |f(t+t_n)-f(t)|<\delta$. - -For example, finite trigonometric polynomials $\sum a_j \sin{(b_j t + c_j)}$ are locally quasiperiodic. -Back to the zeta function: the Hardy $Z$-function is defined as $Z(t)=\pi^{-it/2}\frac{\Gamma(1/4+it/2)}{|\Gamma(1/4+it/2)|}\zeta(1/2+it)$. The functional equation for the zeta function immediately implies that $Z(t)$ is real-valued, and by construction we have $|Z(t)|=|\zeta(1/2+it)|$. One of the nice things about the $Z$-function is that it turns out to be computable in fairly efficient ways (the Riemann-Siegel formula), and it reduces the problem of finding zeros of zeta on the critical line to finding sign changes of the $Z$-function. In fact, the $Z$-function knows about the Riemann hypothesis: If the $Z$-function has a negative local maximum or a positive local minimum, then the Riemann hypothesis is false; see e.g. Section 8.3 of Edwards's book. I don't believe the converse to this is known, so let's call such an extremum a strong failure of the Riemann hypothesis. -Now, I don't believe that the $Z$-function itself is locally quasiperiodic, because the density of its zeros should grow as $t$ grows, and it should wiggle "faster and faster" accordingly; more precisely, the number of zeros in an interval $[t,t+h]$ for $h$ fixed should be $\sim \frac {h}{2\pi}\log{t}$ as $t\to\infty$. However, rescaling in a naive manner, let's consider instead $Z(\frac{t}{\log{t}})$. This should have $\sim \frac{h}{2\pi}$ zeros in an interval $[t,t+h]$ for $h$ fixed and $t \to \infty$, and I see no reason not to believe that - -Conjecture A. The function $Z(\frac{t}{\log{t}})$ is locally quasiperiodic. - -My main reason for enunciating this is that the truth of Conjecture A implies that if there is one strong failure of the Riemann hypothesis, then there are infinitely many strong failures. This is actually pretty evident; take $I$ a small interval containing the relevant bad local extrema and take $\delta$ small enough so the intervals $I+t_n$ contain bad local extrema of the same type. -It's not obvious to me whether Conjecture A is at all accessible by current technology. For example, I don't a single example of an unbounded function which is provably locally quasiperiodic. I would love to see such an example (I've tried and failed to construct one). Also, it seems natural to ask whether there is some simple characterization of locally quasiperiodic functions in terms of properties of their (distributional) Fourier transforms. Is such a characterization reasonable to expect? - -REPLY [4 votes]: Dear David, I want to state first some prelimnary remarks. Do you know about the universality of the Riemann Zeta function? The behaviour in the region $1/2 < \Re s < 1$ of the Riemann zeta function is chaotic. In fact, given $\epsilon >0$, for any compact region $D$ in $1/2 < \Re s < 1$ ans any non vanishing bounded holomorphic function $f$ on $D$, there exists a sequence $T_n = \Omega(n)$ such that -$$ \sup_{z \in D} | f(z) -\zeta(z + iT_n)| < \epsilon,$$ -or even stronger the measure of all such $T$ has lower positive density. Here are the precise statements http://en.wikipedia.org/wiki/Zeta_function_universality. -Perhaps a related fact: The Riemann hypothesis holds if and only if $f$ can be replaced here by the Riemann Zeta function! -Proof for $<=$: Assume that $\zeta$ can be replaced for $f$ anywhere and $RH$ fails once say for a point in some $D$, then $\zeta$ would approximate itself on this $D$ arbitrary good in linear time. This produced every time a zero by Rouche's principle, which are far to many zeros by contradicting density results for zeros. -Poof For $=>$: If $\zeta$ fulfils RH, we can replace $f$ by $\zeta$. -So since the rescaling factor for $Z$ is pretty regular, hence you can deduce this quasiperiodic property in the region $1/2 < Re s <1$ directly from the property known for the Riemann Zeta function. (Approximate a continous function by an entire via the theorem of Mergelyan). On the critical line Joern Steuding & Co. have presented some results last year also for $Re s = 1/2$, which probably imply conjecture $A$.<|endoftext|> -TITLE: Preprint of Hamilton on deformations of foliations -QUESTION [6 upvotes]: Does anyone have access to Hamilton's 1978 Cornell preprint 'Deformation Theory of Foliations'. It is widely quoted but I couldn't find any online copy. - -REPLY [3 votes]: I have a printed copy, but I don't think it was ever published or put online. If you contact me, I can make a copy and send it to you.<|endoftext|> -TITLE: What are the most general classes of simplicial complexes or posets for which the Charney-Davis conjecture is known, and what is the most general setting for which it might expected to be true? -QUESTION [11 upvotes]: What I would like to know is exactly what the title asks: - -What are the most general classes of - simplicial complexes or posets for - which the Charney-Davis conjecture is - known, and what is the most general - setting for which it might expected to be - true? - -I believe it conjectured, for example, that for a flag simplicial sphere (i.e. a flag simplicial complex which is homeomorphic to a sphere) of dimension $2d-1$, -$(-1)^d (1- \frac{1}{2}f_0 + \frac{1}{4} f_1 - \frac{1}{8} f_2 + \dots + (\frac{1}{2})^{2d} f_{2d-1} ) \ge 0$, but that this is still not known. (Here $f_i$ is the number of $i$-dimensional faces.) -I am guessing that the case of simplicial polytopes follows easily from Stanley's $g$-theorem, but what are the most general classes of spheres for which this statement is known? This guess is not correct see this asnwer. -I am aware that there are analogous statements expected to be true for other triangulated manifolds besides spheres, perhaps for more general kinds of posets, etc. But what exactly are these conjectural statements in their most general forms? -Any pointers to survey articles would be greatly appreciated. - -REPLY [9 votes]: Matt also asked: "I am aware that there are analogous statements expected to be true for other triangulated manifolds besides spheres, perhaps for more general kinds of posets, etc. But what exactly are these conjectural statements in their most general forms?" -So let me add (gradually, helped by Eran Nevo) more information regarding the Charney-Davis conjecture and its analogs and extensions. - -The CD-index, order complexes of regular CW spheres and Gorenstein spaces.* - -Starting with a poset P you can define a simplicial complex, called the chain complex of P, whose faces are chains in P. Chain-complexes of posets are always flag. Chain-complexes of graded posets have the property of being "completely balanced" or "colored". If you color each element by its grade, every face is properly colored (it contains no two vertices of the same color). If P is a graded poset and all maximum chains have d elements then its chain complex is (d-1)-dimensional completely balanced simplicial complex. -Especially important are posets of faces of pure regular CW-complexes. (Here we will color an i-face by color 'i'.) We define now in a few steps the CD-index. Let P be a regular CW-complex. -i) For a subset S of indices we let $f_S(P)$ be the number of chains of faces whose dimensions are prescribed by S. (Equivalently, the number of (|S|-1)-dimensional faces of the chain complex whose vertices are collored by the indices of S.) -ii) Define $\beta_S(P)$ (Also denoted by $h_S(P)$) by -$$\beta_S(P)=\sum _{R \subset S}(-1)^{|S\backslash R|}f_T(P).$$ -(This definition extend to completely balanced pure simplicial complexes.) Consider the generating function in two noncommuting variables 'a' and 'b' describing the $2^d$ parameters $\beta_S(P)$ as follows. $$\beta(P)=\sum \beta_S(P)w_S.$$ -where $w_S$ is a word in 'a's and 'b's where you put in the ith place an 'a' if i belongs to S and an 'a' otherwise. -iii) If P is Eulerian than the polynomial $\beta_S(P)$ is symmetric w.r.t. replacing 'a' and 'b'. -(This was proved by Stanley and it extends to Eulerian completely balanced simplicial complexes.) -iv) Bayer and Billera showed that for Eulerian regular CW-complexes (and, more generally, graded Eulerian posts) the linear space spanned by the $f_S$ vectors (or $b_S$ polynomials) has dimension which is the dth Fibonacci number. -v) For Eulerian graded posets Jonathan Fine defined remarkable Fibonacci number of parameters: the coefficients of a certain degree d polynomial Q in two noncommuting variables C (of degree 1) and D (of degree 2). Fine showed that the $\beta(P)$ polynomial can be expressed as a polynomial in the two non-commuting variables C=a+b and D=ab+ba! -vi) The coefficient of $D^{d/2}$ is precisely the Charney-Davis expression. -vii) Kalle Karu proved that for face postes of regular CW spheres of dimension (d-1) (and more generally Gorensetein* regular CW spheres) these coefficients are nonnegative. This supplies important special cases for the Charney Davis conj. -Some link: A blog post on Bayer-Billera's theorem and the CD index; Bayer and Klapper's paper on the CD index' and list of papers citing it (CiteseerX); Karu's paper; - -Gal's $\gamma$-numbers. - -Let $K$ be a simplicial $(d-1)$-dimensional simplicial complex. The $h$-vector of $K$ is defined by -$$ \sum_{0\leq i\leq d}h_i(K)x^{d-i}= \sum_{0\leq i\leq d}f_{i-1}(K)(x-1)^{d-i}.$$ -If $K$ is Eulerian then we have the Dehn-Sommerville relations: $$h_i=h_{d-i}.$$ -The polynomial $$h(K,x)=\sum h_i(K)x^{d-i}$$ can be expressed (uniquely) as a linear combination of monomials $x^i(1+x)^{d-2i}$. -$$h(K,x)= \sum \gamma_i(K)x^i(1+x)^{d-2i}.$$ -Gal Conjecture: If K is a flag Gorenstein* (d-1)-dimensional complex (In particular, a triangulation of the (d-1)-sphere) then $\gamma_i(K) \ge 0$. -When $d$ is even, Gal's top coefficient is the Charney-Davis parameter. -Moreover it is conjectured by Nevo and Petersen that these parameters correspond to face numbers of a completely balanced simplicial complex. -It was also conjectured that the $h$-polynomial for flag triangulated spheres has real zeroes. This strong form of Charney-Davis conjecture was disproved by Światosław Gal as well. -Links: Gal's paper; the paper by Nevo and Petersen.<|endoftext|> -TITLE: Determinants of "almost identity" matrices. -QUESTION [5 upvotes]: Suppose that $A$ is a real square matrix with all diagonal entries $1$, all off-diagonal entries non-positive, and all column sums positive and non-zero. Does it follow that $\det(A)\neq0$? Is this just an exercise? Are these matrices well-known? - -REPLY [3 votes]: I'm answering to the "are these matrices well-known" part. Yes, they belong to at least two classes of widely studied matrices: - -Diagonally dominant matrices, as has been suggested before, i.e., matrices such that $|A_{ii}|>\sum_{j\neq i}|A_{ij}|$ for each $i$. -M-matrices. There are several equivalent definitions of M-matrices, such as matrices in the form $sI-P$, where $P$ is an elementwise nonnegative matrix and $s>\rho(P)$ ($\rho$=spectral radius), or matrices with nonpositive off-diagonal elements and all their eigenvalues in the right half-plane. You can find a comprehensive exposition, including an impressive list of 50 conditions equivalent to "$A$ is a nonsingular M-matrix", on Berman, Plemmons Nonnegative matrices in the mathematical sciences. They have several interesting properties that "look like" those of symmetric positive definite matrices.<|endoftext|> -TITLE: Analytic functions with isotopic x-rays -QUESTION [8 upvotes]: Following Arias-De-Reyna, the x-ray of an analytic function $f$ means markings on the complex plane, with one color showing the real locus of $f$ and another color the purely imaginary locus. -Suppose two analytic (or meromorphic, or analytic with isolated singularities) functions $f$ and $g$ possess isotopic x-rays (so an orientation preserving homeomorphism of the plane moves the x-ray of one precisely onto the x-ray of the other). What can we conclude about $f$ and $g$? -Actually I'm not sure that's the best way to frame the question. Perhaps I should ask for a topological characterization of x-rays, and then ask for the dimension of the space of functions in a given isotopy class, and a method for constructing at least one function with a given x-ray type? For example, an x-ray of a polynomial locates the polynomials roots and their multiplicities from which we can reconstruct the polynomial; so the space of polynomials in the isotopy class of a given polynomial has dimension equal to the number of distinct roots (right?). -Or perhaps one gets a good theory if one studies jointly the x-rays $f$ and $f'$ (with four colors so you can tell them apart)? Indeed, how much does the isotopy-type of the x-ray of $f$ determine about that of $f'$? - -REPLY [4 votes]: Perhaps it is too late (more than few days passed since the questions were asked) -but I will try -to clarify what happens in the non-compact case. Let us restrict ourselves to the fllowing -special case: suppose that we have a meromorphic function which has at most one critical or -asymptotic value on each of the four quadrants, and no other critical or asymptotic values. -In other words, there is a 4 point set $A=(a_j)$ one point in each quadrant and such that -$$f:S\backslash f^{-1}(A)\to S\backslash A$$ -is a covering. -For such functions preimages of the cross can be completely characterized: they are cell decompositions -of the plane, each vertex has degree 4. There can be "infinite faces" which have infinitely many -vertices on the boundary, but the number of vertices on every compact is finite. -Theorem. To each such cell decomposition corresponds a meromorphic function f either in the -unit disc or on the plane. -The free parameters are the 4 ponints $a_j$, plus 2 normalization parameters ($f$ and $f(az+b)$ -will have equivalent cell decompositions). The proof of this theorem is an easy application of the Uniformization theorem. -Now the difficult problem arises: how to tell whether this is a disc or the plane from the decomposition. This is called the Type Problem of a simply connected Riemann surface. -There was a lot of research on this problem in 1950-s. Of more recent works I mention -Peter Doyle, On deciding whether a surface is parabolic or hyperbolic, where a probabilistic -interpretation hinted in Thurston's last comment is established. And also a paper of Merenkov and Schramm who disprove a conjecture of Nevanlinna that the type depends on the growth rate of -concentric balls :-) -Of course there is nothing special with taking preimage of a cross from the beginning. -One can start with any cell decomposition of the Riemann sphere, such that $f$ has one critical -or asymptotic value in each face. The equivalence class of the preimage cell decomposition, -together with the critical and asymptotic values themselves and two normalization constants -define your function uniquely. -But it is hard to tell whether it is meromorphic in the disc or in the plane, except for some -special cases.<|endoftext|> -TITLE: Website to help track recent research level mathematical book publications? -QUESTION [14 upvotes]: I'm responsible for recommending purchases of research level monographs to my university library. My job would be vastly simplified if I could find a website (say) that made it easy to scan a list of recent publications of that sort. At present I find I must visit the websites of dozens of publishers. -I'm open to any suggestions, but let me mention some resources which I already use, but only with effort and partial success: -1) The library catalogs of other institutions. -- Generally by the time another college or university has a book cataloged the book isn't current. Also searching by publication date, to get current books, seems impossible unless I'm already searching under fairly a narrow subject heading such as "algebraic topology," but that would have me making many dozen of searches. -2) Amazon (or the like). Great for currency, but not easy to narrow the search to research level books, and broaden the search to cover all of university level mathematics. -3) Notices of the AMS Certainly current and useful for AMS publications, but the -advertisements would bias me towards big publishers and the particular books they choose to push. -Thanks in advance for whichever way anyone points me. If nothing of the sort exists, feel free to suggests ways to create, or induce the creation, of such a web resource. - -REPLY [3 votes]: In addition to already said, virtually all major publishers (Springer, AMS, Cambridge Univ. Press, World Scientific, ...) have mailing lists announcing the forthcoming titles in the chosen subject area.<|endoftext|> -TITLE: For what subsets S of (Z/nZ)* is there a Euclidean proof that there are infinitely many primes whose residues lie in S? -QUESTION [16 upvotes]: For small values of $n$ and $(a, n) = 1$ it is sometimes possible to give an elementary proof that there are infinitely many primes congruent to $a \bmod n$ along the lines of Euclid's classic proof of the infinitude of the primes. The idea is to find a non-constant integer polynomial $p(t)$ such that the primes dividing an integer value of $p$ are, with finitely many exceptions, congruent to $a \bmod n$ or $1 \bmod n$, and such that the first case occurs infinitely many times; then one proves, just as Euclid did, that the integer values of $p$ are divisible by infinitely many primes, and then one contrives to avoid the residue $1 \bmod n$ (if $a \neq 1$; if $a = 1$ then we take $p(t) = \Phi_n(t)$.) Results of Schur and Murty (see this paper of Keith Conrad) then imply that such a polynomial $p$ exists if and only if $a^2 \equiv 1 \bmod n$. -Say that a subset $S \subset (\mathbb{Z}/n\mathbb{Z})^{\ast}$ has property E if there exists a non-constant integer polynomial $p(t)$ such that the primes dividing an integer value of $p$ are, with finitely many exceptions, congruent to $s \bmod n$ or $1 \bmod n$ for some $s \in S$, and such that the first case occurs infinitely many times. Then we know that any $S$ containing an element squaring to $1$ has property E. -However, there exist subsets $S \subset (\mathbb{Z}/n\mathbb{Z})^{\ast}$ not containing such an element with property E. For example, let $G$ be a proper subgroup of $(\mathbb{Z}/n\mathbb{Z})^{\ast}$, let $a \not \in G$, and consider -$$p(t) = nt + a.$$ -Any prime divisor of $n + a$ is relatively prime to $a$, and at least one of these prime divisors must have residue $\bmod n$ not in $G$. If $p_1, ... p_k$ are finitely many such primes, then $p(p_1 ... p_k) \equiv a \bmod n$ and hence must have a prime factor with the same property which moreover is relatively prime to each $p_i$. It follows that $S = (\mathbb{Z}/n\mathbb{Z})^{\ast} - G$ has property E, and if $n$ does not divide $24$, then taking $G$ to be the subgroup of square roots of $1$ gives the desired example. -Question: What subsets $S$ of $(\mathbb{Z}/n\mathbb{Z})^{\ast}$ have property E and are minimal under inclusion? Are they closed under non-empty intersection? -(If this question is somehow answered in Conrad's paper, my apologies; the material is somewhat beyond me.) - -REPLY [11 votes]: I will try to convince you that no Euclidean proof can possibly show that there are infinitely many primes which are $2 \mod 5$. After giving some definitions, I will explain what I will actually show: -Let $G$ be a finite group. We define an equivalence relation on $G$ by $g \sim h$ if the (cyclic) subgroups generated by $g$ and by $h$ are conjugate. The equivalence classes for this relation are called divisions. Observe that a map of groups always respects this equivalence relation. A more precise (but not yet precise) claim is: - -Consider any Euclidean proof that there are infinitely many primes in some subset $S$ of - $\mathbb{Z}/N^*$. Then the set $S$ contains a complete division. - -Since $\mathbb{Z}/N^*$ is abelian, the conjugacy part of the definition of abelianness doesn't come directly into this theorem, but it will arise in the proof. -So, what is a Euclidean proof? Well, at some step in the proof, I must build a prime $p$ and say "$p$ has the property $P$, and therefore $p$ is in $S$." In every Euclidean proof I have seen, $P$ is one of two things: -(1) There is a number $M$ such that $M$ is not in some subgroup $G$ of $(\mathbb{Z}/N)^*$. So $M$ has a prime divisor which is not in $G$, and we choose $p$ to be that prime factor. So an allowable step in a Euclidean-proof is showing that there are infinitely many primes not in some subgroup $G$ of $(\mathbb{Z}/N)^*$ -(2) Some polynomial $f(X)$ factors in a particular way modulo $p$, and therefore $f$ lies in some class modulo $N$. (Generally, the observation is that $f$ has a root modulo $p$, but I'll allow more general things like $f$ has a quadratic factor" as well. For a polynomial $f$ of degree $n$, and a partition $\lambda$ of $n$, let $D(f, \lambda)$ be the set of primes such that the irreducible factors of $f$ have degrees $(\lambda_1 \lambda_2, \ldots, \lambda_r)$. And let $D(f, \lambda, N)$ be the image of $D(f, \lambda)$ modulo $N$. So an allowable step in an Euclidean-proof will be showing that there are infinitely many primes in some union of $D(f, \lambda, N)$'s. -So, what I will actually be proving is - -Any subgroup $G$ of $(\mathbb{Z}/N)^*$, and any $D(f, n, \lambda)$ in $(\mathbb{Z}/N)^*$, is a - union of divisions. - -So, suppose that $S$ is a set which does not contain any division, and let $T$ be disjoint from $S$, but contain an element representing every division class in $S$. If we have a Euclidean proof, it will construct some prime $p$. We may know that $p$ is not in various subgroups of $(\mathbb{Z}/N)^*$, or that $p$ is in some union of $D(f, \lambda, N)$'s. But that information can't distinguish whether $p$ is in $S$ or in $T$, so our proof can't show that $p$ is in $S$. -OK, my last boxed claim is a precise statement. Let's prove it. -It is easy to see that a subgroup of $(\mathbb{Z}/N)^*$ is a union of divisions (since we are in an abelian group, the conjugacy part of the definition is irrelevant). The complement of a subgroup is likewise such a union. -The interesting thing is $D(f, \lambda, N)$. Let $K$ be a Galois field where $f$ and $x^N-1$ both split. Let $G$ be the corresponding Galois group, so $G$ comes equipped with a map to $(\mathbb{Z}/N)^* \cong \mathrm{Gal}(\mathbb{Q}(\zeta_N)/\mathbb{Q})$. -Now, as you probably know, for $p$ a prime unramified in $K$, the factorization of $f$ modulo $p$ is determined by the Frobenius conjugacy class of $p$ in $G$. What you may or may not know is that it is actually determined by the division class of $p$! (For example, $x^5-1$ has the same factorization modulo primes which are $2 \mod 5$ and primes which are $3 \mod 5$.) To see this, just look at the recipe for reading the factorization off from the Frobenius class. It may or may not help to prove the following lemma: If $g \sim h$, and $X$ is a set with a $G$ action, then there is an order preserving bijection between the $g$ and the $h$ orbits in $X$. -So, the possible Frobenius classes in $G$ of primes in $D(f, \lambda)$ are unions of divisions (I am implicitly using the Cebotarov density theorem here). But then $D(f, \lambda, N)$ is just the projection to $(\mathbb{Z}/N)^*$ of the possible Frobenius classes in $G$ of primes in $D(f, \lambda)$. Since maps of groups take divisions to divisions, this shows that $D(f, \lambda, N)$ is a union of divisions. -Frobenius's density Theorem states that there are infinitely many primes with Frobenius in every division. (And, more precisely, that their Dirichlet density is the size of the division divided by the order of $G$.) It is significantly easier than Cebatarov's, using only the material from a first course in algebraic number theory and a first course in analytic number theory. -Cebatarov's density theorem is the "union" of Frobenius's and Dirichlet's theorems. What I am suggesting is that Euclidean methods, at best, can only get at their intersection. -I am not sure whether or not I think Euclidean proofs can get that far. If you think they can, give me a Euclidean proof that there are infinitely many primes which are in $\{ 3, 5 \}$ mod $7$. I can show infinitely many in $\{ 3,5,6 \}$, and infinitely many in $\{ 2,3,4,5 \}$, but I can't get the intersection.<|endoftext|> -TITLE: Point in Polygon algorithm from the viewpoint of a robot -QUESTION [6 upvotes]: I've come across the following puzzle: - -You're on an island, on which there is - a fence (which is a simple closed - contour). You need to determine - whether you're inside or outside the - fence. - -Now if you had the function defining the contour as well as the point you're in (e.g. you have a GPS), you could calculate the winding number. If you could climb over the fence, you could use ray casting. Both methods are described here. -However, I've come across a claimed solution that presumably does not require any of them: -Traverse the fence - say by keeping the fence stuck to your left side, until you've come back where you started, (assuming you can mark your start position). Repeat the process where the fence is always 1 meter to your left, orthogonally (assuming you can maintain orthogonality, maintain the 1 meter distance and the fence is always wide enough for you to maintain it). -The claim is - if the second trip (the one walking 1 meter away from the fence) took more time (assuming you can measure time and maintain the same exact speed throughout both trips), you're in the exterior. otherwise you're in the interior. -I haven't been able to prove this, and I'm not even sure it's right (couldn't find a counterexample, though). -Any thoughts ? - -REPLY [2 votes]: Lookj up "level sets". -The wikipedia page "Point in Polygon" talks about algorithms that can be used when the polygon's coordinates are known. The proposed solution of the length of a path following the fence along the fence (call it $d_0$) and a path following along the fence but maintaining a constant distance of $x=1$ meter (call it $d_1$) will work for a robot that can do the tasks you're asking of it. However, the answer of the difference in length being $2 \pi \sim 6.28$ meters would only apply if the fence is perfectly circular. -Given a map or diagram of the fence, generate multiple contours or level sets of points which are a constant distance from the fence. You'll end up with something that looks like a contour map or topographical map that the U.S. Geological surveys generates. Notice that for each distance $x$ (up to a certain limiting value), the level sets for $d_x$ may contain points inside the fence as well as outside the fence. Once $x$ is greater than the radius of the circle, the level sets for $d_x$ such that $x \gt r$ will only contain points outside the circle. For fences with concavities (like a pinched figure 8), the inner level set may break up into multiple non-connected paths. -If the fence is square, width edge length $2r$, then the close-fence contour $d_0$ will be $4 \times 2r = 8r$, whereas the 1-meter level set will be - -$4 \times (2r) + 4 \times (\frac{1}{2} \pi) = 8r + 2 \pi $ if the robot's path is 1-meter outside the fence (which consists of the edges translated outward a distance of 1-meter, and of quarter-circle arcs at each of the corner, as correctly pointed out by Mark Bennet's comment below). -$4 \times (2r-2) = 8r - 4$ if the robot path is 1-meter inside the fence. - -Thus for square, non-convex, and pretty much any noncircular fence, the level-set path one meter of the fence will not be $2 \pi \sim 6.28$ meters different from the level-set path of distance $0$ from the fence. -The generalization, however, will still apply that the level set path of distance $x$ away from the fence will be smaller ($d_x \lt d_0$) if the robot follows the level set path within the fence, vs. larger if the robot follows the level set path outside the fence ($d_x \gt d_0$).<|endoftext|> -TITLE: co-$A_\infty$ spaces -QUESTION [10 upvotes]: A co-$A_n$ space is a based space $Y$ equipped with a co-action by the Stasheff associahedron operad $K_\bullet$. This means that $Y$ is comes with certain maps $c_n: Y \times K_n \to Y^{\vee n}$, $n = 2,3,\dots$ that are inductively described (the definition of $c_n$ uses $c_{n-1}$ as input; the map $c_2$ is a co-$H$ structure). The suspension of a based space $X$ has the structure of a co-$A_\infty$ space. -Assume $Y$ is $2$-connected and has the homotopy type of a finite complex. Then -Schwaenzl, Vogt and I showed that a co-$A_\infty$ space $Y$ desuspends to a space $X$ in the -sense that there's a weak equivalence $\Sigma X \simeq Y$. -However we didn't try to check that the given weak equivalence is compatible in the co-$A_\infty$ sense. Part of the problem is that a morphism $f: Y \to Z$ of co-$A_\infty$ spaces should amount a co-$A_\infty$-structure on its mapping cylinder restricting to the given ones on $X \times 1$ and $Y$. However, this doesn't form a category: it's an $\infty$-category. -Now to my questions: -Question 1: is there a documented proof somewhere that the functor which assigns to a based space $X$ its suspension (considered as an co-$A_\infty$ space) induces an equivalence between the homotopy category of $1$-connected spaces and $2$-connected co-$A_\infty$ spaces? -Presumably, such a proof should be Hilton-Eckmann dual to one of the main results in the Book of Boardman and Vogt. -Question 2: Do function spaces coincide up to weak equivalence under this functor? That is, -is the map $$\hom_{\text{Top}_*}(X,X') \to -\hom_{\text{co-}A_\infty}(\Sigma X,\Sigma X')$$ -A weak equivalence under suitable hypotheses on $X$ and $X'$? -By $\hom$ in each case, I mean topologized mapping spaces. -How would one go about proving a result like this? - -REPLY [2 votes]: I have the feeling many of us would agree those statements `should' be true, but I can't think of -anyone who wrote things down publicly. Why not ask at alg-top? overflow requires active logging in -alg-top doesn't -jim<|endoftext|> -TITLE: Is there a complete classification of constant mean curvature surfaces? -QUESTION [13 upvotes]: I'm no expert in this field, but I am familiar with the classification of rotationally symmetric surfaces with constant mean curvature by Delaunay. I am aware that once we drop embeddedness and smoothness, the situation becomes very complicated. -My question is, other than the family of surfaces from Delaunay, is there a classification of (embedded in a Riemannain manifold $M^n$, smooth) surfaces with constant mean curvature? If not, is there a classification known under additional conditions? $M^n = R^n$, closed surfaces, surfaces with boundary, finite total curvature, etc? -I write such a broad question as I find this topic quite interesting and beautiful, and hope that others also share this feeling. - -REPLY [8 votes]: Since my previous answer to this question major progress was made: As Robert Haslhofer already mentioned, Brendle proved the Lawson conjecture that the only minimal CMC torus in the 3-sphere is the Clifford torus. Building on this work, Andrews and Li ("Embedded constant mean curvature tori in the three-sphere." J. Differential Geom. 99 (2015)) classified all embedded CMC tori in $S^3.$ To be more concrete, they used the same two point function as in Brendles proof in order to show that every embedded CMC torus must be rotationally symmetric, and hence is classified and can be written down in terms of elliptic functions explicitly. -The case of higher genus CMC surfaces in space forms is more difficult and unsolved by now. But there has been done computer experiments which suggest that the space of embedded CMC surfaces of higher genus $g\geq2$ with certain symmetries is related to the space of embedded CMC tori, see http://arxiv.org/pdf/1503.07838.pdf and the following image:<|endoftext|> -TITLE: Use of indiscernibles in model theory -QUESTION [12 upvotes]: What is the main use of indiscernibles in model theory? reading through Chang and Keisler's Model Theory it seems that the main motivation for indicernibles is for getting many non-isomorphic models for a theory (like the theory of dense linear order without endpoint). -Also, can you recommend the best source for reading about indiscernibles and their uses? - -REPLY [2 votes]: Indiscernibles are bounds for Skolem functions in all model-theoretic proofs of unprovability.<|endoftext|> -TITLE: Are all topological (finite-dim) real vector spaces homeomorphic to a coordinate space? -QUESTION [5 upvotes]: I know that all real, finite-dimensional topological vector spaces are isomorphic to $\mathbb{R}^n$ for some $n$, but are they also homeomorphic? -The reason I'm asking this is because I was wondering whether or not there were any disconnected real topological vector spaces. - -REPLY [12 votes]: Any (Hausdorff) topological real vector space of dimension $n<\infty$ is homeomorphic to $\mathbf{R}^n$ with the standard topology, see e.g. Rudin, Functional analysis, theorem 1.21. -Here are some comments: - -For some reason it is stated there for complex vector spaces, but, as remarked after the theorem, the proof works for real vector spaces as well. - -Instead of the Hausdorff axiom Rudin uses the (weaker) $T_1$ axiom in the definition of a topological vector space.<|endoftext|> -TITLE: What would you want to see at the Museum of Mathematics? -QUESTION [112 upvotes]: EDIT (30 Nov 2012): MoMath is opening in a couple of weeks, so this seems like it might be a good time for any last-minute additions to this question before I vote to close my own question as "no longer relevant". - -As some of you may already know, there are plans in the making for a Museum of Mathematics in New York City. Some of you may have already seen the Math Midway, a preview of the coming attractions at MoMath. -I've been involved in a small way, having an account at the Math Factory where I have made some suggestions for exhibits. It occurred to me that it would be a good idea to solicit exhibit ideas from a wider community of mathematicians. - -What would you like to see at MoMath? - -There are already a lot of suggestions at the above Math Factory site; however, you need an account to view the details. But never mind that; you should not hesitate to suggest something here even if you suspect that it has already been suggested by someone at the Math Factory, because part of the value of MO is that the voting system allows us to estimate the level of enthusiasm for various ideas. -Let me also mention that exhibit ideas showing the connections between mathematics and other fields are particularly welcome, particularly if the connection is not well-known or obvious. - -A couple of the answers are announcements which may be better seen if they are included in the question. -Maria Droujkova: We are going to host an open online event with Cindy Lawrence, one of the organizers of MoMath, in the Math Future series. On January 12th 2011, at 9:30pm ET, follow this link to join the live session using Elluminate. -George Hart: ...we at MoMath are looking for all kinds of input. If you’re at the Joint Math Meetings this week, come to our booth in the exhibit hall to meet us, learn more, and give us your ideas. - -REPLY [21 votes]: https://www.scribd.com/document/479581247/Letter-to-MoMath-Board -Update: many of us got together to take a stand against unethical practices at the museum. See the above open letter to the Board of Trustees which recommends the replacement of the CEO Cindy Lawrence. -The concerns raised therein are at the intersection of the problems observed by the cosigners and are not comprehensive. - -After serving as Chief of Mathematics at MoMath for the better part of 2 years, I'd like to shed some new light on this. -Firstly, this thread was a beautiful idea, to prompt the community for ideas before the Museum opened. However, at this point, the reality is: the last thing the Museum needs is more math ideas. What they need is proper implementation, and support for education. There is just a huge amount of work that one must do to get from a concept in a mathematician's brain to an interactive exhibit/lesson/activity that will work with kids. That is an ambitious thing to take on even if you don't have any other problems ... which the Museum does (see, for instance, the long history of complaints on Glassdoor -- they are a bit emotional, but having been there, I can say the complaints are well founded). I feel I did some great work there that I'm very proud of, but it was an uphill battle. -So here's what I'd like to see at the Museum: - -proper administrative support for the existing ideas to be correctly -implemented, -a positive change in leadership so that the employees -will be treated with respect, and -for the Board of trustees to take seriously the education standards there should be for a place bearing the name "National Museum of Mathematics."<|endoftext|> -TITLE: Wanted: a graph $G$ without bridges, whose square is not hamiltonian -QUESTION [9 upvotes]: Construct an example of graph $G$ without bridges, such that its square $G^2$ is non hamiltonian. -Note: -Since Fleischner's Theorem (the square of each 2-connected graph is Hamiltonian) and bridges are forbidden, the required graph should have at least one cut-vertex. - -REPLY [10 votes]: You can find an example of a bridgeless graph with cut points, whose square is not hamiltonian in this paper of Fleischner and Kronk. (I know the paper is in German, but the figure of the graph is on the first page.) Fleischner also mentions this example in his paper "The Square of Every Two-Connected Graph Is Hamiltonian".<|endoftext|> -TITLE: Jacobson radical = intersection of all maximal two-sided ideals -QUESTION [9 upvotes]: I'm embarassed to ask this question, but the literature on noncommutative rings seems to give this a berth as if it was absolutely trivial and not worth discussing, and I can't prove it, so all I can do is ask it here... -Let $A$ be a finite-dimensional $k$-algebra, where $k$ is a field. Is it true that the Jacobson radical equals the intersection of all maximal two-sided ideals? (The latter intersection is known as the Brown-McCoy radical of $A$.) -If yes, a short proof (the more self-contained, the better) would be great. -(This is again for use in coalgebra theory.) - -REPLY [5 votes]: Just for the record, here is an example of a (necessarily infinite dimensional) $k$-algebra $A$ where the Jacobson radical is not equal to the intersection of all maximal two-sided ideals. -Let $k$ be a field of characteristic $0$ and let $A = U(\mathfrak{sl}_2) / \langle C \rangle $ where $C = ef + fe + \frac{1}{2}h^2$ is the Casimir element. The image of the augmentation ideal of $U(\mathfrak{sl}_2)$ in $A$ is the unique maximal two-sided ideal of $A$, but $A$ acts faithfully on the Verma module of highest weight $-2$ so $A$ is primitive and its Jacobson radical is zero.<|endoftext|> -TITLE: How to triangulate real projective spaces (as simplicial complexes in Mathematica)? -QUESTION [8 upvotes]: Hello! -I have written a program in Mathematica 7, which calculates for a (finite abstract) simplicial complex all its homology groups. I would really like to test it on the projective spaces, but cannot find a way to triangulate them. There seem to be very few articles on this matter and none of them states directly how the actual triangulations might be achieved, so I turn to MO for help. -How can the real projective spaces -$$RP^n \approx B^n/_{x\:\sim-x;\;\; x\in\partial B^n}$$ -be triangulated as simplicial complexes? A simplicial complex is presented as a list of simplices that aren't a face of any bigger simplex (the "main simplices"). Each k-simplex is just an ordered list of some k+1 integers. -For example, {0,1,2} is a 2-simplex, as is {1,3,15}, etc. Examples of simplicial complexes: {{0,...,n}} is a n-dimensional ball, {{1,2},{1,3},{2,3}} is an empty triangle, skeleton[{Range[n+2]},n] is a n-sphere, etc. -I have already found a concrete triangulation for the real projective plane, but nothing more general. -It would also be appreciated, that the actual triangulation is reasonably small (not necessarily minimal), so that the program calculates homology groups fast enough. Also, answers in code / pseudocode are much desirable (projectiveSpace[n_]:=???). -P.S. I have already written some functions, which can be used (sc...simplicial complex): - -skeleton[sc,k] ...list of all k-faces of all simplices -sum[sc1,sc2]...topological sum (disjoint union) -connected sum[sc1,sc2] ...removes two k-simplicices and adds a tunnel between them -product[sc1,sc2]...staircase triangulation of a product (reorders the vertices) -cone[sc] -suspension[sc] - -REPLY [2 votes]: I've managed to find my mistake and here is the final (if I'm not mistaken) correct solution. -ClearAll[faces, skeleton, subseteq, subset, sBarycentricSubdivision, \ -scBarycentricSubdivision, sphere, projectiveSpace] -faces[s_, k_] := Reverse@Subsets[s, {k + 1}] -skeleton[sc_, k_] := Union[Sort /@ Flatten[faces[#, k] & /@ sc, 1]] - -subseteq[{},A_List] := True -subseteq[{x_,y___}, A_List] := MemberQ[A, x] && subset[{y}, A] -subseteq[A_List,B_List, C__List] := subset[A,B] && subset[B,C] -subset[A_List, B_List] := subseteq[A, B] && Complement[B, A]!={} -subset[A_List, B_List, C__List] := subset[A, B] && subset[B, C] - -sBarycentricSubdivision[{x_Integer}] := {{{x}}} -sBarycentricSubdivision[s_List] := - Subsets[s, {1, Length@s}] //Subsets[#,{Length@s}]& // - Select[#,subset@@#&]& -scBarycentricSubdivision[sc_List] := - Flatten[sBarycentricSubdivision /@ sc, 1] - -sphere[n_Integer] := skeleton[{Range[n+2]},n] - -projectiveSpace[n_Integer] := Module[ - {sc1, vertices1, k, identification, sc2, vertices2, sc3}, - sc1 = scBarycentricSubdivision[sphere[n]]; - vertices1 = - Sort@DeleteDuplicates@ - Flatten[sc1, - 1]; (*the vertices (lists) are ordered antipodally with respect \ -to complement, ie. vertices1[i]={1,...,n+2}-vertices1[-i]*) - k = Length@vertices1; - identification = - Table[vertices1[[-i]] -> vertices1[[i]], {i, 1, k/2}]; - sc2 = DeleteDuplicates[Sort /@ (sc1/.identification)]; - vertices2 = Flatten[sc2, 1] //Sort //DeleteDuplicates; - sc3 = Map[Position[vertices2, #][[1, 1]] &, sc2, {2}] //Sort // - DeleteDuplicates; - sc3 - ] - -And now, the command projectiveSpace[2] indeed returns a triangulation of the projective plane with 7 vertices and 12 faces. Also, the homology groups of $RP^0$,..., $RP^3$ all seem to be correct. -Thank you, mr. Palmieri, greetings from Slovenia.<|endoftext|> -TITLE: A noetherian proof of Zariski's Main Theorem? -QUESTION [9 upvotes]: Recall that Zariski's Main Theorem states that if $f: X \to Y$ is a quasi-finite, separated, and finitely presented morphism into a quasi-compact separated scheme $Y$, then there is a factorization of $f$ into an open immersion followed by a finite morphism. In EGA IV-8, this is proved by reducing to the case of $Y$ the $\mathrm{Spec}$ of a noetherian ring by a finite presentation argument (the general machinery of which is developed in the prior part of that section), then reducing to the case of a local noetherian excellent ring (by again using the finite presentation argument, since by this machinery proving things about the local scheme $\mathrm{Spec}(\mathcal{O}_y)$ is the same as proving things in a neighborhood), and finally by completing and proving the result for $Y$ the spectrum of a complete local noetherian ring, after which it is basically commutative algebra. -This argument is very pretty, but I am curious if there is a more elementary approach in the special case of $Y$ noetherian, or even in the classical case of schemes of finite type over a field (that avoids the general machinery of finite presentation arguments and the descent of properties of morphisms under faithfully flat base-change). Namely, I am curious whether there is an argument that uses less fancy machinery, and could be phrased in the language of varieties. Is there one? - -REPLY [3 votes]: Raynaud and Hochster, and Stacks, give essentially the proof of Peskine -this proof does not use noetherianity -a constructive proof, extracted from Peskine proof is given in -the following paper -{Alonso, M. E. and Coquand, T. and Lombardi, H.}, - TITLE = {Revisiting {Z}ariski main theorem from a constructive point of - view}, - FJOURNAL = {Journal of Algebra}, - VOLUME = {406}, - YEAR = {2014}, - PAGES = {46--68},<|endoftext|> -TITLE: Forbidden mirror sequences -QUESTION [21 upvotes]: Let $\cal{M}$ be a finite collection of two-sided mirrors, -each an open unit-length segment in $\mathbb{R^2}$, -and such that the segments when closed are disjoint. -A ray of light that reflects off the mirrors determines -a mirror sequence or mirror string -consisting of the mirror indices in order of reflection. -Obviously no mirror string can contain $a a$ as a substring. -I am wondering if mirror sequences may be characterized -by a list of forbidden substrings. - -           - -For example, consider the special case of parallel mirrors, -none collinear. -Orient them all vertical, and label them sorted left to right. -Then for labels $a < b < c$, these are forbidden substrings: -$$ b * a c * b $$ -$$ b * c a * b $$ -with $*$ being any string (including the empty string). -So in the above example, after 415 appears, 4 cannot occur again. -Perhaps these are the only forbidden patterns for parallel mirrors? -Edit. Apologies for not initially phrasing a clear question. - -Q. Are there strings of mirror indices that cannot be realized by some ray reflecting - among some collection $\cal{M}$ of mirrors? Is there a list of such strings - that characterize all the realizable sequences? - -In contrast to the parallel-mirrors example above, I am primarily interested in mirrors without -constraints on their placements or orientations. -Perhaps analogous sequences have been studied before, maybe in another context? -Pointers appreciated! - -REPLY [5 votes]: Here is a possible way of producing such forbidden configurations. Suppose you have $(ab)^k$ for some large $k$. Then I'd like to claim that $a$ and $b$ must be nearly parallel (see Thurston's answer). So produce a sequence with three mirrors containing $(ab)^k$, $(bc)^k$, $(ac)^k$ and a forbidden configuration for parallel mirrors such as the one you gave above (which you then have to prove also is forbidden for nearly parallel mirrors).<|endoftext|> -TITLE: Early stabilization in the homotopy groups of spheres -QUESTION [31 upvotes]: Thanks to Freudenthal we know that $\pi_{n+k}(S^n)$ is independent of $n$ as soon as $n \ge k+2$. However, I was looking at the table on Wikipedia of some of the homotopy groups of spheres and noticed that $\pi_2^S$ (the second stable stem) and $\pi_6^S$ are achieved earlier than required by the suspension theorem. So my question is: -What is known about this phenomenon of early stabilization in the homotopy groups of spheres? Does it occur finitely many times or infinitely many times? Also, is there any sort of intuitive reason why stabilization might occur early? -Also, I spotted a few times where stabilization almost appeared, but there was a pesky copy of $\mathbb{Z}$ that appeared and then disappeared right before the stable range. I'm talking about the cases of $\pi_{11+n}(S^n)$, $\pi_{15+n}(S^n)$, and $\pi_{19+n}(S^n)$. I assume this has something to do with Hopf elements showing up somewhere? -I know pretty much nothing about this except the basic homotopy theory, a few very small calculations, and the Pontrjagin construction relating all of this to framed manifolds- so any references or illuminating insights would be helpful! - -REPLY [38 votes]: Nice question! Funny enough, the answer is that you've already found all of the examples of early stabilization (excepting, of course, the fact that you didn't mention $\pi_n S^n$ stabilizing early). This is true even if you ignore odd torsion. -The "pesky copy of ℤ" is indeed related to the Hopf invariant. The fact that these classes exist on the edge of the stable range goes back to Serre's work where he shows exactly which homotopy groups of spheres contain a free summand. -More generally, though, the way one can see what's going on at the edge of the stable range is to use the EHP sequence - see, for example, http://web.math.rochester.edu/people/faculty/doug/mypapers/ehp.pdf. For any integer $n > 0$, there is (2-locally) a homotopy fiber sequence -$$ -S^n \to \Omega S^{n+1} \to \Omega S^{2n+1} -$$ -and this gives rise to a long exact sequence of homotopy groups (after 2-localization) -$$ -\cdots \to \pi_{2n} S^n \mathop\to^E \pi_{2n+1} S^{n+1} \mathop\to^H \pi_{2n+1} S^{2n+1} \mathop\to^P \pi_{2n-1} S^n \mathop\to^E \pi_{2n} S^{n+1} \to 0. -$$ -Here the maps labelled "E" are for suspension, the maps labelled "H" are for the Hopf invariant, and the maps labelled "P" are for something related to a Whitehead product. We know the middle term, so we can rewrite this: -$$ -\cdots \to \pi_{2n} S^n \mathop\to^E \pi_{2n+1} S^{n+1} \mathop\to^H \mathbb{Z} \mathop\to^P \pi_{2n-1} S^n \mathop\to^E \pi_{2n} S^{n+1} \to 0. -$$ -The last map on the right is the "edge" of the stable range and so you have early stabilization if and only if this map is an isomorphism, or equivalently if the map "P" is zero. The map "P" is zero if and only if the map "H" is surjective. -However, "H" in this case really is the classical Hopf invariant: if you have an element $f:S^{2n+1} \to S^{n+1}$ viewed as an attaching map, the element $Hf$ detects which element of $H^{2n+2}$ is the square of the generator of $H^{n+1}$ in the space $Cf$ obtained by using $f$ to attach a cell. J.F. Adams proved that the only time $Hf$ can take the value 1 is if $n+1$ is 1, 2, 4, or 8. So the only time you could have early stability is when the stable stem $n-1$ is -1, 0, 2, or 6. (EDIT: Had an indexing error at the end. Sorry.)<|endoftext|> -TITLE: Degree of Transcendentality and Feynman Diagrams -QUESTION [12 upvotes]: Physicists computing multiloop Feynman diagrams have introduced various -techniques and conjectures that involve the notion of Degree of Transcendentality (DoT). From what I understand one defines -1) $DoT(r)=0$, r rational -2) $DoT(\pi^k)=k$, $k \in {\mathbb N}$, -3) $DoT(\zeta(k))=k$, -4) $DoT( a \cdot b)= DoT(a)+DoT(b)$ -One then proves for example that the $\ell$-loop contribution to a certain scaling function in $N=4$ Supersymmetric gauge theory consists of a sum of terms all of which have DoT equal to $2 \ell-2$. -This can't be rigorous mathematically, since it is not even known that $\zeta(2n+1)$ is transcendental, but is there some circle of ideas, or conjecture in mathematics that if true would give a precise definition to DoT? - -REPLY [9 votes]: Your transcendentality reminds me about the Institute of Algebraic Meditation at Höör (Sweden). To be honest, your definition corresponds to what is known as the weight of a (multiple) zeta value (see Michael Hoffman's http://www.usna.edu/Users/math/meh/mult.html, especially the references on MZVs). These indeed occur in the computation of Feynman's diagrams. As for conjectures related to the transcendental number theory tag, a belief is that $\pi$ and odd zeta values $\zeta(3)$, $\zeta(5)$, etc, are algebraically independent over the rationals.<|endoftext|> -TITLE: Sums of arctangents -QUESTION [9 upvotes]: $$ -\begin{align} -\arctan(x) = {} & \arctan(1) + \arctan\left(\frac{x-1} 2 \right) \\ -& {} - \arctan\left(\frac{(x-1)^2} 4 \right) + \arctan\left(\frac{(x-1)^3} 8 \right) - \cdots -\end{align} -$$ -Is this known? - -REPLY [4 votes]: I've voted up Pietro Majer's incomplete answer and Michael Renardy's incomplete answer in the "comments" section. Here's my own incomplete answer. -Here's how I got this series: start with the identity -$$ -\arctan a - \arctan b = \arctan \frac{a-b}{1+ab}. -$$ -From this we get -$$ -\arctan x = \arctan 1 + \arctan\frac{x-1}{1+x}. -$$ -Substituting 1 for $x$ everywhere in the last expression except the power of $x-1$, we get the 1st-degree term. So we need to replace the last term above by the 1st-degree term plus another arctangent by using the basic identity above, and we get -$$ -\arctan\frac{x-1}{1+x} = \arctan\frac{x-1}{2} + \arctan\frac{-(x-1)^2}{2(1+x) +(x-1)^2}. -$$ -Then again substitute 1 for $x$ everwhere in the last term except in the power of $(x-1)$ in the numerator, to get the 2nd-degree term, and then write the last term above as the sum of the 2nd-degree term and another arctangent of a yet more complicated rational function. And so on. -Does the sequence of arctangents of rational functions go to 0? In some sense? I don't know, nor do I know the general pattern. -I actually tried this first with $x-2$ instead of $x-1$; then I decided that $x-1$ already has enough initial unclarity. -I don't even know whether in some reasonable sense the process goes on forever.<|endoftext|> -TITLE: Negative intersection of symplectic submanifolds -QUESTION [7 upvotes]: For a symplectic 4-manifold, it is possible for two symplectic submanifolds to intersect negatively? Actually it is an exercise in the 4-manifold book by Gompf and Stipsicz to find symplectic planes in $\mathbb{R}^4$ with negative intersection. Now I want to know the closed case. Does this happen also for closed symplectic manifolds? -edited question: -Let $(M, \omega)$ be a closed symplectic manifold. If $A$ and $B$ are different homology classes represented by symplectic submanifolds of complementary dimension, do they always intersect non-negatively? - -REPLY [5 votes]: The answer to this question is YES. I assume you want $A$ and $B$ to be connected. -Already in the case of four manifolds two symplectic surfaces can have negative intersection. -To construct an example, we use that if two symplectic surfaces in a $4$-fold intersect positively, you can always smoothen the neighborhood of their intersection to get a new symplectic surface. (Of course this is not always true in the algebraic case). -Example. Take any symplectic 4-manifold $M$ with a symlectic surface $C$, such that $C^2=-1$ (for example $\mathbb CP^2$ blown up once). Blow up $M$ at a point of $C$ and let $E$ be the exceptional curve, while $C'$ be the proper transform of $C$. Notice that $C'^2=-2$ and -$C'$ intersect $E$ positively in one point. So we can smoothen the union $C'\cup E$ (this is the full preimage of $C$ under the blow up, and we can not smooth this union algebraically) and get a surface that we call $B$. Finally we have $B\cdot C'$=$C'^2+C'\cdot E=-2+1<0$. -ADDED. Smoothing. In order to smoothen $C'\cup E$ symplectically we can first chose coordinates in a local chart $U$ and adjust a bit the symplectic form so the $C'\cup E$ is given in $U$ by $zw=0$, and the symplectic form is $-i(dz\wedge d\bar z+dw\wedge d\bar w)$. Then surely the curve given by $zw=\varepsilon$ is symplectic in $U$ (since the smoothen curve is complex, while the symplectic form is Kahler). But the whole curve is now discontinuous on the boundary of $U$. To cure this we consider $zw=\varepsilon (F(|z|^2+|w|^2))$ where $F(0)=1$ in $U/2$ and $F$ is smooth with compact support. It is symplectic again, since inside $U\setminus (U/2)$ it is a little perturbation of the union of lines $z=0$ and $w=0$<|endoftext|> -TITLE: Generalization of the club filter -QUESTION [7 upvotes]: If $\alpha$ is a (limit) ordinal, then a subset $S\subseteq\alpha$ is club if $\alpha$ is closed as a subset of $\alpha$ under the order topology and unbounded in $\alpha$. The set of all sets containing a club forms a filter on the subsets of $\alpha$, called the club filter. -This definition can be extended in the following way. Let $C$ be any infinite set, and let $P$ be the set of all countable subsets of $C$. Then $S\subseteq P$ is club if $S$ is closed under unions of countable chains (closed), and for all $X\in P$, there is some $Y\in S$ with $X\subseteq Y$ (unbounded). -(One area where this notion of clubness appears is in proofs of a Lowenheim-Skolem theorem for infinitary logic; see e.g. http://www.math.uic.edu/~marker/dwk.pdf.) -Given a set $C$, we can take the set of all subsets of $P$ which contain a club (in the generalized sense); call this the club filter on $P$. -My question is: when is the club filter in this latter case an ultrafilter? -If I understand things correctly, there should be no possibility of the club filter being an ultrafilter if we assume AC. However, in the original sense of the word club, the club filter on $\omega_1$ is an ultrafilter assuming AD, so this leads me to believe that, in ZF + AD, there might be interesting sets $C$ the club filter of which is an ultrafilter. -In particular, what kinds of choice need to fail at $C$ or $P$ in order for the club filter on $P$ to be an ultrafilter? -I hope this question is meaningful; I don't have much background knowledge of models of set theory in which choice fails. - -REPLY [2 votes]: First, in order to avoid Amit's concern, we may as well assume countable choice. Note that if AD (axiom of determinacy) holds in $L(\mathbb{R})$, then DC (axiom of dependent choice) will also be true there, and DC implies countable choice. -What you are asking for then is related to Jech's notion of stationarity. Specifically, we say that $S \subseteq [A]^{\omega}$ ($[A]^{\omega}$ is the set of countable subsets of $A$) is stationary when $S$ meets every club, where club here is in the sense you described. Now you can verify that the club filter on $[A]^{\omega}$ is an ultrafilter if and only if $[A]^{\omega}$ cannot be decomposed into two disjoint stationary sets. -Although not true for arbitrary $A$, you can also verify that $S \subseteq [\omega_1]^{\omega}$ is stationary according to Jech's characterization if and only if $S \cap \omega_1$ is stationary in the usual sense. Since AD implies that the club filter on $\omega_1$ is an ultrafilter ( Model of ZF + $\neg$C in which Solovay's Theorem on stationary sets fails? ), I claim that the club filter on $[\omega_1]^{\omega}$ is actually an ultrafilter if AD is true for suppose not. Then the club filter on $[\omega_1]^{\omega}$ could be decomposed into two disjoint stationary sets $S_1$ and $S_2 = [\omega_1]^{\omega} \setminus S_1$ so that $S_1 \cap \omega_1$ and $S_2 \cap \omega_1$ would be disjoint stationary sets of $\omega_1$. But this is impossible because then $S_1 \cap \omega_1$ and its complement would not be in the club filter on $\omega_1$.<|endoftext|> -TITLE: Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies? -QUESTION [24 upvotes]: Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies? -x,y are integral points on an Elliptic Curve and are easy to find using enumeration of divisors of n (assuming n can be factored). -If yes, will large number of solutions give moderate rank EC? -If one drops $-1$ i.e. $xy(x-y)=n$ the number of solutions can be unbounded via multiples of rational point(s) and then multiplying by a cube. (Explanation): Another unbounded case for varying $a , n$ is $xy(x-y-a)=n$. If $(x,y)$ is on the curve then $(d x,d y)$ is on $xy(x-y-a d)=n d^3$. Find many rational points and multiply by a suitable $d$. Not using the group law seems quite tricky for me. The constant $-1$ was included on purpose in the initial post. -I would be interested in this computational experiment: find $n$ that gives a lot of solutions, say $100$ (I can't do it), check which points are linearly independent and this is a lower bound on the rank. -What I find intriguing is that all integral points in this model come from factorization/divisors only. - -Current record is n=179071200 with 22 solutions with positive x,y. Due to Matthew Conroy. -Current record is n=391287046550400 with 26 solutions with positive x,y. Due to Aaron Meyerowitz -Current record is n=8659883232000 with 28 solutions with positive x,y. Found by Tapio Rajala. - -Current record is n=2597882099904000 with 36 solutions with positive x,y. Found by Tapio Rajala. -EDIT: $ab(a+b+9)=195643523275200$ has 48 positive integer points. – Aaron Meyerowitz (note this is a different curve and 7 <= rank <= 13) -A variation: $(x^2-x-17)^2 - y^2 = n$ appears to be eligible for the same question. The quartic model is a difference of two squares and checking if the first square is of the form $x^2-x-17$ is easy. -Is it possible some relation in the primes or primes or divisors of certain form to produce records: Someone is trying in $\mathbb{Z}[t]$ Can the number of solutions xy(x−y−1)=n for x,y,n∈Z[t] be unbounded as n varies? ? Read an article I didn't quite understand about maximizing the Selmer rank by chosing the primes carefully. -EDIT: The curve was chosen at random just to give a clear computational challenge. -EDIT: On second thought, can a symbolic approach work? Set $n=d_1 d_2 ... d_k$ where d_i are variables. Pick, well, ?some 100? ($d_i$, $y_i$) for ($x$,$y$) (or a product of $d_i$ for $x$). The result is a nonlinear system (last time I tried this I failed to make it work in practice). -EDIT: Related search seems "thue mahler" equation' -Related: unboundedness of number of integral points on elliptic curves? -Crossposted on MATH.SE: https://math.stackexchange.com/questions/14932/can-the-number-of-solutions-xyx-y-1-n-for-x-y-n-in-z-be-unbounded-as-n - -REPLY [10 votes]: Although this is not a mathematical answer I will put the results of my brute force search as an aswer as requested by jerr18. I didn't get anywhere with the thinking part. -Code -You can find the (non-optimal) C-code I wrote under my webpages. The biggest limitation with the program is that it uses 64-bit integers. Feel free to run, test, tweak and/or mutilate the code as you wish. -The program constructs first $n$ with a recursion and then $y$ with a recursion (this way I avoid considering values of $y$ that don't divide $n$). Finally it checks if the positive solution $x$ to the equation $$xy(x-y-1) = n$$ is an integer. -Results -Here are some values found using this program (in roughly 4 hours). -36 positive solutions -$$n = 2597882099904000 = 2^9 · 3^3 · 5^3 · 7 · 13 · 17 · 23 · 29 · 31 · 47$$ -30 positive solutions -$$ n = 34747990981704000 = 2^6 · 3^4 · 5^3 · 7^2 · 11 · 13 · 17 · 19^2 · 29 · 43 $$ -28 positive solutions -$$n = 105140926800 = 2^4 · 3^3 · 5^2 · 7^2 · 13 · 17 · 29 · 31 $$ -$$n = 8659883232000 = 2^8 · 3^3 · 5^3 · 7 · 11 · 13 · 17 · 19 · 31 $$ -$$n = 3783439308448800 = 2^5 · 3^4 · 5^2 · 7^3 · 11 · 13 · 19 · 31 · 43 · 47$$ -$$n = 9928464968822400 = 2^7 · 3^4 · 5^2 · 7^2 · 11^2 · 13 · 17 · 23 · 31 · 41$$ -$$n = 18680310941292000 = 2^5 · 3^4 · 5^3 · 7 · 11^2 · 13^2 · 17 · 19 · 29 · 43$$ -$$n = 88550619849291600 = 2^4 · 3^5 · 5^2 · 7^2 · 11^2 · 13 · 17 · 19 · 23 · 37 · 43$$ -Note that I did not check the results after my program handed them to me... -Edit: Just out of curiosity I tried also with $+1$ instead of $-1$. For example the equation -$$ xy(x-y+1) = 388778796252000 = 2^5 · 3^3 · 5^3 · 7^2 · 11 · 17 · 19 · 23 · 29 · 31$$ -has 38 positive solutions.<|endoftext|> -TITLE: Fully dualizable objects in classical field theories -QUESTION [16 upvotes]: This is a follow up to this MO question: Free symmetric monoidal $(\infty,n)$-categories with duals -Freed-Hopkins-Lurie-Teleman define a classical field theory as a symmetric monoidal functor $I$ from $n$-cobordism to the symmetric monoidal $n$-category $Fam_n(\mathcal{C})$ of $n$-families over a fixed symmetric monoidal $n$-category. This is required to lift a certain "geometric background", i.e., a fixed functor from $n$-cobordism to the $n$-category $Fam_n$. -If $\mathcal{C}$ in an $(\infty,n)$ symmetric monoidal $n$-category with duals, then it is likely that $Fam_n(\mathcal{C})$ has duals, too (I have so far checked this only for $n=1$; in this particular case, the dual of a functor $X\to \mathcal{C}$ from a finite groupoid $X$ to $\mathcal{C}$ is nothing but the dual representation of $X$; and I expect this will still be true for higher $n$'s). Then, by the cobordism hypothesis, the datum of $I$ would be reduced to the choice of a fully dualizable object $I(*)$ in $Fam_n(\mathcal{C})$, i.e., of a functor $X\to \mathcal{C}$ "with a few good properties". The object $I(*)$ -is explicited a couple of times in the paper (namely in the cases $n=1$ and $n=2$), but its fully dualizability is never discussed, nor seems to be claimed that $I(*)$ already contains all the information of $I$. For $n=3$, instead, the natural 3-representation of the groupoid $*//G$ which would starightforwardly generalize the $n=1$ and $2$ cases is not considered explicitely, and only its integrated (or quantized) version apperas. -Therefore, here is the question: -Is it correct that a classical field theory $I$ in the sense of Freed-Hopkins-Lurie-Teleman are equivalent to the datum of a fully dualizable object $I(*):X\to \mathcal{C}$ in $Fam_n(\mathcal{C})$ as claimed above? (with the "background geometry" given by the groupoid $X$) - -REPLY [5 votes]: All of the objects in these iterated span categories are fully dualizable. See this paper by Rune Haugseng.<|endoftext|> -TITLE: stability conditions in the sense of Kontsevich-Soibelman -QUESTION [7 upvotes]: What are the stability conditions in the sense of Kontsevich-Soibelman. -I am reading Bridgeland's stability conditions and I've heard people talking about the Kontsevich-Soibelman Stability. I would appreciate a brief introduction on this, in particular my questions are : - -What are the Kontsevich-Soibelman Stability conditions ? -How is it related to Bridgeland's Stability (or Douglas' $\pi$ - stability on D-branes) ? -Why do we need to consider Kontsevich-Soibelman stability. - -I've to admit my ignorance of the field. Please suggest some references. -Thanks. - -REPLY [7 votes]: Kontsevich-Soibelman's version is a version of Bridgeland's stability given for triangulated $A_\infty$ categories (with a few additional properties; Kontsevich and Soibelman call "non-commutative proper algebraic variety" such an $A_\infty$-category) rather than for triangulated categories as in the original Bridgeland's version. -The main point in Kontsevich-Soibelman definition is that once one thinks of the relevant $A_\infty$ categories as non-commutative analogues of algebraic varieties, one sees that a Bridgeland stability condition can be seen as the datum of a polarization on these non-commutative varieties. The reason for considering $A_\infty$ rather than ordinary categories is that categorical structures arising from branes are naturally $A_\infty$ categories (the most classical example to be done here is probably Fukaya's $A_\infty$ category of a symplectic manifold $X$). -The basic reference for Kontsevich-Soibelman's stability is obviously their paper "Stability structures, motivic Donaldson-Thomas invariants and cluster transformations", arXiv:0811.2435<|endoftext|> -TITLE: Does the etale fundamental group of the projective line minus a finite number of points over a finite field depend on the points? -QUESTION [27 upvotes]: Clearly the etale fundamental group of $\mathbb{P}^1_{\mathbb{C}} \setminus \{a_1,...,a_r\}$ doesn't depend on the $a_i$'s, because it is the profinite completion of the topological fundamental group. Does the same hold for when I replace $\mathbb{C}$ by a finite field? How about an algebraically closed field of positive characteristic? -(note that I'm talking about the full $\pi_1$ and not the prime-to-$p$ part) - -REPLY [27 votes]: It is a result of Tamagawa that for two affine curves $C_1, C_2$ over finite fields $k_1,k_2$ any continuous isomorphism $\pi_1(C_1)\rightarrow \pi_1(C_2)$ arises from an isomorphism of schemes $C_1\rightarrow C_2$. Hence, if $\pi_1( \mathbb{P}^1\setminus\{a_1,\ldots, a_r\})$ were independent of the choice of the $a_i$, then the isomorphism class of the schemes $\mathbb{P}^1\setminus\{a_1,\ldots, a_r\}$ would be independent of the choice of $a_1,\ldots,a_r$. -Tamagawa's result is Theorem 0.6 in this paper: -The Grothendieck conjecture for affine curves, A Tamagawa - Compositio Mathematica, 1997 -http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=298922 -In the case of an algebraically closed field, the answer is also that the fundamental group depends on the choice of the points that are being removed. Again by a theorem by Tamagawa: -If $k$ is the algebraic closure of $\mathbb{F}_p$, and $G$ a profinite group not isomorphic to $(\hat{\mathbb{Z}}^{(p')})^2\times \mathbb{Z}_p$, then there are only finitely many $k$-isomorphism classes of smooth curves $C$ with fundamental group $G$ (the restriction on $G$ excludes ordinary elliptic curves). -This can be found in -Finiteness of isomorphism classes of curves in positive characteristic with prescribed fundamental groups, -A Tamagawa - Journal of Algebraic Geometry, 2004<|endoftext|> -TITLE: Integral cohomology (stable) operations -QUESTION [41 upvotes]: There have been a couple questions on MO, and elsewhere, that have made me curious about integral or rational cohomology operations. I feel pretty familiar with the classical Steenrod algebra and its uses and constructions, and I am at a loss as to imagine some chain level construction of such an operation, other than by coupling mod p operations with bockstein and reduction maps. I am mostly just curious about thoughts in this direction, previous work, and possible applications. So my questions are essentially as follows: -1) Are there any "interesting" rational cohomology operations? I feel like I should be able to compute $H\mathbb{Q}^*H\mathbb{Q}$ by noticing that $H\mathbb{Q}$ is just a rational sphere and so there are no nonzero groups in the limit. Is this right? -2) Earlier someone posted a reference request about $H\mathbb{Z}^*H\mathbb{Z}$, and I am just curious about what is known, and what methods were used. -3) Is there a reasonable approach, ie explainable in this forum, for constructing chain level operations? the approaches I have seen seem to require some finite characteristic assumptions, but maybe I am misremembering things. -4) I am currently under the impression that a real hard part of the problem is integrating all the information from different primes, is this the main roadblock? or similar to what the main obstruction is? -My apologies for the barrage of questions, if people think it would be better split up, I would be happy to do so. -Thanks for your time. - -REPLY [20 votes]: A possibly interesting analogue of the formula $H\mathbb{F}_{2*} H\mathbb{F}_2 = \otimes_{i\ge1} \mathbb{F}_2[\xi_i]$ is $H\mathbb{Z}_{(2)*} H\mathbb{Z}_{(2)} = \bigotimes^\mathbb{L}_{i\ge1} \mathbb{Z}_{(2)*}[\xi_i^2]/(2\xi_i^2)$, where $\otimes^{\mathbb{L}}$ means the derived tensor product. In other words, resolve $\mathbb{Z}_{(2)*}[\xi_i^2]/(2\xi_i^2)$ by (flat or) free $\mathbb{Z}_{(2)}$-modules, tensor the resolutions together, and pass to homology. If I recall correctly, the "first" interesting class $\xi_2^3 + \xi_1^2 \xi_3$ (in degree 9) arises as a torsion product of $\xi_1^2$ and $\xi_2^2$. I needed this for a Shukla homology calculation once. Presumably there is also an odd story.<|endoftext|> -TITLE: Probability that a number and its digit reversal are relatively prime -QUESTION [7 upvotes]: What is the probability that a number and its digit reversal are relatively prime in base b? - -REPLY [3 votes]: I think there are bases $b$ where the probability becomes arbitrarily low. Let $b$ be the product of the first $n$ primes plus one then the difference of $b$ and its palindrome will be divisible by $b-1$ and for them to be relatively prime neither can be divided by the first $n$ primes. So if $n$ is chosen large enough the probability can be made arbitrarily low.<|endoftext|> -TITLE: Minimal polynomial with a given maximum in the unit interval -QUESTION [14 upvotes]: Find the lowest degree polynomial that satisfies the following constraints: -i) $F(0)=0$ -ii) $F(1)=0$ -iii)The maximum of $F$ on the interval $(0,1)$ occurs at point $c$ -iv) $F(x)$ is positive on the interval $(0,1)$ -The answer seems to depend pretty strongly on $c$. It's not difficult to find solutions for all $c$, but the solutions are not minimal. It seems like the solution involves Chebyshev polynomials, but I'm not familiar with them. Can anyone recommended a link? - -REPLY [6 votes]: By replacing $F(x)$ by $F(1-x)$ we may assume that $c\ge1/2$. -The problem is to determine a polynomial $G(x)=G_c(x)$ of minimal -possible degree, say $n$, such that $G(x)>0$ for $0 < x < 1$ and the derivative -of $F(x)=x(1-x)G(x)$ changes sign at $x=c$ and $F(x)\le F(c)$ for $0 \le x \le1$. -Clearly, $G_{1/2}(x)\equiv1$ -and with a very little work $G_c(x)=x+c(2-3c)/(2c-1)$ for $1/2 < c\le2/3$, so $n=1$. -The case $n=2$ produces -$$G(x)=x^2+a\biggl(x-\frac{c(3c-2)}{2c-1}\biggr)-\frac{c^2(4c-3)}{2c-1},$$ -so for $a \ge 0$ we have $G(x) > G(0)=c^2(3-4c)/(2c-1)$ on $x\in(0,1)$, while -the latter expression is non-negative for $2/3 < c\le3/4$. If $a <0$, then -either $G(x)$ is not positive for $0 < x < 1$ or $F(x)$ attains its maximum -at a different point of the interval $0 < x < 1$. This is however a little -bit technical to show. (For example, if we take -$a=1-3c$, then for the corresponding polynomial $G(x)$ we indeed have -$G(x) > 0$, since -$$ -G(x)\ge G\biggl(\frac{3c-1}2\biggr)=\frac{(2c+1)(c-1)^2}{4(2c-1)}. -$$ -But $x=c$ is not the maximum of $F(x)=x(1-x)G(x)$ on the interval.) -If the above pattern remains, then for $n/(n+1) < c \le (n+1)/(n+2)$ -the minimal possible degree of the polynomial $G(x)$ seems to be $n$ -(so that $\deg F=n+2$), with the corresponding choice -$$ -G_c(x)=x^n-\frac{c^n((n+2)c-(n+1))}{2c-1}. -$$ -The limiting case $c=1$ is in favor of this observation: there is -no polynomial $F(x)\not\equiv0$ of the assumed form which attains -its maximum at $x=1$. So, the expected answer to the original -question would be $\deg F=\lceil 1/\min(c,1-c)\rceil$, where -$\lceil x\rceil=n$ when $n-1 < x \le n$. -Edit. With the above choice of $G_c(x)$, $F'(x)=0$ on the interval -$0 < x < 1$ only at $x=c$. Therefore, this choice results in the estimate -$\deg F \le \lceil 1/\min(c,1-c)\rceil$, which is sharp at least for $1/3 \le c \le 2/3$. -Edit 2. Bill's answer gives pretty much evidence for the fact that $1/2 < c < (1+\cos(\pi/n))/2$ gives the estimate $\deg F\le n$ for $n$ even. More remarkably, this is indeed related to the Chebyshev polynomials. The most unpleasant thing is a necessary amount of technical work to be done (but Bill's answer contains all details for such calculations).<|endoftext|> -TITLE: Good ways to engage in mathematics outreach? -QUESTION [66 upvotes]: Greetings all, I have often heard that it would be good if we as a community did more in the way of mathematics outreach: more to explain what it is we do to the community at large, more to expose children and adults to all the fun we are having. -What are some particular ways that people have found to do this? A couple come to mind. There are Math Circles in a number of cities, which welcome volunteers, and these are a fantastic resource. There are also clubs built around preparing middle and high school students for competitions, which are wonderful for many people. -What are some other ways people have found to reach out to the community at large? I would be particularly interested to hear ideas that "anyone can do" -- that don't require special knowledge or connections, or too much commitment (at least at first), and that I could try when I begin my first permanent job. I would also be very interested in any outreach that targeted adults, including faculty outside the math department. -Thank you! - -REPLY [7 votes]: In the UK there are a lot of good programmes for engaging the public with mathematics, some of which have been touched on in other answers. Here is a list of the ones I think have worked the best or been most innovative: -1) Royal Institution Masterclasses: a UK-wide series of masterclasses aimed at age 13 school pupils. Classes are on Saturday mornings for between 5 and 10 weeks. Each class has a different presenter and each class lasts for 2.5 hours with time for lectures, exercises and refreshments. I've been involved in giving the Edinburgh and Glasgow masterclasses and they are a lot of fun for all involved. The kids really appreciate getting to see some maths that they'd never encounter through school, and especially enjoy seeing the diversity and usefulness of the subject. -(In addition, a set of masterclasses gets broadcast on the BBC every Christmas!) -2) FunMaths Roadshow: a collection of 350 maths activities, each suitable for a particular age group between 5 and 20. The resources are cheap and easy to make, as well as being easy to transport, so that they can be taken all around the local area and used with different groups of people. -3) Maths busking: a new phenomenon which has been surprisingly successful! 'Buskers' go out onto the streets and perform games and tricks which appear to be magic but which can be explained by mathematics. It's a great way to engage with people who would otherwise never attend a maths lecture or think of going to a science museum. -4) Mathematical walking tours: do a tour that opens people's eyes to the mathematics around your city. This can be historical curiosities, geometrical constructions in the architecture or optimization in the town planning. Such tours are currently being designed for Oxford and London, or get examples from Manhattan or this maths tourism blog. -Whatever you decide to do, just go ahead and do it with plenty of enthusiasm! Good luck!<|endoftext|> -TITLE: Approximating polynomials in R[x] using integer-valued polynomials -QUESTION [10 upvotes]: An integer-valued polynomial is a polynomial with real coefficients mapping integers to integers. It is well known that all such polynomials $h(x)$ are generated as an additive group by the binomial coefficients $\binom{x}{n}$. My question concerns the problem of approximating an arbitrary polynomial $f$ with real coefficients on a skillfully chosen interval $I$ of length 1 by means of a skillfully chosen non-zero integer-valued polynomial $h$. Specifically, for $f\in\mathbb{R}[x]$ let $$N(f)=\inf_{I,h}\,\,\,\, ||f-h||_{I},$$ where $||\cdot ||_I$ means sup norm, $h$ runs through all non-zero integer-valued polynomials, and $I$ runs through all intervals of length 1. I am looking for a way to compute $N(f)$, and I have a conjecture (a guess really) that makes this computation very simple. -$\textbf{Conjecture:}$ Let $D(x)$ be the distance from the real number $x$ to the nearest integer. Then $$N(f)=\inf_{x\in\mathbb{Z}} \,\,\, D(f(x)).$$ -My question is: Can anyone provide a proof or counter-example or some helpful references? -In particular, as a simple test-case, can anyone compute $N(\frac{1}{2}x)$? Maybe the problem is impossibly difficult or ridiculously easy for some reason I don't see, and someone can put me out of my misery. -$\textbf{Remarks:}$ - -It is clear that $N(f)\ge\inf_{x\in\mathbb{Z}}D(f(x))$, because the closure of every interval of length 1 contains an integer. -I suspect that the conjecture is false if one defines $N$ so that $h$ ranges over polynomials with INTEGER coefficients, but I don't have an example to prove this. -It is easy to check that $N(0)$=0, and in general $N(c)=D(c)$ for any constant $c$, using the fact that the polynomial $\binom{x}{n}$ tends to 0 uniformly on $[0,1]$ as $n$ tends to infinity. -I'm already stuck on the computation of $N(\frac{1}{2}x)$, which is 0 according to the conjecture. One would naturally consider intervals $I$ of the form $[2n-\frac{1}{2},2n+\frac{1}{2}]$, and look at polynomials of the form $h(x):=\sum_{k}c_{k}\binom{x}{k}$, such that all the $c_{k}$ are integers and $h(2n)=n$. - -REPLY [2 votes]: Drawing some consequences from Franz Lemmermeyer's comment above, i.e. that the integer-valued polynomials $f_n:=\frac 1 2(x+x^n)$ converge to $\frac x 2$ uniformly on closed sub-intervals of $]-1,1[$. Then, in the same convergence, we also have that the $k$-fold iteration $f_n\circ\dots\circ f_n$ converges to $x/2^k$ as $n\to\infty$. But then we can reach this way any dyadic rational multiple of $x$; any real multiple of $x$; any polynomial with $0$ constant term; and in turn any continuous function on $]-1,1[$ with an integer value at $0$ is uniform limit on compact sets of a sequence of integer valued polynomials (whence in particular the conjecture follows).<|endoftext|> -TITLE: Is there a "purely algebraic" proof of the finiteness of the class number? -QUESTION [64 upvotes]: The background is as follows: I have been whittling away at my commutative algebra notes (or, rather at commutative algebra itself, I suppose) recently for the occasion of a course I will be teaching soon. -I just inserted the statement of the theorem that the ring $\overline{\mathbb{Z}}$ of all algebraic integers is a Bezout domain (i.e., all finitely generated ideals are principal; note that this ring is very far from being Noetherian). I actually don't remember off the top of my head who first proved this result -- and I would be happy to learn, although that's not my main question here. The reference that sticks in my mind is Kaplansky's 1970 text Commutative Rings, where he proves the following nice generalization: -Theorem: Let $R$ be a Dedekind domain with fraction field $K$ and algebraic closure $\overline{K}$. Suppose that for every finite subextension $L$ of $\overline{K}/K$, the ideal class group of the integral closure $R_L$ of $R$ in $L$ is a torsion abelian group. Then the integral closure $S$ of $R$ in $\overline{K}$ is a Bezout domain. -Neat, huh? Then when I got to deducing the result about $\overline{\mathbb{Z}}$, it hit me: nowhere in these notes do I verify that $\mathbb{Z}$ satisfies the hypotheses of Kaplansky's theorem, namely that the ideal class group of the ring of integers of a number field is always finite. -Don't get me wrong: I wasn't expecting anything different -- I actually don't know of any commutative algebra text which proves this result. Indeed, it is generally held that the finiteness of the class number is one of the first results of algebraic number theory which is truly number-theoretic in nature and not part of the general study of commutative rings. But the truth is that I've been bristling at this state of affairs for some time: I would really like there to be a subbranch of mathematics called "abstract algebraic number theory" which proves "general" results like this. (My reasons for this are, so far as I can recall at the moment, purely psychological and aesthetic: I have no specific ulterior motive here, alas.) To see past evidence of me flirting with these issues, see this previous MO question (which does not have an accepted answer) and these other notes of mine (which don't actually get off the ground and establish anything exciting). -So let me try once again: - -Is there a purely algebraic proof of the finiteness of the class number? - -Unfortunately I don't know exactly what I mean here, because the standard proofs that one finds in algebraic number theory texts are certainly "purely algebraic" in nature or can be made so. (For instance, it is well known that it is convenient but not necessary to use geometry of numbers -- the original proofs of this finiteness result predate Minkowski's work.) Here are some criteria: -1) I want a general -- or "structural" -- condition on a Dedekind domain that implies the finiteness of its class group. (In my previous question, I asked whether finiteness of the residue rings was such a condition. I still don't know the answer to that.) -2) This condition should in particular apply to rings of integers of number fields and also to coordinate rings of regular, integral affine curves over finite fields. -Note that already the standard "purely algebraic proofs" of finiteness of class number in the number field case do not in fact proceed by a general method which also works verbatim in the function field case: additional arguments are usually required. (See for instance Dino Lorenzini's Invitation to Arithmetic Geometry.) As far as number field / function field unity goes, the best approach I know is the adelic one: Fujisaki's Lemma, which is Theorem 1.1 here (see also the theorem on the last page). But this is a topological argument, and the topological and valuation theoretic properties of global fields which go into it are quite particular to global fields: I am (dimly) aware of results of Artin-Whaples which characterize global fields as the ones which have these nice properties: the product formula, and so forth. -It is possible that what I am seeking simply doesn't exist. If you feel like you understand why the finiteness of class number is in some precise way arithmetic rather than algebraic in nature, please do explain it to me! - -Added: here are some further musings which might possibly be relevant. -I like to think of three basic theorems of algebraic number theory as being of a kind ("the three finiteness theorems"): -(i) $\mathbb{Z}_K$ is a Dedekind domain which is finitely generated as a $\mathbb{Z}$-module. -(ii) $\operatorname{Pic} \mathbb{Z}_K$ is finite. -(iii) $\mathbb{Z}_K^{\times}$ is finitely generated as a $\mathbb{Z}$-module. -[Yes, there is also a fourth finiteness theorem due to Hermite, on restricted ramification, which is perhaps most important of all...] -The first of these is acceptably "purely algebraic" to me: it is a result about taking the normalization of a Dedekind domain in a finite field extension. The merit of the adelic approach is that it shows that (ii) and (iii) are closely interrelated: the conjunction of the two of them is formally equivalent to the compactness of the norm one idele class group. So perhaps it is a mistake to fixate on conditions only ensuring the finiteness of the class group. For instance, the class of Dedekind domains with finite class group is closed under localization but the class of Dedekind domains which also have finitely generated unit group is not. However the "Hasse domains" -- i.e. $S$-integer rings of global fields -- do have both of these properties. - -REPLY [38 votes]: Yes, there exist purely algebraic conditions on a Dedekind domain which hold for all rings of integers in global fields and which imply that the class group is finite. -For a finite quotient domain $A$ (i.e., all non-trivial quotients are finite rings), a non-zero ideal $I\subseteq A$ and a non-zero $x\in A$, let $N_{A}(I)=|A/I|$ and $N_{A}(x)=|A/xA|$. Also define $N_{A}(0)=0$. -Call a principal ideal domain $A$ a basic PID if the following conditions are satisfied: - -$A$ is a finite quotient domain, - -for each $m\in\mathbb{N}$,$$\#\{x\in A\mid N_{A}(x)\leq m\}>m$$ -(i.e., $A$ has “enough elements of small norm”), - -there exists a constant $C\in\mathbb{N}$ such that for all $x,y\in A$, -$$N_{A}(x+y)\leq C\cdot(N_{A}(x)+N_{A}(y))$$ -(i.e., $N_{A}$ satisfies the “quasi-triangle inequality”). - - -Theorem. Let $A$ be a basic PID and let $B$ be a Dedekind domain which is finitely generated and free as an $A$-module. Then $B$ has finite ideal class group. -For the proof, see here. -It is easy to verify that $\mathbb{Z}$ and $\mathbb{F}_q[t]$ are basic PIDs, so the ring of integers in any global field satisfies the hypotheses of the above theorem (using the non-trivial fact that rings of integers in global fields are finitely generated over one of these PIDs). -More generally, one can take the class of overrings of Dedekind domains which are finitely generated and free over some basic PIDs. Since it is known that an overring of a Dedekind domain with finite class group also has finite class group, this gives a wider class of algebraically defined Dedekind domains (including $S$-integers like $\mathbb{Z}[\frac{1}{p}]$) with finite class group. -Added: The second condition for basic PIDs can be relaxed to: there exists a constant $c\in\mathbb{N}$ such that for each $m\in\mathbb{N}$, -$$ -\#\{x\in A\mid N_{A}(x)\leq c\cdot m\}\geq m. -$$<|endoftext|> -TITLE: Are there symplectic 4-folds with $b_+>1$, $b_-=0$? -QUESTION [8 upvotes]: This is the question. Is it known that a symplectic $4$-fold with $b_2>1$ should have a homology class $C$ with $C^2<0$? - -REPLY [6 votes]: Symplectic geography in 4 dimensions can be mapped using Chern number coordinates $(c_1^2,c_2)$. The part of the plane where $c_1^2 > 4c_2$ is uncharted. It's unknown whether there are any symplectic 4-manifolds in this region, besides blow-ups of ruled surfaces, though by the Bogomolov-Miyaoka-Yau inequality and the Kodaira-Enriques classification, there are no complex surfaces. -I can't answer the question but I'll point out that a symplectic 4-manifold with $b_-=0$ and $b_+>1$ necessarily lies in this unknown region - in particular, it's not Kaehler. -To see this, rewrite $c_1^2-4c_2$ in terms of Euler characteristic $\chi$ and signature $\sigma$ as $(2\chi+3\sigma)-4\chi$. For a symplectic manifold with $b_-=0$, this quantity equals $4b_1+b_2-4$ and is positive unless $b_1=0$ and $b_2= 1$ or $3$; I use the parity argument mentioned in Paul's comment. If $b_1=0$ and $b_2=3$ then the intersection form on $H_2/tors.$ is $\mathbb{Z}^3$, the unique rank 3 positive-definite unimodular lattice. So $c_1^2=15$ is the sum of three squares; but it's not.<|endoftext|> -TITLE: Optimally directing switches for a random walk -QUESTION [6 upvotes]: If you are sometimes called upon directing a random walk in a directed graph, how should you direct it so as to maximize the probability it goes where you want? -Formal statement -More specifically, suppose you are given a directed graph $G$ with edge weights, two designated vertices $s$ and $t$, and a subset of the vertices $S$. The edges weights represent the transition probabilities of the random walk, the vertex $s$ the start, the vertex $t$ the target, and the set $S$ the set of switches. You are guaranteed that the weights on the out-edges of any node are non-negative and sum to one, that $t$ is absorbing (i.e., $t$ has one out-edge directed towards itself), and that the out-degree of any vertex in $S$ is exactly two. -A random walk is taken on $G$, starting at $s$. For any given vertex not in $S$, the weight on an out-edge is the probability that the walk will travel in that direction. Every time that the walk reaches a switch (a vertex in $S$), you are allowed to choose which of the two edges the walk will travel along (and you are allowed probabilistic strategies). How should you direct the path if you want to maximize the probability that the walk ends up at your target $t$? -Questions -I am most interested in this as an algorithmic question. How fast can you find the optimal strategy with respect to the size of the graph? My specific application has about 100 switches among 200 vertices in a fairly sparse graph (say out-degree bounded above by 6). -But we can also ask purely mathematical questions. For example, my intuition says (and I can hand-wave a proof) that there exists an optimal strategy that is deterministic in the sense that it always chooses the same direction for a given switch and this direction does not depend on the initial vertex $s$. Is this actually true? -Also, is there a sense in which the optimal strategy needs to "coordinate" among the switches? That is, is there a local optimum that is not a global optimum? -Notes -A note on connectivity: we may assume that the graph is sufficiently connected. If not, we can identify all vertices that cannot be reached from the start node, as well as all of those that cannot reach the target node, into a single, absorbing fail state. We may assume the start node is not the fail node. - -REPLY [2 votes]: This is the simple stochastic games problem, but for only one player, and there is a polynomial-time algorithm for it based on linear programming, which is described in Anne Condon's paper "On Algorithms for Simple Stochastic Games." Look in this paper for the linear programming algorithm for SSG's with no min vertices. In one of her papers on simple stochastic games, Condon does indeed prove that the setting of the switches is independent of the start node, and that in the optimal strategy, the switch settings never need to change.<|endoftext|> -TITLE: Comparing fundamental groups of a complex orbifolds and their resolutions. -QUESTION [17 upvotes]: Let $X$ be a complex manifold with quotient singularities, and let $\tilde X$ be its resolution (that exists, for example, by Hironaka). Then I am pretty sure that $\pi_1(X)\cong \pi_1(\tilde X)$. - -Question. Is there a reference for such a statement? At least in dimension 3? - -The reason why this should be true is that it should be possible to find such a resolution of $X$ that the preimage of each point in $\tilde X$ is simply connected ($\mathbb CP^n/G$ is simply-connected for finite $G$). Then everything follows from a standard topological lemma. -I guess this statement should be true as well if $X$ is a complex analytic variety with arbitrary Kawamata log terminal singularities (because Fano manifolds are simply connected). I would be grateful for a reference for any kind of such statement. -ADDED. As Francesco says in his answer, this statement is classical for surfaces. I would like to have a reference what would cover at least the case of $3$-folds with non-isolated quotient singularities (say with abelian stabilisers). -UPD. The question is now 100% settled by the reference provided by the answer of Benoit. - -REPLY [5 votes]: Dear Dmitri, -the result you hope is also true (ie when X has only klt singularities, its fundamental group is isomorphic to the one of any of its desingularization). This theorem is due to Takayama (Local simple connectedness of resolution of log-terminal singularities, International journal of Math 2003). -Bests, Benoît<|endoftext|> -TITLE: Is the ring of all cyclotomic integers a Bezout domain? -QUESTION [17 upvotes]: My previous question about the theorem (apparently due to Dedekind -- thanks, Arturo Magidin!) that the ring $\overline{\mathbb{Z}}$ of all algebraic integers is a Bezout domain got me thinking about rings of integers in infinite algebraic extensions of $\mathbb{Q}$. -As I mentioned here, I was thinking about Kaplansky's formulation/generalization of -Dedekind's theorem: if $R$ is a Dedekind domain with fraction field $K$ such that for all -finite degree field extensions $L/K$ the ideal class group of the integral closure $R_L$ -of $R$ in $L$ is a torsion abelian group, then the integral closure $S$ of $R$ in $\overline{K}$ is a Bezout domain. -Since we know (by nefarious arithmetic arguments...) that the Picard group of the ring of integers of a number field is finite, Kaplansky's theorem applies with $R = \mathbb{Z}$ and gives Dedekind's theorem. -Suppose we now look at Kaplansky's proof and see if it gives more: do we need to go all the way up to $\overline{\mathbb{Q}}$ in order to show that the ring of integers is a Bezout domain? -Indeed not. The key property of the ring $S$ used in the proof is that it is root closed, i.e., closed under extraction of $n$th roots, for all $n \in \mathbb{Z}^+$. There are lots of smaller -- but still infinite -- algebraic extensions of $\mathbb{Q}$ which have this property: the smallest is $\mathbb{Q}^{\operatorname{solv}}$, the maximal solvable extension of $\mathbb{Q}$. - -But what about the ring of integers of $\mathbb{Q}^{\operatorname{ab}}$, the maximal abelian extension of $\mathbb{Q}$? Is this a Bezout domain? If not, what is its Picard group? - -In general, if $R$ is a Dedekind domain with fraction field $K$, $L/K$ is any algebraic field extension and $S$ is the integral closure of $R$ in $L$, then if $[L:K]$ is infinite -$S$ need not be a Dedekind domain but is always a Prufer domain: every finitely generated ideal is invertible. In particular $S$ is a Bezout domain iff $\operatorname{Pic}(S) = 0$. So one may ask: - -What is known about the Picard groups of rings of integers of infinite algebraic extensions of $\mathbb{Q}$? What, if anything, is the connection to more classical algebraic number theory? - -REPLY [5 votes]: Let $K$ be a cyclotomic extension, and $a, b \in O_K$. If the ideal $A = (a,b)$ becomes principal in some cyclotomic extension $L/K$, then clearly you can solve the Bezout problem, -and the converse holds, too: if the Bezout problem is solvable in some cyclotomic extension, then the ideal class of $A$ must capitulate there. Thus the question you are asking is whether the class group of a cyclotomic field capitulates in a cyclotomic extension. For certain pieces of the class group this is believed to be true (Greenberg), but in general I don't know the answer. Georges Gras and Bosca & Jaulent have studied the phenomenon of "abelian capitulation". From a first glance at their article it seems that the answer to your question is yes for the maximal real cyclotomic subfield of all cyclotomic extensions.<|endoftext|> -TITLE: Quadrics containing many points in special position -QUESTION [5 upvotes]: Suppose $n$ quadric hypersurfaces cut - out $2^n$ distinct points - $p_1,\ldots,p_{2^n}$ in - $\mathbb{P}^n$. What is the maximal - number of points $p_i$ a quadric can - contain without containing all of - them? - -For n=2, there is of course a quadric going through any three points and avoiding the last point. -In $\mathbb{P}^3$, it is easy to exhibit a quadric going through 6 points (and avoiding the last two). This is however not possible with 7 points (as can be seen using a projection to $\mathbb{P}^2$). - -REPLY [4 votes]: To complement the answers of Sasha and Damiano I would like to give one (completley non-generic) example of a very specific configuration when a quadric contains exactly -$3\cdot 2^{n-2}$ points. -Take in $\mathbb C^n$ the collection of $2^n$ points $(\pm 1,..., \pm 1)$. It is clear that this collection is the intersection of n quadrics. Indeed, we can obtain these points by taking intersection of $n$ degenerate quadrics $Q_i:=(x_i+1)(x_i-1)=0$, then we perturb a bit each quadric by adding a little multiple of $(\sum_i x_i^2-n)$. This way we get $n$ smooth quadrics that intersect transversally in the above $2^n$ points. -Finally take one more singular quadric $(x_1-1)(x_2-1)=0$ it contains exactly $3\cdot 2^{n-2}$ points.<|endoftext|> -TITLE: is quasicompact part of the definition of etale in EGA -QUESTION [6 upvotes]: Is the projection map $\amalg_{i \in \mathbb{Z}} X \to X$ from an infinite disjoint union of copies of a scheme $X$ to $X$ an etale map, using the definition in EGA IV 17? -According to EGA, an etale map of schemes $f:X \to Y$ is locally of finite presentation and formally etale. Hence I don't think etale maps need to be quasicompact or quasifinite. However, wikipedia claims in property #5 under "Etale morphism" that etale maps are quasifinite, so either I'm missing something or wikipedia is using a different definition. I'm sorry to ask such a dumb question but this has come up repeatedly when I'm trying to read Champs Algebriques. - -REPLY [7 votes]: No, quasi-compact is not part of the definition of etale. Yes, your map from the infinite disjoint union to $X$ is etale.<|endoftext|> -TITLE: What are "classical groups"? -QUESTION [66 upvotes]: Unlike many other terms in mathematics which have a universally understood meaning (for instance, "group"), the term classical group seems to have a fuzzier definition. Apparently it originates with Weyl's book The Classical Groups but doesn't make it into the index there. It was propagated by Dieudonne and others. But I'm never sure exactly what groups are included/excluded by this label. Weyl himself seems to have been interested in general (and perhaps special) linear groups, together with orthogonal (or special orthogonal) and symplectic groups attached to bilinear/quadratic forms. Initially questions were raised mainly in characteristic 0, usually over $\mathbb{C}$ but sometimes other fields as well. -Obviously it helps mathematical communication to have words and symbols which need no further explanation. But ambiguity tends to creep in. For example, what does one mean by "natural numbers" or the symbol $\mathbb{N}$? (Is 0 a natural number or not?) What is a "ring"? (Does it have an identity element or -not?) By now a number of book titles and thousands of research papers refer to -classical groups. But which groups are included? Spin or half-spin groups? Projective versions of the linear groups mentioned above? - -Is there any precise definition of classical groups? - -ADDED: The answers and comments have been enlightening, though like some other people I lean more toward a "no" answer to my basic question. The underlying concern on my part is whether the notion of "classical group" has become too vague to be useful, which I sometimes suspect is the case with newer umbrella terms like "quantum group". It seems that the only safe usage nowadays is "classical groups, by which I mean one of those in the following list ....", at which point the original label has lost most of its purpose. -However ... the careful treatment by Porteous (which I wasn't familiar with) strikes me as well focused even if it omits some groups of interest. Weyl himself wanted a direct and concrete approach to representations and invariants of certain specific matrix groups, mainly over $\mathbb{C}$. That's clearly much too narrow for later purposes, where the geometry of various kinds of forms over various kinds of rings gets more attention, along with internal group structure. But some of the geometric viewpoints might suggest paying more attention to PGL than to GL, contrary to the matrix group emphasis in most other work. -In any case, while the Killing-Cartan classification for Lie algebras still makes it natural to view A-D types as "classical" and the rest as "exceptional", -I'm reluctant to go too far in fitting classical groups into the framework of semisimple Lie or algebraic groups based heavily on differential or algebraic geometry. That framework already has to be stretched to admit general linear groups or rings coming from number theory. And spin or half-spin or adjoint groups, however natural in Lie theory, probably don't fit so well into the familiar world of matrix groups. -One viewpoint I resist is the attempted definition given by Popov in the Springer encyclopedia. This doesn't really cover the ground consistently or comprehensively, besides which the short reference list is totally unbalanced. -P.S. The views expressed in the various answers and comments are mostly quite reasonable, but leave me with the sense that everyday usage won't tend to converge. Maybe I should sum up my lingering uncertainty about the value of the term "classical group" by quoting one of Emil Artin's 1955 papers on the orders of finite simple groups: The notion of classical groups is taken in such a wide sense as to embrace all finite simple groups which are known up to now. - -REPLY [3 votes]: Just a piece of information from the french side. Dieudonn\'e, in ``La g\'om\'etrie des groupes classiques'' (Springer, 1970), takes the definition of a classical group for granted. But browsing through the table of contents, it's clear he means $GL_n(K),SL_n(K),O_n(K,f),U_n(K,f),Sp_{2n}(K)$ plus variants (e.g. the projectivized versions). -In the book ``Groupes de Lie classiques''(Hermann, 1986), R. Mneimn\'e and F. Testard define classical Lie groups in their introduction: same list as in Dieudonn\'e, but assuming of course $K=\mathbb{R}$ or $\mathbb{C}$.<|endoftext|> -TITLE: Happy New Prime Year! -QUESTION [38 upvotes]: It happens that next year 2011 is prime, while outgoing 2010 is -highly composite in the sense that the number of its distinct prime factors -is 4, maximal possible for a year $< 2310$. -Let me denote by $s(n)$ the number of distinct prime factors of $n$ -and note that $s(2011)=1$, $s(2012)=2$ and $s(2013)=3$. I wonder -whether there is a rigorous argument or some heuristic considerations -to show that, for each $k\ge1$, there exist (infinitely many integers) -$n$ satisfying $s(n+1)=1$, $s(n+2)=2$, $\dots$, $s(n+k)=k$. -This can be thought as a generalization of the infiniteness of primes -($k=1$), but I ask this question for curiosity only. -Happy New Prime Year 2011! (Please do not count the exclamation mark as factorial.) - -REPLY [4 votes]: Here is another pattern I learned from Bharath Kumar Annamaneni in his buzz post. -2011= 157 + 163 + 167 + 173 + 179 + 181 + 191 + 193 + 197 + 199 + 211 . -2011, Being A Prime Number Itself, Is Also A Sum Of 11 Consecutive Prime Numbers . -Wow.<|endoftext|> -TITLE: Constructive Bezout cofactors in the ring of algebraic integers -QUESTION [6 upvotes]: We see the following in an answer to This Question : (mangled by me for my purposes, all errors my fault) - -Dedekind stated (in 1871) that the ring of algebraic integers is a Bezout Domain. He calls it an important but difficult theorem, and only presents it as an easy corollary after he has developed all the theory, including (if memory serves) finiteness of the class number. - -That is: Given any algebraic integers $a$ and $b$, there is a common divisor $d$ which is a linear combination of $a$ and $b$. Even more specifically, there exist algebraic integers $x,y,d,a'$ and $b'$ such that $ax+by=d$ $a=da'$ and $b=db'$. (So the common divisor $d$ is a greatest common divisor of $a$ and $b$ in that any common divisor of $a$ and $b$ also divides $d$.) -My question is: How constructively can this be established? - -Is there a satisfying algorithm which given algebraic integers $a$ and $b$ produces algebraic integers $x$ and $y$ such that (the algebraic integer) $d=ax+by$ is a common divisor of $a$ and $b$? - -I hasten to add that of course I know that what works in Euclidean Domains won't work here. The question above is the one I really want the answer to, but I'll throw in the following (which might be very easy) - -Is the answer yes at least for the special case that $a$ is a rational integer and $b$ is a root of a monic quadratic equation? -Can a bound be given for the degree of the minimal polynomials of $x$ and $y$ (and/or $d$) in terms of those for $a$ and $b$? (That is, can the statement be strengthened to something like: "there exist $x,y,d$ all with degree no larger than $\deg(a)\deg(b)$ such that ...") - -I remember that I once found a good $x$ and $y$ for $a=2$ and $b=1+\sqrt{-5}$ and that it involved cube roots of unity (if I recall correctly). I'm sure that that example is well known, but I did it by a very hit and miss process and can't recreate the answer on the spot. - -REPLY [5 votes]: This is completely constructive. Given elements $a, b \in O_K$, the ring of integers in some number field, let $A = (a,b)$ denote the ideal generated by these elements. Compute -the class number $h$ of $K$ (or, if you want, the order you are working in), compute -a generator $c$ with $A^h = (c)$, and set $L = K(\gamma)$ with $\gamma = \sqrt[h]{c}$. Then $A$ becomes principal in $L$, and computing the coefficients $ax + by = \gamma$ is linear algebra -(this should be in Cohen's books).<|endoftext|> -TITLE: "C choose k" where C is topological space -QUESTION [12 upvotes]: One day I read a generating function in a paper. For any "sufficietly nice topological space", $C$: -$$ \sum_{l \geq 0 } q^{2l}\chi(\mathrm{Sym}^l[C]) = (1 - q^2)^{-\chi(C)} = \sum_{l \geq 0} \binom{l + \chi(C)-1}{l} q^{2l} $$ -First of all, I'm not sure what "sufficiently nice" means here. I'm guessing any CW complex will do. I wonder what's an example of a space where this formula doesn't work. - -This formula suggests -$$ \chi(\mathrm{Sym}^l[C]) = \binom{l -1+ \chi(C)}{l} = \chi \binom{l-1 + C }{l} $$ -where the right hand side is some "categorification". Removing the $\chi$'s, is there some sense in which -$$ \mathrm{Sym}^l[C] = \binom{l -1+ C }{l} $$ -is rigorous? - -REPLY [10 votes]: I presume you are asking is whether one can make sense of $\binom{X}{k}$ as a space in such a way that it relates to the formula you are asking about. -The answer is yes. -First let $l = 1$. For a space $X$ define $\binom{X}{k}$ to be the configuration space of subsets of X -having cardinality $k$. Then $\binom{X}{1} = X$. In this case $\text{Symm}^1(X) = X$ -and we have agreement -$$ -\binom{X}{1} = X = \text{Symm}^1(X) . -$$ -Now consider the case $l=2$, and let {1} be the one element set -Then as sets there is an evident bijection -$$ -\binom{X \amalg \text{{1}} }{2} = \binom{X}{2} \amalg (X\times \text{{1}}) -$$ -As a set, the right side is the same thing as $\text{Symm}^2(X) = X\times_{Z_2} X =$ the orbits of the cyclic group of order two acting on $X\times X$ by permutation. To see this, note that $X\times X$ has two kinds of isotropy: one coming from -the diagonal copy of $X$ (with trivial action) and the other being it's complement -which is $X\times X - X$ with free action having quotient $\binom{X}{2}$. -With respect to this identification, your formula makes sense. However, these spaces -have different topology when $X$ isn't discrete. -A similar observation works in the $l >2$ case.<|endoftext|> -TITLE: The closures in $C^0(\mathbb{R}, \mathbb{R})$ of the set of integer valued polynomials, resp, of polynomials with integer coefficients -QUESTION [10 upvotes]: This is a follow up of an interesting recent question on the topic. The answer given there by fedia shows that the matter is rich and complicated, and I can't resist to submit here a further question. - -Q1. What is the closure of the integer valued polynomials in $C^0(\mathbb{R},\mathbb{R})$ - w.r.to the uniform convergence on bounded sets? - -For instance, a nontrivial element in the closure is, by the Euler infinite product -$$\frac {\sin(\pi x)} \pi x = \lim_{n \to \infty}\, \binom {x-1} n \binom {x+n} n $$ -but can one reach all integers-to-integers continuous functions on $\mathbb{R}$? -(After a quick Google search, among the large literature on integer valued polynomials, I was only able to find this book concerning approximation by IVP, that however treats approximation of functions on the p-adic integers rather than on $\mathbb{R}$; I'm not p-adic enough to understand if there are implications to the real case). -[edit] A further question raised by SJR - -Q2. What is the closure of the polynomials with integer coefficients in $C^0(\mathbb{R},\mathbb{R})$ - w.r.to the uniform convergence on bounded sets? - -At a first glance I'd have said $\mathbb{Z}[x]$ is closed in $C^0(\mathbb{R}, \mathbb{R})$ for some elementary reason, but now I don't see it. It is certainly a countable, metrizable commutative group; so it is closed iff it is discrete. It has an additional ring structure and it's a monoid by composition, both structure being compatible with the topology. - -REPLY [4 votes]: I'd say that the closure of the integer-valued polynomials (IVP) by uniform convergence on bounded sets is the whole set $C^0\big((\mathbb{Z},\mathbb{R}),(\mathbb{Z},\mathbb{R})\big)$ of continuous functions mapping integers to integers. -Let $r\le s\in\ \mathbb{N}$. Then $\big|\binom x s \big| < 1 $ holds for all $x \in ]0,r[$. Consider the IVP $f_n(x):=\frac 1 2 (x + x^n)$, that maps $]-1,1[$ into itself, and converges to $\frac x 2 $ uniformly on compact sets of $]-1,1[$. So th $k$-fold iterate $f_n^k:=f_n\circ\dots\circ f_n$ converges to $\frac x {2^k}$, and for all $p\in\mathbb{Z}$ the sequence of IVP $pf^k _n\big(\binom x s \big)$ converges to $\frac p {2^k} \binom x s $ as $n\to \infty$ on $]0,r[$. By density of the dyadic rationals, and by linearity, we get at least any polynomial multiple of $\binom x r$ as a uniform limit of IVP on compact sets of $]0,r[$, and of course, any polynomial taking integer values on $x=0,1\dots, r$ as well. By density, we also get any function $f\in C^0(]0,r[)$ that takes integer values on the integer points; in particular the restriction to $]-1,r[$ of any $f\in C^0\big((\mathbb{Z},\mathbb{R}),(\mathbb{Z},\mathbb{R})\big)$. By translation invariance, and since $r$ is arbitrary, this is true for any interval in place of $]0,r[$, and we conclude by a diagonal argument.<|endoftext|> -TITLE: unboundedness of number of integral points on elliptic curves? -QUESTION [22 upvotes]: If $E/\mathbf{Q}$ is an elliptic curve and we put it into minimal Weierstrass form, we can count how many integral points it has. A theorem of Siegel tells us that this number $n(E)$ is finite, and there are even effective versions of this result. If I'm not mistaken this number $n(E)$ is going to be a well-defined invariant of $E/\mathbf{Q}$ (because different minimal Weierstrass models will have the same number of integral points). -Is it known, or conjectured, that $n(E)$ is unbounded as $E$ ranges over all elliptic curves? -Note: the question is trivial if one does not put $E$ into some sort of minimal form first: e.g. take any elliptic curve of rank 1 and then keep rescaling $X$ and $Y$ to make more and more rational points integral. - -REPLY [26 votes]: I proved that if $E/\mathbf{Q}$ is given using by a minimal Weiestrass equation, then -$ \#E(Z) \le C^{\text{rank} E(Q) + n(j) + 1} $ -where $n(j)$ is the number of distinct primes dividing the denominator of the $j$-invariant of $E$ and $C$ is an absolute constant. This is in J. Reine Angew. Math. 378 (1987), 60-100. -Mark Hindry and I proved that if you assume the abc conjecture, then you can remove the n(j) in the above estimate. This is in Invent. Math. 93 (1988), 419-450. It is a conjecture due to Lang. -The papers contain more general results for (quasi)-S-integral points over number fields.<|endoftext|> -TITLE: A curious construction of a chain complex and its homology -QUESTION [6 upvotes]: ... curious to me, that is. -Suppose two module filtrations $$ \cdots < A_3 < A_2 < A_1 < \cdots $$ and $$ \cdots < B_3 < B_2 < B_1 < \cdots $$ are comparable in the sense that for all $j$, $ B_{j+1} < A_{j} < B_{j-1} $; then there are natural complexes -$$ \cdots \to \frac{A_3}{B_4} \to \frac{A_2}{B_3} \to \frac{A_1}{B_2} \to \cdots $$ -and -$$ \cdots \to \frac{B_3}{A_4} \to \frac{B_2}{A_3} \to \frac{B_1}{A_2} \to \cdots $$ -both of which have homology groups -$$ \frac{A_i\cap B_i}{A_{i+1} + B_{i+1}} .$$ -My question is in two parts: - -this canonical isomorphism $H(A^+/B)\simeq H(B^+/A)$, has it got a name? -is it useful? - -REPLY [7 votes]: I don't know that this qualifies as an answer, but I wanted to point out that your isomorphism arises as the boundary map in a short exact sequence of chain complexes. Note that -\[ -\cdots\to \frac{B_2}{B_4} \to \frac{B_1}{B_3}\to\frac{B_0}{B_2}\to\cdots -\] -is a long exact sequence. When viewed as a chain complex, it fits together with your other two chain complexes (with one of them shifted) into a short exact sequence of chain complexes with columns looking like -\[ -0\to \frac{A_2}{B_3}\to\frac{B_1}{B_3}\to\frac{B_1}{A_2}\to 0. -\] -The induced long exact sequence in homology has one of every three terms zero because the added chain complex is exact. Thus the boundary maps are isomorphisms (I haven't checked that they are the isomorphisms you mention, but it would be somewhat odd if they weren't...). As for question 2, I'm not sure, but at least it's a special case of something useful.<|endoftext|> -TITLE: Stable motivic cohomology with finite coefficients? -QUESTION [10 upvotes]: In this question, which attracted no responces so far, I've asked whether it is possible to extend the Beilinson-Lichtenbaum etale descent rule for motivic cohomology to singular varieties, in particular, by choosing the Grothendieck topologies appropriately. -Subsequently I've looked into Voevodsky's paper "Motives over simplicial schemes" (Journ. K-theory 2010) and found that he seems to explain, in the introduction, that there are two kinds of motivic cohomology for singular varieties: the effective and the stable one. The effective motivic cohomology of a variety $X$ over a field $F$ with coefficients in an abelian group $A$ is given by -$$ - H^i_M(X,A(j))=Hom_{DM(F)}(M(X),A(j)[i])=Hom_{DM(X)}(\mathbb Z,A(j)[i]), -$$ -(see also section 4 of the same paper), while the stable motivic cohomology is -$$ - H^i_{stable}(X,A(j))=\varinjlim\nolimits_n Hom_{DM(X)}(\mathbb Z(n),A(n+j)[i]). -$$ -The reason for the two theories being different is that the cancellation theorem (claiming that the Tate twist is fully faithful) does not hold for motives over a singular variety. -As I tried to explain in my previous question, it seems that the Beilinson-Lichtenbaum etale descent rule for motivic cohomology with finite coefficients -$$ - H_M^i(X,\mathbb Z/m(j)) = H_{Zar}^i(X,\tau_{\le j}R\pi_*\mu_m^{\otimes j}), -$$ -where $\pi\colon Et\to Zar$, breaks down when $X$ is no longer smooth. Can it be true that an isomorphism like -$$ - H_{stable}^i(X,\mathbb Z/m(j)) = H_{Zar}^i(X,\tau_{\le j}R\pi_*\mu_m^{\otimes j}), -$$ -holds for arbitrary singular varieties $X$? Or can it be made to hold by replacing the pair of topologies (Zariski, etale) with a different pair, with one or both of the topologies including resolutions of singularities as covers, perhaps? - -REPLY [3 votes]: I believe you can get results along these lines if you replace Zar with something called the $\ell$dh topology which is a modification of the cdh topology. This topology depends on $\ell$ which is a prime that is not equal to the characteristic you are working over and is a replacement for resolution of singularities over finite fields. The result is originally in Shane Kelly's PhD thesis: http://arxiv.org/abs/1305.5349 -And this is a consequence of corollary 4.25 of http://arxiv.org/pdf/1305.5690v2.pdf which states that every $M\mathbb{Z}_{(\ell)}$-module spectrum satisfies $\ell$dh descent.<|endoftext|> -TITLE: Deformation theory over the field of algebraic numbers -QUESTION [11 upvotes]: Let $X_0$ be a smooth projective variety over $\mathbb{C}$ and let $\Theta_{X_0}$ be the locally free sheaf of $O_{X_0}$-module corresponding to tangent space of $X_0$. -Goal: To find a sufficient condition on $X_0$ so that it admits a model over $\overline{\mathbb{Q}}$ (the field of algebraic numbers). -By spreading out $X_0$ we may choose a proper morphism -$$ -f:X\rightarrow Spec(\overline{\mathbb{Q}}[T_1,\ldots,T_n])=:B, -$$ -where the $T_i$'s are "dependent variables" (i.e. they may satisfy some algebraic relations) such that when we specialize $T=(T_1,\ldots,T_n)$ to the point $P_0=(t_1,\ldots,t_n)\in\mathbb{C}^n$ we recover $X_0$. We may thus view -$X$ via $f$ as a scheme over $Spec(\overline{\mathbb{Q}})$. Using sheaf cohomology, for every $\mathbb{C}$-valued point $p$ of $B$, we get a connecting homomorphism -$$ -\kappa:T_{B/Spec(\overline{\mathbb{Q}}),p} -\rightarrow H^1(X_p,\Theta_{X_p}). -$$ -Note that an element $\partial\in T_{B/Spec(\overline{\mathbb{Q}}),p}$ may be viewed as a derivation of $\mathbb{C}$ over $\overline{\mathbb{Q}}$. -Now if we translate "naively" the Kodaira-Spencer deformation theory to our setting we should have a result which has the follwing flavor: -Tentative theorem: If for all $p\in B$ and all -derivations $\partial\in T_{B/Spec(\overline{\mathbb{Q}}),p}$ one has that $\kappa(\partial)=0$ then $X_0$ admits a model over $\overline{\mathbb{Q}}$. -Question: Do we have such a result and if the answer is yes then what is a good reference where it is proved? -I would like a reference where the proof is as transparent as possible. - -REPLY [2 votes]: The answer is yes, with the modification that the descent may not be from $B$ to $\overline{\mathbb Q}$ but from some etale cover of $B$. I.e. it really is a result about algebraically closed fields of definition. (The intuition being that Kodaira-Spencer trivial implies that a family is isotrivial). -The following was first proved by Buium: -Theorem Let $X$ be a variety, proper over an algebraically closed field $K$. -Then $X$ is defined over the fixed field of the set of all derivations of $K$ which lift to -derivations of the structure sheaf of $X$. -See: -Buium, Alexandru; Differential function fields and moduli of algebraic varieties. Lecture Notes n Mathematics, 1226. Springer-Verlag, Berlin, 1986. -Buium, Alexandru; Fields of definition of algebraic varieties in characteristic zero. Compositio Math. 61 (1987), no. 3, 339–352. -and also: -Gillet, Henri; "Differential algebra - A Scheme Theory Approach", in Differential algebra and related topics: proceedings of the International Workshop, Newark Campus of Rutgers, The State University of New Jersey, 2-3 November 2000, Editors Li Guo, William F. Keigher, World Scientific -The converse statement is of course trivial<|endoftext|> -TITLE: Defining Multiplication in Polynomials over Rings of Matrices -QUESTION [6 upvotes]: More explicitly, if $M_{2 \times 2}(\mathbb{R})[x]$ denotes the ring of polynomials over the ring of 2x2 matrices with real coefficients (with indeterminate x a 2 by 2 matrix with real coefficients), how do i properly define multiplication? e.g. suppose $A_0,A_1,B_0,B_1 \in M_{2 \times 2}(\mathbb{R})$, and let $f[x] = A_0 + A_1x$ and $g[x] = B_0 + B_1x$ be elements of $M_{2 \times 2}(\mathbb{R})[x]$. Then I would assume the product $fg[x]$ would be -$ -fg[x] = (A_0 + A_1x)(B_0 + B_1x) = A_0B_0 + A_0B_1x + A_1xB_0 + A_1xB_1x -$ -But then complications arise due to $M_{2 \times 2}(\mathbb{R})$ being non-commutative. As far as I know (I've only taken one course so far on abstract algebra), this ring is well-defined (in that a polynomial ring can have coefficients in any ring, not just commutative ones). I've checked google, wikipedia, etc and haven't found anything relevant to this topic. Is there any standard literature on this topic? -My plan was to eventually be able to investigate cases in which unique factorizations may hold (if any), or maybe polynomials having unique left or right inverses, etc. - -REPLY [9 votes]: I have two answers for you, depending on what you have in mind. -You want to add an $x$ to the ring of 2x2 matrices, such that while $x$ commutes with multiples of the identity, it doesn't commute with anything else. You can adjoin a noncommuting indeterminate by using what's called the free product. You take the two rings $M_2(\mathbb{R})$ and $\mathbb{R}[x]$, and then you form the free product over $\mathbb{R}$. Wikipedia has an entry on the free product of groups. The construction for rings is fairly similar. -That construction has one weakness: $x$ will not satisfy any relations. There relations that all 2x2 matrices will satisfy, but $x$ in the free product won't. Rings all of whose elements satisfy identities are known as polynomial identity rings. For example, any four 2x2 matrices satisfy an identity of degree four (described in this blog post). So any three elements of $M_2(\mathbb{R})$ and $x$ should satisfy that relation. (I don't know if all possible relations are generated from specializing this one relation. There could be other relations that rely on specific properties of specific matrix elements.)<|endoftext|> -TITLE: Arrangements of points in the plane -QUESTION [18 upvotes]: Let $p_1,\ldots,p_n$ be a collection of distinct points in $\mathbb{R}^2$, no three of which lie on a line. For each $p_i$, let $\omega_i(p_1,\ldots,p_n)$ be the following ordered list (well-defined up to cyclic permutation). Choose some direction $\theta$ such that none of the $p_j$ lie on the ray from $p_i$ going in the direction $\theta$. Rotate $\theta$ clockwise in a full circle, and record the ordered list of the $p_j$ you encounter. The result is $\omega_i(p_1,\ldots,p_n)$. -The ordered lists of points $\omega_i(p_1,\ldots,p_n)$ thus encode some of the combinatorics of how the $p_i$ lie in the plane. -My question is under what circumstances can you go in the other direction? More precisely, assume that you are given $n$ ordered lists $\sigma_1,\ldots,\sigma_n$, where $\sigma_i$ contains exactly the elements of $\{p_1,\ldots,p_n\} \setminus \{p_i\}$ with no repetitions and is well-defined up to cyclic permutations. When can you find points $p_1,\ldots,p_n$ in $\mathbb{R}^2$ such that $\omega_i(p_1,\ldots,p_n) = \sigma_i$ for all $i$? -It is trivial that for $i=1,2,3$, this can always be achieved. However, for $i=4$ it is not hard to find $\sigma_i$ as above that can not be realized. Alas, it is hard to find any general patterns. -Here is a slightly less vague question/guess as to what might be true. I wonder if there might be some kind of "local condition" of the following form. There exists some $N$ such that if $\sigma_1,\ldots,\sigma_n$ is a collection of lists as above, then there exists points $p_1,\ldots,p_n$ as above if and only if for every $m$ element subset $S$ of $\{p_1,\ldots,p_n\}$ with $m \leq N$, the lists obtained from the $\sigma_i$ by deleting the points not in $S$ and throwing away the corresponding lists can be realized? - -REPLY [5 votes]: Goodman and Pollack called sequences that encapsulate the combinatorics -of an arrangement of lines allowable sequences in their classic paper -"Semispaces of Configurations, Cell Complexes of Arrangements" -(J. Combin. Theory Ser. A, -37:257-293, -1984.) -These sequences are equivalent, I believe, to your sequences. -(One of GP's theorems is that a sequence of permutations is realizable by points -iff it is realizable by lines.) -They established the following theorem concerning allowable sequences: - -Every allowable sequence of permutations can be realized by an arrangement of pseudolines. - -Pseudolines are curves each pair of which intersect exactly once. (More precisely, a pseudoline -is a simple closed curve that does not disconnect $\mathbb{P}^2$.) -So I think the answer to your question is that the sequences determine pseudoline arrangements, -which correspond to "generalized configuration of points," rather than line arrangements, which correspond to configurations of points. -Most pseudoline arrangements are not stretchable, i.e., they do not correspond to a line arrangement. So you cannot, in general, reconstruct the point configuration from the sequence. -(However, every arrangement of $\le 8$ pseudolines is stretchable, so your efforts -can be extended to $i=8$.) -There is some work on local conditions with which I am not familar: -Felsner and Weil, "Sweeps, Arrangements and Signotopes" (Discrete Applied Math 109:257-267, 2001). -Perhaps this is related to your local conditions. -There remains no characterization of stretchability. -Peter Shor showed that determining if a pseudoline arrangement is stretchable is NP-hard -("Stretchability of pseudoline arrangements is NP-hard," in Applied Geometry and Discrete Mathematics: The Victor Klee Festschrift, 1991). -This certainly accords with your "Alas, it is hard"! -Another source on this topic is Felsner & Goodman's chapter on "Pseudoline Arrangements" in the Handbook of -Discrete and Computational Geometry (CRC Press, 3rd Ed., 2017). -Here's a near-final PDF of that chapter: Pseudoline Arrangements. They are also mentioned in the -Wikipedia article on arrangements, -but a separate article on pseudoline arrangements is yet to be written.<|endoftext|> -TITLE: How long is the longest path in the game tree of chess? -QUESTION [5 upvotes]: I can only think of an upper bound, which consists of all configurations and so has length $5^{64}$. If the true value is intractable, we may give up solving chess. But if it's small, there still could be a fast algorithm to solve chess. -I say give up because I'm still thinking of traversing the game tree. But this may not be required on a second thought. Anyway I think this question is still interesting theoretically. - -REPLY [4 votes]: Apparently to avoid perpetual check, a rule was initially set such that a sequence of moves that repeats three times will be declared a draw. -A recent nice video by James Grime and Rune Friborg notes that in the 1920's mathematician and chess grandmaster Max Euwe showed that players can engage to create a set of moves corresponding to the Thue-Morse sequence, such that no sequence of moves will be repeated three times. -For example: - -call "white moves a left piece, black moves a left piece" $0$; -call "white moves a right piece, black moves a right piece" $1$; -have the players play a set of moves such as $0110100110010110...$ according to the Thue-Morse sequence. - -Thus there would never be a sequence of moves that repeats three times. Cooperative play would never accidentally lead to a win. Indeed there might be a position wherein the best move for white is to check black with either a left ($a-d$ file) or right ($e-h$ file) rook, and for black to respond by checking white with either a left or right knight. -However, of course there would be a position that repeats three times, even if players cooperate - hence the the change in the rules.<|endoftext|> -TITLE: Derivative of Exponential Map -QUESTION [23 upvotes]: Given a Riemannian manifold $M$, let $\gamma: (a,b) \to M$ be a geodesic and $E$ a parallel vector field along $\gamma$. Define $\varphi: (a,b) \to M$ by $t \mapsto \exp_{\gamma(t)}(E(t))$. Is there a "nice" expression for $\varphi'(t)$? -This question originates in an attempt to understand the proof of corollary 1.36 in Cheeger and Ebin's "Comparison Theorems in Riemannian Geometry." - -REPLY [11 votes]: Let $x(u,t) = \exp_{\gamma(t)}(u E(t))$. For fixed $t$, as $u$ ranges from 0 to 1, the curve $x(\cdot, t)$ is a geodesic segment from $\gamma(t)$ to $\exp_{\gamma(t)}(E(t))$. Then $\phi'(t) = J(1)$ where $J$ is the Jacobi field along this geodesic segment, with the initial conditions $J(0) = \gamma'(t)$ and $(\nabla_{d x/d u} J)(0) = 0$. -Why? As $t$ varies, $x$ is a variation through geodesics, so for fixed $t$, $\frac{\partial x}{\partial t}(u,t)$ is a Jacobi field (call it $J$) along the geodesic $x(\cdot, t)$. Then $\phi'(t) = \frac{\partial x}{\partial t}(1,t) = J(1)$ while $\gamma'(t) = \frac{\partial x}{\partial t}(0,t) = J(0)$. And using the fact that $\nabla$ is torsion-free and $E$ is parallel we have $(\nabla_{\partial x/\partial u} \frac{\partial x}{\partial t})(0,t) = (\nabla_{\partial x/\partial t} \frac{\partial x}{\partial u})(0,t)= \nabla_{\partial x/\partial t}E(t) = 0.$ -(I guess this does not need $\gamma$ to be a geodesic....?)<|endoftext|> -TITLE: Why is Drinfeld's Zastava space called Zastava? -QUESTION [11 upvotes]: I'm trying to get an idea of Drinfeld's Zastava space. It seems to be an infinite-dimensional version of the flag variety, for affine Lie algebras. -But, first of all, why is it called Zastava (Застава)? Sadly I don't understand Russian and I don't understand the connotation. Googling gave me this Wikipedia article about a car company first. But I don't think a car company has much to do with these spaces, or does it? -Apparently it means "outpost" in Russian. But why an outpost? Drinfeld seems to be Ukrainian (again, from Wikipedia) and zastava means "pledge"? -Is it an outpost to a greater understanding of these spaces? Is it a pledge of mathematicians to understand more??? - -REPLY [30 votes]: The term was coined by one Michael de Finkelberg during his visit to Croatia. The word is indeed Croatian and means ``flag''. I was happy to have a Croatian word in mathematics. -The strategy of giving a new notion an old name but in a different language is not perfect.<|endoftext|> -TITLE: Is this Negativstellensatz with uniform denominators known? -QUESTION [7 upvotes]: A theorem of Reznick states that if $f>0$ is a real homogeneous polynomial in several polynomials that is positive away from the origin of ${\mathbb{R}}^n$, then for large $N$, the form $(\sum x_i^2)^N f$ is a sum of squares. -Is there a characterization (or at least some work done) of when $(\sum x_i^2)^N f$ is a sum of squares if we only assume $f\ge 0$? - -REPLY [16 votes]: Hi. I don't usually read this site, but a friend who does told me about your question. Let me give a couple of answers. The first is that my proof completely doesn't work if $f$ - has non-trivial zeros. The second is that the answer depends on n. Hilbert himself proved in 1893 that if $n=3$ and $f(x,y,z)$ is psd with degree m, then $(x^2+y^2+z^2)^N*f(x,y,z)$ is a sum of squares. Landau showed that it is a sum of 4 squares and the bound on $N$ is on the order of $m^2/8$. See my 2000 paper "Some concrete aspects of Hilbert's 17th problem" at -http://www.math.uiuc.edu/~reznick/hil17.pdf -esp. p. 4. On the other hand, if $n \ge 5$, then there are psd forms with "bad points"; that is $f(a) = 0$ and if $h^2 f$ is a sum of squares, then $h(a) = 0$ as well. An example of a form with a bad point can be found in my same paper, p.16. The first such form was found by E. G. Straus, and the expert on the subject is Chip Delzell at LSU. As far as I can remember, the case $n=4$ is still open. You might also be interested in -http://www.math.uiuc.edu/~reznick/aud.pdf -in which is shown that for fixed $(n,m)$, there does not exist a finite set of denominators $\{h_1,\dots, h_T\}$ so that if $f(x_1,\dots,x_n)$ is psd with degree $m$, then $h_j^2f$ is a sum of squares for some $j$.<|endoftext|> -TITLE: Topology of the preimage of a point for degree one holomorphic maps -QUESTION [5 upvotes]: Let $M^n$ and $N^n$ be two compact complex (or complex projective) manifolds. Let $f: M\to N$ -be a holomorphic map of degree one. How to prove that for each $x\in N$ the set -$f^{-1}(x)$ is simply-connected? -Added. The above statement would follow from a different one: -There is an $\varepsilon$-neighbourhood $U_x$ of $x$ such that $f^{-1}(x)$ is the deformation retract of $f^{-1}(U_x)$. This second statement seems very plausible but I don't know how to prove it. -Indeed, since $U_x$ and $f^{-1}(U_x)$ are birational, they have same fundamental group (Griffiths Harris page 474), i.e $f^{-1}(U_x)$ is simply connected. - -REPLY [5 votes]: [This is just a recap/editing of comments that seem to give an answer] -A suggestion about the second statement : follow the gradient of $\phi=d^2_x∘f$, where $d_x$ is the distance to $x\in N$ with respect to a real analytic metric on N, the gradient being taken with respect to another such metric on M. -For this to work, one has first to rule out the possibility that $\phi$ has critical points arbitrarily near $f^{-1}(x)$, but not in it (note that each $z \in f^{-1}(x)$ is a critical point of $\phi$). -But the curve selection lemma (CSL) is precisely adapted to this situation. Namely, if $(z_n)$ is a sequence of critical points of $\phi$ in $f^{−1}(N∖x)$, with $f(z_n)\to x$, one may assume by properness of $f$ that $z_n\to z\in f^{−1}(x)$. Then the curve selection lemma (applied to the semi-analytic set $Z=Crit(\phi)\setminus f^{−1}(x)$, and the point $z$ in its closure) gives an analytic curve $\gamma:[0,\delta[\to M$ of critical points of $\phi$ with $\gamma(0)=z$ and $f(\gamma(t))\neq x$ for $t>0$. -But this is absurd : $t\mapsto \phi(\gamma(t))$ would have to be constant. -To conclude the argument, one may resort to Lojasiewicz inequality (itself a deep consequence of CSL) $|\mathrm{grad} \phi(z)| \geq c \phi(z)^\alpha$ if $\phi(z)<\epsilon$, for some $\alpha\in[0,1[$, $c>0$, $\epsilon >0$. This implies that the gradient trajectories have uniformly bounded lengths since they also are those of $\phi^{1-\alpha}$, which has gradient norm $\geq(1-\alpha)c$ on $\phi^{-1}(]0,\epsilon[)$. -This implies that the neighborhood $\phi^{-1}([0,\epsilon[)$ deformation retracts to $f^{-1}(x)=\phi^{-1}(0)$. -It must be said, however, that this is only a small fragment of the theory of (semi-,sub-) analytic sets developed by Lojasiewicz, Hironaka, etc (with previous work by Whitney and Thom) since the 60's, and that I am not competent to retrace here the story of its development, nor to give proper attributions. In particular, I don't know to what extent it relies on Hironaka's resolution of singularities. -Some references I've found are Milnor's 1969 Singular points of complex hypersurfaces where I first learned -CSL (in the semi-algebraic context), Denef and van den Dries 1988 p-adic and Real Subanalytic Sets, -Bierstone and Milman's 1988 Semianalytic and subanalytic sets, Kurdyka's 1998 On gradients of functions definable in o-minimal structures. Two papers by Lojasiewicz seem also very relevant (but not easily accessible).<|endoftext|> -TITLE: Generalization of a horse-racing puzzle -QUESTION [10 upvotes]: A well-known puzzle goes: -"Suppose that you have 25 horses and a racetrack on which you can race up to 5 horses. If the outcome of each race only tells you the relative speeds of the horses in the race, how many races do you need to determine the fastest 3 horses (and what is the strategy)?" -The solution (look away now if you don't want a spoiler) is to arrange the horses into groups of five and race them, labeling the horses $a_1,\dots,a_5$, ..., $e_1,\dots,e_5$ -- for example, the horse in position 3 in the second race gets the label $b_3$. -Then race horses $a_1, b_1, c_1, d_1, e_1$, and relabel the horses so that all those in the same group as the winner of this race get the label $a_j, j=1,\dots,5$ and so on. Finally, race horses $a_2, a_3, b_1, b_2, c_1$ -- the three fastest horses are now $a_1$ and the two fastest from the final race. -The question: Does this strategy generalize to $m$ horses and $n$ tracks where you want to find the fastest $k$ horses? - -REPLY [8 votes]: Well, this particular strategy generalizes for finding the k best horses when the track size is $n = (k-1)(k+2)/2$ and the number of horses is $n^2$, and it takes n+2 races as in your example: -Split them into n groups of size n, race them in those sets, and label as $a_{11}, a_{12}, \dots, a_{1n}, a_{21}, a_{22}, \dots, a_{2n}, \dots, a_{nn}$ as before (so the horse who came in $j$th place in the $i$th race has label $a_{ij}$. Then race $a_{11}, a_{21},\dots,a_{n1}$, and relabel the first subscripts of all horses using the results of this race. The winner of that race is the best horse. To determine the other k-1 best horses, race the n other horses who have fewer than k horses that are better than them (directly or by transitivity): $a_{12}, a_{13}, \dots, a_{1k}, a_{21}, a_{22}, \dots, a_{2(k-1)}, a_{31}, a_{32}, \dots, a_{3(k-2)},\dots, a_{k1}$. (Note here that conveniently $n = (k-1) + (k-1) + (k-2) + (k-3) + \dots + 2 + 1 = (k-1)(k+2)/2$.) -But this still leaves open the question of what to do for other cases.<|endoftext|> -TITLE: What would one call this generalisation of the moduli space of theta-characteristics -QUESTION [5 upvotes]: A theta-characteristic on a Riemann surface $\Sigma$ is a holomorphic line bundle $L$ such that $L \otimes L \cong \omega$, i.e. a square root of the cotangent bundle. There is a moduli space (stack) of such pairs $(\Sigma, L \in \mathrm{Pic}(\Sigma))$, which I'll denote $\widetilde{M}_g^{1/2}$ when $\Sigma$ has genus $g$, and I believe it is not unreasonable to call this the moduli space of theta-characteristics. This is related to the moduli space $M_g^{1/2}$ of spin Riemann surfaces in that there is a map $M_g^{1/2} \to \widetilde{M}_g^{1/2}$ and this is a $\mathbb{Z}/2$-gerbe. -Now, one can just as well define moduli spaces $\widetilde{M}_g^{1/r}$ and $M_g^{1/r}$ for any $r$, and the natural map is now a $\mathbb{Z}/r$-gerbe. The space $M_g^{1/r}$ classifies families of $r$-spin Riemann surfaces, and $\widetilde{M}_g^{1/r}$ classifies families of Riemann surfaces with a section of the fibrewise Picard which is an $r$-th root of the canonical section. -I am rather uninformed with respect to algebraic geometry, so my question is: What would an algebraic geometer call the moduli space $\widetilde{M}_g^{1/r}$ for $r > 2$? - -REPLY [3 votes]: I've seen many people referring to what you call $r$-spin Riemann surfaces as higher spin curves (sometimes with the $r$, sometimes suppressed), so my thought would be to call it the moduli stack of $r$-theta characteristics or of higher theta characteristics, if the $r$ isn't as important.<|endoftext|> -TITLE: Floating polyhedra with fair equilibria -QUESTION [11 upvotes]: Is there a homogeneous convex polyhedron - which floats so that some subset (perhaps all) of its faces - is distinguished as "up" (above the water line) - in stable equilibrium, each face with equal probability, - and there are no other stable orientations? - -This question was prompted by the recent -MO question on fair polyhedral dice. -My thought is that one might shake a container of liquid and -await the floating polyhedron to stabilize. -Perhaps permitting inhomogeneous polyhedron interiors would -make this more feasible? -I have been unsuccessful in finding much information on floating -polyhedra, perhaps because this is such a bad idea. :-) -I know about the center of boyancy -and the metacenter and -the conditions for stability. -And -I know of Ulam's "Floating Body Problem" from the Scottish Book, -apparently still unsolved -(mentioned in Unsolved Problems in Geometry -by Croft, Falconer, Guy). -None of what I found is shedding much light. -Any pointers would be appreciated! -Edit. -Thanks Igor, sleepless, Douglas for your comments, -and apologies for not being clear! -Let me attempt a sharper question. -Let the density of the polyhedron $P$ be half that of water, -$\rho=\frac{1}{2}$. -I seek a $P$ that has $k>1$ distinct stable floating orientations. -So $P$ is unlike a sphere, which is stable in any orientation -(Ulam's question is: Is the sphere the only such convex body?). -And $P$ is unlike a boat-hull shape that is designed to have a unique stable -orientation, i.e., is monostatic. -In each of $P$'s stable orientations, some face's normal vector $n_i$ -is vertical, perpendicular to the water level, pointing up. -So if you look down from above, you see that face $F_i$. -That is the basic question. -Now some embellishments: - - -It is not essential that $\rho=\frac{1}{2}$ exactly. -Let's say, -$\rho \in [\frac{1}{4},\frac{3}{4}]$. - - -It is not essential that each of the normal vectors $n_i$ -be exactly vertical. -It just needs to be clear which face is "up." -Let's say that $n_i$ should make an angle of $\le \delta$ with -the vertical $(0,0,1)$, and no other face normal $n_j$ -makes an angle of $\le 3 \delta$ with the vertical. - - -Ideally, -if $P$ is placed in the water at an arbitrary orientation, -it should stabilize into one of its $k$ stable positions -with equal probability. -So the total solid angle of orientations that lead to each -of the $k$ distinct stable positions should be $4 \pi / k$. - - -Were these conditions satisfied, one could place $P$ in a glass cylinder half full of -liquid, shake it, wait, and look down through the cylinder top to see the numbered up-face. - -REPLY [14 votes]: Added below: discussion and illustrations for equilibria of floating objects -**Further additions: Plots of potential energy for floating square and cube ** -For the 2-dimensional shape below with 4 long spikes and density .5, it seems clear that the only stable equilibrium is with one spike pointing straight down and another pointing straight up. Two spikes will float to the sides, with their centerlines at the water level. -A slight tilt shifts the center of buoyancy (= mean of points below the waterline) toward the side, and provides a strong righting force. - -(This plot is the locus $|x|^.3 + |y|^.3 = 1$) -This example is easily modified to be a 12-gon in the shape of a cross, with 4 short faces that want to point straight up. -Here's a corresponding 3-dimensional shape, which should float with one of the spikes pointing up: - -Begin additional discussion -For any centrally symmetric object of density .5 floating in water, the plane of the surface of the water passes through the center of mass of the object, since then the volume of water displaced is exactly half the volume of the object. In that case, the potential energy of the object in a given configuration is proportional to depth of the center of mass of the underwater portion of the object: it's as if the underwater portion has negative weight. -To get better intuition for how this works, I tried rotating different shapes using Mathematica, plotting the underwater portion, its center of mass, and watching how the depth of the center of mass (in an amplified view) fluctuated with angle. The red dots in -the pictures below are the centers of mass of the underwater portion. The black dot on -to the left side of the object is at a height that magnifies the height of the underwater center; the higher it is, the more stable the object. (In the live examples, the sliders will adjust the angle, and you can add or remove vertices by clicking). - -It's probably easier to get a good intuition by looking at the relationship of the underwater center of mass (negative weight) versus the above water center of mass (positive weight). -When they are not aligned vertically, then they exert a torque; this torque amounts to the gradient of the potential function. Below, for example, is a square at a skew angle, with the underwater center marked in red, pushing upward, and the abovewater center marked in green pushing downward. This torque rotates the floating square toward a diagonal position where it has a stable equilibrium. -alt text http://dl.dropbox.com/u/5390048/RightingTorque.jpg -For a two-dimensional square (or a long 3-dimensional object with square cross-section), the stable equilibria are when a diagonal of the square is horizontal. Similarly, a hexagon is stable when a diagonal (going from a vertex to its opposite vertex) is horizontal, and otherwise unstable. For polygons with an odd number of vertices, the equilibria are when one face is pointing straight down. -Based on these, I'm confident that an octahedron truncated just slightly will float in equilibrium with the truncated faces pointing upward, because an octahedron with floating with one of its diagonal vertical has a lot of its underwater mass relatively close to the surface. With somewhat less confidence, I think the truncated cube would behave similarly. -It should be straightforward (although tedious to set up) to compute the equilibria in this way for any polyhedron. -Further additions: Plots of potential energy -Here is the plot of potential energy for a floating square of buoyancy .5, as a function of angle. (Units for the vertical scale are arbitrary). The bottoms of valleys are the stable equilibria, at odd multiples of $\pi/4$, when the square is floating with a corner up: - -Here is a similar plot for the cube, when the cube is rotated until the -vector $(x,y,1)$ is vertical. The full graph would have domain $S^2$, tiled by 6 copies of this unit, one for each face of the cube. -From the graph you can see that there is a local maximum when any face is horizontal (the top of the mountain), a saddle point when an edge is horizontal and the diagonal of a face is vertical (the saddle points are the midpoints of the bounding edges in this plot), and a minimum when a diagonal of the cube is vertical (the corners in the plot). Notice how low the mountain passes are between adjacent minima. This indicates when a main diagonal is vertical, rotation any of the three lines perpendicular to any face at its midpoint takes relatively little energy, far less than is required to rotate the cube so that a face is horizontal. (Note: this energy function is highly dependent on the density of the cube, so it is hard to draw a conclusion about a floating ice cube or a floating wooden block). - -Computation is underway for the potential energy for a floating dodecahedron, showing the potential energy when the vector $(x,y,1)$ is rotated vertically, where $(0,0,1)$ is the normal to a face. The image may change as the computation of potential energy is improved. -alt text http://dl.dropbox.com/u/5390048/Dodecahedron.jpg -alt text http://dl.dropbox.com/u/5390048/DodecaFloatEnergy.jpg<|endoftext|> -TITLE: Finite vector bundles over punctured affine spaces -QUESTION [11 upvotes]: Let $X$ be a connected scheme. Recall that a vector bundle $V$ on $X$ is called finite if there are two different polynomials $f,g \in \mathbb N[T]$ such that $f(V) = g(V)$ inside the semiring of vector bundles over $X$ (this definition is due to Nori, if I am not mistaken). For example, any trivial or torsion line bundle is finite in this sense. -Now, let $k$ be a field and $X= \mathbb A_k^n -\{0\}$ with $n\geq 3$. My question is: - -Are there non-trivial finite vector bundles on $X$? If the answer is no, are there elementary ways to see this? - -In the case $k=\mathbb C$, I think the answer is no, as follows: by results of Nori in his thesis, finite bundles gives rise to representations of the fundamental groups scheme, so it is enough to see this group is trivial. But over $\mathbb C$, such group coincides with the etale fundamental group, and $X$ is simply connected. This argument seems to break down over arbitrary fields. -Motivation: I would like to mention this in a talk next week! - -REPLY [5 votes]: If I understand your definition of "finite", then any stably trivial bundle is finite. In particular, if R denotes the real numbers, the bundle over $Spec(R[X,Y,Z]/(X^2+Y^2+Z^2=1)$ defined by the unimodular row (X,Y,Z) is finite. Unless I'm missing something, it should be easy to extend this example to the punctured affine space. $A_R^3-\{0\}$.<|endoftext|> -TITLE: Would an oracle for integral points on elliptic curves be a factoring oracle? -QUESTION [6 upvotes]: Oracle finding all integral points on genus 0 curves is a factoring oracle (e.g. $xy=n$ and $x^2-y^2=n$ -I asked Can the number of solutions $xy(x−y−1)=n$ for x,y,n∈Z be unbounded as n varies? and occurred to me that an oracle giving all integral points may find nontrivial factor of $n$. Drama is this will not work for all $n$. -Would an oracle for finding all integral points on genus 1 curves (in whatever model) be: - -(loosely defined) Weak factoring oracle which finds at least one nontrivial factor -Strong factoring oracle which finds all prime factors? - -The factoring oracle must work for all integers if it exists. -(EDIT): Intuitively if I had genus 0 oracle for integral points I could factor general integers. If the oracle were for genus 1 I don't see a way for general integers but I would be lucky with integers of the form $xy(x-y-1)$ (just an example) - -REPLY [8 votes]: An integral point (actually, a rational point in the affine plane will do) on an elliptic curve $y^2 = x(x^2 + ax + b)$ comes (by the standard technique of simple 2-descent) from a rational point on some quartic -$$ N^2 = b_1M^4 + aM^2e^2 + b_2e^4, $$ -where $b_1b_2 = b$. Thus if you want to factor an integer $N$, ask the oracle for (rational) points on the curve $y^2 = x(x^2 + ax + N)$ for sufficiently many values of $a$ until you find a point that gives you a nontrivial factorization $N = b_1b_2$. If you choose $a$ in such a way that the parity conjecture predicts an odd rank, you will know in advance that such a point exists. -A similar technique works for the Pell equation and shows that solving the Pell equation -$T^2 - dU^2 = 1$ is at least as hard as factoring $d$.<|endoftext|> -TITLE: The sum of integers being a bijection -QUESTION [19 upvotes]: What are the pairs $(P,Q)$ of subsets of $\mathbb N$ for which the map -\begin{eqnarray*} -P\times Q & \rightarrow & {\mathbb N} \\\\ -(p,q) & \mapsto & p+q -\end{eqnarray*} -is a bijection ? -Obvious examples are $P=\mathbb N$ with $Q=\{0\}$, or $P=2\mathbb N$ with $Q=\{0,1\}$. Are there others ? -This question is related to a puzzle given in EMISSARY (fall 2010), asking to find infinite series $f(x)$ and $g(x)$ with coefficients $0$ and $1$, whose product equals $\frac{1}{1-x}$. I suspect that the word infinite was written on purpose, and therefore $P$ and $Q$ must be infinite. -Later. After the answers, I understand that one can find a sequence $(P_j)_{j\ge0}$ of subsets of $\mathbb N$ with $0\in P_j$, such that every $n\in\mathbb N$ writes $\sum_{j\ge0}p_j$ with $p_j\in P_j$, in a unique way. Of course, all but finitely many $p_j$'s are zeros. Now, I feel dumb, because this follows for instance from the writing of integers in some basis. - -REPLY [4 votes]: If you accept that 0 is not a natural number, then there is a very simple answer to your question: take $P$ to be all numbers whose expansions base 4 contain only digits 0 and 1 and $Q$ to contain only digits 0 and 2. Then $P\cap Q=\{0\}$, which we have boldly excluded. -Also, both sets have the lowest possible asymptotic density of order $1/\sqrt n$, which is kinda nice.<|endoftext|> -TITLE: Reasonable "Random" matrices to test numerical algorithms -QUESTION [7 upvotes]: Hello, -in numerical analysis, it is common to compare the behavior of different algorithms, and of different implementation of algorithms. This occurs not only on the theoretical level, but also on the concrete level of implementation - and not to forget, it serves the purpose of demonstration. -A prominent problem is the solution of linear systems, both general as well as various subcases. -To test, benchmark and profile numerical implementations, you run your work on several instances of the problem. However, it is difficult question to obtain a good set of these instances. You want to inspect pathological cases (diff. degrees of ill-conditionedness) as well as "real-life" examples (whatever this may mean). Ideally, you have an algorithm which puts out matrices with certain properties in a "reasonable" probability measure. A good notion of "reasonable" might be accessible, as most such LSE problems from physics or simulations have much more structure as is actually demanded by the algorithms in theory. -In so far, I wonder whether there are works in numerical analysis how to, given $n \in \mathbb N$ randomly produce - -a sequence of ${n \times n}$-matrices -optionally constraint to be symmetric, positive definite, well-conditioned -which is reasonable in whatever sense - -This is probably an interesting topic within the theory of numerical algorithms. -Thanks, -Martin - -REPLY [2 votes]: One way to get useful information on a numerical algorithm is to choose by hand a well-conditioned or even more an ill-conditioned example and perturb it randomly. In fact it is useful to perturb it randomly such that the perturbations form a sequence approaching zero; you can then look at the sequence of solutions that you get and see what you can say about error propagation and the like.<|endoftext|> -TITLE: Is there a systematic method for differentiating under the integral sign? -QUESTION [43 upvotes]: This MO question by Tim Gowers reminded me of a question I've wondered about for some time. In the delightful book Surely You're Joking, Mr. Feynman!, Feynman praises the technique of differentiating under the integral sign (a.k.a. the Leibniz integral rule): - -When guys at MIT or Princeton had trouble doing a certain integral, it was because they couldn't do it with the standard methods they had learned in school. If it was contour integration, they would have found it; if it was a simple series expansion, they would have found it. Then I come along and try differentiating under the integral sign, and often it worked. So I got a great reputation for doing integrals, only because my box of tools was different from everybody else's, and they had tried all their tools on it before giving the problem to me. - -Some examples of this trick are provided on the Wikipedia page that I linked to above. What I am wondering is whether there is a systematic way to attack integrals via the introduction of an extra parameter and applying the Leibniz integral rule. By "systematic" I mean something that could be incorporated into the symbolic integration algorithms of a computer algebra package. -The closest thing I've found in the literature is the paper "The Method of Differentiating under the Integral Sign," by Gert Almkvist and Doron Zeilberger, J. Symbolic Computation 10 (1990), 571–591, which develops an algorithm for finding a differential equation satisfied by the integral -$$R(x) = \int_{-\infty}^\infty F(x,y)\ dy$$ when $F(x,y)$ is holonomic. However, typically the critical step in evaluating an integral "the Feynman way" is to figure out how to introduce an extra parameter in the right way, and the Almkvist–Zeilberger paper does not provide a systematic algorithm for this step. -The Wikipedia examples strike me as ad hoc, so the question I am posing to MO readers is, do you know of any heuristics for introducing extra parameters into integrals, that might form the starting point for a general algorithm? Anything that helps remove the black-magic or rabbit-out-of-a-hat aura of introducing extra parameters would be welcome. - -REPLY [4 votes]: Even this does not address the original problem of differentiating -under the integral sign, there is a partly successful way to -calculate the the loop integrals associated to the Feynman diagrams. -It is hard to describe the strategy, called by the authors method -of brackets, therefore I refer to the original paper - -I. Gonzalez, V.H. Moll, and A. Straub, The method of brackets. - Part 2: examples and applications, Contemp. Math. 517 (2010), 157–171. doi:10.1090/conm/517/10139, arXiv:1004.2062 (pdf) - -Let me cite the short section "Conclusions and future work" in the paper: - -The method of brackets provides a very effective procedure to evaluate definite - integrals over the interval $[0,\infty)$. The method is based on a heuristic - list of rules on - the bracket series associated to such integrals. In particular, a variety of examples - that illustrate the power of this method has been provided. A rigorous validation - of these rules as well as a systematic study of integrals from Feynman diagrams is - in progress.<|endoftext|> -TITLE: Gerbes and Z-graded symmetric monoidal categories -QUESTION [8 upvotes]: Let me first recall a known fact. Suppose $X$ is a complex algebraic variety, $\mathscr{L}$ is a line bundle on $X$ and $L^\times$ is the total space of $\mathscr{L}$ with the zero section removed. Write $\pi:L^\times\longrightarrow X$ for the projection. Then $\mathscr{A}=\pi_*\mathscr{O}_{L^\times}$ is naturally a $\mathbb{Z}$-graded quasicoherent sheaf of commutative $\mathscr{O}_X$-algebras, and $\mathscr{L}$ can be recovered from $\mathscr{A}$ (because we can recover $L^\times$ together with the map $\pi$ and the action of the multiplicative group $\mathbb{G}_m$). -My question is whether this picture can be categorified along the following lines. -On the one hand, instead of line bundles let us look at $\mathbb{G}_m$-gerbes on $X$. -On the other hand, instead of $\mathscr{A}$ let us suppose (for instance) that we have a $\mathbb{Z}$-graded symmetric monoidal $\mathbb{C}$-linear abelian category $\mathscr{M}=\bigoplus_{d\in\mathbb{Z}}\mathscr{M}^d$, so that $\mathscr{M}^0$ is the category of coherent sheaves on $X$ with the usual tensor product. -Is there a setup of this sort where one gets a suitable equivalence between $\mathbb{G}_m$-gerbes on one side and a special class of $\mathbb{Z}$-graded symmetric monoidal categories on the other side? (It is fine with me if the description of the other side needs to be modified.) - -REPLY [7 votes]: This question can be approached abstractly through the general Tannakian formalism, -as laid out e.g. here, or very concretely by hand. You construct maps in both directions. To a $G_m$ gerbe assign its tensor category $QC(Gerbe)$ of sheaves (I'll speak of quasicoherent sheaves out of force of habit - presumably you can work just with coherent). This is a commutative algebra over $QC(X)$, which locally is isomorphic to graded sheaves on $X$ (i.e. to $QC(X)\otimes Rep G_m$). Conversely to such a category assign its spectrum, the stack which to any ring attaches the groupoid of tensor functors from your category to modules. This carries a map to $X$ which is a $G_m$ gerbe. -Put another way, given a sheaf of tensor categories over $X$ (or commutative algebra over $QC(X)$) which is locally isomorphic to $Rep G$, you consider the stack (sheaf of groupoids) of isomorphisms of this sheaf of categories with $QC(X)\otimes Rep G$ [EDIT: Better and more Tannakian to say, the stack of fiber functors to $QC(X)$]. This is a $G$ gerbe (ie 's locally of the form $X\times BG$). This is of course just the usual Tannakian reconstruction as in Deligne, except that we have the base $X$ be a scheme (or geometric stack) instead of the spectrum of a field. -Of course you could also ask to give a more global characterization of such $Z$-graded commutative algebras over $QC(X)$. I think it's equivalent to characterize the module category given by sheaves of degree 1, aka twisted sheaves on the corresponding gerbe. This is your version of a categorified line bundle -- it's a module category locally isomorphic to sheaves. Presumably it can be characterized as an invertible module category -- one for which there exists (or maybe for which you specify - I'm conveniently pretending that everything has been taking place one level of categoricity down, which is fine if you only care about say a class in $H^2(X,G_m)$) an inverse with respect to tensor product of module categories over $QC(X)$. Then the above argument proves that such categorified line bundles (via Spec of the $Z$-graded $QC(X)$-algebra they generate) are equivalent to $G_m$-gerbes. [Edit:] You'll also need to make sure such invertible modules are locally trivial, ie again given by the same cohomology group. You might also want to think of things through a third perspective on this story after the gerby/Tannakian ones, namely that of Azumaya algebras. You want to know that your invertible module category has a generator as an $O_X$-linear category, and thus is equivalent to the category oif modules over a sheaf of algebras, namely the endomorphisms of this generator. And then appeal to a classification of these algebras up to Morita equivalence by the same cohomology group. -[By the way a very interesting recent paper about the derived version of this story is here.]<|endoftext|> -TITLE: Groups with all normal subgroups characteristic -QUESTION [16 upvotes]: Today in my research, I had to use fairly explicitly the rather tautological property of finite cyclic groups that every normal subgroup is characteristic, i.e. fixed by all automorphisms. This got me wondering: - -do (finite) groups with the property that every normal subgroup is characteristic - have a name and/or can they be completely - classified? Generally, has this - property been investigated at all? - -Apart from cyclic groups, some groups possessing the above property that immediately come to mind are simple groups, symmetric groups, semi-dihedral groups, and dihedral groups of twice odd order (however not of twice even order). -This is as far as I got on my short walk home (apart from some false claims, see comments). I suspect that this property might be well studied. -Edit: -The reference that Beren Sanders provided in his answer and the references to and from it all deal with $p$-groups. I still haven't been able to find anything about arbitrary finite groups. Some of the questions that $p$-group theorists ask are just not terribly interesting in the case of arbitrary groups. E.g. the paper that Beren Sanders mentioned proves that every finite $p$-group is contained in another finite $p$-group in which every normal subgroup is characteristic. The same statement for arbitrary finite groups is trivial: just embed your group into a symmetric group. I would still be surprised if nobody had tried to say something reasonably general about finite groups with this property. - -REPLY [9 votes]: Here is a small contribution. If no two factors in a composition series for $G$ are isomorphic, then every normal subgroup of $G$ is characteristic. In fact, in groups of this type, no two distinct normal subgroups can be isomorphic. To see this, suppose $M$ and $N$ are distinct and isomorphic. Look at a composition series obtained by refining the series $1 \subseteq M \cap N < M -TITLE: How many edge-disjoint paths go from upper left to lower right in a $4 \times N$ rectangular gridwork of streets? -QUESTION [12 upvotes]: Background/Motivation -My interest in this problem traces back to an 11 year old girl who really took to one-way path counting problems. After doing several -configurations of streets, she decided to come up with a problem of her own. She presented a $3 \times 3$ gridwork -of two-way streets (forming 4 blocks in a $2 \times 2$ arrangement) and added the condition that a street could be -traversed at most once. She asked how many such paths are there from upper left to lower right? (Answer: 16.) -Stirred by her enthusiasm, we tried generalizing in various directions. If you have 2 long horizontal streets -with $N$ verticals, and let $a_N$ be the number of edge-disjoint paths from upper left to lower right and $b_N$ be -the number of edge-disjoint paths from upper left to upper right, then $a_{N+1} = b_{N+1} = a_N + b_N$ for $N > 1$ and $a_1 = b_1 = 1$. -The $3 \times N$ case is trickier, but the number of edge-disjoint paths from upper left to lower right still satisfies a finite linear recurrence relation. -Naturally, I turned to OEIS and found sequences A013991-A013997, where Dan Hoey gives the number of edge-disjoint paths between opposite corners of $K \times N$ grids for $K = 3, 4, 5, ..., 9$ and small $N$. He also provides the first few values for the $N \times N$ cases (sequence A013990). (Note, his numbering counts blocks, not streets.) For $K=3$, he provides a generating function. In a recent communication, he explained the computer algorithm he used to compute the values but indicates that he did not find a recurrence relation for these sequences, so as far as I know, there is no known way to determine the answer to the title question for large $N$. -I've also spoken with Gregg Musiker, Bjorn Poonen, and Tim Chow about this problem. Although none knew how to do the $4 \times N$ case, Gregg simplified my recurrence relations for the $3 \times N$ case, Bjorn suggested many related questions and suggested an asymptotic formula for the $N \times N$ case, and Tim suggested looking at the related literature on self-avoiding walks, such as the book by Neal Madras and Gordon Slade, though it's not clear to me how related edge-disjoint and self-avoiding are with respect to counting them. -Because there are finite linear recurrence relations for the $2 \times N$ and $3 \times N$ cases, it seems natural to also ask: - -Is there a finite linear recurrence relation for the number of edge-disjoint paths between opposite corners of a $4 \times N$ gridwork of streets? - -Are these problems intractable? - -REPLY [9 votes]: To amplify on Christian's answer, the problem on a $K \times N$ grid for fixed $K$ and varying $N$ admits a finite-state transition model, so in particular is given by a linear recurrence. -The key is to find the right set of states. If you take an edge-disjoint path on a $K \times N$ grid and slice it on a vertical line through the middle passing through a set of horizontal edges, you'll see the path crossing along some odd number of these edges ($\le K$). On both the left and right we'll see a collection of paths with these endpoints. There's another constraint, that we end up with a single, connected path without disjoint loops; to take that in to account, also record a matching: which endpoints are paired up on the right hand side. All but one of the endpoints are paired up in this way. (You could also choose the left, and end up with a slightly different matrix.) -For instance, in the $3 \times N$ case, there are $6$ states. If we record an occupied edge by $\times$ and an unoccupied edge by $\circ$ and turn everything on its side, the states are -$$ -\times\circ\circ\quad\circ\times\circ\quad\circ\circ\times\quad\times_1\times_1\times -\quad\times_1\times\times_1\quad\times\times_1\times_1 -$$ -where the subscript indicates the matching. (In this case, there is at most one matched pair.) -Next consider the transitions. If you consider two adjacent vertical slicings of a path, you'll see two possibly different states. The set of edges that are occupied in the middle is determined by which edges are occupied in the two different states. There is sometimes a choice about how the strands are connected up. However, some of these choices will be ruled out by the constraints on the connectivity; usually you will end up with just $0$ or $1$ possibilities. -For instance, in the $3 \times N$ case, with the states in the order above, I get the following matrix of possibilities: -$$ -M =\begin{pmatrix} -1 & 1 & 1 & 1 & 1 & 0 \\ -1 & 1 & 1 & 1 & 0 & 1\\ -1 & 1 & 1 & 0 & 1 & 1\\ -0 & 1 & 1 & 1 & 0 & 0\\ -0 & 1 & 0 & 0 & 1 & 0\\ -1 & 1 & 0 & 0 & 0 & 1 -\end{pmatrix} -$$ -For the ultimate answer, you want to look at paths that start at the upper-left and go to the lower-right. You can incorporate that nicely by adding an extra slice to the left of the entire diagram, with only the top slot occupied, and another to the right of the diagram, with only tho lower slot occupied. -Concretely, in the $3 \times N$ case, the number of paths is given by the $(1,3)$ entry of $M^N$. -For the $4 \times N$ case, you would get a $16 \times 16$ matrix, which is straightforward but somewhat tedious to work out. As a result, the answer will satisfy a linear recurrence of order $16$. -An interesting variation is to consider only crossingless paths. In this case, the matching must be crossingless, so we only get 5 states in the $3 \times N$ case and $12$ in the $4 \times N$ case. -Update Jan 7: The matrix above is wrong: it should be -$$ -M =\begin{pmatrix} -1 & 1 & 1 & 1 & 1 & 0 \\ -1 & 1 & 1 & 2 & 2 & 2\\ -1 & 1 & 1 & 0 & 1 & 1\\ -0 & 1 & 1 & 1 & 0 & 0\\ -0 & 1 & 0 & 0 & 1 & 0\\ -1 & 1 & 0 & 0 & 0 & 1 -\end{pmatrix} -$$ -Update 2: And here's an image illustrating what is actually being counted: - -I permuted the entries slightly, but they're labelled along the sides. The dotted paths are there to help in the counting: the non-allowed configurations would form a closed loop.<|endoftext|> -TITLE: What is the knot associated to a prime? -QUESTION [25 upvotes]: I can't help but ask this question, having found out about arithmetic topology here on MO. There is a concise description of the MKR dictionary central to this philosophy here. This dictionary is used to translate statements in three-manifold topology into statements in number theory, and vice versa. Often, but not always, the resulting translated statements turn out to be true. -Before asking the following dumb question, let me be absolutely clear that I understand that the general analogy between knots and primes is complicated, and isn't perfect but only fruitful. -That said, I simply cannot help asking the following (certainly stupid) question. -I thank Theo Buehler for his comment below, as it helps to more directly ask the question I wanted to ask. I'll incorporate his comment here as part of the question: - -According to p.24 in Lieven LeBruyn's slides (35 MB!), the trefoil knot is associated to 3. LeBruyn also gives pictures for 2,5 and 7. Is there, associated to the MKR analogy, a natural way to assign a knot to every prime? - -This, I imagine, should not be possible in any reasonable way. I ask with the dim hope that I may be wrong. -Must the knot associated to 3 be the trefoil, as shown in LeBruyn's slides? - -REPLY [50 votes]: Jon, -I should have been more careful in drawing the slide mentioned above (and reproduced below). I didn't want to convey the idea that there is a method to assign to a knot a specified knot (and certainly not one having exactly p crossings...). -I only wanted to depict the result that different primes should correspond to different knots. - (source) -Here's the little I know about this : -Artin-Verdier duality in etale cohomology suggests that $Spec(\mathbb{Z})$ is a 3-dimensional manifold, as Barry Mazur pointed out in this paper - (source) -The theory of discriminants shows that there are no non-trivial global etale extensions of $Spec(\mathbb{Z})$, whence its (algebraic) fundamental group should be trivial. By Poincare-Perelman this then implies that one should view $Spec(\mathbb{Z})$ as the three-sphere $S^3$. Note that there is no ambiguity in this direction. However, as there are other rings of integers in number fields having trivial fundamental group, the correspondence is not perfect. -Okay, but then primes should correspond to certain submanifolds of $S^3$ and as the algebraic fundamental group of $Spec(\mathbb{F}_p)$ is the profinite completion of $\mathbb{Z}$, the first option that comes to mind are circles - (source) -Hence, primes might be viewed as circles embedded in $S^3$, that is, as knots! But which knots? Well, as far as I know, nobody has a procedure to assign a knot to a prime number, let alone one having p crossings. What is known, however, is that different primes must correspond to different knots - (source) -because the algebraic fundamental groups of $Spec(\mathbb{Z})- \{ p \}$ differ for distinct primes. This was the statement I wanted to illustrate in the first slide. -But, the story goes a lot further. Knots may be linked and one can detect this by calculating the link-number, which is symmetric in the two knots. In number theory, the Legendre symbol, plays a similar role thanks to quadratic reciprocity - (source) -and hence we can view the Legendre symbol as indicating whether the knots corresponding to different primes are linked or not. Whereas it is natural in knot theory to investigate whether collections of 3, 4 or 27 knots are intricately linked (or not), few people would consider the problem whether one collection of 27 primes differs from another set of 27 primes worthy of investigation. -There's one noteworthy exception, the Redei symbol which we can now view as giving information about the link-behavior of the knots associated to three different primes. For example, one can hunt for prime-triples whose knots link as the Borromean rings - (source) -(note that the knots corresponding to the three primes are not the unknot but more complicated). Here's where the story gets interesting : in number-theory one would like to discover 'higher reciprocity laws' (for collections of n prime numbers) by imitating higher-link invariants in knot-theory. This should be done by trying to correspond filtrations on the fundamental group of the knot-complement to that of the algebraic fundamental group of $Spec(\mathbb{Z})-\{ p \}$ This project is called arithmetic topology - (source) -(my thanks to the KnotPlot site for the knot-pictures used in the slides) -(my apologies for cross-posting. MathOverflow was the intended place to reply to Jon's question, but this topic was temporarily removed yesterday.)<|endoftext|> -TITLE: Categories of logical formulae -QUESTION [8 upvotes]: Consider the set of formulas of a logic. If there was only one sort of "unary" deduction $\phi \Rightarrow \psi$ - like $(\forall x)\phi(x) \Rightarrow \phi(a)$ - we would immediately have a category of formulas (with deductibility $\Rightarrow$ as morphism). But alas, there are other ("higher") deduction rules, e.g. -$\lbrace p, q \rbrace \Rightarrow p \wedge q$ (rule of conjunction) -$\lbrace p\rightarrow q, p \rbrace\Rightarrow q$ (modus ponens) - -(How) can formulas of "classical logics" - (propositional and FO) be made into a - category despite of those other - relations (= rules)? - -REPLY [2 votes]: I suspect the notion you are looking for is either a multicategory or a polycategory. Multicategories generalise categories by allowing more than one object in the domain of an arrow. These were introduced by Lambek specifically for the study of logical derivations. See - -J. Lambek (1969) "Deductive Systems - and Categories (II)" in Category - Theory, Homology Theory and their - Applications I, Springer Lecture Notes - in Mathematics 87, R. J. Hilton (ed.) - -Polycategories generalise multicategories by allowing multiple objects in the codomain of an arrow. See - -M. E. Szabo (1975) "Polycategories", - Communications in Algebra 3. - -Of course there is a rich history of looking at categories whose objects are formula and whose arrows are proofs, coming mostly from the computer science literature. Lambek and Scott's book may be a good place to start, or for a more modern take try this introductory article http://arxiv.org/abs/1102.1313 However, for classical logics such categories always reduce to boolean algebras, so people tend to work with intuitionistic or linear logics.<|endoftext|> -TITLE: Many flat totally geodesic surfaces ⇒ flat? -QUESTION [41 upvotes]: Let $M$ be a 4-dimensional Riemannian manifold. -Assume there is a huge number (say 100) of flat totally geodesic 2-dimensional surfaces -passing through a point $p\in M$ and assume that their tangent planes at $p$ lie in general position. -Further, assume that sectional curvature is $\ge 0$ (or $\le 0$) in a neighborhood of $p$. -Is it true that $p$ admits a flat neighborhood? -Comments. - -These conditions easely imply that $M$ has zero curvature tensor at $p$. -In dimension 3 the answer is "YES"; it follows since these surfaces have to intersect along geodesics. -If the dimension is ≥ 4, I do not see any reason why the answer should be "YES", but I do not see a counterexample. -For "YES"-answer, you may consider stronger condition, say curvature operator is $\ge 0$ (or $\le 0$) or anything like that. -A related question: assume you have a metric on $\mathbb R^m$ with sectional curvature $\ge 0$ (or $\le 0$). Assume that metric is standard-Euclidean outside of open set $\Omega\subset \mathbb R^m$ and inside $\Omega$ scalar curvature has no zeros. What one can say about $\Omega$? (From the sketch below, it follows that one can not touch $\Omega$ by affine 2-plane in $\mathbb R^m$; it is likely that any component of intersection of $\Omega$ with a 2-plane has to be unbounded; is it all?) - -Sketch for the case $\dim M=3$. -The following observation is due to S. Buyalo. -You can read bit more in our "Sweeping out sectional curvature". -Assume that the number of surfaces is 3. -In general position the surfaces intersect pairwise at 3 distinct geodesics, -say $\alpha_1$, $\alpha_2$ and $\alpha_3$. -Consider a triangle $\triangle x_1x_2x_3$ -such that $x_i\in\alpha_i$ and $x_i\approx p$. -It is easy to see that in $\triangle x_1x_2x_3$ -angles and their model angles are equal. -The curvature condition then implies that this triangle is flat. -Such triangles sweep out a flat octahedral neighborhood of $p$. - -REPLY [28 votes]: Edited 2 January 2011 to clarify: -I don't know if there is an established meaning for general position in this context: it would be nice to have a clarification of terminology (as we've seen from the comments discussion). One possible definition is that a set of subspaces is in topological general position if for any small perturbation, there is a homeomorphism taking one configuration to the other. For 2-planes in $\mathbb R^4$, this just means that any two intersect in a point. If you intersect the subspaces with a small sphere about the origin, this gives a great circle link in $S^3$; for any $n$, there is a finite set of possible link types for these links. - -Example for topological general position - -For topological general position, there is a relatively simple example satisfying the question: take $n$ complex 1-dimensional subspaces of $\mathbb C^2$. The intersection with $S^3$ gives $n$ circles in a Hopf fibration. Now modify the Euclidean metric of $\mathbb C^2$ by -changing the metrics of concentric spheres to homogeneous metrics on $S^3$ obtained by -keeping the Hopf fibers rigid, but uniformly expanding their orthogonal planes by an increasing function of radius $r$. These metrics are invariant by $U(2)$, with nonpositive curvature. (This behaviour is closely related to the geometry of -$\mathbb{CP}^2$ and of complex hyperbolic space $\mathbb {CH}^2$: the family of concentric spheres in either case has this same family of metrics on $S^3$ up to a constant that depends on $r$). - -Possible stronger requirements for general position - -At first it may be tempting to strengthen the condition of topological general position to require that for any small enough perturbation there is a linear map carrying one pattern to another, but this condition is unreasonably strong. The linear equivalence condition is satisfied for almost every triple of 2-dimensional subspaces of a 4-dimensional vector space, since two of the subspaces can be transformed into orthogonal coordinate planes, whereupon the third subspace is the graph of a map from one to the other; by a linear transformation, this can be arranged to map basis vectors to basis vectors. When there is a fourth subspace, the map associated with the third followed by the inverse of the map associated with the fourth is a linear map from a plane to itself, and its characteristic polynomial is an invariant that varies in any open neighborhood of configurations. Note, however, that there are open sets of quadruples of 2-dimensional subspaces for which this self-map has a pair of complex conjugate eigenvalues; when this happens, the quadruple of subspaces is linearly equivalent to a quadruple of complex 1-dimensional subspaces of $\mathbb C^2$. -There are many possible intermediate conditions that could be imposed, depending on what set of special situations you focus on. If you consider all -linear maps defined by triples of planes, as above, you might want to require that the rank of any composition is constant in a neighborhood. However, for many (most) configurations of $n$ planes, when $n$ is high enough, these generate a dense set of linear maps among the planes, and small perturbations can change the rank of something. A more plausible requirement is that the dimension of the Zariski closure of the set of all compositions -of these maps -is constant, or perhaps that the dimension of the Zariski closure of the set of all compositions associated with triples in a subset remains constant. But this definition is special to the setting of half-dimensional subspaces of an even-dimensional space so it's not ideal. It's not hard to modify this for a more general setting, but there are too many possible choices: it's not self-evident that there is one `right' definiton. -One way around the issue is to rephrase the question: is there an open set, or a set of positive measure, or a generic set (intersection of open dense sets) -of $n$ 2-dimensional subspaces of $\mathbb R^4$, such that ___ [a given property is true]. - -Examples for an open set of subspaces - -Now make a random $C^2$-small perturbation of the Hopf fibration to be a foliation of $S^3$ by geodesics, so that the perpendicular plane field $\tau$ is still a contact plane field. -We can construct new metrics in a neighborhood of the origin in $\mathbb R^4$ by modifying -only in the 2-planes orthogonal to the family of 2-planes that are cones on these geodesics, scaling these planes by multiplication by a function $f(r)$ that is concave upward (to aim for negative curvature). If $f$ has $C^\infty$ contact to the constant function 1 at $r = 0$, the resulting metric is smooth. It is flat in each plane that is a cone on leaf of the foliation. -I think that for appropriately chosen $f$ these metrics have nonpositive sectional curvature, but I'll need to come back to this later to back it up (or tear it down), unless someone else will do it.<|endoftext|> -TITLE: switching jobs when you have a tenure-track position -QUESTION [19 upvotes]: I hope this topic is appropriate for MO. I have a tenure-track position at a US university in a location where my spouse is very unhappy, severely underemployed, and has no real prospect of finding suitable employment. I know the economy/job market is bad now and that I will likely have to stay put for a few years. But, hopefully in the near future, things will improve and I would like to test the job market. -Question One: How should I go about obtaining a teaching letter? I work in a small department and I do not think any other faculty member would willingly help me "test the market." (Things could be especially awkward if I cannot find another suitable position and decide to stay in my current position with everyone knowing I tried to leave.) -Question Two: Is it better to apply for other jobs right before or after I get tenure at my current university? (Can having tenure be detrimental in the application process?) -Thanks in advance for any advice ... - -REPLY [16 votes]: 1) It is definitely harder to get a new job if you already have tenure than if you don't. -2) Getting a tenure track position is hard enough. Getting a tenure track position under geographic constraints is really hard. You should be as flexible about the job itself as well as its location as possible. -3) If you're willing to sit tight for a few years, I would advise building as wide a network of professional friends and acquaintances as possible by giving talks at conferences, department seminars and colloquiua. Keep an ear out for job openings for which you are particularly well suited and in departments where you know your candidacy would be supported. You have to be extremely realistic and objective about which schools and departments will be willing to hire you. It will help a lot, if you know people who are willing to help you look and give you frank advice. -4) Get a teaching letter and apply for jobs only when you are really ready to leave. You don't want to do this more than once, if you can help it. But also be ready to do it more than once, if you don't get any or any sufficiently attractive offers the first time. -5) It is indeed a difficult time to do this, so I wish you all the best with this. -ADDED: -6) Be a good citizen in your current department in the sense that you should fulfill all of your teaching and service responsibilities as well as possible. You need as much good will as you can get from everybody including where you are now. -7) If you have any particular talents or skills (outside the usual research credentials) that a math department might find valuable, try to develop them but without undermining your core credentials. One way or another, you want to try to find some way to make yourself stand out relative to everybody else looking for a job.<|endoftext|> -TITLE: Double coset representatives and structure of hecke algebras -QUESTION [6 upvotes]: Let $GL_n(F_q)$ be the general linear group over finite field $F_q$ and $B_n$ be its borel subgroup consisting of all upper triangular matrices. Then the double cosets $B_n\backslash GL_n(F_q)/B_n$ are parametrised by the permutation group $S_n$, which can be viewed as permutation matrices. Further by the theory of representations of symmetric groups, it follows that the number of symmetric permutation matrices is equal to the number of irreducible representations(with multiplicities) of the Hecke algebra $C[B_n\backslash GL_n(F_q)/B_n]$. -My question is: how far is it true? In the sense that suppose I am given a general matrix group (with entries not necessarily from a field) $G$ with a subgroup $B$ such that double cosets $B\backslash G/B$ can be parametrised by symmetric matrices, then is it true that the Hecke algebra $C[B\backslash G/B]$ is commutative? - -REPLY [8 votes]: This question seems slightly strangely phrased to me: the case $\mathbb C[B_n\backslash GL_n(F_q)/B_n]$ gives you a Hecke algebra which is not commutative. It has the same dimension of the group algebra of the symmetric group, and by deformation techniques it's known to have the same number of irreducible representations, but none of this requires commutativity (though that does hold sometimes, and David's answer about the Gelfand trick is a very slick strategy). -If you're interested in where the theory of Hecke algebras relating representations of the symmetric group $S_n$ and the representations of $\text{GL}_n(\mathbb F_q)$ extends to, then there are a number of directions. For a finite group of Lie type $G$, the classification of all irreducible representations hinges on generalizations of the kind of Hecke algebra you mention, where instead of the double-coset algebra, you consider twisted versions of it which are endomorphism algebras of certain "cuspidal" representations of (rational) Levi subgroups of $G$. These turn out to have presentations like the Hecke algebra attached to $S_n$, though sometimes with parameters which are various powers of $q$. One standard reference is Carter's book on finite groups of Lie type, where there is discussion of these endomorphism algebras. (A crucial ingredient is usually that $G$ has a $BN$ pair, so that the $B$--double cosets are indexed by a Coxeter group.) -Similar kinds of Hecke algebras (though now infinite-dimensional "affine" Hecke algebras) arise when studying representations of p-adic groups, Roger Howe has a nice article in Lecture Notes in Math 1804 on the use of Hecke algebras in the p-adic theory -- slight generalisations of the affine Hecke algebras used to study the case of representations generated by an Iwahori fixed vector turn out to allow you to study surprisingly large parts of the representation theory of p-adic groups.<|endoftext|> -TITLE: Integration of differential forms using measure theory? -QUESTION [14 upvotes]: Setup: Let $(M,g)$ be a (possibly non-compact) Riemannian manifold with volume density $d_gV$. Then one may think of $(M,g)$ as a measure space $(\Omega,\mathcal{A},\mu)$, where $\Omega:=M$, $\mathcal{A}:=\sigma(\tau_M)$ is the $\sigma$-Algebra generated by the topology $\tau_M$ of $M$ and for any $A \in \mathcal{A}$, $\mu(A):=\int_M{\chi_A d_gV}$, where $\chi_A:M \to [0,1]$ is the characteristic function of $A$. We obtain $\int_{M}{f d\mu} = \int_M{f d_gV}$, where the left hand side is understood to be an integral in the measure theoretic sense and the right hand side is an integration of a density. This enables us to define the space $L_p(\mu)$ with norm $\|f\|_{L_p(M)}^p = \int_{M}{|f|d_gV}$ on a manifold and apply all the results from integration theory to it, e.g. that it is a Banach space and so on. -My question is: Does this work in the following more general setup: Extend the Riemannian metric on $M$ to a fibre metric in $\bigwedge^k T^{\;*}M$, $0 \leq k \leq m$, (as described in the paragraph below). Then one may define $L_p$-spaces of differential forms by setting $\|\omega\|_{L_p(M)}^p := \int_{M}{|\omega|^pd_gV}$ and setting $L_p^k(M)$ to be the space of all measurable $k$-forms on $M$ (i.e. with Lebesgue measurable coefficient functions in any chart) such that $\|\omega\|_{L_p(M)}<\infty$. Is it possible to construct a measure space $(M,\mathcal{A},\mu)$ such that $L_p^k(M)$ may be thought of as an $L_p(\mu)$ as well?. The problem obviously is the range of a differential form. Formally it is a map -$\omega\colon M \to \bigwedge^k T^*M$, i.e. it takes values in the vector bundle $\bigwedge^kT^*M$. Even if integration theory is available for functions on measure spaces with values in Banach spaces, this does not help since the bundle itself is not a vector space. I am interested in this question, because otherwise I see no alternative but to establish all the results about integration theory for $L_p^k(M)$ again, i.e. that it is a Banach space, Lebesgue Dominated Convergence Theorem, Fubini/Tonelli etc. That seems a bit exaggerated since intuitively this space is not so fundamentally different. -Construction of the fibre metric: For any $0 \leq k \leq m$ the Riemannian metric may be extended canonically to differential forms in $\Omega^k(M)$ in the following way: For one forms $\omega,\eta \in \Omega^1(M)$ define $g(\omega,\eta):=g(\omega^\sharp, \eta^\sharp)$, where $\sharp:T^*M \to TM$ is the sharp operator with respect to $g$. Then define $g$ on decomposable forms by $g(\omega^1 \wedge \ldots \wedge \omega^k, \eta^1 \wedge \ldots \wedge \eta^k):= \det(g(\omega^i, \eta^j))$. - -REPLY [8 votes]: Let $E$ denote a vector bundle over a manifold $M$ equipped with a metric, and $L_p(E)$ the space of measurable sections of $E$ with finite $L_p$ norm. -Obviously, in general, one can't identify $L_p(E)$ -with an $L_p$ space of vector valued functions. -First assume that $M$ is compact. To understand $L_p(E)$, we use a finite set of trivializations $(U_i, h_i)$ of $E$ which cover $M$. Each trivialization identifies $E|_{U_i}$ with $U_i \times\mathbf R^n$ (or $U_i \times\mathbf C^n$). We choose the trivializations such that the bundle norm is equivalent to the euclidean norm, i.e. bounded from above and below. Then an $L_p$ section of $E$ is equivalent to a set of $\mathbf R^n$ (or $\mathbf C^n$)-valued measurable functions ${f_i}$ on $U_i$ satisfying the transition law, such that $\sum \|f_i\|_p$ is finite. -Using this one can easily extend all the basic theorems to $L_p(E)$. In particular, one shows that $L_p(E)$ is equivalent to the completion of $C^{\infty}(E)$ (the space of smooth -sections on $M$) w.r.t. the $L_p$ norm. -If $M$ is noncompact, we write it as the union of locally finite compact subsets $A_i$, such that the intersections of $A_i$ have zero measure. Then the $L_p$ norm of a section -$s$ is given by -$(\sum \int_{A_i} |s|^p)^{1/p}$. -Then the arguments for the compact case can easily be carried over. (We argue on each -$A_i$, and then combine.)<|endoftext|> -TITLE: A geometric Ramsey problem -QUESTION [23 upvotes]: The following problem seems like one to which the answer could well be known: if so, I'd be interested to have a reference. -How large does n have to be such that among any n points in the plane you can find either m points that are collinear or m points such that no three are collinear? The fact that n is finite follows from Ramsey's theorem: colour triples of points according to whether they are collinear. -However, as with many geometric colourings, far better bounds hold than what one can obtain from the abstract Ramsey theorem. Here are what seem to me to be the trivial bounds. In one direction, an m-by-m grid of points does not contain more than m in a line, but if you choose 2m+1 of the points then you must have three that are collinear. So you need at least $cm^2$ points. (It's not quite obvious that you can choose linearly many points in this grid with no three in a line, but an old idea of Erdős does the trick: assume that m is prime and choose all points (x,y) such that $y\equiv x^2$ mod m. It is not hard to check that this set does not contain three points that are in a line even in the mod-m sense, so certainly not in the integer sense. If m isn't prime, then discard a few points until it is.) -In the other direction, you can just greedily pick points such that no three are collinear. If you reach r points and then cannot extend your set, then all subsequent points lie in one of the $\binom r2$ lines defined by the points so far. Therefore, there must be $cn/r^2$ points in a line, by the pigeonhole principle. It follows that it is enough if $n=cm^3$. -My question is, is one of these two bounds known to be correct (up to $n^{o(1)}$), and if so which? It feels quite close to known incidence results: another possibility is that a simple adaptation of a known argument would answer the question. The one thing that suggests that it might be hard is the fact that it takes a slight effort to find that set of points in the grid with no three in a line. - -REPLY [11 votes]: This question has been answered up to a logarithmic factor by Michael Payne and myself ["On the general position subset selection problem"]. We show that $n \leqslant c m^2 \log m$. The proof employs the Szemeredi-Trotter Theorem to bound the number of collinear triples in a point set. Then we apply known results about independent sets in 3-uniform hypergraphs to conclude the result. -I think the answer should be $cm^2$. That is, every set of at least $cm^2$ points contains $m$ collinear points or $m$ points with no three collinear (for some constant $c>0$).<|endoftext|> -TITLE: Continuous bijective way of representing a line on a plane -QUESTION [6 upvotes]: Is there a function $f(a,b)$ which maps ordered pairs to lines in a plane in a continuous, bijective manner? -Here is the definition I am using for the limit with lines: a sequence of lines $L(1), L(2), \dots$ is said to approach another line $L$ if, for any point $p$ on $L$, the limit as $n\to\infty$ of the distance between $p$ and $L(n)$ is $0$. -If there is no such function, can anyone think of a proof? - -REPLY [11 votes]: There is no such bijection. -A line in the plane is almost the same as a plane through the origin in 3-space (by intersecting with the plane at height 1), except there's one plane through the origin that doesn't give you a line (the z=0 plane). So the space of lines in the plane is homeomorphic to $\mathbb{RP}^2$ minus a point: an open mobius strip! So the question is asking if there is a continuous bijection from the open disk $D$ to the open mobius strip $M$. Invariance of domain implies that a continuous bijection between manifolds of the same dimension is a homeomorphism. $D$ and $M$ are not homeomorphic, so there cannot be a continuous bijection between them.<|endoftext|> -TITLE: A non-formal space with vanishing Massey products? -QUESTION [28 upvotes]: Let $X$ be a polyhedron. For each $n$-dimensional face $f$ of $X$ fix a homeomorphism $\sigma_f:\triangle^n\to f$ where $\triangle^n$ is the standard $n-$simplex so that whenever $f$ is a face of $f'$ the map $\sigma_{f'}^{-1}\sigma_f$ is an affine map $\triangle^n\to\triangle^{n'}$ where $n'=\dim \triangle'$. These maps enable one to speak of forms on the simplices of $X$ with rational polynomial coefficients. A degree $n$ Sullivan piecewise polynomial form on $X$ is a choice of degree $n$ rational polynomial forms, one for each face of $X$, that agree on the intersections of the faces. Such forms form a commutative differential graded algebra (cdga) denoted $A_{PL}(X)$. -$X$ is called formal if the cdga's $A_{PL}(X)$ and $H^\ast(X,\mathbf{Q})$ (with zero differential) are quasi-isomorphic i.e. can be connected by a chain of cdga quasi-isomorphisms. One topological consequence of formality is that if $X$ is formal, all higher Massey products on $H^\ast(X,\mathbf{Q})$ vanish (see e.g. Griffiths, Morgan, Rational homotopy theory and differential forms). I vaguely remember having heard that the converse is false: there are non-formal spaces with vanishing Massey products. I would like to ask if anyone knows a (preferably not too expensive) example of such a space. -Remark: this question can of course be stated in terms of cdga's without referring to spaces: by Sullivan's realization theorem (Infinitesimal computations in topology, \S 8) to give an example it would suffice to construct a non-formal cdga $A$ with $A^0=\mathbf{Q},A^1=0$ all of whose Massey products vanish. However, if there is a geometric example, I'd be interested to know. - -REPLY [4 votes]: Many examples of closed highly-connected manifolds with all Massey products vanishing, yet which are not formal, are given in "The rational homotopy type of $(n-1)$-connected manifolds of dimension up to $5n-3$" by Crowley and Nordström. See e.g. Example 1.11: https://arxiv.org/abs/1505.04184.<|endoftext|> -TITLE: On referee-author communications -QUESTION [51 upvotes]: Every time I referee a paper, I dream of a system which would allow me to ask the author a question without troubling the editors. It would save time for everyone involved, most importanly the referee; in my case it would probably half the refereeing time. -Here is the system I envision: together with the paper, the journal would give the referee a link to a secure web form where the referee could type a question which would automatically be emailed to the corresponding author, who in turn could reply in a similar web form. Of course, the editors should have access to the logs of these exchanges, and the system should be set up so that the exchanges (if any) can only be initiated by the referee. -Question. Has this ever been implemented? Are there reasons this cannot (or should not) be implemented in mathematics journals? - -REPLY [5 votes]: The review system for research projects in Italy (during the last few years; it is possibly going to change now) was based on comparing the opinions of two different anonymous referees. At some point the two referees have to exchange opinions with each other and reach an agreement, while remaining anonymous. In practice, this is achieved by channeling communications through a website based at a central location. I post my comments, the other referee can read them and post his comments, etc. Since we both know that our exchange is watched by some official at the Ministry, we feel some pressure as to not revealing our identity. I think this works pretty well, and should work also for peer review. -Actually, several journals have already an electronic system for reports; at the moment it is a little awkward to use it as a communication tool, because of the many steps involved (the messages must be approved through the chain referee->editor->chief editor->editor->author or something like that), but it should be quite easy to make it faster and transform it into a supervised direct channel between referee and author.<|endoftext|> -TITLE: Can Sierpinski's anisotropic bicolouring of the plane, assuming the continuum hypothesis (CH), be extended to three dimensions? -QUESTION [5 upvotes]: Sierpinski showed that, on the assumption of CH (in fact, equivalently to it), each point in the plane can be coloured (say) black or white so that every section of the plane parallel to the $x$ axis is "almost" white $-$ in the sense that all but countably many points of it are white $-$ while every section parallel to the $y$ axis is almost black. Equivalently, given CH, every line through the origin can be almost white, while every circle centred on the origin is almost black. Is there a corresponding result for three-dimensional space? For example, assuming CH, can we bicolour space so that every plane through the origin is almost white, while every sphere centred at the origin is almost black? - -REPLY [6 votes]: Sierpinski himself proved the following striking version of his theorem for the plane: -CH is equivalent to the statement that $\mathbb R^3$ can be written as $A_x\cup A_y\cup A_z$ where each $A_u$, -$u\in\{x,y,z\}$, has a finite intersection with every line parallel to the $u$-axis. -(See http://www.math.wisc.edu/~miller/old/m873-05/setplane.ps, respectively [Simms, John C.; Sierpinski’s theorem. Simon Stevin 65 (1991), no. 1-2, 69–163]. -I believe this theorem is proved in Sierpisnki's 1951 paper in Fund. Math. 38, pp. 1--13, but I would have to actually walk to the library to verify this (and decipher enough of the french in the article).) -There is actually a planar version of this result of Sierpinski, due to Komjath. -A set $A\subseteq\mathbb R^2$ is a cloud if there is $p\in\mathbb R^2$ such that every line through $p$ meets $A$ in only finitely many points. -CH is equivalent to $\mathbb R^2$ being the union of 3 clouds. -In both cases the striking fact is that we go down from countable to finite, which is somehow less arbitrary, in the light of Joel's comment that the previous constructions can be done in ZFC if countable is replaced by $\lneq 2^{\aleph_0}$.<|endoftext|> -TITLE: Automorphic forms and Galois representations over imaginary quadratic fields: generalizing Taylor's theorem? -QUESTION [10 upvotes]: Let $K/\mathbf{Q}$ be an imaginary quadratic field, with $\sigma \in \mathrm{Gal}(K/\mathbf{Q})$ a generator. Suppose $\pi$ is a cuspidal automorphic representation of $GL_2 / K$ with central character $\omega_{\pi}$. If $\pi_{\infty}$ has Langlands parameter $z\to \mathrm{diag}(z^{1-k},\overline{z}^{1-k})$ for $k \geq 2$ an integer, and $\omega_{\pi}^{\sigma}=\omega_{\pi}$, then a well-known theorem of Taylor (proved in the early 90's) yields a compatible system of two-dimensional $\ell$-adic Galois representations of $\mathrm{Gal}(\overline{K}/K)$ attached to $\pi$. My question is simple: given all the recent work on the fundamental lemma and concomitant progress in the field of automorphic Galois representations, it is yet possible to prove this result without any Galois-invariance assumption on the central character $\omega_{\pi}$? - -REPLY [6 votes]: Various groups of people have thought about/are thinking about this. The natural source of the desired Galois reps. is a $U(2,2)$ Shimura variety. The problem is that the cohomology of this variety is not so easy to understand. The fundamental lemma certainly plays some role in controlling it, but I don't think that by itself it overcomes the key difficulties. (My own understanding of the issues is far from perfect, though.)<|endoftext|> -TITLE: Pach's "Animals": What if the genus is positive? -QUESTION [43 upvotes]: Janos Pach asked a deep question 23 years ago (1988) that remains unsolved today: - -Can every animal—a topological ball in $\mathbb{R^3}$ composed of unit cubes glued face-to-face—be reduced to a single unit cube by adding and deleting cubes, while always maintaining the animal (ball) property? - -("Animal" was an apparently original coinage of Janos's.) -I and my students quickly found irreducible animals, i.e. balls of unit cubes from which no cube can be removed without destroying the topological-ball property. -Here is one of 119 cubes due to Tom Shermer (which I exploded vertically for visualization): - - - -Essentially all our irreducible examples are based on Bing's House with Two Rooms (unbeknownst to us at the time). -So if Pach's question has a positive answer, it requires adding cubes as well as deleting. -This history is recounted in Günter Ziegler's Lectures on Polytopes, Springer, 1995, p.276. -His non-shellability Theorem 8.15 (p.243) is based on these irreducible animals. -So, I finally come to my question, which is essentially a question of shellability: - -Can every (embedded) object constructed by gluing unit cubes face-to-face, regardless of genus, be reduced to a single unit cube by adding and deleting cubes, while always maintaining that the surface is a 2-manifold? - -This is exactly Pach's question, but with the ball-requirement removed. All the irreducible animals I know rely on violating the topological-ball requirement for their irreducibility; so it is (remotely) possible that reduction alone suffices(!). -I am tempted to introduce a new genera to encompass Plantae & Animalia; but I resist. -Any pointers that may lead me to information on the generalization of Pach's question would be greatly appreciated. Thanks! -Addendum, 11 May 2011 (original posting on 2 January 2011). -The problem is now solved (positively): Every animal can be reduced by adding and deleting cubes. -The proof is contained in two papers, the second of which appeared as a tech report in May 2010: "A solution to the animal problem," by Akira Nakamura. Here is the PDF. -The first paper, an earlier 2006 tech report, is called simply, "B-Problem," by Akira Nakamura, Kenichi Morita, and Katsunobu Imai. Here is its PDF. -I would summarize but I do not yet understand the papers, which are presented in terms of "digital topology." - -REPLY [6 votes]: Regarding Pach's original question: Bing's house deals with the analogous question of collapsibility. It is an example of a contractible non-collapsible space. For that problem it is known that if you allow both collapses and anti-collapses, every contractable space can be reduced to a point. This follows from "Simple homotopy theory". I dont know if this result of simple homolopy theory extends to animals built from cubes but they might. -The modified higher genus question seems easier since you allow steps that changes the topology of the animal. -One can ask a stronger question if when you fixed the genus g you can pass between two animals of genus g by such steps. The analogous question for collapsibility seems to relate to simple homotopy invariants of surface groups but I dont know what they are.<|endoftext|> -TITLE: How to make Ext and Tor constructive? -QUESTION [50 upvotes]: EDIT: This post was substantially modified with the help of the comments and answers. Thank you! - -Judging by their definitions, the $\mathrm{Ext}$ and $\mathrm{Tor}$ functors are among the most non-constructive things considered in algebra: -(1) Their very definition requires taking an infinite projective or injective resolution; constructing a homotopy equivalence between two such resolutions requires infinitely many choices. -(2) Injective resolutions are rather problematic in a constructive world (e. g. the proof of "injective = divisible" requires Zorn, and as far as I understand the construction of an injective resolution relies on this fact). -(3) Projective/injective resolutions are not really canonical, so $\mathrm{Ext}$ and $\mathrm{Tor}$ are not functors from "pairs of modules" to "groups", but rather functors from "pairs of modules" to some category between "groups" and "isomorphism classes of groups". This is a problem already from the classical viewpoint. -(4) Projective resolutions are not guaranteed to exist in a constructive world, because the free module on a set needs not be projective! In order to avert this kind of trouble, we could try restricting ourselves to very well-behaved modules (such as, finite-dimensional over a field), but even then we are in for a bad surprise: Sometimes, the "best" projective resolution for a finitely-generated module uses non-finitely-generated projective modules (I will show such an example further below). These can be tricky to deal with, constructively. Mike Shulman has mentioned (in the comments) that injective and projective resolutions (and already the proofs that the different definitions of "projective" are equivalent, and that the different definitions of "injective" are equivalent) require choice - maybe the currently accepted notions of injectivity and projectivity are not "the right one" except for finitely-generated modules? (Cf. also this here.) -On the other hand, if we think about the ideas behind $\mathrm{Ext}$ and $\mathrm{Tor}$ and projective resolutions (I honestly don't know the ideas behind injective resolutions, besides to dualize the notion of projective resolutions), they are (at least partially) inspired by some of the most down-to-earth constructive mathematics, namely syzygy theory. So a natural question to pose is: How can we implement the theory of $\mathrm{Ext}$ and $\mathrm{Tor}$, or at least a part of this theory which still has the same applications as the whole theory, without having to extend our logical framework beyound constructivism? -It is not hard to address the issues (1), (2), (3) above one at a time, at least when it comes to the basic properties of $\mathrm{Ext}$ and $\mathrm{Tor}$: -For (1), the workaround is easy: If you want $\mathrm{Ext}^n\left(M,N\right)$ for two modules $M$ and $N$ and some $n\in\mathbb N$, you don't need a whole infinite projective resolution $...\to P^2\to P^1\to P^0\to M$. It is enough to have an exact sequence $P^{n+1}\to P^n\to P^{n-1}\to ...\to P^1\to P^0\to M$, where $P^0$, $P^1$, ..., $P^n$ are projective. (It is not necessary for $P^{n+1}$ to be projective. Generally, $P^{n+1}$ is somewhat like a red herring when $\mathrm{Ext}^n\left(M,N\right)$ is concerned.) -For (2), the only solution I know is not to use injective resolutions. Usually, things that can be formulated with projective resolutions only can also be proven with projective resolutions only. But this is not a solution I like, since it breaks symmetry. -(3) I think this is what anafunctors are for, but I have not brought myself to read the ncatlab article yet. I know, laziness is an issue... At the moment, I am solving this issue in a low-level way: Never speak of $\mathrm{Ext}\left(M,N\right)$, but rather speak of $\mathrm{Ext}\left(M_P,N\right)$ or $\mathrm{Ext}\left(M,N_Q\right)$, where $P$ and $Q$ are respective projective/injective resolutions. This seems to be the honest way to do work with $\mathrm{Ext}$'s anyway, because once you start proving things, these resolutions suddenly do matter, and you find yourself confused (well... I find myself confused) if you suppress them in the notation. [Note: If you want to read Makkai's lectures on anafunctors and you are tired of the Ghostview error messages, remove the "EPSF-1.2" part of the first line of each of the PS files.] -Now, (4) is my main problem. I could live without injective resolutions, without the fake canonicity of $\mathrm{Ext}$ and $\mathrm{Tor}$, and without infinite projective resolutions, but if I am to do homological algebra, I can hardly dispense with finite-length projective resolutions! Unfortunately, as I said, in constructive mathematics, there is no guarantee that a module has a projective resolution at all. The standard way to construct a projective resolution for an $R$-module $M$ begins by taking the free module $R\left[M\right]$ on $M$-as-a-set - or, let me rather say, $M$-as-a-type. Is this free module projective, constructively? This depends on what we know about $M$-as-a-type. Alas, in general, modules considered in algebra often have neither an obvious set of generators nor an a-priori algorithm for membership testing; they can be as complicated as "the module of all $A$-equivariant maps from $V$ to $W$" with $A$, $V$, $W$ being infinite-dimensional. Some are even proper classes, even in the classical sense. The free module over a discrete finite set is projective constructively, but the free module over an arbitrary type does so only if we allow a weaker form of AC through the backdoor. Anyway, even if there is a projective resolution, it cannot really be used for explicit computations if the modules involved are not finitely generated. Now, here is an example of where non-finitely generated modules have an appearance: -Theorem. If $R$ is a ring, then the global homological dimension of the polynomial ring $R\left[x\right]$ is $\leq$ the global homological dimension of $R$ plus $1$. -I am referring to the proof given in Crawley-Boevey's lecture notes. (Look at page 31, absatz (2).) For the proof, we let $M$ be an $R\left[x\right]$-module, and we take the projective resolution -$0 \to R\left[x\right] \otimes_R M \to R\left[x\right] \otimes_R M \to M \to 0$, -where the second arrow sends $p\otimes m$ to $px\otimes m-p\otimes xm$, and the third arrow sends $q\otimes n$ to $qn$. (This is a particular case of the standard resolution of a representation of a quiver, which appears on page 7 of a different set of lecture notes by the same Crawley-Boevey.) -Now, the problem is that even if $M$ is finitely generated as an $R\left[x\right]$-module, $R\left[x\right] \otimes_R M$ needs not be. -Is there a known way around this? -Generally, what is known about constructive $\mathrm{Ext}$/$\mathrm{Tor}$ theory? Are there texts on it, just as Lombardi's one on various other parts of algebra (strangely, this text talks a lot about projective modules, but gives $\mathrm{Ext}$ and $\mathrm{Tor}$ a wide berth)? Do the problems dissolve if I really use anafunctors? Do derived categories help? Should we replace our notions of "injective" and "projective" by better ones? Or is there some deeper reason for the non-constructivity, i. e. is $\mathrm{Ext}$/$\mathrm{Tor}$ theory too strong? - -Endnote for everyone who does not care about constructive mathematics: Even dropping constructivism aside, I believe that there remain quite a lot of real issues with homological algebra. First there is the matter of canonicity, then there is the problem of too-big constructions (proper classes etc.), the frightening unapproachability of injective covers, the idea that one day we might want to work in a topos where even ZF is too much asked, etc. I am changing the topic to a discussion of how to fix these issues (well, it already is such a discussion), constructivism being just one of the many directions to work in. - -REPLY [16 votes]: I have some sympathy for the question, since I have gotten bothered by the noncanoncity -- although less so by the nonconstructivity -- of the usual definition in my youth. -(These days I'm less bothered by such things.) -Here are a few -ways around it for $Ext$. - -There is a definition going back Yoneda, I think, of $Ext^n(M,N)$ as an equivalence -class of exact sequences $0\to N\to \ldots \to M\to 0$ of length $n+2$ (including $M,N$). -Take a look at Hilton and Stammbach's Homological Algebra. This is quite cumbersome but it solves the dependency on, let alone existence of, suitable resolutions. -This was mentioned already by Mike Shulman. You can take $Ext^n(M,N)= Hom_{D(R)}(M,N[n])$ -in the derived category $D(R)$ of $R$-modules. Once you get used to the formalism, this is much more convenient -than 1 (at least for me). -You can choose a canonical resolution initially for the definition of $Ext^n(M,N)$. Here is one general choice. -Let $F^0(M)=F(M)$ be the free module generated by elements of $M$. We have canonical map -$c_M:F(M)\to M$, let $F^{-1}(M)= F(\ker c_M)$. By continuing in this fashion, we build -a canonical free resolution $\ldots F^{-1}(M)\to F^0(M)\to M\to 0$. Dually, you can use the injective hull* to build an injective resolution of $N$. As several people have pointed out, -some important categories have enough injectives but not enough projectives, so it's good to get used to them. - -Addendum Regarding your original question about how constructive this can be made, -I think for certain classes of modules (e.g. finitely presented modules) over certain classes of rings (e.g. polynomial rings), this is not only possible, but it seems to have -been implemented in some computer algebra packages already. -I notice that the latest Macaulay 2 has commands for Ext and Tor. -*If your are unhappy with this, use instead the double character module -$$I(N)= Hom_{\mathbb{Z}}(Hom_{\mathbb{Z}}(N,\mathbb{Q}/\mathbb{Z}), \mathbb{Q}/\mathbb{Z})$$<|endoftext|> -TITLE: Can breadth hurt a job candidate? -QUESTION [28 upvotes]: I am a junior mathematician, currently on the job market. I have been involved in a number of projects spread over several very different topics of mathematics, some being very mathematical projects that appeared in computer science theory forums (I do not consider TCS my specialty). -I had thought that breadth would make me an appealing job candidate (if I work in two areas, I can apply to jobs listed in both!). However, it was recently mentioned to me that this might be viewed negatively as 'lack of focus' and even hurt my job prospects. Another criticism I have heard is that a research group in area X might be reluctant to hire me out of fear that I'm really more interested in area Y. I have been most troubled by recently witnessing a member of the hiring committee at my current university demonstrated a complete ignorance of TCS forums (i.e. not knowing what FOCS is). I assume this means that this individual would value such publications negligibly at best (in contrast, I'd expect/hope that he would be able to identify the a top journal in an area of math outside his expertise). -Can any mathoverflowers either confirm or dispel these concerns? How do committees evaluate candidates with work in multiple areas (and/or reaching into other fields)? I am also interested in hearing how others have marketed themselves in similar situations. - -REPLY [12 votes]: The majority of answers have suggested that breadth doesn't hurt you, but I have a more pessimistic view, based on my personal experience. I find that it is quite common for people to judge your research ability by what they consider to be your best piece of work. So if you're faced with the choice between -focusing on a single area, and publishing two papers in that area, one mediocre and one outstanding; versus -working in two different areas, and publishing one good paper in each area, -then my belief is that you will usually be perceived to be a better researcher in the former case. Now, it's true that the above dichotomy may be a false one. Perhaps your skills and temperament are such that if you choose to focus in just one area, you get bored and can't produce any good papers. But to the extent that you are faced with that choice, then I think choosing to focus on producing one really outstanding piece of work will generally be better for your reputation as a researcher. -Having said that, I don't necessarily advocate shunning breadth (again, assuming you actually have a choice). First of all, as others have pointed out, depending on what job you're after, your perceived status as a researcher may not be the most important factor. Secondly, in the long run you are almost certainly better off working on things you're interested in, rather than pursuing things that you think will impress other people. Of course, some compromise may be necessary since you do need to find employment somehow. But if your natural temperament inclines towards breadth, then ultimately I think you'll be more productive if you pursue your natural inclinations instead of forcing yourself to specialize in a way that you find distasteful. One advantage of breadth is that you have a good chance of finding connections between areas that specialists would never have seen. It may take you a longer time to produce that outstanding paper, but it could very well end up being even more outstanding than any paper you could have published quickly by specializing.<|endoftext|> -TITLE: Non-standard enlargements, $\zeta(s)$ and analytic continuation -QUESTION [10 upvotes]: Consider an extension of the Riemann zeta function $\zeta(s)$ where $s$ now runs over a non-standard enlargement of the complex plane. -Observe that if $s=\sigma + it$ with $\sigma>1$ real and finite (or at least infinitesimally close thereto), but $t$ infinite, then the summands $n^s= n^\sigma(\cos \ln(n)t + i\sin \ln(n)t)$ still have values infinitesimally close to finite complex numbers. Indeed, by fixing an infinite real $T$, we can obtain from $\zeta(s+iT)$, by passing to standard parts, a convergent standard Dirichlet series. -At least with a sufficiently saturated non-standard enlargement, I believe these same Dirichlet series arise from starting with the standard Euler product and shifting, arbitrarily and independently, all the various factors vertically by various amounts. -My questions: What can one say about the possibility of finding analytic continuations for any or all of these Dirichlet series to larger domains? Does the functional equation speak to this matter? If any of these functions have natural boundary at $s=1$ on the standard view, but analytic continuation one the non-standard sense, I would welcome -any insight into how such could happen. -(Corrections welcome if my question betrays any basic misunderstanding!) - -REPLY [8 votes]: I just saw this on arXiv: Nonstandard Mathematics and New Zeta and L-Functions, the PhD thesis of one Benjamin Clare of U. Nottingham. - -This Ph.D. thesis, prepared under the supervision of Professor Ivan Fesenko, defines new zeta functions in a nonstandard setting and their analytical properties are developed. Further, p-adic interpolation is presented within a nonstandard setting which enables the concept of interpolating with respect to two, or more, distinct primes to be defined. - The final part of the dissertation examines the work of M. J. Shai Haran and makes initial attempts of viewing it from a nonstandard perspective. - -(Corrections welcome if my answer betrays any basic misunderstanding!)<|endoftext|> -TITLE: Wall Crossing in Physics and Mathematics -QUESTION [23 upvotes]: This question is motivated by the current interest of Mathematics and Physics community in Wall Crossing. My questions are : - -What is wall crossing in Physics, what are the reasons for current interest in it. -What is wall crossing in terms of mathematics, what is the reason for interest, is it just physics or some mathematical motivation. - -thanks. - -REPLY [42 votes]: Very roughly speaking, "wall-crossing" refers to a situation where you construct a would-be "invariant" $\Omega(t)$, that would naively be independent of parameters $t$ but actually depends on them in a piecewise-constant way: so starting from any $t_0$, $\Omega(t)$ is invariant under small enough deformations, but jumps at certain real-codimension-1 loci in the parameter space (the walls). You might initially think of this as a kind of quality-control problem in your invariant factory, to be eliminated by some more clever construction of an improved $\Omega(t)$; but at the moment it seems that this is the wrong point of view: there are interesting quantities that really do have wall-crossing behavior. -To name one example of such a quantity: suppose you have a compact Kahler manifold $M$ with an anticanonical divisor $D$, and you want to construct the mirror of $M \setminus D$ following the ideas of Strominger-Yau-Zaslow. As it turns out, one of the essential ingredients you will need is a count of holomorphic discs in $M$, with boundary on a special Lagrangian torus $T(t)$ in $M$ (lying in a family parameterized by $t$). The number of such discs in a given homology class exhibits wall-crossing as $t$ varies, and this wall-crossing turns out to be crucial in making the construction work. This story has been developed by Auroux. -In physics, the wall-crossing phenomena that have been studied a lot recently arose in the context of "BPS state counting". If you have a supersymmetric quantum field theory of the right sort, depending on parameters $t$, you can define a collection of numbers $\Omega(\gamma, t) \in {\mathbb Z}$: they are superdimensions of certain graded Hilbert spaces attached to the theory (spaces of "1-particle BPS states with charge $\gamma$"). These quantities exhibit wall-crossing as a function of $t$. -Moreover, $\Omega(\gamma, t)$ are among the relatively few quantities in field theory that we are sometimes able to calculate exactly, so naturally they have attracted a lot of interest. In particular, they are the subject of the Ooguri-Strominger-Vafa conjecture of 2004, which in some cases relates their asymptotics to Gromov-Witten invariants; the investigation of this conjecture (mostly by Denef-Moore) is what triggered the current resurgence of interest in wall-crossing from the physics side. -A particular case is the $4$-dimensional quantum field theory (or supergravity) associated to a Calabi-Yau threefold $X$ (obtained by dimensional reduction of the $10$-dimensional string theory on the $6$-dimensional $X$ to leave $10-6=4$ dimensional space.) In that case the physically-defined $\Omega(\gamma,t)$ are to be identified with the "generalized Donaldson-Thomas invariants" of $X$, studied by Joyce-Song and Kontsevich-Soibelman among others. The mathematical interpretation of $t$ in that case is as a point on the space of Bridgeland stability conditions of $X$. (If $X$ is compact, the last I heard, this space is not known to be nonempty, but the majority view seems to be that this gap will be filled...) -One focal point for the excitement of the last few years is that a pretty remarkable wall-crossing formula has been discovered, the "Kontsevich-Soibelman wall-crossing formula", which completely answers the question of how $\Omega(t)$ depends on $t$, and seems to apply (in some form) to all of the situations I described above. The formula was rather surprising to physicists; the process of trying to understand why it is true in the physical setting led to some interesting physical and geometric spin-offs, some of which seem likely to be re-importable into pure mathematics.<|endoftext|> -TITLE: Flat map with reduced fibers. -QUESTION [5 upvotes]: Hi. -Let $f:X\rightarrow S$ be a flat, surjective morphism of complex spaces with reduced fibers over $S$ reduced. -Q1: Is $X$ reduced too? -Q2: Is the property " reduced fiber" preserved by base change given by the normalization? -Rk: 1) We have some result of this kind in [EGA4], Cor (3.3.5) p.44. -2) We can apply [Matsumara], Cor(ii) p.189 but with the additional asumptions: $S$ is normal and $X$ locally pure dimensional. -Thanks. - -REPLY [4 votes]: This is a consequence of the following result: if $A \to B$ is a flat local homomorphism of local rings, $A$ and $B/\mathfrak{m}_AB$ are reduced, then $B$ is reduced. Keeping in mind that reduced is equivalent to $R_0$ and $S_1$, this follows from Theorem 23.9 in Matsumura's Commutative Ring Theory. -I don't know what the second question means. -[Edit] I just noticed that the hypotheses in the cited theorem are actually stronger, so that it would imply the result for schemes but not for analytic spaces, at least not immediately. I need to think about it some more.<|endoftext|> -TITLE: Limit of a sequence of polygons. -QUESTION [12 upvotes]: Begin with a polygon $P_0$. -Place two points on every edge of the polygon such that they divide each side equally into three parts. Create a new polygon $P_1$ by connecting all new points with lines. -If we begin with a square and iterate this process, what is the limit as the number of iterations approaches infinity? It is clearly not a circle, but what the correct answer is, I do not know. - -REPLY [3 votes]: As noted, goes back to de Rham. Found also in: - -Georges de Rham, "Sur quelques courbes définies par des equations functionnelles". Univ. e Politec. Torino. Rend. Sem. Mat. 16 1956/1957 101–113. - -English tranlation in my book CLASSICS ON FRACTALS.<|endoftext|> -TITLE: Who first found this characterization of Lebesgue integration? -QUESTION [11 upvotes]: Write $L^1$ for the Banach space $L^1([0, 1])$. Given $f \in L^1$, define $f_1, f_2 \in L^1$ by -$$ -f_1(x) = f(x/2), -\qquad -f_2(x) = f((x + 1)/2). -$$ -Let $I = \int_0^1$. Then $I$ is the unique bounded linear functional on $L^1$ satisfying the equations -$$ -I(\text{constant function at }1) = 1, -\qquad -I(f) = (I(f_1) + I(f_2))/2. -$$ -This, then, is a simple characterization of integration. -Presumably this has been known for a really long time -- maybe for a century? -- but I'm having trouble tracking it down in the literature. I simply can't find it anywhere. -Does anyone know anything about the history of this result? - -REPLY [4 votes]: I don't know if this is of help, but I have seen this idea for defining integration elsewhere, specifically on pages 10-11 of Reed and Simon's Functional Analysis. It would go something like this: let $S$ be the space of step functions obtained as linear combinations of characteristic functions of half-open intervals $[k/2^n, (k+1)/2^n)$. Then $S$ is dense in the space $C$ of bounded piecewise continuous functions continuous to the right, with respect to sup norm. By the two conditions, the definition of $I$ is uniquely determined on $S$ and defines a bounded linear functional on $S$ with respect to sup norm, so it uniquely extends to a bounded linear functional on $C$. This in fact defines the Riemann integral $I$, and it's pretty much exactly what Reed and Simon do, except they don't restrict to dyadic rational endpoints.) -(Then, we can go on and define $L^1[0, 1]$ to be the completion of $C$ with respect to the norm given by $f \mapsto I(|f|)$, and since $I$ is a bounded linear functional also with respect to this norm, it extends uniquely to the completion $L^1[0, 1]$.) -Edit:: Ah, Reed and Simon mention that this approach to the Riemann integral can be found especifically in Dieudonné, Foundations of Modern Analysis, or in Loomis and Sternberg, Advanced Calculus, Addison-Wesley, 1968. Perhaps one can consult these for further history.<|endoftext|> -TITLE: Zeros of a holomorphic function -QUESTION [6 upvotes]: Suppose Ω is a bounded domain in the plane whose boundary consist of m+1 disjoint analytic simple closed curves. -Let f be holomorphic and nonconstant on a neighborhood of the closure of Ω such that -|f(z)|=1 -for all z in the boundary of Ω. -If m=0, then the maximum principle applied to f and 1/f implies that f has at least one zero in Ω. -What about the general case? I believe that f must have at least m+1 zeros in Ω, but I'm not able to prove it... -Thank you - -REPLY [9 votes]: Let $C$ be a curve among the m+1 curves defining the boundary. Then since $f$ is non-constant and analytic, $f(C)$ equals the unit circle. Hence $\frac1{2\pi}\Delta_C \arg f(z) \ge 1$. Doing this for all the curves we have $\frac1{2\pi}\Delta_{\partial \Omega} \arg f(z) \ge m+1$ and so $f$ has $m+1$ roots in $\Omega$ by the argument principle.<|endoftext|> -TITLE: Does local triviality in the fppf topology imply local triviality in the etale topology? -QUESTION [7 upvotes]: Given an algebraically closed field $k$, a smooth group scheme $G$ over $k$ -and a principal $G$-bundle $X \rightarrow Y$, which is locally trivial in the fppf topology. -Is this bundle also locally trivial in the etale topology? -Or do we need some extra conditions for this to be true? Literature references about this subject are very welcome. - -REPLY [17 votes]: The answer is yes. Since smoothness is preserved by flat descent, $X$ is smooth over $Y$. This implies that it has sections locally for the étale topology. -No ground field is needed: $Y$ could be any scheme and $G$ any smooth $Y$-group scheme. -[EDIT] to answer a comment by Keerthi Madapusi Sampath: if you don't assume that $X$ is a scheme but only a sheaf on the fppf site of $Y$, the answer is still yes. In fact, since $X$ is fppf-locally isomorphic to $G$, a theorem of Artin implies that $X$ is an algebraic space. In other words, there is a scheme $X'$ and a morphism $X'\to X$ which is representable, surjective and étale. Now $X'$ must be smooth surjective over $Y$, hence has sections étale-locally, and so does $X$. -If $Y$ is the spectrum of a field, group schemes and torsors are always quasiprojective. For any $Y$, if $G$ is affine over $Y$, then so are all $G$-torsors (in particular they are schemes). On the other hand, Raynaud has constructed a regular two-dimensional (local?) scheme $Y$, an elliptic curve $E\to Y$ and an $E$-torsor which is not a scheme.<|endoftext|> -TITLE: Counting characteristic subgroups -QUESTION [15 upvotes]: I cannot seem to find any results at all on characteristic subgroup growth, even of free groups (and even of $F_2$). By contrast, the growth function of all subgroups of finite index is well-understood, as is the growth function of all normal subgroups of finite index (it's the same as enumerating finite groups). The growth function for normal subgroups clearly provides an upper bound, and one can get a truly ghastly lower bound by, eg, intersecting all subgroups of a given index (eg), but I find it hard to believe one can't do better. -For free groups characteristic subgroups are verbal, but it is not clear how this helps: given a collection of "words" computing the index of the corresponding verbal subgroup seems very hard (not recursively computable?). - -REPLY [16 votes]: 4 Jan 2011: Edited to fix discussion of verbal subgroups -I think $F_2$ is expected to behave differently from higher free groups. For a finite simple group $G$, I think it's expected (known?) that all epimorphisms $F_n \rightarrow G$ are equivalent up to $Aut(F_n)$ when -$n > 2$, but there are many orbits for $n = 2$: in particular, the isomorphism class (orbit under the automorphism group of $G$) of the image of the commutator of the generators is an invariant. -Epimorphisms $F_n \rightarrow G$ are equivlaent to $n$-tuples of generators of elements of $G$ that generate. An $n$-tuple that does not generate $G$ must be contained in a proper subgroup; for a finite simple group, there are not many maximal subgroups of small index, so the vast majority of $n$-tuples actually generate. -If there's only one orbit, the orbit size is rougly $|G|^n/|Aut(G)|$; for nonabelian finite simple groups, the outer automorphism group is small, so the orbit size is $|G|^{(n-1)}$ up to a small constant. -The corresponding characteristic subgroup is the kernel of the -homomorphism $F_n \rightarrow \prod G_i$, where the product is over all epimorphisms to $G$ in the orbit, up to automorphism of $G$. A subgroup of the product of two groups is the kernel of the map of the product to a common quotient; using this, since the factors are simple, this homomorphism is an epimorphism. Therefore, -the characteristic subgroup has huge index, -$|G|^{|G|^{(n-1)/a}}$ for some not large $a$ (about 2 or 3). -Even for $n = 2$, the orbits are fairly large, so the corresponding characteristic subgroup have huge index. -I.e. characteristic subgroups with nonabelian simple composition factors are very sparse. Characteristic subgroups with polycyclic factor group are the vast majority, for several reasons: there are a lot of polycyclic groups, they tend to have large automorphism groups, -and different polycyclic quotients of a free group tend to correlate, so the homomorphism to a product of a number of polycyclic quotients is often far from surjective. Thus, -these characteristic subgroups do not have such a huge index. I suspect that in fact characteristic subgroups with abelian quotient should dominate. If so, the answer is boring, since -such subgroups correspond 1-1 with characteristic subgroups of $Z^n$. This may be why there's not much literature on it. - -Verbal Subgroups and Fully Invariant Subgroups - -A word $w$ in letters $X_1, \dots, X_k$ defines a function $f_w: G^k \rightarrow G$, when $k$-tuples of elements of $G$ are substituted for the $X_i$. As explained in comments, a verbal subgroup $H_V \subset G$ is one that is generated by the images of $f_w$, for $w$ in some set $V$. -In other words, $G/H_V$ is what you get by the identities for $w \in V$ that -$\forall (X_1, \dots X_k)\; \; w = 1$ (as well as the relations of $G$ if $G$ is not free). -A fully invariant subgroup $H$ of a group $G$ is one such that every endomorphism -of $G$ takes $H$ to a subgroup of $H$. Fully invariant subgroups are characteristic, but not necessarily vice versa. -In comments, Andy Putnam recalled that fully invariant subgroups of a free group are the ones that are supposed to be verbal, and he was dubious that characteristic subgroups of free groups are verbal. Here is an example to show he is correct: -that characteristic subgroups of $F_2$ are not necessarily verbal: -We'll start from two epimorphisms $f_1$ from $F_2$ to $A_5$ that are in different orbits of the automorphism group of $F_2$: let $f1$ send the generators go to the 5-cycle -$a = (1,2,3,4,5)$ and the order 2 element $b = (1,2)(3,4)$. The commutator of $a$ and $b$ is a 5-cycle. -Let $f_2$ send the generators to $a$ and to the 3-cycle $c = (3,4,5)$. The commutator of $a$ and $c$ is a 3-cycle. The conjugacy class of the commutator of generators is invariant under automorphisms of $F_2$, so the maximal characteristic subgroups $H_1$ and $H_2$ containing the kernels of $f_1$ and $f_2$ are different. -Now consider any word $w(X_1, X_2, \dots, X_N)$ whose image is contained in $H_1$. -Equivalently, in $F_2/H_1$, this word evaluates identically to $1$. This happens if -and only if the projection to each $A_5$ factor evaluates identically to 1. -Therefore, it also evaluates identically to $1$ in $F_2/H_2$. Any verbal subgroup -contained in $H_1$ is in fact contained in the intersection $H_1 \cap H_2$. -To get the actual maximal verbal subgroup contained in the kernel of an epimorphism -such as $f_1$, -first, choose a representative for each orbit of the action of $Aut(A_5)$ on - the product of copies of $A_5$ for each of the $60^2$ -homomorphisms $F_2 \rightarrow A_5$, form the product of these homomorphisms, - and let $P$ be the image. -$P$ comes equipped with a pair of generators; let $R$ be a set of relators. -Then $R$ is a verbal description for the kernel of this map.<|endoftext|> -TITLE: What does log convexity mean? -QUESTION [24 upvotes]: The Bohr–Mollerup theorem characterizes the Gamma function $\Gamma(x)$ as the unique function $f(x)$ on the positive reals such that $f(1)=1$, $f(x+1)=xf(x)$, and $f$ is logarithmically convex, i.e. $\log(f(x))$ is a convex function. -What meaning or insight do we draw from log convexity? There's two obvious but less than helpful answers. One is that log convexity means exactly what the definition says, no more and no less. The other is the more or less circular one that since the Gamma function is so important, any property that characterizes it is also significant. -The wikipedia article http://en.wikipedia.org/wiki/Logarithmic_convexity -point out that "a logarithmically convex function is a convex function, but the converse is not always true" with the counterexample of $f(x)=x^2$. The only logarithmically convex examples in the article come trivially from exponentiating convex functions, and the example $\Gamma(x)$. -Let me say in advance that I'm less interested in the Gamma function than I am in the notion of log convexity, so this question is not a duplicate of -Importance of Log Convexity of the Gamma Function -A thoughtful answer by Andrey Rekalo to that question, is that functions which can be realized as finite of moments of Borel measures are log convex functions. But I'm more interested in things that are implied by (v. imply) log convexity. -My real motivation is the fact that the Riemann Hypothesis implies that the Hardy function is log convex for sufficiently large $t$. (The Hardy function $Z(t)$ is just $\zeta(1/2+it)$ with the phase taken out, so $Z(t)$ is real valued and $|Z(t)|=|\zeta(1/2+it)|$.) This is in Edward's book 'Riemann's Zeta Function' Section 8.3, in the language that RH $\Rightarrow Z^\prime/Z$ is monotonic. This says that between consecutive real zeros, $-\log|Z(t)|$ is convex. -Any insight would be welcome. - -REPLY [8 votes]: In my opinion, some of the reasons that make log-convexity (l.c.) nice / useful are: - -If $f$ and $g$ are l.c., then $f+g$ and $fg$ are both log-convex (note log-concavity on the other hand is not closed under addition, though as Allen commented above, it is closed under convolution (Prekopá's theorem)) - -L.c. functions are related to completely monotonic functions (i.e., functions for which $(-1)^n f^{(n)}(x) \ge 0$ for all $x > 0$, and $n \in \mathbb{N}$), in that any completely monotonic function is also l.c.; If I recall correctly, one place completely monotonic functions come up naturally is when considering Laplace transforms. Another example: the Laplace transform of any non-negative function is l.c. - -L.c. is closely related to multiplicatively convex functions (i.e., $f(\sqrt{xy}) \le \sqrt{f(x)f(y)}$), and it is known that the logarithmic integral is multiplicatively convex. -Many more nice properties may be found in the book: Convex functions and their applications: A contemporary approach by C. P. Niculescu and L.-E. Persson.<|endoftext|> -TITLE: Are there any good websites for hosting discussions of mathematical papers? -QUESTION [106 upvotes]: I was wondering if there are any websites out there which - -systematically provide space for the discussion of mathematics articles (particularly those on the arXiv, though not necessarily just those), and -have a large enough user base to have some hope of having a good discussion (or at least have hope of attracting one in the near future). - -Of course, MO is not so far from such a site, but is organized around questions, not papers, which leads to rather different discussion in practice. Similarly, some blog posts and nLab pages serve this purpose, but are not created systematically. - -REPLY [4 votes]: In Peer review 2.0 thread, from 2021, Martin Sleziak suggested to me this ten years old question. So, there is Papers$^\gamma$ with source code under CC0 Public Domain Dedication. It gains the popularity slowly, in an answer to another MO question I explain possible reasons behind this and give more examples of online platforms devoted to paper discussions. -Disclaimer: I'm the founder of http://papers-gamma.link.<|endoftext|> -TITLE: Open questions in Riemannian geometry -QUESTION [42 upvotes]: What are some major open problems in Riemannian Geometry? I tried googling it, but couldn't find any resources. - -REPLY [7 votes]: Here are two more to add to the listed above: -Conjecture of LeBrun and Salamon : -Quaternionic-Kahler metrics whose universal covers have only discrete isometry groups? -Alekseevsky Conjecture: -http://www2.math.ou.edu/~mjablonski/math/<|endoftext|> -TITLE: Computing homotopies -QUESTION [20 upvotes]: Oftentimes, in the standard algebraic topology books (May, Switzer, Whithead, for instance), there are tricky little proofs that depend on proving that two maps are homotopic. This is comparable to the way we build homotopies, lifts, etc. combinatorially in simplicial homotopy theory, but for some reason I never really acquired the skill-set (maybe the intuition?) to come up with these homotopies in the topological case. I'm just mystified how these little formulas are pulled out of thin air. -Am I missing a key technique that's often taught early-on in an algebraic topology course? Is it tricky even with practice? Have there been any papers that focus on systematic ways of generating these things? -I also noticed that in May's book, he oftentimes writes out explicit formulas for his homotopies, sometimes in a way that obscures the issue at hand (for instance, there is a homotopy that is described by an explicit formula, but it's nothing more than an explicit "representative of the natural homotopy" between the identity map and the constant map on a contractible based space.) How often can these seemingly arbitrary formulas be replaced with more canonical descriptions? (This last question is a soft question to people with experience in topology) - -REPLY [6 votes]: @Harry Gindi: I had the same problem in the 1960s when writing the first 1968 edition of my book now Topology and Groupoids. Then I was looking at papers of Puppe, and found I had no idea how to construct some of their "diagrammatic" homotopies. with different formulae in each bit. This was part of the motivation for finding the gluing theorem for homotopy equivalences, which is now a standard result in abstract homotopy. My answer to another question illustrates how one gets by the methods there explicit homotopies. -There was the same motivation for using double groupoids in homotopy theory. Using the double groupoid defined here one can enrich the category of Hausdorff spaces over the category of these double groupoids with connection (though this has not been written up in detail) and this can be applied to construct homotopies. For some calculations in double groupoids with connections, see Chapter 6 of Nonabelian Algebraic Topology, particularly on rotations (p.169). -One can also enrich the category of filtered spaces over the symmetric monoidal closed category of crossed complexes, and this again enables calculations of homotopies. As said on p.322 of Nonabelian Algebraic Topology (pdf available), this requires further study. -I am not so sure that the work on model categories helps so directly with the calculation of homotopies, unles the cylinder object is used explicit;ly. The use of homotopies is central to Homological Perturbation Theory, and they are used in Graham Ellis' Homological Algebra Programs, in particular for constructing resolutions by induction with a contracting homotopy. -This paper uses higher homotopy groupoids to discuss Toda brackets. -My own work has largely used cubical methods as a background for conjectures and proofs, because of the ease of describing multiple compositions, and homotopies. -June, 2014 I remember a comment of Raoul Bott which I overheard at the 1958 ICM: "Grothendieck was prepared to work very hard to make things tautological!" -March 2016 The following picture - -is part of the argument for proving a formula $\sigma (u +_2 v)= \sigma u +_1 \sigma v$ where $u,v$ are homotopy classes rel vertices of maps $I^2 \to X$ which take the edges to a subspace $A$ and the vertices to a set $C$ of base points, $+_1, +_2$ denote respectively compositions in the vertical and horizontal directions, and $\sigma$ is what is called a "rotation". The diagram is reinterpreted by using the interchange law and rules among the peculiar symbols called "connections". The formula implies the existence of a homotopy and in principal gives it explicitly, but that would be very difficult to write out in terms of the usual type of formulae.<|endoftext|> -TITLE: Flat Module and Torsion-Free Module -QUESTION [27 upvotes]: All rings in this question are integral. -It is known that flat modules are torsion-free. Conversely, torsion-free modules over Prüfer domain (in particular, Dedekind domain) are flat, please see here. My questions are: - - -Is there a general condition under which a torsion free module is flat? -What is the simplest example of torsion-free non-flat module? - - -edit: Since the example in Georges's answer is the coordinate ring of a singular curve, I want to ask the same questions for the coordinate ring $R$ of a smooth algebraic variety: - - -Is there a general condition under which a torsion free module over $R$ is flat? -what is the simplest example of torsion-free non-flat module over $R$ - -REPLY [5 votes]: If the question is only for commutative domains, then yes, torsion-free=flat iff it is Pruefer (see e.g. Fuchs/Salce, Thm. VI.9.10, actually due to Warfield), as was mentioned earlier. If not only domains are in question, then it can be said that Warfield showed that it is true whenever the ring's localizations at maximal ideals are valuation rings (let's call them Pruefer rings). I haven't checked if those are the only commutative rings for which this is true. If one leaves the realm of commutative rings, there is a much bigger class, so-called RD-rings, for which it is true. These are the rings for which Relative Divisibility is the same as purity (just as over the integers). An example is the first Weyl algebra over any field of char 0 (this is even a domain, though not commutative, so possibly not of interest to the poser of the question).<|endoftext|> -TITLE: Collatz conjecture for numbers of th form $2^n +1$ -QUESTION [28 upvotes]: Everybody has heard of the Collatz conjecture and it is a nice programming exercise to write a function, that calculates for a given number $n$ the number of iterations it takes until one reaches $1$. However if one restricts to numbers of the form $2^n+1$ one gets the following sequence of integers (NO matches in oeis). It starts with -7,5,19,12,26,27,121,122,35,36,156,113,52,53,98,99,100,101,102, -72,166,167,168,169,170,171,247,173,187,188,251,252,178,179,317, -243,195,196,153,154,155,156,400,326,495,496,161,162,331,332,408, -471,410,411,337,338,339,340,553,479,480,481,482,483,559,560,561, -562,563,564,565,566,567,568,569,570,571,572,573,574,575,576,626, -578,628,629,630,631,583,584,634,635,636,637,894,895,640,641,898,643 -"Usually" it grows by $1$ and at some positions it takes a completely different value. Then sometimes it jumps back as if there was never a different value involved (like 575,576,626,578) This seemed to me a bit strange/interesting and funny.It there anything known about this special sequence. Maybe there is a characterization of those positions, where this sequence grows by $1$. I am not sure, how to make a well posed question out of this. -EDIT: and there is the same behavior for numbers of the form $2^n-1$ - -REPLY [7 votes]: You can find some closed formulas resembling your problem in -Andrei, S., Kudlek, M., Niculescu, R., S.: Some results on the Collatz problem. Acta Informatica No. 37, Vol. 2, pp. 145-160 (2000) -http://www.springerlink.com/content/aw10yk5gr59l0n1e/ -Hope it helps. -EDIT: (To complement the answer) You should also check the references in the -chapter I.8 ``Consecutive numbers with the same height" in -The Dynamical System Generated by the 3n +1 Function - -Gunther J. Wirsching - -Lec. Notes in Mathematics v. 1681 -There you'll find that there is a infinite number of very long sequences of consecutive numbers for which your Collatz function is constant, which is -also very interesting.<|endoftext|> -TITLE: is $2$ a $2^n$-th power mod $p$ ? -QUESTION [5 upvotes]: Euler's proof that the fifth Fermat number is composite begins with the following argument. If $p$ divides $F_n$, then the order of $2$ in $(\mathbb Z/p\mathbb Z)^*$ is exactly $2^{n+1}$. Hence $p\equiv1$ mod $2^{n+1}$. If $n\ge2$, this implies that $2$ is a square mod $p$, $2=\omega^2$. Then the order of $\omega$ is exactly $2^{n+2}$, hence $p\equiv1$ mod $2^{n+2}$. This limits the set of prime divisors to be tested. For $n=5$, we find $p=257, 641, ...$, in which the first one is impossible because it is $F_3$ and the Fermat numbers are pairwise coprime. Thus the first candidate is $641$, which turns out to divide $F_5$. -Of course, the situation is not so simple for larger values of $n$. But I wander how far this argument could be pushed forward. Because a prime divisor satisfies $p\equiv1$ mod $2^{n+2}$, $2$ should be a $4$-th power if and only if $2^{\frac{p-1}{4}}\equiv1$ mod $p$. If $2=\theta^4$, then ...(bla-bla)... $p\equiv1$ mod $2^{n+3}$. So my question is - -Let $m\ge2$ be given. Assume that $p-1$ is divisible by a large enough power of $2$. Does this imply that $2$ is a $2^m$-th power mod $p$ ? - -Of course, I am not so naive as to look forward for an elementary solution to the long-standing problem of the primality of some $F_n$'s, but such an approach can provide nice exercises in undergraduate classes. - -REPLY [12 votes]: There is no $N$ such that $p \equiv 1 \mod N$ implies that $2$ is a fourth power modulo $p$. -Proof: Suppose otherwise. Without loss of generality, suppose that $4$ divides $N$. -Let $K$ be the field $\mathbb{Q}(2^{1/4})$. If $p \equiv 1 \mod N$, then by hypothesis $x^4-2$ has a root in $\mathbb{F}_p$. Moreover, since $p \equiv 1 \mod 4$, there is a primitive $4$th root of unity in $\mathbb{F}_p$, so $x^4-2$ splits in $\mathbb{F}_p$. -So the prime $p$ splits in the ring of integers of $K$. Recall that, if $K$ and $L$ are two number fields, and every prime which splits in $L$ splits in $K$, then $L$ embeds into $K$. Now, $p \equiv 1 \mod N$ if and only if it splits in the cyclotomic field $\mathbb{Q}(\zeta_N)$. So we would have that $K$ embeds in $\mathbb{Q}(\zeta_N)$. But this contradicts that $\mathbb{Q}(\zeta_N)/\mathbb{Q}$ is abelian, and $K/\mathbb{Q}$ is not Galois.<|endoftext|> -TITLE: What's about "quantum modular forms"? -QUESTION [19 upvotes]: Do you know where one could read on "Modular Forms, K-theory and Knots"? The combination of themes sounds thrilling! -Edit: Zagier's paper on "quantum modular forms" will be published in Clay's volume dedicated to Connes anniversary. -Edit: Copies of the fascinating article circulate in the web. If asked by email, I would help finding it. -Edit: a survey + lecture notes by Ken Ono on harmonic Maass forms. - -REPLY [6 votes]: There is an article Conformal Field Theory and Torsion Elements of the Bloch Group by W. Nahm in the book Frontiers in Number Theory, Physics, and Geometry II. This is a reference for one direction in the abstract of the talk by Zagier.<|endoftext|> -TITLE: Algorithm for summing certain sums involving the floor function -QUESTION [7 upvotes]: I managed to find an efficient algorithm (polynomial time in $P$,$Q$ and $N$) for the following sum: -$$\sum_{i=0}^{N} \lfloor iP/Q \rfloor,$$ -where $P,Q$ are relatively prime positive integers. When $N=Q-1$ it's fairly easy to get a simple closed form, and this is also well-known. The details of the general algorithm can be found here http://mathforum.org/library/drmath/view/73120.html. It is a very nice algorithm that resembles the standard GCD algorithm somewhat. -I'm trying to generalize this algorithm to sums of the following form: -$$\sum_{i=0}^{N} (i^{k0}) \cdot (\lfloor (iP/Q) \rfloor)^{k1},$$ -where $k_0$, and $k_{1}$ are non-negative integers. Does anyone have any good ideas for this? -The algorithm doesn't have to be polynomial in $k_{0}$ and $k_{1}$, and algorithms for small values of $k_{0}$ and $k_{1}$ would also be interesting (such as $k_{0} = 0$). Obviously, $k_{1} = 0$ is not interesting. - -REPLY [2 votes]: It seems to me that at least for $k_0=0$ these sums always satisfy a linear recurrence. Maybe one can guess the general form of this recurrence. For example (using FriCAS): - -(1) -> )expose RECOP -(1) -> floorsum(p, q, k, n) == reduce(+, [floor(i*p/q)^k for i in 0..n]) -(2) -> guessEq(p, q, k) == (N := 50; while empty?(e := guessPRec([floorsum(13, 17, k, n) for n in 0..N], safety==100, maxDegree==0)) repeat N := N+1; getEq first e = 0) -(3) -> guessEq(13,17,0) - (3) - f(n + 1) + f(n) + 1= 0 -(4) -> guessEq(13,17,1) - (4) - f(n + 18) + f(n + 17) + f(n + 1) - f(n) + 13= 0 -(5) -> guessEq(13,17,2) - (5) - - f(n + 35) + f(n + 34) + 2f(n + 18) - 2f(n + 17) - f(n + 1) + f(n) + 338= 0 -(6) -> guessEq(13,17,3) - (6) - - f(n + 52) + f(n + 51) + 3f(n + 35) - 3f(n + 34) - 3f(n + 18) + 3f(n + 17) - + - f(n + 1) - f(n) + 13182 - = - 0 -(7) -> guessEq(13,17,4) - (7) - - f(n + 69) + f(n + 68) + 4f(n + 52) - 4f(n + 51) - 6f(n + 35) + 6f(n + 34) - + - 4f(n + 18) - 4f(n + 17) - f(n + 1) + f(n) + 685464 - = - 0 -(9) -> guessEq(13,17,5) - (9) - - f(n + 86) + f(n + 85) + 5f(n + 69) - 5f(n + 68) - 10f(n + 52) - + - 10f(n + 51) + 10f(n + 35) - 10f(n + 34) - 5f(n + 18) + 5f(n + 17) - + - f(n + 1) - f(n) + 44555160 - = - 0 -(12) -> guessPRec [1,13,338,13182,685464,44555160] - (12) [[f(n): - f(n + 1) + (13n + 13)f(n)= 0,f(0)= 1]] -(13) -> guessPRec [1,18,35,52,69,86] - (13) [17n + 1]<|endoftext|> -TITLE: What are Galois Categories used for? -QUESTION [13 upvotes]: Galois categories are introduced (for the first time?) in SGA1, but here's an English introduction that's available online: http://www.math.uchicago.edu/~may/VIGRE/VIGRE2009/REUPapers/Lynn.pdf -It seems that Galois Categories are a way of axiomatizing all the Galois correspondences in the various fields: Galois theory for fields, Galois theory for covers, Galois theory tame covers and so forth. -What is the benefit, if at all, of this formalism? Is it just to outline the commonalities of these seemingly different topics, or is there some applicable virtue to this language? - -REPLY [16 votes]: (I only just saw this one year on!) I find your question very strange. Grothendieck gives a simple categorical formulation of a situation that encompasses the three main examples of Galois theoretic machines. That means he shows what makes things really tick... isn't that good enough for you! He does this with the clearly stated aim of developing a fundamental group for schemes, and the theory gives that and a lot more. If you go to the slightly wider results on the fundamental groupoid of categories of locally finite sheaves, that is a first step towards his Pursuing Stacks, the letters to Larry Breen, and enroute for his Longue Marche. -In another direction it provides a first step towards the Joyal-Tierney theory of locales etc. and their relation with toposes. It provides a background for all of Jacob Lurie's work on higher toposes, and I could go on with fundamental groups of toposes, homotopy theory of toposes. SGA1 is the key for understanding a large part of modern mathematics. -Grothendieck's methodology was always to seek the clarity that came from abstraction and generalisation. His aim was not only to solve problems (say in algebraic geometry) but to understand as fully as possible their solution and why they worked.<|endoftext|> -TITLE: What objects do the cusps of Modular curve classify? -QUESTION [22 upvotes]: We know that a modular curve has a moduli interpretation, i.e, it classify elliptic curves with additional structure. But after we add cusps on it, is the cusps also has a moduli interpretation? What objects do the cusps classify? - -REPLY [27 votes]: If you just want a classical picture over the complex numbers, the objects lying over the cusp points are Néron polygons equipped with some extra structure. To make a Néron $n$-gon, you take $\mathbb{Z}/n\mathbb{Z} \times \mathbb{P}^1$ ($n$-copies of the 2-sphere), and glue the components into a circular chain, with 0 in one sphere glued transversely to $\infty$ in the next. The free action of $\mathbb{Z}/n\mathbb{Z}$ is part of the structure of the $n$-gon. -The nature of the extra structure depends on the moduli problem describing the modular curve. Here are some examples: - -A Néron 1-gon is a nodal cubic curve, so the object lying over the cusp of $X(1)$ is a nodal projective line equipped with a point in the smooth locus. Silverman gives a description of an analytic neighborhood of the cusp in his Advanced Topics book, in the section on the Tate curve. There is a more formal description in chapter 8 of Katz-Mazur. -For $X_0(N)$, the smooth locus parametrizes elliptic curves equipped with a cyclic subgroup of order $N$, and for each $m|N$ one has cusps that parametrize $m$-gons equipped with a distinguished point (the identity element) and a cyclic subgroup of order $N$ in the smooth locus. -For $X_1(N)$, the smooth locus parametrizes elliptic curves equipped with a point of order $N$, and for each $m|N$ one has cusps that parametrize $m$-gons equipped with a distinguished point (the identity element) and point of order $N$ in the smooth locus. -For $X(N)$, the smooth locus parametrizes elliptic curves equipped with an isomorphism from $(\mathbb{Z}/N\mathbb{Z})^2$ to the $N$-torsion, and for each $m|N$ one has cusps that parametrize $mN$-gons equipped with an isomorphism from $(\mathbb{Z}/N\mathbb{Z})^2$ to the $N$-torsion. - -For each of these moduli problems, there are explicit formulas for the precise number of cusps for each $m$. For $X_0(N)$, you can find it in Shimura, section 1.6. For the others I think it isn't too hard to derive. -As others mentioned, Deligne-Rapoport's compactified moduli problems are described using the notion of generalized elliptic curve, which is a flat proper family of connected curves of arithmetic genus 1 and at most nodal singularities, equipped with a distinguished section and a "group law". Unlike the smooth case, the group law is a necessary extra datum here because we wouldn't have a uniquely defined $\mathbb{Z}/n\mathbb{Z}$-action on the $n$-gon over a cusp without it.<|endoftext|> -TITLE: analogues of Cayley plane as homogenous spaces -QUESTION [5 upvotes]: The Cayley projective plane $\mathbb{OP}^2$ can be defined as a homogenous space $\mathrm{F_4/Spin(9)}$, where $\mathrm{F_4}$ is the compact exceptional simple Lie group. The other possible approach is to define it as equivalence classes of (special) triples of octonions, define a Riemannian metric on it and prove that the group of isometries is compact Lie group of type $\mathrm{F_4}$. This is done in this article. There the authors study also other spaces similar to $\mathbb{OP}^2$, namely they consider the octionionic hyperbolic plane,the octonionic projective plane with indefinite signature and analogue of projective plane made up from split octonions. -My question is: What are the "homogeneous presentations" of these spaces? - -REPLY [3 votes]: Disclaimer: I have not read the paper, but I think that this what's going on. -There are three real forms of the complex simple Lie algebra of type $F_4$. Each such real form will have (real) Lie subalgebras whose complexification is $\mathfrak{so}(9;\mathbb{C})$. Each such pair gives rise to homogeneous 16-dimensional pseudoriemannian manifolds. -This construction certainly accounts for the octonionic projective plane and its noncompact dual, as for the octonionic hyperbolic plane. I have not worked out the other cases mentioned in the question.<|endoftext|> -TITLE: Geometric Realization of a Simplicial Category -QUESTION [6 upvotes]: Let $S:\varDelta^{op}\to (cat)$ be a functor where the category on the right is the category whose objects are categories with cofibrations and morphisms are exact functors(from Waldhausen's paper, Algebraic K-Theory of spaces). Waldhausen talks of the geometric realization of such a simplicial category. In the case of a simplicial set, I know how to construct the geometric realization. However, in this case, $S_n$ is a category. It is not clear to me if he is assuming that this is a small category for each n, in which case we could proceed to construct the geometric realization as in the case of a simplicial set. If each $S_n$ is not a small category, then is it still possible to define the geometric realization of $S$? - -REPLY [6 votes]: As Buehler states in the comments, Waldhausen is taking the nerve degreewise, and then taking the diagonal of the resulting bisimplicial set. This is a model for the homotopy colimit of the simplicial diagram of nerves. -Waldhausen mentions the question of smallness himself in a remark on p. 14 of Algebraic K-theory of spaces. As he observes, it is only necessary to assume that his categories with cofibrations and weak equivalences are small up to weak equivalence.<|endoftext|> -TITLE: Flux through a Mobius strip -QUESTION [20 upvotes]: A friend of mine asked me what is the flux of the electric field (or any vector field like -$$ -\vec r=(x,y,z)\mapsto \frac{\vec r}{|r|^3} -$$ where $|r|=(x^2+y^2+z^2)^{1/2}$) through a Mobius strip. It seems to me there are no way to compute it in the "standard" way because the strip is not orientable, but if I think about the fact that such a strip can indeed be built (for example using a thin metal layer), I also think that an answer must be mathematically expressible. -Searching on wikipedia I found that -http://en.wikipedia.org/wiki/Mobius_resistor -A Möbius resistor is an electrical component made up of two conductive surfaces separated by a dielectric material, twisted 180° and connected to form a Möbius strip. It provides a resistor which has no residual self-inductance, meaning that it can resist the flow of electricity without causing magnetic interference at the same time. -How can I relate the highlighted phrase to some known differential geometry (physics, analysis?) theorem? -Thanks a lot! - -REPLY [8 votes]: David Speyer already explained it very well. But if you want more details of the explicit calculations, you can look here: -http://www.math.uwaterloo.ca/~karigian/teaching/multivariable-calculus/moebius.pdf -I made this handout many years ago for a multivariable calculus class. Enjoy.<|endoftext|> -TITLE: An equation in the free associative ring -QUESTION [5 upvotes]: Let $X$ be an alphabet, $u,v,p,q,r,s$ be words in the alphabet $X$. I am looking for four elements in the free associative ring $R$ (i.e. four linear combinations of words in $X$) $x,y,z,t$ such that $$u-v=x(p-q)y+z(r-s)t.$$ Is this problem decidable? -The problem is motivated by the need to define an analog of Dehn functions for associative rings. In groups, the Dehn function is recursive iff the word problem is decidable. The question (asked by E. Zelmanov) is whether the same is true for rings. - Update: An easier problem: is there an algorithm for solving the equation -$$u-v=x(p-q)y$$ -(one summand)? - -REPLY [4 votes]: If I understand your question, the answer to your easier problem is yes. (I'm assuming that by the free associative ring over $X$ you mean to take coefficients of words over the integers.) -First, you can order words in the alphabet, initially by degree, and then by ordering elements in $X$ and using a lexicographical ordering on words of equal degree. -Case 1: $u=v$. Then if $p=q$ you can let $x,y$ be arbitrary. If $p\neq q$ then either $x=0$ or $y=0$. -Case 2: $u\neq v$. Clearly $p\neq q$. Without loss of generality, we may assume that $u$ is of larger order than $v$, and also that $p$ is of larger order than $q$ (replacing $x$ by $-x$ if necessary). -Let $x'$ be the term from $x$ with largest order, and let $x''$ be the term (with non-zero support) with smallest order (these might agree), and similarly define $y',y''$. We must then have $x'py'=u$ and $x''qy''=v$. There are thus only finitely many choices for $x',x'',y',y''$. -But then there are only finitely many choices for terms between $x'$ and $x''$ if we limit ourselves to words in elements of $X$ appearing in $u,v,p,q$. If $x$ involved a term with an element of the alphabet $X$ not appearing in those four words, a simple argument tells us that $x(p-q)y$ would have a term that cannot cancel involving that variable, hence could not equal $u-v$. Thus, there are only finitely terms to try (and letting the coefficients be arbitrary constants, you get a system of linear equations).<|endoftext|> -TITLE: Splitting infinite sets -QUESTION [21 upvotes]: There are two questions here, an explicit one, and another (more vague) one that motivates it: -I am pretty certain the following should have a negative answer, but at the moment I'm not seeing how to argue about this and cannot locate an appropriate reference. - -In set theory without choice, suppose - $X$ is an infinite set such that for - every positive integer $n$, we can - split $X$ into $n$ (disjoint) infinite - sets. Does it follow that $X$ can be - split into infinitely many infinite sets? What would be a reasonably weak additional assumption to ensure the conclusion. - -("Reasonably weak" would ideally be something that by itself does not suffice to give us that $X$ admits such a splitting, but I am flexible.) -This was motivated by a question at Math.SE, namely whether an infinite set can be partitioned into infinitely many infinite sets. This is of course trivial with choice. In fact, all we need to split $X$ is that it can be mapped surjectively onto ${\mathbb N}$. -However, without choice there may be counterexamples: A set $X$ is amorphous iff any subset of $X$ is either finite or else its complement in $X$ is finite. It is consistent that there are infinite amorphous sets. If $X$ is infinite and a finite union of amorphous sets, then $X$ is a counterexample. The question is a baby step towards trying to understand the nature of other counterexamples. -Note that any counterexample must be an infinite Dedekind finite (iDf) set $X$. One can show that for any iDf $X$, ${\mathcal P}^2(X)$ is Dedekind infinite. For any $Y$, if ${\mathcal P}(Y)$ is Dedekind infinite, then $Y$ can be mapped onto $\omega$ (this is a result of Kuratowski, it appears in pages 94, 95 of Alfred Tarski, "Sur les ensembles finis", Fundamenta Mathematicae 6 (1924), 45–95). As mentioned above, our counterexample $X$ cannot be mapped onto $\omega$, so ${\mathcal P}(X)$ must also be an iDf set. -The second, more vague, question asks what additional conditions should a counterexample satisfy. - -REPLY [8 votes]: Define a permutation model of ZFA as follows. Starting as usual (Ch 4 of Jech's Axiom of Choice) from a well-founded model $\mathcal M$ of ZFAC with infinite set $A$ of atoms, let $G$ be the group of all permutations of $A$; so $G$ can be identified with the group of all automorphisms of $\mathcal M$. For each finite partition $T$ of $A$, let $G_{(T)}$ be the group of permutations in $G$ which fix each element of $T$ (meaning $\sigma\in G_{(T)}$ iff for each $B\in T$ we have that $b\in B$ implies $\sigma b\in B$). Let $\mathcal F$ be the set of subgroups of $G$ which contain $G_{(T)}$ for some finite partition $T$; then $\mathcal F$ is a normal filter of subgroups of $G$, and contains the stabilizer subgroup of each atom in $A$. As usual, a set or atom $x\in\mathcal M$ is called symmetric if its stabilizer subgroup is a member of $\mathcal F$, and we let $\mathcal N$ be the class of hereditarily symmetric elements of $\mathcal M$. -Then $\mathcal N$ is a model of ZFA providing a counterexample. The model $\mathcal N$ has all the finite partitions of $A$ found in $\mathcal M$, but every infinite partition of $A$ into non-singletons would fail to be symmetric.<|endoftext|> -TITLE: Applications of lax 2-limits which are not pseudo 2-limits -QUESTION [6 upvotes]: One application of pseudo 2-limits (bilimits) in algebraic geometry is already found in the definition itself of stacks with values in a 2-category admitting bilimits (i.e. a discrete 1-cell-contravariant pseudo functor defined on a site). Of course there is the general notion of descent, etc. and all of its applications. There are also 2-colimit generalisations of aspects of Grothendieck's Galois theory (from SGA1) and aspects of SGA2. -The question is then whether there are "interesting" situations (not necessarily from algebraic geometry) where one must use lax 2-limits and cannot get away with just using pseudo 2-limits? (Or 2-colimits?) - -REPLY [7 votes]: First I should point out that (as you may know) in most of the 2-categorical literature, "bilimit" and "pseudo (2-)limit" are not used synonymously. The former has a cone which commutes up to specified isomorphism, and has a universal property expressed by an equivalence of categories, while the latter also has a cone which commutes up to specified isomorphism, but has a universal property expressed by an isomorphism of categories. -The nLab uses "2-limit" to mean "bilimit," and so will I from now on. -Now, at least if one allows weighted 2-limits, then there is no situation in which one "must" use lax 2-limits and cannot get away with ordinary ones, since every weighted lax 2-limit can be identified with a non-lax weighted 2-limit which uses a modified weight. -However, lax 2-limits do occur naturally in some places, such as the Eilenberg-Moore object of a monad in a 2-category. Note that weighted 2-limits which are neither unweighted ("conical") 2-limits or lax versions of conical 2-limits do also occur frequently, such as the comma categories mentioned in the comments.<|endoftext|> -TITLE: Are there non-projective normal surfaces which are rational? -QUESTION [9 upvotes]: Every non-singular complete surface is projective. On the other hand, there are non-projective complete surfaces (see e.g. Excercise II.7.13 of Hartshorne) - and there are such examples where the surface is also normal (see e.g. this ). All the examples I have seen of complete normal non-projective surfaces are non-rational. Hence the question: are there (complete) rational non-projective normal surfaces? -Edit: I just saw a previous question which asked for examples of normal non-projective varieties. So I guess this is a sub-question of that one. - -REPLY [5 votes]: Nagata constructs a normal complete rational surface in the paper Existence theorems for nonprojective complete algebraic varieties (see Section 4). His construction uses a blow-up of the plane in 12 points in special position.<|endoftext|> -TITLE: What is the generic poset used in forcing, really? -QUESTION [22 upvotes]: I'm not a set theorist, but I understand the 'pop' version of set-theoretic forcing: in analogy with algebra, we can take a model of a set theory, and an 'indeterminate' (which is some poset), and add it to the theory and then complete to a model with the desired properties. I understand the category theoretic version better, which is to take sheaves (valued in a given category $Set$) over said poset with a given Grothendieck topology (the double negation topology). The resulting topos is again a model of set theory, but now has the properties you want, absent from the original $Set$. -But what is this poset, really? Is it the poset of subobjects of the set you want to append to your theory/model (say a set of a specified cardinality, or some tree with a property)? Is it related to a proof of the property you are interested in? To clarify, I'm not interested in the mechanical definition of an appropriate generic poset, but what it is morally. Bonus points for saying what it 'is' before and after forcing, if this even makes sense. - -REPLY [12 votes]: Let me add some view that hasn't really been mentioned above, namely that of Boolean valued models. Every poset has a completion, which is a complete Boolean algebra. -To each poset we can define a forcing language that essentially looks like first order -logic, only that we use so called names (which depend on the poset and stand for elements of the forcing extension) as constants. -Now the elements of the complete Boolean algebra can be considered as truth values of statements in the forcing language. -Generic filters of the poset and its completion are in 1-1 correspondence to each other. -A filter in the complete Boolean algebra corresponds to a consistent theory in the forcing language. -So, choosing a generic filter for the poset over a model of set theory corresponds to -choosing a "generic" theory in the forcing language, the theory consisting of -statements in the forcing language that are true in the corresponding generic extension. -Actually, with this approach it is not even necessary to really go to a generic extension -at all: If you want to show that CH is not provable in ZFC, all you need to do is construct -a poset so that in the corresponding forcing language all ZFC axioms have truthvalue 1 in -the corresponding complete Boolean algebra and CH does not. -Since everything that follows from ZFC has truthvalue 1 as well, CH does not follow from ZFC. Now, of course, the computations and arguments you have to do to push through this program are just as difficult as in the case of the usual approach.<|endoftext|> -TITLE: Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z[t]$ be unbounded as n varies? -QUESTION [8 upvotes]: We've recently seen this question: Can the number of solutions $ab(a+b+1)=n$ for $a,b,n \in \mathbb{Z}$ be unbounded as $n$ varies? It appears initially plausible that the answer is yes, but evidently there are good reasons to believe that the answer might be no. The same question and lack of certainty applies equally to $$ab(a+b-1)=n.$$ -I wonder about the analogous situation with $\mathbb{Z}$ replaced by $\mathbb{Z}[t]$ - -Is there a known or conjectured upper bound for the number of solutions $ab(a+b-1)=n$ for $a,b,n \in \mathbb{Z}[t]$ as $n$ varies? What is the best known lower bound? - -Meta-reasoning tells us that the number of solutions for this $\mathbb{Z}[t]$ problem is not know or suspected to be unbounded because that would supply us with parametric families of solutions to the problem in $\mathbb{Z}$. -I picked the $-$ version because I was able to find this nice example: -$$[a,b,a+b-1]=[1,2t(t+1)(2t+1),2t(t+1)(2t+1)],[2t^2,2(t+1)^2,(2t+1)^2]$$ and $$[t(2t+1),(t+1)(2t+1),4t(t+1)]$$ are three solutions to $$ab(a+b-1)=\left(2t(t+1)(2t+1)\right)^2.$$ That counts as 3 solutions. (or 6 or 18 if we take $[b,a,a+b-1]$ and $[1-a-b,a,-b]$ as different, though why bother?). -Looking for integer solutions to $ab(a+b-1)=\left(2t(t+1)(2t+1)\right)^2\left(2t(t+1)(2t+1)\right)^2$ most often leads to just the three anticipated solutions but sometimes there are more. Including 6 for $t=55$ and 9 for $t=175$. -It seems reasonable to require that the leading coefficients be positive and that $n=n(t)$ not be an integer (I needed to say that before someone else did!). -The more general question arises as well. But I like concrete examples and don't want to venture a version of minimal Weierstrass form (feel free to enlighten me however) -I've further extended the previous update to this question and made it an answer. Please provide a better one! - -REPLY [2 votes]: The $\mathbb{Z}[t]$ example above with 3 (pairs of) positive solutions for $ab(a+b-1)=n$ has 5 in $\mathbb{Z}[u,v]/(u^2-2v^2+1)$ if $t=\frac{(2v+u)(u-v)}{2}$. This gives us a parametric family of curves in $ab(a+b-1)=n$ in $\mathbb{Z}$ with 5 pairs of positive integer points . We need $u>v$ so first cases are $(u,v,n)=(7,5,21420^2),(41,29,840456540^2),(99,70,32947602728040^2)$. In the first case $n=21420^2$ there is a further sixth pair of solutions but not for the next the next 4 cases. -Suppose that $u^2=2v^2-1$ and let $t=\frac{(2v+u)(u-v)}{2}$. Then we have the same three solutions as before for $ab(a+b-1)=(2t(t+1)(2t+1))^2$. There are also two more. (plus the same things with $a$ and $b$ swapped). These are as follows with the second and third being the new points: -$[{\small 1}, \frac{(2v+u)(2v-u)(u+v)(u-v)uv}{2},\frac{(2v+u)(2v-u)(u+v)(u-v)uv}{2}]$ -$[\frac{(u-v)^2(2v+u)u}{2},{\small (2v-u)(2v+u)v^2}, \frac{(2v-u)(u+v)^2u}{2}]$ -$[\frac{(2v-u)^2(u+v)v}{2}, {\small(u-v)(u+v)u^2}, \frac{(u-v)(2v+u)^2v}{2}]$ -$[\frac{(2v+u)^2(u-v)^2}{2}, \frac{(2v-u)^2(u+v)^2}{2}, { \small u^2v^2}]$ -$[\frac{(2v+u)(u-v)uv}{2}, \frac{(2v-u)(u+v)uv}{2}, {\small (2v-u)(2v+u)(u-v)(u+v)}]$ -note that $\frac{2v-u}{u-v},\frac{v}{u},\frac{2v+u}{v+u}$ are three successive convergents to $\sqrt{2}$. Of course these can be rewritten using the relation $u^2=2v^2-1$ and/or substitutions, for example the product $(2v-u)(2v+u)=1$. -In another direction is again 3 points but in $\mathbb{Z}[t,x]$ -${\small [1+(4t+2)x, 2(t+1)(2t+1)(t+(4t^2+2t-1)x), 2t(2t+1)(t+1+(4t^2+6t-1)x)]}$ -${\small [2t(t+(4t^2+2t-1)x), 2(t+1)(t+1+(4t^2+6t-1)x), (2t+1)^2(1+(4t+2)x)]}$ -${\small [(2t+1)(t+(4t^2+2t-1)x), (2t+1)(t+1+(4t^2+6t-1)x), 4t(t+1)(1+(4t+2)x)]}$ -For a majority of values for $t,x$ the resulting curve over $\mathbb{Z}$ has three positive points. Hower cases of 4 and 5 do happen quite frequently and even 6 sometimes. This suggests that more is there to be found. Replacing $x$ with $-x$ and then multiplying through by $-1$ gives curves $ab(a+b+1)=n$ -The first example represents two iterations of a process: - -Find a family of examples in $\mathbb{Z}$ (say look at ones with at least 3 solutions, one being $[1,b,b]$, from those notice that a few have the other two triples very close to each other. -Find a parametric form in $\mathbb{Z}[t]$. -Put in integers for $t$ to get curves over $\mathbb{Z}$ with at least three points and look for cases with higher rank. -Note that several have 5 solutions and prime factors familiar from convergents to $\sqrt{2}.$ -Find a parametric form of that. - -The second example follows a similar process but this is long enough.<|endoftext|> -TITLE: Winning strategy at chomp (a chocolate bar game)? -QUESTION [15 upvotes]: The game of chomp is an example of a game with very simple rules, but no known winning strategy in general. -I copy the rules from Ivars Peterson's page: - -Chomp starts with a rectangular array of counters arranged neatly in rows and columns. A move consists of selecting any counter, then removing that counter along with all the counters above and to the right of it. In effect, the player takes a rectangular or square "bite" out of the array—just as if the array were a rectangular, segmented chocolate bar. Two players take turns removing counters. The loser is the one forced to take the last "poisoned" counter in the lower left corner. - -A nice non-constructive argument shows that the first player has a winning strategy. The winning strategy can be made explicit in very specific cases. As far as I know, the more general setting for which the winning strategy is known is when we have 3 rows and any number of columns, see this page. -My question is: - -Are there any recent advances on chomp? - -REPLY [2 votes]: The non-constructive proof you refer to is proving a $\Pi_2$ statement, and therefore can be unwound to give an explicit proof. (This was pointed out to me by Mints, in the context of the game Hex, for which the same situation occurs.) -If the argument is what I expect it to be The strategy is to simply produce the tree of all possible moves and then label them as winning or losing (for player 1) by induction: and end-state is winning if player 2 takes the poison, a node where player 1 moves is winning if any of its children are winning, and a node where player 2 moves is winning if all of its children are winning. Roughly the same argument as in the non-constructive proof shows that the root node is winning, and so player 1's strategy is just to always move so they end up on a winning node. -(Of course, this isn't an elegant strategy, so there's still a reasonable open question there, but it is a known winning strategy in any formal sense of the term.)<|endoftext|> -TITLE: Borsuk--Ulam question -QUESTION [6 upvotes]: A version of the Borsuk--Ulam theorem states that a continuous antipodal map from the M-sphere into euclidean N-space has a zero -provided that M is at least N. Clearly the general case follows from the case when M = N. But is the case when M >> N any easier to prove than the equidimensional case? - -REPLY [4 votes]: I don't think so, since any antipodal (non-existent) map $S^n\to S^{n-1}$ would easily be "suspended" to an antipodal map $S^{n+1}\to S^n$. Iterating and composing would then yield antipodal maps $S^m\to S^{n-1}$ with arbitrarily large $m$. The $n=2$ case is somewhat easier though.<|endoftext|> -TITLE: Can anyone calculate KK(A,B) when neither A or B are the complex numbers? -QUESTION [6 upvotes]: Here I am referring to Kasparov's KK-theory, a bivariant functor on the category of separable C* algebras. It is well known that $KK(A, \mathbb{C})$ is K-homology and $KK(\mathbb{C}, B)$ is K-theory, both of which can be calculated for a huge collection of C* algebras (often by topological methods). -I am wondering if anybody knows how to actually calculate $KK(A,B)$ when it is not simply isomorphic to K-theory or K-homology (so I guess I also have to exclude $C_0(\mathbb{R})$. It seems that most of the time the interest in KK groups is not actually calculating them but in constructing specific elements (such as the Dirac / dual Dirac elements in proofs of the Baum-Connes conjecture). -It occurred to me that I have never actually seen anybody explicitly calculate $KK(A, B)$ in nontrivial cases. I am left wondering if this is because nobody knows how to do it or if it just isn't particularly useful to do so. Explanations or references are both appreciated. - -REPLY [3 votes]: Hi Paul, I do not know if this is interesting (enough) for you, but in 'The operator K-functor and extensions of C*-algebras' Kasparov shows that for suitable A and B (I guess A,B seperable, A nuclear) there exists an isomorphism from $\mathrm{KK}(A,B)$ onto a group $\mathrm{Ext}(A,C_0(\mathbb{R})\otimes B)$ of 'stable equivalence classes of extensions' -$$0\xrightarrow{}{} C_0(\mathbb{R})\otimes B\otimes\mathcal{K}\xrightarrow{}{} C \xrightarrow{}{} A \xrightarrow{}{} 0\text{ .}$$. -Maybe the RHS is not concrete enough for this to qualify as an answer.<|endoftext|> -TITLE: Computer Science for Mathematicians -QUESTION [50 upvotes]: This is a big-list community question, so I'm sorry in advance if it is deemed too soft but I haven't seen anything similar yet. -I've seen computer scientists post questions looking to learn things from pure maths. This is basically the other way around... My ignorance may prevent me from being as specific as I think I would like to be and so I have separated my main question into two. - -What good books - readable - introductions - are there for - mathematicians to learn about computer - science? - -By this I really mean the science of how computers work. There are perhaps some books out there which are written in a style which mathematicians can relate to - e.g. not practicality-focused, starting from the more abstract fundamentals and building up (I may be wrong but I'm under the impression that a lot of books in other disciplines shy away from presenting things this way around, whereas mathematicians (for better or worse) are accustomed to it). Partly to illustrate what the first question is not asking, the second question is - -What good books - readable - introductions - are there for - mathematicians looking to learn about - theoretical computer science, as it is as a - subfield of maths? - -Here is where my ignorance prevents me from explaining the question any more because I can only assume these two aren't the same thing... -It seems quite frustrating that I have made it to grad school and know very little about computers and theoretical CS. -Standard "one recommendation per post" is probably appropriate, + a few sentences about what the books did for you. Also, maybe I should say that I'm not looking to ditch my current interests and become a computer scientist, so things being readable is a fairly strong condition. I'm not looking to become an expert, just to deal with my own ignorance. Thanks in advance. - -REPLY [3 votes]: You could also be interested in the dragon book: "Compilers: principles, techniques, and tools", by Aho, Lam, Sethi and Ullman, which is considered a classic in computer science, from the angle of compilation theory and practice.<|endoftext|> -TITLE: What is the probability that the range of a set of N randomly chosen real numbers in [0, 1] is less than the reciprocal of N? -QUESTION [6 upvotes]: (Random number with uniform distribution over [0, 1]) -For clarification, in the case where N = 2, we can use geometric probability. On the coordinate plane consider points with 0<=x,y<=1. The condition is satisfied on a diagonal band of area 3/4 from the origin to (1,1). Similarly, with N = 3 the volume of the space in which the condition is satisfied is 7/27. - -REPLY [14 votes]: The keyword is the order statistics. The distributions of the maximum and minimum values of a sample of $n$ independent uniformly distributed random variables are given respectively by the laws -$$U_{max}\sim \mbox{Beta}(n,1),\qquad U_{min}\sim \mbox{Beta}(1,n).$$ -The range $U_{max}-U_{min}$ has a $\mbox{Beta}(n-1,2)$ distribution (see, e.g., Section 2.5 of A First Course in Order Statistics) so -$$\mathbb P\{U_{max}-U_{min} < a\}=\frac{1}{B(n-1,2)}\int_{0}^{a}x^{n-2}(1-x)dx=na^{n-1}-(n-1)a^n.$$<|endoftext|> -TITLE: Neron models of elliptic curves with level N structure? -QUESTION [8 upvotes]: In the Deligne-Rapoport paper entitled "Les schemas de modules de courbes elliptiques" the following is written (I translated in english): -Let $E$ be an elliptic curve with $\Gamma(N)$-level structure defined over -$\mathbb{C}((T))$. Let $E'$ be the minimal model of $E$ over $\mathbb{C}[[T]]$. -It may happen that $E'$ has bad reduction (i.e. when one reduces modulo $T$). So let -$A$ be the neron model of $E'$ over the d.v.r. $\mathbb{C}[[T]]$. Then it seems that the special -fiber of $A$ (i.e. when $T=0$) is always isomorphic to -$\mathbb{C}^{\times}\times\mathbb{Z}/kN$ for some suitable integer $k$. -Q: How come this $N$ shows up in the special fiber of $A$ ? -This is a little bit strange since in the definition of a Neron model no such $N$ appears. - -REPLY [9 votes]: Roughly speaking, the N-torsion defined over the base injects into the Neron model (in characteristic 0), so the special fiber of the Neron model needs to have a subgroup isomorphic to Z/NZ x Z/NZ, since by assumption there is full level N structure. The special fiber (since there's bad reduction) has the form C^* x F, where F is a finite cycle group. So F contains an N-torsion element, which means F = Z/kN.<|endoftext|> -TITLE: Primes in arithmetic progressions -QUESTION [7 upvotes]: Denote by $\pi(x,a,q)$ the number of primes $p\le x$ of the form $p=qk+a$ -and $E(x,a,q)=\phi(q)^{-1}\mathrm{Li}(x)-\pi(x,a,q)$. -What is the strongest conjectured bound on $E(x,a,q)$ in terms of $x,q$? - -REPLY [12 votes]: As Felipe notes, the main term should be li(x)/\phi(q). Replacing this in your definition of $E(x,a,q)$ above we have that, the best that is know even on the GRH is that $E(x,a,q) = O(x^{1/2}\log x)$. This estimate doesn't get better when $q$ is large (compared to $x$) and a lot more is believed to be true. For instance Montgomery conjectures that $E(x,a,q) = O_{\epsilon}( x^{1/2+\epsilon}/ q^{1/2})$ (see Conjecture 13.9 in his book on Multiplicative Number Theory). -Edit: This is really is the best one can hope for, as Friedlander and Granville have shown (using Maier's Matrix Method) that the $O_{\epsilon}(x^{1/2+\epsilon})$ cannot be replaced with a term of the form $O_{A}(x^{1/2} \log^{A}(x))$. -See: Friedlander and Granville, Limitations to the equi-distribution of primes, III, Compositio Math. 81 (1992), 19-32.<|endoftext|> -TITLE: Can ZFC → NBG be iterated? -QUESTION [29 upvotes]: von Neumann-Bernays-Gödel set theory (NBG) is a conservative extension of ZFC which contains "classes" (such as the class of all sets) as basic objects. "Conservative" means that anything provable in NBG about sets can also be proven in ZFC. The essential properties which make this true (as opposed to, say, Morse-Kelley set theory, which is not conservative) are that classes cannot be elements of other classes (in particular, powerclasses and function classes do not exist), and class comprehension is "predicative", i.e. quantified variables can range only over sets. -My question is, can the operation "ZFC → NBG" be iterated? Can we add to NBG new basic objects called (say) "2-classes" (such as the 2-class of all classes), which can contain classes as elements (but not other 2-classes), and where 2-class comprehension can use quantified variables that range over classes, but not over 2-classes? (Well, obviously, we can, but the real question is whether it would be conservative over NBG, hence also over ZFC.) -Background: I am wondering about this as a foundation for category theory. Most "large" categories which arise in mathematics, outside of category theory itself, can be defined in NBG (or even in ZFC, sort of, using the usual trick of representing proper classes by the first-order formulas defining them). But in category theory, we sometimes want to study things like "the category of large categories" or "the functor category between two large categories," which cannot be defined in NBG or ZFC. The usual solution is to assume a Grothendieck universe (or inaccessible cardinal), but this seems somewhat extravagant, since most of these new beasts only live "one more level up" from the classes in NBG. So I wonder whether we can get away with a further conservative extension of this sort. - -REPLY [17 votes]: There is a cheap way of doing this, which may not be the optimal approach when a subtle task (such as the foundational question you have in mind) is the goal. But, then again, this may suffice. -Working in an appropriately strong theory, to simplify, the standard way to check that NBG is conservative over ZFC is to see that any model $M$ of ZFC can be extended to a model $N$ of NBG in such a way that the "sets" of $N$ give us back $M$. Again to simplify, assume the model $M$ is transitive. The model $N$ we associate to it is Gödel's $\mathop{\rm Def}(M)$, the collection of subsets of $M$ that are first order definable in $M$ from parameters (The proper classes are the elements of $\mathop{\rm Def}(M)\setminus M$.) -This suggests the simple solution of defining the models of "iterated-NBG" as the result of iterating Gödel's operation. So, given a transitive model $M$ of ZFC, the $\alpha$-th iterate would simply be what we usually denote $L_\alpha(M)$. -I am restricting to transitive models here, but there is a natural first order theory associated to each stage of the iteration just described (at least, for "many" $\alpha$), and I guess one could try to axiomatize it decently if enough pressure is applied. -There are some subtleties in play here. One is that most likely we want to stop the iteration way before we run into serious technicalities ($\alpha$ would have to be a recursive ordinal, for one thing, but I suspect we wouldn't want to venture much beyond the $\omega$-th iteration). Another is that the objects we obtain with this procedure would have wildly varying properties depending on specific properties of $M$. -For example, if $M$ is the least transitive model of set theory, then we "quickly" add a bijection between $M$ and $\omega$. In general, if $M$ is least with some (first order in the set theoretic universe) property, then we "quickly" add a bijection between $M$ and the size of the parameters required to describe this property (this is an old fine-structural observation. "Quickly" can be made pedantically precise, but let me leave it as is). -So you may want to work not with ZFC proper but with a slightly stronger theory (something like ZFC + "there is a transitive model of ZFC" + "there is a transitive model of "ZFC+there is a transitive model of ZFC"" + ...) if you want some stability on the theory of the transitive models produced this way. (Of course, this is an issue of specific models, not of the "iterated-NBG" theory per se). - -I should add that I do not know of any serious work in the setting I've suggested, with two exceptions. One, in his book on Class Forcing, Sy Friedman briefly mentions a version of "Hyperclass forcing" appropriate to solve some questions that appear in a natural fashion once we show, for example, that no class forcing over $L$ can add $0^\sharp$. The second is by Reinhardt in the context of large cardinals and elementary embeddings, and is described by Maddy in her article "Believing the axioms. II". As far as I remember, neither work goes beyond hyperclasses, i.e., classes of classes.<|endoftext|> -TITLE: Are there homology groups for cosimplicial groups? -QUESTION [6 upvotes]: Hi, -Assume you have a cosimplicial group $G$, so that for each $n \ge 0$ there is a group $G_n$, and you have the usual cofaces and codegeneracies. - -Is there a known way to associate to this a collection of homology/homotopy groups in a sensible way? - -"Sensible" means at least that it should provide a generalisation of the following two particular cases: -(1) When each $G_n$ is abelian, one can get a cochain complex (I believe this is called the Dold-Kan construction), and one can consider its cohomology groups. -(2) (I apologize for the vagueness of this one.) In low degrees, one can sometimes use a couple of tricks. I have, for example, come across the following situation: the map $x\mapsto d^0(x) d^2(x)$ was a homomorphism $G_2 \to G_3$, as was the map $x\mapsto d^3(x)d^1(x)$; so I could consider their equalizer. Moreover the map $x\mapsto d^0(x)d^1(x)^{-1}d^2(x)$ was a homomorphism $G_1 \to G_2$, whose image was normal in the preceding equalizer. Taking the quotient gave a generalisation of $H^2$ in this lucky situation. The "general" definition which I'm asking for, should it exist, would hopefully coincide with this equalizer trick whenever it makes sense. -Let me also point out that, in the example above, I had originally started with some bigger cosimplicial group $(\Gamma_n)$ and then decided to restrict to the smaller $(G_n)$ precisely so that I could use this little trick. I do believe that the details of this example are completely irrelevant to the general discussion; I mention it because someone might know how to go from a cosimplicial group to a "nicer" one somehow. -Thank you for reading this. -Pierre - -REPLY [4 votes]: You will probably find interesting the paper by Mariam Pirashvili called "Second cohomotopy and nonabelian cohomology". Cohomotopy groups $\pi^nG^\bullet$ of cosimplicial groups $G^\bullet$ have been considered in very low dimensions $n=0,1$ since long time ago, e.g. in the context of non-abelian cohomology. Actually, the only group is $\pi^0G^\bullet$, which is the equalizer of the cofaces -$$G^0\mathop{\rightrightarrows}\limits^{d^0}_{d^1}G^1.$$ -Then $\pi^1G^\bullet$ is just a pointed set and there seems to be no way of going higher in general (unless of course $G^\bullet$ is abelian, which is a very strong restriction). Nevertheless, M. Pirashvili has found a nice condition, satisfied by many examples of interest, under which the previous $\pi^0G^\bullet$ and $\pi^1G^\bullet$ are an abelian group and a group, respectively, and one can define a pointed set $\pi^2G^\bullet$. There are also nice exact sequences associated to cosimplicial subgroups, etc.<|endoftext|> -TITLE: Infinitely many solutions of a diophantine equation -QUESTION [44 upvotes]: If $P(x,y,...,z)$ is a polynomial with integer coefficients then every integer solution of $P=0$ corresponds to a homomorphism from $\mathbb{Z}[x,y,...,z]/(P)$ to $\mathbb{Z}$. So there are infinitely many solutions iff there are infinitely many homomorphisms. If $P$ is homogeneous, we consider solutions up to a scalar factor. -Now if $G$ is a finitely generated group and $\Gamma=\langle X\mid r_1,...,r_n\rangle$ is another group, then any solution of the system of equations $r_1=1,...,r_n=1$ in $G$ corresponds to a homomorphism $\Gamma\to G$. We usually consider homomorphisms up to conjugation in $G$. Now it is well known that if there are infinitely many homomorphisms from $\Gamma$ to $G$, then $\Gamma$ acts non-trivially (without a globally fixed point) by isometries on the asymptotic cone of $G$ (i.e. the limit of metric spaces $G,G/d_1, G/d_2,...$ where $d_i\to \infty$, $G/d_i$ is the set $G$ with the word metric rescaled by $d_i$; here $d_i$ is easily computed from the $i$-th homomorphism $\Gamma\to G$: given a homomorphism $\phi$, we can define an action of $\Gamma$ on $G$: $\gamma\circ g=\phi(\gamma)g$; the number $d_i$ is the largest number such that every $g\in G$ is moved by one of the generators by at least $d_i$ in the word metric). - Question. Is there a similar object for Diophantine equations, that is if $P$ has infinitely many integral solutions, then $\mathbb{Z}[x,y,...,z]/(P)$ "acts" on something, not necessarily a metric space. -The reason for my question is that several statements in group theory and the theory of Diophantine equations assert finiteness of the number of solutions. For example, Faltings' theorem states that if the genus of $P$ is high enough, then the equation $P=0$ has only finite number of solutions. Similarly, if $\Gamma$ is a lattice in a higher rank semi-simple Lie group, then the number of homomorphisms from $\Gamma$ to a hyperbolic group is finite (the latter fact follows because the asymptotic cone of a hyperbolic group is an $\mathbb{R}$-tree, and a group with Kazhdan property (T) cannot act non-trivially on an $\mathbb{R}$-tree). - Update: To make the question more concrete, consider one of the easiest (comparing to the other statements) finiteness results about Diophantine equations. Let $P(x,y)$ be a homogeneous polynomial. If the degree of $P$ is at least 3 and $P$ is not a product of two polynomials with integer coefficients, then for every integer $m\ne 0$ the equation $P(x,y)=m$ has only finitely many integer solutions. It is Thue's theorem. Note that for degree 2 the statement is false because of the Pell equation $x^2-2y^2=1$. The standard proof of Thue's theorem is this. -Let the degree of $P$ be 3 (the general case is similar). Represent $P$ as $d(x-ay)(x-by)(x-cy)$ where $a,b,c$ are the roots none of which is rational by assumption. Then we should have -$|(x/y-a)|\cdot |(x/y-b)|\cdot |(x/y-c)|=O(1)/|y^3|$ for infinitely many integers $x,y$. Then the right hand side can be made arbitrarily small. Note that if one of the factors in the left hand side is small, the other factors are $O(1)$ (all roots are different). Hence we have that $|x/y-a|=O(1)/y^3$ (or the same with $b$ or $c$). But for all but finitely many $x,y$ we have $|x/y-a|\ge C/y^{5/2+\epsilon}$ for any $\epsilon>0$ by another theorem of Thue (a "bad" approximation property of algebraic numbers), a contradiction. -The question is then: is there an asymptotic geometry proof of the Thue theorem. - -REPLY [18 votes]: Revised -Interesting question. -Here's a thought: -You can think of a ring, such as $\mathbb Z$, in terms of its monoid of affine endomorphisms -$x \rightarrow a x + b$. The action of this monoid, together with a choice for 0 and 1, give the structure of the ring. However, the monoid is not finitely generated, since the -multiplicative monoid of $\mathbb Z$ is the free abelian monoid on the the primes, times -the order 2 group generated by $-1$. -If you take a submonoid that uses only -one prime, it is quasi-isometric to a quotient of the hyperbolic plane by an action of $\mathbb Z$, which is multiplication by $p$ in the upper half-space model. To see this, place a dots labeled by integer $n$ at position $(n*p^k, p^k)$ in the upper half plane, for every pair of integers $(n,k)$, and connect them by horizontal line segments and by vertical line segments whenever points are in vertical alignment. The quotient of upper half plane by the hyperbolic isometry $(x,y) \rightarrow (p*x, p*y)$ has a copy of the Cayley graph for this monoid. This is also quasi-isometric to the 1-point union of two copies of the hyperbolic plane, one for negative integers, one for positive integes. It's a fun exercise, -using say $p = 2$. Start from 0, and recursively build the graph by connecting $n$ to $n+1$ by one color arrow, and $n$ to $2*n$ by another color arrow. If you arrange positive integers in a spiral, you can make a neat drawing of this graph (or the corresponding graph for a different prime.) The negative integers look just the same, but with the successor arrow reversed. -If you use several primes, the picture gets more complicated. In any case, one can take rescaled limits of these graphs, based at sequence of points, and get asymptotic cones for the monoid. The graph is not homogeneous, so there is not just one limit. -Another point of view is to take limits of $\mathbb Z$ without rescaling, but -with a $k$-tuple of constants $(n_1, \dots , n_k)$. The set of possible identities among polynomials in $k$ variables is compact, so there is a compact space of limit rings for $\mathbb Z$ with $k$ constants. Perhaps this is begging the question: the identitites that define the limits correspond to diophantine equations that have infinitely many solutions. -Rescaling may eliminate some of this complexity. -A homomorphism $\mathbb Z[x,y,\dots,z]/P$ to -$\mathbb Z$ gives a homomomorphism of the corresponding monoids, so an infinite sequence of these gives an action on some asymptotic cone for the affine monoid for $\mathbb Z$. -With the infinite set of primes, there are other plausible choices for how to define length; what's the best choice depends on whether and how one can prove anything of interest.<|endoftext|> -TITLE: Where and when did "transition to abstraction" courses start? -QUESTION [18 upvotes]: I often find myself debating the content and structure of such courses and I would find it useful to know the basic history. -I don't remember any such offerings during my own undergraduate days in the '70s. I have always supposed these courses appeared as compensation for a decline in high school Euclidean geometry teaching, but I would call my understanding anecdotal. -Some old-timers may have the whole history in their heads, if so thanks. Otherwise it would be useful to me to hear if you had such a course a long time ago (and where, and from who), but please only comment if your date trumps the earliest previously posted date. In any case I feel sure such courses were popular by the 1980s. -Polya's How To Solve It dates to 1945 and roughly addresses these needs, but I have always understood it as a popular book rather than a a text. So I wonder what were the first texts written to support such courses? -Please refrain, of course, from opining about the efficacy or effectiveness of such courses, but feel free to cite any published research addressing the same. Thanks. - -REPLY [15 votes]: I assume that this question is asking about Mathematics Education courses in the United States. A short answer to this question is to read a nice piece on the history of the mathematics major (in the U.S.) by Alan Tucker. The citation is: -Tucker, A. The History of the Undergraduate Program in Mathematics in the United States. The American Mathematical Monthly, 120(8), pp. 689-705. -(I believe a free version is available here.) -Here are a couple relevant pieces to your specific question from the Tucker article. The first one references the sort of class that you mention and which was available prior to the 1970s. The latter one notes that it was primarily the universities themselves, rather than, e.g., recommendations from mathematicians (an example being the CUPM = the MAA's Committee on the Undergraduate Program in Mathematics) that were the driving forces behind the institution of such courses. - - - -Separately, I was recently reading through a Mathematics Education thesis from Columbia University Teachers College that discusses the history of the mathematics major at liberal arts institutions. You can find a copy through DigitalDissertations; the citation is: -George. M. The history of liberal arts mathematics. Teachers College, Columbia University, ProQuest, UMI Dissertations Publishing, 2007. 3288599. -Besides being of general interest, this thesis also led me to work by a mathematician who worked at the University of Chicago by the name of E.P. Northrop. Already by the 1940s Chicago was instituting a course quite similar to the sort that you have described. More information can be found in a couple of Northrop's papers: -Northrop, E. P. (1945). Mathematics in a liberal education. The American Mathematical Monthly, 52(3), 132-137. -Northrop, E. P. (1948). The Mathematics Program in the College of the University of Chicago. The American Mathematical Monthly, 55(1), 1-7. -Here is a relevant excerpt from the former citation: - -As requested, you can see in the above a reference to a specific textbook (Fundamental Mathematics) that was used for purposes similar to those you outlined, and which dates back to the 1940s. -I am sorry I cannot respond to your question in more detail; a full answer is probably fodder for an entire thesis in the history of Mathematics Education. Hopefully, though, the few references provided here can get you started if you wish to read further. - -REPLY [6 votes]: I took such a course at Temple University in the spring of 1972. The professor was Ray Coughlin, and the book was Fundamental Concepts of Mathematics by Foulis, which I fell in love with. I still keep a few copies around and occasionally give them away to students at about that stage.<|endoftext|> -TITLE: On the fundamental group of hypersurfaces -QUESTION [6 upvotes]: Let $H$ be a smooth projective hypersurface in $\mathbb{P}^n(\mathbb{C})$ where $n\geq 3$. Then by the Lefschetz hyperplane theorem we have that $H^1(H,\mathbb{C})= -H^1(\mathbb{P}^n(\mathbb{C}),\mathbb{C})=0$. It thus follows that $\pi_1(H)^{ab}$ is a finite -abelian group. -Q.1 Do we have an example of an $H$ such that $\pi_1(H)$ is infinite? -Q.2 Is it possible to compute explicitly $\pi_1(H)$ (or more modestly -$\pi_1(H)^{ab})$ in term of the defining equation of -$H$ ? - -REPLY [7 votes]: Let me supplement Algori's answer a bit. The statement for fundamental groups goes -back to Zariski, I believe. Standard Morse theory proofs yield this and more ( Milnor's -book gives a nice account). -Over any field, there are similar results using the etale fundamental group. -See for example SGA2, exp XII, cor. 3.5.<|endoftext|> -TITLE: commuting matrices -QUESTION [7 upvotes]: I have a set $\{A_1, A_2, .. A_k\}$ of $n$ by $n$ real matrices and I know that they are 'perturbated versions' of a set of commuting matrices : $\{P_1,..,P_k\}$, by perturbated versions I mean that I have a bound on -$\sum_i\|A_i-P_i\|\leq \epsilon$. My problem is that I want to find an epsilon perturbation of $\{A_1,...,A_k\}$ so that they commute. It is clear that at least one such perturbation exists since they are perturbed versions of a commuting set. But my question is if there is any algorithm to find a permutation that makes the matrices in $A$ commute. - -REPLY [6 votes]: I think it really comes down to looking at the lattice of invariant subspaces for each $A_i$. I believe this is what Igor basically is saying. -For a single real linear transformation $V \rightarrow V$ with distinct characteristic roots, $V$ decomposes as a sum of 1 and 2-dimensional invariant subspaces, and these generate the lattice of invariant subspaces. When you perturb the matrix by a known amount, the subspaces can wander around over a certain range --- how much they shift depends on how close together are the characteristic roots, and also on the angles between the subspaces. For simplicity, -think of the case with all real distinct eigenvalues. The projectivized action, on -$\mathbb{RP}^{n-1}$, then has $n$ fixed points. The first derivative of the projectivized map is simple to determine, given the eigenvectors and eigenvalues --- the eigenvalues of the derivative at a point for eigenvalue $\lambda$ are the ratios of the other eigenvalues to $\lambda$. Therefore, if you change the map by adding multiples of the other eigenvectors as a linear function of the projection to the eigenspace under consideration -you shift your given eigenvector by a known linear function of the coefficients. -If you have several matrices with all real distinct eigenvalues that are near enough to a commuting set of matrices, then presumably the eigenvectors are also near to each other, so you can tell which eigenvectors need to line up with each other. You have a good measurement of how hard it is to move each eigenvector in a given direction, so you can -find a point that's approximately equally hard for each (do this separately, for each matching collection of eigenvectors.) Then adjust each of the eigenvectors to go to that point, and repeat on the other eigenspaces. -When there are complex roots, there are 2-dimensional invariant subspaces, which appear as invariant circles in $\RP^{n-1}$. You can do much the same thing with these. -When there are multiple real or complex roots, then it's necessary to look at larger-dimensional invariant subspaces. One strategy is to first fix up invariant subspaces -associated with the product of all factors of the characteristic polynomial whose roots are in a small neighborhood, or a pair of complex conjugate neighborhoods. The goal is to get such subspaces to agree with, to contain, or to be contained in similar suspaces for the other matrices. If there are repeated roots, then first see if there's a small perturbation that makes multiple eigenvectors (or multiple 2-dimensional invariant subspaces) --- when this is possible, it gives a bigger target for matching commuting matrices. If not, look at the largest invariant subspace it contains. -I doubt if there's an easy way other than looking at these invariant subspace structures. Perhaps in the actual applications you have in mind, they're not so complicated. - -REPLY [4 votes]: First, let us consider a simpler setting. Let $A_1$ and $A_2$ be given as inputs. The aim is to find a matrix $B$ such that $A_1+B$ commutes with $A_2+B$, and that $B$ has small norm. -The key-constraint is thus: $(A_1+B)(A_2+B) = (A_2+B)(A_1+B)$, which simplifies to the worst-case (since no unique solution is possible) Sylvester equation -$$(A_1-A_2)B + B(A_2-A_1) = A_2A_1-A_1A_2$$ -Thus, a formulation as an optimization problem could be -$$ - \min\quad \|B\|^2\qquad - \text{s.t.}\quad (A_1-A_2)B + B(A_2-A_1) = A_2A_1-A_1A_2. -$$ -In principle (if we use the Frobenius norm for $B$), this is just a least-norm problem (similar to least-squares). -In general, if we have $A_1,\ldots,A_k$, then we will have a huge-number of constraints (for all pairs of commutativity requirements), but the overall problem should still be essentially a least-norm problem, and can be thus solved ``easily.''<|endoftext|> -TITLE: Moving one family of commuting self-adjoint operators to another without losing commutativity on the way -QUESTION [37 upvotes]: This is actually not a question of mine, so I'll be short on motivation and say nothing beyond that if this were true, a few fancy harmonic analysis techniques that a colleague of mine used in proving his recent results could be replaced by the mean value theorem. -Suppose that $A_1,\dots,A_n$ and $B_1,\dots,B_n$ are two commuting families of self-adjoint operators in a Hilbert space $H$ (that is all $A$'s commute, all $B$'s commute, but $A$'s may not commute with $B$'s). Assume that $\|A_k-B_k\|\le 1$ for all $k$. Is it true that there exists a one-parameter family $C_k(t)$ of self-adjoint commuting (for each fixed $t$) operators such that $C_k(0)=A_k$, $C_k(1)=B_k$ and $\int_0^1\left\|\frac d{dt}C_k(t)\right\|dt\le M(n)$ where $M(n)$ is a constant depending on $n$ only? In other words, is the set of commuting $n$-tuples of self-adjoint operators a "chord-arc set"? - -REPLY [5 votes]: I am not sure it is relevant exactly, or if you still have interest in this problem, but perhaps you might find this comment useful or interesting. -$\mathrm{Hom}(\mathbb{Z}^r,G)$ is generally NOT path-connected. For instance, if $r\geq 3$ and $G=\mathsf{SO}(n)$, over $\mathbb{R}$ or $\mathbb{C}$, for $n\geq 4$ then it is disconnected. Thus, in this setting, there are commuting collections $\lbrace A_i\rbrace$ and $\lbrace B_j\rbrace$ that are NOT connected by a path through commuting elements.<|endoftext|> -TITLE: Simplified treatment of resolutions of complex analytic varieties? -QUESTION [8 upvotes]: According to the article of Hauser: -The Hironaka theorem on resolution of singularities http://www.ams.org/journals/bull/2003-40-03/S0273-0979-03-00982-0/home.html -The existence of resolution of complex analytic varieties was proven in: - -Aroca, J.-M., Hironaka, H., Vicente, J.-L.: The theory of the maximal contact. Memorias -Mat. Inst. Jorge Juan, Madrid 29 (1975). -Aroca, J.-M., Hironaka, H., Vicente, J.-L.: Desingularization theorems. Memorias Mat. -Inst. Jorge Juan, Madrid 29 (1975). - -Question 1. I would like to know if since then some alternative more simple proofs were found? And especially are there some (relatively) recent books, lecture notes or exposition articles covering this topic? -The result of Hironaka on resolution of algebraic varieties is now exposed pedagogically in the book of Kollar: "Lectures on resolution of singularities", but as far as I can judge (please correct me if I am wrong), Kollar does not treat the case of complex analytic varieties. -Question 2. Suppose I want to use this result on resolution of singularities of complex analytic varieties (and the answer to question 1 is NO). Should I cite these two articles, or cite nothing, pretending that everyone is aware of these two aricles? I have an impression that nowadays people tend not to cite these results on resolution unless they work exactly on this problem. - -REPLY [7 votes]: I had a chance to check with Jarek Włodarczyk, who points out -that the analytic case is not much harder (to him), but that it does require some extra care. -In addition to the above references, the resolution of singularities of -analytic spaces is treated in - -Bierstone, Milman, Canonical desingularization in characteristic zero by blowing up the maximum strata of a local invariant. Invent. Math. (1997) -Włodarczyk, Resolution of singularities of analytic spaces. Proceedings of Gökova Geometry-Topology Conference 2008<|endoftext|> -TITLE: Generalized complex groupoid -QUESTION [5 upvotes]: Is there any nontrivial example of Generalized complex groupoid? -By trivial, I mean all the classes of symplectic groupoids/ Abelian varieties as well as their products. -What I mean is that, is there any example of a groupoid in the category of generalized complex (GC) manifolds, in the sense of Hitchin. GC geometry is a way to unify complex geometry and symplectic geometry in to one realm. The definition could be found in wiki. -The space of arrows is a generalized complex manifold, and the source and target maps are generalized complex morphisms. -I can only come up with trivial examples, which are symplectic/complex, so are not real examples reflecting the nature of the theory. --Any complex manifold is a GC manifold, so any abelian variety is a GC groupoid. --Any symplectic manifold is a GC manifold, so any symplectic groupoid is also a GC groupoid. --any product of these examples is also a GC groupoid. -But I can not find out any other example. - -REPLY [5 votes]: There are three examples which I'm aware of. - -GC Lie groups: these are Lie groups equipped with GC structures compatible with the multiplication map. Holomorphic Poisson Lie groups are an example of this, but there are others. For example, the known examples of generalized Kahler structures on compact even-dimensional semisimple Lie groups (we just need a bi-invariant metric, not all hypotheses are necessary) consist of two commuting GC structures, one of which is multiplicative in the above sense, and the other of which is a GC homogeneous space over the GC Lie groupd defined by the first. This situation will be familiar to those in Poisson Lie Group theory, and this is joint work in progress with Jiang-Hua Lu. David is of course correct in his statement that GC actions of GC Lie groups would then define GC action groupoids. -B-symplectic groupoids as described in http://arxiv.org/abs/math/0412097. These are, first and foremost, symplectic groupoids, but they have an extra 2-form making them GC Lie groupoids. -Any holomorphic Poisson groupoid is an example of a generalized complex groupoid. For example, if Z is a Poisson manifold, then $Z\times Z$ is a Poisson groupoid and hence a generalized complex groupoid.<|endoftext|> -TITLE: Implication of Polignac's conjecture on prime distribution in models of PA -QUESTION [6 upvotes]: Polignac's conjecture (PC) is that there exists infinitely many pairs of consecutive prime numbers that are a distance $d$ apart for some natural number $d$. The twin prime conjecture is the particular instance of this conjecture for $d = 2$. The fact that this conjecture remains open has some interesting implications on nonstandard models of Peano Arithmetic (PA). Specifically, it is a standard exercise to show that every model of PA has order type $\mathbb{N} + \mathbb{Z} \cdot A$ for some dense linear order without endpoints. Thus every nonstandard model has an initial segment of Natural numbers followed by nonstandard numbers all appearing in an unbounded dense linearly ordered collection of what are called integer blocks or $\mathbb{Z}$-blocks. What I realized (someone else must've realized this too so please mention references if you know of any) is that if Polignac's conjecture turned out to be false, then we'd have the following strong limitation on the number of primes appearing in $\mathbb{Z}$-blocks. - -($\mathbb{N} \vDash \lnot PC$) If $M$ is a model of PA and is $\Sigma^0_1$-equivalent to the theory of $\mathbb{N}$, then $M$ can have at most one prime appearing in any $\mathbb{Z}$-block: If not, then for some $d \in \mathbb{N}$, there would be a pair of nonstandard numbers that $M$ would view as two consecutive primes a distance $d$ apart. Since this occurs in a $\mathbb{Z}$-block, for any true Natural number $n$, the model $M$ thinks that there is a pair of consecutive primes greater than $n$ a distance $d$ apart. Then by $\Sigma^0_1$-elementarity, $\mathbb{N}$ would think the same thing so $\mathbb{N}$ would have unboundedly many pairs of consecutive primes a distance $d$ apart, making Polignac's conjecture true (in the standard model). - -My question concerns the other models of PA: - -Can we prove that there is a nonstandard model of PA having a $\mathbb{Z}$-block with at least two primes? Even better, can we prove that there is a model of PA with unboundedly many $\mathbb{Z}$-blocks having at least two primes? - -REPLY [3 votes]: For simplicity, take $d=2$. Then the the existence of a model of PA with at least one non-standard pair of twin primes is equivalent to the assertion -For all (fixed, standard) primes $p$ the sentence $$\phi_p:\forall x>p,\,\,\, x \textrm{ is not prime or } x+2 \textrm{ is not prime }$$ -is not provable in PA. -Is it reasonable to spend lots of time hunting for a proof of this before the twin prime conjecture is resolved? My guess is "no" and here is why. -All known constructions of non-standard models of PA depend in some way on an oracle who can determine which sentences are consistent with PA, or who can determine which sets are and are not members of some non-principal ultrafiter. I have no objections to such arguments, but it is well to be clear about the awesome powers of such an oracle: Relative to where we are, this is the point of view of eternity. It would be remarkable indeed if any statement of elementry number theory could be derived from such general constructions, and to my knowledge, none ever has. -There is an analogy here with a paragraph in Hardy and Wright's number theory text. Let $c$ be the constant $.020300500000007\ldots$. The point is that the definition of $c$ depends on foreknowledge of the sequence of primes. Using $c$ we can give a very simple formula for the $n$th prime, which is, as Hardy remarks, completely useless for proving things about primes. -In the same vein, there is a frontspiece to a book, I think by Mahler, that says: - If you want to make sausage you have to put some pork in the grinder. - -It is dangerous to say never, and in spite of Tennenbaum's theorem, there might come a day when some representation of models of PA is discovered that allows one to get information about these models from some source other than the Peano axioms... but as far as I know, that day has not arrived. The closest thing to such a representation theorem I have ever seen is an unpublished theorem of Tennenbaum, proving that every countable model of PA can be embedded in $\mathbb{R}^{\omega}$ modulo the cofinite filter: In other words, you can think of the elements of models of PA as germs at infinity of sequences of real numbers. I once asked Greg Cherlin about the usefulness of this representation. His response was "There is no free lunch."<|endoftext|> -TITLE: Prerequisites for P-adic Representations -QUESTION [6 upvotes]: I wonder what the prerequisites for learning the representation theory of reductive groups over a p-adic field are? Can someone recommend me any book or article for learning this theory if I wanna clearly know what this theory is about and what kind of applications this theory has? -I just know some fundamental concepts, like Part 1 of Serre's "Linear rep'ns of finite groups", of the rep'ns of finite groups and the structure of p-adic fields. Thanks! - -REPLY [6 votes]: When I learned this material, 10 or so years ago, I read Cartier's article in Corvalis (linked to by Thomas in the comments above), -and then Casselman's notes (mentioned by Ramin). My experience has been that all of this material is a little dry and difficult to learn (due to lack of apparent motivation) if it is not coupled with an understanding of its relationship to the global theory of automorphic forms. In turn, I find automorphic forms difficult to think about without having the link to the classical case of modular forms. -Unfortunately, while there is a lot of literature available which explains the relations between the classical theory of modular forms and the more representation-theoretic and more general theory of automorphic forms, I found a lot of it unsatisfying when I was learning this material for the first time. One article that I like is Deligne's expository article in LNM 349, in which he concretely passes from the theory of modular forms on the upper half-plane to the representation-theoretic view-point of functions on $GL_2$ of the adeles. It might at least serve as a helpful supplement to more comprehensive treatises such as those of Gelbart or Bump.<|endoftext|> -TITLE: Is there a mathematical justification/proof for the claim that the universe contains a finite amount of information? -QUESTION [5 upvotes]: Since I was first introduced to it, I've been intrigued by the claim that the universe contains a finite amount of information. (That link is not where I first encountered the concept; it is simply the first example of this claim I could find from a quick Google search.) -Basically, the argument seems to be that if there is a finite amount of matter in the universe, that matter can only store a finite amount of information. On the surface, I have to concede that this makes a lot of sense. After all, if I'm thinking in terms of bits (for example), I might visualize a hypothetical "infinite hard disk drive" that could store unlimited data. This device would presumably have to be infinite in size, since it stores information on a physical platter that obviously occupies some space. -Digging a little deeper, however, I start to doubt this presumption. After all, information can be compressed according to a system of encoding information in a particular set of symbols. Then as long as the system provides a way of decoding that information, you could effectively increase the capacity of any storage mechanism by encoding its contents using said system (analogous to converting every file on a hard disk using some compression algorithm such as LZMA). -But, there's still more to it than that. It goes without saying that any system of compression like what I just described comprises its own information, and therefore needs to be stored somewhere itself. -Since the universe is "all there is" (?), a system of encoding the information contained within it would have to be a part of that very information. This is where I think I hit a mental wall. On the one hand, it seems that you could extract a seemingly unlimited amount of information from finite data—by using a system to encode that data, another system to encode the encoded data, and so on and so forth—whereas on the other, intuition tells me that there must come a point where, if the data as well as the system must share a space, there is no longer any room for either more data or another system of encoding it. The available space becomes too "crowded," so to speak. -Is there a mathematical principle or theorem that answers this question? Is the problem I'm describing (determining a limit on the capacity of material data to store information) defined, analyzed, and/or illuminated by any particular concept(s) in mathematics? - -REPLY [4 votes]: The holographic principle has been mentioned so I'll just add a little more info from the physics perspective. -The Bekenstein entropy bound implies that there is a finite number of information (entropy) in a finite volume of space with finite energy. -Speaking in terms of entropy, one can see for example that there must be a limit on how many fundamental particles there are. The reasoning is that given a particle made up of sub particles, the total number of degrees of freedom of the particle is the product of the degrees of freedom of each sub particle (are there sub particles with only one degree of freedom?). Since the total number of degrees of freedom must be finite, this implies that one cannot subdivide particles forever. -There are some particularly striking consequences of these entropy bounds. For example, Verlinde argues that the force of gravity between particles is a result of the holographic principle. This can be thought of as (indirect) physical evidence of the holographic principle at work.<|endoftext|> -TITLE: Is it possible to show that an infinite set has a countable (infinite) subset, without using the Axiom of Choice? -QUESTION [11 upvotes]: Let X be an infinite set. -Is it possible to show the existence of a countably infinite subset of X without using the Axiom of Choice? - -REPLY [20 votes]: Short answer: No. -By countably infinite subset you mean, I guess, that there is a 1-1 map from the natural numbers into the set. -If ZF is consistent, then it is consistent to have an amorphous set, i.e., an infinite set whose subsets are all finite or have a finite complement. If you have an embedding of the natural numbers into a set, the image of the even numbers is infinite and has an infinite complement. -So the set cannot be amorphous. - -REPLY [9 votes]: No. A set which has a countably infinite subset is called Dedekind-infinite. Clearly every Dedekind-infinite set is infinite; the statement that every infinite set is Dedekind-infinite is not provable in ZF (assuming ZF is consistent, of course). You don't need full AC, though. In fact, the equivalence isn't even as strong as countable choice.<|endoftext|> -TITLE: Relationship between quasicrystals and PV numbers -QUESTION [6 upvotes]: Freeman Dyson writes in his article "Birds and Frogs", that a unique quasicrystal -exists corresponding to every Pisot-Vijayaraghavan number. Is this statement a theorem? -p.s.I couldn't find about this in any paper. - -REPLY [11 votes]: In her book Quasicrystals and geometry Majorie Senechal discusses the relationship between quasicrystals and PV number on pp. 126-128. She cites Pisot (1946) and Cassels (1965). -First she presents an alternative characterization of PV numbers: -Theorem 4.1 Let $\mu_1>1$ be a real algebraic integer. Then $\mu_1$ is a PV number if and only if there exist nonzero $q \in \mathbb{R}$ such that $$\lim_{m\rightarrow \infty} \mu_1^m q = 0 \mod \mathbb{Z}.$$ -Thus it follows that -Theorem 4.2 The diffraction condition is satisfied for an $\mathcal{A}$ sequence if and only if the leading eigenvalue $\lambda_1$ of $\mathcal{A}$ is a PV number. -I note that she defined an $\mathcal{A}$ sequence to be "any sequence of points $\Lambda = \{x_n\}$ such that, for all $n,$ $x_n - x_{n-1} \in \{\alpha_1, \ldots, \alpha_n\}$ and $\Lambda$ has suitably defined predecessors of all orders with respect to the [linear map which may be represented by a primitive matrix] $\mathcal{A}.$" -It's a nice book, easy to read. Highly recommended. -Oh, the references are: -C. Pisot (1946), Repartition (mod 1) des puissances successives des nombres reels, Commentarii Mathematici Helvetici, Vol. 19, 153-60. -J.W.S. Cassells (1965), Diophantine Approximation, Cambridge Tracts in Mathematics and Mathematical Physics No. 45, Combridge University Press.<|endoftext|> -TITLE: Pushing Complex Structure Forward -QUESTION [5 upvotes]: Let $p: E\to B$ be a covering map of $C^\infty$ manifolds, where $E$ has a complex structure. There are many cases when we want to know whether $B$ has a complex structure (which is obviously unique) making $p$ an analytic map, for example in the construction of families of elliptic curves. -The difficulty is that given a small open set $U\subset B$, and pulling back the covering map to $$p|_{p^{-1}(U)}: \bigsqcup_\alpha U_\alpha\to U,$$ the $U_\alpha$ may induce incompatible complex structures on $U$. The easiest example of this phenomenon is the covering map $\mathbb{CP}^1\simeq S^2\to \mathbb{RP}^2$, where $\mathbb{RP}^2$ does not admit any complex structure, as it is not orientable. This case is not too badly behaved, however--in particular, given $U_\alpha, U_\beta\subset \mathbb{CP}^1$ over some $U\subset \mathbb{RP}^2$, the transition map $U_\alpha\to U\to U_\beta$ seems to me to be antiholomorphic. -So I have two questions about this general situation: - -1) Is there an example of a covering map $p: E\to B$ of $C^\infty$ manifolds with $E$ complex, such that $B$ admits some complex structure, but none making $p$ analytic? -2) Given a covering map $p: E\to B$ with $E$ complex, is there an algebra-topological obstruction to the existence of a complex structure on $B$ making $p$ analytic? - -REPLY [4 votes]: This seems to be a question about holomorphicity of diffeomorphisms in a given complex structure. Replace your covering map $E \to B$ by its Galois closure (= frame bundle) $X \to B$. Now by construction $X \to B$ is a covering space which is Galois with Galois group $G$ (= groups of self bijections of a fixed fiber of $E \to B$). Since $X \to B$ factors through $E$, every complex structure on $E$ will induce a complex structure on $X$, and a complex structure on $B$ makes $E \to B$ holomorphic if and only if it makes $X \to B$ holomorphic. But the later question is just the question of whether all elements of $G$ which act as diffeomorphisms of $X$ will preserve the complex structure. Some of them preserve it automatically, e.g. the elements of the subgroup $H \subset G$ for which $E = X/H$. But for the rest it is an actual condition. If all those diffeomorphisms preserve your complex structure, then the quotient exists as a complex manifold. If one of them doesn't, then your are out of luck. -I don't think you can get more concrete obstructions.<|endoftext|> -TITLE: Reciprocals of Fibonacci numbers -QUESTION [18 upvotes]: Is the sum of the reciprocals of Fibonacci numbers a transcendental? - -REPLY [12 votes]: To supplement Joseph's answer, I add my review MR2354148 on [C. Elsner, S. Shimomura and I. Shiokawa, Acta Arith. 130:1 (2007), 37--60]. -Let $\lbrace F_n\rbrace _{n\ge0}$ and $\lbrace L_n\rbrace _{n\ge0}$ be Fibonacci and Lucas numbers, -respectively, -$F_0=0$, $F_1=1$, $F_{n+2}=F_n+F_{n+1}$ for $n\ge0$, and -$L_0=2$, $L_1=1$, $L_{n+2}=L_n+L_{n+1}$ for $n\ge0$. -Using Nesterenko's theorem -[Yu.V. Nesterenko, Sb. Math. 187:9 (1996), 1319--1348. MR1422383] -and expressing the series -$$ -\zeta_F(2s)=\sum_{n=1}^\infty\frac1{F_n^{2s}} -\quad\text{and}\quad -\zeta_L(2s)=\sum_{n=1}^\infty\frac1{L_n^{2s}}, -\qquad s=1,2,\dots, -\qquad (1) -$$ -via the Eisenstein series -$$ -E_{2s}(q)=1-\frac{4s}{B_{2s}}\sum_{n=1}^\infty\sigma_{2s-1}(n)q^n, -\qquad \sigma_k(n)=\sum_{d\mid n}d^k, -$$ -where $B_{2s}\in\mathbb Q$ are Bernoulli numbers, -the authors prove the algebraic independence of -the numbers in the collections $\zeta_F(2)$, $\zeta_F(4)$, $\zeta_F(6)$ -and $\zeta_L(2)$, $\zeta_L(4)$, $\zeta_L(6)$ as well as express -algebraically even "zeta values" $\zeta_F(2s)$ (and $\zeta_L(2s)$) -for $s\ge4$ in terms of the three algebraically independent numbers -in the corresponding collection. Similar algebraic independence results -are shown for the alternating versions of (1). Known irrationality -results for $\zeta_F(k)$ and $\zeta_L(k)$ with odd $k$ -(when the series have no known relations with the modular world) are indicated. -It is worth mentioning that these results go in a natural parallel -with the ones for the so-called $q$-zeta values defined in -[W. Zudilin, Math. Notes 72:5-6 (2002), 858--862. MR1964151] -and [C. Krattenthaler, T. Rivoal, and W. Zudilin, -J. Inst. Math. Jussieu 5:1 (2006), 53--79. MR2195945]. -In particular, it is natural to expect a "Fibonacci" analogue of Rivoal's theorem -[T. Rivoal, C. R. Acad. Sci. Paris Ser. I Math. 331:4 (2000), 267--270. MR1787183] -on the infiniteness of -irrational numbers in the set $\zeta_F(1),\zeta_F(3),\zeta_F(5),\dots$ -(or $\zeta_L(1),\zeta_L(3),\zeta_L(5),\dots$), -based on the techniques developed in the paper under review and -in the joint paper of Krattenthaler, Rivoal and the reviewer cited above. -Reviewed by Wadim Zudilin - -To summarize, the difficulty of proving the transcendence for odd $\zeta_F(s)$ with $s$ odd is similar to the one for odd zeta values. The irrationality of $\zeta_F(1)$ is known but already its non-quadraticity remains an open problem.<|endoftext|> -TITLE: Generalization of Raynaud's (p, p, ... p) result -QUESTION [5 upvotes]: Does Corollary 3.4.4 in Raynaud's paper ``Schemas en Groupes de Type (p, ..., p)'' apply also to the case where G is quasi-finite? If not, what is the more general statement? -The corollary states: -``Soit G un K-schéma en groupes fini, commutatif, annulé par une puissance de p et qui se prolonge en un R-schéma en groupes fini et plat. Soit H un quotient de Jordan-Hôlder de G. Alors H est un schéma en F-vectoriels, pour un corps fini convenable F. Si F à $p^r$ éléments, -le caractère $\psi : I_t \to F^*$, qui décrit l'action du groupe de Galois $Gal(\bar{K}/K)$ sur le F-vectoriel $G_i(K)$ est alors de la forme -$\psi = \psi_{i + 1}^{n_1} \cdots \psi_{i + r}^{n_r}$ avec $0 \leq n_j \leq e$ pour tout j.'' -(Here, R is a strictly Henselian discrete valuation ring, K is its fraction field, char K = 0, and p is the residue characteristic of R.) - -REPLY [8 votes]: As James Borger writes in his comments above, the answer is surely no. The restrictions on the characters appearing in the Galois representation attached to $G$ are being forced by the fact that $G$ has a finite flat model. If you throw that assumption away, you lose any control on the ramification of the Galois action on $G(\overline{K})$, and it can be whatever you like. -(The point is that any representation of $Gal(\overline{K}/K)$ on a $p$-power order abelian group, or on an $F$-vector space for some char. $p$ finite field $F$, is the group -of $\overline{K}$-bar points of a finite group scheme, or $F$-vector space scheme, over $K$. But if the ramification conditions in the conclusion of Raynaud's theorem aren't met, then this group scheme won't have a finite flat prolongation over $R$.) -If you want to have a theory generalizing Raynaud's theory of finite flat group schemes -which applies to other representations of $Gal(\overline{K}/K)$ (i.e. ones that don't come -from finite flat group schemes over $R$) then you will need to learn integral $p$-adic Hodge theory, which is a highly developed theory. The first step is to learn something about Fontaine--Lafaille theory (which deals with the case when $e = 1$). The step after this is to learn about Breuil modules and/or Kisin modules. But all of this is rather technical and involved, and a serious investement to learn. So you might be better off explaining your motivation and goals, so that people can give more specific and appropriate advice.<|endoftext|> -TITLE: Bertrand theorem - central forces -QUESTION [7 upvotes]: Here is a version of Bertrand theorem. Let us consider a force $F(r)$ which depends only on the distance to a given point. If all trajectories which remain bounded are closed, then either $F(r)=ar$ either $F(r)=b/r^2$. -Here are my questions : is this statement (or a related one) a mathematical theorem ? If yes, is there a nice reference ? -Thank you for any answer. - -REPLY [5 votes]: Bertrand's theorem is essentially correct as stated, but most of the proofs are a little weak at one point - the assumption of stability of "nearly circular orbits" - that the apsidal angles are continuously dependent and have a finite limit as the perturbations approach 0. This is the argument that Bertrand used to assert that, since they were always rationally commensurate with one revolution, the angle had to be constant. - In general this may not be true, because perturbations may be spirals instead of rosettes (the apsidal angle integral may diverge). However, if one imposes the "Lagrange stability condition" (that r^3 times the force law increases with r), then the orbits must be rosettes, continuous dependence can be established, and the power rules can be "calculated". - The existence of non-circular bounded orbits can then be used to show the restriction of the Lagrange stability condition can be removed from the statement of the result. - A final remark: this can all be shown rigorously following Bertrand's brilliant argument using mathematical techniques known at the time (Riemann integration theory) his original paper was published(1853).<|endoftext|> -TITLE: Tate models for semistable algebraic varieties with mixed reduction over a local field -QUESTION [6 upvotes]: It's known that if $A$ is an abelian variety of totally multiplicative reduction over a p-adic field K, then, after taking a finite field extension, it becomes isomorphic, as a rigid analytic group, to $G_m^g/\prod_{i=1}^g q_i^{\mathbb{Z}}$ where $q_i$ are points (after a finite field extension) of $G_m^g$ which generate a discrete subgroup. -My question is, what can be said when $A$ is not totally multiplicative or good reduction, but is some semistable abelian variety of mixed multiplicative-good reduction. I would guess that instead of being a quotient of $G_m$ by a discrete subgroup, $A$ will be (as a rigid analytic variety) a quotient of a good-reduction semi-abelian variety (i.e. a variety which is an extension by $G_m^m$ of an abelian variety of good reduction) by a free discrete subgroup. Does anyone know whether this is true, or whether there's something else replacing the Tate uniformization in this case? - -REPLY [10 votes]: If $A$ has semi-abelian reduction, then $A$ is uniformized by a semi-abelian variety $G_A$, -namely there is an exact sequence -$$0 \to \Gamma_A \to G_A \to A \to 0,$$ -where $\Gamma_A$ is free of finite rank, $G_A$ is semi-abelian, and the maximal abelian variety quotient of $G_A$ has good reduction. -This is due to Raynaud, I believe, and is also discussed in SGA 7. -For precise references, allow me (out of laziness) to cite one of my papers --- see the -discussion at the beginning of Section 3.<|endoftext|> -TITLE: Group ring computation -QUESTION [5 upvotes]: Let $G$ be a finite abelian group. Is it true that the following element of the group ring ${\mathbb Z}[G]$: -$$ -\prod_{g\ne 1}(1-g) -$$ -is non-zero? - -REPLY [15 votes]: Counterexample : $G=(\mathbb{Z}/2)^2$. The product is $1-a-b+ab -ab +a^2b+ab^2 -a^2b^2$, with $a^2=b^2=1$, $ab=ba$. -PS: on the other hand, this is true if (and only if?) $G$ cyclic, since then you have an injective character $\chi : G\to \mathbb{C}^\times$, whose linear extension to $\mathbb{Z}[G]$ has nonzero value on you element. -PPS: your expression is indeed $0$ in $\mathbb{Z}[G]$ whenever $G$ isn't cyclic. It is enough to shows it's zero in $\mathbb{C}[G]$, which (by elementary representation theory) is a product of fields $\mathbb{C}_\chi$ over the characters $\chi\in \hat{G}$. The result follows, since no character is injective when $G$ isn't cyclic.<|endoftext|> -TITLE: Prime numbers dividing the orders of the sporadic groups -QUESTION [13 upvotes]: When we consider the list of the prime numbers that divide the order of the 26 (or 27 if you include Tits group T) sporadic groups, we find that they all are among the 20 smallest prime numbers. -In fact, the order of the monster sporadic group M is divided by the 15 following primes: -2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 41, 47, 59 and 71. -Moreover, the order of the sporadic group J4 is divided by: 2, 3, 5, 7, 11, 23, 29, 31, 37 and 43, -and the order of the sporadic group Ly is divided by 2, 3, 5, 7, 11, 31, 37 and 67. -So that, because no other prime is dividing the order of another sporadic group, we find that -considering the list of the 20 smallest prime numbers, only the 16th prime, 53, and the 19th prime, 61, are omitted of the list of the divisors of the orders of the sporadic groups. -Do we know an explanation for this curious fact ? -Gérard Lang - -REPLY [2 votes]: A trivial remark: Since the number of these primes is less than the number of sporadic groups, the multiplicative subgroup of ${\mathbb Q}^{\times}$ generated by the orders of sporadic groups is not free on 26 (or 27) generators. :-).<|endoftext|> -TITLE: Which categorical (coproduct-like) operation captures integration of measures? -QUESTION [7 upvotes]: Suppose we have a measure space $(X,a)$, a measurable space $Y$ and for every $x\in X$ we have a measure $b_x$ on $Y$. Suppose that $(Z,c)$ is a measure space such that as a measurable space $Z=X\times Y$ and the measure $c$ is the integral of measures $b_x$ with respect to the measure $a$. -I have a vague intuition that $(Z,c)$ is a coproduct of the family $(Y,b_x)$ along $(X, a)$ - -Question: Is there a category of measure spaces and a categorical construction in it which captures this intuition? - -One could ask a similar question about a category of Hilbert spaces and integrals of Hilbert spaces over measure spaces, and preferably the "categorical construction" in question should also answer this problem. -The categorical construction in question should preferably be similar to coproduct, and it would be very nice if it actually was a coproduct in some category. -On the other hand, maybe my vague intuition is wrong, and comments on that would also be appreciated. - -REPLY [7 votes]: Yes, there is such a category, namely the category of measurable spaces and morphisms of measurable spaces equipped with fiberwise measures (also known as operator valued weights). -See this answer for an introduction and this answer for a list of further references. -Intuitively, a morphism f: X→Y in this category is a morphism of measurable spaces -together with a measure on each fiber of f that varies measurably with respect to Y. -In particular, if Y is the point, then a morphism f: X→pt is simply a measurable space equipped with a measure, -i.e., a measured space. -As David Roberts has already pointed out, the categorical construction that captures the intuition of the question is called the dependent sum. -In the notation of the nLab artcle we have B=Z, A=X, I=pt, -the morphism f: A→I is the measure a, the morphism g: B→A is the projection map Z=X×Y→X equipped with the family of measures b_x. -The dependent sum of g indexed by f exists and is the composition fg, -which in our case is the measure c.<|endoftext|> -TITLE: Can we characterize the spatial tensor product of von Neumann algebras categorically? -QUESTION [9 upvotes]: The tensor product of commutative algebras is exactly their coproduct -in the category of commutative algebras. -In other words, if A and B are two commutative algebras, -then the covariant functor that represents A⊗B assigns -to an algebra Z the set of all pairs of morphisms f: A→Z and g: B→Z. -Tensor product of noncommutative algebras also admits a categorical characterization. -Namely, if A and B are two noncommutative algebras, -then the functor that represents A⊗B assigns -to an algebra Z the set of all pairs of morphisms f: A→Z and g: B→Z -whose images commute in Z, i.e., m(f⊠g)=ms(f⊠g), -where m is the multiplication Z⊗Z→Z, s is the symmetry Z⊗Z→Z⊗Z, -and ⊠ is the external tensor product: f⊠g: A⊗B→Z⊗Z. -The category of commutative von Neumann algebras also admits a coproduct, -which therefore can be thought of as the categorical tensor product of von Neumann algebras. -This tensor product can be extended to noncommutative von Neumann algebras in the same -way as described above. -Apparently this product was first described by Alain Guichardet in his 1966 paper. -The categorical tensor product is much bigger than the spatial tensor product. -The difference for commutative algebras is explained in this answer: -Is there a category structure one can place on measure spaces so that category-theoretic products exist? -Is there a categorical characterization of the spatial tensor product of von Neumann algebras? -By the universal property of the categorical tensor product for any two von Neumann algebras A and B -there is a canonical morphism Q: C→S of von Neumann algebras -from the categorical tensor product C to the spatial tensor product S. -This morphism is an epimorphism, i.e., it is surjective. -However, unless one of the algebras is finite-dimensional, it has non-trivial kernel. -Hence, the algebra S is represented by -a subfunctor of the covariant functor that assigns to a von Neumann algebra Z -the set of all pairs of morphisms f: A→Z and g: B→Z with commuting images. -Can we characterize categorically the pairs (f,g) that belong to this subfunctor? -Alternatively, the kernel of the morphism Q is a σ-weakly closed -two-sided ideal of C, which corresponds to a central projection of C. -Can we characterize this central projection categorically? - -REPLY [4 votes]: I am not sure whether this helps, but along the lines of the comment of Theo Johnson-Freyd, one can say the following: -If $A,B,C$ are von Neumann algebras and $Mod(A),Mod(B),Mod(C)$ their categories of representations, then there exists a natural isomorphism between morphisms $A \to B \bar{\otimes} C$ and $Mod(B)\times Mod(C) \to Mod(A)$. Here, the representation categories are viewed as categories ${\cal D}$ fibered over (equipped with forgetful functor $U_\cal D$ to) the category of Hilbert spaces, and the product ${\cal D} \times {\cal E}$ of two such fibered categories is the product category, equipped with the forgetful functor $U_{\cal D} \times U_{\cal E}$ followed by the Hilbert space tensor product functor. -This statement depends on the fact that the categories of representations have generators and extends from the spatial tensor product to fiber products of von Neumann algebras relative to some subalgebra. Categorically, this has to be made more precise and may not be quite what you're looking for. -Details for the fiber product (which reduces to the spatial tensor product if the subalgebra is just complex numbers) may be found in Section 4.3 of link text<|endoftext|> -TITLE: A question about non-norm-euclidean real quadratic fields -QUESTION [22 upvotes]: After lurking MO for a while, I decided I'd jump in. Why not start with a question? -Specifically, I have a question about a natural set that sort of "measures" the failure of a real quadratic field to be norm-Euclidean. -Let $K$ be a real quadratic field and let $V$ denote the two-dimensional real vector space $K\otimes_{\mathbb Q}\mathbb{R}$. The ring of integers $\mathcal{O}_K$ embeds as a lattice in $V$, and the quotient is a real torus $T$ equipped with an automorphism of infinite order induced by a fundamental unit. The field norm on $K$ extends in a natural way to a map $V\to \mathbb{R}$ (extend the galois automorphism and use the usual definition of norm, or just work in coordinates in the obvious fashion). -Now let $U$ be the open set in $V$ consisting of all $y$ such that $|Nm(y-x)|<1$ for some $x\in \mathcal{O}_K$. Thus, for example, $K$ is norm-Euclidean if $U=V$. In general the complement is some closed set in $V$ that is $\mathcal{O}_K$-translation invariant and descends to a closed set $X$ of the torus $T$. -Here's where I wish I knew how to make one of those nifty shaded boxes to highlight the question... - -What is the nature of the set $X$? - -Though I know basically nothing of ergodic theory, it seems that $X$ must have measure zero since it's preserved by the (ergodic) automorphism induced by the fundamental unit. It seems to look like some sort of Cantor-like set, but I have no idea how to really get my hands on it. -Two specific questions: -1) Is there some reasonable way to parameterize or otherwise describe $X$ somewhat explicitly (I ran across something called a "binary arithmetic coding" once that purports to be able to do this, but it seemed ill-suited to this particular set to me). -2) What is the Hausdorff dimension of $X$? Does it have any relation to the arithmetic of $K$? (Wouldn't it be cute if it had something to do with the class number, etc.) -Of course, one could ask a similar question about any number field other than $\mathbb{Q}$ and quadratic imaginary fields, but this seemed like the place to start. - -REPLY [9 votes]: As it stands, I guess your question is hopeless. There are some answers if you replace the condition $|N(x-y)| < 1$ by $|N(x-y)| < \kappa$, where $\kappa$ is chosen as the infimum over all constants $\lambda$ for which $|N(x-y)| < \lambda$ satisfies $U=V$. This infimum is called "Euclidean minimum" (or, classically, the inhomogeneous minimum of the norm-form of the corresponding quadratic number field). -Barnes and Swinnerton-Dyer (The inhomogeneous minima of binary quadratic forms, Acta Math. 87, 88, 92, 1952--1954) have studied the set X for various families of number fields (and $\kappa$ as above). The set is sometimes finite and sometimes countably infinite. Recent results in this direction have been obtained by Jean-Paul Cerri, Inhomogeneous and Euclidean spectra of number fields with unit rank strictly greater than 1 [J. Reine Angew. Math. 592 (2006), 49-62] using ergodic theory.<|endoftext|> -TITLE: 2nd Incompleteness and Model Theory -QUESTION [7 upvotes]: In the presence of Godel's Completeness Theorem, the 2nd Incompleteness Theorem has -the following strictly model theoretic interpretation: if there exists any model at all of (say) ZFC, there also exists a model M of ZFC such that M contains no (internal) model of ZFC. -That is the direct model-theoretic translation of: if ZFC is consistent, then ZFC + (ZFC is inconsistent) is consistent. -My question(s): Either -1) Does this model theoretic statement possibly enjoy a "purely" model-theoretic proof? or -2) Is there a good reason not to expect a purely model-theoretic proof of this statement? -Since category theory affords a rich context for studying internalization, perhaps I should -substitute "category-theoretic" for "model-theoretic." -[As a graduate student years ago I wondered whether "every model of ZFC contains an internal model of ZFC" would simply run directly afoul of the Axiom of Foundation until someone refined my understanding of "internal model" and convinced me that the story couldn't be that simple.] - -REPLY [7 votes]: Hi David. The following is ("probably", he says) not what you want, but I am leaving it here, as it may explain some of the context. It is an argument of Woodin, similar to one of Jech, which though mathematically alike, is more "proof theoretic" in nature. I wrote some notes on it; they are here. -The key fact you refer to is however not proved model theoretically, but by an appeal to the arithmetic fixed point lemma, also a key fact in Gödel’s approach. -The idea is the following (the note provide the missing details and makes this precise): A property $P$ of models of set theory is hereditary iff whenever $P(M)$ and $N\in M$ is a model of set theory such that $M$ thinks that $P(N)$, then in fact $P(N)$. -Then the following holds: - -For any hereditary $P$, either $P(N)$ fails for all $N$, or else there is an $M$ such that $P(M)$ but $P(N)$ fails for all $N\in M$. - -This is proved by an application of the fixed point lemma: Take $\phi$ such that (ZFC proves tha) $\phi$ holds iff for all $N$, $P(N)$ implies $N\models\lnot\phi$. -It is immediate that if $P(M)$ holds for some $M$, then it holds for some $M$ such that $M\models\phi$. But then $M$ thinks that there is no $N$ satisfying $P$. -Ok. If $P(N)$ is "$N$ is a model of set theory", then $P$ is hereditary, and we have the second incompleteness theorem. As shown in the note, quite a few similar results follow all at once by considering appropriate properties $P$.<|endoftext|> -TITLE: Analytic vs. formal vs. étale singularities -QUESTION [16 upvotes]: There are multiple definitions of when points $P$ and $Q$ on schemes $X$ and $Y$ are equisingular. One can require that either $P$ and $Q$ have isomorphic analytic (i.e. complex analytic), or formal, or étale neighborhoods. My main question is how are these three definition related. I know that for simple singularities they are equivalent, and for difficult ones I guess they are non-equivalent. Does anyone know precise statements, references, proofs, examples when these are non-equivalent notions? -The analytic and the étale case is a little bit subtler. One can ask for the adequate local rings to be isomorphic and also for isomorphic analytic/étale neighborhoods, as above. Are these two approaches equivalent? - -REPLY [5 votes]: None of these notions is the usual definition of equisingular. For example, any choice of four distinct lines though some given point in the plane is usually considered equisingular to any other, however, the analytic isomorphism class depends on the cross-ratio of the slopes.<|endoftext|> -TITLE: Theorems that are 'obvious' but hard to prove -QUESTION [90 upvotes]: There are several well-known mathematical statements that are 'obvious' but false (such as the negation of the Banach--Tarski theorem). There are plenty more that are 'obvious' and true. One would naturally expect a statement in the latter category to be easy to prove -- and they usually are. I'm interested in examples of theorems that are 'obvious', and known to be true, but that lack (or appear to lack) easy proofs. -Of course, 'obvious' and 'easy' are fuzzy terms, and context-dependent. The Jordan curve theorem illustrates what I mean (and motivates this question). It seems 'obvious', as soon as one understands the definition of continuity, that it should hold; it does in fact hold; but all the known proofs are surprisingly difficult. -Can anyone suggest other such theorems, in any areas of mathematics? - -REPLY [6 votes]: If $T_1, \dots, T_k$ are subtrees of a tree $T$ which pairwise intersect, then $\cap_{i=1}^k T_i$ is non-empty. This is a standard fact in graph theory and almost completely obvious, but it is surprisingly difficult to give a correct proof.<|endoftext|> -TITLE: Looking for a particular family of C.Y quintics -QUESTION [7 upvotes]: It is possible to construct (in many ways) a family of Calabi-Yau quintics $\mathcal{X}\rightarrow \Delta$, over disk, such that the fiber over $0$ has a singularities locally given by the equation $x_1^2+x_2^2+x_3^2+x_4^2=0$ and the others are smooth quintics. In such families there is vanishing cycle in generic fiber which is a Lagrangian isomorphic to $S^3$ and collapses to the node at central fiber. This Lagrangian $S^3$ is obtained by the deformation of the local equation: $$x_1^2+x_2^2+x_3^2+x_4^2=0 \rightarrow x_1^2+x_2^2+x_3^2+x_4^2=\epsilon$$ -Here is my questions: Consider the Fermat quintic $z_1^5+\cdots+z_5^5=0$ and its real locus which is a Lagrangian isomorphic to $S^3 /\mathbb{Z}_2$. Is there any one parameter family of C.Y 3-folds with one fiber isomorphic to Fermat quintic, such that real locus of quintic apears as vanishing cycle for this family ? -Comment: I think if there is such family then the singular fiber has to have orbifold singularities given by local equation: ${x_1^2+x_2^2+x_3^2+x_4^2=0}/\mathbb{Z}_2$ and I think its impossible to write a family of degree 5 equations with this type local singularities! - -REPLY [6 votes]: Let me say first, that I really like this question. Very unusual question about such well known things (in fact I did not know even that the real quintic $\sum_i x_i^5=0$ is $\mathbb RP^3$). -This is not an answer, but more like my interpretation of this question (hopefully correct one). The first comment is about vanishing cycles. Usually, when we speak about vanishing cycles in symplectic geometry, we speak about lagrangan spheres in a fiber of a Lefshetz fibraton constructed using parallel transport given by the symplectic connection on the fibers. But Lefshetz pencil is just the most simple object of algebraic geometry. We can consider instead other one-parameter families of algebraic varieties, that can have singularities more complicated than double points $A_1$. In this case again we can ask what will be the shape of the vanishing cycle constructed via symplectic parallel transport? How does it look like? The point that I want to make is that in a large case of situations, this vanishing cycle will not look at all like a manifold. -Namely, we will consider the example coming from a function on $\mathbb C^n$ with isolated singularity. I.e., we have an analytic function $F:\mathbb C^n\to \mathbb C$ with isolated singularity at $0$ and consider its level sets: hyper-surfaces $F_t:=F^{-1}(t)$ intersected with $B_\varepsilon(0)$ (the ball of radius $\varepsilon$). Then we know that the homotopy type of $F_t\cap B_{\varepsilon}$ is a bucket of $k$ spheres, where $k$ is the Milnor number of the singularity. Now, I think (and here I can not provide the proof), that the vanishing cycle is a deformation retract of $F_t\cap B_{\varepsilon}$. Hence it can not be diffeomorphic to a manifold unless $k=1$. In this case it has to be homotopy equivalent to a sphere, i.e., homeomoerphic to it (by Poincare). -In the case when the singularity of $F$ is non-degenerate (double point), the vanishing cycle is of course diffeomorphic to $S^n$, but maybe if we consider Brieskorn singularities we will be able to get vanishing cycles that are exotic spheres as well? -The above situation concerns the case when the total space of the fibration is smooth. In this case we does not seem to be able to get $\mathbb RP^{n-1}$ as a vanishing cycle. But if we allow singularities in the total space of the fibration, we can get it. Indeed if we consider the function $\sum_i x_i^2=0$ on the total space $\mathbb C^n/\mathbb Z_2$, where $\mathbb Z_2$ is acting by $z\to -z$, the vanishing cycle will be diffomorphic to $\mathbb RP^{n-1}$. In sum, it seems to me that the following question is interesting: can we describe the class of manifolds what appear as vanishing cycles of algebraic one-parameter families? -As for the question itself, I don't know what is the answer...<|endoftext|> -TITLE: Perturbation in C*-Algebra -QUESTION [7 upvotes]: Let x be an element in a C*-algebra A, is it true that if x approximately commute with every element in A, then x is near the centre of A? More precisely, I want to know whether the following is true： Let x be an element in a C*-algebra A with norm 1. Then for any $\epsilon>0$, there exist a $\delta>0$ such that the following holds: $\forall y\in A,||y||=1,||xy-yx||<\delta\Rightarrow dist（x, Centre(A)）<\epsilon$. This is true if A is the matrix algebra, but I was wondering whether it can be generalized to any C*-algebra - -REPLY [4 votes]: In general, the answer to this has to be "no". To construct a counterexample, let $S$ be a dense subset of the unit interval (0,1], and let $\mathcal{H}=\ell^2(S)$ be the Hilbert space with orthonormal basis $(e_s)_{s\in S}$. For each positive integer $n$, let $\mathcal{H}_n$ be the subspace of $\mathcal{H}$ generated by linear combinations of $e_s$ over $1/(n+1) < s\le 1/n$. We can define a $C^*$ sub-algebra $\mathcal{A}$ of the bounded linear operators $B(\mathcal{H})$ as follows: $A\in B(\mathcal{H})$ is in $\mathcal{A}$ if and only if the following are satisfied - -$A$ maps $\mathcal{H}_n$ into $\mathcal{H}_n$. -There exists a sequence of complex numbers $a_n$ such that, for each fixed $n$, -$$\Vert Ae_s-a_ne_s\Vert+\Vert A^*e_s-\bar a_ne_s\Vert\to0$$ -as $s\to 1/n$. - -It is easily seen that this is a sub-$C^*$-algebra. I'm thinking of $\mathcal{A}$ intuitively as being those elements of the direct product (sum?) $\prod_nB(\mathcal{H}_n)$ "joined continuously at the edges", and roughly as continuous functions $S\to\mathbb{C}$ but allowing some non-commutativity within the intervals $(1/(n+1),1/n]$. -If operator $A$ is in its center then it must restrict to the center of each $B(\mathcal{H}_n)$, so must be a constant on each of these. That is, $Ae_s=\lambda_se_s$ for some $\lambda_s\in\mathbb{C}$, which is independent of $s$ over $1/(n+1) < s\le 1/n$. Also, $s\mapsto\lambda_s$ must be continuous at $s=1/n$, from which we see that $s\mapsto\lambda_s$ is constant. So, $\mathcal{A}$ has trivial center. -Fixing a positive integer $m$, define an operator $A\in\mathcal{A}$ by $Ae_s=\min(ms,1)e_s$. Then, $\Vert A\Vert = 1$ and ${\rm dist}(A,{\rm Center}(\mathcal{A}))=1/2$. On the other hand, the restriction of $A$ to each $\mathcal{H}_n$ is within a distance $1/(2(m+1))$ from its center. This gives $\Vert AB - BA\Vert \le \Vert B\Vert / (m+1)$ for all $B\in\mathcal{A}$. -Choosing $m$ as large as we like, this contradicts your claim.<|endoftext|> -TITLE: Fibrewise homotopy-equivalence of unit sphere bundles vs isomorphism of tangent bundles -QUESTION [9 upvotes]: Let $M$ be a smooth $m$-dimensional manifold, $TM$ its tangent bundle and $SM$ its unit sphere bundle. -Are there some simple examples where $SM$ is fibrewise homotopy-equivalent to the trivial bundle $S^{m-1} \times M$, yet $TM$ is not trivial as a vector bundle? Does it ever happen for $M$ a sphere? -Via classifying space machinery this amounts to comparing the orthogonal group $O_m$ to the space of homotopy-equivalences of $S^{m-1}$, $HomEq(S^{m-1})$, in particular its asking for tangent bundle classifying maps $M \to BO_m$ such that the composite $M \to BO_m \to BHomEq(S^{m-1})$ is null. -As far as I know I've never come across examples of this sort, but then again I haven't studied the homotopy-properties of the map $O_m \to HomEq(S^{m-1})$ in much detail. Are there many canonical references on this topic? -This is related to a math.stackexchange question: https://math.stackexchange.com/questions/16779/conditions-for-a-smooth-manifold-to-admit-the-structure-of-a-lie-group - -REPLY [9 votes]: It is proved in [Kaminker, J., Proc. Amer. Math. Soc. 41 (1973), 305–308] that the tangent sphere bundle of a closed smooth H-manifold is (unstably) fibre homotopy trivial. On other other hand, surgery theory allows to construct H-manifolds with non-trivial rational Pontrjagin classes, see e.g. [Victor Belfi, Pacific J. Math. vol 36, Number 3 (1971), 615-621] but this is a standard surgery-theoretic argument. -Finally, in [Milnor-Spanier, Pacific J. Math. vol 10, Number 2 (1960), 585-590] it is shown that the tangent bundle to $S^n$ is fiber homotopy trivial if and only if $n=1,3,7$, -which are exactly the parallelizable spheres (by Bott-Milnor). -Thus the above is a complete answer to your question. This phenomenon you ask for never occus for spheres but occurs for lots of H-manifolds.<|endoftext|> -TITLE: Can Lie algebra cohomology prove Cartan's Semisimplicity Criterion? -QUESTION [8 upvotes]: Here is what I mean by "Cartan's semisimplicity criterion": -Let $\mathfrak g$ be a finite-dimensional Lie algebra over a field of characteristic $0$. Assume that the center of $\mathfrak g$ is trivial. Then, the following three assertions are equivalent: -(1) The Killing form on $\mathfrak g\times \mathfrak g$ is nondegenerate. -(2) Every short exact sequence of finite-dimensional representations of $\mathfrak g$ splits. -(3) Every subrepresentation of the adjoint representation of $\mathfrak g$ has a complementary subrepresentation. -What I am looking for is a slick proof for this equivalence (although the only thing I really need is a proof of (1) $\Longrightarrow$ (3)). I am aware of the proof in Fulton-Harris Appendix C, but this could fill an hour of talking and seems to involve many unmotivated ideas. Is there something more explanatory? Using cohomology perhaps? Is the whole thing obvious from an advanced viewpoint? (I don't mean using the classification of simple Lie algebras, of course...) Maybe newer ideas such as Lie algebroids, algebraic groups etc. can help? - -REPLY [5 votes]: My favorite proof is showing that the trivial module is projective in the category $\mathcal R$ of finite dimensional reps; you can do this using cohomology and the quadratic Casimir (so here the Killing form comes in). After that, showing that all fin. dim. reps. are semisimple amounts to showing that they are all projective in $\mathcal R$, and this follows from the identity $\hom_\mathfrak g(M,N)=\hom_\mathfrak g(k,\hom(M,N))$ and a little cohomological yoga. - -REPLY [4 votes]: My favorite proof is the Harish-Chandra isomorphism. -Choose the PBW isomorphism $U(\mathfrak g)\cong U(\mathfrak n_-)\otimes U(\mathfrak h)\otimes U(\mathfrak n_+)$ and let $p: U(\mathfrak g)\to U(\mathfrak h)\cong \mathbb k[\mathfrak h^*]$ be the projection killing all higher order terms in the other factors. Then -Theorem. The restriction of $p$ to the center $Z(U(\mathfrak g))$ is an injection whose image is the invariant functions for the $\rho$-shifted Weyl group action $w\bullet \lambda =w(\lambda+\rho)-\rho$ on $\mathfrak h^*$. -This means that the center acts in the same way on two different highest weight representations if and only if $\lambda+\rho$ and $\mu+\rho$ are in the same orbit of the Weyl group (consider the action on the highest weight vector). This can't happen for two different dominant weight vectors ($\lambda+\rho$ is in the interior of the dominant Weyl chamber), so the action of the center distinguishes different simple representations. -Now we need to show that these can't glue to each other; you can easily reduce to the case of an extension $V\to W \to V$ for a simple $V$. In this case, just pick a splitting on the highest weight space, and let $V'\subset W$ be the sub-rep generated by the preimage of the highest weight in the quotient copy of $V$. By PBW, the intersection of this with the highest weight space is 1-dimensional, so it cannot have any intersection with the sub-module, but it maps surjectively to the quotient. Thus it is a complement and the sequence splits. -I think this shows that there really is something delicate about this semi-simplicity; this isn't the lowest tech proof, but they're always a little subtle. You can also see this from how wildly semi-simplicity fails in the infinite-dimensional case; look at some literature on Verma modules and category $\mathcal O$ to see how complicated this can get.<|endoftext|> -TITLE: Shot in the dark: Is there an english translation of Deligne-Rapoport "Les schemas de modules..." anywhere? -QUESTION [7 upvotes]: Extensive googling (and searching here) has yielded nothing, unfortunately. -I knew a language genius once who offered to translate it for me as a favor, but I turned him down because it seemed like too much to ask someone to do for free. -At some point I suppose I'm going to just have to learn french. - -REPLY [2 votes]: https://math.stackexchange.com/questions/8854/has-deligne-rapoport-been-translated<|endoftext|> -TITLE: ultrapowers and higher order logic -QUESTION [12 upvotes]: One of the reasons I think ultrapowers are interesting is the following corollary of Łoś's theorem: - -Let $V$ be a relational structure and $^*V$ an ultrapower of $V$. Then a first order statement about $V$ is true if and only if the corresponding statement for $^*V$ is true, where the relations $R$ on $V$ are replaced by their enlargements $^*R$. - -This suggests the following question: what construction should replace the ultraproduct if I want a similar result for 2nd or higher order statements? A little googling suggests that a book by van Benthem called Modal Logic and Classical Logic might contain an answer, but I don't have access to a copy. - -REPLY [3 votes]: As pointed out by Matthew Towers, any ultrapower construction that preserves second order statements must be trivial at least in those cases where the original structure satisfies a categorical second order statement. -But there may be fragments of second order logic that can be preserved by ultrapowers and/or ultraproducts. - -As a trivial example, any purely existential second order statement which is true in all factors will be true in the ultraproduct (since an expansion of the original structure by new predicates induces an expansion of the ultraproduct). -As a trivial non-example: already the purely universal and monadic second order statement "every subset containing 0 and closed under successor is the whole set" is not preserved in any nontrivial ultrapower of the natural numbers, nor can it be preserved in any (nontrivial) proposed "advanced ultrapower construction".<|endoftext|> -TITLE: How to un-pixelate pixelated regions in films? -QUESTION [11 upvotes]: Everybody knows the effect of pixelated objects (e.g. faces) on TV. Is there a way - and which mathematical method lies behind it - to un-pixelate the region? Beware: I am not talking about smoothing it out (unblur) image by image - but about using the information of a whole sequence of pixelated pictures where the object which is pixelated naturally changes the location within the overall pixelated region. I guess it must be some kind of "un-averaging over time" (e.g. differentiation over time comes into mind). -Has anybody tackled this problem yet? Are there references, demos, software...? -Thank you. - -REPLY [14 votes]: This is a problem that has been addressed by the computer vision community in the past few years. You may want to look into upsampling and super-resolution techniques as well as vectorization methods. Here is a recent paper on the topic in ICCV 2009 which gives a good summary of the recent efforts in this field: http://www.wisdom.weizmann.ac.il/~vision/SingleImageSR.html<|endoftext|> -TITLE: Restrictions of perfect Morse functions to submanifolds -QUESTION [9 upvotes]: A Morse function $f: M \rightarrow \mathbb R$ on a connected closed manifold $M$ is called $\mathit{perfect}$ with respect to the field $\mathbb F$ if all of the Morse inequalities are equalities, i.e. the number of critical points of $f$ with index $k$ coincides with the $k$-th Betti number of $M$ with respect to $\mathbb F$-coefficients for all $k$. -Now assume that $f: M \rightarrow \mathbb R$ is a perfect Morse function on a closed connected Riemannian manifold and that $N \subset M$ is a closed submanifold, such that $N$ contains all the critical points of $f$ and the restriction $f|_N$ is a Morse function on $N$. If in addition the gradient $\nabla f$ is tangent to $N$ along $N$, i.e. $\nabla f|_N \in \Gamma(TN)$, then $N$ is a union of flow lines and the critical points of $f|_N$ are exactly the critical points of $f$. -Now my question is: Is it possible to deduce perfectness of $f|_N$ from perfectness of $f$ in this setting, or are there additional conditions under which this is possible? - -REPLY [8 votes]: Isn't there a counterexample with $(M,N)=(\mathbb CP^2,\mathbb RP^2)$?<|endoftext|> -TITLE: For which $c$ is $\mathbb{Z}[\sqrt{c}]$ a unique factorization domain? a Euclidean domain? -QUESTION [20 upvotes]: Let $c$ be an integer, not necessarily positive and $|c|$ not a square. Let $\mathbb{Z}[\sqrt{c}]$ be the set of complex numbers $$a+b\sqrt{c}, \quad a, b\in \mathbb{Z},$$ -which form a subring of the ring $\mathbb{C}$ under the usual addition and multiplication. -Are the following questions completely solved? - -For what $c$ is $\mathbb{Z}[\sqrt{c}]$ a Euclidean domain? -For what $c$ is $\mathbb{Z}[\sqrt{c}]$ a UFD (unique factorization domain) but not Euclidean ? -For what $c$ is $\mathbb{Z}[\sqrt{c}]$ not a UFD ? - -I know that for $c=-1$, Question 1 has a positive answer; for $c=-5$, Question 3 has a positive answer. - -REPLY [22 votes]: A quadratic order has unique factorization (or is Euclidean) only if it is the maximal order of a number field (it must be integrally closed); for your examples, this holds if and only if $c$ is a squarefree integer congruent to $2$ or $3$ modulo $4$. - -It is known, as others have pointed out, which quadratic fields are Euclidean with respect to the absolute value of the norm. In the complex quadratic case, every Euclidean ring is norm-Euclidean (Motzkin); in the real quadratic case, there are examples (by students of Murty, namely Clark and Harper) of Euclidean rings that are not norm-Euclidean. -By a result of Weinberger, every real quadratic field with unique factorization is Euclidean assuming the generalized Riemann hypothesis. -The ring of integers in a quadratic number field is not a UFD if its class number is nontrivial; it is easy to construct examples by making $c$ a product of at least three primes. It is believed but not known that infinitely many quadratic number rings have class number $1$. - -See Simachew or my -own survey for references. - -REPLY [7 votes]: This is more a comment to the question (which I cannot do). -As written (intentionally?) [ADDED: Apparently, it was intentional.] the specified rings are not always the (full) ring of algebraic integers of the field $\mathbb{Q}(\sqrt{c})$ (see, e.g., https://en.wikipedia.org/wiki/Quadratic_integer for details). -In these cases the rings in question are not integrally closed and thus not UFDs, even if the class number of the field is one and thus the (full) ring of algebraic integers would be a UFD (see, e.g., https://en.wikipedia.org/wiki/Class_number_problem ). -Possibly, one needs to take this into account too, when using the list, mentioned in another answer, where the rings are Euclidean. -ADDED: Franz Lemmermeyer's answer is considerably more complete than this remark. - -REPLY [6 votes]: Question 1 is slightly ambiguous. Does "euclidean" mean euclidean with respect to the obvious norm, or does it mean euclidean with respect to some norm? -For example, Z[$\sqrt{14}$] is not euclidean w.r.t. the usual norm, but I'm not sure it's known whether there's some other norm that makes it euclidean.<|endoftext|> -TITLE: topology on the automorphism group of a C* algebra -QUESTION [9 upvotes]: Let $A$ be a $C^*$-algebra. The group $Aut(A)$ of $\ast$-automorphisms of $A$ is usually equipped either with the pointwise norm topology, i.e. the topology generated by the semi-norms $\lVert \varphi \rVert_a = \lVert \varphi(a) \rVert$ or with the uniform topology, i.e. the one which it earns by inclusion into the Banach space of bounded linear operators from $A$ to $A$. Moreover, let $U(A) = U(M(A))$ be the unitary group of the multiplier algebra equipped with the strict topology. Now, the map $Ad : U(A) \to Aut(A)$ is continuous if $Aut(A)$ carries the pointwise norm topology. It induces a continuous bijection $U(A) / Z(U(A)) \to Inn(A)$ onto the inner automorphisms, which is not a homeomorphism unless the $C^\ast$-algebra is a continuous trace $C^{\ast}$-algebra. So, my questions are - -Is the induced bijection $U(A)/Z(U(A)) \to Inn(A)$ a homeomorphism, if $Aut(A)$ carries the uniform topology? -Is there a natural topology on $Aut(A)$, which induces a homeomorphism -$U(A)/Z(U(A)) \to Inn(A)$? - -REPLY [6 votes]: If $Aut(A)$ carries the uniform topology but $U(A)/Z(U(A))$ the topology induced by the strict one, then the bijection is not continuous. -For example, let $A=C_0(N,M_2(C))$ be the $C^*$-algebra of all $2\times 2$-matrix-valued functions on the naturals vanishing at infinity, and define for each $n \in N \cup \{\infty\}$ a unitary $U_n \in U(A)$ such that for $k \leq n$, $U_n(k)$ is $0$ on the diagonal and $1$ off the diagonal and for $k > n$, $U_n(k)$ is the identity.Then the sequence of the $U_n$ converges strictly to $U_{\infty}$. But, if $\chi_{m} \in A$ denotes the function that is equal to the identity in $M_2(C)$ on $\{0,\ldots,m\}$ and $0$ thereafter, then $\|(Ad_{U_n}- Ad_{U_{n+1}})(\chi_m)\|$ is constant non-zero for $m>n$ independently of $n$.<|endoftext|> -TITLE: What is the closure of product loci in A_g? -QUESTION [6 upvotes]: Let $A_g$ denote the moduli space of principally polarized abelian varieties of dimension $g$. For any partition of $g$, one can consider the corresponding locus inside $A_g$ of products of lower-dimensional abelian varieties. -My question is: can one describe explicitly the closure of this locus in (some) compactification? -I guess part of this question is what compactification one should use. For instance, if one simply chooses a toroidal compactification $\widetilde{A}_g$, one can also extend the universal family to a family of semistable abelian varieties. Can one describe necessary and sufficient conditions for a semistable abelian variety at the boundary for it to be in the closure of such a product locus? Or perhaps if one restricts attention to rank-one degenerations, so it does not depend on a choice of toroidal compactification? -Alternatively, one could choose a point in the closure of $A_g$ inside Alexeev's space of stable semi-abelic pairs -- are there necessary and sufficient conditions in this situation? - -REPLY [2 votes]: This isn't really an answer, but perhaps a suggestion for where to get some ideas. Michael van Opstall studied moduli spaces of products of stable curves (of general type) in this and this papers. While his focus is on curves of genus $g>1$, some of the general deformation theoretic arguments might help you. Also, after all, a polarized abelian variety is of log general type, so possibly the arguments in this paper can be extended to pairs and then applied à la Alexeev.<|endoftext|> -TITLE: A question on the singular series and singular integral in Hardy-Littlewood Circle Method -QUESTION [5 upvotes]: Given a set $A$ of positive integers, set $r_{A,h}(N)$ to be the number of $h$-tuples $(a_1, \cdots, a_h)$ such that $N = a_1 + \cdots + a_h$. Set $f(z) = \sum_{a \in A} z^a$. Then by Cauchy's Theorem we have that $r_{A,h}(N) = \frac{1}{2\pi i} \int_{|z| = r} \frac{f(z)^h}{z^{N+1}}dz$ for $0 < r < 1$. This can be simplified if we replace $f$ with the trigonometric polynomial $f_N(z) = \sum_{a \in A, a \leq N} z^a$. The key is to estimate the integral, and to do this one breaks the circle into major and minor arcs and to show that the integral over the minor arcs is negligible compared to the integral over the major arc. A key component is to show that the integral over the major arc can be factored into a singular integral, usually denoted $J(N)$ or $J_A(N)$ if the set $A$ is unclear, and a 'singular series' $\mathfrak{S}(N)$ or $\mathfrak{S}_A(N)$ with some small error term. The Circle Method has been applied fruitfully to provide an asymptotic formula for the classical Waring's problem and Vinogradov's Theorem giving an asymptotic formula for sum of three primes (see for example Nathanson's Additive Number Theory - Classical Bases). -My question is, if $A$ is an arbitrary set of positive integers, how does one go about constructing $\mathfrak{S}_A(N)$ or $J_A(N)$? If this is impossible in general or far too intractable, what properties of the integer powers (less interesting) and the primes (more interesting) allows us to find explicit singular integrals and singular series? Can anyone give an example of an 'arbitrary' (meaning not given by a specific formula, somewhat like the primes) set of positive integers with a nice singular series and a singular integral? - -REPLY [2 votes]: $A$ should have a fairly regular distribution in arithmetic progressions to yield a reasonable singular series and singular integral. This is why Vinogradov published his solution to the ternary Goldbach problem only after the Siegel-Walfisz theorem was available. The idea is that the relevant trigonometric polynomial has its peak at $z=1$ (obviously) and similar looking but smaller peaks at roots of unity (which is not obvious at all). Now the extent of success depends in large on the extent of "similarity" in this statement.<|endoftext|> -TITLE: Amenability of groups in terms of a perturbation condition -QUESTION [15 upvotes]: Let $G$ be a countable group and $\lambda \colon G \to U(\ell^2 G)$ its left-regular representation. Suppose that there exists a constant $C>0$ such that for all $T \in B(\ell^2 G)$ -$$\inf \lbrace\|T-S\| \mid S \in \lambda(G)' \rbrace \leq C \cdot \sup\lbrace \|\lambda(g)T- T \lambda(g) \| \mid g \in G\rbrace.$$ -(Here $\lambda(G)'$ denotes the commutant of $G$ in $B(\ell^2 G)$.) - -Question: Is $G$ amenable? - -It is fairly easy to see that amenability of $G$ implies the existence of such a constant. Indeed, one may take $S$ to be some fixed point for the conjugation action on $\overline{\rm conv}\lbrace \lambda(g)T\lambda(g)^* \mid g \in G\rbrace$. I am asking for the converse of this statement. -EDIT: Since the derivation problem came up in Kate's comment, I want to clarify to what version of it my question is related. The inequality above holds for some $C$ if and only if the first bounded cohomology of $G$ with coefficients in $B(\ell^2 G)$ (with the conjugation action induced by $\lambda$) is reduced. This is a straightforward application of the open mapping theorem. Now, two things are unclear: - -Question: Can $H^1_b(G,B(\ell^2 G))$ be reduced without being zero? - -and - -Question: Can $H^1_b(G,B(\ell^2 G))$ be zero without $G$ being amenable? - -REPLY [3 votes]: Let $M=C^*_\lambda(G)''$ be group von Neumann algebra of $G$. The the condition above implies: -$d(T, M')\leq C ||ad(T)|_{M}||$ for every $T\in B(l^2 G)$. -The last inequality is equivalent to saying that every derivation of $M$ into $B(l^2 G)$ is inner. -Edit: the above inequality is satisfied automatically (was clarified to me by Stuart White). -It is known that if $M\subset B(H)$ has a cyclic vector, then every bounded derivation -from $M$ into $B(H)$ is inner [E. Christensen, Extensions of -derivations II, Math. Scand., 1982]. Thus [Christensen, op cit, Cor -5.4] we have -$d(T,M')\leq 3/2\|(\mathrm{ad}| T)_{M}\|$ -for every $T\in B(l^2 G)$.<|endoftext|> -TITLE: Row reduction of sparse matrices -QUESTION [5 upvotes]: Let $p$ be prime (of size roughly $100$, say). Suppose that $M$ is a matrix with coefficients in $\mathbf{F}_p$ with roughly $An$ rows and $n$ columns, where $A>1$ is some fixed small constant. Suppose, moreover, that - -Every row of $M$ has at most $B$ non-zero entries, which are all $\pm 1$, for some small $B$. -The rank of this matrix is small, say, of order $O(\sqrt{n})$ (this may or may not be relevant.) - -I am interested in computing the rank of $M$. Does the fact that $M$ is sparse imply that there are relatively efficient ways of computing this rank? How large should I expect to be able to take $n$ and still be able to compute the rank? The current program I am running (which was kindly written [very quickly without a view to optimization] for me by someone else) chokes up with -$n \sim 5000$ or so, and for $n \sim 4000$ seems to take around $20$ minutes. - -REPLY [4 votes]: The LinBox project provides a C++ library which can effectively compute the rank of sparse matrices over finite fields. -You could also look at Sage which should also allow this, but I am not sure whether they have effective implementations for this particular case. If you have access to it, then Magma can also do this.<|endoftext|> -TITLE: How did Bernoulli prove L'Hôpital's rule? -QUESTION [17 upvotes]: To prove L'Hôpital's rule, the standard method is to use use Cauchy's Mean Value Theorem (and note that once you have Cauchy's MVT, you don't need an $\epsilon$-$\delta$ definition of limit to complete the proof of L'Hôpital). I'm assuming that Cauchy was responsible for his MVT, which means that Bernoulli didn't know about it when he gave the first proof. So what did he do instead? - -REPLY [4 votes]: I think the important reference in which its author describes in detail the proof of L'Hospital's rule done by l'Hospital in his book but with todays language is the following -Lyman Holden, The March of the discoverer, Educational Studies in Mathematics, Vol. 5, No. 2 (Jun., 1973), pp. 193-205.<|endoftext|> -TITLE: Is Robinson Arithmetic biinterpretable with some theory in LST? -QUESTION [16 upvotes]: Let ZFC$^{\text{fin}}$ be ZFC minus the axiom of infinity plus the negation of the axiom of infinity. It is well-known that ZFC$^{\text{fin}}$ is biinterpretable with Peano Arithmetic. In this sense one could argue that ZFC$^{\text{fin}}$ is PA couched in the language of set theory (ie one nonlogical binary relation, $\in$) rather than the language of arithmetic ($+$, $\cdot$, $0$, $S$). This gives us some confidence that "there exists an infinite set" -- and the hierarchy of large cardinal axioms beyond -- is an at least somewhat-natural extension of arithmetic. -In precise terms, every theory in this hierarchy proves the consistency of all those before it. In vague terms, each theory in this hierarchy adds "more infiniteness" than those before it. -Does the hierarchy start at PA, or is there a step below it? Robinson Arithmetic is a theory in the language of arithmetic; among its properties are: - -Robinson Arithmetic is essentially undecidable (as PA and all stronger theories are) -PA proves the consistency of Robinson Arithmetic -Robinson Arithmetic is finitely axiomatizable - -The first point might be considered an argument for why Robinson Arithmetic is part of the same hierarchy as PA/ZFC$^{\text{fin}}$ -- it has enough coding power to express primitive recursion. The second point shows why Robinson Arithmetic is strictly below PA/ZFC$^{\text{fin}}$ on this hierarchy. The third point explains -- in vague terms -- what sort of "infiniteness" PA/ZFC$^{\text{fin}}$ add to Robinson Arithmetic: it adds infinite collections of axioms. -From PA on up, all theories on the hierarchy are biinterpretable with some theory in the language of set theory. - -Question: is Robinson Arithmetic biinterpretable with some theory in the language of set theory? - -REPLY [5 votes]: Since the other (excelent) answers didn’t touch the biinterpretability aspect of the question, let me comment on that. -First, a caveat: arbitrary structures (say, in a finite language) can be encoded by directed graphs. Thus, it might very well be the case that $Q$ is biinterpretable with a theory in a language with one binary predicate (i.e., in the language of set theory) which looks nothing like a “set theory” at all, and is not included in or consistent with any recognizable theory of sets like ZFC. -So, a more natural question is if $Q$ is biinterpretable with any theory that contains at least a modicum of set theory. While I didn’t exactly specified what this means, I propose that the answer is negative because of the following result of Albert Visser: $Q$ is not biinterpretable with any theory which includes the axiom of pairing $\forall x,y\,\exists z\,\forall t\,(t\in z\leftrightarrow t=x\lor t=y)$. See On $Q$.<|endoftext|> -TITLE: Coproducts of complete Boolean algebras -QUESTION [12 upvotes]: Does the category of complete Boolean algebras have binary coproducts? -Note that this category does not have countable coproducts. Indeed, the coproduct of countably many copies of the four element complete Boolean algebra would be the free complete Boolean algebra on countably many generators, and such an object does not exist. - -REPLY [16 votes]: Chris Heunen's comment under the OP can be turned into a proof. Suppose the category of compact Hausdorff extremally disconnected spaces has binary products. Let $X \times Y$ denote the product in that category. If $|X|$ denotes the underlying set, then of course the canonical map -$$|X \times Y| \to |X| \times |Y|$$ -is an isomorphism, because $|X| \cong \hom(\ast, X)$ where $\ast$ is the one-point space, i.e., the underlying set functor is representable and representables preserve products. -Chris observes that the ordinary product space $X \times_{Top} Y$ of two compact Hausdorff extremally disconnected spaces need not be extremally disconnected. However, under our supposition we would have a continuous comparison map -$$X \times Y \to X \times_{Top} Y$$ -in $Top$ which is a bijection at the level of the underlying sets. Being a continuous bijection between compact Hausdorff spaces, it is a homeomorphism, and this contradicts Chris's observation.<|endoftext|> -TITLE: About an exercise in Serre's "Trees" -QUESTION [16 upvotes]: Following section 1.4, which is entitled "Constructions Using Amalgams" is the innocent-enough looking exercise: -Show that the group defined by the presentation (presumably on the generators $x_1, x_2, x_3$) -$x_2x_1x_2^{-1}=x_1^2 $ -$ x_3x_2x_3^{-1} = x_2^2$ -$ x_1x_3x_1^{-1} = x_3^2$ -is trivial. -Now, if one simply (well, not simply in practice but simply as in I could write code to do it much more easily than I can actually do it myself...one quickly loses track of inverses etc. and discovers new magical relations that are, in fact, too good to be true) manipulates relations, writes certain words and then tries to move generators past each other, one can obtain this result. Which leaves me to wonder: why precisely is this exercise here? -Guess 1: This exercise stands in contrast to an example in this section where we have the same set of relations except on 4 generators, so 4 relations, and so the same kind of proof does not work out at all (you can no longer move all the generators past one-another); in fact Serre uses amalgams to demonstrate that the aforementioned group contains the free group on two generators. One may perhaps be tempted (as I was, indeed, for way too long) to attempt to use amalgams to solve the exercise which is the topic of this post, to no avail. Maybe the point here is that sometimes amalgams get you nothing; that some types of questions lend themselves to proof by generator and relation manipulation but others to more "abstract" diagram chasing? -Guess 2: There is an elegant proof somehow using amalgams and it escaped me. -Guess 3: Some other reason. -Anyway, perhaps from my years of secondary, college, and finally graduate math textbook experience, I feel somehow uneasy solving an exercise at the end of a section without employing any tools at all from that section. I worry I've missed the point. -Any ideas? (Excluding perhaps that I am overthinking this and need to let it go?) - -REPLY [8 votes]: I don't have the book next to me but I'm quite sure I remember the exercise and, as far as I was able to determine, your first guess is the correct one. This isn't an amalgam exercise in any natural way (or any way at all), but rather an example to put the case of 4 generators in perspective.<|endoftext|> -TITLE: References on Moishezon spaces in English/French -QUESTION [8 upvotes]: I'm looking for references on the proof (due to B. Moishezon, I guess) that any Moishezon space becomes a projective smooth complex variety after a finite number of blow-ups (called a modification?) His own articles on this that I could find are mainly in Russian, except a few survey papers (in English) without too much proof. I don't read Russian. -Could anyone give some references in English/French? Also, since I didn't read Moishezon's original paper, I don't know the precise statement for this result (e.g. if one can impose some constraints on the subspaces that we blow up). Can someone give the precise statement? - -REPLY [9 votes]: Dear Shenghao, -I) I am happy to report that according to Ueno's Classification of Algebraic Varieties and Compact Complex Spaces, Springer LNM 439, 1975, Moishezon's papers have been translated into English : AMS Translation Ser.2, 63 (1967), 51-177. -II) Here is the precise statement you request. Given a Moishezon variety $X$, you can obtain a smooth projective manifold $\tilde X$ from it by a finite succession of blowing-ups and the canonical morphism between their meromorphic function fields -$\mathcal M (X) \to \mathcal M (\tilde {X})$ is an isomorphism of fields. If $X$ is smooth, the blow-ups can be taken with smooth centers. -III) And finally two facts you probably already know. -a) A Moishezon manifold is projective algebraic iff it is Kähler. -b) A smooth Moishezon surface , i.e. manifold of dimension two, is automatically projective. This is a theorem by Chow and Kodaira ( proved long before Moishezon introduced his general definition ).<|endoftext|> -TITLE: Presheaves are locally sheaves? -QUESTION [7 upvotes]: On nlab it says that a presheaf is locally isomorphic to a sheaf. What do they mean by locally isomorphic? Their definition of locally isomorphic is given in terms of Grothendieck topologies which i think is overkill. -When I first read the nlab page, I thought that it might mean that every presheaf, when restricted to a small enough open set is a sheaf, but I have doubts now because I can't find a proof in the literature and I can't prove it myself. - -REPLY [4 votes]: Arrgh, I wish to delete this, but do not know how. So i will make it into a question. Georges' nice answer is a presheaf that violates the existence axiom s2 on every nbhd of a point. Is there an example that violates the uniqueness axiom s1 on every nbhd of some point?<|endoftext|> -TITLE: Perron number distribution -QUESTION [65 upvotes]: A Perron number is a real algebraic integer $\lambda$ that is larger than the absolute value of any of its Galois conjugates. The Perron-Frobenius theorem says that any -non-negative integer matrix $M$ such that some power of $M$ is strictly positive has a -unique positive eigenvector whose eigenvalue is a Perron number. Doug Lind proved the converse: given a Perron number $\lambda$, there exists such a matrix, perhaps in dimension -much higher than the degree of $\lambda$. Perron numbers come up frequently in many places, especially in dynamical systems. -My question: - -What is the limiting distribution of Galois conjugates of Perron numbers $\lambda$ in - some bounded interval, as the degree goes to infinity? - -I'm particularly interested in looking at the limit as the length of the interval goes to -0. One way to normalize this is to look at the ratio $\lambda^g/\lambda$, as $\lambda^g$ -ranges over the Galois conjugates. Let's call these numbers \emph{Perron ratios}. -Note that for any fixed $C > 1$ and integer $d > 0$, there are only -finitely many Perron numbers $\lambda < C$ of degree $< d$, since there is obviously a bound on the discriminant of the minimal polynomial for $\lambda$, so the question is only interesting when a bound goes to infinity. -In any particular field, the set of algebraic -numbers that are Perron lie in a convex cone in the product of Archimedean places of the field. For any lattice, among lattice points with $x_1 < C$ that are within this cone, the projection along lines through the origin to the plane $x_1 = 1$ tends toward the uniform -distribution, so as $C \rightarrow \infty$, the distribution of Perron -ratios converges to a uniform distribution in the unit disk (with a contribution for each complex place of the field) plus a uniform distribution -in the interval $[-1,1]$ (with a contribution for each real place of the field). -But what happens when $C$ is held bounded and the degree goes to infinity? This question seems -related to the theory of random matrices, but I don't see any direct translation from -things I've heard. Choosing a random Perron number seems very different from choosing -a random nonnegative integer matrix. -I tried some crude experiments, by looking at randomly-chosen polynomials of a fixed degree whose coefficients are integers in some fixed range except for the coefficient of $x^d$ -which is $1$, selecting from those the irreducible polynomials whose largest real root is Perron. This is not the same as selecting a random Perron number of the given degree -in an interval. I don't know any reasonable way to do the latter except for small enough $d$ and $C$ that one could presumably find them by exhaustive search. -Anyway, here are some samples from what I actually tried. -First, from among the 16,807 fifth degree polynomials with coefficients in the range -3 to 3, there are $3,361$ that define a Perron number. Here is the plot of the Perron ratios: -alt text http://dl.dropbox.com/u/5390048/PerronPoints5%2C3.jpg -Here are the results of a sample of 20,000 degree 21 polynomials with coefficients -between -5 and 5. Of this sample, 5,932 defined Perron numbers: -alt text http://dl.dropbox.com/u/5390048/PerronPoints21.jpg -The distribution decidedly does not appear that it will converge toward a uniform distribution on the disk plus a uniform distribution on the interval. Maybe the artificial bounds on the coefficients cause the higher density ring. - -Are there good, natural distributions for selecting random integer polynomials? Is there a - way to do it without unduly prejudicing the distribution of roots? - -To see if it would help separate what's happening, -I tried plotting the Perron ratios restricted to $\lambda$ in subintervals. For -the degree 21 sample, here is the plot of $\lambda$ by rank order: -alt text http://dl.dropbox.com/u/5390048/CDF21.jpg -(If you rescale the $x$ axis to range from $0$ to $1$ and interchange $x$ and $y$ axes, -this becomes the plot of the sample cumulative distribution function of $\lambda$.) -Here are the plots of the Perron ratios restricted to the intervals $1.5 < \lambda < 2$ -and $3 < \lambda < 4$: -alt text http://dl.dropbox.com/u/5390048/PerronPoints21%281.5%2C2%29.jpg -alt text http://dl.dropbox.com/u/5390048/PerronPoints21%283%2C4%29.jpg -The restriction to an interval seems to concentrate the absolute values of Perron ratios even more. The angular distribution looks like it converges to the uniform -distribution on a circle plus point masses at $0$ and $\pi$. -Is there an explanation for the distribution of radii? Any guesses for what it is? - -REPLY [11 votes]: There are many contexts in which one can establish that a family of polynomials, or even a single polynomial, has roots which are distributed according to some measure which approximates the uniform measure on the unit circle. -Let $f(x) = a_N x^N + \ldots + a_1 x + a_0$ be a polynomial with real coefficients, and assume that $a_N \ne 0$. Consider the quantity: -$$L_N(f) = \log \left( \sum |a_k| \right) - \frac{1}{2} \log |a_N| - \frac{1}{2} \log |a_0|.$$ -A theorem of Hughes and Nikeghbali (The zeros of random polynomials cluster uniformly near the unit circle, Compositio 144 (1998)) says (informally) that if $L_N(f)$ is small compared to $N$, then the roots are distributed uniformly along the unit circle. An ingredient of their result is a theorem of Erdos-Turan (On the distribution of roots of polynomials, Ann. of Math. (2) 51 (1950)) which (under the same hypothesis) shows that the arguments of the roots are distributed uniformly. Using this, we can answer: - -Is there an explanation for the distribution of radii? Any guesses for what it is? - -If one takes the coefficients to lie in some bounded range, say $[-5,5]$, then $L_N(f)$ has order $O(\log(N))$, which is certainly $o(N)$. Hence the random polynomials generated in this way have roots which are distributing in the unit circle. On the other hand, the polynomials being chosen are those which have a unique largest root of size at most five and usually around $5$, and hence the renormalized roots will appear to cluster around the circle of radius $1/5$. The clustering becomes more pronounced when one restricts to polynomials where the largest root lies in $[5,5-\lambda]$ for smaller $\lambda$ (Hence Thurston's graph with more restricted intervals are more "ring-like." (There is no point mass along the real axis by the theorem of Erdos-Turan.) - -Are there good, natural distributions for selecting random integer polynomials? Is there a way to do it without unduly prejudicing the distribution of roots? - -Given a large dimensional probability space, a natural way to generate random elements according to the distribution is to use a random walk Metropolis-Hastings algorithm. Concretely, one can start with a random Perron polynomial with largest root $< 5$, and then perturb the coefficients according to some distribution (say a normal distribution with small variance). If the new polynomial is also Perron with largest root $< 5$, choose this polynomial, or else keep the same polynomial. This Markov process should --- under suitable conditions --- generate random polynomials (in the aggregate) according to the required distribution. For example, let's run this algorithm and compare it to the roots of random monic polynomials with coefficients in $[-5,5]$ which have a unique largest root. The roots of polynomials with coefficients in $[-5,5]$ are illustrated in the first graph, and the result of the Metropolis-Hastings algorithm for all Perron polynomials with largest root less than $5$ is given in the second graph: - -Notice that the roots of the first graph cluster around the unit circle, as explained above. However, the roots of the second graph are clustering around the boundary. Indeed, this turns out to reflect reality (see the theorem below). - -What is the limiting distribution of Galois conjugates of Perron numbers $\lambda$ in some bounded interval, as the degree goes to infinity? - -Let's first consider the easier problem of describing "real Perron polynomials" --- that is, monic polynomials with real coefficients which have a unique largest root $\le r$ (necessarily real). The actual Perron numbers form integral lattice points in this space (although not all lattice points, just the ones corresponding to irreducible polynomials; however, the reducible lattice points form a thin subset). If one fixes the degree $N$ and increases the radius $r$, then (because the regions involved are suitable nice) the lattice points are distributed more or less uniformly inside the space (Davenport's Lemma). However, if one fixes $r$ and lets $N \rightarrow \infty$, it is no longer so clear whether the distribution of lattice points can be approximated by the corresponding real region. Studying lattice points in non-convex regions (or even convex ones) often leads to pretty thorny number theoretic issues, but one can at least hope (and compare with experiment) that the real geometry gives some indication of the truth of the matter. To this end, one has the following: -Theorem: Fix a real positive integer $r > 0$. Let $\Omega_N$ denote the space of monic polynomials with real coefficients which have a real root $\alpha$ such that $r > \alpha > |\sigma \alpha|$ for all conjugates $\sigma \alpha \ne \alpha$. Then, as $N \rightarrow \infty$, the roots of a random polynomial in $\Omega_N$ are distributed uniformly along the circle $|z| = r$. -The proof of this theorem uses the theorem of Hughes and Nikeghbali mentioned above. The point is that one has to obtain estimates of how the quantities $|a_i|$ vary over the space $\Omega_N$, which reduces, in the end, to evaluating and/or estimating various integrals over $\Omega_N$. For example, consider the example above of degree $21$ polynomials with a largest root $< 5$. The model given by the OP choose $a_{21} \in [-5,5]$. It turns out, however, that the expected value of the constant term $a_{21}$ over $\Omega_{21}$ (with $r = 5$) is -$$ \frac{3^2 \cdot 5^{21} \cdot 7 \cdot 13 \cdot 17 \cdot 19}{2^{17} \cdot 11} \sim 8.748 \times 10^{13}.$$ -This is pretty big compared to $[-5,5]$! -Other statistics -There are a few other interesting probabilities one can compute using integrals over spaces like $\Omega_N$ and related spaces. For example, one can consider all monic polynomials of degree $2N$ with the property that their roots all have absolute value at most one. This defines a compact region of (the coefficient space) $\mathbf{R}^N$, and so it has a natural uniform measure. Then the probability that a random such polynomial has no real roots in the interval $[-1,1]$ (equivalently, is positive on $[-1,1]$, equivalently, has no roots on all of $\mathbf{R}$) is equal to -$$\sim \frac{2C}{\sqrt{2\pi} \cdot (2N)^{3/8}},$$ -where the constant $C$ is equal to -$$C = 2^{-1/24} e^{- 3/2 \cdot \zeta'(-1)} = 1.24514 \ldots.$$ -Curiously enough, there is a theorem of Dembo, Poonen, Shao, and Zeitouni that a random polynomial (in the more usual sense) whose coefficients are chosen -with (say) identical normal distributions with zero mean is positive in $[-\infty,\infty]$ - with probability $N^{-b + o(1)}$ and positive in $[-1,1]$ with probability $N^{-b/2 + o(1)}$ - for some universal constant~$b/2$, which - they estimate be $0.38 \pm 0.015$. On the other hand, -the exponent occurring above is $3/8 = 0.375$. Is there any direct relationship between -these theorems? For example, does this suggest that $b = 3/4$? (The result of DPSZ actually holds for a very wide range of distributions with the same undetermined $b$, but it does not apply to our very rigid context.) -Further Remarks -There's some further analysis one can make of the space $\Omega_{N}$, or more generally the space of all monic polynomials with real coefficients whose roots have absolute value at most $r$. For example, one can show that a "random" polynomial of degree $N$, subject to the constraint that all its roots have absolute value at most $r$, will be a Perron polynomial (i.e. have a unique largest root) with probability exactly $1/N$ if $N$ is odd and $1/(N+1)$ if $N$ is even. Some of this analysis will appear in my thesis (I will add a link here when it is written!). As my advisor said, "Any question that Thurston asks is probably worth thinking about."<|endoftext|> -TITLE: Functorial way of showing that the Segre (or Plucker) morphism is a closed embedding? -QUESTION [12 upvotes]: Let $\mathcal{F}, \mathcal{G}$ be quasi-coherent sheaves on a base scheme $S$. Then the $S$-schemes $\mathbb{P}(\mathcal{F}), \mathbb{P}(\mathcal{G})$ are defined by a universal property: to map from an $S$-scheme $T$ (with structure map $g: T \to S$) into $\mathbb{P}(\mathcal{F})$ is the same thing as giving a line bundle $\mathcal{L}$ on $T$ and a surjection $g^*(\mathcal{F}) \to \mathcal{L}$. The Segre embedding $\mathbb{P}(\mathcal{F}) \times_S \mathbb{P}(\mathcal{G}) \to \mathbb{P}(\mathcal{F} \otimes \mathcal{G})$ is defined in a canonical manner using this universal property. Similarly, the Plucker embedding of the Grassmannian into a projective scheme is done by giving a morphism of the functors they represent. -The Segre and Plucker embeddings are closed immersions, as is checked locally using the definitions. However, the properness of projective space $\mathbb{P}^n_{\mathbb{Z}}$ can be easily proved using only the valuative criterion, which seems easier and more elegant (though it can also be done directly). Is there a functorial way to see that the Segre and Plucker embeddings are closed immersions? - -REPLY [4 votes]: I think this is a very interesting question. It gives another proof of the Segre embedding, makes it somehow more natural and puts it into a more general context, which I explain below. -I will just talk about the Segre embedding $\mathbb{P}^n \times \mathbb{P}^m \to \mathbb{P}^{n+m+nm}$. The general case of $\mathbb{P}(F) \times \mathbb{P}(G) \to \mathbb{P}(F \otimes G)$ is similar. -Let $x_0,...,x_n$ be the homogeneous coordinates of $\mathbb{P}^n$, $y_0,...,y_m$ the ones of $\mathbb{P}^m$ and $z_{00},...,z_{nm}$ the ones of $\mathbb{P}^{n+m+nm}$. The idea of the Segre embedding is $z_{ab} = x_a y_b$. Observe that with this equation we can interchange $a,c$ and $b,d$ in $z_{ab} z_{cd}$ without changing the result. Thus, consider the homogeneous ideal generated by the $z_{ab} z_{cd} - z_{cb} z_{ad}, z_{ab} z_{cd} - z_{ad} z_{cb}$ and let $I$ denote the corresponding ideal of the structure sheaf of $\mathbb{P}^{n+m+nm}$. Now I claim that $V(I)$ and $\mathbb{P}^n \times \mathbb{P}^m$ satisfy the same universal property. So these schemes are isomorphic, and we get a closed immersion $\mathbb{P}^n \times \mathbb{P}^m \to \mathbb{P}^{n+m+nm}$. -Let $X$ be a test scheme. Then $\hom(X,\mathbb{P}^n \times \mathbb{P}^m)$ is the set of tuples $(\mathcal{L}_1,\rho,\mathcal{L}_2,\tau)$, where $\mathcal{L}_1$ (resp. $\mathcal{L}_2$) is a line bundle on $X$ and $\rho_0,...,\rho_n$ (resp. $\tau_0,...,\tau_m$) are global generators of $\mathcal{L}_1$ (resp. $\mathcal{L}_2$). -On the other hand, $\hom(X,V(I))$ is the set of all tuples $(\mathcal{L},\sigma)$, where $\mathcal{L}$ is a line bundles on $X$ and $\sigma_{00},...,\sigma_{nm}$ are global generators of $\mathcal{L}$, which satisfy the relations $\sigma_{ab} \sigma_{cd} = \sigma_{cb} \sigma_{ad}$ and $\sigma_{ab} \sigma_{cd} = \sigma_{ad} \sigma_{cb}$. -With these descriptions, there is a very natural map $\hom(X,\mathbb{P}^n \times \mathbb{P}^m) \to \hom(X,V(I))$ given by $\mathcal{L} = \mathcal{L}_1 \otimes \mathcal{L}_2$ and $\sigma_{ab} = \rho_a \otimes \tau_b$. Actually this map makes sense in every abelian tensor category, but its bijectivity seems to be a special property of the abelian tensor category of quasi-coherent sheaves on a scheme (more generally, sheaves of modules on a ringed space). The usual proof uses the local explicit description of the classifying projective schemes, but you can do it directly by writing down an inverse map: -Let $\mathcal{L},\sigma$ be given as above. Construct a line bundle $\mathcal{L}_1$ as follows: For every $i$, put $X_{\rho_i} = \cup_j X_{\sigma_{ij}}$. On this open set, let $\mathcal{L}_1$ be generated freely by a symbol $\rho_i$. The cocycles $\rho_a / \rho_i$ are defined as $\sigma_{aj} / \sigma_{ij}$ for some $j$; the choice of this $j$ does not matter because of the relations for $\sigma$. Similarily we may define $\mathcal{L}_2$ with global generators $\tau_j$. It's easy to show that $\mathcal{L} \cong \mathcal{L}_1 \otimes \mathcal{L}_2$ with $\rho_i \otimes \tau_j$ corresponding to $\sigma_{ij}$.<|endoftext|> -TITLE: Clearing misconceptions: Defining "is a model of ZFC" in ZFC -QUESTION [41 upvotes]: There is often a lot of confusion surrounding the differences between relativizing individual formulas to models and the expression of "is a model of" through coding the satisfaction relation with Gödel operations. I think part of this can be attributed to the common preference for using formulas over codings. For example, a standard proof showing that $V_{\kappa} \models ZFC$ for $\kappa$ inaccessible will appeal to the fact that all of the ZFC axioms relativized to $V_{\kappa}$ are true. But then one learns about the Lévy Reflection Theorem scheme which allows every (finite) conjunction of formulas to be reflected to some $V_{\alpha}$. Perhaps this knowledge is followed by a question of whether the Compactness theorem can be used to contradict Gödel's Second Incompleteness Theorem. -Specifically, consider the following erroneous proof that ZFC + CON(ZFC) proves its own consistency: -Introduce a new constant $M$ into the language of set theory and add to the axioms of ZFC all of its axioms $\varphi_n$ relativized to $M$, denoted $\varphi_n^M$. Provided that ZFC is consistent, every finite collection of this theory is consistent by the Lévy Reflection Theorem whereby the Compactness Theorem tells us that the entire theory ZFC + "$M \models ZFC$" will be consistent. Consequently, this theory has a (ZFC) model $N$ so in this model, there exists a model $M$ of ZFC. To summarize then, arguing in ZFC + CON(ZFC), we've seemingly proven that we have a ZFC model $N$ modeling the consistency of ZFC by virtue of it having the model $M$ (i.e., seemingly $N \models ZFC + CON(ZFC)$ so we would have a proof of CON(ZFC + CON(ZFC)). -The misstep in this proof is of course a misuse of the conclusion of the Compactness theorem, mainly the assumption that such an $N$ will think that $M$ is a ZFC model. With some enumeration of the formulas of the axioms $\{\varphi_n| n \in \mathbb{N}\}$ of ZFC, it is clear that $N$ will certainly think that $M \models \varphi_n$ for any particular $n \in \mathbb{N}$ analogous to how a nonstandard model of Peano arithmetic has an element $c$ satisfying $c > n$ for any particular $n \in \mathbb{N}$. The problem of course in the case of $N$ is that there may be formulas with nonstandard indices not accounted for just as there will definitely be nonstandard numbers greater than $c$ in the PA example. -If one were to carry out the same proof with the more tedious arithmetization of syntax, then this link may be more apparent. -To a lesser extent, there may also be confusion with the fact that $0^{\sharp}$ provides us with a proper class of $L(\alpha) \preceq L$. This may lead to the question of whether $L$ has its own truth predicate, contradicting Tarski's Theorem. But of course $L$ will only realize that each of these $\varphi^{L(\alpha)}$ is true for any ZFC axiom $\varphi$, and if one attempts to appeal to the arithmetization of syntax, one can begin to see the problem that these $\alpha$ may not (and of course will not) be definable (without parameters) in the constructible universe L. -Since these types of misconceptions can be common among logicians and non-logicians alike, I thought I would ask the highly intelligent mathematicians who have worked through such problems or helped illuminate them to others if they would do so here as well. I think compiling a collection of tidbits of wisdom in this area from the collective perspectives of the MO Community can be illuminating to all. As such, my question is as follows: - -What insights can you share regarding the questions of formalizing "is a model of ZFC" in ZFC and the various "paradoxes" that arise? - -For example, maybe you can show a related seemingly paradoxical problem and resolve it, or simply share your thoughts on how to avoid such traps of logic. - -REPLY [58 votes]: Here is a result along the lines you are requesting, which -I find beautifully paradoxical. -Theorem. Every model of ZFC has an element that is a -model of ZFC. That is, every $M\models ZFC$ has an element -$m$, which $M$ thinks is a structure in the language of set -theory, a set $m$ and a binary relation $e$ on $m$, such -that if we consider externally the set of objects $\bar -m=\{\ a\ |\ M\models a\in m\ \}$ with the relation $a\mathrel{\bar e} -b\leftrightarrow M\models a\mathrel{e} b$, then $\langle -\bar m,\bar e\rangle\models ZFC$. -Many logicians instinctively object to the theorem, on the -grounds of the Incompleteness theorem, since we know that -$M$ might model $ZFC+\neg\text{Con}(ZFC)$. And it is true -that this kind of $M$ can have no model that $M$ thinks is -a ZFC model. The paradox is resolved, however, by the kind -of issues mentioned in your question and the other answers, -that the theorem does not claim that $M$ agrees that $m$ is -a model of the ZFC of $M$, but only that it externally is a -model of the (actual) ZFC. After all, when $M$ is -nonstandard, it may be that $M$ does not agree that $m$ -satisfies ZFC, even though $m$ actually is a model of ZFC, -since $M$ may have many non-standard axioms that it insists -upon. -Proof of theorem. Suppose that $M$ is a model of ZFC. Thus, -in particular, ZFC is consistent. If it happens that $M$ is -$\omega$-standard, meaning that it has only the standard -natural numbers, then $M$ has all the same proofs and -axioms in ZFC that we do in the meta-theory, and so $M$ -agrees that ZFC is consistent. In this case, by the -Completeness theorem applied in $M$, it follows that there -is a model $m$ which $M$ thinks satisfies ZFC, and so it -really does. -The remaining case occurs when $M$ is not -$\omega$-standard. In this case, let $M$ enumerate the -axioms of what it thinks of as ZFC in the order of their Goedel numbers. An initial segment of -this ordering consists of the standard axioms of ZFC. Every -finite collection of those axioms is true in some -$(V_\alpha)^M$ by an instance of the Reflection theorem. -Thus, since $M$ cannot identify the standard cut of its -natural numbers, it follows (by overspill) that there is -some nonstandard initial segment of this enumeration that -$M$ thinks is true in some $m=(V_\alpha)^M$. Since this -initial segment includes all actual instances of the ZFC -axioms, it follows that $m$ really is a model of ZFC, even -if $M$ does not agree, since it may think that some -nonstandard axioms might fail in $M$. $\Box$ -I first learned of this theorem from Brice Halimi, who was visiting in New York in 2011, and who subsequently published his argument in: - - -Halimi, Brice, Models as universes, Notre Dame J. Formal Logic 58, No. 1, 47-78 (2017). ZBL06686417. - - -Note that in the case that $M$ is $\omega$-nonstandard, then we actually get that a rank initial segment $(V_\alpha)^M$ is a model of ZFC. This is a very nice transitive set from $M$'s perspective. -There are other paradoxical situations that occur with countable computably saturated models of ZFC. First, every such M contains rank initial segment $(V_\alpha)^M$, such that externally, $M$ is isomorphic to $(V_\alpha)^M$. Second, every such $M$ contains an element $m$ which $M$ thinks is an $\omega$-nonstandard model of a fragment of set theory, but externally, we can see that $M\cong m$. Switching perspectives, every such $M$ can be placed into another model $N$, to which it is isomorphic, but which thinks $M$ is nonstandard.<|endoftext|> -TITLE: Do elongated convex objects all have long simple geodesics? -QUESTION [10 upvotes]: Let $S$ be a closed convex surface, the boundary of a compact -convex body in $\mathbb{R}^3$. -I am interested in whether there are conditions on its shape -that ensure that it supports a long, simple (non-self-crossing) geodesic. -The length of a geodesic for my purposes is the longest distance -you can travel along the geodesic before returning to your starting -point. Some condition is necessary for the type of result I seek, -for all the geodesics on a sphere have the same length. -Define the elongation $L$ of $S$ as the largest height to diameter -ratio, $h/d$, of a cylinder of height $h$ and diameter $d$ in -which $S$ is tightly inscribed. By tightly inscribed I mean -that $S$ touches the top, bottom, and sides of the cylinder in -such a manner that neither the height nor diameter can be reduced. -I could use a theorem of this type: - -If $S$ has elongation $L \ge k$, then there is a simple - geodesic on $S$ of length $\ge f(k)$, where $f(k)$ is some - increasing function of $k$, e.g., $c k$ for a constant $c > 0$. - -Perhaps such a theorem cannot exist. -Or maybe a theorem of this ilk exists, but only with -certain smoothness assumptions? -There are always at least three simple closed -geodesics on $S$, by a theorem of Lyusternik and Schnirelmann, but perhaps -they might all be short? -For an ellipsoid, the three simple closed geodesics -follow the major and minor axes, and the longest of those -satisfies the type of relationship I seek. -(Elongation could as well be defined in terms of an enclosing ellipsoid rather than cylinder.) -And a cylindrical $S$ supports a long spiral geodesic: -           -Such spirals are exactly the type of geodesic I seek. Thanks for any ideas or pointers! -Edit. -This may not add much, but here is how I view a long geodesic on a cylinder: starting at $a$, -crossing the bottom in a segment $x x'$, crossing the top in $y y'$, and stopping at $b$ just before it is about to cross itself. - -REPLY [5 votes]: There is a highly relevant paper of Gene Calabi and J. Cao, where they show that there always exists a geodesic of length at least twice the diameter (in the metric space sense) of your surface. I think this answers your question.<|endoftext|> -TITLE: Easier induction proofs by changing the parameter -QUESTION [11 upvotes]: When performing induction on say a graph $G=(V,E)$, one has many choices for the induction parameter (e.g. $|V|, |E|$, or $|V|+|E|$). Often, it does not matter what choice one makes because the proof is basically the same. However, I just read the following ingenious proof of König's theorem due to Rizzi. -König's theorem. For every bipartite graph $G$, the size of a maximum matching $v(G)$ is equal to the size of a minimum vertex cover $\rho(G)$. -Proof. Induction on $|V|+|E|$. Base case is clear. Now if $G$ has maximum degree 2, then $v(G)=\rho(G)$, so we may assume that $G$ has a vertex $x$ of degree at least 3. Let $y$ be a neighbour of $x$ and let $Y$ be a minimum vertex cover of $G - y$. Evidently, $Y \cup y$ is a vertex cover of $G$. But, by induction $|Y|=v(G-y)$, so we are done unless $v(G)=v(G-y)$. Thus, $G$ has a maximum matching $M$ avoiding $y$. Let $e \in E - M$ be incident to $x$ but not to $y$. By induction, -$v(G)=|M|=v(G-e)=\rho(G-e)$. -Let $Z$ be a vertex cover of $G-e$ of size $|M|$. Note that $y \notin Z$, since $M$ does not cover $y$. This implies $x \in Z$, since $xy \in E$. But then $Z$ also covers $e$ and hence is a vertex cover of $G$. - -Question. What are some other instances (not necessarily in graph theory), where simply changing the induction parameter results in a nice shorter proof? - -REPLY [6 votes]: Gauss' "second" (1815) proof of the fundamental theorem of algebra (Werke, Volume 3, 33-56, or see Paul Taylor's translation, currently available here) follows an interesting pattern, similar to the one in Cauchy's proof of the AM-GM inequality mentioned in Pietro's answer. (It does more than this: It introduces the discriminant, for example.) -Gauss shows that a polynomial with real coefficients can be factored into real polynomials of first and second degree. We have that a polynomial of odd degree has a root. From this, he argues by assigning to a polynomial $p$ of degree $n$ a new polynomial $p^+$ of degree $n(n-1)/2$, in such a way that pairs of (possibly complex) roots of $p^+$ determine (possibly complex) roots of $p$ via quadratic equations. - -So the pattern is induction not on the degree $n$ of the polynomial, but on the largest power of 2 dividing $n$. - -I first encountered this neat idea not through Gauss work, but through a proof by Derksen of the fundamental theorem of algebra via linear algebra (Harm Derksen, "The fundamental theorem of algebra and linear algebra", American Mathematical Monthly, 110 (7) (2003), 620–623.) -The skeleton of Derksen's proof is as follows: One actually shows that: - -If $V$ is a complex finite dimensional vector space, and ${\mathcal F}$ is a (possibly infinite) family of pairwise commuting linear operators on $V$, then the operators in ${\mathcal F}$ admit a common eigenvector. - -For this, one considers the statement $E(K,d,\kappa)$: If $V$ is a vector space over $K$ of finite dimension, and $d\not{\mathrel{|}}{\rm dim}(V)$, then any family ${\mathcal F}$ with $|{\mathcal F}|=\kappa$ of pairwise commuting linear maps from $V$ to itself admits a common eigenvector. -One easily checks that the case $\kappa$ infinite follows from the finite one, and this follows by a straightforward induction, so it is enough to show $E({\mathbb C},d,1)$. -For this, one first shows $E({\mathbb R},2,1)$: A linear map from ${\mathbb R}^n$ to itself, say, with $n$ odd, admits a real eigenvalue. This follows from odd degree real polynomials having roots. -Then, one shows $E({\mathbb C},2,1)$. For this, let $V$ be a ${\mathbb C}$-vector space of odd degree n, and let $L(V)$ be the space of ${\mathbb C}$-linear transformations of $V$ to -itself. Given a linear $T:V\to V$, one associates to it a real-vector subspace of $L(V)$ of real dimension $n^2$, and a pair of commuting linear maps there, in a way that from any common (real) eigenvalue we can reconstruct a complex eigenvalue of $T$. Then one uses $E({\mathbb R},2,2)$. -One then argues by induction on $k$ that $E({\mathbb C},2^k,1)$ holds. As before, to a $T:V\to V$ with $V$ of appropriate dimension $n$, one associates a complex subspace of $L(V)$ of dimension $n(n-1)/2$ and a pair of commuting linear maps there, so the result follows from $E({\mathbb C},2^{k-1},2)$. -(I must confess I haven't worked through the details enough to comment on whether this is essentially Gauss' proof in a different language.)<|endoftext|> -TITLE: Nagata's bizzare examples -QUESTION [6 upvotes]: Hi, -due to Nagata and his clever and bizzare examples I'm unsure in this: -1) Is there a regular ring of infinite Krull dimension? -2) Is it true that: Regular ring of finite Krull dimension = Commutative Noetherian ring of finite global dimension? (answer is yes if R is local by Serre's Theorem) -Regards, -David - -REPLY [9 votes]: For 1), Nagata's counterexample goes as follows: -Let $\mathbb{N}=A_1\cup A_2 \cup \cdots$ be a partition such that $|A_i|<|A_{i+1}|$. Take $S=k[x_1,x_2,\ldots]$, prime ideals $I_i=\langle x_i | i\in A_i \rangle$ and consider the localization $S[U^{-1}]$ in the multiplcative set $U=S \setminus \bigcup_i I_i$. Then the height of each of the $I_i$ is $|A_i|$ and so $S[U^{-1}]$ has infinite Krull dimension. It not so hard to show that the ring is regular, but a more surprising point is that it is also Noetherian. -For 2) I believe the answer is yes by this reference.<|endoftext|> -TITLE: Is an A-infinity thing the same the same as strict thing viewed through a homotopy equivalence? -QUESTION [16 upvotes]: If I have a topological monoid ($X,\mu$), a space $Y$ and a homotopy equivalence $f,g$ between them, then $Y$ has the structure of an $A_\infty$ space defined 'pointwise' by -$$ y_1 * y_2 := g \left(\mu(f(y_1),f(y_2))\right) $$ -Also, I have heard someone say the reverse as "Every $A_\infty$-space is (weakly?) homotopy equivalent to a topological monoid." -My question is this: In how far is this the general case? Can I define/think of $A_\infty$ structures as strict structures as viewed through the distorting glasses of a homotopy equivalence? When is an $A_\infty$ structure of this type - i.e. is there always an equivalent strict version? (I guess no in general. Can you tell me more?) -FYI: I started out with the question: Is the based loop space of a space $X$ always on the other side of a homotopy equivalence of a (strict) topological group? -thanks, -a. - -REPLY [6 votes]: Some ancient history: The first proof that the loop space of any reasonable space is equivalent to a topological group is due to John Milnor, in his 1956 Annals of Math. paper ``Construction of universal bundles, I'' (I'm sure its hypotheses are removable nowadays.) -Some slightly less ancient history is Theorem 13.5 in my 1972 Springer Volume ``The Geometry of Iterated Loop spaces'', http://www.math.uchicago.edu/~may/BOOKS/geom_iter.pdf on my web page. An $A_{\infty}$-space is a $\mathcal{C}$-space $X$ for any $A_{\infty}$ operad $\mathcal{C}$. In the notation there, I construct a topological monoid $B(M,C,X)$, a $\mathcal{C}$-space $B(C,C,X)$ and a pair of $\mathcal{C}$-maps $B(C,C,X) \longrightarrow X$ and $B(C,C,Z)\longrightarrow B(M,C,X)$. The first is always a homotopy -equivalence and the second is always a weak homotopy equivalence, although I only proved that there when $X$ is connected. This is the standard construction that Tyler described in more detail. (I don't think I was yet on Overflow when the question was first asked.)<|endoftext|> -TITLE: Positive integers $n$ that divide $\sigma_2(n)$ -QUESTION [5 upvotes]: For a positive integer $n$ let -$$ -\sigma_2(n) = \sum_{d \mid n} d^2. -$$ -There are many positive integers $n$ for which -$$ -n \mid \sigma_2(n). -$$ -But, when $n$ has the particular form -$$ -n=pq^2 -$$ -where $p, q$ are distinct odd prime numbers and -$$ -p \equiv 1 \pmod{4} -$$ -there seems to be none. -Question: What happens in these case. -It is easy to see that the condition is equivalent to -$$ -pq^2 \mid 1+p^2+q^2+q^4. -$$ - -REPLY [4 votes]: As already mentioned in other posts, $p\equiv1\bmod3$ and $q\equiv1\pmod4$. -Since $p\mid q^4+q^2+1=(q^2+q+1)(q^2-q+1)$, the prime $p$ divides (at least) -one of the two factors; in particular, $p\le 1+q+q^2<(1+q)^2$ implying -$q>\sqrt p-1$. It happens very rare that $p^2+1$ is divisible by a square -$>(\sqrt p-1)^2$. Excluding the case $p=7$ and $q=5$ (when we indeed have -$q^2\mid p^2+1$ and $p\mid q^4+q^2+1$), there are only 19 such cases -for $p<10^8$; in all them the corresponding factor $q^2>(\sqrt p-1)^2$ -involves $q$ prime such that $p\nmid q^4+q^2+1$. Even more, -the residue $q^4+q^2+1\pmod p$ seems to be a completely random number... -It looks like the two conditions $p^2\equiv-1\bmod{q^2}$ and $q^4+q^2+1\equiv0\bmod p$ -do not "feel each other". Therefore, it is quite unlikely that a solution -to the system of congruences in prime $p>7$ and integer $q>5$ (not necessarily prime!) -exists. I even doubt about integer solutions $p$, $q$...<|endoftext|> -TITLE: Generalizing Eichler-Shimura to higher dimension, again -QUESTION [5 upvotes]: This question is related to -Intuition behind the Eichler-Shimura relation? -and -L-functions and higher-dimensional Eichler-Shimura relation -Answering the first question above, Matt Emerton gives a sketch of a proof of the Eichler-Shimura congruence relation which tells that the Hecke correspondence mod p is a sum of the graph of the Frobeniusm and it's transpose. -I am wondering how the statement and the proof can be generalized to moduli of higher dimensional abelian varieties with level structure (and maybe some more structure, like PEL). -It seems like the reason for having only two isogenies $E \to E'$ in Matt's answer is that p-isogenies correspond to subgroups of order p in E[p] as a scheme, and E[p] (if we assume E to be ordinary) in char p is a product of $Z/p$ and the dual group, $\mu_p$, and apparently these two groups: $Z/p$ and $\mu_p$ are the only nontrivial subgroups in E[p], so taking quotients we come up with the Frobenius and Verschiebung (dual isogeny). -Now let's say A is an abelian variety of dimension g in char p, which has maximal p-rank, -i.e. $A[p] = Z/p^g \times \mu_{p^g}$. What are the subgroups of order $p^g$ of such a group? I am not familiar with how local groups behave, but I can assume there will be g+1 isogenies $A \to A'$, each one having the Kernel of the kind $H_1 \times H_2$, where $H_1$ and $H_2$ are subgroups in $Z/p^g$ and in $\mu_{p^g}$ respectively. -Also one probably needs a statement that abelian varieties with maximal p-rank are Zarisky open and dense in the moduli space, which is true in dimension 1. -Then the reduction mod p of the Hecke correspondence $T_p$, appropriately defined will be a sum of these g+1 cycles? Is that making any sense? -Thanks. - -REPLY [6 votes]: Dear Evgeny, -The paper Congruence relations on some Shimura varieties of Wedhorn (and some follow-up papers) proves what is (I think) the state of the art on Eichler--Shimura relations on higher dimensional Shimura varieties. Namely, he establishes the congruence relation for a wide class of PEL Shimura varieties. Just as you surmise, the density of the ordinary locus plays an important role in his argument. -Regards, -Matt<|endoftext|> -TITLE: When do we need the axiom of compact support for a homology theory to be uniquely defined? -QUESTION [6 upvotes]: Looking over the treatment of the Eilenberg-Steenrod axioms in a few of my favorite introductory algebraic topology texts, I see that some include an "axiom of compact support", while others do not. Whether or not one needs such an axiom for the homology theory to be uniquely defined (assuming it satisfies the dimension axiom) seems to depend on the category of pairs for which one is defining the homology theory. For example, if one is only looking at the homology of compact pairs, the axiom of compact supports is certainly unnecessary. Spanier invokes the axiom to handle arbitrary polyhedral pairs, and this also seems to be the situation in Munkres. But then Hatcher (if I'm reading it correctly) proves uniqueness (assuming the dimension axiom) on CW pairs, without any mention of the compact support axiom, which I find somewhat surprising without limiting to compact pairs. -So for which categories containing possibly non-compact spaces does one need the compact support axioms (along with the dimension axiom) in order to know that the homology theory is unique? In particular, what is the status for polyhedral pairs, CW pairs, and arbitrary topological pairs? And if I'm reading Hatcher correctly, is there an intuitive reason why the compact support axiom isn't necessary for CW pairs but is for polyhedral pairs? -It's amazing the things one is forced to think about when teaching a graduate algebraic topology class! - -REPLY [6 votes]: The axioms that Hatcher uses (on page 160 of the current online version of his book) include an axiom about arbitrary wedges going to direct sums. This has the same effect (in the presence of his other axioms and in the CW setting) as an axiom about compact support (by which I assume you mean a statement that the homology of a space or pair coincides with the direct limit of homology of compact spaces or pairs mapping to it).<|endoftext|> -TITLE: Under what circumstances do morphisms on the stalks of a sheaf induce a sheaf morphism -QUESTION [14 upvotes]: It is very well known that if $\alpha : \mathscr{F} \to \mathscr{G}$ is a morphism of sheaves, then it induces homomorphisms on the stalks. I have been wondering for a while if given a collection of homomorphisms between the stalks of two sheaves and some suitable patching condition, can we construct a homomorphism of sheaves? -I have been thinking about this problem a little bit lately, but I have not been able to come up with a "canonical" solution. In other words I have found ways of producing sheaf morphisms from stalk morphisms but they have always felt very unnatural. -Let me phrase the question a little bit more formally. Let $X$ be a topological space and $\mathscr{F}, \mathscr{G}$ sheaves on $X$. Let $\alpha_p : \mathscr{F}_p \to \mathscr{G}_p$ be a homomorphism for each $p \in X$. Is there a suitable condition on the $\alpha_p$s such that there exists a sheaf morphism $\alpha : \mathscr{F} \to \mathscr{G}$ whose induced maps on the stalks are exactly the $\alpha_p$s? - -REPLY [14 votes]: There is a simple condition. Stalk maps $\alpha_p : F_p \to G_p$ yield a map of sheaves $\alpha : F \to G$ if and only if for all open $U \subseteq X$ and sections $s \in F(U)$ there is an open covering $U = \cup_i U_i$ and sections $t_i \in G(U_i)$ such that $\alpha_p(s_p) = (t_i)_p$ for all $p \in U_i$ and all $i$. This can also be stated as continuity of the $\alpha_p$ (see the answer of David Roberts). -This cirterion is sometimes useful, for example in the construction of fibered products of locally ringed spaces (German: http://maddin.110mb.com/pdf/faserprodukte.pdf).<|endoftext|> -TITLE: The straightest possible path embeddable in a path of polygons -QUESTION [6 upvotes]: I'm studying a problem involving the sets of discrete curves that can be embedded in a non-trivial polygon, from a source to a target point, as shown below. - -Initially my interest was limited to the set containing discrete curves with the possible smallest segment count and within this set the discrete curve with the shortest length. I've been able to find this discrete curve (not shown in the above graphic). -Now I'm considering the problem of the straightest-possible discrete curve that can be embedded in the polygon, and I suspect, or rather I'm hoping, that both curves, the one with the smallest segment count and the straightest one are identical. I'd appreciate help showing that is or isn't the case. -Note: I'm aware of the concept of geodesics as shortest possible paths. However, the curve I'm studying isn't the shortest path but the path with the smallest segment count. In the example provided above, the shortest path would cut around corners as it seeks to minimize the path length. Interestingly, this curving of the path increases its segment count. -Clarification of the question (courtesy of arex) -Suppose a simple polygon is given, along with two points, s and t, within the polygon. Is there a polygonal path from s to t that simultaneously minimizes the number of segments and amount of turning. Simple means homeomorphic to a disk; a polygonal path is a piecewise-linear curve; and turning is measured by the sum of the absolute values of the turning angles. - -I'm making the following restatement of the initial problem, now solved, because it describes the sort of path that I'm investigating. Also the description may trigger insight into the question of bounds on straightness: -The polygon can be considered as a tunnel in which there is no line of sight between two points, which we can imagine to be the source and destination of a light beam. We seek to install the smallest number of light beam relays required to forward a signal as quickly as possible though the tunnel from source to the destination. The locations of the relays are the equivalent to the intermediate vertices of the shortest discrete curve with the minimum possible segment count embeddable in a non-trivial/non-convex polygon. - -REPLY [4 votes]: The natural generalization of your "straightest-possible" constraint is a curvature-constrained path. -I know this is not what you asked, but I wanted to mention that even finding a path with maximum -curvature 1 inside a convex polygon is already difficult. This was studied by -Agarwal et al. in a paper, -"Curvature-Constrained Shortest Paths in a Convex Polygon," -(SIAM Journal on Computing, Volume 31, Issue 6, 2002) -Here is one of their figures, which hints at why it is difficult to find a path between initial $I$ and final $F$ -positions/orientations: - -Returning to the question (or questions) you did ask, -I think looking at the paper -"Polygonal path simplification with angle constraints," by Chen et al. -(Computational Geometry, Volume 32, Issue 3, November 2005, Pages 173-187), might help. -They explicitly consider the "tunnel" version of your revised question, phrased in terms of an -error tolerance $\epsilon$. Perhaps most usefully for your purposes, they describe all the related -work in this domain, which has considered many variations. -Usually those variations start with a path $P$, and then try to find another path (a "simplification") -with certain properties: (1) The min-# problem: Find a path that remains within $\epsilon$ of $P$ but has the fewest -segments; (2) The min-$\epsilon$ problem: Given a fixed number $m$ of segments, find a path using -no more than $m$ segments that minimizes $\epsilon$. This specific paper solves the min-# problem -with the addition of angular constraints, which seems close to your initial formulation (as clarified in the comments). -Edit. Here is a preliminary version of the "Simplification" paper: simplification.pdf.<|endoftext|> -TITLE: The Klein bottle and the Heawood Conjecture -QUESTION [16 upvotes]: Let $\Sigma_g$ be a surface of genus $g$. The Heawood Conjecture gives a closed formula in one variable, $\chi$ (the Euler characterstic of $\Sigma_g$), for the minimal number of of colors needed to color any graph drawn on $\Sigma_g$. -Ringel-Youngs, 1968: The Heawood Conjecture is true, except for $g=0$ (where we don't know the answer - this would be the Four-Color Conjecture) and the Klein bottle (where a graph was shown to be colorable with fewer than predicted by the formula). -Of course, the Four-Color Conjecture was eventually proven (not by Ringel or Youngs, although I think they were both involved to some extent), leaving the Klein bottle as the odd-man out. - -(a) What is the "reason" for the Klein bottle's exceptionality here? (b) Does the answer to part (a) manifest in any other way - meaning, are there other theorems, patterns, etc to which the Klein bottle is an exception that can be traced back to (a)? - -REPLY [8 votes]: Even though there is always a preference towards uniformity in understanding of general results such as Heawood's theorem, this somehow feels like the wrong question to ask. While I am not familiar all the way with all the proofs involved (the full proof is massive, and involves the four color theorem as well) it is obvious that there are many special cases treated separately, with the most difficult cases being the planar case and $g\in \{59, 83,158,257\}$. So even if the resulting formula is uniform, the behavior of each case is not obviously so. -In the end, the answer to your question boils down to the reason that Francesco Polizzi mentions in his answer, and can perhaps be elaborated a bit more by mentioning the improvement of the Heawood result by Dirac, Albertson and Hutchinson that every graph in $\Sigma$ is actually $H(\Sigma)-1$ colorable unless it has a subgraph isomorphic to a complete graph on $H(\Sigma)$ vertices. You can see here that the Klein bottle is expected to behave differently as $K_7$ doesn't embed in it, but suddenly we get the planar case as another exception and the projective plane was dealt separately too.<|endoftext|> -TITLE: Is the classification of finite p-groups a smooth problem? -QUESTION [7 upvotes]: Fix a prime $p$. Then the question is about the difficulty of classifying finite $p$-groups. Originally I was going to ask if this was a "wild" problem but thanks to Joel David Hamkins' answer to question -10481 I now know to ask if it is a smooth problem. -At one time there was a minor industry producing papers which showed that a particular list of invariants did not classify finite $p$-groups so my understanding is that it would be remarkable if this problem was smooth. -Also for each prime $p$ and each $n$ the number of groups of order $p^n$ (up to isomorphism) is finite. This gives a sequence of integers for each $p$. For $p=2,3,5,7$ these sequences appear in OEIS as sequences A000679, A090091, A090130, A090140. -A supplementary question is: -Does knowing if the classification is or is not smooth have any bearing on the complexity of these sequences? - -REPLY [3 votes]: 1) The classification of p-groups (even p-groups of class 2) is indeed wild over $\mathbb{F}_p$ [V. Sergeichuk, The classification of metabelian p-groups (Russian), Matrix problems, Akad. -Nauk Ukrain. SSR Inst. Mat., Kiev, 1977, pp. 150-161.] -(It's not clear to me why Hamkins's answer on your other question led you away from asking about wildness...) -2) Rather than smoothness - since as pointed out by others, classification problems on finite spaces are trivially smooth - in addition to wildness, you might consider computational complexity as a measure of difficulty of a finite classification problem. In particular, the existence of a polynomial-time algorithm for a classification problem is some evidence that the classification is "easy." -There is currently no known poly-time algorithm to test isomorphism of finite groups, given by their multiplication tables, and it is widely believed that p-groups (even those of class 2) are the hardest cases. (Beware of potential circularity in beliefs though: the wildness of classifying p-groups of class 2 is one of the pieces of evidence for the preceding belief, although there are others as well.) -Furthermore, despite all of the exciting recent progress on classifying groups by coclass, as far as I know this hasn't yet been used to make any progress towards a polynomial-time algorithm for testing isomorphism of p-groups. Furthermore, isomorphism of p-groups of class 2 are believed to be the hardest cases, and these are exactly the groups of maximal coclass, for which the recent classification results say the least. -(Polynomial-time as a version of "easiness" is a little subtle compared to mathematical experience: for example, the fact that finite simple groups are all generated by 2 elements (a consequence of CFSG) implies that there is a relatively simple polynomial-time algorithm to test isomorphism of finite simple groups (find a generating pair of elements in the one group, try all possible maps of that pair into the other group and check if each is an iso), but that algorithm relies on a massive, difficult theorem for its correctness. However, I still think the computational complexity of a classification problem is one useful gauge of its difficulty, particularly when it comes to finite problems.)<|endoftext|> -TITLE: Norms of higher derivatives of mappings between Riemannian manifolds -QUESTION [6 upvotes]: Let $M, N$ be Riemannian manifolds and $f: M \to N$ be a smooth map -(I'm actually only considering diffeomorphisms (flows) -$\Phi^t: M \to M$, but just for the sake of generality). -The first derivative of $f$ can be understood as its tangent map -$T f: T M \to T N$. Higher derivatives can abstractly be viewed as -maps between higher order tangent bundles. -I want to make estimates on the (operator norm) size of these higher -derivatives. In the higher order tangent spaces (see also the recent -question In how many ways can an iterated tangent bundle (T^k)M be viewed as a fibre bundle over (T^(k-1))M?) I'd have to use -induced metrics, which I don't readily know how to work with, and besides, -I think these would include the base, lower order derivatives as well. -I would prefer to keep things defined on the tangent/tensor bundle, in -a similar way as taking covariant derivatives for vector fields, but I -don't know how to do this for for maps $f: M \to N$. -So my question roughly is: are there natural/practical representations -of norms of higher order derivatives of maps between manifolds? -One thing I did come up with is representing $f$ in normal coordinates, as these are the most canonical charts and then use the norms in the tangent spaces at the argument and image points $x$ and $y = f(x)$. -(The basis for this question is that I want to obtain a Gronwall-like -growth estimate for the higher derivatives of a flow $\Phi^t$ in -terms of the exponential growth of its tangent flow $D \Phi^t$.) - -REPLY [7 votes]: You could indeed pull back each tensor bundle of $N$ via the map $f$ to $M$ and use the naturally induced metric and connection to define norms of higher covariant derivatives of $f$. You can do this for every order greater than or equal to $1$. Depending on your needs, this might work fine. -If you need something that will work under weaker a priori assumptions, then you might need to embed $N$ isometrically into a higher dimensional Euclidean space (via Nash's theorem) and treat $f$ as a vector-valued function. This allows you to work with maps $f$ that are not necessarily continuous and define its weak derivatives. This is often used in the study of harmonic maps.<|endoftext|> -TITLE: Does a compact semilocally simply connected geodesic space have the homotopy type of a CW complex? -QUESTION [5 upvotes]: Does a compact semilocally simply connected geodesic space have the homotopy type of a compact CW complex? Actually what I'd like to know is whether the fundamental group of such a space is finitely presented. -Edit: As Bruno Martelli notes, this is obviously false, but the question of whether the fundamental group is finitely presented, which is what I really want to know, is still open. - -REPLY [7 votes]: I hope I have not goofed, but I think the answer to your modified question is yes: - -The fundamental group of a semi-locally simply connected, compact and geodesic space is finitely presented. - -Here are the ingredients - all numbers in parentheses refer to Bridson-Haefliger, Metric spaces of non-positive curvature, Springer Grundlehren, 1999, Part I. - -The universal covering space $\widetilde{X}$ equipped with the length metric induced by the covering projection is a length space (3.25). -The fundamental group $\pi_{1}(X)$ acts on $\widetilde{X}$ properly and cocompactly by isometries (8.3 (2)), see also (8.5). -If a length space $\widetilde{X}$ admits a proper and cocompact action by isometries then it is locally compact (8.4 (1)) and hence proper and geodesic (3.7). -A group is finitely presented if and only if it acts properly and cocompactly by isometries on a simply connected geodesic space (8.11). - -All this taken together yields that $\widetilde{X}$ is a simply connected geodesic metric space and $\pi_{1}{(X)}$ acts properly and cocompactly by isometries, hence $\pi_{1}{(X)}$ must be finitely presented.<|endoftext|> -TITLE: Distribution of the spectrum of large non-negative matrices -QUESTION [15 upvotes]: This question is related to that of Thurston. However, I am not interested in algebraic integers, and I wish to focus on random matrices instead of random polynomials. -When considering (entrywise) non-negative matrices $M$, a natural probability measure seems to be -$$\prod_{i,j=1}^ne^{-m_{ij}}dm_{ij}.$$ -By Perron-Frobenius theorem, the spectral radius $\rho(M)$ is an eigenvalue, associated to a non-negative eigenvector. Almost surely, $M$ is positive and therefore this eigenvalue is simple and its eigenvector is positive. - -What is the distribution of the eigenvalues of $M$ as the size $n$ goes to infinity ? What is the relevant normalization ? Should we consider $\lambda/\rho(M)$ or $\lambda/\sqrt{\rho(M)}$ or something else ? Is it the same asymptotics as in the case of the conjugates of the algebraic Perron numbers considered by Thurston ? - -Note that because of the constraint $m_{ij}\ge0$, an exponential law looked more natural to me than a Gaussian. Has anyone an other suggestion of probability over non-negative matrices ? - -REPLY [11 votes]: This is a non-centered iid random matrix whose entries have mean one and variance one (and decay exponentially at infinity), and as such, is subject to the circular law with one outlier. Thus, there will be one eigenvalue roughly near n, and the rest will be more or less uniformly distributed in the complex disk of radius $\sqrt{n}$. The latter result is due to Chafai (at least for almost all of the eigenvalues), and the former is due to Silverstein. I discuss some finer aspects of the outlier eigenvalue (and show that all the other eigenvalues are nearly contained in the disk) in a more recent paper. (See in particular Phillip Wood's Figure 3 in that paper for an example of the eigenvalue distribution of a similar matrix model to the one you propose.)<|endoftext|> -TITLE: Arbitrarily thin additive bases of the natural numbers -QUESTION [6 upvotes]: A subset $A$ of $\mathbb{N}$ is called a basis of order $k$ if the set $kA$ = {$a_1 + \cdots + a_k | a_1, \cdots, a_k \in A$} $= \mathbb{N} \setminus C$, where $C$ is a finite set of positive integers. Let $r_{k,A}(N)$ be the number of representations of $N$ as a sum of $k$ (not necessarily distinct) elements of $A$. Erdos proved the following theorem in 1956: -There exists a basis $B \subset \mathbb{N}$ such that $r_{2,B}(N) = \Theta (\log(n))$ for every sufficiently large $n$. -In 1990, together with Tetali, he proved the following generalization. -For any fixed $k \in \mathbb{N}$, there exists a basis $B \subset \mathbb{N}$ such that $r_{k,B}(N) = \Theta (\log(n))$. -Now my question is, can there be any 'thinner' bases? That is, can we improve on the $\log(n)$ in the above theorems? Erdos and Turan conjectured that $\limsup_{n \rightarrow \infty} r_{2,B}(n) = \infty$ and later Erdos conjectured that $\limsup_{n \rightarrow \infty} r_{2,B}(n)/\log(n) > 0$ for any basis $B$ of order 2, suggesting that the answer is that one cannot really improve on the $\log(n)$. Are there are progress in this direction, either positive or negative? - -REPLY [3 votes]: As far as I know, one can achieve $r_{2,B}(n) \sim c \log(n)$, which is stronger than big-theta. I'm certain that nobody has beaten the logarithmic barrier in this problem, though. -Regarding the Erdos-Turan conjecture, essentially nothing is known except for a few factoids that rule out certain approaches. For example, there are some results showing that that $\limsup$ must be at least 6, and I wouldn't be surprised if a reasonable computation could push that up to 10 or 12 (the name Grekos comes to mind, but I'm not certain of his involvement). There has also been progress of the sort of proving and disproving various analogs (sets of integers, for example, instead of sets of naturals, or replace addition with some other binary function), Nathanson, Hegarty and others have worked on this sort of analog. - -Edit: The result I was thinking of was one of Cassels', which isn't what I recalled. Cassels proved that there is a basis $\{b_1 -TITLE: What is the state of our ignorance about the normality of pi? -QUESTION [51 upvotes]: Famously, it is not known whether $\pi$ is a normal number. Indeed, there are far weaker statements that are not known, such as the statement that there are infinitely many 7s in the decimal expansion of $\pi$. I'd like to have some idea of where the boundary lies between what we know and what we do not know. For example, I would guess that it is not even known not to be the case that all decimal digits from some point on are 0s and 1s. Am I right about this? -A partial answer to this question was given by Timothy Chow in a discussion of another question: Is pi a good random number generator?. He pointed out that some very weak facts can be deduced from known results about how well $\pi$ can be approximated by rationals. I suppose I could ask whether that is essentially the only technique we have. Could it be, for instance (as far as what is proved is concerned -- obviously it isn't actually the case) that the digits of $\pi$ are all 0s and 1s from some point on and that there is a constant $C$ such that the number of 1s in the first $n$ digits is never more than $C\sqrt{n}$? - -REPLY [18 votes]: I believe that Gerry's answer to the earlier question on MO -is exhaustive enough for this question as well. The normality of $\pi$ (which would imply the -irrationality measure 2) is not known yet, and even the questions like "all decimal digits from some point on are 0s and 1s" -cannot be shown unconditionally. -The best known estimate for the irrationality measure of $\pi$, due to V. Salikhov -[Russ. Math. Surv. 63:3 (2008), 570--572] -(see also MR2483171 (2010b:11082)) -reads $|\pi-p/q| > q^{-7.6063\dots}$ for all integers $p$ and $q>q_0$. The improvement -of Hata's record is achieved by constructing rational approximations to $\pi$ alone -(Hata's result is in fact about $\mathbb Q$-linear independence of $1$, $\log 2$ and $\pi$). -An interesting approach of attacking the normality of numbers like $\pi$, as well as problems like 0s--1s, is related -to the so-called BBP formulas. -However all such formulas for $\pi$ correspond to its binary (also hexadecimal) expansions. -The last piece of news is a base 10 formula for $1/\pi^2$ due to G. Almkvist and J. Guillera -(arXiv:1009.5202 [math.NT]), -$$ -\sum_{n=0}^\infty\frac{(6n)!}{n!^6} -(532n^2+126n+9)\frac1{10^{6n}}=\frac{375}{4\pi^2}, -$$ -which is not of BBP type but close to it in a certain sense. -A nice exposition of The Life of $\pi$ is J. Borwein's public -lecture given at the 2010 Australian Math Society meeting.<|endoftext|> -TITLE: Model categories of simplicial objects -QUESTION [10 upvotes]: If $\mathcal{C}$ is a category, then surely the category of simplicial -objects $s\mathcal{C}$ is not automatically a model category. What conditions -must $\mathcal{C}$ satisfy in order for $s\mathcal{C}$ to have a reasonable model structure? - -REPLY [13 votes]: Quillen gives a couple of sets of sufficient conditions for $s\mathcal{C}$ to be a model category, in his Homotopical Algebra book. Notably, this includes the case when $\mathcal{C}$ is a complete and cocomplete category with a small set of projective generators. This gives model structures for categories of simplicial groups, simplicial rings, simplicial lie algebras, etc. - -REPLY [2 votes]: It always has a model structure using Kan's theory of Reedy categories. For a proof, see Hirschhorn Model Categories and their Localizations 15.3. -This is because $\Delta$ and $\Delta^{op}$ are both Reedy categories. -I will address the more general question as well: -If $C$ is any small category, the condition for $M^C$ to be a model category is that $M$ be cofibrantly generated. However, it is not necessarily true that $M^C$ is itself cofibrantly generated. In general, the condition we need for that is called combinatoriality. However, the existence of such a cofibrantly generated model structure that is not combinatorial would disprove Vopenka's principle (a large cardinal axiom), so I cannot give an example of one.<|endoftext|> -TITLE: Does Riemann map depend continuously on the domain? -QUESTION [17 upvotes]: I've always taken this for granted until recently: -In the simplest case, given Jordan curve $C \subseteq \mathbb{C}$ containing a neighborhood of $\bar{0}$ in its interior. Given parametrizations $\gamma_1:S^1 \rightarrow C$. -Is it true that for all $\varepsilon >0$, there exists $\delta >0$ s.t. any Jordan curve $C'$ with a parametrization $\gamma_2:S^1 \rightarrow C_2$ so that $||\gamma_1-\gamma_2||<\delta$ in the uniform norm implies the Riemann maps $R, R'$ from $\mathbb{D}$ to the interiors of $C, C'$ that fixes the origin and have positive real derivatives at $\bar{0}$ would be at most $\varepsilon$ apart? - -REPLY [5 votes]: This fact is precisely Theorem 2.11 in the book Boundary Behaviour of Conformal Maps by Pommerenke. It follows from Caratheodory's Kernel Theorem.<|endoftext|> -TITLE: Non-diagonalizable doubly stochastic matrices -QUESTION [15 upvotes]: Are there constructive examples of doubly stochastic matrices (whose rows and columns all sum up to $1$ and contain only non-negative entries) that are not diagonalizable? - -REPLY [9 votes]: Here's a 2-parameter family of examples. -Let $$A=\pmatrix{a&b&1-a-b\cr2-2a-3b&3a+4b-2&1-a-b\cr a+3b-1&3-3a-5b&2a+2b-1\cr}$$ If $a$ and $b$ are chosen so that all the entries are between 0 and 1, then this is a doubly-stochastic matrix. It has the repeated eigenvalue $3a+3b-2$, which is not 1 if $a+b\ne1$. The matrix $A-(3a+3b-2)I$ has rank 2, provided $a+2b\ne1$. So if $a$ and $b$ are chosen in accordance with the stated conditions, this is a non-diagonalizable doubly-stochastic matrix. -I found this by letting $A(1,1)=a$, $A(1,2)=b$, $A(2,1)=a-\lambda$, $A(2,2)=b+\lambda$, letting the other entries be what they have to be to make all row- and column-sums 1, and then choosing $\lambda$ to make the trace $1+2\lambda$, so the eigenvalues have to be $\lambda$ repeated and 1. -EDIT: If, for some reason, you want the matrix to be singular, then just take $\lambda=0$. Then you get the one-parameter family $$A=\pmatrix{a&(2/3)-a&1/3\cr a&(2/3)-a&1/3\cr1-2a&2a-(1/3)&1/3\cr}$$ Now if you choose $a=1/4$ you get (essentially) David Speyer's example.<|endoftext|> -TITLE: Teaching undergraduate students to write proofs -QUESTION [43 upvotes]: In my experience, there are roughly two approaches to teaching (North American) undergraduates to write proofs: - -Students see proofs in lecture and in the textbooks, and proofs are explained when necessary, for example, the first time the instructor shows a proof by induction to a group of freshman, some additional explanation of this proof method might be given. Also, students are given regular problem sets consisting of genuine mathematical questions - of course not research-level questions, but good honest questions nonetheless - and they get feedback on their proofs. This starts from day one. The general theme here is that all the math these students do is proof-based, and all the proofs they do are for the sake of math, in contrast to: -Students spend the majority of their first two years doing computations. Towards the end of this period they take a course whose primary goal is to teach proofs, and so they study proofs for the sake of learning how to do proofs, understanding the math that the proofs are about is a secondary goal. They are taught truth tables, logical connectives, quantifiers, basic set theory (as in unions and complements), proofs by contraposition, contradiction, induction. The remaining two years consist of real math, as in approach 1. - -I won't hide the fact that I'm biased to approach 1. For instance, I believe that rather than specifically teaching students about complements and unions, and giving them quizzes on this stuff, it's more effective to expose it to them early and often, and either expect them to pick it up on their own or at least expect them to seek explanation from peers or teachers without anyone telling them it's time to learn about unions and complements. That said, I am genuinely open to hearing techniques along the lines of approach 2 that are effective. So my question is: -What techniques aimed specifically at teaching proof writing have you found in your experience to be effective? -EDIT: In addition to describing a particular technique, please explain in what sense you believe it to be effective, and what experiences of yours actually demonstrate this effectiveness. - -Thierry Zell makes a great point, that approach 1 tends to happen when your curriculum separates math students from non-math students, and approach 2 tends to happen math, engineering, and science students are mixed together for the first two years to learn basic computational math. This brings up a strongly related question to my original question: -Can it be effective to have math majors spend some amount of time taking computational, proof-free math courses along with non-math majors? If so, in what sense can it be effective and what experiences of yours demonstrate this effectiveness? -(Question originally asked by Amit Kumar Gupta) - -REPLY [9 votes]: I taught the `transitions' course at a large state university a number -of years ago, with reasonable success. -The clientele of this (purely elective) course was mainly B students -in calculus who would likely have done poorly in real analysis -or abstract algebra, and would have had difficulty completing a math major. -To maximize the impact on students' ability to understand and produce -proofs, several things were important: -a) The text was Velleman's How to Prove It: A structured approach, -which is readable by average students, clearly delineates the structure and -construction of typical proofs, and is full of problems which are -elementary but not boring. (For a regular beginning analysis class -I just ask the students to read this book---esp. chapter 3 as mentioned -by Jon Bannon---and I discuss the basics of this material -for a few lectures.) -b) The format of most class sessions was discussion not lecture. To have these students passively listen, like in their -previous courses which they demonstrably failed to master, would be -useless. Discussion was structured like in a humanities or language -course, led by the instructor with specific goals in mind and calling -on individual students to involve everyone and make sure they get it. -The 22 students were informed that it was essential that they come to -class prepared, having read the day's material and having worked the -relevant problems, laid out in each week's syllabus. -c) Why we insist on ``proof beyond unreasonable doubt'' was explained, -referring to the great discoveries of 19th and early 20th century -analysis (especially regarding infinite sets and fractals) -that demand the enormously skeptical approach to establishing truth -which now dominates much of modern mathematics. -Many of the students were weak at the start and apparently benefited -from all this. For one, this course was a big step in eventually -changing his career from fisherman to gaining a Masters and working in a scientific software company. Another later did A work in a senior-level ODE course I taught. -But I did not conduct a randomized controlled study.<|endoftext|> -TITLE: Convergence of alternating harmonic sums -QUESTION [10 upvotes]: I owe the idea of asking this question to Max Muller and -his curiosity. -What is the set of $\alpha$ in the interval $0\le\alpha < 1$ for which -the alternating sum -$$ -\sum_{n=1}^\infty\frac{(-1)^{n+[n^\alpha]}}n -$$ -converges? Here $[\ \cdot\ ]$ denotes the integral part of a number. -It clearly converges when $\alpha=0$ and -my post -to Max's answer implies the convergence when $\alpha=1/2$. -What about more general $\alpha$? Of course, the question is meaningful -for positive $\alpha\notin \mathbb Z$ as well, but then it seems to be much harder. - -REPLY [4 votes]: Let me try and give an answer for $\alpha>1$ also. This one uses some technology from a -paper of mine with Boshernitzan, Kolesnik and Wierdl. -I want to use exponential sums. Using the notation of that paper we take $a(n)=n+n^\alpha$. That paper lets us control $\hat A_t(1/2)=(1/t)\sum_{n\le t}e([a(n)]/2)=(1/t)\sum_{n\le t}(-1)^{n+[n^\alpha]}$ where $e(x)=e^{2\pi i x}$ (see after Lemma 7.2). -The proofs of Theorem 3.4 and Theorem 7.1 give (if you read carefully) the existence of an $\epsilon>0$ and a $C$ such that $|\hat A_t(1/2)| < Ct^{-\epsilon}$ for all $t$. This says that the difference between the number of $+1$'s and the number of $-1$'s (the discrepancy) for $n\le t$ is at most $t^{1-\epsilon}$ (ignoring constants from now on). -Now let $K>2/\epsilon$ and divide the integers into blocks $I_j=(j^K,(j+1)^K]$. The discrepancy up to $j^K$ is at most $j^{K-2}$ by the above. Similarly the discrepancy up to $(j+1)^K$ is also at most $j^{K-2}$. So the discrepancy in the $I_j$ block is at most $j^{K-2}$. -We now have $\sum_{n\in I_j}(-1)^{n+[n^\alpha]}/n = \sum_{n\in I_j}(-1)^{n+[n^\alpha]}/j^K + \sum_{n\in I_j}(-1)^{n+[n^\alpha]}(1/n-1/j^K)$. -By the above comment, the first term contributes $j^{-2}$. In the second term, there are $j^{K-1}$ terms, each contributing in absolute value at most $j^{-K-1}$ giving a maximum total contribution of $j^{-2}$. -It follows that the contributions from the $I_j$-blocks are absolutely summable and we're done. -Of course $\alpha<0$ is trivial so this is good for all $\alpha$ except the positive integers.<|endoftext|> -TITLE: Does one of the hexagon identities imply the other one? -QUESTION [6 upvotes]: Suppose we have a monoidal category equipped with additional data that almost makes it a braided monoidal category except that only one of the hexagon identities -is satisfied. -Can we then prove the other hexagon identity? -If not, is there an explicit counterexample and can we prove the other identity under the additional -condition that the braiding is symmetric? - -REPLY [11 votes]: Consider the category of $A-$graded vector spaces (here $A$ is an abelian group) with -obvious tensor product and trivial associator. Then each isomorphism $a\otimes b\to b\otimes a$ can be specified as a nonzero complex number $B(a,b)$. Now the first hexagon axiom says that the function $B(a,b)$ is linear in the first variable and the second axiom says that $B(a,b)$ is linear in the second variable. Clearly these conditions are independent. The symmetric braiding would correspond to skew-symmetric function $B(a,b)$ -and linearity in one variable would imply bilinearity.<|endoftext|> -TITLE: How to picture $\mathbb{C}_p$? -QUESTION [63 upvotes]: I hope this is appropriate for mathoverflow. Understanding $\mathbb{C}_p$ has always been something of a stumbling block for me. A standard thing to do in number theory is to take the completion $\mathbb{Q}_p$ of the rationals with respect to a $p$-adic absolute value. The resulting field is then complete, but has no good reason to be algebraically closed. You can take its algebraic closure, but that is not complete, so then you take the completion of that, and get a field which is both complete, and algebraically closed, denoted by $\mathbb{C}_p$. -I understand that it is a reasonable desire to have a field extension of $\mathbb{Q}_p$ that is both complete and algebraically closed; my trouble, however, is getting some sort of grasp on how to picture this object, and to develop any intuition about how it is used. Here are my questions; I'd imagine the answers are related: - -Am I even supposed to be able to picture it? -Is there some way I ought to think of a typical element? -Is it worth it, in terms of these goals, to look at the proofs of the assertions in my first paragraph? -How is $\mathbb{C}_p$ typically used? (this question may be too vague, feel free to ignore it!) - -Please feel free to answer any or all of these questions. - -REPLY [11 votes]: Since there are already several very good answers, I just discuss question 4 (how is ${\mathbb C}_p$ typically used?) with one example of use which made a great impression on me when I learnt it, and made me think that ${\mathbb C}_p$ was something deep and serious, and not only a very amusing curiosity. This example is a theorem of Tate and Sen, which states that if $V$ is a finite-dimensional over $\mathbb{ Q_p}$ vector space with a continuous linear action of $G=$Gal$(\overline{\mathbb Q_p}/{\mathbb Q_p})$ (that is, $V$ is a $p$-adic representation of $G$), then the following are equivalent: -(1) $\dim_{\mathbb Q_p} (V \otimes {\mathbb C_p})^{G} = \dim_{\mathbb Q_p} V.$ (Here, G acts on $\mathbb{C_p}$ by extending by continuity its action on $\overline{\mathbb Q_p}$ -and it acts on $V \otimes {\mathbb C_p}$ by acting on both factors.) -(2) The inertia subgroup of $G$ acts on $V$ through a finite quotient (in more knowledeable words, $V$ is potentially unramified). -To appreciate this theorem, it may be useful to solve for oneself the following elementary exercise: if in (1), ${\mathbb C}_p$ is replaced by $\overline {\mathbb Q_p}$, then (2) should be replaced by "$G$ acts on $V$ through a finite quotient". Somehow, replacing -$\overline {\mathbb Q_p}$ by its completion allows (1) to see inside the group $G$ -and detect the behaviour of the inertia subgroup in it. -I believe that someone who understands the proof of this theorem has necessarily -a good understanding of $\mathbb{C}_p$, and this will be my answer to question 3 as well: -knowing the proof of the basic assertions on $\mathbb{C}_p$ given in the questions is a first step into a good understanding of that field and its elements, but won't take you very far. Learning the proof of the above theorem will let you get a much deeper look inside $\mathbb{C}_p$ -- and in addition you will learn a nice result, which is a first step in the fundamental $p$-adic Hodge Theory.<|endoftext|> -TITLE: "Bad" reduction of Shimura curves via dual graphs -QUESTION [6 upvotes]: I have the following naive (and inexpert) question about the -reduction of Shimura curves at primes dividing the discriminant -of the underlying quaternion algebra. It requires some background -to state. That is, let $F$ be a totally real field of degree $d$. -Fix a real place $\tau_1$ in the set of real -places $\lbrace \tau_1, \ldots, \tau_d \rbrace$ of $F$. Let -$B$ be a quaternion algebra over $F$ that is split at $\tau_1$ -and ramified at $\tau_2, \ldots, \tau_d$. Let $H \subset -\widehat{B}~(= B \otimes \widehat{Z})$ be a compact open -subgroup. Let $M_H$ denote the Shimura curve over $F$ of -level $H$, with complex points given by \begin{align*} -M_H({\bf{C}}) &= B^{\times}\backslash \widehat{B}^{\times} -\times \left({\bf{C}}-{\bf{R}} \right)/H.\end{align*} -Fix a prime $v \subset \mathcal{O}_{F}$. Assume that -$H$ can be factored as $H^v \times H_v$, with $H_v \subset -B_v^{\times}~(= B^{\times} \otimes F_v)$ maximal. If $v$ does -not divide the discriminant of $B$, then it is known by -work of Morita and Carayol that $M_H$ has good reduction -over $v$, hence that there exists a smooth model ${\bf{M}}_H$ -of $M_H$ over $\mathcal{O}_{(v)}$. If $v$ divides the discriminant -of the quaternion algebra $B$, then it is known by work of Varshavsky -for instance that there exists an integral model -${\bf{M}}_{H}^V$ of $M_H$ over $\mathcal{O}_{(v)}$. (N.B. there -is apparently also a model due to Drinfeld, described extensively -in the literature for the case of $F={\bf{Q}}$, though it -is not clear to me why Drinfeld's work, which seems -to require a moduli theoretic description of $M_H$, extends -to the general totally real fields setting). Anyhow, let -$F_v$ denote the completion of $F$ at $v$, with $\kappa_v$ the -residue field and $\pi_v$ a uniformizer. By -Cerednik's theorem, the completion of ${\bf{M}}_H^V$ -along its closed fibre is canonically isomorphic to the product -\begin{align*} GL(F_v)\backslash \widehat{\Omega}^{unr} \times -D^{\times}\backslash \widehat{D}^{\times}/\overline{H}^v. -\end{align*} Here, $\widehat{\Omega}^{unr}$ denotes the product -$\widehat{\Omega} \times_{\operatorname{Spf}\mathcal{O}_{F_v}} \operatorname{Spf} -\mathcal{O}_{F_v}^{unr}$, where $\widehat{\Omega}$ denotes the $v$-adic -upper half plane (viewed as a formal scheme), and -$\mathcal{O}_{F_v}^{unr}$ the ring of Witt vectors with coefficients -in $\overline{\kappa}_v$. The action of $\gamma \in GL(F_v)$ on -$\widehat{\Omega}^{unr}$ is via the image of $\gamma$ in $PGL(F_v)$ -on the component $\widehat{\Omega}$, and via multiplication by -$\operatorname{Frob}_v^{n(\gamma)}$ on -$\widehat{\mathcal{O}}_{F_v}^{unr}$, where $n(\gamma) = - -ord_v \left( \det(\gamma) \right)$. As well, -$D$ denotes the totally definite quaternion algebra over $F$ obtained -from $B$ by switching invariants at $v$ and $\tau_1$, with -$\overline{H}^v$ the compact open subgroup of $\widehat{D}^{\times v}$ -corresponding to $H^v$ under a fixed isomorphism $B^{\times v} \cong -D^{\times v}$. The theory of Mumford-Kurihara unifomization then gives -the following information about this curve ${\bf{M}}_{H}^V$: - -The curve ${\bf{M}}_{H}^V$ is an admissible curve -over $\mathcal{O}_{F_v}$ in the sense of Jordan-Livne, i.e. - -(i) ${\bf{M}}_H^V$ is a flat, proper curve over $\mathcal{O}_{F_v}$ -with a smooth generic fibre. -(ii) The special fibre of ${\bf{M}}_H^V$ is reduced; the normalization -of each of its irreducible components is isomorphic to ${\bf{P}}^1_{\kappa_v}$, -and its only singular points are $\kappa_v$-rational, ordinary double points. -(iii) The local ring ${\bf{M}}_{H, x}$ at any singular point $x$ of -the special fibre is isomorphic as an $\mathcal{O}_{F_v}$-algebra to -$\mathcal{O}_{F_v}[[X,Y]]/(XY - \pi_v^{m(x)})$, for $m(x) \geq 1$ a uniquely -determined integer. - -The dual graph $\mathcal{G}({\bf{M}}_H^V) - = (\mathcal{V}({\bf{M}}_H^V),\mathcal{E}({\bf{M}}_H^V))$ -of the special fibre of ${\bf{M}}_H^V$ is -isomorphic to $GL(F_v)^{+} \backslash \left( -\Delta \times D^{\times}\backslash \widehat{D}^{\times}/\overline{H}^v -\right)$, minus any loops. Here, $GL(F_v)^+ \subset GL(F_v)$ denotes the -collection of matrices with determinant having even $v$-adic valuation, -and $\Delta = (\mathcal{V}(\Delta), \mathcal{E}(\Delta))$ the -Bruhat-Tits tree of $SL(F_v)$. - -My question is the following: why is the edgeset -$\mathcal{E}({\bf{M}}_H^V)$ nonempty? The dual graph -$\mathcal{G}({\bf{M}}_H^V)$ is clearly disconnected, and -seen easily to be given by the disjoint union of connected -graphs \begin{align*} \coprod_i \mathcal{G}_i &= -\coprod_i \overline{\Gamma}_i \backslash \Delta. -\end{align*} Here, each $\overline{\Gamma}_i$ denotes -the image in $PGL(F_v)$ of a suitable arithmetic subgroup -$\Gamma_i \subset D^{\times} \cong D_v^{\times} \cong GL(F_v)$. -Each component graph $\mathcal{G}_i = (\mathcal{V}_i, -\mathcal{E}_i)$ is connected. Now, since $\Delta$ is a tree, -each component graph $\mathcal{G}_i = \overline{\Gamma}_i -\backslash \Delta$ is a tree. It is then well known that -each (first) Betti number $\beta(\mathcal{G}_i) := -\vert \mathcal{E}_i \vert - \vert \mathcal{V}_i \vert + 1$ must -vanish, i.e. $\vert \mathcal{E}_i \vert = -\vert \mathcal{V}_i \vert -1 = 0$. If so, then the cardinality -of the edgeset $\mathcal{E}({\bf{M}}_H^V)$ must also equal -zero. i.e. the special fibre of ${\bf{M}}_H^V$ would have no -singular points ... what have I missed here? - -REPLY [8 votes]: Inkspot is indeed correct that the component graphs are indeed not generally trees. -As you seem to have deduced for yourself, Cerednik-Drinfeld uniformization is a highly nontrivial concept, and it really helps to have some examples to set it in your mind. Most helpful in this direction is Ogg's "Mauvaise réduction des courbes de Shimura" where he draws out a few of these dual graphs. -In general the dual graphs have $2h$ vertices $x$ where $h$ is the class number of $\mathcal{O}_x$, a level $H$ Eichler order in $D$ (your totally definite quaternion algebra, so note there's a choice of which $x$ to make here, but as long as the level is squarefree all orders are hereditary and it doesn't make a difference). -An orbit-stabilizer theorem computation then shows that for instance when $B$ is a quaternion algebra over $\mathbf{Q}$, $p+1$ (the size of the set of edges $y$ stemming from a particular vertex $x$ in the Bruhat-Tits tree) is equal to $\sum_{e(y) = x} \frac{\mathcal{O}_x^\times}{\mathcal{O}_y^\times}$. So it's not just that there are edges, but we know exactly how many there are! (For a readable account of details of this, see Kurihara's paper on Equations defining Shimura Curves) -Also, if I may take issue with 1.(ii) and 1.(iii), you've given a good description of the special fiber over $\overline \kappa_v$ (which is what I'm taking the dual graph to represent the data for), not necessarily $\kappa_v$. What you've claimed is that the special fiber is a Mumford Curve, that is, the transverse union of a number of copies of $\mathbb{P}^1$'s. The truth is that the special fiber is a quadratic twist of a Mumford curve, where the Galois action is not simply given by the $| \kappa_v|$-Frobenius, but where the action of Frobenius is identified with the action of the Atkin-Lehner operator $w_p$, which interchanges some of the components (if you want to think about the graph, its vertex set can be partitioned into $ \{x_1, \dots , x_h, x_1', \dots, x_h'\}$ where $w_p(x_i) = x_i'$). -All that said, some of the best advice I've heard for trying to understand this stuff is to first completely understand what happens when $v$ divides the LEVEL because in that case the moduli problem is much easier (if an abelian variety here is isogenous to a product of supersingular elliptic curves, it's isomorphic to a product of supersingular elliptic curves) -Here are a few additional references: -http://www.math.mcgill.ca/cfranc/documents/bctranslation.pdf (a translation of Boutot-Carayol) -http://math.berkeley.edu/~ribet/Articles/bimodules.pdf (this includes a somewhat more intuitive description of the components of the Mumford curve) -http://math.uga.edu/~pete/thesis.pdf (a comprehensive introduction to Shimura Curves and the action of the Atkin-Lehner group) -http://www.springerlink.com/content/gj8365486214l141/ (this is Oort's "which abelian surfaces are products of supersingular elliptic curves", see also his book on moduli of supersingular abelian varieties, as when $v$ ramifies in $B$, you're asking a question about moduli of supersingular abelian varieties, see the appendix on Honda-Tate theory to the thesis above)<|endoftext|> -TITLE: What kind of structures allow Galois descent? -QUESTION [7 upvotes]: EDIT: Question solved. - -Let me explain what I mean. -The classical formulation of Galois descent, e. g. in Crawley-Boevey's "Cohomology and central simple algebras", uses the following notion: -Definition: A $K$-vector space with additional structure is one of the following: - -a $K$-vector space; -a $K$-algebra; -an $A$-module, for some fixed $K$-algebra $A$; -an $A$-$B$-bimodule, for some fixed $K$-algebras $A$ and $B$; - -etc. -My question is what exactly the "etc." means, or how to precise this definition. I tried to do this the following way: A $K$-vector space with additional structure is given by a $K$-vector space $X$ and a family of linear maps of the forms $U_1\otimes U_2\otimes ...\otimes U_m\otimes X^{\otimes a}\to X^{\otimes b}$ (with $U_1$, $U_2$, ..., $U_m$ being fixed $K$-vector spaces) that satisfy certain polynomial equations (for example, in the case of a $K$-algebra, our maps are the multiplication map $X^{\otimes 2}\to X^{\otimes 1}$ and the unity map $X^{\otimes 0}\to X^{\otimes 1}$, and the equations they must satisfy are associativity and unitality). -This is similar to the definition in Gille and Szamuely's "Central Simple Algebras and Galois Cohomology", except I have the $U_i$ and they don't (but this doesn't make a difference, because they are fixed and thus a map $U_1\otimes U_2\otimes ...\otimes U_m\otimes X^{\otimes a}\to X^{\otimes b}$ is just a family of maps $X^{\otimes a}\to X^{\otimes b}$). -Now I've got a problem. It's the proof of Theorem 5.9 in 1 rsp. that of Theorem 2.3.3 in 2. The part where you are given a cocycle in $H^1\left(G,\mathrm{Aut}\left(X^L\right)\right)$ (in the notation of 1) rsp. $H^1\left(G,\mathrm{Aut}\left(\Phi\right)\right)$ (in the notation of 2) and you want to show that the invariant subspace $X_{\rho}$ (in the notation of 1) rsp. $\left(\ _a V_K\right)^G$ (in the notation of 2) is a $K$-vector space (in the notation of 1) rsp. $k$-vector space (in the notation of 2) with the same additional structure. It works well for the standard cases ($K$-vector space, $K$-algebra, $A$-module, $A$-$B$-bimodule), but all of these cases have a peculiar property: that each of the maps $U_1\otimes U_2\otimes ...\otimes U_m\otimes X^{\otimes a}\to X^{\otimes b}$ has $b\leq 1$. There are structures for which this doesn't hold ($K$-coalgebras, $K$-Hopf algebras, separable $K$-algebras with separability idempotent, and many more), and I would like to know how to do Galois descent (viz., prove that elements of $H^1\left(G,\mathrm{Aut}\left(X^L\right)\right)$ yield isomorphism classes of twisted forms of $X$ split by $L$, to use the notations of 1). The proof from 1 and 2 doesn't seem to work in this case; at least I don't see why every $G$-invariant map $\left(X\otimes L\right)^{\otimes 2}\to \left(X\otimes L\right)^{\otimes 2}$ must come from a map $X^{\otimes 2}\to X^{\otimes 2}$. - -REPLY [3 votes]: If you allow tensor powers of duals of $X$, then all the things you are talking about are just elements of certain vector spaces ($X\to X$ is just an element of $X^∗\otimes X$, for example). And any Galois invariant element of $V\otimes L$ is an element of $V$ (where $V$ is a vector space over $K$).<|endoftext|> -TITLE: Is there any transitivity for separable algebras? -QUESTION [9 upvotes]: If $R$ is a commutative ring (with $1$), then an $R$-algebra $A$ is said to be separable if $A$ is projective as an $A$-$A$-bimodule. (The notion of an "$A$-$A$-bimodule" includes the requirement that $R$ acts the same way from the left and from the right. The notion of "projective $A$-$A$-bimodule" is defined in the same way as the notions of "projective $A$-left module" and "projective $A$-right module". If you are unhappy with this definition, you can rewrite any $A$-$A$-bimodule as an $A\otimes_R A^{\mathrm{op}}$-left module, and then use the notion of a projective left module.) -There is a criterion stating that an $R$-algebra $A$ is separable if and only if there is an element $e\in A\otimes_R A$ such that the multiplication map $A\otimes_R A\to A$ sends $e$ to $1\in A$, and such that $ae=ea$ for all $a\in A$. Equivalently, an $R$-algebra $A$ is separable if and only if the $A$-$A$-bimodule epimorphism $A\otimes_R A\to A$ given by multiplication of the two tensorands has a section in the category of $A$-$A$-bimodules. (This is both in Crawley-Boevey, chapter 4. I have difficulties finding other literature which does the notion of separability in full generality. For some reason, most books consider it enough to talk about separable $k$-algebras with $k$ a field.) -Now my question is, is there a transitivity theorem like this: -If $B$ is a separable commutative $R$-algebra, and $A$ is a separable $B$-algebra, then $A$ is a separable $R$-algebra as well? -Maybe some conditions like projectivity (of $B$ as $A$-module and $A$ as $R$-module) must be added; that would be ok for me. -If something like this holds, then the proof that separability of a field extension as algebra is just Galois-theoretical separability could be simplified (most importantly, the ugly Primitive Element Theorem would not be needed anymore). - -REPLY [3 votes]: You do not even need that the intermediate algebra be commutative: - -Let $R$ be commutative and let $A\subseteq C\subseteq B$ be a chain of $R$-algebras. If $B$ is a separable extension of $C$ and $C$ of $A$, then $B$ is a separable extension of $A$. - -This is one of the exercises in Pierce's Associative Algebras, §10.8. - -Later: Here is an answer to darij's question in the comments below: -If two $R$-algebras $A$ and $B$ are Morita equivalent, so that their categories of left modules ${}_A\mathrm{Mod}$ and ${}_B\mathrm{Mod}$ are equivalent (as $R$-linear categories) then one can see that their categories of bimodules ${}_A\mathrm{Mod}{}_B$ and ${}_B\mathrm{Mod}{}_B$ are also equivalent (as $R$-linear categories, again) by an equivalence which maps the $A$-bimodule $A$ to the $B$-bimodule $B$. -Now $A$ being a separable $R$-algebra means that the $A$-bimodule $A$ is a projective object in ${}_A\mathrm{Mod}{}_A$. Since projectivity of an object is preserved by equivalences of categories, we conclude that - -an $R$-algebra which is Morita equivalent to a separable $R$-algebra is itself separable. - -Now $A$ and $M_n(A)$ are Morita equivalent $R$-algebras for all $n\geq1$, so an affirmative answer to the question follows. -One can even be bolder: if $A$ and $B$ are now only derived equivalent (but now I think we need to assume they are both flat over $R$...) then it also follows that they are simultaneously separable, by a similar reasoning.<|endoftext|> -TITLE: Analogue of Smith normal form for matrices over $\mathbb Z[t]$ -QUESTION [19 upvotes]: Let $R$ be a principal ideal domain and $A \in M_n R$. It is well known that there exist invertible matrices $Q$ and $S$ and a diagonal matrix $D= {\rm diag}(a_1,\dots,a_n)$ such that - -$a_i \mid a_{i+1}$ for all $1 \leq i \leq n-1$, and -D=QAS. - -The matrix $D$ is called Smith normal form of $A$ and is unique. Obviously, this applies to both $\mathbb Z$ and $\mathbb R[t]$. - -Question 1: Is there any analogue for $R= \mathbb Z[t]$? Is there any classification of matrices over $\mathbb Z[t]$ up to equivalence? - -A related question is the following: - -Question 2: Is there any classification of $n \times n$-matrices over $\mathbb Z$ up to conjugation? - -(The relation comes from looking at the matrix $t\cdot 1_n - A$ for $A \in M_n \mathbb Z$. Then, classification of $A$ up to conjugation is the same as classification of $t\cdot 1_n-A \in M_n \mathbb Z[t]$ up to equivalence.) -An obvious first invariant is the characteristic polynomial. Even if the matrix is assumed to be symmetric, it is unclear to me what kind of additional information could be added. -In that respect I know of a theorem of Latimer and MacDuffee which says that if the characteristic polynomial $f$ of $A \in M_n \mathbb Z$ is irreducible, then conjugacy classes of integer matrices with the same characteristic polynomial are in bijection with $\mathbb Z[\alpha]$-ideal classes in ${\mathbb Q}(\alpha)$, where $\alpha$ is a root of $f$. However, this seems to depend on the irreducibility of $f$ and I do not know of an extension to the general case. (This is nicely explained in notes by Keith Konrad.) - -Question 3: Are there any other positive results going in the direction of the Theorem of Latimer-MacDuffee? What if the characteristic polynomial is a product of two irreducible polynomials? - -Related but maybe easier: - -Question 4: Is there some characterization (in terms of the characteristic polynomials + additional invariants) of pairs of matrices in $A,B \in M_n\mathbb Z$, such that $A$ and $B$ are conjugate in $M_n \overline{\mathbb F}_p$ for all primes $p$ and in $M_n \mathbb C$? - -And finally - -Question 5: Is there some characterization (in terms of the characteristic polynomials + additional invariants) of pairs of matrices in $A,B \in M_n\mathbb Z$, such that $A$ and $B$ are conjugate in $M_n \overline{\mathbb F}_p$ for a fixed prime $p$ and in $M_n \mathbb C$? - -(Of course, Question 4 and 5 can be answered by looking at the Jordan decomposition for each of the fields separately. However, the question is, can we do something more conceptual?) - -REPLY [19 votes]: For Q1 the problem is that one invariant of the matrix is the (isomorpism class -of the) cokernel and any $\mathbb Z[t]$-module generated by $n$ elements and -$n$-relations appears as such an invariant. There simply are too many modules -over a $2$-dimensional ring such as $\mathbb Z[t]$. -As for Q2 you cannot really hope for a classifiction even for the conjugacy classes -of matrices of finite order. In fact already for matrices of order $p^m$, $p$ prime -and $m>2$, the problem is wild which essentially means that any complete -classification is hopeless. -Finally (as far as these comments go) for the part of Q3 where the characteristic -polynomial is the product of two distinct irreducible polynomials $f$ and $g$: -If $M$ is $\mathbb Z^n$ as a module of $\mathbb Z[t]$ through the matrix, then -we have a direct sum part $M'\bigoplus M''\subseteq M$, where $M'$ is the -annihilator of $f$ and $M''$ of $g$. Hence, $M'$ and $M''$ are given by ideals -as in Latimer-McDuffee and can be considered classified. Then $M$ is given by a -submodule $\overline M$ of $M'\bigotimes\mathbb Q/\mathbb Z\bigoplus -M''\bigotimes\mathbb Q/\mathbb Z$. As $M'$ and $M''$ are the kernels of -multiplication by $f$ resp.\ $g$ we also get that $\overline -M\cap M'\bigotimes\mathbb Q/\mathbb Z=0$ and the same for $M''\bigotimes\mathbb -Q/\mathbb Z$. Hence $\overline M$ is the graph of an isomorphism between -submodules $\overline M'\subseteq M'\bigotimes\mathbb Q/\mathbb Z$ and -$\overline M''\subseteq M''\bigotimes\mathbb Q/\mathbb Z$. This means that -$\overline M'$ and $\overline M''$ are killed by both $f$ and $g$ and as they -are relatively prime, the kernel of $g$ on $M'\bigotimes\mathbb Q/\mathbb Z$ (as -well as the kernel of $f$ on $M''\bigotimes\mathbb Q/\mathbb Z$) are finite (and -usually quite small). Hence one can (in principle) determine the possible -$\overline M$ and they have to be considered modulo automorphisms of $M'$ and -$M''$ which are given by units in their endomorphism rings (which are overorders -of $\mathbb Z[t]/(f)$ resp. $\mathbb Z[t]/(g)$). -This sometimes works very well. For instance this is exactly one way of doing -the classification of matrices of order $p$ (where $f=t-1$ and -$g=t^{p-1}+\cdots+t+1$). On the other hand I am pretty sure it is as -hopeless as the general conjugacy problem for general $f$ and $g$.<|endoftext|> -TITLE: Derivation of von Neumann algebra which is zero on MASA -QUESTION [5 upvotes]: Are there any example of $II_1$-factor $M$ with maximal abelian von Neumann subalgebra $A$ and non-zero derivation $\delta:M\rightarrow B(H)$ such that $\delta(a)=0$ for every $a\in A$? - -REPLY [2 votes]: Sorry to be naive but why can't you just represent $M$ on a separable Hilbert space and take the inner derivation induced by any non-central element in the masa? This maps into $M\subset B(H)$, vanishes on the masa, but is non-trivial on $M$ because the element is non-central.<|endoftext|> -TITLE: Is there a quaternionic algebraic geometry ? -QUESTION [44 upvotes]: Let $\mathbb{H}$ be the skew-field of quaternions. I'm aware of the -Theorem 1. A function $f:\mathbb{H}\to\mathbb{H}$ which is $\mathbb{H}$-differentiable on the left (i.e. the usual limit $h^{-1}\cdot (f(x+h)-f(x))$, for $h\to 0$, exists for every $x\in \mathbb{H}$) is a quaternionic affine function on the right (i.e. of the form $x\mapsto x\cdot \alpha + v$). -This means that there are no interesting smooth quaternionic funcions, hence no interesting "quaternionic-smooth manifolds" (which is not the same as the quaternionic-Kahler or hyperkahler structures you encounter in differential geometry and complex analytic geometry). -I think I can also recall the -Theorem 2. If a function $f:\mathbb{H}\to\mathbb{H}$ is locally $\mathbb{H}$-analytic (i.e. it can be locally developped in power series, for the suitable noncommutative notion of "power series"), than it corresponds to a real-analytic function $f:\mathbb{R}^4\to \mathbb{R}^4$, and any real-analytic funcion $f:\mathbb{R}^4\to \mathbb{R}^4$ can be obtained in this way. -That says that $\mathbb{H}$-analytic functions are essentially the same as quadruples of real-analytic functions of 4 variables. Hence there is no "quaternionic-analytic geometry" distinguishable from $4n$-dimentional real-analytic geometry. -I think the same happens with quaternionic (noncommutative) polynomials: they're just 4-tuples of real polynomials in 4 variables. -But, is it reasonable that the zero locus on $\mathbb{H}^n$ of a "noncommutative polynomial" with $\mathbb{H}$-coefficients doesn't have any further mathematical structure than it's real-algebraic variety structure? -It would be nice to be able to see things such as $\mathbb{HP}^1$ as a "quaternionic curve ", and to speak of a point " $\mathrm{Spec}(\mathbb{H})$ " (whatever it means) if it possible... - -Is there a theory of "quaternionic algebraic geometry", maybe as a branch or particular case of some noncommutative (algebraic) geometry theory? - -Of course, if such a theory has some sense, it cannot be the "obvious analog" of complex algebraic or analytic geometry, as theorems 1. and 2. above show. - -REPLY [4 votes]: You may perhaps be also interested in the quite recent paper -Gentili, Graziano; Stoppato, Caterina Zeros of regular functions and polynomials of a quaternionic variable. Michigan Math. J. 56 (2008), no. 3, 655–667 -(and other papers by the same authors).<|endoftext|> -TITLE: Why does (Ribbon) Graph (co)Homology Compute (co)Homology of MCG? -QUESTION [13 upvotes]: The title says it all. I am looking for an explanation or reference for why the homology of the ribbon graph complex computes the cohomology of the mapping class groups of surfaces. -I've seen explanations of this using operads, but my understanding is that the operad viewpoint is more recent and not how the above question was originally understood. - -REPLY [11 votes]: My favorite way to 'see' the connection between ribbon graphs and mapping class groups is to use the contractibility of the complex of arcs in a surface. Given a surface $S$, the arc complex $\mathcal{A}(S)$ has vertices given by isotopy classes of arcs with endpoints lying on the boundary of the surface, and simplices given by disjoints collections of such arcs. Hatcher has a rather attractive way to construct a contraction of this complex down to a single vertex. -By an easy inductive argument one can deduce from the contractibility of the arc complex that the poset $\mathcal{A}_0(S)$ of simplices for which the arcs cut the surface into pieces that are all discs is also contractible. Such a collection of arcs has a dual ribbon graph. -The mapping class group $MCG(S)$ acts on the poset $\mathcal{A}_0(S)$, and the homotopy quotient is equivalent to the classifying space of the mapping class group since the poset is contractible. On the other hand, one can build a model for the homotopy quotient by taking the category $MCG(S) \int \mathcal{A}_0(S)$ in which objects are objects of $\mathcal{A}_0(S)$ and a morphism $x \to y$ is a mapping class group element $\alpha$ such that $\alpha\cdot x \subset y$ (it is a theorem of Thomason that this models the homotopy quotient). By sending a collection of arcs to its dual ribbon graph one sees that this category is equivalent to the category of ribbon graphs of type $S$.<|endoftext|> -TITLE: Which types of Diophantine equations are solvable? -QUESTION [14 upvotes]: Is there a list somewhere of which types of Diophantine equations are solvable, which types are not solvable, and which types are not known to be solvable or not? (When I say solvable, I mean that we can determine in a finite number of steps whether or not there exist solutions.) For instance, we know that linear Diophantine equations are solvable. But we know by Matiyasevich's theorem that there are some Diophantine equations that are not solvable. -I am interested in seeing what the borderline between solvable and unsolvable looks like. - -REPLY [30 votes]: Craig: -For a while, there was some research on improving bounds on the number of variables or degree of unsolvable Diophantine equations. Unfortunately, I never got around to cataloging the known results in any systematic way, so all I can offer is some pointers to relevant references, but I am not sure of what the current records are. -Perhaps the first paper to consider along these lines is - -Y. Matiyasevich, J. Robinson, "Reduction of an arbitrary Diophantine equation to one in 13 unknowns", Acta Arithmetica (27), (1975), 521-553 - -The most significant paper in the area is undoubtedly - -J.P. Jones, "Universal Diophantine equation", Journal of Symbolic Logic (47), (1982), 549-571. - -Let me quote from Matiyasevich's review for MathReviews: - -It has been known since 1970 that every recursively enumerable set $W$ has a representation of the form $$ x\in W\Leftrightarrow\exists w_1\cdots w_\nu P(x,w_1,\cdots,w_\nu)=0, $$ - where $P$ is a polynomial with integer coefficients and $x,w_1,\cdots,w_\nu$ are nonnegative integers. J. Robinson and the reviewer [Acta Arith. 27 (1975), 521--553; MR0387188 (52 #8033)] showed that one can always take $\nu=13$. Later the reviewer claimed [see, for example, Logic, foundations of mathematics and computability theory (London, Ont., 1975), Part 1, 121--127, Reidel, Dordrecht, 1977; MR0485685 (58 #5508)] that it suffices to have $\nu=9$; however he never wrote anything but terse notes with limited circulation. In the paper under review the author first performs the hard work of developing and making public all the cumbersome technical details of the proof. Then he develops what is known as a universal Diophantine equation, namely, an equation $U(x,z,u,y,w_1,\cdots,w_{28})=0$ such that for any recursively enumerable set $W$ a corresponding equation $P=0$ in the representation above can be obtained from $U=0$ just by substituting particular numerical values for the parameters $z$, $u$ and $y$. While the mere existence of universal equations has also been known since 1970, the author here manages to exhibit $U$ almost explicitly in only 7 printed lines! - -Here is a link to Jones's abstract; of note is the following part: - -In these equations there are 28 unknowns $a,b,c,d,e,f,g,h,i,j,k,l, m,n,o,p,q,r,s,t,w,A,B,C,D,E,F,G$ and four parameters $z,u,v,x$. The degree is $5^{60}$. -The degree can be reduced to 4 at the expense of the number of unknowns. The following pairs $(n,d)$ where $n$ is the number of unknowns and $d$ is the degree, are sufficient for all r.e. sets: -$(58, 4),(38, 8),(32, 12),(29, 16),(28, 20),(26, 24),(25, 28),(24, 36),(21, 96)$, $(19, 2668),(14, 2.0\times10^5),(13, 6.6\times10^{43}),(12, 1.3\times10^{44}),(11, 4.6\times10^{44})$, $(10, 8.6\times10^{44}),(9, 1.6\times10^{45})$. -The last pair $(9, 1.6\times10^{45})$ is the 9 unknowns theorem. This theorem, due to Matijasevich, is proved in this paper. - -There are two other, more recent, papers I'm aware of: - -Zhi Wei Sun, "J. P. Jones's work on Hilbert's tenth problem and related topics", Adv. in Math. (China) 22 (1993), no. 4, 312–331. - -I haven't seen this paper, but I believe it is in Chinese. Here is the Abstract: - -This paper is a survey of modern results on Hilbert's tenth problem (especially the work of J. P. Jones). It consists of six sections: 1. Hilbert's tenth problem; 2. The nine unknowns theorem; 3. Universal Diophantine equations; 4. Classification of quantifier prefixes over Diophantine equations; 5. Diophantine representations; 6. Applications of Hilbert's tenth problem. Some new results due to the author, such as the undecidability of $\exists^{11}$ over ${\mathbb Z}$, are also mentioned in the survey. - -Finally, there is the following (quoting from MathReviews): - -Fritz Grunewald, Dan Segal, "On the integer solutions of quadratic equations", J. Reine Angew. Math. 569 (2004), 13–45. -In the paper under review the authors construct an algorithm to determine whether an arbitrary quadratic equation in several variables has solutions in positive integers. In a 1972 paper, C. L. Siegel [Nachr. Akad. Wiss. Göttingen Math.-Phys. Kl. II 1972, 21-46; MR0311578 (47 #140)] constructed an algorithm to determine whether an arbitrary quadratic equation had integer solutions. The transition from integer solutions to positive integer solutions in the general context of Diophantine equations is usually done using Lagrange's theorem concerning representation of integers by the sums of four squares. Unfortunately, replacing all the variables by sums of squares clearly changes the degree of the equation, and therefore this method will be of no use if the degree of the equation is of concern. -To construct an algorithm for positive integer solutions, the authors solve a more general problem. They give a decision procedure to solve a system of quadratic equations in integers subject to finitely many specified congruences and linear inequalities. En route to constructing this algorithm, the authors also produce some interesting number-theoretic results concerning the distribution of integral points on quadric hypersurfaces at infinity. -While of independent number-theoretic interest, this paper is also a valuable contribution to our understanding of the boundary of Diophantine undecidabilty with respect to the degree of the equation. J. P. Jones [Bull. Amer. Math. Soc. (N.S.) 3 (1980), no. 2, 859-862; MR0578379 (81k:10094)] has shown that the degree of the equations with the decidable Hilbert's tenth problem for positive integer solutions has to be less than 4, and the results in this paper indicate that the only remaining question pertains to the degree 3 equations. -Reviewed by Alexandra Shlapentokh. - -Let me close by mentioning that work on the tenth problem is still very active, although it has moved from just the setting of ${\mathbb Z}$ to more general number rings and beyond. A great reference is the recent book by Shlapentokh, "Hilbert's tenth problem. Diophantine classes and extensions to global fields". New Mathematical Monographs, 7. Cambridge University Press, Cambridge, 2007. - -[Edit (April 12, 2017)] The talk by Zhi-Wei Sun mentioned in the comments, "On Hilbert's tenth problem and related topics", a talk given at the City University of Hong Kong on April 14, 2000, is at his page. -Also, Sun has just posted to the arXiv a paper on precisely this topic, Further Results on Hilbert's Tenth Problem. Here is the abstract: - -Hilbert's Tenth Problem (HTP) asked for an algorithm to test whether an arbitrary polynomial Diophantine equation with integer coefficients has solutions over the ring $\mathbb Z$ of the integers. This was finally solved by Matijasevich negatively in 1970. In this paper we obtain some further results on HTP over $\mathbb Z$. We show that there is no algorithm to determine for any $P(z_1,\dots,z_9)\in\mathbb Z[z_1,\dots,z_9]$ whether the equation $P(z_1,\dots,z_9)=0$ has integral solutions with $z_9\ge0$. Consequently, there is no algorithm to test whether an arbitrary polynomial Diophantine equation $P(z_1,\dots,z_{11})=0$ (with integer coefficients) in $11$ unknowns has integral solutions, which provides the best record on the original HTP over $\mathbb Z$. We also show that there is no algorithm to test for any $P(z_1,\dots,z_{17})\in\mathbb Z[z_1,\dots,z_{17}]$ whether $P(z_1^2,\dots,z_{17}^2)=0$ has integral solutions, and that there is a polynomial $Q(z_1,\dots,z_{20})\in\mathbb Z[z_1,\dots,z_{20}]$ such that - $$ \{Q(z_1^2,\dots,z_{20}^2): z_1,\dots,z_{20}\in\mathbb Z\}\cap\{0,1,2,\dots\} $$ - coincides with the set of all primes.<|endoftext|> -TITLE: Representability on the big étale site and base change -QUESTION [5 upvotes]: I am reading M. Artin's treatment of the proper base change theorem for étale cohomology in his "Théorèms de représentabilité pur les espaces algébriques", and I have trouble understanding the following remark on page 222: -If $f:X\rightarrow S$ and $g:S'\rightarrow S$ are morphisms of algebraic spaces (or schemes, if you prefer), and if $f':X'\rightarrow S'$, $g':X'\rightarrow X$ denote the base changes of $f$ and $g$, then one can construct for any abelian sheaf $F$ on the big étale site of $X$ the base change morphism -$g^*R^qf_*F\rightarrow R^q f'_*(g'^*F)$ (the higher direct images also computed on the big sites). If I understand correctly, Artin claims that if $F$, $R^q f_*F$ and $R^qf'_*(g'^*F)$ are representable on the big étale site of $X$, resp. $S$, resp. $S'$ (i.e. locally constructable), then the base change morphism is an isomorphism. -Why is that? Is that an easy fact? - -REPLY [3 votes]: With help from Milne's book on étale cohomology, I figured out how answer the question, although I am not sure that this argument is what Artin had in mind, and I still think that there's an easier argument. -There are morphisms of topoi $\pi_X: X_{ET}\rightarrow X_{et}$ from the topos associated -to the big étale site of $X$ to the topos of the small étale site. Similarly for $S$, and -$f:X\rightarrow S$ induces $f^s:X_{et}\rightarrow S_{et}$ and $f^b:X_{ET}\rightarrow -S_{ET}$, and the obvious diagram commutes, i.e. $f^s\pi_X=\pi_Sf^b$. -Given a sheaf $F$ in $S_{ET}$ we get a base change morphism -$$ \pi_S^*R^qf^s_*F\rightarrow R^qf_*^b\pi_X^*F$$ -Milne calls this "universal base change morphism", for good reasons: Given any morphism $g:S'\rightarrow S$, you also get morphisms of topoi $g^b:S'_{ET}\rightarrow S_{ET}$ and ${g'}^b:X'_{ET}\rightarrow X_{ET}$. Using this to restrict the universal base change morphism to $S'_{ET}$, we get the usual base change morphism for $g$ and $F$. (For this one has to check the commutativity of a few diagrams. All the ingredients can be found, e.g., in great detail in the Stacks Project) -Now, if $F$ is locally constructible, i.e. if the adjunction map $F\rightarrow \pi_X^*\pi_{X,*} F$ is an isomorphism, then it is not hard to check that the universal base change morphism is an isomorphism, and thus every base change morphism is an isorphism.<|endoftext|> -TITLE: Euler characteristic of orbifolds -QUESTION [9 upvotes]: Hello, -Suppose $M$ is a compact oriented smooth manifold and $G$ is a finite group acting on it. Then it is well-known, although I have yet to find a proof or derivation of it, that the (normal topological) Euler characteristic of the orbifold $M/G$ is $\chi(M/G) = \frac1{|G|}\sum_i\sum_{g\in G} (-1)^i\mathrm{Tr}_{H^i(M)}(g^*)$, where $H^i(M)$ is the $i$-th (De Rham) cohomology group and $g^*$ is the map on it induced by $g$. -Everywhere where I have looked, this is then said to equal $\frac1{|G|}\sum_{g\in G}\chi(M^g)$ (where $M^g$ is the set of points that $g$ leaves fixed) because of the Lefschetz formula. Not being familiar with this formula, I've looked up several versions of it. Especially the one for compact oriented manifolds seems very useful, but it (along with a number of other versions) holds only when all the fixed points are isolated. In particular, the set of fixed points should be countable. However, I have seen this formula being used used in situations in which this does not hold. For example, let the permuation group $S_n$ act on $M\times\dots\times M = M^n$ by permuting the factors. The set of fixed points of this action is certainly not countable. -So how does this work? Is this Lefschetz formula applicable to this situation after all, or is there another usable version of it that should be used here? Also, is there perhaps a book or arXiv document that shows how to calculate the first expression of the Euler characteristic? - -REPLY [21 votes]: As far as I understand your question, you want to see a derivation of the formula for $\chi(M/G)$. Here it is: - -The difficult part of the argument os to show that there is an isomorphism $H^* (M/G; \mathbb{Q}) \cong H^* (M; \mathbb{Q})^G$ (the $G$-invariants). It is induced from the quotient map $M \to M/G$, but that is not so important. In the following, all cohomology have rational coefficients. -If that is done, the argument is easy. By elementary representation theory of finite groups: - -$$ -dim (H^i (M)^G) = \frac{1}{|G|} \sum_{g \in G} Tr_{H^i (M)} (g). -$$ - -We show the difficult part in two steps. Consider the Borel construction $EG \times_G M$. There is a fibre bundle $\pi:EG \times_G M \to BG$ with fibre $M$ and a map $f: EG \times_G M \to M/G$. -The Leray-Serre spectral sequence of $\pi$ begins with $E_{2}^{pq}=H^p (G;H^q (M))$. If $p > 0$, this group is zero and hence $H^n(EG \times_G M) \cong H^0 (G, H^n(M))=H^n (M)^G$. -$f$ induces an isomorphism in cohomology: -Let $x \in M$ and let $H$ be the stabilizer subgroup at $x$. Pick a $G$-equivariant Riemann metric on $M$. Consider $V=\bigcup_{g \in G} B_{\epsilon}(x)$. If $\epsilon$ is small enough, then $V$ is a disc bundle a $G$-equivariant vector bundle on $G/H$. Clearly, $EG \times_G V \simeq EG \times_G G/H \simeq BH$. Moreover, $V/G\simeq \mathbb{R}/H$ is contractible. -The consequence of this discussion is that there exists a finite cover of $M/G$ by open contractible sets, such that all intersections are again contractible or empty. Also, the preimages of the covering sets and their intersections have trivial rational cohomology, because the cohomology of $BH$ is trivial. By the Mayer-Vietoris sequence, induction on the number of covering sets and repeated application of the 5-lemma, it follows that $f^* :H^* (M/G) \to H^* (EG \times_G M)$ is an isomorphism in rational cohomology. - -REPLY [7 votes]: Just to add to Johannes answer, when you have a map $f:M\to M$ and you want to compute $\sum_i (-1)^i \textrm{Tr}_{H_i(M)}(f_\ast)$, one may proceed as follows : imagine $i$-cycles in $M$, i.e., elements in $H_i(M)$, arising as either $i$-cycles in $M^f$ which are fixed by $f_\ast$ or they are not fixed by $f_\ast$. One can do this rigorously by choosing an appropriate triangulation of $M$, homotoping $f$ to be a simplicial map and choosing a basis for $H_i(M)$ for all $i$ simultaneously. It then follows that in this basis, the trace picks up $1$ for each element in the basis of $H^i(M^f)$, which when summed up with alternating signs produces $\chi(M^f)$. The classical Lefschetz fixed point formula can be proven in exactly the same way.<|endoftext|> -TITLE: Is there a smooth free circle action on the Klein bottle? -QUESTION [6 upvotes]: Can the circle group $S^1$ act smoothly and freely on the Klein bottle? I'm sure there is some obvious reason why the answer is no, which eludes me right now. -We can view $K$ as the quotient of $S^1\times S^1\subset\mathbb{C}\times\mathbb{C}$ by the involution $(z_1,z_2)\to (-z_1,z_2^{-1})$. Then we get an almost-free $S^1$-action $(z,[z_1,z_2])\to [zz_1,z_2]$ with $\mathbb{Z}_2$ isotropy ($-1$ fixes the circles $[z_1,1]$ and $[z_1,-1]$). - -REPLY [11 votes]: No. If you had a smooth free action, the quotient would be a compact connected 1-manifold, so a circle. So the Klein bottle would be an orientable circle bundle over the circle, but there's only one and that is a torus. -So the tools I'm using are (1) when the quotient of a manifold by a free action of a compact Lie group is another manifold, (2) classification of 1-manifolds, (3) classification of circle bundles.<|endoftext|> -TITLE: Is being finitely generated a local property? -QUESTION [11 upvotes]: I am trying ot figure out a proof of the following fact, that I believe is true, but it seems to me that something is lacking. -Suppose we have commutative, unitary rings $A,B$ and a (unit preserving) homomorphism $\varphi \colon A \rightarrow B$ which makes $B$ an $A$-algebra. Suppose also we have elements $f_1,\dots,f_n \in B$ which generate the unit ideal and such that $B_{f_i}$, namely the localisation of $B$ with respect to $f_i$, is a finitely generated $A$-algebra. Show that then $B$ is a finitely generated $A$-algebra. -Could someone give me a rigorous proof of this fact (or a counterexample, if this is false)? -Thank you! - -REPLY [3 votes]: To supplement these expert answers, I want to give one that at least to me shows how to use geometry to think of the algebraic proof. Not being an algebraic thinker this sort of thing is crucial for me, and indeed by thinking in these terms I was able to do this initially puzzling, although elementary, computation in my head. The key idea is that of a partition of unity, that allows one to decompose a global element into a sum of local elements. -I apologize for revisiting this old and elementary question, but it is basic and I think it is worth trying again, as Georges did, to expose the geometric idea behind the algebra. -To prove: if R is an S algebra and spec(R) has an open covering by basic open sets Ufj j=1,...m, such that each affine ring Rfj is finitely generated as S algebra, where the functions fj generate the unit ideal in R, then R is also finitely generated as S algebra. -proof: To take advantage of the hypothesis, we want to restrict an element h of R, thought of as a function on spec(R), to each open set Ufj, then write each restriction in terms of a finite number of elements of R, and add back up to get h itself in terms of a finite number of elements of R. But since the sets Ufj overlap, the naive restrictions will not add back to h. Thus we want to use the more subtle restriction obtained from a “partition of unity”. I.e. if we write the constant 1 as a sum of functions each supported in the Uj, then by multiplying h by these functions, we can write h also as such a sum. -Since the fj generate the unit ideal we have functions gj such that f1g1+...+fmgm = 1. Thus given h = h.(f1g1)+...+h.(fmgm), it suffices to write each “restriction” h.(fjgj) as an S linear combination of monomials in terms of a finite number of elements of R, which are independent of h. -If we restrict h to Uj in the naive sense, we have by hypothesis that h is an S linear combination of monomials involving a fixed finite number of elements of form r/fj^s, with r in R; hence for some t, h.fj^(s+t) is an S linear combination of monomials in a fixed finite set of elements of R, if we add fj to the finite set of elements r. -Now using the fact that 1^N = 1, for any N, we can form our partition of unity using any power of the fj. Thus for any n, we can write h = h.(f1^n.G1)+...+h.(fm^n.Gm) where each h.(fj^n) is an S linear combination of monomials in a given finite number of elements of R, hence (after including the Gj among the finite set of elements) also h.(fj^n.Gj), hence also h.<|endoftext|> -TITLE: Simple description of a Chow ring of blow-ups. -QUESTION [7 upvotes]: Is there a simple description of a Chow ring of a blow-up of a point on a smooth projective variety? Or at least of successive blow-ups of $\mathbb{P}^n$? - Maybe something like $A(\tilde{X})=f^*(A(X))\oplus\mathbb{Z}(E)$, where $f\colon\tilde{X}\to{}X$ is a blow-up, E is an exceptional divisor, with multiplication given by $E\cdot{}E_k=-E_{k-1}$, $E_0=f^*(P)$, where $E_k{}$ is a k-dimensional linear subspace of an exceptional divisor $E(=E_{n-1})$, and $P$ is a point we are blowing up. What I'm suggesting is true for surfaces (exercise 6.5 in appendix A of Hartshorne), and seems geometrically plausible in the case $X=\mathbb{P}^n$. - Also, it'd be great to know what cycles are effective. - I'm afraid all this is really trivial for someone understanding Fulton's book, but I'm not at that level yet. - -REPLY [7 votes]: The general formula about the intersection ring of blow-ups is discussed in Fulton's book. In your case you want to study the intersection ring of a smooth algebraic variety $V$ blown up at a point $Z$. There is a simple formula for this situation by Keel. You can find it in his paper: Intersection Theory of Moduli Space of Stable N-Pointed Curves of Genus Zero. -Another nice reference is the paper "A compactification of configuration spaces" by Fulton-MacPherson. In section 5 of this paper they mention the Keel's formula and state the facts needed in the computation of the Chow ring. I summarize it below. -The key fact is that the restriction map from the Chow ring of the variety $V$ to the Chow ring of the point $Z$ is surjective. The intersection ring of the blow-up $\widetilde{V}$ is generated over $A(V)$ by the class of the exceptional divisor $E$ with the ideal $I$ of relations described bellow: -1) Let $J_{Z/V}$ be the kernel of the restriction map from $A(V)$ to $A(Z)$. -It contains all elements in $A(V)$ of positive degree, for example. -2) Assume that you can write $Z$ as a transversal intersection $\cap_{i=1}^r D_i$ of the divisor classes $D_i$. Define the polynomial $P_{Z/Y} \in A(V)[t]$ by the rule $P(t)=\prod_{i=1}^r(t+D_i)$. This polynomial is called a Chern polynomial of $Z$. It -depends on the choice of the divisor classed $D_i$. It means that it is not unique and is determined up-to an element in $J_{Z/V}$. -The ideal $I$ is generated by $J_{Z/V}\cdot E$ and $P_{Z/V}(-E)$. The Chow ring of $\widetilde{V}$ is therefore equal to $\frac{A(V)[E]}{I}$.<|endoftext|> -TITLE: Is there a Poincare-Hopf Index theorem for non compact manifolds? -QUESTION [16 upvotes]: Does Poincare-Hopf index theorem generalizes in any way to non compact manifolds ? In particular, I am interested in the case of a smooth vector field on a cylinder $\mathbb{T}_1\times\mathbb{R}$? If so, are there some additional assumption that one has to impose on a vector field considered (maybe it should vanish outside some compact set or decay very fast "at infinity"?). Sorry if the question is silly - I know the Hodge index theorem only from very elementary sources (Arnold's book on ODEs and Wikipedia). -Motivation (physical digression): A friend of mine tries to model action of a cardiac tissue in a heart and its soundings. One of his aims is to understand phenomena called "spiral waves" (they are believed to be partially responsible for hearth attacks). I don't know the details but those "spiral waves" can be described by some ODE defined on a domain which is closely related to the real geometry of considered tissue. From the information about indexes of singular points of a corresponding vector field it is possible to deduce some qualitative information about occurrence of this phenomenon. - -REPLY [13 votes]: Suppose $M$ has empty boundary. Let $U\subset M$ be an open set with compact closure whose topological boundary contains no zero of the continuous vector field $X$ on $M$. Suppose $X$ is smooth and hence generates a local semiflow $f_t$, $t \geq 0$. -For sufficiently small $t>0$ the map $f_t$$\colon U \to M$ is defined and has a "fixed point index" $I(f_t, U)$ (see A. Dold, ``Lectures on Algebraic Topology,'' Die - Grundlehren der matematischen Wissenschaften Bd. 52. Springer, New - York (1972)). - It can be shown that the integer $i(X,U):=I(f_t, U)$ is independent of $t$ and $U$, and is stable under perturbation of $X$. -If $X$ is not smooth, approximate it by a sequence of smooth fields $X_j$ and define $I(X,U):= \lim_{j\to \infty} I(X_j,U)$. -If $X$ has only finitely many zeros in $U$ and none on $U\cap \partial $, then $I(X, U)$ is the sum of their Poincare-Hopf indices. -If $M$ has nonempty boundary, this work if at every boundary point $p$ there is an integral curve $u\colon [0,\epsilon)\to M$ with initial point $u(0)=p$.<|endoftext|> -TITLE: Examples of non-metrizable spaces -QUESTION [34 upvotes]: I want to know some examples of topological spaces which are not metrizable. Of course one can construct a lot of such spaces but what I am looking for really is spaces which are important in other areas of mathematics like analysis or algebra. I know most spaces arising naturally in other areas of mathematics are metrizable because of the Urysohn metrization theorem. But still there must be some examples of non-metrizable spaces.So far I know the following examples: - -Zariski topology -Weak* topology on $X^{*}$ if X is an infinite dimensional Banach space -The topological vector space of all functions $f:\mathbb{R}\rightarrow\mathbb{R}\ \ $ under pointwise convergence. - -Your help is appreciated. - -REPLY [6 votes]: Sorgenfrey line is not metrizable, since it is separable but does not satisfy the second axiom of countability.<|endoftext|> -TITLE: Old question of Serre on discriminants of a sequence of polynomials -QUESTION [8 upvotes]: Let $P_n(t)$ be polynomials with integer coefficients with $d_n = \deg(P_n(t))$ going to infinity when $n$ goes to infinity -and with nonzero discriminants $disc(P_n(t)) \neq 0$. -Question: Is -$$ -\lbrace disc({P_n(t)})\rbrace ^{\frac{1}{d_n}} -$$ -bounded when $n$ goes to infinity ? - -REPLY [4 votes]: To reiterate Serre's question (Open Problem 2.5 in Odlyzko's survey): Must there be only finitely many polynomials having root discriminant below a given bound? -With this answer I just want to note that the much stronger statement formulated in the last paragraph of user631's answer, which concerned the existence of a $c > 1$ such that all irreducible polynomials of discriminant $\leq (cd)^d$ are essentially cyclotomic, is false without additional assumptions. Since we may take $c = M(f)^2$ as Mahler proved, this is a natural statement to consider: its truth would have strengthened Lehmer's conjecture. -A counterexample is given by $x^d-x-1$, whose discriminant has absolute value $d^d + (-1)^{d}(d-1)^{d-1}$. We may take more generally any sequence of irreducible trinomials with $\pm 1$ coefficients and degree going to infinity. This follows by the explicit calculation of the discriminant of the general trinomial; the statement and simple derivation of the formula is given as Theorem 2 in Swan's 1962 paper Factorization of polynomials over finite fields in the Pacific Journal of Mathematics. (Another reference is Prasolov's book Polynomials, which reproduces the same calculation). -A counterexample of a rather different kind (in particular, having unbounded Mahler measure under all integer translations) is provided by the minimum polynomial of the generator $\zeta_n+\zeta_n^{-1} = 2\cos(2\pi/n)$ of the integer ring of the maximal totally real subfield of $\mathbb{Q}(\zeta_n)$. Further counterexamples would include the minimum polynomials of the generators over $\mathbb{Z}$ of the integer rings of other bounded index monogenic subfields, if such exist, of either $\mathbb{Q}(\zeta_n)$ or the splitting fields of the $\pm 1$ trinomials considered in the previous paragraph. -However, one may wish to restrict to reciprocal polynomials; for Lehmer's problem this would be sufficient. For these the statement seems very interesting, and could well be true; by the formula in Swan's article, it certainly holds for all trinomials. The above examples are of no use here: the only reciprocal $\pm 1$ trinomials are $x^{2n} \pm x^n +1$, and those have only cyclotomic factors.<|endoftext|> -TITLE: Does "finitely presented" mean "always finitely presented", considered in general -QUESTION [9 upvotes]: I'm wondering about the question - -"If we have a finitely presented __, is it necessarily finitely presented with respect to any finite generating set for it?" - -I know this is true for groups and for $R$-modules. Does anyone know whether this is true for $A$-algebras? Commutative $A$-algebras? Other things people might happen to know about it for? - -REPLY [9 votes]: Yes in general. -See Adámek and Rosicky, Locally Presentable and Accessible Categories Cambridge University Press, Cambridge, (1994). -T. 3.12 p. 143. -Of course "in general" I mean: every "algebraic theory" (many sorted) Set models. -For topological algebraic structures, (like profinite groups) this equivalence isn't true for the Yiftach Barnea example<|endoftext|> -TITLE: Weak form of the $abc$-conjecture? -QUESTION [6 upvotes]: The $abc$-conjecture implies that the equation $a+b=c$ has only finitely many primitive solutions in the multiplicative semigroup generated by any particular finite set of primes. -I would appreciate any information about the status of this a priori weaker statement, -and citations to the literature if any exist. - -REPLY [10 votes]: The fact that the S-unit equation has finitely many solutions is due to Siegel and Mahler. The statement has been generalized quite a bit (for example to the fact that $u_1+\cdots +u_n=1$ has finitely many solutions, which uses some generalization of Schmidt's subspace theorem). You can find a proof and proper references in "Diophantine Geometry: An introduction" by M. Hindri, J.H. Silverman, among other places.<|endoftext|> -TITLE: Supersingular elliptic curve dilemma -QUESTION [8 upvotes]: Let $E$ be a supersingular elliptic curve over a finite field of -characteristic $p$, and $\mathbb{F}_q\supset \mathbb{F}_{p^2}$ be a finite field large enough such -that all (absolute) endomorphisms of $E$ is defined over $\mathbb{F}_q$. -We write $G$ for the absolute Galois group of $\mathbb{F}_q$. It is well known that the Frobenius automorphism $\varphi$ is a (topological) generator of $G$. Let us fix a prime $\ell\neq -p$. By the Tate Conjecture, -$$\mathrm{End}(E)\otimes\mathbb{Q}_\ell=\mathrm{End}_{\mathbb{Q}_\ell}(V_\ell(E))^G=\mathrm{End}_{\mathbb{Q}_\ell}(V_\ell(E))^\varphi. $$ -Since $E$ is supersingular, $\mathrm{End}(E)$ is an order in a quaternion -algebra. In particular, $\mathrm{rank}_\mathbb{Z}(\mathrm{End}(E))=4$. So -$$ \mathrm{End}(E)\otimes\mathbb{Q}_\ell=\mathrm{End}_{\mathbb{Q}_\ell}(V_\ell(E))$$ -It follows that $\varphi$ is in the center of -$\mathrm{End}_{\mathbb{Q}_\ell}(V_\ell(E))$. In other words, $\varphi$ is a -scalar. This clearly leads to a contradiction (say, with the Riemann -Hypothesis). Where did the argument go wrong? - -REPLY [10 votes]: There is nothing wrong. A typical case will be -$\mathbb F_q = \mathbb F_{p^2}$ and $\varphi = -p$. -Since $q^{1/2} = (p^2)^{1/2} = p$, this is consistent with RH.<|endoftext|> -TITLE: Inconsistent theory with long contradiction -QUESTION [40 upvotes]: What can one say about an inconsistent theory $T$ which has no contradictions (i.e. deductions of $P \wedge \neg P$) of length shorter than $n$, where $n$ is some huge number? -There have been some discussions about the consistency of ZFC, for instance, where it has been asserted that it would be OK for ZFC to be inconsistent as long the contradiction was enormous. However, this still seems like a bad situation to me since there are constructions which depend on consistency but don't care about deduction per se. For example, constructing a model of a theory. -Can someone explain the consequences of this? -EDIT: After hearing some good feedback, I think I can phrase this question in a more concrete way: -To what extent, and in what situations, is it possible to work consistently with a theory that is inconsistent? - -REPLY [3 votes]: Let me expound on a somewhat plausible scenario of an inconsistency in the large cardinal hierarchy that may take a very long time to appear. -A rank-into-rank embedding is an elementary embedding $j:V_{\lambda}\rightarrow V_{\lambda}$. -Lemma If $j$ is a rank-into-rank embedding, then $(j*j)(\alpha)\leq j(\alpha)$ for all ordinals $\alpha$. -Proof Suppose that $\alpha<\lambda$. Let $\beta$ be the least ordinal such that $j(\beta)>\alpha$. Then -$$V_{\lambda}\models\forall x<\beta,j(x)\leq\alpha.$$ -Therefore, by elementarity, -$$V_{\lambda}\models\forall xo_{n}(2)$, then the least natural number $n$ such that $o_{n}(1)>o_{n}(2)$ is greater than $f_{9}^{\text{Ack}}(f_{8}^{\text{Ack}}(f_{8}^{\text{Ack}}(254)))$. -Therefore if one tries to prove that rank-into-rank cardinals are inconsistent by exhibiting an $n$ such that $o_{n}(1)>o_{n}(2)$, then one would need to take more than $f_{9}^{\text{Ack}}(f_{8}^{\text{Ack}}(f_{8}^{\text{Ack}}(254)))$ steps. -Of course, there may be short-cuts in showing that $o_{n}(1)>o_{n}(2)$ for some $n$ which take much less time than actually calculating the least $n$ such that $o_{n}(1)=5$. Or there could be an entirely different sort of contradiction with the assertion that there exists a rank-into-rank cardinal. -For the record, here are the first few values of $o_{n}(1)$ and $o_{n}(2)$. -$o_{1}(1)=0,o_{1}(2)=1;o_{2}(1)=1,o_{2}(2)=1;o_{3}(1)=2,o_{3}(2)=2;o_{4}(1)=2,o_{4}(2)=2;o_{5}(1)=3,o_{5}(2)=3;o_{6}(1)=3,o_{6}(2)=3;o_{7}(1)=3,o_{7}(2)=4;o_{8}(1)=3,o_{8}(2)=4;o_{9}(1)=4,o_{9}(2)=4$. -If such an inconsistency were to pop up only very far away, then such an inconsistency will not have any effect on the mathematics that mathematicians here on Earth will do because nobody will live long enough to observe such a contradiction. -This being said, set theorists generally do not think that there is any contradiction anywhere in the large cardinal hierarchy up until say $I0$ no matter how distant the contradiction is located.<|endoftext|> -TITLE: Solution of Plateau Problem for a simple, smooth closed curve on a Riemannian Manifold (Kahler) gives a surface that can be parametrized by a closed disk? -QUESTION [5 upvotes]: Hi, -Perhaps it's a stupid question, in that case i'll delete it. -Let M be a compact orientable smooth (Kahler if changes things) manifold of dimension $dim_{\mathbb{R}}(M)=2n$ with $n\geq1$, let $\gamma$ be a simple smooth closed curve that lies in a (holomorphic) coordinate chart and that can be taken as small as necessary (one can choose $\gamma$ such $diam(\gamma)<\epsilon$ with $\epsilon>0$). If i want to solve Plateau -problem for a $\gamma$ so small such that the solution is contained in the coordinate chart, -do i get something that can be parametrized by closed disc? -Thank you in advance. -Edit: -I'll try to clarify my question, this is what i wanted to know. Let $U$ be a sufficiently small geodesically convex set of a manifold $M$ and $\gamma$ a smooth simple closed curve lying in $U$ (no other assumptions on $\gamma$). -1) Can $\gamma$ be the boundary of an embedded closed disk? -2) If $\gamma$ can be the boundary of a closed disk, then can it be the boundary of a minimal (as a surface, not only among the disks that it bounds) embedded closed disk? -I anticipate that i couldn't see works of Douglas so i don't know if the answer to my question is there. -My suspect was that the answer could be yes for dimension 2 (i think about jordan curve theorem), Professor Thusrston example of the knotted curve suggests me that in dimension 3 i need additional assumptions on the curve not only on the linking number. But what happens for dimension $n\geq4$? - -REPLY [12 votes]: I've been waiting for someone with more expertise than me to answer, but since they haven't (so far) I'll say something. -There are different versions of Plateau-like problems; I'm not sure if there's a specific single one that's generally accepted as "the Plataeu problem". One can ask for a mapped-in disk with minimal area having the given boundary, a mapped-in surface of minimal area, a current of minimal area, an integral current of minimal area, an embedded "minimal surface" meaning that it's just a critical point for area among embedded (or mapped in if you prefer) surfaces, a minimal disk ... A lot is known about these different questions, and the answers aren't the same. -First: even for a curve in Euclidean space, there might not be an embedded minimal disk. -The easiest examples are for a knotted curve in R^3. However, there are also unknotted -curves in R^3 that do not bound a disk in their convex hull, so they do not bound any embedded minimal disk. The minimum area of a disk bounding an unknotted curve grows exponentially in an appropriate measure of the complexity of the curve; the genus of a minimum area surface also grows exponentially. (Fred Almgren and I once wrote a paper about this). The same examples can be transported to any higher dimension, e.g. taking -the product with another manifold. -Second: Jesse Douglas showed how to find mapped-in minimal disks in great generality. -This will work locally within any manifold of dimension. The basic technique is that for any parametrization of the curve by the boundary of a disk, first find an energy-minimizing map of the disk that extends this parametrization --- a harmonic map. Now consider the energy as a function of the parametrization (which is a kind of Teichmuller space). The critical points of the harmonic energy with respect to the Teichmuller space -are minimal surfaces: the basic insight is that critical points of the harmonic maps within the Teichmuller space are when the harmonic map is conformal. When it is not conformal, you can change the conformal structure on the disk (which is equivalent to giving a reparametrization, by the uniformization theorem) and reduce energy. -Third: minimal surfaces of any type (inclding currents) whose boundary is inside a convex set are always contained in the convex set. Here convex means that geodesic arcs between nearby points on the boundary -are contained within the set. Even weaker conditions are sufficient, but this is good enough for your purposes --- a metric ball of small radius in any Riemannian manifold is convex in this sense. -Fourth: there always exist minimizing objects of some sort, but they might not be things that you are happy with: one thing that happens is that $k$ times a curve might -bound a surface of area less than $k$ times that for the original curve (in dimensions > 3). A limit -of $1/k$ times the minimum area for $k$ times the curve might be a diffuse current, -spread out in an entire region. I think this can happen even locally in a Kahler manifold, but I'm not sure. Even without using fractional weights, the minimizing object -would in general be an integral current that is like a surface but with singularities. -I don't know the classification for the 2-d case, but i'm sure the experts do. -On the other hand, if you take a nearly round circle in a small coordinate chart, there should be an embedded minimal disk --- basically because nearly flat minimal surfaces -are stable, so changing the metric a little bit gives you a new minimal surface by implicit function theorem type arguments (in the space of surfaces, which in this case can be described as graphs of functions). In R^3, there's a theorem that any curve whose -total curvature is less than 4$\pi$ bounds an embedded minimal disk; I suspect there are known estimates likke this in Riemannian manifolds of higher dimension, using curvature -assumptions about some convex set contining the curve, but I don't actually know. -Is there an expert who can correct or extend what I've said?<|endoftext|> -TITLE: cohomology theory for algebraic groups -QUESTION [16 upvotes]: Is there a cohomology theory for algebraic groups which captures the variety structure and restricts to the ordinary group cohomology under certain conditions. - -REPLY [4 votes]: This post came up on my RSS feed, and I didn't see that it had been answered in 2011, but since I've now written the following, you may as well have it ... -One other point, which I think is relevant to your question. If you take $G$ over $k=\bar k$, char $k=p$ with a Frobenius map $F$ whose fixed points $G^F$ is a split Chevalley group $G^F=G(q)$ with $q=p^r$, then CPSvdK in their 1977 Inventiones paper explain a strong connection between the rational cohomology $H^i(G,M)$ and $H^i(G(q),M)$ for $M$ a finite-dimensional rational $G$-module. As other posters have pointed out, there is divergence between the two, even for $q$ sufficiently large. One has to perform sufficiently many Frobenius twists to $M$ to make the two sides agree. The first example is for $G=SL_2$ and $k$ has characteristic $2$. Then if the natural module is $V=L(1)$, this has $H^1(SL_2,V)=0$ in the rational setting but for all $r>1$, $H^1(SL_2(2^r),V)$ is $1$-dimensional. The 'generic cohomology' is achieved as the stable limit $H^i(G(q),M)$ for $q=p^r$ as $r\to \infty$. So $H_\mathrm{gen}^1(SL_2,V)\cong k$. -What's the difference between $H^i$ and $H^i_\mathrm{gen}$? Well, with a little computation you can see that in order to get the non-trivial cocycles of $H^1_\mathrm{gen}(SL_2,V)$ in $H^1(SL_2,V)$ you would have to be able to take square roots. Well, of course, you can't do this in the category of algebraic morphisms. -But there is also another cohomology coming from forgetting all variety structure and allowing maps to be non-morphisms. So we're just looking at $G$ as a(n infinite) group. We could denote this $H^i_{\mathrm{abs}}(G,M)$ (where the 'abs' is for abstract). In the category of abstract $G$-modules, of course you can take any set theoretic map you like, so square roots are fine. And in fact one has an isomorphism $H^i(G_\mathrm{abs},M)\cong H^i(G_\mathrm{gen},M)$. I am led to believe that there is an article by Brian Parshall, [EDIT: with thanks to Wilberd van der Kallen] 'Cohomology of algebraic groups' in [The Arcata conference on Representations of Finite Groups, Proc Symp Pure Math 47 Part 1] which gives a proof of this for finite-dimensional rational modules $M$ using an argument he attributes to van der Kallen.<|endoftext|> -TITLE: Longest coinciding pair of integer sequences known -QUESTION [21 upvotes]: There are arbitrarily many pairs of integer sequences (of arbitrary origins) that coincide upto an $N$ but differ for an $n > N$. I assume, the coincidence will be considered accidentally then by default, but I may be mistaken about that. -One is disadviced to draw any conclusions from coincidences of integer sequences unless its proven, that they coincide for all $n$. (Even then there may be no sensible conclusions, as I have learned here: Equivalence of families of objects with the same counting function.) -In any case, it is hard not to be entrapped to draw a conclusion when $N$ is very large. But what is "very large"? Thus my question: - -What is the largest $N$ with two known - integer sequences coinciding upto $N$ - but differing for an $n > N$? - -(Can this information be captured from OEIS by an intelligent query?) -(I am aware of the fact that one can trivially define pairs of integer sequences which conincide for all $n$ but a single and arbitrarily large one. It should be clear that I am not interested in those but in pairs that are not adjusted to each other this way.) - -REPLY [9 votes]: A possibly unbeatable example: the sequence $a(n)$, where $a(n)$ is the periodicity of the first row in the Laver table based on the set $\left\lbrace 1,\ldots, 2^n\right\rbrace$: -The first values are $$1, 1, 2, 4, 4, 8, 8, 8, 8, 16, 16,\ldots$$ -and then it stays at $16$ for ages: -Randall Dougherty showed that the first $n$ for which $a(n)$ can possibly differ from $16$ is A(9,A(8,A(8,255))), where A denotes the Ackermann function, a recursive function whose first values are already huge. -However, it can be shown that the sequence actually tends to infinity, under some additional axiom that is independent to the usual ones from ZFC.<|endoftext|> -TITLE: Are the axioms for higher category-theory effectively computable? -QUESTION [14 upvotes]: I ask this, although I don't conduct any research in the area, or even plan to. -- There seems to be general agreement that the axioms for higher categories grow very rapidly in complexity as the order $n$ of the category grows to infinity. I am referring to the ''algebraic style'' axioms here, axiomatising a ''maximally weak'' notion of $n$-category. I assume this makes enough sense to make the following question meaningful: - -is it known (or guessed at) whether there is a Turing-machine that computes these axioms (taking the order $n$ of the category as input)? - -REPLY [5 votes]: Yes. -…at least, for Leinster’s reformulation of Batanin’s definition of globular operadic weak ω-category (and hence also for the finite-dimensional versions of this). Showing this is essentially a matter of repeatedly applying one lemma: if $\mathbf{T}$ is an essentially algebraic theory with a computable presentation, then the free $\mathbf{T}$-structure on a computably presented object is again computably presented. By “computably presented”, I mean essentially that the sets of operations and axioms are all computably enumerable. -In the Leinster/Batanin definition, one starts with strict $\omega$-categories (certainly a computably presentable theory, by the standard explicit axiomatisaion); by their observation above, their monad $T$ is computably presentable; from this, one can show that the theory of $T$-operads is computably presentable; similarly, then, the theory of $T$-operads-with-contraction; so the free $T$-operad-with-contraction $L$ is computably presentable. -But now the operations of the theory of weak $\omega$-categories are the elements of $L$; and the axioms are given by elements of “powers” of $L$, in the monoidal structure $\otimes$ built by $T$ and pullbacks; so these sets are all computably enumerable, so we’re done. - -From here on I’m a little beyond my comfort zone, and wouldn’t want to swear that the details hold up: someone who knows realizability toposes better than I do can probably tell better whether I’ve missed some subtlety. -A nice way to look at the above argument could be to say: develop the theory of weak $\omega$-categories in $\newcommand{\Eff}{\mathcal{E}\textit{ff}} \Eff$, the effective topos — that is, repeat all the normal definitions in the internal logic of $\Eff$, to get an internal theory $\mathbf{T}^{\Eff}_\omega$. (Possibly $\mathbf{PER}$ or some other category of ‘computably presented sets and functions’ might work better than $\Eff$.) -Now, the global sections functor $\Gamma \colon \Eff \to \mathbf{Sets}$ is a left exact left adjoint, so in particular, it will commute with pullbacks and with most ‘free object’ constructions — so, with all the ingredients used in the definition of the theory of weak $\omega$-categories. So when we hit $\mathbf{T}^{\Eff}\omega$ with $\Gamma$, we just recover the original external theory $\mathbf{T}\omega$. That is, $\mathbf{T}^{\Eff}\omega$ is a computable presentation of $\mathbf{T}_\omega$ -Intuitively, we’re ‘shadowing’ every construction we do in $\mathbf{Sets}$ with a computable presentation, by performing the same constructions in parallel up in $\Eff$. -This approach should also work for most other theories of higher categorical structures — power-sets and non-finite exponentiation are the main logical constructions not preserved by $\Gamma$, and off the top of my head, only the definitions of higher categories which involve topological constructions will require these.<|endoftext|> -TITLE: The concept "conjugate class" in monoids. -QUESTION [11 upvotes]: Is there any concept in monoids that is similar to the concept "conjugate class" in groups? For example, are there any such similar concept in symmetric inverse monoids? Thank you very much. - -REPLY [3 votes]: This seems to be a relevant paper: Kudryavtseva and Mazorchuk - On three approaches to conjugacy in semigroups.<|endoftext|> -TITLE: How to get explicit unramified covers of an elliptic curve? -QUESTION [11 upvotes]: Say we begin with an explicit elliptic curve over $\mathbb{C}$, say: $y^2=x(x-1)(x-2)$. According to abstract reasoning this elliptic curve has an (several, in fact) unramified cover with group $C_n$. My question is, how can one find the equations that define this cover? To be very precise: how can one describe the function field extension of $Quot(\mathbb{C}[x,y]/(y^2-x(x-1)(x-2)))$ as the quotient field of a polynomial ring modulo an explicit ideal? -It would be nice to have a method that can generate all such unramified $C_n$ covers, but a method that generates some unramified $C_n$ cover would be helpful too. - -REPLY [6 votes]: Let me expand Felipe's answer a bit. -Vélu's formulae given in the very readable short paper [1] are very easy to use. For instance if you are given a $n$-torsion point one can immediately write down the Weierstrass equation for the quotient by this point. -To get such a point, one can use the division polynomials which can be computed recursively very fast. The zeroes of this polynomial are exactly the $x$-coordinates of the $n$-torsion points. Oover the complex number it is not difficult to get them also using the parametrisation by the Weierstrass $\wp$-function. -There are quite a few paper improving Vélu's formula. Most importantly, there is Kohel's thesis [2]. Or for instance [3]. These were used in the implementation for isogenies in sage [sage] and in [magma]. -[1] MR0294345, Jacques Vélu, Isogénies entre courbes elliptiques. C. R. Acad. Sci. Paris Sér. A-B 273, 1971. -[2] MR2695524, David Kohel, Endomorphism rings of elliptic curves over finite fields. Thesis (Ph.D.), University of California, Berkeley. 1996. -[3] MR2398793, Bostan, A.; Morain, F.; Salvy, B.; Schost, É. Fast algorithms for computing isogenies between elliptic curves. Math. Comp. 77 (2008). -[magma] http://magma.maths.usyd.edu.au/magma/handbook/text/1321 -[sage] http://www.sagemath.org/doc/reference/sage/schemes/elliptic_curves/ell_curve_isogeny.html<|endoftext|> -TITLE: Each element of fundamental group of a topological group represented by homomorphism? -QUESTION [13 upvotes]: This may be a fairly simple question. Suppose G is a (T0) topological group. Assume that G is path-connected, locally path-connected, and semilocally simply connected, so that covering space theory applies. -Question: Is it true that for any element of $\pi_1(G,e)$ (where e is the identity element of G), there exists a [ADDED: continuous] homomorphism from $S^1$ to $G$ having that element of $\pi_1(G,e)$ as its homotopy class? -Another way of formulating this is that there is a set map: -$$\operatorname{Hom}_{cts}(S^1,G) \to \pi_1(G,e)$$ -The subscript cts is to indicate continuous. -(when G is abelian, the left side has a group structure too [ADDED: under pointwise multiplication], and the Eckmann-Hilton principle tells us that we get a group homomorphism). - -Is the set map surjective in all cases (regardless of whether G is abelian)? -Does the image of $\operatorname{Hom}(S^1,G)$ generate $\pi_1(G,e)$ as a group (this is equivalent to surjectivity when $G$ is abelian)? -Does surjectivity work for Lie groups? Compact Lie groups? -Does the weaker formulation (2) work for Lie groups? - -I have a sketch of an argument/proof that may show (4) (basically, using properties of one-parameter subgroups), but I'm hoping somebody will have a clean proof that works in general for topological groups. - -REPLY [11 votes]: Here's an abelian counterexample: Let $G$ be the following topological group homotopy equivalent (as a space) to $\mathbb RP^\infty$. It is the realization of the simplicial group that corresponds by Dold-Kan to the chain complex in which the only nontrivial chain group, the first, has order $2$. To put it another way, if a group $A$ is abelian then the usual simplicial model for $BA$ is itself a group (in fact, abelian); let $G$ be $BA$ where $A$ has order $2$. -In this topological group, every nontrivial element has order $2$. Therefore there is no nontrivial homomorphism (continuous or not) from $S^1$. -Note that the reasoning here is very different from what it was in my other example. It is not that there is no homotopically nontrivial map $BS^1\to BG$, and it is not (as in Allen Hatcher's comment) that there is no nontrivial $H$-space map (homomorphism up to homotopy) $S^1\to G$.<|endoftext|> -TITLE: Picard group of scheme over DVR -QUESTION [9 upvotes]: Let $A$ be a DVR and let $X/A$ be a smooth, proper scheme with geometrically integral fibers. Is there an easy way to see that the Picard group of $X$ is isomorphic to the Picard group of the generic fiber $X_\eta$ of $X$? - -REPLY [9 votes]: This is too long to be a comment to Matt Emerton's answer. I think his conclusion is correct. Namely, if $X$ is smooth and proper over $A$, then $\mathrm{Pic}(X)\to \mathrm{Pic}(X_{\eta})$ is an isomorphism (whenever the generic fiber is geometrically connected or not). As explained by Matt, one can suppose $X$ is connected and we have only to show the injectivity. Clearly the kernel is generated by the irreducible components of the special fiber $X_s$. So it is enough to show that each of these irreducible components is a principal (Cartier) divisor. -Frist consider $A'=H^0(X, \mathcal O_X)$. It is finite and flat over $A$. As $X_s=X\otimes_{A'} (A'/\pi A')$ where $\pi$ is a uniformizing element of $A$, is smooth over $A/\pi A$, this force $A'/\pi A'$ to be separable over $A/\pi A$, hence $A'$ is étale over $A$. -The scheme $X$ is also a smooth proper $A'$-scheme with geometrically connected generic fiber. Let $s_1,\dots, s_n$ be the points of Spec$(A')$. Then $A'/\pi A'=\oplus_i k(s_i)$ and $X_s$ is just the disjoint union of the $X_{s_i}$. By Zariski, $X_{s_i}$ is connected, smooth, hence irreducible. Let $\pi_i\in A'$ be a uniformizing element at $s_i$ and a unit at $s_j$ for all $j\ne i$. Then the divisor div$(\pi_i)$ of $\pi_i$ on $X$ is $X_{s_i}$. -In Laurent's example, each copy of $\mathbb P^1_k$ is actually a principal divisor. So there is no more divisors on $\mathbb P^1_{A'}$ than on $\mathbb P^1_{\mathrm{Frac}(A')}$. Hope I don't miss something.<|endoftext|> -TITLE: Invariants that might determine graph up to isomorphism -QUESTION [10 upvotes]: Are there any graph invariants which have a reasonable chance of capturing the graph up to isomorphism? In other words, some candidates for a function $f$ such that $f(G)=f(H)$ if and only if $G$ and $H$ are isomorphic. -For instance, in the case of trees, weighted graph polynomial ($U$-polynomial) of Welsh/Noble 1999 is a candidate because no counter-example has been found. Are there such candidates for general graphs? - -Clarification: I'm interested in examples of functions which - capture some graph invariant, are practical to compute, and are not - yet proven to assign the same value to a pair of non-isomorphic graphs - -REPLY [7 votes]: In some ways, provably, no (assuming the graphs are infinite). See MR1011177 (91f:03062) Friedman, H; Stanley, L; "A Borel reducibility theory for classes of countable structures." J. Symbolic Logic 54 (1989), no. 3, 894–914. -This paper shows (although the argument is terse, and at least some is older folklore) that any Borel (in an appropriate sense) function f mapping graphs to any thing else with an equivalence relation E in such a way that G is isomorphic to H iff f(g) E f(H) must be at least as complicated as the graphs themselves. -For a similar result on finite graphs, see MR2135387 (2006e:03049) Calvert, Cummins, Knight, and Miller, Comparing classes of finite structures. (Russian) Algebra Logika 43 (2004), no. 6, 666--701, 759; translation in Algebra Logic 43 (2004), no. 6, 374–392.<|endoftext|> -TITLE: Roadmap to Computer Algebra Systems Usage for Algebraic Geometry -QUESTION [10 upvotes]: I've decided it's time to start learning how to use a computer to do calculations... I've used Singular to some small extent so far, but I want to start relying on computer algebra systems more. -Question -Which computer algebra system is best for what, and what is the easiest/most fun(?) way to learn how to deal with them? -I should mention that I do (arguably) Arithmetic Geometry. So, ideally I should invest my time in learning something that is capable of making the abstract existence theorems in that field explicit (see for example: How to get explicit unramified covers of an elliptic curve?, or another example is normalizations). -P.S. I don't have access to Magma. How powerful is their "online calculator", and is it worthwhile to learn Magma just to use the online version? - -REPLY [17 votes]: Your question is "Which computer algebra system is best for what, and what is the easiest/most fun(?) way to learn how to deal with them?" - -Regarding community, I think Sage (http://sagemath.org) is the best CAS, since Sage is completely open and free, and there are about a dozen mailing lists, and thousands of subscribers and messages a month. There is also a forum like mathoverflow for Sage: http://ask.sagemath.org/questions/. Also, Sage builds heavily on Singular, so it may feel familiar. See this 1-page article I published in the Notices for more about motivation: http://www.ams.org/notices/200710/tx071001279p.pdf For a sense of the capabilities of Sage, skim the reference manual: http://sagemath.org/doc/reference/. There is also an optional package of more unstable code for arithmetic geometry here: http://code.google.com/p/purplesage/ -Regarding raw interpreted speed, it's been claimed that the Magma interpreter is often faster than the Sage interpreter (=Python). However, Sage has Cython http://cython.org, which can produce code that is much faster than anything one can produce using the Magma (or any other CAS) interpreter. -Magma is the only existing CAS that has well developed capabilities for computing with function fields of transcendance degree 1 over $\mathbf{F}_p$, i.e., the function field analogue of algebraic number theory. Chris Hall and I are working on something similar for Sage right now, but this will take a long time. -Magma is also the only CAS that can compute fairly general spaces of Hilbert Modular Forms (again, work is also under way to add this to Sage, but it is a nontrivial project). Someday you might care about this. -Regarding elliptic curves, Sage has much more related to $p$-adic $L$-functions and Heegner points, but Magma has 3 and 4 descent. - -The answer to your question: "Is it worth learning Magma?" is definitely still yes, since there are still many algorithms today in arithmetic geometry that are only implemented in Magma, and available nowhere else (definitely not in Macaulay2, Singular, Mathematica, Sage, etc.). It will only take a few days, and you will have a better sense of what is possible. The exact same argument applies to Sage as well. Learn both. For Sage, you basically should: 1. learn Python, and 2. go through the Sage tutorial, which takes 2-3 hours. -Another point: Sage and Magma have a huge overlap in functionality related to arithmetic geometry, but the overlap in code bases is almost zero (and in many cases the people who implemented the algorithms in both systems are disjoint too). Thus there is some nontrivial value in comparing the output of both systems. See, e.g., this trac ticket about implementation of a function for computing all integral points on an elliptic curve for a nice example of this http://trac.sagemath.org/sage_trac/ticket/3674 This ticket also illustrates how the code that goes into Sage is all publicly peer reviewed (unlike this case with Magma).<|endoftext|> -TITLE: Realizing Baumslag-Solitar groups as functions of the $n$-adic integers -QUESTION [5 upvotes]: Let $\mathbb{Z}_n$ denote the ring of the $n$-adic integers. I recently read a paper which used the fact that the Baumslag-Solitar groups BS($\pm$1,n) and BS(n,$\pm$1) can be realized as functions $\mathbb{Z}_n \rightarrow \mathbb{Z}_n$. Can BS(m,n) (for m and n arbitrary) be realized as a group of functions $\mathbb{Z}_r \rightarrow \mathbb{Z}_r$ for some $r$? Thanks! - -REPLY [6 votes]: If you mean action by automorphisms, then the answer is "no" since the Baumslag-Solitar groups $BS(m,n)$, $|m|\ne |n|\ge 2$ are not residually finite. The groups $BS(m,n)$ do act nicely on the products of a tree and the Hyperbolic space: http://www.emis.de/journals/JLT/13-2/galpl.ps.gz .<|endoftext|> -TITLE: Criteria for coherence of rings -QUESTION [16 upvotes]: I'm trying to collect pointers into the literature about coherent rings. Recall that a ring is left coherent if its finitely generated left ideals are finitely presented. -This condition was introduced by Stephen Chase in [Stephen U. Chase, Direct products of modules, Trans. Amer. Math. Soc. 97 (1960), 457–473. MR0120260] as a characterization of rings whose class of flat right modules is closed under arbitrary direct products. -There are many results out there giving criteria for coherence. In particular, providing conditions which, when satisfied, ensure that a ring constructed in some way (polynomials, formal series, &c) from another ring is coherent. There are also interesting negative results: the nicest one I can find is [Jean-Pierre Soublin, Anneaux et modules cohérents, J. Algebra 15 (1970), 455–472 MR0260799] which shows that the Hilbert basis theorem is false for coherence. -Coherence is a more or less technical thing, so it tends to appear a bit 'hidden' in various contexts, so I am pretty sure I am missing a big part of the literature in spite of having spent a non-negligible time following links on MathSciNet. For example, it appears that model theorist have (had?) an interest in coherence. - -May the amazing collective MO erudition help me find relevant results? - -REPLY [2 votes]: Here are some further interesting references: - -C. Faith, Coherent rings and annihilator conditions in matrix and polynomial rings, Handbook of algebra, Vol. 3 (2003). -M. E. Harris, Some results on coherent rings, Proc. Amer. Math. Soc. 17 (1966), 474-479. -M. E. Harris, Some results on coherent rings II, Glasgow Math. J. 8 (1967), 123-126. -E. Matlis, Commutative semi-coherent and semi-regular rings, J. Algebra 95 (1985), 343-372. -G. Sabbagh, Coherence of polynomial rings and bounds in polynomial ideals, J. Algebra 31 (1974), 499-507.<|endoftext|> -TITLE: Where can you find Grothendieck's "Récoltes et Semailles"? -QUESTION [10 upvotes]: Where can you find Grothendieck's "Récoltes et Semailles"? -Is it available anywhere? - -REPLY [7 votes]: Update: Récoltes et Semailles has just been published in two volumes by Gallimard.<|endoftext|> -TITLE: Adjacency matrices of graphs -QUESTION [29 upvotes]: Motivated by the apparent lack of possible classification of integer matrices up to conjugation (see here) and by a question about possible complete graph invariants (see here), let me ask the following: - -Question: Is there an example of a pair of non-isomorphic simple finite graphs which have conjugate (over $\mathbb Z$) adjacency matrices? - -It is well-known that there are many graphs which have the same spectrum. This implies that their adjacency matrices are conjugate over $\mathbb C$. -In Allen Schwenk, Almost all trees are cospectral. New directions in the theory of graphs (Proc. Third Ann Arbor Conf., Univ. Michigan, Ann Arbor, Mich., 1971), pp. 275–307. Academic Press, New York, 1973 it was shown that almost all trees have cospectral partners. Maybe $\mathbb Z$-conjugate graphs can be found among trees? - -REPLY [40 votes]: Yes. -Consider the adjacency matrices -$$ A = \left[\begin{array}{rrrrrrrrrrr} -0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\ -1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\\\ -0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\\\ -0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\\\ -0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\\\ -0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\\\ -0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\\\ -0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\\\ -0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\ -0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\\\ -0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 -\end{array}\right] $$ -and -$$ B = \left[ \begin{array}{rrrrrrrrrrr} -0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\ -1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\ -0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\\\ -0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\\\ -0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \\\\ -0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\\\ -0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\\\ -0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\\\ -0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\ -0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\\\ -0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 -\end{array}\right]. $$ -These are both the adjacency matrices of trees, and both have characteristic polynomial -$$\lambda^{11}-10\lambda^9+34\lambda^7-47\lambda^5+25\lambda^3-4\lambda.$$ -Each tree -has exactly two vertices of degree 3, separated by a path of length 1 in the case of $A$ but length 2 in the case of $B$. In particular, the trees are not isomorphic. -Now consider the [EDIT: improved, much nicer] matrix -$$ C = \left[\begin{array}{rrrrrrrrrrr} -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & -1 \\\\ -0 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\\\ -0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\\\ -0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\\\ -0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\\\ -0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\\\ -0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\\\ -0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\\\ -0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\\\ -0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\\\ -0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\\\ -\end{array}\right] $$ -with determinant $-1$. -Since $C^{-1}AC = B$, the two trees (on 11 vertices) are non-isomorphic but -have adjacency matrices that are conjugate over $\mathbb Z$. -Now to explain the where the example comes from. The pair of graphs -was constructed by a method, attributed to Schwenk, that I found in Doob's -chapter of Topics in algebraic graph theory (edited by Beineke and Wilson). -The first 9 rows and columns of $A$, in common with $B$, come from a particular tree on 9 vertices that has a pair of attachment points such that extending the tree in the same way from either point gives isomorphic spectra. -Adding a single pendant vertex cannot work for this problem, as I found using Brouwer and van Eijl's trick, mentioned -by Chris Godsil, of comparing the Smith normal forms of (very) small polynomials in $A$ and $B$, in this case $A+2I$ and $B+2I$. When a path of length two is added at either of the two special vertices, however, there doesn't seem to be any obstruction of this type. -I then set about trying to conjugate both $A$ and $B$, separately, to the companion matrix of their mutual characteristic polynomial, by looking for a random small integer vector $x$ for which the matrix $X_A = [ x\ Ax\ A^2x\ \ldots\ A^{10}x]$ has determinant $\pm 1$, and similarly $y$ giving $Y_B$. (The fact that I succeeded fairly easily may have something to do with the fact that $A+I$ is invertible over $\mathbb Z$.) The matrix -$X_AY_B^{-1}$ then acts like the $C$ above. -[EDIT: The actual matrix $C$ I found at random and first posted was not nearly so pretty, with a Frobenius norm nearly ten times the current example. But taking powers 0 to 10 of $A$ times $C$ gave a $\mathbb Q$-basis for the full space of conjugators, whose Smith normal form (as 11 vectors in $\mathbb R^{121}$) was all 1's down the diagonal, so in fact it was a $\mathbb Z$-basis. Performing an LLL reduction on this lattice basis then gave a list of smaller-norm matrices, the third of which is the more illuminating $C$ given above, of determinant $-1$. The other determinants from the reduced basis were all $0$ and $\pm 8$.] -Taking rational $x$ and not restricting the determinant of $X_A$ gives a space of possible rational matrices $C$ of dimension 11, which are generically invertible; varying $y$ gives the same space [EDIT: as does multiplying on the left by powers (or in the more general case commutants) of $A$]. Since the spectrum of $A$ has no repeated roots, this is also the dimension of the commutant of $A$, and every matrix conjugating $A$ to $B$ lies in this space. Starting with a rational basis, it is not hard to find an exact basis for the integer sublattice, and taking the determinant of a general point in the integer lattice gives an integer polynomial in 11 variables which takes the value $1$ or $-1$ if and only if the matrices $A$ and $B$ are conjugate over $Z$. If there are repeated roots, you have to work a little harder; in general the full space has dimension the sum of the squares of the multiplicities, and is generated by multiplying on the left by a basis for the commutator space of $A$. A basis for the commutant can be produced (for a diagonalizable matrix) by first conjugating $A$ to a direct sum of companion matrices for the irreducible factors of the characteristic polynomial, and then one at a time, for each $k$-by-$k$ block corresponding to a $k$-times repeated factor of degree $m$, replacing each of the $k^2$ blocks with powers $0$ to $m-1$ of the companion matrix for that factor, with $0$ everywhere elsewhere.<|endoftext|> -TITLE: How to interpret topologically that the equalizer in Groupoids of ${\rm id}, {\rm id}: BG \rightrightarrows BG$ is $G/G$ (adjoint action)? -QUESTION [11 upvotes]: Let $G$ be a (discrete) group, and $1/G$ the corresponding groupoid with one object. Consider the diagram in (the 2-category) Groupoids with one vertex, labeled $1/G$, the one arrow from that vertex to itself, given by the identity map. -$$ \begin{matrix} 1/G \\ {\huge \circlearrowleft} \\ \scriptstyle \mathrm{id} \end{matrix}$$ -(This diagram is equivalent to the pair of parallel arrows $1/G \overset{\rm id}{\underset{\rm id}\rightrightarrows} 1/G$. Note that I am not filling in the loop with a 2-cell.) -A cute fact is that the ("2-") limit of this diagram in Groupoids is the action groupoid $G/G$ of the adjoint action of $G$ on itself. (See e.g. 2 limit in nLab or HTT Chapter 4 for a definition of limits.) -Now, in homotopological terms, the groupoid $1/G$ looks like the classifying space ${\rm B}G$, and the above diagram looks like ${\rm B}G \times S^1$. I have the possibly-mistaken impression that limits are supposed to look like topological cones (but maybe this is because we use words like "cone" when talking about limits). -Question: In terms of homotopy, how should I visualize the limit cone -$$ \lim\left( \begin{matrix} 1/G \\ {\huge \circlearrowleft} \\ \scriptstyle \mathrm{id} \end{matrix} \right) \quad \begin{matrix} {\huge \to} \\ {\large \circlearrowleft \!\!\!\!\!\! \circlearrowleft} \end{matrix} \quad \begin{matrix} 1/G \\ {\huge \circlearrowleft} \\ \scriptstyle \mathrm{id} \end{matrix} $$ -? -(Edits: per Quid's request in the comments, I replaced some broken images with diagrams, trying to reconstruct them from memory. $\circlearrowleft \!\!\!\!\! \circlearrowleft$ is my attempt at a doubled circle arrow, i.e. a 2-cell filling in the cone walls.) - -REPLY [11 votes]: To slightly amplify what Ben wrote, the diagram is precisely a presentation of $Map(S^1,BG)=L(BG)$ rather than of $BG\times S^1$. More generally the loop space of a space $X$ can be presented as the homotopy fiber product $LX= X\times_{X\times X} X$, the self-intersection of the diagonal, which is a slightly different way (which I find more convenient) to present self-homotopies of the identity map of $X$. In the case of a groupoid (or a stack) this results in the inertia groupoid, i.e., objects (points) together with automorphisms. Again in the case of $BG$ we have one object (the trivial $G$-torsor on a point, in one presentation) and its automorphisms form a $G$, with automorphisms given by $G$ acting adjointly. -On the level of functions/chains (interpretation depending on your context), rather than points, you get a formula that looks more like what you wrote, i.e. $$F(X) \otimes S^1= F(X) \otimes_{F(X)\otimes F(X)} F(X),$$ aka the Hochschild homology (or chains) of functions on $X$.<|endoftext|> -TITLE: Self-dual Complete Category -QUESTION [7 upvotes]: Is there an example of a self-dual complete category that is not a partially-ordered set? - -REPLY [11 votes]: Mike has already answered the question, but I thought I would add some assorted remarks on some of the discussion above. -Mike gave an example of what is called a $\ast$-autonomous category, as first defined by Michael Barr in Springer LNM 752 (1979). The field has been very heavily investigated, partly because $\ast$-autonomous categories are the categorical models of Girard's celebrated linear logic. In brief, a $\ast$-autonomous category is a symmetric monoidal closed category $V$ equipped with a "dualizing object" $D$, which means, letting $\hom$ denote the internal hom of $V$, that the canonical natural transformation -$$A \to \hom(\hom(A, D), D)$$ -is a natural isomorphism. This implies in particular that -$$\hom(-, D): V^{op} \to V$$ -is an equivalence, so that $V$ is self-dual. Complete $\ast$-autonomous categories are certainly of considerable interest, and many examples are known. Some (including Mike's example) are given in this nLab article. -There are also various constructions which allow one to embed sufficiently nice instances of duality, such as the Pontryagin duality alluded to by Martin, into complete $\ast$-autonomous categories. One of the most potent general constructions is called the Chu construction, which takes an input any symmetric monoidal closed category with pullbacks and a preassigned object $d$, and produces as output a $\ast$-autonomous category $Chu(C, d)$ together with a coreflective (in particular full) symmetric monoidal embedding -$$i: C \to Chu(C, d)$$ -whose dualizing object is manufactured from $d$ in a canonical way. If $C$ is complete and cocomplete, then so is $Chu(C, d)$. For a sample theorem which shows how certain nice topological dualities embed into Chu-type dualities, including the case of Pontryagin duality, see this paper by Barr. -Incidentally, with regard to Martin's answer, it is indeed the case that the category $LCHAb$ of locally compact Hausdorff abelian groups is not complete. We know for example that the topological product $\mathbb{R}^{\mathbb{N}}$ of countably many copies of $\mathbb{R}$ is not locally compact; it remains to see that there is no such product in $LCHAb$ (i.e., even if we try to strengthen the product topology on the set $\mathbb{R}^\mathbb{N}$ to some locally compact topology in some way, there still exists no solution to the universal mapping problem). If there were, then using the universal property, we could show that the scalar product $\mathbb{R} \times \mathbb{R}^\mathbb{N} \to \mathbb{R}^\mathbb{N}$ is continuous, making $\mathbb{R}^\mathbb{N}$ a Hausdorff topological vector space. But it is well-known that a locally compact Hausdorff TVS over the real numbers is finite-dimensional (reference; see theorem 4.3 on page 5). As arsmath suggests, this implies that the category of locally compact Hausdorff spaces cannot be complete (if it were, then so would be the category of abelian group objects, which we just proved is not the case).<|endoftext|> -TITLE: Higher homotopy algebraic structure on the homology of an operad -QUESTION [18 upvotes]: Given a DGA $A$, then by standard techniques such as homological perturbation theory, the ring structure on the homology $H(A)$ extends to a minimal $A_\infty$-algebra structure such that $H(A)$ is quasi-isomorphic to $A$, and moreover, this structure on the homology is unique up to isomorphism. The $A_\infty$ algebra structure is describe explicitly, as usual, by the collection of higher multiplications $m_n: H(A)^{\otimes n} \to H(A)$ (which are closely related to Massey products). -An operad in the category of chain complexes can be thought of as a generalisation of a DGA - we now have a sequence of complexes $P(n)$, and the $\circ_i$ compositions are associative products between these. The homology $H(P(n))$ is again an operad in chain complexes (with zero differential), and the above fact for DGAs should generalise to say that $H(P)$ carries the additional structure of a strongly homotopy operad for which it is quasi-isomorphic to $P$. Has this structure been describes explicitly in terms of higher $n$-ary analogues of the $\circ_i$ compositions? I would be very grateful if someone could point me towards an appropriate reference. - -REPLY [15 votes]: You are completely right Jeffrey. Since an operad is (strictly speaking) a generalization of the notion of associative algebra, there exists a homotopy transfer of structure through homotopy equivalences. This has been written, with explicit tree formulae, by Johan Granåker in "Strong homotopy properads" available at http://arxiv.org/abs/math/0611066 and published in IMRN. To get the statement on the level of operads, remove the two first letters in the word "properad" and consider only rooted trees instead of graphs in loc. cit. (More seriously, an associative algebra is an operad concentrated in arity 1. Properads model operations with several inputs and several outputs. So an operad is a properad concentrated in arity (many [inputs], 1 [outputs])). -In the paper with Merkulov, we use the homotopy transfer theorem for co(pr)operads to understand the minimal model of the properad encoding associative bialgebras (we apply it to the bar construction). And in a forthcoming paper with Gabriel Drummond-Cole, we use the same idea to make the minimal of the operad encoding Batalin-Vilkovisky algebras explicit.<|endoftext|> -TITLE: Is a fibration in algebraic geometry a fibre bundle? -QUESTION [9 upvotes]: Let $X \to B$ be a smooth, proper, dominant map of schemes over $\text{Spec }k$ an algebraically closed field of characteristic zero with $B$ integral. We have the generic fibre $\overline{F}$ defined over $\text{Spec }\overline{K(B)}$ and by base-changing along $\text{Spec }\overline{K(B)} \to \text{Spec }k$, we obtain a map $\overline{X} \to \overline{B}$ such that we can now write down the sequence $\overline{F} \to \overline{X} \to \overline{B}$. To what extent is this a fibre bundle? To ask a definite question, is there some etale map $\overline{B}' \to \overline{B}$ such that further pulling back will yield an isomorphism $\overline{X}' \simeq \overline{F}' \times \overline{B}'$? -This question is closely related to Flatness in Algebraic Geometry vs. Fibration in Topology and Is an algebraic geometer's fibration also an algebraic topologist's fibration?. In particular, it is motivated by (1) Ehresmann's theorem that locally analytically such a morphism should be a (topological) fibre bundle and (2) the fuzzy thinking that "locally analytically" should mean "after an etale base change", but I feel like the answer to the question I posed it above is probably in the negative. For example, it seems unlikely to me that two smooth hypersurfaces of degree $d$ in $\mathbb{P}^n$ which are (automatically) diffeomorphic but not isomorphic should suddenly become isomorphic after an etale base change. However, I don't know of any weaker way to algebro-geometrically state the condition that some map be a fibre bundle, however -- is there anything then that we can say algebro-geometrically with respect to the above maps, or do we have to be content with the differential-geometric statement that it's a fibre bundle in that category? - -REPLY [17 votes]: The answer here is a resounding no. I think the most important point is that you're applying the wrong topological intuition here. A variety shouldn't be thought of as like a manifold, but as like a complex manifold, and the corresponding theorem to the "submersion=fiber bundle" theorem in smooth manifolds is just false for complex manifolds. Just as an example, all elliptic curves are topologically the same, so the solutions to $x(x-1)(x-a)=y^2$ are a smooth fiber bundle over (most of) $\mathbb{C}$ with coordinate $a$, but all the fibers which aren't in the same orbit of $SL_2(\mathbb Z)$ on $\mathbb{C}$ are not isomorphic as complex manifolds. -The other way of saying this is that complex structures can exist in families; they have moduli. Moduli spaces exactly measure how theorems like "submersion=fiber bundle" fail since they measure continuous variation of structure.<|endoftext|> -TITLE: Is Krull dimension non-increasing along ring epimorphisms? -QUESTION [11 upvotes]: Let $f \colon R \to S$ be an epimorphism of commutative rings, where $R$ and $S$ are integral domains. Suppose that $\mathfrak{p} \subset S$ is a prime such that $f^{-1}(\mathfrak{p}) = 0$. Does it follow that $\mathfrak{p} = 0$? - -If answer is "yes", then it follows that for any epimorphism of commutative rings $R \to S$, strictly increasing chains of prime ideals in $S$ lift to strictly increasing chains in $R$; hence, Krull dimension is non-increasing along ring epimorphisms. -All of the above hold for quotients and localizations, which are the only examples of ring epimorphisms that come readily to my mind. - -REPLY [17 votes]: Yes. Letting $k$ be the field of fractions of $R$, we have the following commutative diagram. -$$ -\begin{array}{ccc} -R&\stackrel{f}{\rightarrow}&S\\\\ -\downarrow\scriptstyle{}&&\downarrow\scriptstyle{}\\\\ -k&\stackrel{g}{\rightarrow}&S_{\mathfrak{p}} -\end{array} -$$ -However, $f$ and the localization $S\to S_{\mathfrak{p}}$ are epimorphisms, so $g$ is an epimorphism with domain a field. This means that it is surjective, so $S_{\mathfrak{p}}$ is a field, and $\mathfrak{p}=0$. -To see that an epimorphism $g\colon k\to A$ of commutative rings with domain a field $k$ is surjective, consider the morphisms $u,v\colon A\to A\otimes_k A$ given by $u(a)=a\otimes1$ and $v(a)=1\otimes a$. Then, $u\circ g=v\circ g$ and, from the definition of epimorphism, $u=v$, in which case $g(k)=A$. - -REPLY [5 votes]: Let's make several reductions. First, the condition implies $f$ is injective, so letting $K=\operatorname{Frac}(R)$, we have that $f: K\to S'$ is an epimorphism, letting $S'$ be the localization of $S$ at the zero ideal of $R$; it is easy to check that this is an epimorphism via the universal property of localization. -But as George Lowther points out, an epimorphism from a field to an integral domain must be surjective. To see this, assume to the contrary that $K\to S'$ is not surjective; thus $K'=\operatorname{Frac}(S')$ is not equal to $K$. But $S'\to K'$ is an epimorphism, so composing, we've reduced to the case of a map between two fields $f: K\to K'$. But $K'$ admits many embeddings into (say) its algebraic closure which agree on $K$, contradicting that $f$ was an epimorphism. -EDIT: As the commenters point out, the algebraic closure doesn't quite work in the case $K'/K$ is purely inseparable, but this case is not difficult; see George Lowther's answer for an easy general argument. - -REPLY [2 votes]: First, pick a maximal chain of prime ideals in $S$ and mod out by the minimal one. Now $S$ is an integral domain of the same dimension. Similarly, you might as well assume $f$ is injective, since that can only decrease the Krull dimension of $R$. -So, now, we have a map, which must induce an isomorphism on fraction fields, and both algebras inject into their fraction fields. Now, take an ideal $I\subset S$ such that $I\cap R=0$, and let $s\neq 0$ be an element of $I$. Then $s=r'/r''$ for $r',r''\in R$. Thus $sr''\in R\cap I$, and we have arrived at a contradiction.<|endoftext|> -TITLE: A combinatorial approximation functor sSet->qCat -QUESTION [24 upvotes]: Let $sSet_J$ denote the category of simplicial sets equipped with the Joyal model structure. Simply by the fact that $sSet_J$ is locally presentable and its class of anodynes ($\neq \mathbf{Cof} \cap \mathbf{W}$ (the mid-anodynes are properly included in the class of trivial cofibrations)) has a small generating set with accessible source and target, Quillen's small object argument allows us to replace any simplicial set by a Joyal-equivalent quasicategory (and functorially so!). -However, as is often (would it be ungentlemanly for me to say "always"?) the case with factorizations constructed using the small object argument, it is extremely difficult to say anything concrete at all about the resulting approximations, which are typically immense (as they are constructed by a transfinite recursion). -The classical model structure on simplicial sets (denoted just as $sSet$) has an extremely elegant combinatorial fibrant replacement functor due to Dan Kan, called $\mathbf{Ex}^\infty$. The $n$-simplices of $\mathbf{Ex}^\infty S$ are exactly the k-fold subdivided n-simplices of $S$ for $k\geq 0$. -This tells us a lot of concrete information about the fibrant replacement, which we simply can't get from those approximation functors arising from the small object argument. The difference: The $k$-th stage of the transfinite composition does not depend on the previous terms. This is similar to presentations of sequences by direct (is that the right word?) formulae vs recursive formulae. -Question - -Does there exist anything similar to $\mathbf{Ex}^\infty$ for quasicategories? How about for the other widely-used simplicial models for $(\infty,1)$-categories: complete Segal spaces and Segal categories? -(Incidentally, I think that there is an analogue of $\mathbf{Ex}^\infty$ for simplicial categories gotten by applying $\mathbf{Ex}^\infty$ on hom-objects. However, this is not nearly as powerful, since not every object in $sCat$ is cofibrant). - -REPLY [8 votes]: $\def\Cnec{{\frak C}^{\rm nec}} -\def\Choc{{\frak C}^{\rm hoc}} -\def\C{{\frak C}} -\def\N{{\rm N}} -\def\id{{\rm id}} -\def\Exi{{\rm Ex}^∞} -\def\sCat{{\sf sCat}} -\def\sSet{{\sf sSet}_{\sf Joyal}}$ -Such a fibrant replacement functor can be constructed by composing three existing constructions: - -The Dugger–Spivak functor $$\Cnec: \sSet→\sCat$$ -that sends a simplicial set $X$ to the simplicial category $\Cnec(X)$ -whose set of objects is $X_0$ (the set of 0-simplices of $X$) -and the simplicial set of morphisms from $x∈X_0$ to $y∈X_0$ -is the nerve of the category of necklaces from $x$ to $y$. -Objects of this category are necklaces from $x$ to $y$, -i.e., maps of simplicial sets $N_k:=Δ^{k_1}∨⋯∨Δ^{k_m}→X$ -($m≥0$, $k_i≥1$) -that map the initial vertex of $Δ^{k_1}$ to $x$, -the terminal vertex of $Δ^{k_m}$ to $y$, -and $∨$ denotes the operation that fuses the terminal vertex -of the simplex to its left with the initial vertex of the simplex to its right. -Morphisms of necklaces from $N_k→X$ to $N_l→X$ are simplicial maps $N_k→N_l$ -that preserve the initial and terminal vertices and make the obvious triangle commute. -This functor is a Dwyer–Kan equivalence of relative categories. - -Kan's $\Exi$ functor, applied to each mapping simplicial set in a simplicial category: -$$\Exi:\sCat→\sCat.$$ -This is a fibrant replacement functor in the Dwyer–Kan and Bergner model structures. - -Cordier's homotopy coherent nerve functor -$$\N:\sCat→\sSet.$$ -This is a right Quillen equivalence. - - -By construction, the composition -$$\N∘\Exi∘\Cnec:\sSet→\sSet$$ -is a Dwyer–Kan equivalence of relative categories that lands in quasicategories. -Furthermore, we have a zigzag of natural weak equivalences -$$\id→\N∘\Exi∘\C←\N∘\Exi∘\Choc→\N∘\Exi∘\Cnec$$ -in the model category $\sSet$, where $\Choc$ is another functor constructed by Dugger and Spivak. -This construction satisfies the stated criterion in the original post, namely: - -The $k$-th stage of the transfinite composition does not depend on the previous terms. - -One could also use the functor $\C$ (the left adjoint of $\N$) instead of $\Cnec$. -The functor $\C$ has a concrete and explicit (but slightly more complicated) -description in terms of necklaces, similar to $\Cnec$. -The advantage of using $\C$ is that one has a genuine natural weak equivalence -$$\id→\N∘\Exi∘\C,$$ -as opposed to a zigzag of natural weak equivalences.<|endoftext|> -TITLE: 3D conformal mappings -QUESTION [8 upvotes]: Are there analogues to conformal mapping in 3 dimensions? -I have a specific example I am trying to solve.. Laplace's equation in 3D with slightly complicated rectilinear boundaries. (Think of solving a harmonic function over a 3D boundary which is a cube but with a sub-cube "bitten" out of one corner.) -Laplace's equation is still valid under conformal transformations, so for example in 2D I could take a square domain with a subsquare bitten out of a corner, and apply an inverse tranformation like some of these and solve the equation in a simple square domain. -Are there similar conformal-like transformations in 3D? Perhaps they wouldn't be called conformal maps, but maybe something exists which would work similarly for my 3D Laplace equations. - -REPLY [4 votes]: (If I understand the question correctly) there is a theorem by Liouville stating that starting from dimension 3 the only conformal transformations are the (analogs of) the Möbius ones - composites of reflections in hyperspheres and hyperplanes. It is e. g. in Wikipedia. (Separate Wikipedia entry) -There are more general quasiconformal maps, maybe they are useful here, this I don't know. -PS Just notices that this has been already said in comments to the OP actually, so I probably do not deserve these upvotes...<|endoftext|> -TITLE: Effective Chebotarev Density -QUESTION [14 upvotes]: Let $K$ be a number field, and $p$ be a rational prime. Then the Chebotarev Density Theorem implies we can find primes $v$ and $w$ of $K$ of degree 1 which are split and nonsplit respectively in $K[\sqrt{p}]$. What is the best known effective (upper) bound for the norms of the least such primes (not assuming GRH)? In particular, is there a bound which is asymptotically strictly less than $\sqrt{p}$ (times a constant coming from the field $K$)? -EDIT: I'd like to clarify, in response to the comments below. The situation I'm wondering about is when we fix K, and let p vary. So when K is a cyclotomic field (adjoin, say, the qth root of unity for a prime q), I'm asking about the least prime which is a quadratic nonresidue (resp. residue) mod p, which is 1 mod some fixed prime q, and I'm hoping that there is a bound of the form $\sqrt{p}$ times (something in terms of q). Under GRH this is true --- in fact under GRH, we can get a bound of the shape $(\log p)^2$ times constants coming from K. - -REPLY [11 votes]: I assume $\sqrt{p}$ is not contained in $K$, then the bound you are looking for is available. -Let $\chi$ be the ray class character attached to the quadratic extension $K(\sqrt{p})/K$, then the $L$-function $L(s,\chi)$ has conductor essentially $p$. By a recent result of Venkatesh (Theorem 6.1 in Annals of Math. 172 (2010), 989-1094) we have the subconvex bound $L(s,\chi)\ll |s|^N p^{1/4-1/200}$ on the criticial line $\Re s=1/2$, where $N>0$ is a constant. It follows, by a simple Mellin transformation technique, that for any fixed smooth function $V:(0,\infty)\to\mathbb{C}$ of compact support we have -$$\sum_{\mathfrak{m}\subset\mathcal{O}_K}\chi(\mathfrak{m})V(N\mathfrak{m}/X)\ll p^{1/4-1/200} X^{1/2}.$$ -Therefore the absolute value of the left hand side is smaller than $X$ for some $X\gg p^{1/2-1/100}$, where the implied constant depends only on $K$ and $V$. This implies that $\chi$ takes both values $\pm 1$ on prime ideals with norm $\ll p^{1/2-1/100}$. -Perhaps one can complement this with Vinogradov's trick, see Corollary 9.19 in Montgomery-Vaughan: Multiplicative number theory I. -EDIT: As the OP pointed out, all the $p$'s above should be replaced by $p^{(K:\mathbb{Q})}$. - -REPLY [9 votes]: The paper: -J. C. Lagarias, H. L. Montgomery and A. M. Odlyzko, A bound for the least prime ideal in the Chebotarev density theorem, Invent. Math. 54 (1979) 271-296 -gives a bound of the form $c \sqrt p$ for some unspecified $c$. -My paper with J. Vaaler: -The least nonsplit prime in Galois extensions of Q, J. Number Theory, 85 (2000), 320-335. -gives an effective constant (when $K=\mathbb Q$, but the argument should generalize). Actually, the quadratic field case of our argument is already in Gauss. Our paper has a bunch of other references, including what you can get with GRH. Improving the square root bound without GRH is a big open problem. -The paper also gives me Erdos number 2 :-) -EDIT: As in GH's answer, the natural quantity for the bounds is the discriminant, so $p$ needs to be replaced by $p^n, n=[K:\mathbb Q]$, in the case of $K(\sqrt p)$. Here is an example where this will make a big difference. Take $K$ to be the cyclotomic field of $p$-th roots of unity where $p \equiv 3 \mod 4$, so the quadratic extension is non trivial. The OP asks for degree one primes, these are primes above rational primes $l \equiv 1 \mod p$, so they have norm $l > p$ and you can't expect a $\sqrt p$ bound.<|endoftext|> -TITLE: Does the Borel functor take equivariant fibrations to fibrations? -QUESTION [10 upvotes]: Let $p\colon X\to B$ be a fibration. Let $G$ be a topological group acting continuously on $X$ and $B$, and assume that the map $p$ is $G$-equivariant. -We can apply the Borel functor $EG\times_G-$ for a contractible, free $G$-space $EG$. This gives a map $1\times_G p\colon EG\times_G X\to EG\times_G B$. - -Under what conditions on the group $G$ and the actions is the map $1\times_G p$ guaranteed to be a fibration? - -References appreciated! - -REPLY [11 votes]: I believe what you are asking is true whenever $X\to B$, considered as an unequivariant map, -is a Serre fibration. - First some definitions: -Call a map of $G$-spaces $E \to B$ a $G$-Serre fibration if and only if for all subgroups $H\subset G$, the map of fixed points $E^H \to B^H$ is a Serre fibration. In particular, -$EG \times E \to EG \times B$ is a $G$-Serre fibration if and only if it is a Serre fibration of unequivariant spaces. It is known this notion of fibration arises from a model structure on $G$-spaces, in which a map $X\to Y$ is a $G$-weak equivalence iff each map of fixed point sets $X^H \to Y^H$ is a weak homotopy equivalence. A map $X\to Y$ is a $G$-cofibration iff $Y$ is obtained from $X$ by attaching cells of the form $D^n \times (G/H)$ where $H$-varies through subgroups and the attaching maps are $G$-maps, or more generally if the pair $(Y,X)$ is a retract of a relative $G$-cell complex. -I do not know a reference for the above, but I am confident it's in the literature. -(Added Later: see the comment below for two references.) - Now the argument: -Suppose that $A\to U$ is an acyclic cofibration in the Serre model structure on spaces. Without loss in generality, we can assume that $U$ is obtained from $A$ by cell attachments. Suppose we are given a lifting problem: -$A \to X \times_G EG $ -$\downarrow\qquad \qquad \downarrow$ -$U \to B\times_G EG$ - We need to find a map $U \to X\times_G EG$ making the diagram commute. -Here's how: pull back the above to an equivariant lifting problem -$\tilde A \to X \times EG $ -$\downarrow\qquad \qquad \downarrow$ -$\tilde U \to B\times EG$ -where $\tilde A$ for example is given by the pullback of $A \to BG \leftarrow EG$ -(the map $A\to BG$ is the composite $A\to X\times_G EG \to BG$). It is relatively straightforward to check that the inclusion $\tilde A\to \tilde U$ is an acyclic $G$-cofibration, -where the cells that are being attached are of the form -$D^n \times G$, i.e., they're free. -It follows from the model category structure on $G$-spaces that there's an equivariant lift $\tilde U \to X\times EG$ making the diagram commute. Now take orbits to get the a lift -$$ -U \to X\times_G EG -$$ -solving the original lifting problem.<|endoftext|> -TITLE: What fails when using call/cc as realizer of the Peirce formula -QUESTION [7 upvotes]: Define the axiom constants $p_{A,B}^{((A\rightarrow B)\rightarrow A)\rightarrow A}$ as realizers of the Peirce formula, and $f_A^{\bot\rightarrow A}$ as realizers of the Ex Falso Quodlibet. Then $p_{A\vee\lnot A,\bot}^{(\lnot (A\vee\lnot A)\rightarrow(A\vee\lnot A))\rightarrow(A\vee\lnot A)}\lambda_{u^{\lnot(A\vee\lnot A)}} . f_{A\vee\lnot A}^{\bot\rightarrow (A\vee\lnot A)} u (\vee_+^{\lnot A\rightarrow A\vee\lnot A} \lambda_{v^A}.u(\vee_+^{A\rightarrow A\vee\lnot A} v))$ proves the law of the excluded middle $A\vee\lnot A$. -Now, the function call-with-current-continuation, or short call/cc, in scheme, has the signature of $p$ (in Haskell, this one is replaced by its negative translation). The rest of the above term is covered by the Curry Howard Isomorphism, and therefore, using an instance of call/cc as a realizer of $p$ gives a program of type $A\vee\lnot A$. -Of course, something that one expects in programs extracted from constructive proofs must fail here, otherwise one could decide every proposition $A$. I was just wondering, what exactly it is. -Scheme's call/cc will fail with this term, since the continuation is called twice - but that is not really a problem, since one could just copy the whole context of the continuation and bind it into a new function which can then be called multiple times - that would produce a lot of overhead, but still would provide an effective algorithm which must not exist. -And having a call/cc in the beginning of a term does also not seem very useful, but inside another term, this should not be a problem. -Then, maybe in this context we must not use $f$. Normally, instances $f$ of the Ex Falso Quodlibet must never be called, and in this term, it could be called (and would give an instance of $A\vee\lnot A$) - but on the other hand, the continuation is called twice before ever coming to call $f$, so I am not sure whether this is actually the problem. -And then of course, $u$ is called twice, once with an instance of $A\vee\lnot A$ that eliminates to $A$, and once with an instance that eliminates to $\lnot A$. So probably calling $u$ with both outcomes of $A\vee\lnot A$ may produce endless loops or strange behaviour. -As I said, I am not sure what exactly fails here. Does anybody know? - -REPLY [11 votes]: I'm not sure what you mean by 'fail' here. -It is true, as you say, that classical proofs can exhibit behaviour that constructive proofs can't. In general, it's no longer true that a term of type A reduces to a constructive proof of the proposition A -- but that's only to be expected. Proof-terms in classical logic are strongly normalizing, though (if you're sensible about it), so there are no 'endless loops'. The difference is that, in general, such proof-terms can have more than one distinct normal form, and there is no good reason to pick one over any other. -You say - -Scheme's call/cc will fail with this term, since the continuation is called twice - -which is not true. Scheme continuations can be called any number of times. Also - -Normally, instances f of the Ex Falso Quodlibet must never be called - -is not true either. The ex falso sequitur quodlibet term discards the current continuation and returns its argument to the top level. -In terms of behaviour, functional programs use their continuations exactly once. Allowing EFSQ means that a program can discard, but not copy, its continuation. Admitting the double-negation rule duplex negatio affirmat allows programs to use their continuations an arbitrary number of times, including zero (or not, for Peirce's law). This corresponds to allowing exactly one, at most one and arbitrarily many formulas on the right-hand side of the turnstile in sequent calculus. -What the excluded-middle term does is this: it pretends to be a proof of not-A until such time as it is presented with a proof of A, whereupon it travels back in time (so to speak) and returns that proof of A instead (Wadler has a paper about this, I think). In general, classical proofs behave more like systems of communicating processes than functional programs. There are a couple of papers by Barbanera and Berardi that go into this in detail.<|endoftext|> -TITLE: Formal group laws and L-series -QUESTION [46 upvotes]: Let E be an elliptic curve, let $L(s) = \sum a_n n^{-s}$ -denote its L-function, and set -$$ f(x) = \sum a_n \frac{x^n}{n}. $$ -Then Honda has observed that -$$ F(X,Y) = f^{-1}(f(X) + f(Y)) $$ -defines a formal group law. -The formal group law of an elliptic curve has applications to -the theory of torsion points, apparently because formal groups -are useful tools for studying such objects over discrete valuation -domains. -Nevertheless I would appreciate it if someone could point out -the intuition behind this approach. What is the connection -between the L-series and the group law on the curve given by -the formal group law? Do formal group laws just give a -streamlined proof of basic properties of the elliptic curve -over $p$-adic fields, or is there more to them? -I've also seen the work of Lubin-Tate in local class field theory, and -I do remember that I found the material as frightening as cohomology -at first. It would be nice if the answers had something from a salesman's -point of view: why should I buy formal group laws at all? - -REPLY [48 votes]: Okay, here's a few words about the relation between the $L$-series and the formal group. In general, if $F(X,Y)$ is the formal group law for $\hat G$, then there is an associated formal invariant differential $\omega(T)=P(T)dT$ given by $P(T)=F_X(0,T)^{-1}$. Formally integrating the power series $\omega(T)$ gives the formal logarithm $\ell(T)=\int_0^T\omega(T)$. The logarithm maps $\hat G$ to the additive formal group, so we can recover the formal group as -$$ F(X,Y) = \ell^{-1}(\ell(X)+\ell(Y))$$. (See, e.g., Chapter IV of Arithmetic of Elliptic Curves for details.) -Now let $E$ be an elliptic curve and $\omega=dx/(2y+a_1x+a_3)$ be an invariant differential on $E$. If $E$ is modular, say corresponding to the cusp form $g(q)$, then we have (maybe up to a constant scaling factor) $\omega = g(q) dq/q = \sum_{n=1}^{\infty} a_nq^{n-1}$. Eichler-Shimura tell us that the coefficients of $g(q)$ are the coefficients of the $L$-series $L(s)=\sum_{n=1}^\infty a_n n^{-s}$. Integrating $\omega$ gives the elliptic logarithm, which is the function you denoted by $f$, i.e., $f(q)=\sum_{n=1}^\infty a_nq^n/n$, and then the formal group law on $E$ is $F(X,Y)=f^{-1}(f(X)+f(Y))$. -To me, the amazing thing here is that the Mellin transform of the invariant differential gives the $L$-series. Going from the invariant differential to the formal group law via the logarithm is quite natural.<|endoftext|> -TITLE: Questions on standard (motivic) conjectures -QUESTION [7 upvotes]: Over an (algebraically closed) characteristic $p$ field, is it known that the cohomological equivalence of cycles relation (with respect to $\mathbb{Q}_l$-adic \'etale cohomology) does not depend on the choice of a prime $l$ (distinct from $p$). At least, is it known that: if the numerical equivalence of cycles relation coincides with the cohomological one for one value of $l$, this is also true for all other $l$'s? -I believe that in order to have Kunneth decompositions for cohomological (pure) motives and the Standard Lefshetz conjecture is suffices for the numerical equivalence of cycles relation to coincide with the cohomological one. Is this true, or does the Hodge Standard conjecture play some role here (or in the first question)? - -REPLY [5 votes]: The answer to both questions in 1 is NO. For example, Clozel has shown that for an abelian variety over the algebraic closure of a finite field, there are infinitely many l for which numerical and homological equivalence coincide, but this doesn't help with proving the statement for all l (or even the independence of homological equivalence from l, for all l). -The answer to question 2 can be found in Kleiman's articles.<|endoftext|> -TITLE: Pseudogroups and the flavors of the topology of manifolds -QUESTION [12 upvotes]: Topologists study topological manifolds, differentiable manifolds and PL-manifolds (and some other flavors), with each class distinguished by the selection of a pseudogroup that restricts the transition functions that may occur in an atlas. Other pseudogroups -exist and do receive study, but in many cases the extra structure they imbue has a geometrical rather than a topological character, e.g. on account of local invariants or continuously varying moduli. -The TOP, DIFF and PL pseudogroups each have their intrinsic importance for applications and their historical caché, so I understand why they attract so much attention. But I ask: do there exist theorems that characterize abstractly those pseudogroups that lead to reasonable theories of manifolds, theories with a topological character? In particular, does any such theorem explain the de facto privileged status of TOP, DIFF and PL (and a few others)? If no, then how big is the zoo of exotic topologically-flavored pseudogroups? -================================================================================== -Addendum: -I see an analogy (though that word might be too weak?) between arithmetic geometry and manifolds-with-pseudogroup-structure. In arithmetic geometry one asks whether a variety has a definition over a field, and also whether distinct varieties defined over one field become isomorphic over some larger field. This reminds one of how, say, passing from DIFF to TOP one meets new manifolds that didn't admit a DIFF-structure, but also sees new TOP-isomorphisms between distinct DIFF-manifolds. -The big difference, methodologically, seems that in arithmetic geometry the varieties (~manifolds) and the fields (~pseudogroups) will often enter on an equal footing. For example, one constructs important fields in class field theory as the fields of definition certain varieties. But in topology I have the impression that one always chooses at the start one or more pseudogroups, and then develops their theory; I don't know anywhere that the pseudogroups emerge naturally out the topological phenomena. -Perhaps I should start over and shape this into an autonomous question? - -REPLY [11 votes]: Contact structures have a strong topoological character --- there are no local invariants -for them, and no local moduli. They especially have a lot of known interaction with "standard" topology in dimension 3. They are quite flexible, with a rich group of automorphisms. -Symplectic structures do have local moduli, but the local parameter is just a cohomology -class, which is itself straightforward topology. Symplectic structures have a rich group -of automorphisms; it's a matter of word choice if you want to say they don't have -a "topological" character. Obviously they're have undergone a vast amount of study, -and they're very important. -Other "topological" examples: the pseudogroup of local bilipschitz homeomorpisms, -and the pseudogroup of local quasiconformal homeomorphisms. -But really, for many purposes, I don't see the need to segregate the different kinds of -pseudogroups: the relationships among them often tell us something about topology as -well as the particular structure. When there is occasion to study actual diffeomorphisms and homeomorphisms between manifolds, rather than just isotopy classes of -homeomorphisms, the subject is very closely related to foliations, which is very closely -related to flat bundles with fiber a manifold. There is a good deal of -machinery, classifying spaces and the like, that works in similar ways for many different -cases.<|endoftext|> -TITLE: smoothly varying smooth structures -QUESTION [12 upvotes]: Can one vary smooth structures on $\mathbb R^4$ smoothly/continuously? -This question popped out of Ben's answer here. - -REPLY [8 votes]: Yeah, sure. A $1$-parameter family of smooth structures on a topological manifold $M$ can be taken to be a smooth structure on $M \times I$ such that the projection map $M \times I \to I$ is a submersion. Similarly for higher families. -To relate it to your comments on Ben's thread, you can (apparently) find a $1$-parameter family of smooth structures on $\mathbb R^4$ such that all pairs of fibres $\mathbb R^4 \times \{a\}$ and $\mathbb R^4 \times \{b\}$ for all $a \neq b$ are not diffeomorphic. -That 2nd paragraph is really Larry Siebenmann talking. I don't believe I've ever seen such a construction.<|endoftext|> -TITLE: Ising model on groups -QUESTION [18 upvotes]: Can anything interesting be deduced about the properties of a group from the behavior of the Ising model on its Cayley graph? (i.e. existence and character of phase transitions, critical behavior) I'm not sure, though, if one should expect any general results (link between large-scale geometry and universality class, maybe?), since even very simple geometry of the group (say, $\mathbb{Z^2}$) can give highly nontrivial statistical properties. - -REPLY [6 votes]: Hi Marcin, -recently writing a paper about phase transition on the Ising model with positive non-uniform magnetic field in infinite graphs I discovered that joining some results in the literature we can relate amenability and Phase transition in the Ising model with positive magnetic field: - -Theorem 1: If $G$ is a non-amenable infinite connected graph then there is a ferromagnetic Ising model on $G$, with constant positive magnetic field having phase transition. - -There is a partial converse of this result for quasi-transitive amenable graphs. A infinite graph $G=(V,E)$ is quasi-transitive if there exist a finite number of vertices $x_1,\ldots,x_k$ such that for any $x\in V$, there is an automorphism of $G$ taking $x$ to some $x_i$. - -Theorem 2: If $G$ is a amenable quasi-transitive infinite connected graph then all ferromagnetic Ising model on $G$, with constant positive magnetic field has no phase transition. - -Theorem 2 is interesting converse because of the quasi-transitive hypothesis can not be removed since Bausev shown that we have phase transition in ferromagnetic Ising model with magnetic field being constant at all sites of the lattice $\mathbb{Z}^2 \times \mathbb{Z}_+$ . - Definitions. - -An Ising model on a graph $G=(V,E)$ is defined as follows: -Let $\mathcal{L}$ be the set of finite parts of $V$ and suppose that $\Lambda_n\in\mathcal{L}$ is such that $\cup_{n\in\mathbb{N}}\ \Lambda_n=G$. The Hamiltonian of the Ising model in $\Lambda_n$ with a boundary condition $\omega\in \{-1,1\}^{V}$ is given by -$$ -H_{\Lambda_n}(\sigma|\omega)= --\sum_{\substack{i,j\in E:\\ i,j\in\Lambda_n}} J\ \sigma_i\sigma_j --\sum_{i\in \Lambda_n}h_i \ \sigma_i - \sum_{\substack{i,j\in E:\\ i\in\Lambda_n,j\in\Lambda_n^c}} J\ \sigma_i\omega_j, -$$ -where $\sigma=(\sigma_i)_{i\in V}\in\{-1,1\}^{V}$, $J\in\mathbb{R}$ (the model is called ferromagnetic of $J>0$) and $h_i\in\mathbb{R}$ is said the magnetic field. Finally we say that this Ising model has phase transition if the closed convex hull of the set -$$ -\left\{w-\lim_{\Lambda_n\uparrow G}\ \mu_{\Lambda_n}^{\beta,\omega}:\omega\in\{-1,1\}^{V} \right\} -$$ -is singleton for all $\beta>0$. The measures $\mu_{\Lambda_n}^{\beta,\omega}$ are defined by -$$ -\mu_{\Lambda_n}^{\beta,\omega}(\sigma)= -\left\{ -\begin{array}{rl} -\frac{\exp(-\beta H_{\Lambda_n}^{\omega}(\sigma))}{Z_{\Lambda_n}^{\omega}},&\text{if}\ \sigma_i=\omega_i\ \forall i\in\Lambda_n^c;\\ -0,& \text{otherwise}, -\end{array} -\right. -$$ -One nice reference about the results I stated above is Jonasson, J. and Steif, J. E.: Amenability and Phase Transition in the Ising Model. J. Theor. Probab. 12, 549-559 (1999).<|endoftext|> -TITLE: Cross correlation detection in binary Hamming distance -QUESTION [7 upvotes]: Given two long binary strings of length N, it's easy to find the Hamming distance between them. If you're allowed to cyclically shift one of the strings, you'll get N different Hamming distances when comparing the two. What is an efficient way to find the maximum Hamming distance over all N shifts? -This question is motivated by a sensor which tends to emit streams of "random" bits. There's potential slight correlations at unknown delays of millions of bits, a kind of slight bias of a ghost echo. I'm looking for a test to see if these correlations can be detected. -I think the problem can be solved with application of the discrete Fourier transform, but I'm not sure if there's a "binary" Fourier transform analogue and how it could identify the maximum Hamming over all the circular shifts. -Thanks! - -REPLY [6 votes]: I'd suggest that you start by encoding your signal in terms of the symbols +1 and -1 rather than 0 and 1. If you have two signals x and y, then take elementwise products of x and y and sum to get a measure of the distance between x and y. If the signals are identical, then the sum will be n. If the signals differ in each position, the sum will be -n. If there are matches in k positions, then the sum will be k-(n-k)=2k-n. -e.g. (using MATLAB notation) -x=[1; 1; 1]; -y=[-1; 1; -1]; -sum(x.*y)=-1 -Once you know this, then you can consider taking the circular convolution of two signals x and y. The elements of the circular convolution of x and y are these sums for various circular shifts of z. e.g. -x=[+1, -1, +1] -y=[+1, +1, -1] -let -z=circconv(x,y)=ifft(fft(x).*fft(y)) -z=[3, -1, -1] -This tells us that shifting y circularly to the left by one position results in a perfect match, and that the other two circular shifts of y result in largest mismatches. -This algorithm takes O(n*log(n)) time to compute z and then O(n) time to find the max (and/or min) elements of z.<|endoftext|> -TITLE: Applications of the Theorem of Gelfand-Naimark -QUESTION [8 upvotes]: Hi, -I am interested in the correspondence of algebraic results about C(X) (the space of continuous functions $X\to {\mathbb C}$(complex numbers) or $X\to {\mathbb R}$(real numbers) and topological properties of X,for example results like this .(Does by the way someone know what's the "deep result by bkouche" mentioned in the paper?) -Can you by the way use this result to prove interesting theorems with this translation -(like: --a manifold(or even a CW-complex) is paracompact --the theorem of Tietze, etc?) -There are many such correspondences which are obtained by using Gelfand-Naimark but I couldn't find literature where you can find full details with all needed definitions and proofs.(I couldn't even find a proof of the categorical Gelfand-Naimark theorem in the nonunital case,only some sketches.) Does such literature exist for a beginner in this topic? The book "Basic Noncommutative Geometry" written by Khalkhali is a good source but omits most details (see page 16 for a little list). -So I would be glad if you can recommend to me a good book/link, or write a nice result here if it's not too complicated. -Example: X is connected iff C(X) has no idempotents because direct sums of subalgebras correspond to disjoint union of closed subspaces,since C is an equivalence. - -REPLY [4 votes]: Dear trew, here is an application I like of Gel'fand-Naimark's representation theorem . I'm not sure it answers your question in a strict sense but I hope it is sufficiently close... . -Consider a completely regular topological space $X$ and its algebra of continuous and bounded functions $A= \mathcal C_b (X) $. By Gel'fand-Naimark's representation theorem, the algebra $A$ is isomorphic to to $ \mathcal C (\bar X)$ for a uniquely (up to homeomorphism ) defined topological space $\bar X$. Well, this $ \bar X$ is the -Stone–Čech compactification $\bar X=\beta (X)$ of $X$. There are other definitions of that compactification but I find this one appealing to those who (like me) are more familiar with spectra of rings than with ultrafilters. -Technical note Completely regular (= $T \;\; 3\frac{1}{2}$) means Hausdorff and global continuous functions ( not necessarily bounded ! ) numerous enough to separate a closed set from an exterior point. This is needed in order that $X$ embed into $\bar X=\beta (X)$.<|endoftext|> -TITLE: The Dual Abelian Variety -QUESTION [9 upvotes]: Let $A$ be an abelian variety defined over an algebraically closed field, say over $\mathbb{C}$. There is a dual abelian variety $\hat{A}$, along with a Poincare line bundle $L$ on $A\times \hat{A}$. Is there any relation between $\widehat {A\times A} $ and $\hat{A}\times \hat{A}$, for instance are they isogenous. What happens when $A$ is principally polarized, can we say relate the Poincare bundles in this case. - -REPLY [18 votes]: You can define $\hat{A}$ as $\underline{\mathrm{Pic}}^0(A)$. Now, over any algebraically closed field $k$, let $V$ and $W$ be proper (irreducible, reduced) varieties. Then $\underline{\mathrm{Pic}}^0(V)\times \underline{\mathrm{Pic}}^0(W)\to\underline{\mathrm{Pic}}^0(V\times W)$ is an isomorphism. Injectivity is immediate, and surjectivity follows from the theorem of the cube (see e.g. Mumford, Abelian Varieties, section 6 in chapter II).<|endoftext|> -TITLE: Is H^2(W_p,C^times) well-known? -QUESTION [8 upvotes]: Let $W_p$ be the Weil group of $\mathbf{Q}_p$. What is the Galois cohomology group $H^2(W_p,\mathbf{C}^{\times} )$ (with trivial action)? Is it zero, or something huge and complicated? -(This group comes up, at least for me, when you want to compare two Weil group representations whose projectivizations agree.) - -REPLY [11 votes]: It is known that $H^2(W, C^\times)$ is trivial, when $W$ is the Weil group of a global or local field, with the trivial action on $C^\times$, and the cohomology is taken in the sense of Moore (measurable cochains). This is the main result of C.S. Rajan, "On the vanishing of the measurable cohomology groups of Weil groups", Compositio Math. 140 (2004) 84-98 (also easy to find online). -Is this well-known? I don't know. I only found this paper recently, and hopefully now it will become better-known!<|endoftext|> -TITLE: Does generator of continuous time random walk map heat kernel from L^2 to L^2? -QUESTION [5 upvotes]: Let $\Gamma = (G,E)$ be an undirected, infinite, connected graph with no multiple edges or loops. We equip $\Gamma$ with a set of edge weights $\pi_{xy}$, where, given $e=\{x,y\}\in E$, we write $\pi_{\{x,y\}} = \pi_{xy}=\pi_{yx}>0$. If $\{x,y\}\not\in E$, we set $\pi_{xy}=0$. We write $\pi_x$ for $\sum_{y\sim x} \pi_{xy}$. We also assign a set of positive vertex weights $(\theta_x)_{x\in G}$. -We consider the continuous time random walk on $\Gamma$ with generator $\mathcal{L}_\theta$ given by -\begin{equation*} -(\mathcal{L}_\theta f)(x) = \frac{1}{\theta_x}\sum_{y\sim x}\pi_{xy}(f(y)-f(x)), -\end{equation*} -and denote the resulting Markov process by $(X^\theta_t)_{t\geq 0}$. Roughly speaking, at a vertex $x$, this process waits an exponential time with mean $\theta_x/\pi_x$, and then jumps to one of its neighbors $y$ with probability $\pi_{xy}/\pi_x$. It's not hard to see that if one chooses the $\theta$ s and $\pi$ s in certain ways, the process can behave quite pathologically and run away from its starting point very quickly (e.g., the process may have a finite lifetime). -The function $p_t(x,y) := \frac{\mathbb{P}^x(X^\theta_t=y)}{\theta_y}$ is the called the heat kernel of the random walk. -I'm interested in some analytic aspects of the generator $\mathcal{L}\_\theta$. For example, $\mathcal{L}\_\theta$ is a bounded operator from $L^2(\theta)$ to $L^2(\theta)$ if and only if -\begin{equation*} -\sup_{x\in G} \frac{\pi_x}{\theta_x}< \infty. -\end{equation*} -Set $u(x) := p_t(x_0,x)$. Note that $u$ is always in $L^2(\theta)$, as -\begin{equation*} -\sum_{x\in G} u^2(x)\theta_x = \sum_{x\in G} p_t(x_0,x)p_t(x,x_0)\theta_x = p_{2t}(x_0,x_0) \leq \frac{1}{\theta_{x_0}} -\end{equation*} -Is it the case that $\mathcal{L}_\theta u$ is always in $L^2(\theta)$ also? - -REPLY [2 votes]: I was able to answer this question, although it didn't turn out to be useful in the way I thought it would be. -Let $P^\theta_t$ be the transition operator, $(P^\theta_tf)(x) = \sum_{y\in G} p_t(x,y)f(y)\theta_y$. Then a fairly easy computation shows that $P_t$ maps $L^2(\theta)$ to $L^2(\theta)$, and one can also show that if $v\in C_c(G)$, then $P_t(\mathcal{L}\_\theta v) = \mathcal{L}_\theta (P_tv)$, basically from a self-adjointness calculation. If one applies this to the function $u=\theta^{-1}_{x_0}\delta_{x_0}$, then we get $\mathcal{L}\_\theta u = \mathcal{L}_\theta (P_tv) = P_t(\mathcal{L}_\theta v)$, and the right-hand side of this is in $L^2(\theta)$.<|endoftext|> -TITLE: How are the classifying space of $E_8$ and $K(\mathbb{Z},4)$ related? -QUESTION [40 upvotes]: I recently heard the following fact : -Up to the $15$th skeleton, the classifying space $BE_8$ and $K(\mathbb{Z},4)$ are homotopy equivalent? -I have two questions on this : -(1) Is there any easy way to see this? Of course, knowing the first fourteen homotopy groups of $E_8$ is enough but then the question is how does one compute them? -(2) Is there any feasible explanation that suggests that $4$th cohomology classes (possibly related to gerbes), i.e., elements in $H^4(X;\mathbb{Z})$, arise from physical considerations and if $X$ is of dimension $14$ or less then we're classifying $E_8$-bundles on $X$, thereby suggesting that $E_8$ arises out of physical considerations? -The last question is a little vague but any pointers would be great! - -REPLY [10 votes]: re 1): André's answer is superb, but just for the record, original references determining the homotopy groups of E8 are here: http://ncatlab.org/nlab/show/E8#HomotopyGroupsReferences -re 2): -One way to think about the phenomenon $B E_8 \simeq_{15} B^3 U(1) \simeq K(\mathbb{Z},4)$ from the point of view of string theory is to compare it to -a) the equivalence $B \mathrm{PU}(\mathcal{H}) \simeq B^2 U(1) \simeq K(\mathbb{Z},3)$ that controls Freed-Witten anomaly cancellation over D-branes -b) generalized complex geometry and exceptional generalized geometry that controls various other geometric structures in string theory. -In all these cases, one is looking at geometry which arises from reduction of structure groups along maps $G \to K$ of groups with the property that they are weak homotopy equivalences. -This is true for the inclusions of maximal compact subgroups that control generalized complex and exceptional generalized geometry, hence the U-duality symmetry of supergravity theories in various dimensions. -Notice that these inclusions are far from being equivalences as morphisms of Lie groups. But they are equivalences of the underlying topological spaces. -This situation now has a good analog in higher smooth geometry, which "explains" the role of $E_8$. -Namely, there is a smooth 3-group $\mathbf{B}^2 U(1)$ (a smooth group 2-stack) and the universal degree 4-class on $B E_8$ has a smooth refinement to a morphism of smooth 3-groups (group 2-stacks) -$$ - \Omega \mathbf{a} : E_8 \to \mathbf{B}^2 U(1) . -$$ -There is a higher analog of the notion of "reduction of structure groups" along such higher maps, and this controls the geometry of the supergravity C-field. For comparison, there is similarly a morphism of smooth 2-groups (smooth group stacks) -$$ - \Omega \mathbf{dd} : \mathrm{PU}(\mathcal{H}) \to \mathbf{B}U(1) -$$ -and its induced "generalized geometry" by "reduction of higher structure groups" controls precisely the Chan-Paton bundles on D-branes twisted by the $B$-field. -Both of these morphisms of smoth higher stacks become equivalences of topological spaces under geometric realization (the first on 15-coskeleta, hence over the relevant spacetimes). So we may think of this as saying that: -"The Lie group $E_8$ is a generalized maximal compact subgroup of the smooth 3-group $\mathbf{B}^2 U(1)$. The geometry of the $C$-field is the 'generalized geometry' controled by this 'inclusion'." -For more details on all this, see around section 4.3 of -http://arxiv.org/abs/1201.5277 -and the big overview tables in section 4.4.1 of -http://ncatlab.org/schreiber/files/cohesivedocumentv032.pdf<|endoftext|> -TITLE: Detecting a cover of the figure-8 knot complement -QUESTION [13 upvotes]: I have a specific concrete example of a complete, finite volume, cusped hyperbolic 3-manifold $M$, and I am trying to determine whether or not $M$ is a cover of the figure-8 knot complement (call this $F_8$). -The manifold $M$ is described combinatorially, and all the numbers work out; for instance I know $M$ and $F_8$ are commensurable. But $M$ is big: it is built out of 630 regular ideal cubes, or equivalently, 3150 regular ideal tetrahedra, so it would have to be a 1575-sheeted cover. This is of course a finite problem, but too big for SnapPea (or any other software, I imagine) to handle. By the way, I know that $M$ is not a link complement in $S^3$. -I am looking for ideas for (a) how to prove $M$ IS NOT a cover of $F_8$, if indeed it is not; and (b) how to prove $M$ IS a cover of $F_8$, if indeed it is. -For (b) I am looking for symmetries of $M$ to mod out by, and I am making slow progress, but for (a) I have no ideas. - -REPLY [19 votes]: Edited, in light of description of manifold in comments (at end) -Added #2: The answer is no, details at the end -In principle this is doable, but caution is necessary. Given a manifold tiled by -ideal simplices, its fundamental group is a subgroup $H$ of the full group $G$ of isometries of finite index in the tiling, as is the fundamental group $H_0$ of the figure eight knot complement. You can determine whether $H$ is conjugate in $G$ to a subgroup of $H_0$ by -a finite check: label the two simplices of the figure eight knot complement, assigning -them vertex orderings, and just try developing this pattern into the pattern for -triangulation of $H$. If you transcribe the information suitably into a computer program, the -check should be instantaneous (on human timescale). Snappea has a good system for -handling the combinatorics of how simplices are glued together; this could be copied or used. -(or you could do it by hand). -But a caution: $H_0$ might be conjugate within $PSL(2,\mathbb C)$ but not within $G$. -To start, the tiling of $H^3$ by regular ideal cubes has two subdivisions into tilings -by regular ideal simplices, obtained by inscribing a tetrahedron in the cube in one of two -ways. More generally, since these particular groups are arithmetic, they have a large -commensurability group, which is $PGL(2, \mathbb Q(\sqrt{-3}))$. If your manifold -is not a covering of the figure eight knot, to prove it you need to consider -conjugation within this larger group. It's known how to understand all the maximal -lattices commensurable with an arithmetic group. This translates into a countable sequence of patterns, one for each prime ideal in the Eisenstein lattice - $O_{-3}=\mathbb Z[1/2 + \sqrt{-3}/2$, given the triangulation of $H^3$ by ideal regular tetrahedra, of grouping them into a repeating pattern of larger polyhedra which can be retriangulated a second way. -The retriangulation of the cube is the one associated with 2. -Whether or not these retriangulations are compatible to your given manifold depends on -congruence conditions. I.e. find generators for the group in $PSL(2,O_{-3})$, and -look what subgroup they generate mod various prime ideals. This gives a finite set of ways to retriangulate --- then you need to continue, obtaining a finite list of subgroups -of $PSL(2,O_{-3})$ that are conjugate to $H$ within $PSL(2, \mathbb Q(\sqrt(-3)))$. -For each of these, you would need to check whether they are contained in $H_0$. -It's reasonably likely your manifold is a finite-sheeted covering of the figure eight knot complement, in which case, most of this isn't necessary---you'll find the covering without such an elaborate search. If it's not a covering, you might be able to prove that it's -not more easily by looking at the length spectrum. -If you have a reasonably conceptual description of your manifold, then it might also be -possible to work out the answer by pure thought, without having to do a lot of combinatorial -searching. -Added -I didn't notice your comment giving a concrete description when I wrote the above. -It's a nice manifold. To recap the description: for each triple of disjoint edges on -the complete graph K7, there is -a cube. The faces of the cube correspond to the ends of edges; each face is glued to -the cube obtained by moving the corresponding endpoint to the unoccupied vertex. -It's easy to check that the edges have order 6. There are $7!/2^3$ cubes. -First question: can the cubes be consistently divided into 5 tetrahedra, by choosing -for each cube a set of 4 vertices at distance 2 apart (on the 1-skeleton of the cube), -so that the subdivisions of the faces match up? No: think of the operation -of reflecting a cube in a pair of opposite faces. This corresponds to walking -a single edge around a triangle in K7. It comes back after 3 times, with its -ends reversed; the reflected subdivision does not match. This means there is - a 2-fold cover of the manifold where the subdivision would work consistently. -This 2-fold covering space appears more likely to be a covering of the figure -eight knot complement, but it requires checking. -I think it should be possible to prove from this that your manifold does not cover the figure eight -knot complement, by showing that your group is conjugate to a subgroup of -the stabilizer of an -edge but not the stabilizer of -a vertex in the action of $PSL(2, \mathbb Q(\sqrt{-3}))$ on the 5-way-branching tree -(Tits building) corresponding to the prime $\left <2\right >$, but I haven't thought through the details. -Additional details -The prime 2 does not split in the ground field, so $O_{-3}/\left <2\right >$ is the field with 4 elements. The cusps of the Bianchi group are at elements of the projective line -of $\mathbb Q(\sqrt{-3})$; reduction mod $\left <2\right >$ maps them to the 5 elements of the -projective line of $F_4$. You can think of the structure of the corresponding Tits building -or Bass-Serre tree as follows: from a tiling by regular ideal tetrahedra, if you -pick a tetrahedron, it and its four facial neighbors make an ideal cube; by reflections, -this extends to a grouping of the tiling by tetrahedra into cubes. There are 5 -ways to do this, depending which tetrahedron in the original cube you pick to be the -central one, and these can be distinguished algebraically by naming which of the -5 elements of $\mathbb P^1F_4$ is omitted form the vertices of the central tetrahedron. -For any tiling by cubes, you can subdivide it into regular ideal tetrahedra in two ways. -Thus, the full group of automorphisms of the cubical tesselation intersects the -full group of automorphisms of the tetrahedral tesselation in index 2 in the first, -and index 5 in the second (a parabolic subgroup mod $<2>$). To get from a parabolic subgroup mod $\left <2\right >$ to the edge stabilizer, you add an element of $PGL(2,O_{-3})$ -that is not conjugate to $PSL$, something like diagonal matrix $(2,1)$. -So: it's clear that the group as described stabilizes an edge of this tree, but does -not stabilize a vertex. Since the full commensurability group, -$PGL_2(\mathbb Q(\sqrt{-3}))$, acts on the tree, -this implies it is not conjugate to a subgroup that stabilizes a vertex. Since -the figure eight knot group stabilizes a vertex, it is not conjugate to a subgroup -of the figure eight knot group, i.e. it is not a covering space.<|endoftext|> -TITLE: Holomorphic function with a.e. vanishing radial boundary limits -QUESTION [8 upvotes]: Hello everybody. -I'm looking for an "easy" example of a (non-zero) holomorphic function $f$ with almost everywhere vanishing radial boundary limits: $\lim\limits_{r \rightarrow 1-} f(re^{i\phi})=0$. -Does anyone know such an example. -Best -CJ - -REPLY [13 votes]: According to a footnote in the famous Hardy-Ramanujan paper "Asymptotic formulae in combinatory analysis", the function $f(q)=\prod_{n=1}^{\infty}\frac{1}{1-q^n}$ vanishes like $f(re^{i\theta})=o((1-r)^{1/4-\varepsilon})$ for almost all $\theta$. No proof is given, though I can't imagine Hardy would have made a statement like this without a proof in his pocket. -Edit: This isn't actually hard to guess at. By Euler's pengatonal number theorem, we have $f(q)^{-1}=\sum_{n\in \mathbf{Z}}(-1)^{n}q^{n(3n-1)/2}$, so Plancherel gives -$\int_{0}^{2\pi}|f(re^{i\theta})|^{-2}d\theta=2\pi\sum_{n\in \mathbf{Z}}r^{n(3n-1)} \sim 2 \pi^{3/2}3^{-1/2}(1-r)^{-1/2}.$<|endoftext|> -TITLE: Perimeters of random-walk polygons -QUESTION [13 upvotes]: I have a random walk on $\mathbb{Z}^2$ that takes a step -with equal probability in the three directions that avoid -retracing the previous step. -The walk proceeds until it returns to a lattice point -previously visited, at which time it pinches off a simple, -closed loop or polygon: - - - -I'd like to know the distribution of the perimeters of -these polygons. -(In the example above, the perimeter is 10.) -Simulations show that the average perimeter is about 5.6, -with perimeter 4 the overwhelming favorite, -as one would expect: - - - -I feel this distribution must be known to the experts and -not difficult to explicitly detail, but -after looking at hitting times, first-passage times, self-avoidance times, -and various other frequently studied random walk quanities, -I am not finding a close-enough analog to help. -Thanks for any pointers you might provide! -Addendum. -Here is log-plot of the probability of a perimeter of length $L$, based on a simulation of $10^6$ walks. The first point represents 642,225 perimeters for $L=4$, the second point 176,043 perimeters for $L=6$, etc. The last point plotted is 135 instances of $L=38$. (There is one polygon of length $L=74$ in these million trials.) The average perimeter length is 5.62, which occurs after an average of 8.46 steps. - -REPLY [3 votes]: I believe the paper below is on this exact question: -http://pre.aps.org/abstract/PRE/v55/i3/pR2093_1<|endoftext|> -TITLE: Is fuzzy mathematics useful in pure mathematics ? -QUESTION [11 upvotes]: Fuzzy sets and logic seem to be mostly used for applying to real-world situations, control-theory, game-theory, economics, statistics, data management, artificial intelligence, automated reasoning etc -Are there any proofs of theorems in pure mathematics of a non-fuzzy nature that make use of fuzzy concepts ? -Fuzzy set theory may be defined axiomatically and therefore be "pure" however here are some quotes from the Fuzzy logic article at Scholarpedia which highlight the applied nature: -"Humans have a remarkable capability to reason and make decisions in an environment of uncertainty, imprecision, incompleteness of information, and partiality of knowledge, truth and class membership. The principal objective of fuzzy logic is formalization/mechanization of this capability." -"During much of its early history, fuzzy logic has been an object of skepticism and derision, in part because fuzzy is a word which is usually used in a pejorative sense. Today, fuzzy logic has an extensive literature and a wide variety of applications ranging from consumer products and fuzzy control to medical diagnostic systems and fraud detection" -If you're thinking that the idea of fuzzy proofs of nonfuzzy theorems is strange, then I would say that it doesn't, on the face of it, seem to me to be any less strange than proofs by the probabilistic method. - -REPLY [7 votes]: Fuzzy measure theory has applications in pure measure theory. The Choquet capacity theorem is a standard tool for showing the universal measurability of analytic sets. The theory of capacities (or fuzzy measures) is fairly well developed and strongly related to "normal" analysis. -The theory of capacities was not created in the context of fuzzy mathematics, but M. Sugeno developed a form of fuzzy integration in his PhD thesis that shares many formal similarities with the Choquet integral and some work on the Sugeno integral carried over to the Choquet integral. -A rather extensive introduction to these topics is given in the book Generalized Measure Theory by Wang and Klir.<|endoftext|> -TITLE: De Rham cohomology of formal groups -QUESTION [19 upvotes]: Let $G$ be some (dimension $1$, to simplify) formal group over a characteristic $0$ field $K$. The law of $G$ is denoted by $\oplus$. If $w(X) \in K[[X]] dX$ is a differential form, let $F_w(X)$ be the unique power series such that $dF_w=w$ and $F_w(0)=0$. Let $F_w^2(X,Y) = F_w(X \oplus Y) - F_w(X) - F_w(Y)$. Say that $w$ is second kind if $F_w^2$ has bounded coefficients and that $F_w$ is exact if $F_w$ has bounded coefficients. The 1st de Rham cohomology group of $G$ is defined by -$$H^1_{dR}(G)= \text{\{second kind forms\}} / \text{\{exact forms\}}.$$ -Theorem: the group $H^1_{dR}(G)$ has dimension $h$, the height of $G$. -Question: where can I find a proof of this? -The above definitions and theorem are in pages 633-634 of Colmez' "Periodes $p$-adiques des varietes abeliennes" for example, and he refers to Fontaine's book "Groupes $p$-divisibles sur les corps locaux", but without giving a precise reference. Iovita also uses these definitions in "Formal sections and de Rham cohomology of semistable abelian varieties" and refers to chapter V of Katz' "Crystalline cohomology, Dieudonne modules and Jacobi sums". In either case, I can't say that the references have been very helpful. - -REPLY [18 votes]: I have put an updated copy of my formal groups notes here: -http://neil-strickland.staff.shef.ac.uk/courses/formalgroups/fg.pdf -They are not really finished, but the relevant material is discussed in Section 18.<|endoftext|> -TITLE: Consistent r.e. extensions of non r.e. theories. -QUESTION [5 upvotes]: Let $\mathcal{L}$ be some first-order language, and $T$ be a consistent set of formulas of $\mathcal{L}$ which is not recursively enumerable. -Under what conditions will there be $T'\supset T$ such that $T'$ is consistent and recursively enumerable? - -REPLY [2 votes]: Let $A = T \cup R$ where $R$ is any set of refutable statements from $T$ (i.e., $T \vdash \lnot \varphi$ for all $\varphi$ in $R$). A simple necessary condition for such a $T^{\prime}$ to exist is that $A$ is not recursively (computably) enumerable. If it were, then given a r.e. (c.e.) $T^{\prime} \supseteq T$, we could recursively enumerate all statements of $T$ by listing only the ones in $T^{\prime}$ that also appear in $A$ (i.e., $T = T^{\prime} \cap A$). The reason for this equality is that $T^{\prime}$ is an extension of $T$ so it must contain all the statements in $T$ but then it cannot include any in $R$ by its consistency. -Now let $I$ be the set of all statements that are independent of $T$ (i.e. all $\varphi$ such that $T \nvdash \varphi$ and $T \nvdash \lnot \varphi$). In the case that $T$ is deductively closed meaning $T \vdash \varphi$ implies that $\varphi \in T$, a corollary of the above result is that $I$ cannot be the complement of a recursively enumerable set. This is the instance of letting $R$ be the set of all statements refutable from $T$.<|endoftext|> -TITLE: Proving that every graph is an induced subgraph of an r-regular graph -QUESTION [5 upvotes]: How would you prove that every graph $G$ is an induced subgraph of the $r$-regular graph, where $r\geq \Delta(G)?$ -I can picture the answer for when $G$ itself can be turned into a $\Delta$-regular graph: make a union of $G$ with a copy of itself and then connect the vertices across the two vertex sets $U$ (from $G$) and $W$ (from the copy of $G$) such that $u_i$ and $w_j$ are connected if and only if $v_i$ and $v_j$ would be connected in the original graph in order to turn it (the original graph) into a $\Delta$-regular graph. -However, I cannot figure out how to do it in the general case where, for instance, the order of $G$ may be even or odd (and, thus, may not be made into an $r$-regular graph if r is odd as well) or for when $r>\Delta$. (I am also having trouble with the just language of graph theory and how to write proofs for it if you couldn't tell.) - -REPLY [5 votes]: Take two copies $G_1$ and $G_2$ of $G$ and add and edge between each vertex $v$ of $G_1$ and every vertex of $G_2$ corresponding to non-neighbours of $v$. Then $G$ is obviously an induced subgraph, the obtained graph is $|G|-1$ regular and has order $2|G|$.<|endoftext|> -TITLE: Stiefel–Whitney classes in the spirit of Chern-Weil -QUESTION [62 upvotes]: Chern-Weil theory gives characteristic classes (e.g. Chern class, Euler class, Pontryagin) of a vector bundle in terms of polynomials in the curvature form of an arbitrary connection. There seems to be no hope in getting Stiefel-Whitney classes from this method since Chern-Weil gives cohomology classes with real coefficients while Stiefel-Whitney classes have $\mathbb Z/2$ coefficients. Further, since any vector bundle over a curve has vanishing curvature, classes obtained by Chern-Weil can't distinguish, for example, the Mobius bundle from the trivial bundle over the circle (while Stiefel-Whitney classes do). -Nonetheless, I am wondering if there is a more general or abstract framework that allows one to define the Stiefel-Whitney classes in the spirit of Chern-Weil. For example, maybe this is done through a more abstract definition of a connection/curvature. - -REPLY [4 votes]: There exist flat manifolds (i.e., closed Riemannian manifolds with vanishing sectional curvature) which do not admit any spin or spin${}^c$-structure. But since the existence of spin, resp. of spin${}^c$-structures is detected by the second Stiefel-Whitney class, I am strongly led to believe that there is no way of defining these à la Chern-Weil. The reason is that the phrase 'à la Chern-Weil' entails for me that you use in an essential way the curvature tensor, but in these examples it vanishes.<|endoftext|> -TITLE: Online video of some courses -QUESTION [6 upvotes]: Who knows online video of Riemannian Geometry and Commutative Algebra? If you know, please recommend them to me. I am really eager to learn these courses. - -REPLY [3 votes]: There is now a great course on Commutative Algebra by Richard Borcherds on Youtube, it follows the book of David Eisenbud : -Commutative Algebra with a view toward Algebraic geometry -https://www.youtube.com/playlist?list=PL8yHsr3EFj53rSexSz7vsYt-3rpHPR3HB<|endoftext|> -TITLE: Wanted: chain of nowhere dense subsets of the real line whose union is nonmeagre, or even contains intervals -QUESTION [8 upvotes]: Let $X$ be a topological space. When I call a set nowhere dense, meagre or similar without qualification, I mean that it has this property as a subset of $X$. Call a subset of $X$ weager (for weakly meagre) if it is the union of a chain (wrt containment) of nowhere dense sets. Using that finite unions of nowhere dense sets are nowhere dense, it is easy to see that meagre implies weager. Call $X$ an Astaire space (for a stronger Baire space) if every weagre subset of $X$ has empty interior. Obviously Astaire implies Baire. Two natural, but rather silly (not just because of the terminology) questions are: - -Does weagre imply meagre? If not, does Baire imply Astaire? - -Unsurprisingly, the 2nd (and hence also the 1st) question has a negative answer. Let $X$ be uncountable. In fact, for convenience, take $X$ to be the well-ordered set of all countable ordinals. Topologize $X$ by putting open all sets which are either empty or have countable complement. Then $X$ is a Baire space - in fact the notions countable; closed and not $X$; nowhere dense; and meagre all coincide for subsets of $X$. However, $X$ is the union of the chain of all its countable initial segments so $X$ is not an Astaire space. -The above example is somewhat unsatisfactory since the space is far from Hausdorff, but the ease with which it arose made me wonder whether my question had a positive answer even when $X = \mathbb{R}$. Adapting my example, it is at least possible to express an uncountable subset of $\mathbb{R}$ as the union of a chain of countable subsets of $\mathbb{R}$ but this is quite unhelpful because, in this context, there is no guarantee that countable implies nowhere dense, or that uncountable implies nonempty interior (or even nonmeagre for that matter). So that I don't spend too much more time today thinking about things I know nothing about and/or dreaming up silly names for concepts that probably already have much more respectable names - I pose to you the following question: - -Is the real line an Astaire space? If not, are there at least weagre subsets of $\mathbb{R}$ which are not meagre? - -Or, in plain English for those of you who only skimmed this nonsense: - -Does there exist a chain of nowhere dense subsets of $\mathbb{R}$ whose union has nonempty interior? If not, is there such a chain whose union is not meagre? - -Thank you, Michael. - -REPLY [15 votes]: Theorem. There is no chain of nowhere dense subsets of $\mathbb{R}$ whose union contains an interval. -Proof. Suppose there was such a chain $\{\ B_i \mid i\in I\ \}$, where $\langle I,\lt\rangle$ is a linear order and $i\lt j$ implies $B_i\subset B_j$. First, I claim that this chain cannot have countable cofinality, since then we could find a countable cofinal subset of $I$, and the union of the $B_j$ from this cofinal subset would also contain an interval, violating the Baire category theorem. So every countable subset of $I$ is bounded. In this case, consider the set $Q$ of rational numbers $q$ in the interval from the union $\bigcup_i B_i$. Each of them appears in some $B_{i_q}$, and the set of all $i_q$ for $q\in Q$ is a countable set and hence bounded in $I$. Thus, there is some $j\in I$ beyond all $i_q$. So $B_j$ contains all those $q$ and thus is not nowhere dense. QED -Edit. As George pointed out in the comments, essentially the same argument establishes the full property: -Theorem. There is no chain of nowhere dense subsets of $\mathbb{R}$ whose union is non-meager. -Proof. Suppose that $B_i$ for $i\in I$ is a chain of nowhere dense sets whose union $\bigcup_i B_i$ is non-meager. Again, we see that every countable subset of $I$ must be bounded, for otherwise the union would be a countable union of nowhere dense sets and hence meager. Since $\bigcup_i B_i$ is non-meager, it must be dense on an interval, and so it must have a countable subset $Q$ that is also dense on an interval. By the argument above, since $I$ has uncountable cofinality, this set $Q$ must be in some $B_j$ for large enough $j\in I$, contradicting that $B_j$ is nowhere dense. QED - -REPLY [3 votes]: (Joel's answer appeared as I was typing this.) -I think the answer is no. -Suppose to the contrary there exists a nonmeager set $A \subset \mathbb{R}$ which is the union of some chain $\{K_i\}_{i \in I}$ of nowhere dense sets. $A$ is separable, so we may enumerate a countable dense set $\{x_n\} \subset A$. Then we can find an increasing sequence $\{K_{i_n}\}$ with $x_n \in K_{i_n}$. Set $K = \bigcup_n K_{i_n}$. Since $K$ is meager $K \ne A$, so there exists $x \in A \backslash K$. Now there must be some $K_j$ with $x \in K_j$. Now for each $n$ we certainly don't have $K_j \subset K_{i_n}$, so we must have $K_{i_n} \subset K_j$ since the $K_i$ are a chain. Thus $K \subset K_j$, but then $K_j$ contains all the $x_n$ and so is not nowhere dense. -Added: This indeed shows that a second countable Baire space cannot be the union of a chain of nowhere dense subsets. Here is a stab at a counterexample in the non-second countable case. -Consider the non-separable complete metric space $\ell^\infty = \ell^\infty(\mathbb{N})$. I claim its Hamel dimension $\dim \ell^\infty$ is $\mathfrak{c}$. First, we have the natural inclusion $\ell^1 \subset \ell^\infty$; $\ell^1$ is a separable Banach space, so it is known that $\ell^1$ has Hamel dimension $\mathfrak{c}$, and thus $\dim \ell^\infty \ge \mathfrak{c}$. On the other hand, $\ell^\infty$ is the continuous dual of $\ell^1$, and thus is naturally included into the algebraic dual of $\ell^1$, which must also have Hamel dimension $\mathfrak{c}$; thus $\dim \ell^\infty \le \mathfrak{c}$. By Schroeder-Bernstein, $\dim \ell^\infty = \mathfrak{c}$. -Now suppose we believe the continuum hypothesis $\mathfrak{c} = \aleph_1$. Pick a Hamel basis $B$ for $\ell^\infty$; since it is in bijection with the least uncountable ordinal, we can well-order it in such a way that for any $x \in B$, $B_x = \{y \in B : y < x\}$ is countable. Note $B$ has no greatest element, so $\bigcup_{x \in B} B_x = B$. Let $E_x = \mathrm{span } B_x$; clearly $\{E_x\}$ is a chain, and $\bigcup_{x \in B} E_x = \ell^\infty$. But each $E_x$ has countable Hamel dimension and therefore is separable, so it must be nowhere dense in $\ell^\infty$.<|endoftext|> -TITLE: topological type of smooth manifolds with prescribed homotopy type and pontryagin class -QUESTION [8 upvotes]: Can someone help explain the following result: -If the dimension is at least five, there are at most finitely many different smooth manifolds with given homotopy type and Pontryagin classes. -Thank you so much! - -REPLY [11 votes]: In the $1$-connected case, one may argue as follows: -Let $X$ be a closed $1$-connected smooth $n$-manifold, $n \ge 5$. The theory of the Spivak fibration shows that any homotopy equivalence $f: M^n \to X$ with $M$ smooth -is covered by a stable fiber homotopy equivalence of underlying stable tangent spherical fibrations of $M$ and $X$. Call $f$ stably tangential if this -equivalence of stable spherical fibrations lifts to an isomorphism of stable tangent vector bundles. -Then the surgery exact sequence -shows that any stable tangential homotopy equivalence $f: M \to X$ -is homotopic to a diffeomorphism $f': M \sharp \Sigma \to X$, where $\Sigma$ is a homotopy sphere, and $\sharp$ means connected sum. -(You can either quote here Corollary II.3.8 of Browder's book, or you can deduce it directly from the surgery exact sequence. The point is that connected sum gives an action of the homotopy $n$-spheres on the the structure set of $X$, and one can compare the surgery exact sequence for $M$ and the sphere to deduce the above statement.) -To finish the proof of what you want, notice: - -Kervaire and Milnor showed that there are only finitely many homotopy spheres in each dimension $\ge 5$. -The obstruction to the homotopy equivalence being stably tangential is given by its normal invariant, which is an element of $[X,\text{G/O}]$ (it's given by the "difference" between the stable tangent bundles of $M$ and $X$, i.e., $(f^{-1})^*\tau_M - \tau_X$, appropriately interpreted). -The map $\text{G/O} \to B\text{O}$ is a rational homotopy equivalence (by Serre). The image of the normal invariant in $[X,B\text{O}]$ is rationally detected by the difference of the Pontryagin classes of $M$ and $X$ using the fact that $H^*(B\text{O}; \Bbb Q)$ is a polynomial algebra on the $p_i$.<|endoftext|> -TITLE: Eigenvalues of an "oblique diagonal" matrix -QUESTION [16 upvotes]: I am looking for guidance about the behavior of powers of a particular matrix (call it $A_n$ for $n\ge2$), which has come up in a counting problem about quantum knot mosaics (a good reference for quantum knot mosaics is here ). Here's a description of the matrix. It has $4^n$ rows and columns. Instead of a traditional diagonal matrix with its non-zero entries on the main diagonal, the non-zero entries of $A_n$ are on an ``oblique diagonal" of slope $4$, modulo the size of the matrix. More precisely, the non-zero entries occur where -$\left\lfloor\frac{\text{column}}4\right\rfloor=\text{row}\text{, mod}\;4^{n-1}$. -The matrix has sixteen possibly non-zero values $a_1,\ldots,a_{16}$, arranged as follows (with boxes for visual clarity): -$\begin{array}{l|lll|lll|} -\text{row}\backslash\text{column}&0&4&\ldots&4^n/2&4^n/2+4&\ldots\\ \hline -0&\boxed{a_1\,a_2\,a_1\,a_2}\\ -1&&\boxed{a_1\,a_2\,a_1\,a_2}\\ -\vdots&&&\ddots\\ \hline -4^n/8&&&&\boxed{a_3\,a_4\,a_3\,a_4}\\ -4^n/8+1&&&&&\boxed{a_3\,a_4\,a_3\,a_4}\\ -\vdots&&&&&&\ddots\\ \hline -2\cdot4^n/8&\boxed{a_5\,a_6\,a_5\,a_6}\\ -2\cdot4^n/8+1&&\boxed{a_5\,a_6\,a_5\,a_6}\\ -\vdots&&&\ddots\\ \hline -3\cdot4^n/8&&&&\boxed{a_7\,a_8\,a_7\,a_8}\\ -3\cdot4^n/8+1&&&&&\boxed{a_7\,a_8\,a_7\,a_8}\\ -\vdots&&&&&&\ddots\\ \hline -4\cdot4^n/8&\boxed{a_{9}\,a_{10}\,a_{9}\,a_{10}}\\ -4\cdot4^n/8+1&&\boxed{a_{9}\,a_{10}\,a_{9}\,a_{10}}\\ -\vdots&&&\ddots\\ \hline -5\cdot4^n/8&&&&\boxed{a_{11}\,a_{12}\,a_{11}\,a_{12}}\\ -5\cdot4^n/8+1&&&&&\boxed{a_{11}\,a_{12}\,a_{11}\,a_{12}}\\ -\vdots&&&&&&\ddots\\ \hline -6\cdot4^n/8&\boxed{a_{13}\,a_{14}\,a_{13}\,a_{14}}\\ -6\cdot4^n/8+1&&\boxed{a_{13}\,a_{14}\,a_{13}\,a_{14}}\\ -\vdots&&&\ddots\\ \hline -7\cdot4^n/8&&&&\boxed{a_{15}\,a_{16}\,a_{15}\,a_{16}}\\ -7\cdot4^n/8+1&&&&&\boxed{a_{15}\,a_{16}\,a_{15}\,a_{16}}\\ -\vdots&&&&&&\ddots\\ \hline -\end{array}$ -Since $A_n$ has a straightforward geometrical description with non-zero entries only on a diagonal (albeit an oblique one), the following question seems reasonable: - -Is there an elementary way to compute the eigenvalues of this matrix in terms of $a_1,\ldots,a_{16}$ and $n$? - -I'm no expert in tensor algebra, but it seems that $A_n$ might be expressed as the tensor product of $A_2$ with $n-1$ copies of another transformation. Even an approximation of the largest eigenvalue would be useful, but it would be best to avoid the power method with the Rayleigh quotient since I'm trying to analyze the powers of the matrix in terms of the eigenvalues, not vice versa. Any insight into the computation of the eigenvalues of $A_n$ would be greatly appreciated and would go a long way in answering a question in the paper referenced above. - -REPLY [2 votes]: Not a complete answer but a heuristic argument that there is little hope for a good estimation, let alone an exact solution. Take the matrix $$\pmatrix{a&b&.&.&.&.&.&.\\ .&.&a&b&.&.&.&.\\ .&.&.&.&a&b&.&.\\ .&.&.&.&.&.&a&b\\ a&b&.&.&.&.&.&.\\ .&.&a&b&.&.&.&.\\ .&.&.&.&c&d&.&.\\ .&.&.&.&.&.&c&d }$$ which is a special case of an $8\times 8$ "equivalent" (sort of) and has the characteristic polynomial $$x^3\left[x^5-(a+d)x^4+a(d-b)x^3+b(a+d)(ad-bc)x-b(ad-bc)^2\right].$$ Now, the factor in brackets is of form $x^5+px^4+qx^3+rx+s$ with linearly independant coefficients (in terms of $a,b,c,d$), i.e. a quintic in almost general form, with just the $x^2$ term missing. AFAIK this has no closed form solution, and I wonder if there are reasonable estimates of its maximal root in terms of $p,q,r,s$.<|endoftext|> -TITLE: What is the wild fundamental group? -QUESTION [12 upvotes]: In the abstract of - -Singularités irrégulières Correspondance et documents - Pierre Deligne, Bernard Malgrange, Jean-Pierre Ramis - Documents mathématiques 5 (2007), xii+188 pages (link) - -there is a reference to 'wild' fundamental groups, in contrast to Galois fundamental groups. What is the wild fundamental group? It is in the setting of analysis of singularities of differential equations and Stokes phenomena. -(If there are any tags that do/don't belong, don't hesitate to edit). - -REPLY [20 votes]: Consider the category of algebraic integrable connections on a smooth connected algebraic variety $X$ with a base point $x$ (over a field of characteristic zero). This is a Tannakian category with a fibre functor given by taking fibres over $x$ and is hence equivalent to the representations of some pro-algebraic group. This group is the wild fundamental group. One may consider the subcategory of integrable connections with regular singularities (which is a condition at infinity wrt $X$). This subcategory (when the base field is the complex numbers) is equivalent to the category of finite dimensional complex representations of the fundamental group and is hence described by the "Tannakian hull" of the fundamental group, in general a very large pro-algebraic group. In any case, the fact that the regular connections form a Tannakian subcategory makes this Tannakian hull a quotient of the wild fundamental group. The kernel somehow reflects the possible irregularities of general connnections on $X$. These can also be described exactly in terms of Stokes phenomena. -The reason why one would be interested in the wild fundamental group and not just in irregular connections per se (and their associated Stokes phenomena) is on the one hand that it essentially gives a description of the differential Galois group of integrable connections, on the other hand that there is a very visible but still imperfectly understood connection (sic!) between irregular singularities and wild ramification in positive characteristic. This incidentally is the reason for the terminology "wild". -Addendum: As for references there are several books by N. Katz (in the Annals of Mathematics Study series) that take up both the differential Galois group side and characteristic $p$ side of (what should be) the same systems. They give, I feel, a good feeling for how these ideas can be used in concrete situations. There are of course also several relevant books on differential Galois theory. However, many of them tend to be less geometric (in particular the (sometimes only implicitly chosen) base point is usually the generic point of $X$ which gives a less geometric flavour to the whole business). The book by van der Put and Singer; "Galois theory of linear differential equations" tries to find some middle road.<|endoftext|> -TITLE: Intuition for the Hardy space $H^1$ on $R^n$ -QUESTION [15 upvotes]: the standard intuition for Lebesgue spaces $L^p(\mathbb R^n)$ for $p \in [1,\infty]$ are measurable functions with certain decay properties at infinity or at the singularities. -In particular, a typical $L^p$ function is a function like $|x|^{-\alpha}$ with $\alpha > \frac{n}{p}$ for $p$ < $\infty$, or simply $1$ for $p = \infty$. Furthermore, functions which are almost everywhere absolute-value-dominated by an $L^p$ function are elements of $L^p$, too. This is useful for approximation arguments, as the pointwise error has just to be dominated by an $\epsilon$-multiple of another $L^p$ function. -In contrast to that, hardy spaces seem to be less intuitive due to cancellation properties. Hence I wonder: - -How do "typical" $H^1$ functions look like? -In particular, what can you typically "do" with Hardy functions? - -For example, I guess convolution arguments take a more prominent role in approximation arguments than perturbation arguments do, but I am not sure about that. -The hardy space $H^1$ shall be defined via -$f \in H^1 $ if $f \in L^1_{\mathrm{loc}}$ and $Mf \in L^1$, where -$$Mf(x) := \sup \limits_{B_r(x_0) \ni x, \phi \in \mathcal L(B_r(x_0))} \int f \phi dx$$ -$$\mathcal{L}(B_r(x_0)) = \left\{ \phi \in C(B_r(x_0)) s.t. |\phi(x)| < \dfrac{\max(r-|x-x_0|,0)}{r|B_r(x_0)|} , \mathrm{Lip}(\phi) < \frac{1}{r|B_r(x_0)|} \right\}$$ - -REPLY [16 votes]: In many ways $H^1$ is just a natural substitute for $L^1$. -A typical $H^1$ function is a $1$-atom, i.e. a function $\phi\in L^1(\mathbb R^n)$ such that the support of $\phi$ is contained in some ball $B(a,r)$, the bound -$$\sup|\phi(\cdot)|\leq \frac{1}{m(B(a,r))}$$ -holds true, and -$$\int_{\mathbb R^n} \phi(x)dx=0.$$ -In fact, following a suggestion by Fefferman, one can use $1$-atoms to characterize the whole Hardy space $H^1$. More precisely, the following is true. - -Theorem. A function $f$ belongs to $H^1(\mathbb R^n)$ if and only if there exist $1$-atoms $\phi_k$ and complex numbers $c_k$ such that - $$f=\sum_{k=1}^{\infty}c_k\phi_k,$$ - where the convergence is in $L^1$ and $\sum_{k}|c_k|<\infty.$ - -One should note that the mean-value condition in the definition of $1$-atom plays a key role here. If it is removed from the definition, the above description with sums would result in the standard Lebesgue space $L^1$. -$H^1$ is a good testing field to study various classes of multiplier transformations. For instance, the singular integral operators of the form -$$Tf=\lim_{\epsilon\to0}\int_{|y|>\epsilon}\frac{K(x-y)}{|x-y|^n}f(y)dy$$ -where $K$ is a smooth homogeneous function of degree $0$ such that $\int_{|y|=1}K(y)dy=0$, extend to bounded operators on $H^1$. In particular, the Riesz transforms are bounded on $H^1$. Moreover, an integrable function $f$ belongs to $H^1$ if and only if its Riesz transforms $R_jf$ are also in $L^1$.<|endoftext|> -TITLE: Given a prime $p$ how many primes $\ell2$, how many primes $\ell < p$ there exist which are quadratic residues mod $p$? -(2) Given a prime $p>2$, how many primes $\ell < p$ there exist which are quadratic nonresidues mod $p$? -As for (1) I can prove $\gg\log p/\log\log p$ by an elementary argument. Indeed, put $p':=(-1)^{(p-1)/2}p$ and observe, by quadratic reciprocity, that a prime $\ell\neq p$ divides some value $x^2-p'$ for $x\in\mathbb{Z}$ if and only if $\ell$ is a quadratic residue mod $p$. Now consider $|x^2-p'|$ for $0 < x < \sqrt{p}$: these are integers in $(0,p)$ or $(p,2p)$ depending on $p$ mod $4$. At any rate, these numbers are built up from the $k$ primes enumerated under (1), and their number is $\gg\sqrt{p}$. As each of the $k$ prime exponents is $\ll\log p$, we conclude $\sqrt{p}\ll(\log p)^k$ and my claim follows. -Added 1. As Anonymous pointed out, we should restrict to odd $0 < x < \sqrt{p}$, and talk about the odd part of $|x^2-p'|$. In addition, using the upper bound part of (7.16) on p. 203 of Montgomery-Vaughan: Multiplicative Number Theory (proof on pp. 204-208), we can see $k>(\log p)^{2-o(1)}$ for the number of primes under (1). -Added 2. Regarding (2), Lucia pointed out that $\gg p^\delta$ follows with a decent $\delta>0$ from a result of Bourgain and Lindenstrauss. I found this response very satisfactory, and I accepted it officially. Still, I would welcome any further developments regarding the above questions (1) and (2). -Added 3. The recent preprint of Paul Pollack contains several nice new results and valuable historic references regarding the above two questions. An even more recent preprint by him and Kübra Benli settles (1) in the sense that there are $\gg p^{1/9}$ prime quadratic residues $\ell0$ such that ($\ell$ runs over primes below) -$$ -\sum_{N^{\alpha} <\ell < N; (\frac{\ell}{p}) = -1} \frac{1}{\ell} \ge \frac 12 -\epsilon. -$$ -In particular the number of primes $\ell$ with $(\frac{\ell}{p}) =-1$ is trivially at least $(1/2-\epsilon)N^{\alpha}$. Now use this with $N=p$, and we deduce the result mentioned above. I didn't check the details, but I think one can get a pretty decent value of $\delta$ above -- maybe even as big as $3/8$ (the level of distribution is like $p^{\frac 34}$ and the sifting limit should be $1/2$).<|endoftext|> -TITLE: Are there topological restrictions to the existence of almost quaternionic structures on compact manifolds? -QUESTION [21 upvotes]: I start with some background, but people familiar with the subject may jump directly to the question. -Let $M^{4n}$ be a compact oriented smooth manifold. Recall that an almost hypercomplex structure on $M$ is a 3-dimensional sub-bundle $Q\subset End(TM)$ spanned by three endomorphisms $I$, $J$ and $K$ satisfying the quaternionic identities: $I^2=J^2=-Id$, $IJ=-JI=K$. -An almost quaternionic structure on $M$ is a 3-dimensional sub-bundle $Q\subset End(TM)$ which is locally spanned by three endomorphisms with the above property. -In both cases one may assume (by an averaging procedure) that $M$ is endowed with a Riemannian metric $g$ compatible with $Q$ in the sense that $Q\subset End^-(TM)$, i.e. $I$, $J$ and $K$ are almost Hermitian. Using this one sees that an almost hypercomplex or quaternionic structure corresponds to a reduction of the structure group of $M$ to $\mathrm{Sp}(n)$ or $\mathrm{Sp}(1)\mathrm{Sp}(n)$ respectively, but this is not relevant for the question below. -Notice that in dimension $4$ every manifold has an almost quaternionic structure (since $\mathrm{Sp}(1)\mathrm{Sp}(1)=\mathrm{SO}(4)$), but there are well-known obstructions to the existence of almost hypercomplex structures. For example $S^4$ is not even almost complex. Finally, here comes the question: - -Are there any known topological obstructions to the existence of almost quaternionic structures on compact manifolds of dimension $4n$ for $n\ge 2$? - - -EDIT: Thomas Kragh has shown in his answer that there is no almost quaternionic structure on the sphere $S^{4n}$ for $n\ge 2$. I have found further obstructions in the litterature and summarized them in my answer below. - -REPLY [8 votes]: I know this is a bit late, but as you mentioned Cadek's and Vanzura's paper, I'd like to point out (selfishly?) that there's also my paper which uses a bit of their work and gives some integrality conditions on the existence of quaternionic structures on closed manifolds -- an example is stated below. I should emphasize that I really mean honest quaternionic not just almost quaternionic here, although the referee believed that the same should hold for only almost quaternionic structures too. -Theorem: Let M be an 8-dimensional compact quaternionic manifold with Pontryagin classes $p_1(TM)$ and $p_2(TM)$ and a fundamental class $[M]$. Then the following expressions are integers $$\biggl(\frac{143}{960}p_{1}^{2}-\frac{89}{240}p_{2}\biggr)[M], \quad \biggl( -\frac{17}{480}p_{1}^{2}+\frac{71}{120}p_{2}\biggr)[M].$$<|endoftext|> -TITLE: Are there (-2)-curves on an Enriques surface? -QUESTION [6 upvotes]: Let $X$ be an Enriques surface. A $(-2)$-curve is an irriducible rational curve on X such that $C^2 = -2$. By Proposition [VIII,16.1] from Barth-Peters-Van de Ven, we have that if $D^2 = -2$, then it is a $(-2)$-curve, but do such curves exist? - -REPLY [14 votes]: As explained in J.C. Ottem's answer, the generic Enriques surface contains no smooth rational curves at all. -However, it can happen that some special Enriques surface $X$ contains $(-2)$-curves, and also infinitely many of them (see this paper by Cossec and Dolgachev). The maximal number of disjoint $(-2)$ curves on $X$ is eight, and Enriques surfaces with eight disjoint $(-2)$-curves are classified in the article -Mendes Lopes, Margarida; Pardini, Rita -Enriques surfaces with eight nodes -Math. Z. 241 (2002), no. 4, 673–683. -The authors first show that, setting $C_1, \dots,C_8$ to be the exceptional $(-2)$-curves of $X$, the divisor $C_1+\dots+C_8$ is divisible by $2$ in the Picard group of $X$, or equivalently there exists a double cover $\widetilde{X} \to X$ branched exactly over them. -The main theorem then states that an Enriques surface with eight disjoint $(-2)$-curves is isomorphic to $X=D_1\times D_2/G$, where $D_1,D_2$ are elliptic curves and $G$ is either $\mathbb{Z}_2^2$ or $\mathbb{Z}_2^3$. - -REPLY [10 votes]: It is well-known that (at least over $k=\mathbb{C}$) that the generic Enriques surface does not contain any smooth rational curves at all. This can be seen, for example, using the global Torelli theorem for Enriques surfaces. For a complete proof, see -Barth, W., Peters, C.: Automorphisms of Enriques surfaces. Invent. Math. 73, 383-411 (1983). -However, as Fransesco's answer shows, there are Enriques surfaces containing rational curves. Moreover, it is also known that once $S$ contains a rational curve, then generically it contains infinitely many. The reason for this is basically since the automorphism groups of Enriques surfaces tend to be very large. -In fact, Cossec and Dolgatchev proved the following surprising result about rational curves on an Enriques surface: - -Let $S$ be an Enriques surface of degree -$d$ in a projective space $\mathbb{P}^n$. Assume -that S contains a smooth rational -curve, then it contains such a curve -of degree less or equal to $d$. - -This implies for example that the subset of the Hilbert scheme parametrizing Enriques surfaces of degree $d$ in $\mathbb{P}^n$ containing smooth rational curves is a constructible subset. - -REPLY [4 votes]: Since an Enriques surface is elliptic, it may have (-2)-curves as a component of singular fibers. You may refer S. Kondo, Enriques surfaces with finite automorphism groups. Japan. J. Math.(N.S.) 12 (1986), no. 2, 191--282. In the paper Kondo constructed explicitly many examples of Enriques surfaces with finitely many (-2)-curves.<|endoftext|> -TITLE: Locally constant sheaves for the étale topology, lack of intuition about "étale-localness" -QUESTION [18 upvotes]: I have started studying some étale cohomology and I am trying to build up some intuition about the concept of local for the étale topology. I can understand some nice examples (like Kummer exact sequence) but I am still quite confused by some "easy" notions such as locally constant sheaves. -I believe that an étale sheaf which is étale locally isomorphic to the same constant sheaf should be also globally isomorphic to that constant sheaf if the isomorphisms verify some cocycle condition, but here is a toy example which seems to contradict this: -Let $k$ be a field, $n$ an integer invertible in $k$ and assume that $k$ does not contain all $n$-th roots of unity. Now consider the two following étale sheaves on $X=Spec\; k$: - -The sheaf of n-th roots of unity $\mu_n$; -The constant sheaf $\mathbb Z/n \mathbb Z$. - -They are not isomorphic since their sections on $Spec\; k$ are different, but they become isomorphic after some finite separable extension of scalars so they are isomorphic étale locally. To be precise, $U=Spec(k[T]/(T^n-1))$ is an étale cover of $X$ such that the pullbacks of the two sheaves are isomorphic. -Why are this two sheaves locally isomorphic but not isomorphic? -Is it normal that this isomorphism doesn't "patch"? (which would imply that the sheaves over the small étale site on $Spec\; k$ don't form a prestack) - -If I try to think to all this "stalkwise", changing to the point of view of topoi, (I'm not very familiar with the theory of topoi so please correct me if I am writing nonsense) I believe that: -the topos of sheaves over $Spec\;k$ with the small étale site has enough points, a family of conservative points consisting of just one element (the étale local ring is some separable closure $k^{sep}$ of $k$); and on this local ring the two sheaves above coincide. -It should follow that as soon as we have a morphism of sheaves inducing this isomorphism on the stalk the two sheaves should be isomorphic, which is not the case. -Is it just because we don't have such a morphism or am I missing something more fundamental here? - -REPLY [5 votes]: The short answer is that locally isomorphic things needn't be globally isomorphic, and this -isn't specific to the etale topology. Let me spell it out for locally -constant sheaves of vector spaces on an ordinary (sufficiently nice) topological space $X$. -Such sheaves correspond to -representations of the fundamental group (see Why are local systems and representations of the fundamental group equivalent). Two locally constant -sheaves $F$ and $G$ of the same rank are locally isomorphic, and in fact they pullback to -to isomorphic sheaves on the universal cover $\tilde X\to X$. -However, they won't be isomorphic unless the corresponding representations match. -This is entirely analagous to the example of the nonisomorphic sheaves -$\mathbb{Z}/n\mathbb{Z}$ and $\mu_n$ pulling -to isomorphic sheaves on $Spec( k^{sep})$. -(As I was writing this, I realize that Emerton has already given an answer, but perhaps two is better than none.)<|endoftext|> -TITLE: Is there Harer stability for moduli of curves with level structure? -QUESTION [13 upvotes]: The famous Harer stability theorem asserts that the homology group $H_d(\mathcal{M}_{g,n},\mathbf{Z})$ is independent of g and n in the range $0 \leq 2d < g-1$. This is proven by analyzing the maps of mapping class groups $\Gamma_{g,n}\to \Gamma_{g+1,n}$ given by gluing a torus with a disk removed to a boundary circle (when $n \geq 1$), and $\Gamma_{g,n} \to \Gamma_{g,n-1}$ by gluing in a disk, and showing that these maps induce an isomorphism on homology in low dimensions (regardless of the choices involved in writing down such maps). -By considering curves with level structures, one obtains finite covers of $\mathcal{M}_{g,n}$, or equivalently, finite index subgroups of the mapping class group. So let's consider a finite group G, and denote by $\mathcal{M}_{g,n}[G]$ the moduli space parametrizing n-pointed smooth curves of genus g equipped with an étale G-torsor. Is the corresponding statement for $H_d(\mathcal{M}_{g,n}[G],\mathbf{Z})$ known? It is not hard to write down analogues in this context of the maps of mapping class groups above. -Remark: The corresponding statement for moduli of spin curves is known (and is also a theorem of Harer), so one might hope for a statement like this because of the similarities between the spaces of r-spin curves and the spaces of curves with $G=\mathbf{Z}/r\mathbf{Z}$ level structure. - -REPLY [15 votes]: This is a hard open problem. Essentially nothing is known except for linear congruence subgroups. Denoting by $Mod_{g,n}(L)$ the level $L$ linear congruence subgroup, the desired result is only known for $H_1(Mod_{g,n}(L);\mathbb{Q})$ (which is due to Hain) and for $H_2(Mod_{g,n}(L);\mathbb{Q})$ (which is due to me). See my paper "The second rational homology groups of the moduli space of curves with level structures" (available on my webpage). -I should also remark that the corresponding result is false if you replace $\mathbb{Q}$ with $\mathbb{Z}$, even for $H_1$. See Theorem F of my paper "The Picard group of the moduli space of curves with level structures". -One observation that is worth making (and I make it in my paper "The second rational homology...") is that if $H_k(Mod_{g,n}(L);\mathbb{Q})$ stabilizes, then we have an isomorphism $H_k(Mod_{g,n}(L);\mathbb{Q}) \cong H_k(Mod_{g,n};\mathbb{Q})$. Indeed, since we are dealing with finite-index normal subgroups, the Hochschild-Serre spectral sequence collapses and gives that $H_k(Mod_{g,n};\mathbb{Q})$ is isomorphic to the co-invariants of the action of $Mod_{g,n}$ on $H_k(Mod_{g,n}(L);\mathbb{Q})$. However, stability implies that any Dehn twist acts trivially on $H_k(Mod_{g,n}(L);\mathbb{Q})$ (draw the picture -- the homology is entirely supported "away" from the simple closed curve), so we get the desired isomorphism. - -REPLY [6 votes]: This is not an answer to your question, but is directly related to your remark so I thought I should mention it. -I have recently proved, though I am afraid that it has not appeared yet, that moduli spaces of $r$-Spin curves exhibit homological stability. However, the truth of this statement depends sensitively on what one means by "moduli spaces of $r$-Spin curves": -If one takes the moduli stack $\mathcal{M}_{g}^{1/r}$ that represents families of Riemann surfaces equipped with a line bundle $\ell$ on the total space and a chosen isomorphism $\ell^{\otimes r} \cong \omega$ to the fibrewise cotangent bundle, then all is well and one has integral homology stability. However, if one takes the "rigidification" $\widetilde{\mathcal{M}}_{g}^{1/r}$ obtained by killing the natural $\mathbb{Z}/r$-worth of automorphisms of every object, the homology does not stabilise integrally, though it does over $\mathbb{Z}[1/r]$. In fact, even the first homology of $\widetilde{\mathcal{M}}_{g}^{1/r}$ jumps around all over the place. -The (orbifold) fundamental group of $\widetilde{\mathcal{M}}_{g}^{1/r}$ at some $r$-Spin surface $(\Sigma, \ell)$ may be identified with the subgroup $\widetilde{\Gamma}_g^{1/r} \leq \Gamma_g$ of those mapping classes which preserve the $r$-Spin structure $\ell$ up to isomorphism. Consequently, the groups $\widetilde{\Gamma}_g^{1/r}$ do not have integral stability. It is only a certain extension of these groups by $\mathbb{Z}/r$ which has integral stability.<|endoftext|> -TITLE: Consecutive numbers with n prime factors -QUESTION [10 upvotes]: Let $P(m,n)$ mean that there is a number, $M$, such that starting with $M$ there are $m$ consecutive numbers each having exactly $n$ distinct prime factors. Is it obvious that $P(m,n)$ is true for all $m$ and $n$? My gut says "obviously" and $P(4,4)$ and $P(5,5)$ are definitely true (for 134043 and 129963314 respectively). It seems like some sort of pigeonhole proof based on the number of factors available might work, but upon reflection, I'm not so sure. Maybe I'm missing something obvious. - -REPLY [3 votes]: Check out the related question Happy New Prime Year! . I have some code posted there which tracks constant sequences as well as increasing and decreasing sequences. (Check out 2302 to 2308.) -A comment made by someone else and then deleted contained the observation that multiples m of the nth primorial had s(m) >= n, so that runs of values less than n must have length less than the nth primorial. Also, if you look at multiples of 6, you get that s(m)=2 for at most 11 consecutive values instead of at most 29, so there is room for improvement in -the upper bound to such lengths. -Gerhard "Reduce, Reuse, Recycle for Rep" Paseman, 2011.01.18<|endoftext|> -TITLE: Rational curves on varieties of general type -QUESTION [14 upvotes]: Let $S$ be a complex surface of general type. Are there infinitely many smooth rational curves on $S$? And more general, what if $V$ is a variety of general type? - -REPLY [2 votes]: Some related additional comments: -Siu-Yau showed that when the bisectional curvature of a compact K\"ahler variety is positive then there exists a rational curve. -Birkar, et al, showed that when $X$ is a Moishezon manifold which is not projective. Then $X$ contains a rational curve. Moreover we have the same result when $X$ -is a normal Moishezon variety with analytic $\mathbb Q$-factorial singularities. -Existence of rational curve is related to Gromov-Witten Invariants de to Ruan-Tian. -Let $X$ be a smooth projective variety over $\mathbb C$. Fix a point $x\in X$, a homology class $A\in H_2(X(\mathbb C),\mathbb Z)$ and very ample divisors in general position $H_i\subset X$, $i=1,...,k$. -Let $y_0,...,y_k\in \mathbb CP^1$ be general points. Take maps $f:\mathbb CP^1\to X$ such that $f_*[\mathbb CP^1]=A$, $f(y_0)=x$ and $f(y_i)\in H_i$ -Ruan-Tian, introduced an invariant -$$\tilde F_{A,X}(x,H_1,...,H_k;y_0,...,y_k):={\text{number of such maps}}$$ -Gromov-Witten theory of pseudo-holomorphic curves shows that one can make a similar definition where $X$ is replaced by a symplectic manifold $(M,\omega)$. Then the corresponding invariant is denoted by $$\tilde \Phi_{A,X}(x,H_1,...,H_k;y_0,...,y_k):={\text{number of such maps}}$$ -Ruan-Tian, showed that if $\tilde \Phi_{A,X}(x,H_1,...,H_k;y_0,...,y_k):={\text{number of such maps}}$ then there is a rational map $f:\mathbb CP^1\to X$ such that $f_*[\mathbb CP^1]=A$, $f(y_0)=x$ and $f(y_i)\in H_i$ -Moreover, $\tilde \Phi_{A,X}(x,H_1,...,H_k;y_0,...,y_k)=\tilde F_{A,X}(x,H_1,...,H_k;y_0,...,y_k)$ if the following conditions are satisfied. - -If $C_1,...,C_m\subset X$ are rational curves such that $\sum [C_i]=A$ and $x\in C_1$, then $m=1$ -If $g:\mathbb CP^1\to X$ is any map such that $g_*[\mathbb CP^1]=A$, $g(y_0)=x$ then $H^1(\mathbb CP^1,g^*TX)=0$ - -Existence of rational curves are very important in the classification of the solution of K\"ahler Ricci flows . -i.e an effective way to consider Green-Griffiths-Lang conjecture is using classification of the solutions of relative Kahler Ricci flow and study of existence of rational curve appears,<|endoftext|> -TITLE: Theory of addition and a predicate that recognizes powers of 2 -QUESTION [5 upvotes]: What is the complexity of the theory of addition (Presburger arithmetic) augmented by a unary predicate that recognizes powers of 2? - -REPLY [7 votes]: The theory of the natural numbers with addition and $x\mapsto 2^x$ is decidable. -One reference is the Cherlin-Point paper "On extensions of Presburger arithmetic". -It can be found on Francoise Point's webpage: -http://www.logique.jussieu.fr/~point/papiers/cherlin_point86.pdf<|endoftext|> -TITLE: Zeroes of complete L-functions -QUESTION [9 upvotes]: Hello, -Let $F$ and $G$ be two functions belonging in the Selberg class, $\Lambda_{F}$ and $\Lambda_{G}$ the complete L-functions associated to $F$ and $G$. I would like to know whether this assertion is true or not: -"$\Lambda_{F}$ and $\Lambda_{G}$ have the same zeroes if and only if $F=G$ or $F=\overline{G}$." -Thank you in advance. - -REPLY [7 votes]: Expanding on Stopple's comment above, I believe the following argument based on Landau's explicit formula answers the question. -Here is a generalization of Landau's explicit formula for the zeros of the Riemann zeta-function which is exercise 8.4.8 in M. Ram Murty's book Problems in Analytic Number Theory: Let $F$ be in the Selberg class, $n>1$ be a positive integer, and $T>1$. Then -$$ \sum_{|\gamma|\leq T} n^\rho = -\frac{T}{\pi}\Lambda_F(n) + O( n^{3/2}\log T )$$ -where $\rho=\beta+i\gamma$, $\beta>0$, runs over the non-trivial zeros of $F(s)$. Here the coefficients $\Lambda_F(n)$ are defined by -$$ -\frac{F'}{F}(s) = \sum_{n=1} \frac{\Lambda_F(n)}{n^s}.$$ -Now suppose that $F$ and $G$ are in the Selberg class and have the same zeros (with multiplicity). Then we deduce from Landau's formula that -$$ |\Lambda_F(n) - \Lambda_G(n)| \ll \frac{n^{3/2}\log T}{T} $$ -for all $n>1$. Letting $T\rightarrow \infty$, it follows that -$\Lambda_F(n) = \Lambda_G(n)$ for all $n>1$. This implies that $F=G$.<|endoftext|> -TITLE: All polynomials over a finite field are sums of $2$ square-free polynomials -QUESTION [9 upvotes]: I quote Proposition 2.3, page 14 lines -3 and -4 of Michael Rosen's book -Number Theory in Function Fields: -Let $b_n$ be the number of square-free monics in $A= \mathbb{F}_q[t]$ of degree $n.$ -Then $b_1=q$ and for $n>1$, $b_n=q^n-q^{n-1}.$ -Using the proposition it is easy to prove that any polynomial $P \in A$ can be written -$$ -P = S_1+S_2 -$$ -where $S_1$ and $S_2$ are square-free, provided -$$ -q \neq 2 -$$ -Question: What happens when $q=2.$ - -REPLY [6 votes]: Alright, putting together all the comments it appears to still be true when $q = 2$, depending on how you like your square-free polynomials. -First, to quote Sonia's comment, If $P$ is not sq.free, then for each sq.free polynomial $Q$ of degree $n$, we have a non-zero $P−Q$, of degree $\leq n−1$, $2^{n−1}$ in all. The number of sq.free polynomials of degree $\leq n−1$ is $2^{n−1}$. So there must be nonempty intersection. – Sonia Balagopalan -Now, suppose $P$ is square-free and has degree $n$. For any square-free polynomial of lesser degree, $Q$, $P-Q$ is a polynomial of degree $n$. There are $2 + 2 + 2^4 + \ldots + 2^{n-2} = 2^{n-1}$ choices for such $Q$, but there are only $2^{n-1}$ non-square-free polynomials of degree $n$ and we certainly did not land on $0$, so at least one of these must be another square-free polynomial of degree $n$. -So any polynomial of degree $n \geq 3$ can be written as the sum of two square free polynomials of degrees $n$ and $d$ where $d < n$. What about when $n = 2$? The only polynomials to consider are -$P_1 = t^2,$ -$P_2 = t^2 + 1,$ -$P_3 = t^2 + t$ and -$P_4 = t^2 + t + 1$. -If you want to add the restriction that our two square free polynomials MUST have degree at most 2, as Sonia claims, $P_3$ and $P_4$ are counterexamples. Otherwise we may write -$P_1 = (t^2 + t) + (t)$, -$P_2 = (t^2 + t) + (t + 1)$, -$P_3 = (t^2 + t^7 + 1) + (t + t^7 + 1)$ and -$P_3 = (t^2 + t^7 + t) + (t^7 + 1)$. -(Note: There is nothing too special in picking the $7$.) -Upon further thought, you can do what you want with the degree 1 polynomials. A similar trick using a larger degree polynomials will work, or you can use those as counterexamples to a restricted degree version if you like. Although trivial, you may also want to express $0$ and $1$ as sums, which can certainly be done as $z$ and $z+1$ are both square-free.<|endoftext|> -TITLE: Topology of black holes -QUESTION [10 upvotes]: I've asked this question of some physicist friends of mine and I've never gotten a satisfactory answer: What is topologically possible for a neighborhood of a black hole? To clarify, I'm curious about the topology as a 4-manifold, although I'd also be interested to hear about time-like and space-like slices as well. I've heard that a time-like slice of the event horizon can be a torus or sphere, but this isn't really what I'd like to know, although I imagine that there is a close connection between the topology of the event horizon (as a 3-manifold) and the topology of a neighborhood of the black hole. Please ask if any clarifications are needed. - -REPLY [15 votes]: So far J Verma and RBega provided two succint descriptions on topology of the apparent horizon itself (for any space-time admitted trapped regions), and so by association, the topology of the event horizon in a stationary asymptotically flat black-hole space-time. -I'll try to provide an answer to an interpretation of the question you asked for: namely that of the topology of a neighborhood of the event horizon. That I have to interpret the question is because you have not actually provide a description of what you mean by a neighborhood. Using the Hawking topology theorem you can easily manage that the topology of the event horizon is $\mathbb{S}^2\times \mathbb{R}$, and is an embedded null hypersurface. So a tubular neighborhood of it necessarily has the topology $\mathbb{S}^2\times \mathbb{R}^2$, which is, for one thing, simply connected. But there is absolutely nothing to prevent you from choosing a neighborhood to be some arbitrary open set containing the event horizon with complicated topology. Indeed, you can easily imagine removing a four dimensional tube disjoint from the event horizon from the $\mathbb{S}^2\times\mathbb{R}^2$ set to get something that is not simply connected. -Because of this freedom to choose subsets, the naive reading of you question leads to the answer that "pretty much as bad as you want". -But that answer is rather physically useless: it doesn't capture anything essential about black holes. In fact, the answer given above is identical to the answer to the following question: let $U$ be an open connected subset of $\mathbb{R}^4$, what kinds of topology can $U$ admit? -A more useful question to ask is: given an isolated gravitating body (such as a black hole), what is the topology of the space-time outside of it? And that question is one admitting a good answer. The content is the topological censorship theorem. In physicist's language, to quote Friedman, Schleich, and Witt, - -general relativity does not allow an observer to probe the topology of spacetime: Any topological structure collapses too quickly to allow light to traverse it. - -An early version of this is due to Gannon, who showed that Cauchy hypersurface with non-trivial topology will necessarily generate a development which is null geodesically incomplete. The FSW paper showed that under some restrictions, all the non-trivial topology must be hidden behind the event horizon. -A stronger generalisation of the topological censorship theorem is due to Galloway, who later, with Schleich, Witt, and Woolgar, extended the result from asymptotically flat space-times to also asymptotically anti-deSitter ones. One interesting crucial assumption of these theorems is the requirement for null or time-like "Scri". -A somewhat related paper is this one by Schleich and Witt which I didn't read in detail so cannot say more about.<|endoftext|> -TITLE: Vanishing Trace -QUESTION [7 upvotes]: Let $\mathcal H$ be a Hilbert space, and let $a \in \mathcal B(\mathcal H)$ satisfy $\mathrm{Tr}(a)=0$. If $a$ is self-adjoint, then we can find a vector $\xi \in \mathcal H$ such that $\langle \xi | a \xi \rangle=0$. Is this true in general? Is it always possible to find an orthonormal basis of such vectors? - -REPLY [6 votes]: The answers to both your questions are yes. (ie for any trace class operator $a$ on a Hilbert space $H$ with $Tr(a)=0$, there exists an orthonormal basis of $H$ consisting of vectors $\xi$ such that $\langle a \xi,\xi\rangle=0$). -First note that the fact that there exists an orthonormal basis of vectors $\xi$ such that $\langle a \xi,\xi\rangle=0$ is immediate (by something like "transfinite induction") from the fact that for any $a$ with trace zero, there exists one non-zero vector $\xi$ with $\langle a \xi,\xi\rangle=0$. Indeed, Zorn's Lemma implies that there exists a maximal closed subspace $H$ of $K$ such that there is an orthonormal basis of $K$ consisting of vectors such that $\langle a \xi,\xi\rangle=0$. If $K$ were different from $H$, consider $P$ the orthonormal projection on the orthogonal of $K$. The restriction of $Pa$ to the orthogonal of $K$ is still of trace class and zero trace, so there exists a unit vector in the orthogonal of $K$ such that $\langle a \xi,\xi\rangle=0$, which contradicts the maximality of $K$. -Now to the first point. You want to prove that is $Tr(a)=0$, zero belongs to the numerical range $W(a)$ of $a$, which is the set of $\langle a \xi,\xi\rangle$ for $\xi$ in the unit sphere of $H$. $W(a)$ is a convex subset of $\mathbb C$ for any operator $a$. This is usually stated for matrices (see this page), but since everything happens in two dimensions this remains true in general. -But the assumption $Tr(a)=0$ implies that there is a sequence $(\lambda_n)_n$ in $W(a)$ such that $\sum_n \lambda_n=0$. This clearly implies that $0$ belongs to the convex hull of the $\lambda_n$'s, which is contained in $W(a)$. -Edit: To answer Bill's objection, here is a proof that if $\sum_n \lambda_n=0$, then $0$ belongs to the convex hull of the $\lambda_n$'s. (I agree that for a general compact operator, $0$ does not always belong to $W(a)$ but only to its closure, but my point was that here the assumption $Tr(a)=0$ makes it different). -Assume, by contradiction, that $0$ does not belong to the convex hull of the $\lambda_n$'s. By (some form of) Hahn-Banach, the $\lambda_n$'s all belong to some closed half-plain, say for simplicity $Im(\lambda_n)\geq 0$ for all $n$. The equality $\sum_n Im(\lambda_n)=0$ then implies that the $\lambda_n$'s are all real. They cannot be all positive (or all negative), otherwise $\sum_n \lambda_n \ne 0$. This is a contradiction.<|endoftext|> -TITLE: Finite groups generated by pseudo-reflections -QUESTION [6 upvotes]: A theorem of Chevalley, Shepard, and Todd states that if $G$ is a finite group and $\rho: G \rightarrow GL_n(\mathbb{C})$ a representation so that $\rho(G)$ is generated by pseudoreflections, then $\mathbb{C}[z_1,\dots,z_n]^G$ (the subring of $G$-invariant polynomials) is again a polynomial ring. -From my understanding, a pseudoreflection is diagonalizable with diagonal form $\text{diag}(1,\dots,1,\zeta)$ where $\zeta$ is a root of unity. -My question is whether the property of being generated by pseudoreflections is a property of the representation or of the group itself, and if so what is the equivalent abstract group-theoretic property. If it is not the case, could someone point me towards a group which furnishes a counterexample? - -REPLY [13 votes]: Regarding a comment on Anton's answer: -Todd and Shephard classified in [Shephard, G. C.; Todd, J. A. Finite unitary reflection groups. Canadian J. Math. 6, (1954). 274--304. MR0059914 (15,600b)] all finite groups generated by pseudo-reflections. Not all groups are isomorphic to one in their list. -For example, the center of a group generated by pseudo-reflections which is irreducible (i.e., not a direct product of two smaller groups of the same kind) is cyclic. So to find an example, it is enough to find a group which is not a direct product and which has non-cyclic center. -Restricting to $p$-groups with even prime $p$, GAP tells me there is such a group of order $2^4$, the semi-direct product of $C_4$ by itself, and another, a semi-direct product of $C_4\times C_2$ by $C_2$, both with center isomorphic to $C_2\times C_2$. -Another source of examples: if $G$ is a finite group with a faithful representation generated by pseudo-reflections, then the corresponding determinant representation is non-trivial (as the determinant of a pseudo-reflection is not $1$) It follows that $G$ is not simple.<|endoftext|> -TITLE: Comprehensive and self-contained treatment of Algebraic Geometry using Functor of Points approach -QUESTION [15 upvotes]: The book everyone seems to use to study Algebraic Geometry is Hartshorne's book. However, I hear a good number of people saying that this book totally misses the functorial point of view. Hence, could you please recommend a good source to learn AG using the Functor of Points approach? Thanks!! - -REPLY [8 votes]: Dear Brian, it seems that algebraic geometers who adopt your favoured approach are essentially specialists in algebraic groups. -My favourite example would be Jantzen's Representations of Algebraic groups", Academic Press 1987, in which all of Chapter 1 (18 pages) is devoted to the functor approach you require. Let me emphasize that Jantzen doesn't limit himself to affine schemes nor to group schemes. He considers completely general schemes defined as local functors admitting an open covering (in the functor sense!) consisting of affine schemes . I am sure you'll love the ingenious but natural definitions of open subfunctor, closed subfunctor, base change ... introduced in this meaty chapter: good luck!<|endoftext|> -TITLE: A question about primes as an additive basis -QUESTION [10 upvotes]: Let $\mathcal{P}$ denote the set of primes. Define the function $r_2(N)$ to be the number of ways to write $N$ as a sum of two not necessarily distinct primes (where order matters). Then the famous Goldbach Conjecture can be phrased as following: -$r_2(2N) > 0$ for all $N \geq 1$. -A related function, $r_3(N)$ which is the number of ways $N$ can be written as the sum of three not necessarily distinct primes (order matters), is used in the following classic result by I.M. Vinogradov: -Vinogradov's Theorem: Define $\mathfrak{S}(N) = \displaystyle \sum_{q=1}^\infty \frac{\mu(q)c_q(N)}{\varphi(q)^3}$ where $\mu$ is the Mobius function, $c_q$ is the Ramanujun sum, and $\varphi$ the Euler totient function. Then for odd $N$, we have: -$r_3(N) = \displaystyle \mathfrak{S}(N) \frac{N^2}{2(\log(N))^3}\left(1 + O\left(\frac{\log \log(N)}{\log(N)}\right)\right)$. -This clearly shows that $r_3(N) > 0$ for all sufficiently large odd $N$, which is to say that all sufficiently large positive odd integers $N$ can be written as the sum of three primes. Goldbach's Conjecture would be proved if we can find some arithmetic function $f(N)$ which tends to infinity such that $r_2(N) >> f(N)$. -It has been confirmed that the Goldbach Conjecture holds for the first $10^{18}$ even integers (according to Wikipedia) by exhaustive computer search. But my question is, is there anything known about $r_2(N)$ for these cases? In particular I am asking whether or not there are any examples where $r_2(N)$ is unexpectedly small. If this happens infinitely often, for instance, then attempting to prove Goldbach's Conjecture by finding a lower bound that tends to infinity would be infeasible. -Another formulation would be, is there any heuristic evidence to suggest that there exist a positive integer $M$ such that for infinitely positive integers $N$ we have $r_2(N) \leq M$? - -REPLY [6 votes]: Considering the density of primes in combination with Brunn's sieve (which shows that the number of weighted solutions to $p+q=n$ can't be more than a constant times the expected value) should show that $r_{2}(n) \leq M$ can't hold for all (or even almost most) primes. In fact more is known, Montgomery and Vaughan (The exceptional set in Goldbach's problem. Collection of articles in memory of Juriĭ Vladimirovič Linnik. Acta Arith. 27 (1975), 353–370.) have shown that the number of exceptional values in $[0,N]$ where $r_{2}(n)=0$ is much smaller, at most $N^{1-\delta}$ for $\delta >0$. This probably can be modified to give a similar bound on the frequency of $r_{2}(n) \ll M$ holding. Of course, it is strongly believed that this can't happen.<|endoftext|> -TITLE: On the non-rigorous calculations of the trajectories in the moon landings -QUESTION [14 upvotes]: In a paragraph written by a person emphasizing that rigour is not everything in mathematics (I wish I had written down the details), it was stated that the moon landings would have been impossible without the use of nonrigorous algorithms (I believe he was referring to the calculations of the trajectories of the spacecraft). What was the nonrigorous algorithm that was used? And why were rigorous algorithms insufficient? Was it because they were too slow on 60's hardware or was it because rigorous methods could not provide the answers required? And if such computations were to be done today, could they be done rigorously? - -REPLY [29 votes]: The existence of the Arenstorf Orbits was discovered in 1963 on the basis of numerical computations. The Arenstorf orbits appear as periodic solutions to the equations for the plane restricted three body problem (see the original paper by R. Arenstorf): - -$$x''+2ix'-x=-\frac{(1-\mu)(x+\mu)}{|x+\mu|^3}-\frac{\mu(x+\mu-1)}{|x+\mu-1|^3},$$ - where $x=x_1+ix_2$ is the complex position vector of the infinitesimal body referred to a co-system rotating with angular velocity $1$ about the center of gravity of the two attracting bodies of masses $1-\mu$ and $\mu$ ($0\leq\mu\leq 1$) as origin. - -When $\mu=0$, the solutions describe the classical Keplerian motion. There are two types of the closed orbits which can be obtained by perturbation methods from $\mu=0$. The closed orbits of the first kind were discovered by Poincaré. They are close to circular Keplerian orbits. The closed orbits of the second kind (i.e. the Arenstorf 8-shaped orbits) are close to elliptic Keplerian orbits of arbitrary eccentricity. -The estimates showing that the Arenstorf orbits can pass very near the Moon and the Earth probably would not have been found without extensive numerical computations.<|endoftext|> -TITLE: Expected second moment for random points on a circle -QUESTION [11 upvotes]: Let $S$ be a circle with unit circumference. Suppose that $n$ random points are chosen independently uniformly from $S$; choosing one arbitrarily as $x_1$, label the rest $x_2, \dots, x_n$ in clockwise order. What is the expected value of -$$ -\max |x_{i}-x_{i-1}| -$$ -(where $x_0$ is interpreted as $x_n$)? Or even more, what is the distribution of this maximum? -(As written, the expression $|x_{i}-x_{i-1}|$ represents distance in the Euclidean plane; I'd prefer to use the distance along the circumference of the circle, which isn't that different when $n$ is large.) -Trivially the maximum (using distance along the circumference of the circle) is at least $1/n$; I'm mostly interested in whether this is the correct order of magnitude (although the exact constant is an interesting question as well). - -REPLY [5 votes]: The problem is equivalent to choosing $n-1$ points at random on the unit interval and considering the length of the longest resulting subinterval. Given that, the distribution of the maximum and its expected value are in my answer to this recent math.SE question on Average Length of the Longest Segment. It's closely related to the one given in the comments above by Shai Covo. I'll reproduce the answer here (which is given in terms of making $n-1$ cuts randomly chosen on a rope of unit length). - -If $X_1, X_2, \ldots, X_{n-1}$ denote the positions on the rope where the cuts are made, let $V_i = X_i - X_{i-1}$, where $X_0 = 0$ and $X_n = 1$. So the $V_i$'s are the lengths of the resulting pieces of rope. - -The key idea is that the probability that any particular $k$ of the $V_i$'s simultaneously have lengths longer than $c_1, c_2, \ldots, c_k$, respectively (where $\sum_{i=1}^k c_i \leq 1$), is $$(1-c_1-c_2-\ldots-c_k)^{n-1}.$$ This is proved formally in David and Nagaraja's Order Statistics, p. 135. Intuitively, the idea is that in order to have pieces of size at least $c_1, c_2, \ldots, c_k$, all $n-1$ of the cuts have to occur in intervals of the rope of total length $1 - c_1 - c_2 - \ldots - c_k$. For example, $P(V_1 > c_1)$ is the probability that all $n-1$ cuts occur in the interval $(c_1, 1]$, which, since the cuts are randomly distributed in $[0,1]$, is $(1-c_1)^{n-1}$. - -If $V_{(n)}$ denotes the largest piece of rope, then -$$P(V_{(n)} > x) = P(V_1 > x \text{ or } V_2 > x \text{ or } \cdots \text{ or } V_n > x).$$ This calls for the principle of inclusion/exclusion. Thus we have, using the "key idea" above, -$$P(V_{(n)} > x) = n(1-x)^{n-1} - \binom{n}{2} (1 - 2x)^{n-1} + \cdots $$ -$$+ (-1)^{k-1} \binom{n}{k} (1 - kx)^{n-1} + \cdots,$$ -where the sum continues until $kx > 1$. -Therefore, -$$E[V_{(n)}] = \int_0^{\infty} P(V_{(n)} > x) dx = \sum_{k=1}^n \binom{n}{k} (-1)^{k-1} \int_0^{1/k} (1 - kx)^{n-1} dx $$ $$= \sum_{k=1}^n \binom{n}{k} (-1)^{k-1} \frac{1}{nk} = \frac{1}{n} \sum_{k=1}^n \frac{\binom{n}{k}}{k} (-1)^{k-1} = \frac{H_n}{n},$$ -where the last step applies a known binomial sum identity. - -For much more on this problem, see Section 6.4 ("Random Division of an Interval") in David and Nagaraja's Order Statistics, pages 133-137, and the corresponding exercises on pages 153-155. -As a side note, the distribution of $V_{(n)}$ was apparently first obtained by Ronald Fisher in "Tests of Significance in Harmonic Analysis," Proceedings of the Royal Society of London, Series A, 1929, pp 54-59. (Sorry for the JSTOR link.)<|endoftext|> -TITLE: "Strøm-type" model structure on chain complexes? -QUESTION [14 upvotes]: Background -The Quillen model structure on spaces has weak equivalences given by the weak homotopy equivalences and the fibrations are the Serre fibrations. The cofibrations are characterized by the lifting property, but in the end they are those inclusions which are built up by cell attachments (or are retracts of such things). -The Strøm model structure on spaces has weak equivalences = the homotopy equivalences and fibrations the Hurewicz fibrations. Cofibrations the closed inclusions which satisfy the homotopy extension property. -The projective model structure on non-negatively graded chain complexes over $\Bbb Z$ has fibrations given by the degreewise surjections and weak equivalences given by the quasi-isomorphisms. Cofibrations are given by the degree-wise split inclusions such that the quotient complex is degree-wise free. -From the above, it would appear to me that the projective model structure on chain complexes is analogous to the Quillen model structure on spaces. - Question - Is there a model structure on (non-negatively graded) chain complexes over $\Bbb Z$ for which the weak equivalences are the chain homotopy equivalences? -(Extra wish: I want cofibrations in the projective model structure to be a sub-class of the cofibrations in the model structure answering the question. Conjecturally, they should be the inclusions satisfying the chain homotopy extension property.) - -REPLY [5 votes]: Very recently a paper appeared on the arxiv by Barthel-May-Riehl which addresses this question in a very complete way. It discusses the three model structures on DG-algebras (answering the OPs question and covering the mixed model structure as well), then goes on to give six model structures on DG modules over a DGA. This paper generalizing the references in Peter May's answer above and gives model category foundations for some classical constructions in differential graded algebra.<|endoftext|> -TITLE: Subspace of $L^2$ that lies in $L^\infty$ -QUESTION [21 upvotes]: Let $E$ be a closed subspace of $L^2[0,1]$. Suppose that $E\subset{}L^\infty[0,1]$. Is it true that $E$ is finite dimensional? -PS. This is actually a question from the real analysis qualifier. I came across it as I was teaching qualifier preparation course, and was solving problems from old qualifiers. So, though it might follow from some advanced theory of Banach spaces, I am most interested in the 'elementary' solution, using only methods from standard real analysis course. Note: if $E\subset{}C[0,1]$, then it is a problem from Folland, and there is a solution there. However, it does not work for $L^\infty$, not without some trick. - -REPLY [30 votes]: Another solution: as Mikael wrote, $||f||_{\infty} \leq C ||f||_2$ for every $f \in E$. -Let $f_1,\ldots,f_n$ be an orthonormal family in your subspace. -Then for every $x \in [0,1]$, $f_1(x)^2+\ldots+f_n(x)^2 \leq ||f_1(x)f_1+\ldots+f_n(x)f_n||_{\infty} \leq C \|f_1(x)f_1+\ldots+f_n(x)f_n\|_2$ $$=C \sqrt{f_1(x)^2+\ldots+f_n(x)^2},$$ -and by squaring we get $f_1(x)^2+\ldots+f_n(x)^2 \leq C^2$, and integrating gives $n \leq C^2$.<|endoftext|> -TITLE: Recommended page layout settings for latex -QUESTION [15 upvotes]: I would appreciate input and thoughts on what a good page layout for a mathematical article looks like in latex. -What would you consider good practice? What is your own personal preference? Which page layouts annoy you as a reader? -11pt or 12pt? Standard latex fonts, or do you prefer another (commonly available) choice? What's a good compromise between traditional typographic rules (no more than 60 characters per line) and not wasting too much empty space on a page? -To make it really concrete, let's say I am specifically asking for guidance for articles to be posted on the arXiv, i.e. the article will be printed both on letter and A4 paper. And unlike journals, you don't have the luxury of choosing your page size appropriately. -(P.S.: What are good examples of layout for math content on A4-sized pages done by professionals?) - -REPLY [7 votes]: From arXiv, it is possible to download the source of any paper. You could check the page layout from a paper you like. - -REPLY [5 votes]: This is undoubtedly not the answer you want, but I just use the amsart style with 12 point type. I see no reason to trust my own typographical taste above that of the people who created the style file. Journals will apply their own styles anyway, and any fancy stuff I did in my preprints would just interfere with this re-styling.<|endoftext|> -TITLE: Calculating the decomposition of a vector bundle over rational curve -QUESTION [10 upvotes]: Consider the rational curve (conic) given by image of the map -$$ u([z,w])=[z^2,-z^2,w^2,-w^2,zw] \in \mathbb{P}^4 $$ -which lies in quintic 3-fold $X: x_1^5+\cdots+x_5^5- x_1\cdots x_5=0$. -By Grothendick theorem and the fact that $X$ is Calabi-Yau, we know that $u^*N_C^X= O(a) \oplus O(b)$, for some $a+b=-2$, where $N_C^X$ is the normal bundle of $C=imgae(u)$ in $X$. -How should I calculate $a,b$ for this (or any other) explicitly given map? - -REPLY [4 votes]: Since you know the explicit equation of the conic, you can compute everything by using Macaulay2. -The following script should be clear: -i1 : k=ZZ/32003; - -i2 : ringP1=k[x, y]; - -i3 : ringP4=k[z1, z2, z3, z4, z5]; - -i4 : I= ideal(z1^5+z2^5+z3^5+z4^5+z5^5-z1*z2*z3*z4*z5); - -o4 : Ideal of ringP4 - -i5 : ringQuintic=ringP4/I; - -i6 : conicMap=map(ringP1, ringQuintic, {x^2, -x^2, y^2, -y^2, x*y}); - -o6 : RingMap ringP1 <--- ringQuintic - -i7 : conic=image conicMap; - -i8 : IC=ideal conic; - -o8 : Ideal of ringQuintic - -i9 : ConormalModuleConic = IC/IC^2; - -i10 : ConormalSheafConic= sheaf ConormalModuleConic; - -i11 : NormalSheafConic= dual sheaf ConormalModuleConic; - -i12 : HH^0(ConormalSheafConic) - - 4 -o12 = k - -o12 : k-module, free - -i13 : HH^1(ConormalSheafConic) - -o13 = 0 - -o13 : k-module - -i14 : HH^0(NormalSheafConic) - -o14 = 0 - -o14 : k-module - -i15 : HH^1(NormalSheafConic) - -o15 = 0 - -o15 : k-module - -The output reads -$h^0(X, N_{C|X}^{*})=4, \quad h^1(X, N_{C|X}^{*})=0, \quad h^0(X, N_{C|X})=0, \quad h^1(X, N_{C|X})=0$, -hence $N_{C|X}=\mathcal{O}_{\mathbb{P}^1}(-1) \oplus \mathcal{O}_{\mathbb{P}^1}(-1)$, according to Sasha's and Sandor's answers.<|endoftext|> -TITLE: Hard Cube Puzzle -QUESTION [12 upvotes]: You are and your friend are given a list of $N$ distinct integers and are told this: -Six distinct integers from the list are selected at random and placed one at each side of a cube. The cube is placed in the middle of a rectangular room in front of its only door, with one face touching the floor, its 6 sides parallel to the walls of the room. Your friend must enter the room and is allowed to alter the orientation of the cube, with the restriction that afterwards it's in the same place with one face touching the floor and its 6 sides kept parallel to the walls of the room. Your friend will then be sent away, after which you can enter the room and are allowed to observe the 5 visible sides of the cube. -What is the largest $N$ that guarantees you to be able to determine the number on the bottom of the cube and what should you instruct your friend to do with the cube for that $N$? -I don't really know how to approach this, the only result I have so far: -An upper bound of $N=29$ and a trivial strategy for $N<10$. -I'm also looking for a solution for the general case of an S-sided dice. -Update: a lower bound of $N=18$ is given here: Guessing a hidden number on a cube. - -REPLY [28 votes]: This is a variation on a classic card trick (audience pick 5 cards, magician A removes a card of his choosing and hands the rest to magician B, who then names the missing card.) -There are two ways you can view the process - from the point of view of the cube-orienter, or from the point of view of the final guesser. -It turns out to be more useful to consider the cube-orienter's job. -Given a cube, he has to pick an orientation to leave it in. Let's view this as a function from the set of possible cubes to the set of 'visible orientations', i.e. we ignore what's on the bottom face. Then this function must be a bijection, and it must satisfy the constraint that the image of a cube under this function does give a valid 'visible orientation' of that cube. -This constraint can be represented as a bipartite graph, where there are $\frac{1}{24}n(n-1)(n-2)(n-3)(n-4)(n-5) = 30\binom{n}{6}$ vertices on the left representing the different cubes, $n(n-1)(n-2)(n-3)(n-4)$ vertices on the right representing the 'visible orientations', and each cube has $24$ edges linking it to its valid 'visible orientations'. Conversely, each 'visible orientation' has $n-5$ edges linking back to the cubes it could have arisen from. In this context, the bijective function we seek is a matching in this graph. -Hence the original problem is equivalent to the existence of a matching in this bipartite graph. A necessary and sufficient condition for such a matching to exist is given by Hall's Marriage Theorem. We need to check that for any set of $k$ cubes, the cardinality of the union of 'visible orientations' linked to these cubes is greater than or equal to $k$. But $k$ cubes give rise to $24k$ edges, which give rise to at least $24k/(n-5)$ 'visible orientations', which is greater than or equal to $k$ for $n \leq 29$. -So there is a strategy for $n = 29$, and conversely this is the best possible just by comparing the number of vertices on either side. -Update: An explicit strategy was requested, so here is an adaptation of the standard card trick strategy: -Let's assume we're working with numbers from $0$ to $28$ for convenience. -The cube-orienter adds up the numbers on all the faces of the cube modulo $6$. Call the result $i$, with $0 \leq i \leq 5$. Now if $i=0$, put the smallest face face-down, if $i=1$, put the second smallest face-down, and so on. Later on, the guesser will be able to add up all the visible faces modulo $6$, and a little thought shows that the hidden face will be congruent to the negative of this modulo $6$, provided the guesser renumbers the unseen numbers from $0$ to $23$. This means the guesser will know the answer is one of $4$ cards, and these remaining $4$ degrees of freedom can be communicated by the $4$ possible rotations the cube-orienter can leave the cube in (with a fixed face down). E.g. of the side-faces, the largest can be pointing left/forward/right/backward.<|endoftext|> -TITLE: Stable base loci cannot contain isolated points -QUESTION [6 upvotes]: Let $X$ be a normal projective complex variety. -A theorem of Fujita-Zariski says that if $L$ is a Cartier divisor on $X$ such that -the base locus $Bs(|L|)$ is a finite set then $L$ is semiample. -It seems to me that by using this theorem it is possible to prove that the stable base locus $\mathbb{B}(L):=\bigcap_{m \in \mathbb{N}} Bs(|mL|)$ of a Cartier divisor on a variety as above cannot contain isolated points. -Do you know a reference for this (or a counterexample)? - -REPLY [8 votes]: Yes, you are right, and the result is almost immediate using the FZ theorem. A proof can be found in the recent paper Restricted volumes and base loci of linear series. by Ein, Lazarsfeld,Mustaţă, Nakamaye and Popa. For the sake of completeness I sketch the argument here. -Just to clarify: The Fujita–Zariski Theorem says that if a line bundle $L$ is ample on its base locus, then it is is semiample. If the base-locus is finite then this condition is automatically satisfied. Here semiampleness means that a multiple $L^{\otimes n}$ is globally generated. -Since $X$ is normal, I'll switch from a line bundle $L$ to a divisor $D$ (just for the sake of notation). Suppose that $x$ is an isolated point in $\mathbb{B}(D)$. We have $Bs(mD)\supset Bs((m+1)D)\supset \cdots $ so by Noetherianness we can choose a large $m$ so that that the stable base locus $\mathbb{B}(D)$ equals $Bs(|mD|)$. Let $X'$ be the blow-up of $X$ with center $Z=Bs(mD)\setminus \{x\}$. The total transform of $mD$ is can be decomposed as $E+M$ where $E=\pi^{-1}(Z)$ and $M$ is a divisor with a base locus at $f^{-1}(x)$ (with some multiplicity). Now, since the base locus of $M$ is finite, Fujita–Zariski implies that $pM$ is base-point free for some large $p$. But then $x\not\in Bs(mpD)$, which contradicts $x\in \bigcap_{m\ge 1} Bs(mD)$.<|endoftext|> -TITLE: Log structure and degeneration -QUESTION [5 upvotes]: I am interested in compactification of the moduli space of elliptic curves, and I heard that Log geometry is very important for the problem. -I am developping the same technique for quantum geometry. -My question was that: -1-Why Log structure can give us a better way to understand degeneration of Elliptic curves? What's the motivation behind? -Is that right that log structure gives us a way to embed the scheme locally into affine space, and the degeneration happens in the space? -Like the log structure on $N^2\rightarrow k[x,y]/x.y=0.$, which is fibered over $N\rightarrow k$. In that case, the fourier transform maps $N^2$ into $A^2$, which is the embeding of the nodal curve. -2-Does it make sense to define log-group, i.e. a group with a log structure. -3-Does it make sense to define log structure on a stack which is not algebraic? -Reference for Log geometry: http://www-personal.umich.edu/~satriano/logcurves.pdf - -REPLY [3 votes]: Read the introduction in the Kato-Usui book. It's got some nice motivating examples, including a degenerating family of elliptic curves with pictures. -There is no problem with defining a group object in the category of log schemes. -This depends on what you want to do with such a structure, i.e., what properties you need. Maybe you should familiarize yourself with the algebraic case first (e.g., in Olsson's paper Logarithmic geometry and algebraic stacks). - -Finally, Matt's notes are a nice introduction, but I hesitate to call them a reference. You might want to look at Ogus's book or Kato's original paper.<|endoftext|> -TITLE: Nontrivial finite group with trivial group homologies? -QUESTION [70 upvotes]: I stumbled across this question in a seminar-paper a long time ago: -Does there exist a positive integer $N$ such that if $G$ is a finite group with $\bigoplus_{i=1}^NH_i(G)=0$ then $G=\lbrace 1\rbrace$? -I believe this to still be an open problem. For $N=1$, any perfect group (ex: $A_5$) is a counterexample. For $N=2$, the binary icosahedral group $SL_2(F_5)$ suffices (perfect group with periodic Tate cohomology). And I found in one of Milgram's papers a result for $N=5$, the sporadic Mathieu group $M_{23}$. Note that this question is answered for infinite groups, because we can always construct a topological space (hence a $BG$ for some discrete group $G$) with prescribed homologies. -Is there another known group with a larger $N\ge 5$ before homology becomes nontrivial? -Are there any classifications of obstructions in higher homology groups? -[[Edit]]: Another view. A group is $\textit{acyclic}$ if it has trivial integral homology. There are no nontrivial finite acyclic groups. Indeed, a result of Richard Swan says that a group with $p$-torsion has nontrivial mod-$p$ cohomology in infinitely many dimensions, hence nontrivial integral homology. - -REPLY [26 votes]: Here is an approach to try to answer this question. -To show the affirmative side -[there is an N] look at the prime two and at swans argument that for a finite group with a two primary subgroup the cohomology mod two must be nonzero in infinitely many dimensions. I tried but didn't find it. if that results in a finite N with nonzero cohomology for all finte groups with a two primary part the problem is solved. because if the two primary component is absent, the group is solvable and N=1 results in non zero homology. -On the other hand, to show the negative side, there is no such N. if swans argument does not yield such a concrete N for the prime two then one might well believe there is no such N, and that a string of examples might be constructed by thinking about the proof of swan's theorem.<|endoftext|> -TITLE: On the determinant of an odd, continuous Galois representation. -QUESTION [9 upvotes]: In his paper, Duke paper, Serre consider continuous, odd Galois representation -$\rho: G_{\mathbb{Q}}\longrightarrow GL_{n}(\overline{\mathbb{F}}_{p})$ where $p$ is a rational prime. -Roughly, (I don't understand much French except for the help from Google translation) Serre claims (section 1.3) that -$\det(\rho(Frob_{l})) = \epsilon(Frob_{l})\omega^{k}(Frob_{l})$ for all prime $l\nmid pN$ where $N$ is defined as the level of the representation (with an explicit formula given in the paper) and $\epsilon$ is a Dirichlet character and $k$ is some positive integer. -This seems to be standard since other papers cited it without reproving and I could not find any reference for the proof. In particular, my questions are: -1) Where can I find a proof for this. -2) What exactly is $\epsilon$, in some paper, there is the claim that $\epsilon$ is the unique quadratic character mod $p$ ramified only at $p$, and I do not understand where this comes from? -3) How can one finds $k$. -For motivation, I think $\det(\rho(Frob_{l}))$ is an important invariant to compute since, for example, it appears in the attachment equation that associates these representations with modular forms. -Thanks in advance for any insight. - -REPLY [6 votes]: $\det (\rho)$ is a one dimensional rep of the absolute Galois group of the rationals, i.e., it is a character. All such characters can be described by class field theory or, more simply, by the Kronecker-Weber theorem. So is a Dirichlet character and, by the hypotheses, its conductor divides $pN$. Factor it as a character of conductor $p$ (that will be $\epsilon$) times a character of conductor $N$. The latter is a power of the cyclotomic character and $k$ is defined to be that power. The bit about quadratic character must be under additional hypotheses. -Edit: I got $p$ and $N$ switched above. The character of conductor $N$ is $\epsilon$. The character of conductor $p$ is a power of the cyclotomic character because $(\mathbb{Z}/p)^*$ is cyclic.<|endoftext|> -TITLE: Zeta Function: Zero Density Theorems. -QUESTION [14 upvotes]: I was reading about some of the zero density Theorems for my Analytic Number Theory Topics course. While looking over some more complicated results and proofs a few simple questions came up: -First, let $$N(\sigma,T)=|\{ \rho=\beta+i \gamma \text{ }:\text{ } \zeta(\rho)=0, \text{ } 0< \gamma < T, \text{ } \sigma\leq\beta<1 \}|$$ be the cardinality of the set of zeros of the zeta function real part greater than $\sigma$ and imaginary part between 0 and $T$. -We call theorems pertaining to the size of $N(\sigma,T)$ zero density theorems. (Complicated bounds on the size of the empty set) These estimates are usually written in the form $N(\sigma , T)\ll T^{A(\sigma)(1-\sigma)+\epsilon}$ where the $\ll$-constant is uniform in $\sigma$. (So the smaller $A(\sigma)$ is, the better, and all such theorems concern the size of $A(\sigma)$) -My questions are: -1) Why do we have a factor of $(1-\sigma)$ in the exponent $T^{A(\sigma)(1-\sigma)}$? It is clear to me what this implies about the density of the zeros, but why does it arise so naturally? -2) The so called density hypothesis is that $A(\sigma) \leq 2$. The best known bound is $A(\sigma) \leq 2.4$ (or possibly 2.3) What is so special about $A(\sigma)\leq 2$? Are there links between this value and certain theorems one may hope to prove? Why does this in particular deserve such a name as "the density hypothesis?" -Thanks a lot!! -(Also, if you have a good reference book or paper that talks about these issues I would be happy to know about it!) - -REPLY [11 votes]: To resonate with John's answer, it is in the application where exponents of the form $c(1-\sigma)$, with $c$ constant, are natural and important. I recommend you read page 265 of Iwaniec-Kowalski: Analytic number theory, after which you will see this connection very clearly. Let me refine this answer. -For $\sigma$ close to $1/2$ Ingham (Quart. J. Math. 8 (1937), 255-266) proved that the exponent $2(1-\sigma)+\epsilon$ follows from the Lindelöf Hypothesis, so the Density Hypothesis was born. On the other hand, Turán (Acta Math. Hung. 5 (1954), 145-163) conjectured that for $\sigma\geq 1/2+\delta$ it should be possible to derive from the Lindelöf Hypothesis the much stronger exponent $\epsilon$. He accomplished this derivation for $\sigma\geq 3/4+\delta$ in a joint paper with Halász (J. Number Theory 1 (1969), 121–137.). In the same paper they also proved unconditionally that $(1-\sigma)^{3/2}\log^3(1-\sigma)^{-1}$ is an admissible exponent when $1-\sigma$ is sufficiently small. -This shows that, from the point of our current understanding, the dependence $1-\sigma$ is rather natural when $\sigma$ is close to $1/2$, but less so when $\sigma$ is close to $1$. There is a definite turning point at $\sigma=3/4$: if one is very-very optimistic, current technology (Halász' inequality and their refinements due to Montgomery, Huxley, Jutila, Bourgain and others) might lead to a proof of the Density Hypothesis for $\sigma>3/4$, but certainly some very new ideas will be needed to make an improvement for $\sigma\leq 3/4$.<|endoftext|> -TITLE: Eigenvalues of sum of a non-symmetric matrix and its transpose $(A+A^T)$ -QUESTION [9 upvotes]: Suppose we have a matrix $M$ such that $M$ is non-symmetric real and has positive eigenvalues. Do we have a relation between eigenvalues/eigenvectors of $(M+M^T)$ and those of $M$? -What if $M$ and $(M+M^T)$ both are of low rank? -Suppose, $M = AP$ where $A$ is a positive semi-definite matrix and $P$ is an orthogonal projection matrix of the form $UU^T$ ($U$ being an orthogonal matrix). $M$, $A$, $P$ are all of size $n\times n$, $U$ is of size $n\times k$, where $k \ll n$. We know that $M$ will have real and non-negative eigenvalues. The question is, how are the eigenvectors/eigenspaces of $M$ and $(M+M^T)$ are related? - -REPLY [12 votes]: Let $N:=(M+M^T)/2$. besides the obvious equality $Tr(N)=Tr(M)$ which is an equality of the sums of eigenvalues, you have the following. Let $\lambda_\pm$ be the smallest/largest eigenvalues of $N$. Then every eigenvalue of $M$ satisfies $\Re\lambda\in[\lambda_-,\lambda_+]$. In addition, if $w(M):=\max\{\lambda_+,-\lambda_-\}$ is the numerical radius of $M$, then -$$w(M)\le\|M\|\le2w(M),$$ -in operator norm. This implies that the singular values, hence the moduli of the eigenvalues of $M$, are not greater than $2w(M)$.<|endoftext|> -TITLE: On a schoolchild puzzle of V.I. Arnold (re: toric varieties) -QUESTION [33 upvotes]: When reading the interview with Vladimir Arnold in the April 1997 edition of the Notices, I came across the following anecdote. - -Many Russian families have the -tradition of giving hundreds of such -problems to their children, and mine -was no exception. The first real -mathematical experience I had was when -our schoolteacher I. V. Morozkin gave -us the following problem: Two old -women started at sunrise and each -walked at a constant velocity. One -went from A to B and the other from B -to A. They met at noon and, continuing -with no stop, arrived respectively at -B at 4 p.m. and at A at 9 p.m. At what -time was the sunrise on this day? -I spent a whole day thinking on this -oldie, and the solution (based on what -is now called scaling arguments, -dimensional analysis, or toric variety -theory, depending on your taste) came -as a revelation. - -I found the solution in a rather straightfoward fashion, but I was curious as to the parenthetic remark. So, can anybody tell me (as a total outsider to algebraic geometry), what does this problem have to do with toric varieties? - -REPLY [5 votes]: I think the problem cannot be solved because the sun rises at different times at A and B in general -(say A = Vladivostok and B = Moscow). All what can be said is that (12-tA)(12-tB) = 4x9 = 36. -If by chance the sun rise times are identical then tA = tB = 6. -This is obtained by using the similitude of triangles in the space-time diagram.<|endoftext|> -TITLE: Basis-free definition of Casimir element? -QUESTION [17 upvotes]: Let $V$ be a finite-dimensional vector space and let $\mathfrak g \subset \mathfrak{gl}(V)$ be a representation of a semisimple Lie algebra on $V$. Let $e_1, \dots, e_n$ be a basis for $V$. Let $e_1', \dots, e_n'$ be the dual basis of $\{e_i\}$ under the Killing form $B_V(X,Y) = \mathrm{trace}(X \circ Y)$. The Casimir element of the universal enveloping algebra $U(\mathfrak g)$ (or $\mathrm{Sym}^2(V)$, if you prefer) is defined by -$c_V = \sum_i e_i \cdot e_i'$ -One then proves that this definition is independent of the choice of basis. -My question is: is there a good definition of $c_V$ that can be stated without referring to a basis? -A perfunctory glance at Bourbaki, Humphreys, Fulton and Harris, and Wikipedia turned up none but the above definition. -I'm mostly interested in representations of ordinary finite-dimensional Lie algebras (over $\mathbb{C}$, even), but if the issue is more naturally addressed in the context of some more general sort of algebra, then I'd be interested to learn about it, even though I don't really know much about, say, Hopf algebras. -EDIT: -Qiaochu's answer below looks like just what I'm looking for. I wrote this clarification and will leave it up in case anyone' interested. But Qiaochu also improves on most of what I say here. -Peter McNamara gives a sensible definition below which I ought to have been aware of: the Casimir is the unique element of degree 2 in the center of the universal enveloping algebra, up to some normalization. I suppose this definition will probably work with some modification in cases where $\mathfrak g$ is not the adjoint rep. This is a nice definition: it sheds some light on what makes the Casimir tick, and points our attention to the center of the unversal enveloping algebra as an object of interest. Where is a good reference to read about the center of a universal enveloping algebra systematically? -I'm not entirely satisfied with this definition, though, and I'd like to illustrate why with a pair of examples: coordinate-free definitions of the determinant, and of the trace. -The determinant on an $n$-dimensional vector space $V$ may be defined (up to normalization) as the unique element of $\Lambda^{n}(V)$ (the highest-dimension grading of the alternating algebra). This definition is highly analogous to Peter McNamara's definition of the Casimir. In either case, we define widget by finding an algebra $A$ of which widget is naturally an element, and by careful analysis of $A$, we show that widget is uniquely characterized by the features of $A$. One of the benefits of this sort of definition is that it points our attention towards the algebra $A$ as a worthy object of study. On the other hand, one might wish to avoid using $A$ altogether. -In contrast, here is a definition of the trace on a finite-dimensional vector space $V$. Consider the isomorphism $ \mathrm{End}(V) \cong \mathrm{End}(V)^{*} $ given by viewing $\mathrm{End}(V)$ as $V \otimes V^{*} $, swapping the factors, and using the canonical isomorphism $V \cong V^{**}$. The trace is the image of the identity under this map. It might be said that the tensor algebra is used in an auxiliary manner, but there is (I think) a difference: only the most general, "hands-off" properties of tensor products are used. The heaviest lifting comes from the isomorphism $V^{**} \cong V$ (in order to show that it is actually an isomorphism, one must keep track of dimensions of a finite-dimensional vector space under dualization). So the only hard work comes on the original object of interest: the vector space $V$ itself. In contrast, the definition of the determinant required some detailed understanding of the alternating algebra, which was not originally in question. -There might be a logical way to state this: something along the lines of whether the definition can be given in an extremely weak fragment of logic (on the order of Horn logic?). It would have to be a fragment which doesn't easily allow definitions. More likely there's no real difference between the definition of the trace and the determinant beyond the former being "easier" than the latter. Nevertheless, if there were such an "easy" definition of the Casimir, it ought to be enlightening. - -REPLY [7 votes]: I can't help but add: consider the chain of $G$-equivariant natural maps, using the identification of the Lie algebra $g$ with its dual via the non-degenerate $G$-equivariant Killing form: -$$ -End(g) \approx g\otimes g^* \approx g\otimes g \subset \bigotimes{}^\bullet g \rightarrow Ug -$$ -The identity endomorphism of $g$ commutes with the (Adjoint) action of $G$, so its image on the far end does, as well, and is Casimir. The often-seen expressions in coordinates (chosen depending on the circumstance), derive from the expression for the identity map in $g\otimes g$. -(I also can't help but add that trying to understand where these quadratic expressions "live", it was a great relief at some point to understand that such computations take place in the enveloping algebra... so, even if it is possible to talk about Casimir without the enveloping algebra, it may be easier to explain the context with the enveloping algebra.)<|endoftext|> -TITLE: sharper minkowski bound -QUESTION [9 upvotes]: If we want to bound the norm of the smallest ideal which generates a nontrivial ideal class, is there a better bound than Minkowski's bound? -(Note that Minkowski's bound is to guarantee something much stronger, namely that every ideal class has a representative whose norm is less than the given bound. Here, we just want that some nontrivial ideal class has such a representative.) - -REPLY [6 votes]: The classical reference to an improvement of Minkowski bounds is - -R. Zimmert, Ideale kleiner Norm in Idealklassen und eine Regulatorabschätzung, Invent. Math. 62 (1980), 367-380<|endoftext|> -TITLE: Smooth and analytic structures on low dimensional euclidian spaces -QUESTION [7 upvotes]: So it is relatively easy to show that there exists only one smooth structure on -the real line $\mathbb{R}$. So here are 2 natural questions: -Q1: Up to equivalence, is there only one real analytic structure on $\mathbb{R}$? If so, -then do we have a simple proof of that? -Q2: Where can I find the simplest proofs that there exists only one smooth structure -on $\mathbb{R}^2$ and $\mathbb{R}^3$? -So I've heard that on $\mathbb{R}^4$ there are infinitly (in fact uncountably) many non-equivalent smooth structures, so what about real analytic strucutres? Is there some kind of moduli space of smooth structures on $\mathbb{R}^4$. if so, in how many ways is it possible to deform -a smooth structure into a real analytic one? - -REPLY [2 votes]: Regarding Q1, put an analytic Riemann metric on your 1-manifold. Integrating a unit speed vector field gives an analytic diffeomorphism to $\mathbb R$. Another way to prove analytic structures are unique is to notice the same argument that one uses to prove that the group of $C^k$-diffeomorphisms of $\mathbb R$ has the homotopy type of $\mathbb Z_2$ works for analytic diffeomorphisms -- simply take the straight-line homotopy between your original diffeomorphism and either the identity or the negative identity, appropriately. -Regarding Q2, I don't know much in the way of really simple proofs. But when $n=2$ you've got the Uniformization Theorem from complex analysis. That's relatively simple.<|endoftext|> -TITLE: Tamely ramified p-adic Galois representations -QUESTION [12 upvotes]: The following question came up in a discussion with a colleague about local Galois representations: - - To what extent is the classification of continuous $p$-adic representations - of $G_{\mathbf{Q}_{\ell}}$ for $\ell\neq p$ similar to the classification of - tamely ramified $p$-adic representations for $\ell=p$? - -More precisely, let $\rho: G_{\mathbf{Q}_{\ell}}\rightarrow \mathrm{GL}_n(\mathbf{Q}_p)$ be a (continuous) $p$-adic representation. If $\ell\neq p$, then Grothendieck proved (using the observation that such a representation kills an open subgroup of wild inertia) that $\rho$ -is determined by the associated Weil-Deligne representation (see, for example, -the notes of Brinon and Conrad, pg. 111, or Taylor's 2002 ICM article). - -When $\ell=p$ and $\rho$ is trivial on the wild inertia subgroup, is it the -case that $\rho$ is necessarily de Rham? - -What seems clear to me is that if one assumes that $\rho$ is Hodge-Tate, then the only Hodge-Tate -weight is zero. If indeed $\rho$ were de Rham = pst, then the associated filtered $(\phi,N)$-module -would have trivial filtration, and so one ``ought" to be able to recover it from -the attached Weil-Deligne representation. In other words, the classification of $p$-adic -representations of $G_{\mathbf{Q}_{\ell}}$ for $\ell\neq p$ is literally the same -as the case $\ell=p$, provided one throws in the (rather drastic) condition that wild inertia -is killed (or at least some open subgroup of it is killed). -Does this sound correct? - -REPLY [16 votes]: It's true that if $\rho$ is tamely ramified, then $\rho$ is de Rham. In fact, it's even potentially crystalline with all Hodge-Tate weights equal to 0. -First, note that $\rho(I_{\mathbb Q_p})$ is finite. The reason is that the image of $\rho$ lands in $GL_n(\mathbb Z_p)$, which has a pro-$p$ subgroup of finite index, namely the principal congruence subgroup $1+pM_n(\mathbb Z_p)$. Since $I_{\mathbb Q_p} \to GL_n(\mathbb Z_p)$ factors through a prime-to-$p$ group (tame inertia) by assumption, $\rho(I_{\mathbb Q_p})$ injects into $GL_n(\mathbb F_p)$, hence it is finite. -It follows that there is a finite extension $K/\mathbb Q_p$ such that $\rho|_{I_K}$ is trivial. (The kernel of ($\rho$ restricted to $I_{{\mathbb Q}_p}$) corresponds to a finite (tame) extension of $\mathbb Q_p^{nr}$ and we can choose $K$ such that that extension is contained in $\mathbb Q_p^{nr} \cdot K$.) -It's a general fact that $\rho|_{I_K}$ crystalline implies $\rho|_{G_K}$ crystalline, hence $\rho$ is potentially crystalline. (Added: this follows from Hilbert's theorem 90. If $L/K$ is a Galois extension and $Gal(L/K)$ acts semilinearly and continuously on a finite-dimensional $L$-vector space $V$, then $V$ has a basis that is $Gal(L/K)$-invariant.) -In particular, you only get WD representations with $N = 0$.<|endoftext|> -TITLE: Converse to Bishop-Gromov Inequality -QUESTION [12 upvotes]: Is the converse to the Bishop-Gromov Inequality true? -In other words, if, for a complete $n$-dimensional Riemannian manifold $M$, there is $k \in \mathbb{R}$, such that defining $V_k(R)$ to be the size of a ball of radius $R$ in the standard space $S^n_k$, for any $p \in M$ and $R\geq 0$ we have that -$$ -vol_M(B_R(p)) \leq V_k(R) -$$ -then is it true that on $M$, $Ric \geq (n-1)k$? - -REPLY [11 votes]: This is false for $n=3$, and for $k<0$. One can get a different pinching condition, $R(g)g-Ric \geq -4g$ which is weaker than $Ric \geq -2g$ ($R(g)$ is the scalar curvature), and which gives the same upper bound on volume as Gromov's inequality (for a manifold with negative curvature or no cut points). The idea is that in the proof of Gromov's inequality, one can integrate out the curvature tangent to the sphere of radius $r$ using the theorem egregrium and Gauss-Bonnet, giving a weaker curvature condition. There are metrics which satisfy the first curvature condition but not the Ricci lower bound. For example, there are homogeneous negatively curved metrics on $R^3$ which in one direction are foliated by totally geodesic hyperbolic planes of curvature $-K_1^2$, and in a perpendicular direction are foliated by totally geodesic hyperbolic planes of curvature $-K_2^2$ for $0 < K_1\neq K_2>0$ (this is a double warped product, and is Thurston's "ninth geometry" of the form $dr^2+e^{2K_1r}dx^2+e^{2K_2r}dy^2$). One obtains a counterexample for the choice of $K_1 =5/4,K_2=1/2$. -Added computations: The curvature operator of this metric may be normalized to have sectional curvatures $-K_1^2, -K_2^2, -K_1K_2$. If we assume that $K_1\geq K_2$, then the minimal curvature is $-K_1^2$. Under the same assumption, the minimal eigenvalue of the Ricci operator is $-K_1^2-K_1K_2$. The minimal eigenvalue of $Rg-Ric$ is $-2K_1^2-K_1K_2-K_2^2$. We then see that for $K_1=5/4, K_2=1/2$, the minimal eigenvalue of $Ric$ is $-35/16<-2$, and the minimal eigenvalue of $Rg-Ric$ is $-4$.<|endoftext|> -TITLE: Shannon's communication paper and finite differences -QUESTION [14 upvotes]: In Shannon's 1948 paper "A Mathematical Theory of Communication", early on he derives the equation $$N(t)=N(t-t_1)+N(t-t_2)+\ldots+N(t-t_n).$$ -He then says "according to a well-known result in finite differences, $N(t)$ is then asymptotic to $X_0^t$ where $X_0$ is the largest solution to the equation $X_0^{-t_1}+\ldots X_0^{-t_n}=1$." -He does not cite a reference. Obviously if the $t_i$ are commensurable, this reduces to a standard constant coefficient recurrence relation, but Shannon does not explicitly make this assumption (in his examples all the $t_i$ are rational). -The result seems to be true in the case of general positive $t_i$ also if you assume some kind of regularity of $N(t)$, but here is my question: -Can anyone suggest a reference that treats this? (the books on finite differences I've seen seem to deal with the commensurable case). - -REPLY [9 votes]: When the $t_i$ are incommensurable in the sense that they generate a dense subgroup, $N(t)=CX_0^t+o(X_0^t)$ for a given constant $C$. This is a consequence of the standard renewal theorem and needs no hypothesis on the monotonicity of the function $t\mapsto N(t)$. -To see this, let $(\xi_k)$ denote some i.i.d. random variables such that $P[\xi_k=t_i]=X_0^{-t_i}$ for every $k$ and $i$. Introduce $M(t)=N(t)/X_0^t$. Then -$$ -M(t)=E[M(t-\xi_1)]. -$$ -Fix $t_0$ such that $t_0\ge t_i$ for every $i$. For every positive $k$, let $S_k=\xi_1+\cdots+\xi_k$. For every $t > t_0$, consider the first time $T(t)$ such that $S_{T(t)}\ge t-t_0$. Since $T(t)$ is a stopping time, the martingale property yields -$$ -M(t)=E[M(t-S_{T(t)})]. -$$ -Reversing the time axis, $t_0-(t-S_{T(t)})$ becomes the overshoot over $t-t_0$ for the renewal process based on the sequence $(\xi_k)$ and starting from $0$. In the non lattice case, the renewal theorem asserts that $t_0-(t-S_{T(t)})$ converges in distribution to a random variable $\xi_0$ when $t\to+\infty$. Being lattice means that there exists a nonzero $a$ such that the random variables $a\xi_k$ are almost surely integer valued, hence the non lattice case corresponds to non commensurate parameters $t_i$. -Thus, when the $t_i$ are non commensurate, $N(t)/X_0^t=M(t)\to C$ wih -$$ -C=E[M(t_0-\xi_0)]=X_0^{-t_0}E[N(t_0-\xi_0)X_0^{\xi_0}]. -$$ -Finally, $\xi_0$ is distributed like $u\xi'$ where $u$ and $\xi'$ are independent, $u$ is uniform on $[0,1]$ and the distribution of $\xi'$ is the size-biased distribution of the distribution of $\xi_k$, given by $P[\xi'=t_i]=t_iX_0^{-t_i}/E[\xi_k]$. Hence one can write $C$ as an explicit integral of the function $N$ over $[0,t_0]$. -A reference is Applied Probability and Queues by Søren Asmussen.<|endoftext|> -TITLE: Can a homotopy inverse of the map from a Lie group to loops on its classifying space be given by holonomy? -QUESTION [13 upvotes]: Let $G$ be the compact Lie group $SO(n)$. There are some classical constructions of the classifying bundle of $G$ based upon on direct limits of Grassmann and Stiefel manifolds: -$$BG \simeq \underset{m \to \infty}{\lim} SO(m+n)/SO(m) \times SO(n)$$ -$$EG \simeq \underset{m \to \infty}{\lim} SO(m+n)/SO(m)$$ -with the evident $G$-bundle projection $EG \to BG$. One may give an explicit map $i: G \to \Omega BG$ which is a weak homotopy equivalence, by comparison of long exact homotopy sequences. Since $G$ and $\Omega BG$ have the homotopy types of CW complexes, the map $i$ is in fact a homotopy equivalence. -I am interested in whether an explicit homotopy inverse $h: \Omega BG \to G$ to $i$ can be given by taking holonomy of loops with respect to some suitably chosen connection on the universal bundle. Naturally the "manifold" $BG$ is infinite-dimensional, but I'm thinking it would suffice to work with the finite-dimensional $G$-bundles $V_m \to G_m$ (Stiefel to Grassmann) which approximate to the direct limits above, provided that the connections on each of the approximating bundles are compatible with respect to inclusion into the next, "compatible" here having an obvious sense in terms of $G$-valued holonomy. -Has anyone seen this idea worked out? Naturally I'm curious also about whether a similar idea works out for a general compact Lie group $G$. - -REPLY [12 votes]: Yes, certainly. The model for BG as you described it has a canonical, universal connection -for its $SO(n)$ bundle: just the induced Riemannian connection from Euclidean space. -As you -move an $n$-dimensional plane in $\mathbb E^{n+m}$, the induced connection -is the limit of compositions - of orthogonal projections between nearby planes. In the limit, these become isometries. -It doesn't matter what is the dimension of the ambient space, as long as the projection is -defined. So, a loop in the Grassmanian gives an element of $G$. One example: if you -hold a bicycle wheel by its axle, and move it around in a loop, when it comes back, it has -rotated by some angle. That's the map. -It's a homotopy inverse of the map going the other way, namely, the classifying map for -the bundle obtained by the suspension of an element of $G$. The suspension of a homomorphism -has a canonical connection; its holonomy is $G$, so one composition of these two maps -equals the identity. The other composition is homotopic to the identity, basically -because the space of connections is contractible, and the bundle $EG$ over this model -of $BG$ has a universal connection: Every connection on every $SO(n)$ bundle over -a $CW$ complex $X$ is induced from a map $X \rightarrow BG$ (and this is also true -in a relative form). This is a standard fact; - I don't have a handy reference, but the proof is "soft". -However, to get a more immediate classifying space for connections that works -for all Lie groups, just make -a simplicial complex whose simplices are connections for a $G-bundles$ over the simplex. -The $G$-bundle is specified by in terms of a trivialization associated with each vertex; -the data needed is the chart-transition cocyle. In addition, give a connection; -this amounts to specifying a connection form. Glue these simplices with $G$-bundles -and connections together, to make a model for $BG$. Since the space of connections -is contractible, this has the same homotopy type as the more usual model for $BG$ where just -the cocycle is specified. -The same construction works to give an explicit homotopy inverse -in a much more general context, e.g. the group of diffeomorphisms for a manifold. - -REPLY [5 votes]: I'm going to change my original answer, since I interpreted the question wrongly. -(I hope that's alright.) -Suppose $p: E\to B$ is a Hurewicz fibration, where $F = p^{-1}(\ast)$ is the fiber over the basepoint and $B$ is connected. Then one can cook up a map $\Omega B \times F \to F$ which might be called a "holonomy" in the algbraic topology sense. -The idea is this: Let -$$ -\Lambda_p = E\times_B B^I -$$ -be the space of path lifting problems for $p$ (this is the space of pairs $(e,\lambda)$ where -$e\in E$ and $\lambda$ is a path starting at $p(e)$. There is a map -$$ -q: E^I \to \Lambda_p -$$ -by sending path $\lambda$ in $E$ to $(\lambda(0), p\circ \lambda)$. Then the condition that -$p$ be a Hurewicz fibration is tantamount to saying that $q$ has a section. A choice of section might be regarded as parallel transport along a path in the algebraic topological sense. Choose such a section. This gives a way of associating to each -path in $B$, starting at $x$ and ending at $y$, a map $E_x \to E_y$, where $E_x$ is -the fiber at $x$. This map is a homotopy equivalence. (When $p$ is a fiber bundle, -one can choose the section in such a way that each parallel transport is a homeomorphism of fibers.) -Evaluating the section when $x=y$ is the basepoint gives the holonomy operation -$\Omega B \times F \to F$, or adjointly as $\Omega B \to G(F)$, where $G(F)$ is -the topological monoid of self homotopy equivalences of $F$. If $p$ is a fiber bundle with structure group $G$, then -the transport operation described above can be factored as -$$ -\Omega B \to G\to G(F) . -$$ -If we choose a basepoint in $F$, then the value of the operation on the basepoint gives a map -$$ -\Omega B \to F . -$$ -This map is well-known: it's the map sitting in the homotopy fiber sequence -$$ -\Omega B \to F \to E . -$$ -(this should be in any reasonable text on the subject). -So, in the particular case when $p: EG \to BG$ and $F = G$, then map $\Omega BG \to G$ will be a homotopy equivalence, using the above homotopy fiber sequence, since $E = EG$ is contractible. We have also seen this map as decribed by the orbit of a point in -$G$ under the holonomy operation $\Omega BG \times G \to G$ as given above. - -REPLY [2 votes]: Let $\pi:P \to M$ be a smooth principal $G$-bundle on a Hilbert manifold. There exist smooth connections in this situation, so pick one. Pick a point $p \in P$, $x:=\pi(p)$. Let $\Omega_{\infty} M$ be the space of smooth loops based at $x$. Consider the lifting problem -$$ -\xymatrix{ -\Omega_{\infty} \times \{0\} \ar[d] \ar[r] & P \ar[d] -\Omega_{\infty} M \times [0,1] \ar[r] \ar[ur]^{l} & M -} -$$ -The bottom map is the evaluation, the top map is the constant map $p$. Now parallel transport along curves defines the lift $l$. Restriction of $l$ to $\Omega_{\infty} M \times \{1\}$ defines the holonomy $hol:\Omega_{\infty} M \to \pi^{-1}(x) =G$, the last identification depending on the point $p$. -Now specialize to a compact Lie group $G$. Take the Grassmann manifold $Gr_n$ of $n$-dimensional subspaces of the Hilbert space $\ell^2$. This is a model for $BSO(n)$ as a Hilbert manifold, the corresponding model for $EO(n)$ is the Stiefel manifold $V_n$ of orthonormal $n$-frames in $\ell^2$; this is a Hilbert manifold as well. If $G$ is compact, we can embed $G$ into $O(n)$ for some $n$ (Peter-Weyl Theorem). Then $V_n \to V_n/G=BG$ is a model for the universal $G$-bundle, in the context of Hilbert manifolds. -Now I claim that $hol: \Omega_{\infty} BG \to G$ is a weak homotopy equivalence. -Let $f:E \to B$ be a Hurewicz fibration with fibre $F=f^{-1}(x)$. There is the fibre transport map $T:\Omega B \to F$ obtained by lifting the paths. It is not hard to see that $\pi_{n+1}(B) \cong \pi_n (\Omega B) \stackrel{T}{\to} \pi_n (F)$ is the same as the connecting homomorphism in the long exact homotopy sequence of $F \to E \to B$. If the fibration is $EG \to BG$, you get a weak homotopy equivalence $\Omega BG \to G$. The fibre transport is defined only up to homotopy, but above, we have constructed one using a connection. Thus the holonomy is homotopic to the fibre transport and hence a weak homotopy equivalence. -I am running out of steam, but I think it can be shown along these lines that $hol$ is also a homotopy inverse to the natural map $G \to \Omega_{\infty} BG$ (which does not look so natural in this setting).<|endoftext|> -TITLE: Estimating the Volume of the Metric Polytope -QUESTION [8 upvotes]: A metric on $n$ points $N$ can be represented as a vector $x \in \mathbb{R}_+^{n \choose 2}$. -For each pair of distinct $i, j \in N$, we have $d(i,j) = d(j,i) = x_{i,j}$. The set of all metrics is the set of points which lie inside the cone defined by the (triangle) inequalities: -$$x_{i,j} - x_{i,k} - x_{j,k} \leq 0$$ -for each distinct triple $\{i,j,k\} \subset N$. -If we bound the cone by the inequalities $x_{i,j} + x_{i,k} + x_{j,k} \leq 2$, then we have the metric polytope. - -Question: Are there any (lower or upper) bounds known on the volume of the metric polytope, for general n? - -In the end, I am interested in estimating the size of the smallest $\epsilon$-net for the set of bounded metrics on $n$ points, which I asked about here. I would also be interested in estimates for polytopes that result in other ways of bounding the metric cone -- for example, by including the inequalities $x_{i,j} \leq 1$ for all $i,j$. - -REPLY [7 votes]: Some googling reveals: - -What you call a metric polytope is also called a semimetric polytope (see for example the standard reference: Geometry of cuts and metrics ) -In the same book, see here, the authors say that the volume of the rooted semimetric polytope in dimension $n+1$ is: $$\frac{n!}{(2n)!}2^n$$ - -EDIT: Further web searching (not googling) showed up the paper: On Skeletons, Diameters and Volumes of Metric Polyhedra. In this paper, the authors give explicit volumes for $n=3$ to $n=6$, and given those volumes it seems that the volume goes very rapidly to zero (e.g., $vol_4=2/45$, $vol_5=4/1701$, $vol_6=71936/1477701225$). -At this point, my current knowledge of the volume ends!<|endoftext|> -TITLE: Survey article on Intersection Theory -QUESTION [16 upvotes]: Does anybody knows about good overview on intersection theory. -The book of Fulton has very hard language. Does there exist simple overview on this topic with many examples? - -REPLY [23 votes]: Eisenbud and Harris are coming out with a book on intersection theory, "3264 and all that", and if you know Harris's style at all, you'll know it's chock full of down-to-earth examples that should be right along the lines of what you're looking for. (Sorry to recommend a book that's not strictly speaking published yet, but it does sound like exactly what you're asking!) - -REPLY [6 votes]: This is certainly not an overview of intersection theory as a whole, but for its classical roots I highly recommend: - -Steven L. Kleiman, Problem 15: rigorous foundation of Schubert's enumerative calculus. Mathematical developments arising from Hilbert problems (Proc. Sympos. Pure Math., Northern Illinois Univ., De Kalb, Ill., 1974), pp. 445–482. Proc. Sympos. Pure Math., Vol. XXVIII, Amer. Math. Soc., Providence, R. I., 1976. - -I found this article to be both completely lucid and completely fascinating -- and I am someone who, in general, has no great interest in intersection theory or (especially) Schubert calculus.<|endoftext|> -TITLE: Chern classes in flat families -QUESTION [5 upvotes]: Given a smooth projective variety $X$ over an algebraically closed field $k$. Now given a another projective variety $T$ and a coherent $O_{X\times T}$-module $F$, which is flat over $T$. -Given $r,s \in T$, and let $F_r$ and $F_s$ be the pullback of $F$ to the fibers over $r$ and $s$. -Do we have $ch(F_r)=ch(F_s)$ in this case? I.e. is the Chern character constant in flat families? -What about the Chern classes $c_i$? -Is there some literature about the behaviour of these characteristic classes in (flat) families? - -REPLY [14 votes]: The answer depends on the version of the Chern classes you are using. -For example, if you consider Chern classes with values in the Chow ring then the answer is negative. The simplest example is the following, take $X = T$ to be an elliptic curve, and $F$ to be the line bundle corresponding to the diagonal. Then $c_1(F_r)$ is the point $r$ itself, but different points are not rationally equivalent. -On the other hand, if you consider Chern classes with values in cohomology with rational coefficients then the answer is positive (if $T$ is connected!). Indeed, since $X$ is smooth and projective any flat sheaf has a resolution by vector bundles. Because of the flatness its restriction gives a resolution of the restricted sheaf. Since Chern classes are multiplicative in exact seqeunces, the question reduces to the case of a vector bundle. Further, because of the splitting principle we can assume we can assume that $F$ is a line bundle. And in this case we can deduce everything from Riemann--Roch. -Note by the way that the same argument shows that Chern classes are also constant in non-flat families, if you only replace the restriction $F_r$ by derived restriction. - -REPLY [6 votes]: As soon as ${\rm Pic}^\circ(X)$ is non-trivial there is a counter-example. Take a non-trivial line bundle $\mathcal L$ that deforms to the trivial one. Then $c_1(\mathcal L)$ goes from non-zero to zero.<|endoftext|> -TITLE: Weight filtration and Hodge theory for tropical varieties -QUESTION [12 upvotes]: Many concepts is algebraic geometry have tropical analogues. -Question: Is there an analogue of the weight filtration or Hodge filtration for tropical varieties? -A tropical curve ends up being essentially a metric graph. The tropical genus is the first Betti number of the graph. There is a period mapping (analogous to the classical Abel-Jacobi period map) from the moduli space of tropical curves of genus $g$ to the space $GL_g(\mathbb{R})/O_g(\mathbb{R})$. Can this period map be interpreted as a classifying map for variation of tropical Hodge structure? - -REPLY [7 votes]: Itenberg, Kazarkov, Mikhalkin and Zharkov have formulated a definition of tropical $H^{p,q}$. You can read about it in this preprint. -Their definition is restricted to the case that $\mathrm{Trop} \ X$ locally looks like a tropical linear space. (For example, if $X$ is a curve then, at a vertex of degree $d$, the directions of the incoming edges must span a space of dimension $d-1$ and the unique relation between the minimal lattice vectors on these directions must be that their sum is $0$.) This can be thought of as a tropical "smoothness" condition. -Roughly speaking, $H^{0,q}$ is related to the topological cohomology of $\mathrm{Trop} \ X$ which should, in this context, be viewed as the cohomology of the sheaf of locally constant functions on $\mathrm{Trop} \ X$. $H^{p,q}(X)$ is related to the cohomology of a sheaf on $\mathrm{Trop} \ X$ which is related to the degree $p$ part of Orlik-Solomon algebras of the matroids locally describing the relevant tropical linear spaces. (In other words, to $H^p$ of the corresponding hyperplane arrangement complement.) - -The authors have been talking about these results for long before the above linked preprint was made public. Here are some earlier references: -A 2009 lecture by Zharkov at MSRI. Section 2.2 of Kristin Shaw's thesis (a student of Mikhalkin).<|endoftext|> -TITLE: Interesting Calculus Questions/Exercises -QUESTION [52 upvotes]: I am in the process of redesigning the calculus course that I have taught five or six times. What I would like to know is if anyone has some really good examples or exercises that I could either do in class or give as a project. In particular, I've found that I don't have many good examples/exercises that illustrate the awesomeness of the main theorems (Intermediate Value Theorem, Mean Value Theorem, etc.). All levels of difficulty are certainly appreciated. The intent is to have material that I can present or assign here and there throughout the course that goes beyond basic calculus and will challenge even those to whom math comes naturally. -An example of what I'm looking for is something like showing a continuous function on $S^1$ has to map two antipodal points to the same value. -EDIT: In response to Qiaochu Yuan, Calc I and II together form all of single variable calculus. For Calc I: limits, differentiation, Riemann integration (improper as well). For Calc II: sequences, series, polar coordinates, parametric coordinates. The old book for this course was Stewart's "Calculus: Early Transcendentals", but I don't follow any book when I teach. - -REPLY [2 votes]: I think the "period three implies chaos" theorem can be proved at the level under discussion. This is the result that says that if $f:I\to I$ is a continuous map from an interval to itself with a point of period 3 then for every positive integer $k$, $f$ has a point of period $k$. -There's an exposition by Li and Yorke in the Monthly, 82 (Dec. 1975) 985-992, and I'm sure there are expositions elsewhere.<|endoftext|> -TITLE: Can surfaces be interestingly knotted in five-dimensional space? -QUESTION [22 upvotes]: It's possible this question is trivial, in which case it will be answered quickly. In any case, I realized that it's a basic question the answer to which I should know but do not. -Everybody loves knots — one-dimensional compact manifolds mapped generically into three-dimensional compact manifolds — and it's natural to ask about "knots" in higher dimensions. Of course, the space of generic maps of a one-dimensional compact manifold into a four-dimensional compact manifold is connected, so there is no interesting "knotting". Instead, people usually think about "surface knots in 4d", which are usually defined as embedded compact 2-manifolds in a 4-manifold. -But surfaces can map into 4-space in much more interesting ways. In particular, whereas a generic map from a 1-manifold to a 3-manifold is an embedding, two generic surfaces in 4-d can be "stuck" on each other: the generic behavior is to have point intersections. So a richer theory than that of embedded surfaces in 4-space is one that allows for these point self-intersections — it would be the theory of connected components of the space of generic maps. -Still, though, thinking about these self-intersections is hard, and their existence is part of what makes 2-knot theory hard (for instance, it interferes with developing a good "Vassiliev" theory for 2-knots). If you really want to reproduce the fact that generic maps have no self intersections, you should move the ambient space one dimension higher. -Hence my question: - -Can compact 2-manifold embedded into a compact 5-manifold be interestingly "knotted"? I.e. let $L$ be a compact 2-manifold and $M$ a compact 5-manifold; are there multiple connected components in the space of embeddings $L \hookrightarrow M$? - -I expect the answer is "no", else I would have heard about it. But my intuition is sufficiently poor that I thought it best to ask. - -REPLY [6 votes]: I think the first reference in the literature for the result you want is Wen-Tsun Wu, On the isotopy of $C^r$-manifolds of dimension $n$ in Euclidean $(2n+1)$-space. Sci. Record (N.S.) 2 1958 271--275. -In it he proves any two $C^r$-embeddings $M^n \to \mathbb R^{2n+1}$ are isotopic, provided $n>1$. $M^n$ is any compact $n$-manifold (the title does not mention he also assumed connected, but it is). The techniques are now standard -- your two embeddings $f_0, f_1 : M \to \mathbb R^{2n+1}$ are homotopic $f_t : M \to \mathbb R^{2n+1}$ so you look at the "graph", $(x,t) \longmapsto (f_t(x), t)$ as a map $M \times [0,1] \to \mathbb R^{2n+!} \times [0,1]$ and then approximate by an immersion and consider appropriate usage of the Whitney trick.<|endoftext|> -TITLE: Number Theory Representation of Primes -QUESTION [5 upvotes]: For a primes $p$ sufficiently large, does there always exists -positive integers $k,a,b\in\mathbb{N}$ such that $p=(k+1)(ab)+k(a+b)$ or equivalently -$p\equiv (ab)\bmod ((a+b)+ab)$? -Please note that from Frobeneius $p=(k+1)X+kY$ is solvable in non-negative integers $X,Y$ -provided p>XY-(X+Y), what I am looking for is the existence of a sub-class of positive intger solutions where X=ab and Y=a+b. - -REPLY [3 votes]: Thanks for your interesting comments particularly about the Sophie Gemaine Primes. It is interesting that when $2p+1$ is composite with $2p+1=cd$ where $c\equiv 1(\bmod 4)$ and -$d\equiv 3\bmod 4)$ there is always a solution. Another avenue of exploration I have been thinking about is to use some argument around constructing Covering Systems of Congruences, but this is a vague idea at best.<|endoftext|> -TITLE: Generalized Euler phi function -QUESTION [18 upvotes]: Let $n$ be an integer, there is a well-known formula for $\varphi(n)$ where $\varphi$ is the Euler phi function. Essentially, $\varphi(n)$ gives the number of invertible elements in $\mathbb{Z}/n\mathbb{Z}$. My questions are: -1) Since Dedekind domains have the same factorization theorem for ideals analogous to that of the integers, can one define a generalized Euler phi function type for an ideal of a Dedekind domain, i.e, $\varphi(I)$ shall give the number of invertible elements in $R/I$, and is there a nice formula for it? It makes sense to me that perhaps the formula should resemble that of the integer, using the factorization of $I$ into prime ideals. But I do not have a concrete idea of what it should be. -2) What about domains that are not Dedekind, more specifically, what are the minimum hypotheses that one can impose on a domain so that one can have perhaps a formula for Euler phi function type on the ideals? I am not sure if this even makes sense at this point. - -REPLY [2 votes]: An old and two new references : Page 13 of W. Narkiewicz, Elementary and Analytic Theory of Algebraic Numbers, third edition, Springer Monogr. Math., Springer-Verlag, 2004. -C.Miguel, Menon’s identity in residually finite Dedekind domains, Journal of Number Theory 137 (2014) 179–185. DOI: 10.1016/j.jnt.2013.11.003 -A. Kobin, A Primer on Zeta Functions and Decomposition Spaces, arXiv:2011.13903v1.<|endoftext|> -TITLE: On locally convex (and compactly generated) topological vector spaces -QUESTION [6 upvotes]: Part 1: -How big is the category $TVS_{loc.conv.}$ of locally convex topological vector spaces (and continuous maps)? -In other words (and less cheekily), is there a free locally convex TVS having any given set as basis? This would imply the functor $TVS_{loc.conv.} \to Set$ is essentially surjective and has an adjoint. -Part 2: -Consider now the intersection $T$ of $TVS_{loc.conv.}$ (as a subcategory of $Top$) with $CGWH$, the subcategory of $Top$ of compactly generated weak Hausdorff spaces. -How big is $T$? (Or, is $T$ essentially small?) -Note that a Banach space is locally compact iff it is finite dimensional, but I am being stupid and not remembering the relationship between local compactness and compact generation, so I can't immediately use this fact. - -REPLY [4 votes]: Since the first question doesn't seem to have been addressed directly in the answers so far, here are some suggestions. Firstly, since we are discussing topological vector spaces, I think the most natural question is to consider the forgetful function onto the category of topological spaces and since functional analysts are interested in function spaces, to completely regular spaces (of course, the case of sets can be incorporated by regarding a set as a discrete topological space). One then has a natural construction of the free locally convex space---one takes the free vector space generated by the topological space $X$ and provides it with the finest locally convex topology which agrees on $X$ with the original one. In our situation, this will be Hausdorff and will contain $X$ as a closed topological subspace. It is simple and natural to carry this one stage further and consider the completion of this space. It will have the corresonding univeral property, now for functions with values in a complete locally convex space. This space has a natural explicit representition, e.g., if we start with $[0,1]$, we get the space of Radon meassures on the interval. One of the nice things about this construction is that it can be varied almost infinitely and provides a unified approach to many spaces whose initial development was slow and painful---some of which are again forgotten lore. -As examples, we can consider spaces with the universal property for bounded functions and replace continuity by other smoothness conditions---uniform continuity if $X$ is a uniform space, $C^\infty$ if $X$ is an open subset of some euclidean space, holomorphicity (subsets of the complex plane or its higher dimensional analogues), measurablility if $X$ is a measure space and so on---similarly for functions on suitable manifolds. This provides a unifying approach to such topics as uniform measures, distributions, analytic functionals and so on. -As regards the second question, I have the feeling that functional analysts and topologists use the term compactly generated with different meanings. For the former, a locally convex space (in particular, a Banach space) is compactly generated if it contains a compact subset whose span is dense. For the latter, a topological space is compactly generated if it has the finest topology which agrees with itself on compact sets (otherwise known as a $k$-space or a Kelley space). As remarked above, metric spaces have the latter property and have the former one if they are separable. Further examples of spaces which have the latter property without being metrisable are the so-called Silva spaces, i.e. countable inductive limits of sequences of Banach spaces with compact interconnecting mappings. Many of the important spaces of distributions belong to this class, as do spaces of analytic functionals.<|endoftext|> -TITLE: What is the relationship between modular forms and Maass forms? -QUESTION [21 upvotes]: Modular forms are defined here: -http://en.wikipedia.org/wiki/Modular_form#General_definitions -Maass forms are defined here: -http://en.wikipedia.org/wiki/Maass_wave_form -I wonder if modular forms can be transfered into Maass forms. -Or they two are different categories of automorphic forms. - -REPLY [31 votes]: Automorphic forms correspond to representations that occur in $L^2(G/\Gamma)$. In the case when $G$ is $SL_2$, holomorphic modular forms correspond to (highest weight vectors of) discrete series representations of $G$, while Maass wave forms correspond to (spherical vectors of) continuous series representations of $G$.<|endoftext|> -TITLE: On the structure of the maximal abelian Galois group of a number field -QUESTION [8 upvotes]: Let $K$ be a number field. I am wondering if the following exact sequence $$1 \longrightarrow[\widehat{\mathcal O}_K^\times] \longrightarrow Gal(K^{ab}/K) \overset{\pi}{\longrightarrow} Cl_K \longrightarrow 1$$ splits, i.e. if there is a homomorphism $s:Cl_K \to Gal(K^{ab}/K)$ such that $s \circ \pi = id$. Here $[\widehat{\mathcal O}_K^\times]$ stands for the image of $\widehat{\mathcal O}_K^\times$ in $Gal(K^{ab}/K)$ under Artin's reciprocity morphism and $Cl_K$ denotes the ideal class group of $K$. -Equivalently we can rewrite the above sequence idelically as -$$1 \to \widehat{\mathcal O}_K^\times / \overline{\mathcal O _{K,+} ^\times} \to \widehat{\mathcal O}_K^\natural / \overline{\mathcal O _{K,+}^\natural} \to \widehat{\mathcal O}_K^\natural / (\widehat{\mathcal O}_K^\times \cdot \mathcal O _{K,+}^\natural) \to 1,$$ -where $\mathcal O _K ^\natural = \mathcal O _K - \{ 0 \} $, $\widehat{\mathcal O}_K^\natural = \mathbb A _{K,f}^\times \cap \widehat{\mathcal O}_K$. Further, subscript $+$ denotes totally positive elements and over lined objects are meant to be closures in the idele topology. (I don't know if this is helpful.) - -REPLY [8 votes]: Let $K = {\mathbb Q}(\sqrt{-5})$ and let $P = (3+2\sqrt{-5})$ denote a prime ideal of norm $29$ in the ring of integers of $K$. There does not exist a quadraic extension of $K$ ramified exactly at $P$, but there is one over the Hilbert class field $K(i)$ of $K$. This means that the ray class group modulo $P$ does not split into the product of the ideal class group, which has order $2$, and another group. In particular, the maximal abelian extension unramified outside of $P$ does not split into a compositum of an unramified extension and a purely ramified extension, and neither does $K^{\rm{ab}}$.<|endoftext|> -TITLE: Equidistribution relative to the Zariski topology -QUESTION [17 upvotes]: I'm interested in measuring the distribution of an ordered set of points relative to the Zariski topology. Possibly this is a standard idea (with different terminology?) in algebraic geometry. Any pointers to the literature would be appreciated. Here's what I have in mind. -Let $X\subset\textbf{P}^N$ be a projective variety defined over an algebraically closed field $k$. For any finite set of points of $X$, define the $X$-degree of the set to be the minimal degree of a homogeneous polynomial $F\in k[X_0,\ldots,X_n]$ that does not vanish identically on $X$, but vanishes at all of the points in the set. (In other words, the minimal degree of a hypersurface that goes through the points, but does not contain $X$.) -Now take a sequence of points $S=(P_1,P_2,P_3,\ldots)\subset X$, and for each $n\ge1$, let $S[n]=\{P_1,P_2,\ldots,P_n\}$. Intuitively, if $\deg_X S[n]$ grows quickly, then the set $S$ is well distributed relative to the Zariski topology. A rough guess is that one might define equidistribution by the condition -$$ - \lim_{n\to\infty} \frac{\log\deg_XS[n]}{\log n} = \frac{1}{\dim X}. -$$ -(It is not hard to check that the limsup of the left-hand side is at most $\frac{1}{\dim X}$.) -[Edit: value of limit fixed as suggested by JSE] - -REPLY [5 votes]: First of all, I think the limit above would like to be 1/(dim X) rather than 1/N -- given d^N points on X, there will indeed be a degree-d polynomial passing through them, but this polynomial might well vanish on all of X (e.g. consider the case where X is a line in P^N -- in order to pass through d points on the line and not vanish on the line, your polynomial had better have degree d.) -I would say this is closer to a weak notion of "general position" than of "equidistribution" -- for points on the plane, aren't you asking for some coarse version of "not too many of the first few points on a line, not too many of the first few points on a conic, ..... etc?" -Anyway, as for where this kind of thing appears in the literature, maybe the argument of Heath-Brown about uniform bounds for points on curves? (Huayi Chen has some appealing-looking recent papers extending this approach to more general varieties.) You could describe these results as saying that if you have a long list of rational points whose height grows very slowly, they DO tend to line up on unexpectedly low-degree hypersurfaces; so in your context, this would be the opposite of the usual situation in which slow growth of height enforces equidistribution!<|endoftext|> -TITLE: Using zero-sharp to characterize L -QUESTION [6 upvotes]: Hi all, -In most set-theory accounts dealing with $ 0 ^ \sharp $ and its related effects on the constructible sets, a proof of the following theorem is crucial: -(*) Assuming there is a measurable cardinals, for uncountable limit cardinals $ \alpha , \beta $ $ L_\alpha $ is elementary equivalent to $ L_\beta $. -Few accounts are Jech, Kanamori, Devlin and Silver's papers. -The proof goes through a heavy abstract machinery of the Silver Indiscernibles in model theory, relying on a Ramsey cardinal to find a "remarkable well founded E.M set". -My question - does anyone know of a shorter proof for (*)? something along the line of Godel condensation theorem for L? - -REPLY [8 votes]: A short proof of $(*)$ uses the characterization of $0^\sharp$ in terms of mice. A nice account of this version is in Ernest Schimmerling's paper "The ABC of mice". -The idea is that a sharp is a kind of "local measurable cardinal". More formally, we can think of a set equivalent to $0^\sharp$ (in a strong sense) to be of the form $M=(L_{(\kappa^+)^L},\in,{\mathcal U})$ for some $L$-cardinal $\kappa$ and some $L$-measure ${\mathcal U}$ on $\kappa$. We ask some definability requirements of $M$ that imply that $M$ is countable (in $V$), and we require that we can iterate ultrapowers by ${\mathcal U}$ and obtain well-founded models. -(I say "local measurable" because as soon as we continue the constructible hierarchy over $M$, we are able to define a map from $\omega$ onto $L_{(\kappa^+)^L}$, so not only is $M$ countable, but ${\mathcal U}$ is very far from being a measure in $L[M]$. On the other hand, ${\mathcal U}$ measures all the subsets of $\kappa$ in $L$, so ultrapowers using ${\mathcal U}$ of $L$ or initial segments of $L$ make sense.) -It follows at once from continuity properties of embeddings that if $\alpha<\beta$ are uncountable cardinals in $V$, then iterating the ultrapower embedding by ${\mathcal U}$ will send $L_\kappa$ eventually to $L_\alpha$ and later to $L_\beta$, and so in particular $L_\alpha\equiv L_\beta$. -What we are doing here is casting Kunen's characterization of $0^\sharp$ as the key defining property. All of this, and the way to recover the usual version of $0^\sharp$ from this one, is explained in very nice detail in Ernest paper, that I recommend you study. -The presentation itself is part of the folklore of the subject. I learned it from John Steel. It is essential to understand this presentation if one is to make sense of current fine structure or inner model theory. -That being said, the EM-blueprints approach is not outdated. It is essentially the only way we can make sense of sharps of sets that are not 'internally well-ordered'. This idea was first developed by Robert Solovay when he introduced ${\mathbb R}^\sharp$ in the course of his work on determinacy. It has been much refined by Hugh Woodin during his study of the Chang model. - -REPLY [5 votes]: I think you are just asking for a direct proof from a measurable cardinal. In this case, here is one way to do it. If there is a measurable cardinal $\kappa$ with normal measure $\mu$, then by considering the collapse of a countable elementary substructure $X\prec V_\theta$ one finds a countable transitive set $M_0$ with a measurable cardinal $\kappa_0$ and normal measure $\mu_0$. Since the iterated ultrapower of $M_0$ by $\mu_0$ maps directly into the iteration of $V$ by $\mu$, we see that all such iterations are well-founded. In other words, $\langle M_0,\in,\mu_0\rangle$ is iterable. Let $j_\alpha:M_0\to M_\alpha$ be the $\alpha$-th iteration, where at successor steps we take the ultrapower by the image of $\mu_0$, and at limit stages we take the direct limit. Since the internal ultrapowers do not increase cardinality and the maps are continuous at limits, it follows that the corresponding images $\{j_\alpha(\kappa_0) \mid \alpha\in ORD\}$ is a closed unbounded class of ordinals, and contains all uncountable cardinals of $V$, since the process is continuous at limits. -Now, just restrict to $L$ and observe that $M_{\kappa_\alpha}\prec M_{\kappa_\beta}$ whenever $\alpha\lt\beta$, because the iteration between $\alpha$ and $\beta$ has critical point $\kappa_\alpha$. In particular, $L_{\kappa_\alpha}\prec L_{\kappa_\beta}$, and this includes all the uncountable cardinals, so your conclusion is obtained. -One can weaken measurability just to having a countable iterable structure and use the same argument. -(But I see now that Andres has just posted an answer with the same idea.)<|endoftext|> -TITLE: Reference for combinatorics of cell decomposition of the Hilbert scheme of points in the plane -QUESTION [13 upvotes]: It is known from either Morse theory or Bialynicki-Birula decomposition that the fixed points of a ${\mathbb{C}}^*$ action on a smooth algebraic variety over $\mathbb{C}$ determine a cell decomposition of this variety (cells correspond to fixed points by sending each point to a fixed point by the ${\mathbb{C}}^*$ action). Such cell decomposition of the Hilbert scheme of points on the plane $({\mathbb{C}}^2)^{[n]}$ is described in several sources, notably the original papers by G. Ellingsrud and S.A. Stromme and Chapter 5 of the book about Hilbert schemes of points on surfaces by H.Nakajima. -It follows that the fixed points of the torus action (and hence the cells) on $({\mathbb{C}}^2)^{[n]}$ are indexed by partitions of $n$. However, it is hard for me to infer from the above references what is the combinatorial order of these cells (i.e. which cells lie on the boundary of which other cells) -- this should induce a certain order on the partitions of $n$. Is there a reference which would state what this order is explicitly? - -REPLY [9 votes]: It might be useful to see quickly why Jeremy's answer, although a very reasonable guess, is wrong. Consider the two partitions $(2,2)$ and $(3,1)$. In refinement order, neither one is greater than the other. -Consider the subscheme of $\mathbb{P}^1 \times \mathbb{C}^2$ cut out by -$$x^2=u xy + v y^2=y^3=0$$ -where $(x,y)$ are the coordinates on $\mathbb{C}^2$ and $(u: v)$ are the homogenous coordinates on $\mathbb{P}^1$. Every fiber over $\mathbb{P}^1$ has length $4$, so this is a flat family, and we get a map $\mathbb{P}^1 \to \mathrm{Hilb}_4(\mathbb{C}^2)$. -The image of this map is a torus invariant curve in the Hilbert scheme. Its two torus fixed points are the monomial ideals $\langle x^2, y^2 \rangle$ and $\langle x^2, xy, y^3 \rangle$, corresponding to the partitions $(2,2)$ and $(3,1)$. -So, in any Bialynicki-Birula decomposition, one of these partitions must dominate the other.<|endoftext|> -TITLE: Do isomorphic semi-direct products correspond to conjugate automorphisms? -QUESTION [5 upvotes]: Let $H$ and $N$ be two groups with $H$ cyclic. Let $f,g:H \rightarrow \mathrm{Aut}(N)$ be homomorphisms such that $N\rtimes _f H \cong N\rtimes _g H$. Then does that mean $f(H)$ and $g(H)$ are conjugate in $\mathrm{Aut}(N)$? - -REPLY [7 votes]: The answer is no. You can easily have a situation where $f$ is the trivial map, while $g$ makes $H$ act through an inner automorphism of $N$, so that in both cases $N\rtimes H\cong N\times H$. For concreteness, let $N=D_8$ be the dihedral group of order 8, let $\sigma$ be a non-central involution in $N$, let $H=\langle h\rangle\cong C_2$, and define $g(h)(n)=\sigma^{-1} n\sigma\;\forall n\in N$. Then $G=N\rtimes_g H\cong N\times H$, since the subgroup $\langle h\sigma\rangle$ of $G$ is of order 2, intersects $N$ trivially, and commutes with it.<|endoftext|> -TITLE: Self Avoiding Walk Enumerations -QUESTION [5 upvotes]: Let $c(n)$ be the number of Self avoiding walks (SAW) of length $n$ on an infinite lattice $L$. Are there any known non-geometric interpretations of $c(n)$?. For example, is there a number theoretic version of SAW's? We know for example that Catalan numbers count a myriad of things so perhaps SAW appear elsewhere? For reference, $c(n)$ for the usual 2-D integer lattice looks like 1,4,12,36,100,284,780,,... for $n=0,1,2,... -The online integer sequence library gives no results other than the number of SAWs. -Note: I'm not exactly interested in characterizations, such as the ones in this question. Something like the Hammersley and Welsh characterization in terms of excursion width is closer to what I'm looking for. In fact one sees partition functions crop up naturally, with theorems of Hardy and Ramanujan used for growth estimates. - -REPLY [6 votes]: I strongly doubt that there exists a non-trivial bijection between self-avoiding walks on Z^d and other combinatorial objects. I don't have any intuition as to whether the generating function may have some other number theoretic interpretation, but I haven't seen or heard anything in this regard. -SAWs are sufficiently complicated that it seems almost certain that the generating function for SAWs even on the square lattice them is non-holonomic (see e.g. Andrew Rechnitzer discussing the related model of self-avoiding polygons, "Haruspicy 2: The anisotropic generating function of self-avoiding polygons is not D-finite", link), and the model is unlikely to be "solvable" in the conventional sense. -FYI, Iwan Jensen has enumerated SAWs on Z^2 to 71 steps here, and higher dimensional enumerations due to Gordon Slade, Richard Liang and I are available here.<|endoftext|> -TITLE: Generalization of Curl to higher dimensions -QUESTION [7 upvotes]: In terms of vector field analogies to closed and exact differential forms, conservative and incompressible vector fields (gradient and divergence) generalize to higher dimensions, but curl and irrotational fields do not. Why? -Cross product doesn't generalize either but one can use exterior products and hodge duals to fullfill the need. In differential geometry, there is a duality between the boundary operator on chains and the exterior derivative as expressed by the general Stokes' theorem. -By a theorem of De Rham, the exterior derivative is the dual of the boundary map on singular simplices. From this perspective, there should be a generalized curl. Perhaps, there is an explanation in terms of the Poincare Lemma where for n>3 (but perhaps not 7), curl fails for higher dimensions? - -REPLY [3 votes]: The curl of a vector field $X=P\partial_x+Q\partial_y+R\partial_z$ is equal to -$$ -\mathrm{Curl}(X)= (R_y-Q_z)\,\partial_x +(P_z-R_x)\,\partial_y+ (Q_x-P_y)\,\partial_z -$$ -For the moment we replace $\partial_x,\partial_y,\partial_z$ with $dy\wedge dz,\ dx\wedge dz$ and $dx\wedge dy$ respectively (In fact we apply the Hodge star operator to the dual of the basis $\partial_x,\partial_y,\partial_z$). -So actually the component of $\mathrm{Curl}(X)$ is identical to the components of the $2$ form -$$ -\alpha=(R_y-Q_z)\,dy\wedge dz +(P_z-R_x)\,dx\wedge dz +(Q_x-P_y)\,dx\wedge dy -$$ -On the other hand the vector field $X$, being a section of the tangent bundle $T\mathbb{R}^3$, can be considered as a map -\begin{align} -X\colon \mathbb{R}^3& \to \mathbb{R}^3\times \mathbb{R}^3\\ -(x,y,z)&\mapsto (x,y,z, P(x,y,z),Q(x,y,z),R(x,y,z)). -\end{align} -Note that the following equality holds: -$$ -\alpha =X^* \omega, -$$ -where $\omega$ is the natural symplectic structure of $\mathbb{R}^3 \times \mathbb{R}^3$ with $\omega= dx\wedge dp +dy\wedge dq+dz \wedge dr$. -The situation described above is a motivation to consider the following generalization of the concept of the curl of a vector field on an arbitrary Riemannian manifold. - -A generalized curl: Let $(M,g)$ be a Riemannian manifold. The metric $g$ gives an isomorphism (hence diffeomorphism) between $TM$ and $T^* M$. So the standard intrinsic symplectic structure of the cotangent bundle is carried to a symplectic structure $\omega$ on $TM$. Now assume that $X:M \to TM$ is a vector field. We define the curl of $X$ as a $2$-form with the following formula: - $$ -\mathrm{Curl}(X):=X^* \omega. -$$ - -This was already mentioned at the MO question A generalization of Gradient vector fields and Curl of vector fields.<|endoftext|> -TITLE: Curvature and Riemannian metric -QUESTION [7 upvotes]: Hi all, -I am going to give a talk in a seminar about the general theme 'sum of squares'. My interests lie in Differential Geometry, so I recalled that the curvature is a local invariant, an obstruction to the metric being locally a sum of squares of coordinate differentials. -Can some people suggest me some good books which clearly illustrate the relation between the curvature and the metric? Also, do you guys have other suggestions about 'sum of squares' in Geometry? Thanks a lot! - -REPLY [2 votes]: This is maybe late for your seminar, but a classical textbook about Riemannian geometry, including relations between curvature and metric tensor, with any signature, is -Riemannian geometry by L.P.Eisenhart -Another very interesting book for you could be -Spaces of constant curvature by J. A. Wolf -Indeed, it seems that you are searching for the Riemannian manifolds whose metric element can be written as sum and/or difference of squares of coordinate differentials. This implies that the curvature is constant and equal to 0. As shown in Wolfs'book, this can be locally realized by several manifolds with different "global geometry". As example in dimension 2, $ds^2=dx^2+dy^2$ can be realized on the (Euclidean) plane, on the cylinder, on the torus, on the Moebius strip and on the Klein bottle, while $ds^2=dx^2-dy^2$ on the (Minkowski) plane, on the torus and on the Klein bottle.<|endoftext|> -TITLE: Algorithm for least distance of powers of integers -QUESTION [7 upvotes]: From Michailescu's theorem (Catalan's conjecture) we have that the only $a,b,m,n \in \mathcal{Z}^{+}$ with $m,n>1$ such that $a^{m} - b^{n} = 1$ are: $a=3$, $b=2$, $m=2$, $n=3$. -1) Is there an algorithm which, for any $a,b \in \mathcal{Z}^{+}$ finds the minimum of $|a^{m} - b^{n}|$ $\forall m,n \in \mathcal{Z}^{+}$ (with $m,n>1$ )? -2) Do we know for how many different values of $m,n$ this minimum distance can be achieved? -3) If we do not have such an algorithm, do we know if this problem is decidable? - -REPLY [5 votes]: I'd be curious to know how often the closest pair has $\min(m,n)>2$. Bennett shows that $|a^m-b^n|<\frac{\max(\sqrt{a^m},\sqrt{b^n})}{4}$ happens at most once. However this might not happen for most pairs $a,b$ and even when this happens, it might not be a min. (And it seems rare for it to happen with $\min(m,n)>1$.) A rather spectacular (to my mind) case is $13^3-3^7=10$. The first few convergents of the continued fraction for the ratio of the logs are $1/2,2/5,3/7,239/558$ This shows that 3/7 is an extremely good approximation (of course). But the closest pair is $13^1-3^2=4$. -later I looked for instances of $|a^m-b^n| -TITLE: What is Symplectic Area? -QUESTION [6 upvotes]: In classical Mechanics, momentum and position can be paired together to form a symplectic manifold. If you have the simple harmonic oscillator with energy $H = (k/2)x^2 + (m/2)\dot{x}^2$. In this case, the orbits are ellipses. How is the vector field determined by the (symplectic) gradient, then? -Also, does anyone know an interpretation for the area inside a closed curve in phase space? - -REPLY [2 votes]: You can view $\mathbb{R}^{2n}$ as a quotient of the real Heisenberg group $\mathcal{H}^{2n+1}$ modulo its center. For a closed loop $\alpha$ in $\mathbb{R}^{2n}$ and a point in $\mathcal{H}^{2n+1}$ over $\alpha(0)$ there's unique lift $\tilde{\alpha}$ of $\alpha$ to $\mathcal{H}^{2n+1}$ going through this point. The symplectic area enclosed by $\alpha$ expresses the signed distance from $\tilde{\alpha}(0)$ to $\tilde{\alpha}(1)$ with respect to a left invariant Riemannian metric on $\mathcal{H}^{2n+1}$.<|endoftext|> -TITLE: On rational functions with rational power series -QUESTION [9 upvotes]: Let $f(z)=\sum_{n\geq 0}a_n z^n$ be a Taylor series with rational coefficients with infinitely non-zero $a_n$ which converges -in a small neighboorhood around $0$. Furthermore, assume that -\begin{align*} -f(z)=\frac{P(z)}{Q(z)}, -\end{align*} -where $P(z)$ and $Q(z)$ are coprime monic complex polynomials. By developing $\frac{P(z)}{Q(z)}$ as a power sereis around $0$ and comparing it with $f(z)$ we obtain infinitely many polynomial equations in the roots of $P(z)$ and $Q(z)$ which are equal to rational numbers so this seems to force the roots of $P(z)$ and $Q(z)$ to be algebraic numbers. -Q: How does one prove this rigourously? - -REPLY [4 votes]: Well, I think there is a simpler argument. For a power series $g(x)\in\mathbb{C}[[x]]$ -and $\sigma\in Aut(\mathbb{C})$ (note that except for the complex conjugation or the identity $\sigma$ is not continuous!) we may define define the power series with coefficients twisted by $\sigma$ which we denote by $g^{\sigma}(x)$. Now an element in $Aut(\mathbb{C})$ respect finite sum and products so it follows from that, -that for all $\sigma\in Aut(\mathbb{C})$ one has -$$ -f^{\sigma}(z)=\frac{P^{\sigma}(z)}{Q^{\sigma}(z)}. -$$ -From this (and the unique factorization of $\mathbf{C}[x]$) it follows that $P(z)$ and $Q(z)$ have rational coefficients.<|endoftext|> -TITLE: The geometry of Nadirashvili's complete, bounded, negative curvature surface -QUESTION [17 upvotes]: I would like to understand the geometric structure of -a surface that Nadirashvili constructed which resolved what -was known as Hadamard's Conjecture. -Perhaps in the 15 years since his construction, others have -redescribed the example, and perhaps even made a graphics image of it? -Background. -Hilbert's theorem that implies that the hyperbolic plane cannot -be realized as a surfaces in $\mathbb{R}^3$ is well known. -Perhaps less well known is Hadamard's Conjecture, which -asked if there is a complete negative curvature surface -in a bounded region of $\mathbb{R}^3$. -This is discussed at some length in Burago and Zallgaller's -book Geometry III: Theory of Surfaces. The problem was -solved after that 1989 book was written, as Berger explains -in A Panoramic View of Riemannian Geometry (p.135): - -(Incidentally, the answer to this related MO question on Compact Surfaces of Negative Curvature -does not resolve my question, as it relies on Burago and Zallgaller.) -Here is the citation: - -Nikolaj Nadirashvili, - "Hadamard's and Calabi-Yau's conjectures on - negatively curved and minimal surfaces." - Invent. Math. 126(3) (1996), 457–465. - -The main theorem is this: - -Theorem. There exists a complete surface of negative Gaussian curvature - minimally immersed in $\mathbb{R}^3$ which is a subset of the unit ball. - -I have studied the paper, but my grasp of the underlying -mathematics is not strong enough to convert his description -into a geometric picture. -If anyone knows of later discussions that might help, I would -appreciate pointers or references. Thanks! -Edit. I was not able to access MathReviews until now. This is from the review by M. Cai (MR1419004 (98d:53014)): - -For the proof, the author starts with a minimal immersion of the unit disk into a fixed ball in $\mathbb{R}^3$ with the Gaussian curvature of the immersed surface being negative, then he inductively defines a sequence of minimal immersions of negative curvature into the fixed ball in such a way that the sequence converges to a complete immersion. - -This helps. - -REPLY [7 votes]: The conjecture attributed to Hadamard, if one regards that as being concerned with the existence of a complete embedded negatively curved surface in a ball, is still open. I have read the corresponding papers of Hadamard, and I think that it is fairly clear that he was concerned mainly with embedded surfaces. -The book Geometry III, edited by Burago and Zalgaller, contains articles by Rozendorn where the Hadamard conjecture for embedded negatively curved surfaces is discussed. It is also proposed there what the surface should look like: a coral with a fractal structure, as shown below. -                                      -The basic idea here is that the surface (or the coral) trifurcates at a sufficiently rapid rate as it approaches the boundary of the sphere, thus each point of the surface will be infinitely far from the boundary of the ball as measured within the surface. The tricky thing here, which has not been verified, is that these pieces can be glued together without the creation of flat points. It is known that the individual constituent pieces can be negatively curved.<|endoftext|> -TITLE: Does a Trivial Tangent Bundle Induce a Multiplication? -QUESTION [6 upvotes]: Let $M$ be a connected smooth manifold, and assume that it is parallelisable; that is, its tangent bundle is trivial. Does $M$ admit an H space structure? That is, does there exist a smooth map $\mu:M\times M\to M$ and an identity element $e$ satisfying $\mu(e,x)=\mu(x,e)=x$? -The motivation for asking is the following: given any Lie group $G$, its tangent bundle is trivial. What about the converse? It's hard enough coming up with a parallelisable manifold that is not a Lie group ($\mathbb{S}^7$ is such an example). The best idea I've heard is to think about quotients of Lie groups by discrete subgroups, but the few examples I've tried weren't parallelisable in the end. - -REPLY [12 votes]: Ryan Budney's comment pretty much killed the question, but anyway... -Let $X_{m,n} = S^{2m}\times S^{2n+1}$, with $m\le n$ (strictly) positive integers. -Lemma: $X=X_{m,n}$ is parallelisable. -Proof: this follows from playing around with vector bundles, the key facts being that $TX = \pi^*(TS^{2m}) \oplus \pi^*(TS^{2n+1})$ and that trivial bundles are natural. -More precisely, the second factor of $X$ has Euler number 0, so one can split off a trivial 1-bundle from its tangent bundle, pull it back through the projection and see it as the pull-back of a trivial bundle over the first factor. So $TX = \pi^*(TS^{2m}\oplus \varepsilon) \oplus V$, and the first summand is trivial. Now one can split off a trivial bundle of rank 2 from the first factor and use it to trivialise the second factor. -Lemma: $X$ is not an $H$-space. -Proof: the cohomology ring (say with $\mathbb{F}_2$ coefficients) of an $H$-space is a finite-dimensional commutative Hopf algebra, therefore it's generated in odd dimensions. But, in our case, $H^*(X)$ has $H^{2m}(X)$ as the first nontrivial group. -This gives a nice family of simply connected counterexamples to your question, the smallest of which is $S^2\times S^3$. Notice how, actually, there is no simply connected counterexample in dimension one, two, three (thanks to Perelman) and four (e.g. because the Euler number of a simply connected 4-manifold is strictly positive).<|endoftext|> -TITLE: Statistics of Extended GCD -QUESTION [13 upvotes]: Consider a coprime pair of integers $a, b.$ As we all know ("Bezout's theorem") there is a pair of integers $c, d$ such that $ac + bd=1.$ Consider the smallest (in the sense of Euclidean norm) such pair $c_0, d_0$, and consider the ratio $\frac{\|(c_0, d_0)\|}{\|(a, b)\|}.$ The question is: what is the statistics of this ratio as $(a, b)$ ranges over all visible pairs in, for example, the square $1\leq a \leq N, 1 \leq b \leq N?$ -Experiment shows the following amazing histogram: -EDIT by popular demand: the histogram is for an experiment for $N=1000.$ The $x$ axis is the ratio, the $y$ axis is the number of points in the bin. The total number of points is $1000000/\zeta(2),$ so there are $100$ bins each with around $6000$ points. -But no immediate proof leaps to mind. - -REPLY [3 votes]: Here is an attempt to give a somewhat finer grained view of the distribution. The set of ratios $\sqrt{\frac{s^2+t^2}{a^2+b^2}} \subset(0,\frac{1}{2})$ are essentially the values in the first half of the Farey sequence $\lbrace \frac{p}{q} | \gcd(p,q)=1,\ 2p \le q -TITLE: Is anything known about the "closure" of an additive category by adding all the images and kernels? -QUESTION [12 upvotes]: It is possible well-known: is there a "minimal" pre-abelian (or abelian) category, containing the given additive category, or which conditions should be performed for this property? -For example, let us take a category of vector bundles over a given manifold. Then, we can include it into a category of sheaves on this manifold, and close it there taking kernels and cokernels, finally getting the category of perfect sheaves (sheaves with finite resolution). -Overlooking the size of the category of vector bundles and all the sheaves, we can see that this construction somehow is similar as trying to find $\mathbb{Q} \subset \mathbb{R}$, knowing $\mathbb{Z} \subset \mathbb{R}$ inclusion. -Maybe any abstract procedure for this "closure" exists, or there are many "minimal" pre-abelian categories? -Idea 1. -It may be possible to define an universal property of this category. For example, this might work: "after gluing all the isomorphic objects together, if $V$ is our additive category, $V \hookrightarrow A$ is it's closure and $B$ is other pre-additive category, equipped by morphism $V \hookrightarrow B$ there are $A \longrightarrow B$, commuting with previous two morphisms". -Idea 2. -I've tried to construct this category directly. If we know a left resolution for an object $X$, we know $Hom(X, A)$. If we know right resolution, we know $Hom(A, X)$. Left resolution and right resolution together is an exact sequence $0 \rightarrow L_n ... \rightarrow L_1 \rightarrow R_1 ... \rightarrow R_n \rightarrow 0$. -Now let us assume that we actually know what an exact sequence is. Than we can add to our category a new object, corresponding to the one of the morphisms of a given exact sequence, and define Hom's to this object and from this object. -What's with the composition? -Let us work with a sequence of length four (which will simplify our calculations). I don't know anything about the case of long sequences. -So, having a sequence $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow D \rightarrow 0$ we imeddiately add a new object $X$, equipped with -$0 \rightarrow A \rightarrow B \rightarrow X \rightarrow 0$ -and -$0 \rightarrow X \rightarrow C \rightarrow D \rightarrow 0$ -Then, watching the properties of $Hom$, for any object S we define $Hom(S, X)$ by an exact sequence $0 \rightarrow Hom(S, X) \rightarrow Hom(S, C) \rightarrow Hom(S, D)$ (and $Hom(X, S)$ dually). -It's easy to see that $Hom(X, X)$ is well-defined. -There are also no problems with compisition of $Hom(X, S)$ and $Hom(S, T)$ for any $S$ and $T$. We just lift $Hom(X, S)$ to the $Hom(B, S)$, compose them and check that it's well-defined. -But how can we compose morphism of $Hom(S, X)$ and $Hom(X, T)$? Both parts of our exact sequence participate in the definition of this sets! So we need a some kind of condition on our "exact sequence". -I think I know what to do with it. -But maybe this is a well-known construction? Thanks for answering. - -REPLY [6 votes]: Note: I'm only addressing David's question how one one can "add cokernels" to an additive category and how one can use this in order to embed an additive category into an abelian one, even in a universal way. -Concerning the request on minimality in the question, it is unclear to me what exactly it is, Lev wants to achieve. What happens in the mentioned example of sheaves that one identifies an interesting category (vector bundles = finitely generated projective $\mathcal{O}_{X}$-modules) and ends up with the category coherent sheaves. In this case, the category constructed below embeds into the category of sheaves just because one started out with a class of projective objects in an abelian category already. - -The idea is to embed the additive category $\mathcal{A}$ into the category of morphisms $\mathcal{A}^{\to}$. Note that a morphism in $\mathcal{A}^{\to}$ corresponds to a commutative square. More precisely, the idea is that a morphism $(A \to B)$ of $\mathcal{A}$ should represent its cokernel. -Let's pretend this works. Since the cokernel of $0 \to A$ in $\mathcal{A}$ is $A$, we see that we must embed $\mathcal{A}$ via $A \mapsto (0 \to A)$. This embedding of $\mathcal{A} \to \mathcal{A}^{\to}$ is clearly fully faithful. But we're not quite there, yet. There are nonzero morphisms of morphisms $(A^{-1} \to A^{0}) \to (B^{-1} \to B^{0})$ that should induce the zero map on the (putative) cokernels, namely precisely those for which the map $A^{0} \to B^{0}$ factors over $B^{-1}$: - -It is easy to see that these morphisms form an ideal $\mathcal{J}$ in $\mathcal{A}^{\to}$ (they are closed under composition and sums), so we may factor this ideal out. In other words, a morphism of $\mathcal{A}^{\to}$ is identified with zero if and only if it lies in $\mathcal{J}$. The resulting category $\text{fp}(\mathcal{A}) = \mathcal{A}^{\to}/\mathcal{J}$ is what we want because we have: - -Theorem (Freyd, 1965; Beligiannis, 2000) - -The category $\text{fp}(\mathcal{A})$ has cokernels. The functor $A \mapsto (0 \to A)$ is fully faithful and universal among functors to additive categories with cokernels: more precisely, every functor $F: \mathcal{A} \to \mathcal{C}$ to a category with cokernels extends uniquely to a cokernel-preserving functor $\text{fp} (\mathcal{A}) \to \mathcal{C}$. -Moreover, $\text{fp}{(\mathcal{A})}$ is abelian if and only if $\mathcal{A}$ has weak kernels (a weak kernel has the factorization property of a kernel but uniqueness of the factorization is not required). - - -So if $\mathcal{A}$ has weak kernels, we're already done. If not, we may play the same game again, using this theorem. We first embed the category with kernels -$(\text{fp} \mathcal{A})^{\text{op}}$ into the abelian category -$\text{fp}\left( (\text{fp} \mathcal{A})^{\text{op}} \right)$ and then pass to the opposite category $\mathcal{F}{(\mathcal{A})} = \left(\text{fp}\left( (\text{fp} \mathcal{A})^{\text{op}} \right)\right)^{\text{op}}$. The diagrams get a bit unwieldy, but one may check that: - -Theorem (Adelman, 1971) - The embedding $\mathcal{A} \to \mathcal{F}(\mathcal{A})$ is fully faithful and universal among all functors to abelian categories (every functor to an abelian category extends to an exact functor on $\mathcal{F}(\mathcal{A})$). - -The category $\text{fp}(\mathcal{A})$ is sometimes used in connection with the derived category/triangulated categories (the inclusion $\mathcal{T} \to \text{fp}(\mathcal{T})$ is the universal homological functor on the triangulated category $\mathcal{T}$. It already appears in Verdier's thesis - attributed to Freyd). Closely related are also the hearts of $t$-structures (perverse sheaves). -The most important references are: - -Peter Freyd, Representations in abelian categories, Proc. Conf. Categorical Algebra (La Jolla, Calif., 1965), Springer, New York, 1966, pp. 95–120. -A. Beligiannis: ''On the Freyd Categories of an Additive Category'', Homology, Homotopy and Applications Vol. 2, No.11 (2000), pp. 147-185.<|endoftext|> -TITLE: Sums of subsets of $\mathbb{Z}/n\mathbb{Z}$ -QUESTION [11 upvotes]: I have encountered a problem that I suspect has been thoroughly studied but I have not been able to find references. Can anyone point me to a published reference dealing with this or a closely related problem? -Here is the problem: -Let $A$ and $B$ be disjoint $k$-subsets of $\mathbb{Z}/n\mathbb{Z}$. Consider $S(A,B)=\sum_{x\in A}x - \sum_{y\in B}y$. As $(A,B)$ ranges over all possible ordered pairs of disjoint $k$-subsets of $\mathbb{Z}/n\mathbb{Z}$, how are the sums $S(A,B)$ distributed over the elements of $\mathbb{Z}/n\mathbb{Z}$? More precisely, for how many of the $\binom{n}{k}\binom{n-k}{k}$ choices of $(A,B)$ is $S(A,B)$ equal to each of the elements of $\mathbb{Z}/n\mathbb{Z}$? -Again I am just looking for references. I actually have a solution in the case that n is prime, but I assume the result is known for more general n. I would be interested in any leads. - -REPLY [6 votes]: For convenience I will first consider how many sums $S(A,B)$ are equal -to 0. I might have made some computational errors, but I think that -the method is correct. Regard $n$ as fixed. Let $f_k$ be the number of -pairs $(A,B)$ of $k$-element subsets (not necessarily disjoint) of -$\mathbb{Z}/n\mathbb{Z}$ such that $\sum_{x\in A}x-\sum_{y\in -B}y=0$. Let $g_k$ be the same, except that $A$ and $B$ must be -disjoint. It is easy to see that - $$ f_k = g_k+{n-2(k-1)\choose 1}g_{k-1}+{n-2(k-2)\choose - 2}g_{k-2}+\cdots. $$ -These equations for $0\leq k\leq n$ can be inverted to express -the $g$'s in terms of the $f$'s, so it suffices to find $f_k$. -Let $[x^iy^j]P(x,y)$ denote the coefficient of $x^iy^j$ in the -polynomial $P(x,y)$. By standard properties of roots of unity, we have - \begin{eqnarray*} f_k & = & [x^ky^k] \frac 1n \sum_{\zeta^n=1} - \prod_{j=0}^{n-1} (1+x\zeta^j) (1+y\zeta^{-j})\\ & = & - \frac 1n\sum_{d|n}\phi(d)(1-(-x)^{n/d})^d (1-(-y)^{n/d})^d. - \end{eqnarray*} -It is routine to extract the coefficient of $x^ky^k$, -Completely analogous reasoning works for sums equal to any $m\in -\mathbb{Z}/n\mathbb{Z}$, though the computation becomes messier. A -related computation appears in -http://math.mit.edu/~rstan/pubs/pubfiles/8g.pdf. See also Enumerative -Combinatorics, vol. 1, Exercise 1.105.<|endoftext|> -TITLE: Is it possible to construct (without choice, even?) a non-finitely-generated group with no proper non-finitely-generated subgroup? -QUESTION [5 upvotes]: Is there a non-finitely-generated group each of whose proper subgroups is finitely generated? If so, what form of choice (if any) is required to construct such a group? - -REPLY [12 votes]: (CW since this is just expanding on George Lowther’s comment to the question, which could really have been an answer in the first place; if George L wants to convert his answer to a comment himself, I can delete this one.) -For any prime $p$, the Prüfer $p$-group is as desired. -There are several constructions of this; a good one for present purposes is -$$\mathbb{Z}[1/p]\ /\ \mathbb{Z}$$ -i.e. rationals with denominator a power of $p$, modulo the integers. -To see that this works, note that it is the union of the linearly ordered chain of finitely generated (indeed, cyclic) subgroups $H_i := \{ [a / p^i]\ |\ 0 \leq a < p^i \}$, over $i \in \mathbb{N}$. -Now any element of $H_{i+1}$ not in $H_{i}$ must be of the form $[a/p^{i+1}]$ with $a$ coprime to $p$, and hence generates the whole of $H_{i+1}$. So any subgroup is either equal to some $H_i$, or else contains them all and is the whole group. -On the other hand, the entire group is clearly not finitely generated since any finite set of elements is contained in some $H_i$.<|endoftext|> -TITLE: Singular Cardinals, and A Strange Question. -QUESTION [8 upvotes]: Let $\mu$ be any infinite cardinal, and define a collection $N\subset[\mu]^\mu$ to be, maximal almost disjoint (MAD) over $\mu$, iff - -$\forall\{A,B\}\in[N]^2$ $( A\cap B \in [\mu]^{<\mu})$ -$\forall X\in[\mu]^\mu \exists A\in N$ $( X \cap A \in [\mu]^\mu)$ - -My questions are as follows: when $\mu$ is singular - -What is known about MAD families (or any other combinatorial structures, like, towers, SFIP families without pseudo-intersection, etc) over $\mu$? -Are such families degenerate in the sense that an infinite family can have cardinality below $\mu$? -Is there any connection between such constructs on $\mu$ and the corresponding constructs on $cf(\mu)$? - -The main point I really want to know is this: Is it possible to add new subsets to an arbitrary singular cardinal without adding new subsets the the cardinals below it? -Side Request: I've been told there are forcing constructions which will add an order type $\omega$ cofinal sequence to a cardinal with cofinality $\omega$, can anyone point me in the correct direction with a book or article? - -REPLY [6 votes]: Theorem If $0^{\sharp}$ does not exist and $\lambda$ is a singular cardinal, then any forcing adding subsets to $\lambda$ necessarily adds subsets to a cardinal below $\lambda$. - -Proof: Let $\mathbb{P}$ be a partial order in the ground model and $G \subseteq \mathbb{P}$ be $V$-generic. Without loss of generality, we may assume that $\mathbb{P}$ is a partial order on a cardinal so that its elements are all ordinals. Also let $\vec{s} = \langle s_{\alpha}| \alpha < \text{cof}(\lambda)\rangle$ be a cofinal sequence in $\lambda$ in the ground model and $\dot{A}$ a $\mathbb{P}$-name for a subset of $\lambda$ in $V[G]$. Now suppose that $V$ and $V[G]$ agree on the bounded subsets of $\lambda$. Then for all $\alpha$, we have $a_{\alpha} = s_{\alpha} \cap \dot{A}_G \in \mathcal{P}(\lambda)^{V}$ so for every $\alpha < \text{cof}(\lambda)$, there will be: -$p_{\alpha} \in G$ such that $p_{\alpha} \Vdash \dot{A} \cap \check{s_{\alpha}} = \check{a_{\alpha}}$ -Because $V$ is a definable class in $V[G]$, the forcing relation for $V$ is definable in $V[G]$, and we may therefore choose such a $\text{cof}(\lambda)$-sequence of conditions $p_{\alpha}$ below a condition forcing that $\dot{A}$ is a name for a subset of $\lambda$. Let $S_A = \{p_{\alpha}| \alpha < \text{cof}(\lambda)\} \in V[G]$ be such a set of ordinals. Now $S_A$ is a set of ordinals having size $\text{cof}(\lambda)$ in $V[G]$ so by the nonexistence of $0^{\sharp}$, it follows from Jensen's covering lemma that there is a constructible set $C$ of size $\theta = \max\{\omega_1, \textrm{cof}(\lambda)\}$ in $V[G]$ covering $S_A$. Then since $\lambda$ is singular, $\theta < \lambda$ whereby $C \in L \subseteq V$ will also have size $\theta$ in $V$ by virtue of the fact that a poset adding no new subsets to $\theta$ cannot collapse any cardinals below $\theta^{++}$. But now $S_{A} \subseteq C$ must also be in $V$ because otherwise $V[G]$ would be adding a subset of $\theta$ induced by $f''S_{A}$ where $f: C \rightarrow \theta$ is a bijection in the ground model. However, then $V$ can construct $A$ from $S_A$ in the ground model since -$A = \bigcup\{a \in \mathcal{P}(\lambda)| p \in S_{A} \land s_{\alpha} \in \text{range}(\vec{s}) \land p \Vdash \dot{A} \cap \check{s_{\alpha}} = \check{a}\}$. $\Box$ -In particular, this shows that if $V$ is a forcing extension of $L$, then we cannot add a subset to a singular cardinal without adding a subset to a cardinal below it. I don't have an answer for what happens when $0^{\sharp}$ does exist, but at least this shows that your very interesting question is closely tied to the existence of certain large cardinals.<|endoftext|> -TITLE: Free subgroups vs law -QUESTION [17 upvotes]: Consider the following two conditions for a group $G$: -(1) $G$ does not satisfy a nontrivial law. -(2) $G$ contains a non-abelian free subgroup. -Obviously (2) implies (1) and it is easy to construct torsion groups that do not satisfy any law (e.g., the direct product of all finite groups). Thus (1) does not imply (2) in general. However the following question seems open: - -Are (1) and (2) equivalent for profinite groups? - -Here is a similar question: - -Suppose that a residually finite group does not satisfy a law. Does its profinite completion contain a nonabelian free subgroup? - -These questions may be thought of as possible generalizations of the Tits alternative to residually finite (or profinite) groups. -The answer to the second question is positive for finitely generated p-groups. This follows from [Wilson, Zelmanov, Identities for Lie algebras of pro-p groups, JPAA 81(192), 103-109], where the authors prove that if $G$ is a finitely generated residually finite p-group, then either it is finite (and hence satisfies a law), or its profinite completion contains a nonabelian free group. In the general case they only prove that if the profinite completion of a residually finite group does not contain a nonabelian free subgroup, then some Lie algebra associated to the group satisfies a law. - -REPLY [6 votes]: Since Henry asked, here is a reference: Aner Shalev in the first chapter (Lie Methods in the Theory of pro-$p$ Groups) in New Horizons in pro-$p$ Groups posed 4 conjectures in decreasing order of strength: - -Let $G$ be a finite $p$-group satisfying some identity $w$ with probability $\epsilon>0$. Then $G$ satisfies some identity depending only on $w$, $p$, and $\epsilon$. -Let $G$ be a finitely generated pro-$p$ group satisfying some identity with positive probability. Then $G$ satisfies some identity. -Let $G$ be a finitely generated pro-$p$ group satisfying some coset-identity. Then $G$ satisfies some identity. -Let $G$ be a $k$-generated pro-$p$ group satisfying some identity on all generating $k$-tuples. Then $G$ satisfies some identity. - -There is a discussion there about what is known. I am not sure what developments occurred since the book appeared and what is known about the profinite case in general. I think it may have been mentioned by Shalev or Mann in other surveys.<|endoftext|> -TITLE: Why does the map $BG\to A(*)$ fail to split? -QUESTION [7 upvotes]: There is a map $BG \to A(\ast)$ where $BG$ classifies stable spherical fibrations and $A(\ast)$ is -Waldhausen's algebraic $K$-theory of a point. The map is induced by applying Quillen's plus construction to the inclusion -$$ -BGL_1(S^0) \to BGL_\infty(S^0) -$$ -where $BGL_1(S^0)$ is $BG$. Here $BGL_\infty(S^0)$ can be defined as the homotopy colimit -over $k$ and $n$ of $BG(\vee_k S^n)$, where $G(\vee_k S^n)$ is the topological monoid of homotopy automorphisms of a $k$-fold wedge of $n$-spheres. (Note: $A(\ast) = \Bbb Z \times BGL_\infty(S^0)^+$.) - Question 1: I've heard it mentioned that there can be no retraction $A(*) \to BG$ to the map $BG \to A(\ast)$. Can someone please explain to me why there is no such splitting, and if possible, give a reference? -More generally, - Question 2: If $R$ is a structured ring spectrum, is there a reasonable set of conditions which guarantees that the map $BGL_1(R)\to K(R)$ admits a retraction? -(A related question is under what conditions does an $R$-module map -$f: R^n \to R^n$ which is a weak homotopy equivalence -admit a determinant $\text{det}(f) \in GL_1(R)$). - -REPLY [5 votes]: Question 1: There are several arguments. -In degree 2 there is a reference: the proof of corollary 3.7 of Waldhausen's "Algebraic K-theory of spaces, a manifold approach". See http://www.math.uni-bielefeld.de/~fw/ for a copy. Consider the maps $BG \to A(\ast) \to K(Z)$ and apply $\pi_2$. Here $\pi_2 BG = Z/2$, the composite is zero (restrict to $BSG$) and $\pi_2 A(\ast) \to K_2(Z)$ is an isomorphism. Hence the first homomorphism cannot be injective. -In degree 3 there is another argument. Right composition with the unstable map $\eta : S^3 \to S^2$ takes the generator of $\pi_2(BO)$ to zero in $\pi_3(BO) = 0$, so by naturality with respect to $BO \to BG$, composition with $\eta$ takes the generator of $\pi_2(BG) = Z/2$ to zero in $\pi_3(BG)$. On the other hand, composition with $\eta$ takes the generator of $\pi_2 Q(S^0)$ to a nonzero element in $\pi_3 Q(S^0)$, so by naturality with respect to $Q(S^0) \to A(\ast)$, composition with $\eta$ is nonzero from $\pi_2 A(*) = Z/2$ to $\pi_3 A(\ast)$. Therefore $BG$ cannot split off $A(*)$. -In degree 4 there is your argument. -Question 2: So far, I only know that $BGL_1(R) \to K(R)$ admits a retraction if $R$ is -the realization of a commutative simplicial ring. There is no retraction for $R = ku$ by Corollary 2.3 of "Divisibility of the Dirac magnetic monopole as a two-vector bundle over the three-sphere" by Ch. Ausoni, B. I. Dundas and J. Rognes, Documenta Mathematica 13 (2008) 795-801, and ku is already pretty close to a commutative simplicial ring.<|endoftext|> -TITLE: Extreme rigidification of homotopy self-equivalences -QUESTION [9 upvotes]: Suppose $X$ is a CW-complex. The monoid of homotopy self-equivalences $M = hAut(X)$ is the subspace of $Map(X,X)$ consisting of those maps with a homotopy inverse. It is a union of path components. It obviously acts on $X$, and the homotopy type only depends on the homotopy type of $X$. -It is known that we can find maps $G \leftarrow M' \to M$ of topological monoids, all homotopy equivalences, with $G$ a topological group. -It is also known that we can use this to find a $G$-space $Y$ and maps of $M'$-spaces $Y \leftarrow X' \to X$ which are all homotopy equivalence. In other words, this rigidifies the action of $hAut(X)$ to an honest action of a topological group. -However, even in this situation we have a composite map $G \to Aut(Y) \to hAut(Y)$ that we know is a homotopy equivalence, but it is unlikely to be the case that $Aut(Y)$ is homotopy equivalent to $hAut(Y)$. -Can we rigidify this and find an $Y$ whose automorphism group is equivalent to its homotopy automorphism group? Or does there exist a space for which the map $Aut(Y) \to hAut(Y)$ is never a homotopy equivalence for any space homotopy equivalent to $X$? - -REPLY [4 votes]: This is a substantial revision of my original post. It shows that if we replace the "equivalence" Tyler is asking for by a "retract" then the answer is yes. - -Given a CW space $Y$, we can take $G(Y) =$ the topological monoid of homotopy automorphisms of $Y$. The Borel construction -$$ -EG(Y) \times_{G(Y)} Y \to BG(Y) -$$ -is then a quasifibration. Let $U \to BG(Y)$ be the effect of converting it into a fibration. -Let $G$ be a topological group with a chosen homotopy equivalence $$BG\simeq BG(Y). $$ For example, we can do what -Tyler does, or we can simply take $\Omega BG(Y)$, where this means the realized Kan loop of the total singular complex of $BG(Y)$. -Let $EG \to BG$ be a universal $G$-principal bundle, and set -$$ -Z \quad := \quad \text{pullback}(EG \to BG \simeq BG(Y) \leftarrow U) -$$ -Then $Z \subset EG \times U$ inherits a $G$-action and its underlying homotopy type is that of $Y$. Then the Borel construction -$$ -EG\times_G Z \to BG -$$ -is a fiber bundle which is weak fiber homotopy equivalent to $U \to BG(Y)$. -Step 3 implies that $BG(Y)$ is a retract up to homotopy of $B\text{homeo}(Z)$. -This will imply that $G(Y)$ is a homotopy retract of $\text{homeo}(Z)$ in the $A_\infty$ sense, with $Z \simeq Y$.<|endoftext|> -TITLE: Proof theoretic ordinal -QUESTION [16 upvotes]: In Ordinal Analysis, Proof-theoretic Ordinal of a theory is thought as measure of a consistency strength and computational power. -Is it always the case? I. e. are there some general results about formal theories which say that if we know the ordinal measure of a formal theory than we would know its consistency strength and computational power to some extent? - -REPLY [16 votes]: This is a very interesting question (and I really want to see what other answers you receive). I do not know of any general metatheorems ensuring that what you ask (in particular, about consistency strength) is the case, at least under reasonable conditions. -However, arguments establishing the proof theoretic ordinal of a theory $T$ usually entail this. You can find a nice summary in the paper "A Model-Theoretic Approach to Ordinal Analysis" by Jeremy Avigad and Richard Sommer, The Bulletin of Symbolic Logic, Vol. 3, No. 1, (Mar., 1997), pp. 17-52, available here. -What follows is their description of these arguments. Avigad and Sommer are assuming that $T$ is a theory in the language of arithmetic, but very little of the description needs to be modified if that is not the case. - -Saying that the proof-theoretic ordinal of a theory $T$ is less than or equal to $\alpha$ usually entails all of the following results: -(1) There is some formula $\varphi (y)$ such that $T$ doesn't prove ${\rm TI}(\alpha, \varphi(y))$, where ${\rm TI}(\alpha, \varphi(y))$ formalizes transfinite induction up to $\alpha$ for the formula $\varphi(y)$. -(2) Over a weak base theory, ${\rm PRWO}(\alpha)$ proves the 1-consistency of $T$. Here ${\rm PRWO}(\alpha)$ is a scheme which asserts that there are no primitive recursive descending sequences beneath $\alpha$, and "the 1-consistency of $T$" is the formalized $\Pi^0_2$ assertion that if $T$ proves any $\Sigma^0_1$-formula (possibly with parameters) then that formula is true. -(3) If $T$ proves a recursive function $f$ to be total, then $f$ is $\prec\alpha$-recursive. [Where $\prec$ is the ordering on ordinal notations induced by the intended interpretation.] By "$T$ proves the recursive function $f$ to be total" we mean that $T$ proves $$\forall x\exists! y \varphi(x,y)$$ for some $\Sigma^0_1$ formula $\varphi$ that defines the graph of $f$ in the standard model. -(4) If $\lt$ is any recursive ordering and $T$ proves $(+)$: $$\forall X\hspace{1mm} {\rm TI}({\lt}, y \in X)$$ then the order-type of $\lt$ in the standard model is less than $\alpha$. (If $T$ doesn't allow for quantification over sets of numbers, we replace $(+)$ by $${\rm TI}({\lt}, X(y)),$$ where $X$ is a new predicate symbol that we allow to appear in the axiom schemata of $T$.) - -Now, if the program above can be carried out, general theorems of speed-up of proofs (see for example Samuel R. Buss, "On Gödel's Theorems on Lengths of Proofs I: Number of Lines and Speedup for Arithmetics", The Journal of Symbolic Logic, Vol. 59, No. 3, (Sep., 1994), pp. 737-756) should guarantee that if $T_1$ has larger proof theoretic ordinal than $T_2$, then in fact the "computational power" of $T_1$ outruns that of $T_2$ in significant and quantifiable ways.<|endoftext|> -TITLE: Point on a line nearest a point in Banach space -QUESTION [12 upvotes]: I have a Banach space geometry question (a curiosity-driven spin-off from a research topic). Given a point $x$ on the unit sphere of a Banach space and a vector $y\ne 0$, there is a multiple $t_0y$ of $y$ for which $\|t_0y-x\|$ is minimized (this will be unique if the norm is strictly convex). -My question is this: -For which Banach spaces $X$ is it guaranteed that $\|t_0y\|\le \|x\|$? -My "Euclidean intuition" suggested that this should be the case for all Banach spaces, but a little experimentation showed that this is not the case. You quickly see this is really a question about two dimensions. In fact it seems to fail for every $\ell^p$, $p\ne 2$ (see the attached figure in $p=1.2$). - -Could it be true that this property characterizes Hilbert space? (I looked at the obvious sources: (MO 11192 and papers mentioned in there and didn't find anything of the sort). - -REPLY [14 votes]: The answer is no in dimension 2 and yes in dimension 3 and higher. The property that the nearest-point projection to a line does not increase the norm is equivalent to the symmetry of orthogonality relation defined as follows: $x$ is orthogonal to $y$ iff $\|x+ty\|\ge\|x\|$ for all $t\in\mathbb R$. -It is well-known that symmetry of this orthogonality relation in dimension $\ge 3$ implies that the norm is Euclidean, see e.g. Thompson's "Minkowski geometry", Theorem 3.4.10. -This is not the case in dimension 2. There are many counter-examples (I believe they are called Radon planes). Basically you only need to ensure that every unit vector with its unit orthogonal one span a constant parallelogram area, this is easy to satisfy and is equivalent to the symmetry of orthogonality. For a simple explicit example (although non-smooth), consider a norm on the plane whose unit ball is a regular hexagon.<|endoftext|> -TITLE: Are strict pushout squares in Cat exact squares? -QUESTION [5 upvotes]: Let $C,C',D\in \operatorname{Cat}$, and consider the following strict pushout square -$$\begin{matrix} -C&\overset{f}{\to} &D^{op}\\ -\downarrow^\pi&\swarrow&\downarrow^{\iota_2}\\ -C'&\underset{\iota_1}{\to}&D'^{op} -\end{matrix} -$$ -where $D':=C'^{op}\coprod_{C^{op}} D$ and the 2-cell (denoted by $\swarrow$) is the identity. -Is this square exact (or at least $Set$-exact) in the sense that the canonical Beck-Chevalley transformation -$$\pi_!f^*\to \pi_!f^*\iota_2^*\iota_{2!} = \pi_!\pi^*\iota_1^*\iota_{2!}\to \iota_1^*\iota_{2!}$$ -is an isomorphism? -Note that $(-)_!$ denotes the left kan extension (that is, the left adjoint of the canonical pullback $(-)^*$ (looking at diagram categories over an arbitrary base category $M$ (resp. looking at diagram categories into $Set$))). -If it is true, does it hold for simplicial categories (resp. for the diagram categories targeting $sSet$)? - -REPLY [7 votes]: No, it's false for general categories; it's true for groupoids, however. -A crucial test case for Beck-Chevalley is often called (by categorical logicians especially) Frobenius reciprocity; it's the case where one forms the pullback -$$\begin{matrix} -C & \overset{\Delta}{\to} & C \times C \\ -\downarrow^\Delta & & \downarrow^{1 \times \Delta} \\ -C \times C & \underset{\Delta \times 1}{\to} & C \times C \times C -\end{matrix}$$ -and asks whether the canonical map $\Delta_! \Delta^\ast \to (\Delta \times 1)^\ast (1 \times \Delta)_!$ is invertible. You probably recognize this sort of thing from other contexts; in the monoidal category as opposed to monoidal bicategory context, where this arrow is replaced by an equation, this is the defining equation for a Frobenius monoid, the one which relates a monoid structure to a comonoid structure. It also crops up in representation theory, and in any context where one has ambidextrous adjunctions (where one functor is simultaneously a left and right adjoint to another). Most famously, it arises in the monoidal category of 2-cobordisms between oriented compact 1-dimensional manifolds. (You can see the cobordisms by drawing the string diagrams for both sides of the equation, where you get the famous "I = H" equation, and then thickening it by taking the boundary of small $\varepsilon$-neighborhoods of these string diagrams in 3-space.) -Anyway, it's fun to calculate the canonical transformation directly and see what happens. -My preferred formalism is composition of profunctors, because it basically behaves like matrix multiplication. On the "I" side of the equation, one is composing the profunctor -$$C^{op} \times C \times C \to Set: (d, c_1, c_2) \mapsto (C \times C)(\Delta d, \langle c_1, c_2 \rangle)$$ -with the profunctor -$$(C \times C)^{op} \times C \to Set: (c_3, c_4, d') \mapsto (C \times C)(\langle c_3, c_4 \rangle, \Delta d')$$ -(This should be liked to multiplying a $1 \times 2$ matrix by a $2 \times 1$ matrix.) Officially, the result is a profunctor -$$(C \times C)^{op} \times (C \times C) \to Set$$ -$$(\langle c_3, c_4 \rangle, \langle c_1, c_2 \rangle) \mapsto \int^d (C \times C)(\langle c_3, c_4 \rangle, \Delta d) \times (C \times C)(\Delta d, \langle c_1, c_2 \rangle)$$ -where the coend can be rewritten -$$\int^d C(c_3, d) \times c_4, d) \times C(d, c_1) \times C(d, c_2)$$ -This may look complicated, but it's not. The elements of the coend at a particular 4-tuple are equivalence classes of diagrams which look like this: -$$\begin{matrix} -c_3 & & & & c_4 \\ - & \searrow^{\phi_3} & & \swarrow^{\phi_4} & \\ - & & d & & \\ - & \swarrow^{\phi_1} & & \searrow^{\phi_2} & \\ -c_1 & & & & c_2 -\end{matrix} -$$ -where the equivalence relation is generated by stipulating that two 4-tuples (one factoring through $d$, another through $d'$) are equivalent if both arise from a diagram of shape -$$\begin{matrix} -c_3 & & & & c_4 \\ - & \searrow & & \swarrow \\ - & & d & & \\ - & & \downarrow & & \\ - & & d' & & \\ - & \swarrow & & \searrow & \\ -c_1 & & & & c_2 -\end{matrix}$$ -(which looks like an "I"). Now for the "H" side; it's actually easier because there are various Yoneda lemma reductions (which I'll skip) that allow one to boil down the profunctor composite to -$$(C \times C)^{op} \times (C \times C) \to Set$$ -$$(\langle c_3, c_4 \rangle, \langle c_1, c_2 \rangle) \mapsto C(c_3, c_1) \times C(c_4, c_1) \times C(c_4, c_2)$$ -whose elements are diagrams in a kind of "H" shape: -$$\begin{matrix} -c_3 & & c_4 \\ -\downarrow & \swarrow & \downarrow \\ -c_1 & & c_2 -\end{matrix}$$ -And now you should be able to see that the two profunctor composites are definitely distinct in general. For example, in the I-composite, $c_3$ is connected by a morphism to $c_2$, but not so in the H-composite. Of course, in groupoids they are connected because we can travel along inverses, and the isomorphism between the diagram classes is easy to establish. -There are probably less elaborate calculations which arrive at the same point, but it is often the case (for example in certain simple type-theoretic contexts) where verification of the Beck-Chevalley condition boils down to Frobenius reciprocity. I may come back to edit if I think of a clearer way to say it, but anyway Frobenius is often a crucial test case, as I said at the beginning. -Finally: Harry, please don't delete this question. Very often I find that you have deleted questions that I was considering or preparing an answer for. In the present case, the question is good, and I hope you find my answer helpful.<|endoftext|> -TITLE: partition functions and Galois representations? -QUESTION [13 upvotes]: The recent answer's link to Ono's work makes me ask and wonder if his new results on partition functions tell something about Galois representations? (Hoping that question is not a case of this) - -REPLY [20 votes]: Dear Thomas, -As far as I know, this work is not related directly to Galois representations, but is rather a particular calculation in the theory of $p$-adic modular forms (although it is not really described this way explicitly in the paper). -The $p$-adic theory of modular forms of half-integral weight was developed in the 2004 Ph.D. thesis and subsequent papers of Nick Ramsey. (See Nick's comment below this answer, and the several papers available on his web-page.) The deduction of the results of Folsom--Kent--Ono from Ramsey's thesis is explained in a short note recently written by my colleague Frank Calegari. -The key idea is that iterating the $U_{\ell^2}$ operator on a space of ($\ell$-adic) modular forms -of half-integral weight projects to the ordinary part of the space, which is finite-dimensional and more-or-less explicitly computable. Applying this procedure to the modular form $1/\eta$ of weight $-1/2$ (recall that $1/\eta = q^{-1/24}\prod_{n=1}^{\infty}(1-q^n)^{-1}$ is the generating -function for partitions) gives the results of Folsom--Kent--Ono. -[Added: The shift from $p$-adic to $\ell$-adic in the second paragraph is made just because in the work of Folsom--Kent--Ono, and so also in Calegari's note, the distinguished prime is called $\ell$. On the other hand, when talking about this area in general, people normally speak of $p$-adic modular forms rather than $\ell$-adic ones.]<|endoftext|> -TITLE: Which almost complex manifolds admit a complex structure? -QUESTION [29 upvotes]: I was reading Yau's list of problems in geometry, and one of them is to prove that any almost complex manifold of complex dimension $n \geq 3$ admits a complex structure. It's been some time since Yau's list was published, so what is the status of this problem today? -Obviously it isn't hasn't been shown to be true, because we're still looking for complex structures on the six-sphere, but I have a vague feeling of having read that this doesn't hold. So do we know any counterexamples to this question? If not, then is anyone working on this problem? -Also, Yau only stated the problem for manifolds of dimension $n \geq 3$. We know this is true in dimension one, because there we have isothermal coordinates which give complex structures, but why didn't Yau mention almost complex surfaces? Do we know this holds there, or are there counterexamples in dimension 2? - -REPLY [18 votes]: In complex dimension 3 or more it is still an open conjecture -(which was re-stated Yau a couple of years ago in his UCLA lectures). -There is not a single known example of an almost complex manifold -of dimension $\geq$ 3 not admitting a complex structure. -In dimension 2 it is easy, of course, because -the non-Kähler complex surfaces are understood -much better than Kähler ones: every non-Kähler -surface with $b_1 >1$ is diffeomorphic to a blow-up of a -locally trivial elliptic fibration over a curve. Hence any -4-dimensional compact almost complex manifold -with odd $b_1 >1$ and a fundamental group not virtually -isomorphic (*) to a cross-product of a fundamental group of a curve -and $\mathbb{Z}$, cannot be a complex surface. -(*) Here "virtually isomorphic" means "isomorphic up to a -finite index subgroup".<|endoftext|> -TITLE: Orderings of ultrafilters -QUESTION [6 upvotes]: Let $U$ is a set. I will speak about filters on this set. -If $f$ is a function and $a$ is a filter then I define $f \left[ a \right]$ as -the filter whose base is $\lbrace f[A] | A \in a \rbrace$. -I will call super-embedding-1 of filter $a$ into filter $b$ a function $f$ -such that $f \left[ a \right] \subseteq b$ and super-embedding-2 of filter $a$ -into filter $b$ a function $f$ such that $f \left[ a \right] = b$. -Let define preorders $\leqslant_1$ and $\leqslant_2$ on the set of filters: -$b \leqslant_1 a$ if there are super-embedding-1 from $a$ to $b$ and $b -\leqslant_2 a$ if there are super-embedding-2 from $a$ to $b$. -Question 1: $\leqslant_1$ is the same as $\leqslant_2$? -Filters $a$ and $b$ are isomorphic if exists a bijective super-embedding-2 $f$ -from $a$ to $b$ such that $f^{- 1}$ is super-embedding-2 from $b$ to $a$. For -two other equivalent characterizations of isomorphic filters see -this -blog post and -this -blog post (the second blog post requires this article). -Being isomorphic is an equivalence relation. I will call classes of filters -equivalence classes under the being isomorphic relation. I will call classes -of ultrafilters these classes of filters which contain ultrafilters. -Further I will denote $i = 1, 2$. So every open problem below is in fact two -problems. -Question 2: Is $\leqslant_i$ for ultrafilters the same as Rudin-Keisler order -(paragraph 9 of Comfort and Negrepontis ``The Theory of Ultrafilters'') of -ultrafilters? If not, how they are related? -Question 3: Is the preorder of classes of filters induced by $\leqslant_i$ a -partial order? -Question 4: Is the preorder of classes of ultrafilters induced by -$\leqslant_i$ a partial order? -Question 5: If it is a partial order, is it a linear order? -Question 6: If it is a linear order, is it a well-order (or maybe -anti-well-order)? -Question 7: If in the above definition of isomorphic filters super-embedding-2 -is replaced with super-embedding-1, does it remain equivalent to the above -definition? - -REPLY [8 votes]: I understand your question better now. -First, in your general context of filters the relations -$\leq_1$ and $\leq_2$ are not the same. To see this, let -$G=\{I\}$ be the trivial filter on a set $I$ with at -least two points, and let $\mu$ be any nonprincipal -ultrafilter on $I$. Since $G\subset \mu$, we see that -$\mu\leq_1 G$ as witnessed by the identity function $i$ on $I$. (Details: since $i[I]=I$, it follows that $i[G]$ is the filter with base $\{I\}$, which is the same as $G$. So $i[G]=G$, which is a subset of $\mu$, and so $\mu\leq_1 G$.) Meanwhile, I claim that $\mu\not\leq_2 G$. To see this, observe that for any function $f:I\to I$, we have $f[G]$ is the filter with base $\{f[I]\}$, and so $f[G]\neq\mu$ since $\mu$ is nonprincipal. -So the relations are -different. -Note also that if $\mu$ is an ultrafilter on $I$ and $F\leq_1 \mu$ via the function $f$ for a filter $F$, then $F$ is an ultrafilter. The reason is that if $Y\notin F$, then $f^{-1}Y\notin\mu$ and so $f^{-1}(I-Y)\in\mu$, which implies $f[f^{-1}(I-Y)]\in F$, which implies $I-Y\in F$, so $F$ is an ultrafilter. -Next, I claim that for ultrafilters, the relations are the -same. -Theorem. If $\nu$ is an ultrafilter, then $F\leq_1\nu\iff F\leq_2\nu$. -Proof. It suffices to prove the forward direction. Suppose $\nu$ is an ultrafilter on a set $J$ and $F$ is a filter on $I$ and $F\leq_1\nu$ as witnessed by $f:J\to I$. So $f[\nu]\subset F$. Consider any $X\in F$. If $f^{-1}X\in\nu$, then we get $X\supset f[f^{-1}X]\in f[\nu]$ and so $X\in f[\nu]$. Otherwise, since $\nu$ is an ultrafilter, we have $f^{-1}(I-X)\in\nu$ and so $I-X\supset f[f^{-1}(I-X)]\in f[\nu]\subset F$, which would put disjoint sets in $F$, a contradiction. QED -Finally, I claim that for ultrafilters, the relation $\leq_2$ is the same as the -Rudin-Keisler order. The usual definition of this order is -that if $F$ is a filter on $J$ and $f:J\to I$ is any -function, then one we may define a filter $G=f*F$ on $I$ by -$X\in G\leftrightarrow f^{-1}X\in F$. The Rudin-Keisler -order is defined so that $G\leq_{RK} F$ if and only if -there is $f$ for which $G=f*F$. -Suppose $F$ is a filter on $J$ and $f:J\to I$. I claim -generally that $f*F=f[F]$. This is because $Y\subset -f^{-1}f[Y]$ for $Y\subset J$ shows that $f[F]\subset f*F$; -and conversely $f[f^{-1}X]\subset X$ for $X\subset I$ shows -$f*F\subset f[F]$. -It follows that $\leq_2$ is the same as the Rudin-Keisler -order.<|endoftext|> -TITLE: Integer Points on the Elliptic Curve $y^2=x^3+17$. -QUESTION [13 upvotes]: I came across the problem "find all integer solutions to $y^2=x^3+17$." -I've tried several things, without any success, and I was hoping that someone could help out. (Some ideas or a reference for where to find it are both appreciated) -By numerical calculation I have found that the following integer points $(x,y)$ lie on the curve -$(-1,4)$, $(-2,3)$, $(2,5)$, $(4,9)$, $(8,23)$, $(43,282)$, $(52,375)$, $(5234,378661)$ -and this is probably all of them. -Thanks - -REPLY [7 votes]: Uspensky and Heaslet, Elementary Number Theory, published in 1939, credits Delaunay (on page 400) with showing that $y^2=x^3+17$ has only the eight solutions, and goes on to say, "Whether his method will always work is still an open question, and the problem, despite its simple appearance, is a very difficult one." No reference is cited<|endoftext|> -TITLE: What is known about the birational involutions of P^3? -QUESTION [18 upvotes]: Describing the group of birational automorphisms of $\mathbb{P}^n$, $\mbox{Bir}(\mathbb{P}^n)$, for $n\ge 3$ is a fundamental open problem in birational geometry. For $n=2$, the classical theorem of Noether says that this group is generated by linear transformations and the Cremona transformation, which is given by -$$ -\phi:(x_0 : x_1 : x_2) \to (x_0^{-1} : x_1^{-1} : x_2^{-1}). -$$For $n\ge 3$, there is an analogous Cremona transformation, but it is known that the group $\mbox{Bir}(\mathbb{P}^n)$ is no longer generated by this and $\mbox{PGL}_{n+1}(\mathbb{C})$. My question is therefore - -Are there examples of other birational - involutions of $\mathbb{P}^3$? - -In case the answer is yes, have these been classified? I'm also interested in the analog of Noether's theorem in this case: Are there examples of birational transformations of $\mathbb{P}^3$ that can not be written as a composition of birational involutions? - -REPLY [6 votes]: There are plenty of other examples of birational involutions of $\mathbb{P}^3$ which are known. For the moment, no precise classification as in dimension $2$. However, the following preprint of Yuri Prokhorov gives a first nice step toward a classification: -Yuri Prokhorov, "On birational involutions of $\mathbb{P}^3$" -http://arxiv.org/abs/1206.4985 -In dimension $2$, we can say if two involutions are conjugate only by looking at fixed points. When the curves pointwise fix a curve of positive genus, the curve has to be the same (easy argument), but this condition is in fact sufficient (hard argument, only follows from classification). The same works for any order, looking carefully at the fixed points of all powers (see J. Blanc, "Elements and cyclic subgroups of finite order of the Cremona group." -Comment. Math. Helv. 86 (2011), no. 2, 469-497, http://arxiv.org/abs/0809.4673). -In dimension $3$, we can have the same discussion, replacing fixed curves of positive genus with fixed surfaces which are not birationally ruled (other surfaces can be contracted by birational maps). The article of Yuri Prokhorov studies the case where such fixed surface exists.<|endoftext|> -TITLE: palindromic subsequences -QUESTION [11 upvotes]: I'd like any insight or references to the following two conjectures (see the glossary below for definitions): -Conjecture 1: For any string $x$, there exists a longest common subsequence of $x$ and its reversal $x^R$ that is a palindrome. -Conjecture 2: For any string $x$ over a two-letter alphabet, all longest common subsequences of $x$ and $x^R$ are palindromes. -Conjecture 2 is not true for strings over a three-letter alphabet, a counterexample being $abacbab$, which has $abcab$ and $bacba$ as longest common subsequences. -Glossary: -A string (or word) is any finite sequence of objects ("letters") drawn from some finite set (the "alphabet"). -For any string $x = x_1x_2\cdots x_{n-1}x_n$ of length $n$, the reversal of $x$ is $x^R := x_nx_{n-1}\cdots x_2x_1$. -A string $x$ is a palindrome if $x = x^R$. -A string $x$ is a subsequence of a string $y$ if $x$ results from $y$ by removing zero or more letters (in arbitrary locations, closing up any gaps that result). -A longest common subsequence (LCS) of two strings $x$ and $y$ is a string $z$ that is a subsequence of both $x$ and $y$ such that no string longer than $z$ has this property. Generally, $x$ and $y$ may have several different LCSs. There is a well-known algorithm to find an LCS of two given strings that runs in quadratic time (see e.g., Cormen, Leiserson, Rivest, and Stein, Introduction to Algorithms). - -REPLY [3 votes]: This problem is discussed and solved here .<|endoftext|> -TITLE: Modular forms of fractional weight -QUESTION [27 upvotes]: Modular forms of integral weight are prominent in number theory. -Furthermore, there are $\theta$-functions and the $\eta$-function, having weight 1/2, -which also have a rich theory. -But I have never seen a modular form of weight e.g. 1/3. -I have been wondering about this for a long time. Are there -examples of modular forms of fractional weights other than multiples of 1/2? -And if yes, is there are reason why they are poorly studied? - -REPLY [16 votes]: Automorphic forms naturally live on the adelic points of a reductive algebraic group (modulo rational points and a compact subgroup giving the level). One may interpret automorphic forms -of fractional weights to be automorphic forms which live on a topological cover of the adelic group as in the classical metaplectic works mentioned earlier. In this vein, the most modern treatment is in the paper of Brylinski-Deligne. The only arithmetically related work seems to be due to Marty Weissman. The Brylinski-Deligne paper works equally well over function fields and it would be interesting to see its connections with Lafforgue's works.<|endoftext|> -TITLE: Cohomology of Structure Sheaves: Algebraic, Constructible and more -QUESTION [19 upvotes]: I am not an algebraic geometer, but I am a topologist who uses sheaves. I have studied some algebraic geometry and am interested in what happens as I reduce the amount of rigidity in the structure sheaves on a space. Specifically I want to know what the cohomology of the following structure sheaves tell you. Please do things over characteristic zero. -If $X$ is a topological space then the natural structure sheaf of continuous functions has no interesting cohomology because of the existence of partitions of unity. Consequently, $C^k$, smooth and topological manifolds have structure sheaves with no interesting cohomology. -To contrast, schemes ($X,\mathcal{O}_X$) with their structure sheaves of regular functions have lots of interesting information. In particular, there is non-trivial higher cohomology. However, I am still unsure what these groups tell you (aside from what all sheaf cohomology means - obstructions to extending sections). For example Hartshorne exercise 4.3 tells you that $H^1(U,\mathcal{O}_U)$ is infinite dimensional (spanned by $x^iy^j|i,j<0$) where $U=\mathbb{A}^2_k-(0,0)$. For $X$ a curve then the dimension of $H^1(X,\mathcal{O}_X)$ tells you the genus. For affine pieces this cohomology is trivial, so the cohomology of the structure sheaf detects "non-triviality" of a space. Are there any other characterizations of the higher cohomology groups of the structure sheaf? -I am actually interested in the definable/o-minimal/constructible setting. So I want to consider a constructible space $X$ along with it's structure sheaf of ($\mathbb{R}$ or $\mathbb{Z}$-valued) constructible functions as a ringed space. Since one implementation of definable spaces is the semi-algebraic (or semi-analytic) setting, I would like to know that the cohomology of the structure sheaf here tells you. So if someone could address any of the following: - -a real analytic space with the sheaf of analytic functions (no partitions of unity, so potential higher cohomology?) Question: Since regular functions in AG are defined as locally being the quotient of polynomials, would regular for analytic be locally the ratio of analytic? (EDIT: I am interested primarily in the real case since GAGA shows in some cases complex analytic spaces are "as rigid as" complex algebraic ones.) -a semi-analytic space with the above structure sheaf(ves) -a semi-algebraic space with its structure sheaf (Which is? Do Nash functions come into play here?) -for a cell complex, is there a natural way of considering it as a ringed space? If so what would that cohomology tell you? - -I apologize for the wide spread of questions. Partial answers will be voted up. - -REPLY [7 votes]: Dear Justin, let me address your first question. First of all, the local quotient of two analytic functions is called meromorphic. In dimension one you can see meromorphic functions as regular functions with codomain $\mathbb P^1$. However in dimension $\geq 2$ this won't work : for example at the origin of $\mathbb C^2$ the function $z_1/z_2$ approaches any complex number according to the path you take (you have to blow-up $\mathbb C^2$ if you want a well-defined holomorphic map to $\mathbb P^1$). -There is a related interesting question, the Poincaré problem, for a complex manifold $X$: is a meromorphic map globally -the quotient of two holomorphic ones, i.e. do we have $\mathcal M(X)=Frac (\mathcal O(X))$ ? Here are answers in two extreme cases. -Compact manifolds Since holomorphic functions are constant, the answer to Poincaré's problem is "no", unless the manifold has only constants as meromorphic functions (this happens for some tori for example) -Stein manifolds The answer to Poincaré's problem is "yes" if the Stein manifold $X$ satisfies the purely topological condition $H^2(X,\mathbb Z)=0$. In this case we have indeed $\mathcal M(X)=Frac (\mathcal O(X))$ . -Interestingly the proof uses that $H^1(X, \mathcal O)=H^2(X, \mathcal O)=0$ for the Stein manifold $X$. -The result $\mathcal M(X)=Frac (\mathcal O(X))$ is due to Weierstrass in the particular case $X=\mathbb C$ -and to Poincaré for $X=\mathbb C^2$. You can read the proof of the general result in Grauert-Remmert's Theory of Stein Spaces.<|endoftext|> -TITLE: How do I split a homotopy idempotent? -QUESTION [5 upvotes]: I want to check that the homotopy category of cochain complexes of an idempotent splitting, preadditive category is idempotent splitting. -Let $a\xleftarrow{e}{}a$ be an idempotent chain map up to chain homotopy, $e^2\sim e$; that is, there exists maps $a_{i-1}\xleftarrow{h_i}a_i$ with $e_i^2-e_i=h_{i+1}d_i+d_{i-1}h_i$. Assuming that idempotents split in the underlying category, how can I construct a chain complex $im\left(e\right)$ with chain maps $im\left(e\right){\xleftarrow{p}\atop \xrightarrow[i]{}}a$ such that $ip\sim e$ and $pi\sim1_{im\left(e\right)}$? - -REPLY [6 votes]: Let me re-denote your chain complex $a$ by $C$. You can define a chain complex $D$ as the mapping telescope of the infinite sequence -$$ -\cdots\overset e\to \quad C \quad \overset e\to \quad C \quad \overset e\to \cdots -$$ -This can be constructed as follows: Form the homotopy coequalizer of the pair of maps -$$ -1,S_a: \bigoplus_{\Bbb N} \quad C \quad \to \quad \bigoplus_{\Bbb N} \quad C -$$ -where $1$ is the identity and $S_a$ is given by applying $a$ and then shifting by one unit to the right in the index. (The homotopy coequalizer is gotten from this diagram by replacing the target -$\oplus_{\Bbb N} C$ with its cylinder $\oplus_{\Bbb N} C \otimes I$ and forming the coequalizer of the two inclusions given by $1$ and $S_a$ on each end.) -The effect of this construction is to homotopically invert the map $e$, giving you a model for -$C[e^{-1}]$. There is an evident inclusion -$i: C \to D$. There is a map $D \to C$ which is defined on the $k$-th summand using the -map $e^{\circ k}$. This will do what you want it to.<|endoftext|> -TITLE: In which ways can the isogeny theorem fail for local fields? -QUESTION [35 upvotes]: Fix a field $K$ with absolute Galois group $G$. By an isogeny theorem over $K$, I mean the statement that the map $\operatorname{Hom}(A,B)\otimes\mathbb{Z}_l \to \operatorname{Hom}_G(T_l A, T_l B)$ is an isomorphism, where $A,B$ are abelian varieties over $K$, and $T_l A$ is the Tate module of $A$. Such a statement was proved for finite fields by Tate, for global function fields by Zarhin and for number fields by Faltings. -I'm interested in the case where $K$ is a $p$-adic field. The statement is then generally false, but is sometimes true. It holds e.g. when $A,B$ are elliptic curves with bad reduction and $l = p$ (Serre) or when $A,B$ have the same (good) reduction and again $l=p$ (Serre-Tate). It certainly fails if $A,B$ have good reduction and $l \ne p$. -What I don't have is a clear picture of the various possibilities for the reductions and for which cases the statement holds or fails. So that's the question. - -REPLY [23 votes]: I think the statement can fail in the case of elliptic curves of good reduction even when $l = p$. But then your comment on -Serre-Tate theory confused me for a little while! (A discussion with Jared Weinstein helped me clear it up.) -The post ended up getting long as a result; it's partly for my own reference. -Recently FC pointed out the following to me: if $E$ over $\mathbb Q_p$ is an elliptic curve of good supersingular reduction -and $p > 3$, then $T_p(E)$ (up to isomorphism) does not depend on $E$. -First, $a_p = 0$ by the Weil bounds. The $p$-adic Galois representation $\rho := T_p(E) \otimes \mathbb Q_p$ is crystalline -(due to good reduction) of Hodge-Tate weights 0, 1 with (crystalline) $a_p = 0$. In particular, it is non-ordinary so $\rho$ -is (even residually) absolutely irreducible. It's a basic fact in $p$-adic Hodge theory that any 2-dim. absolutely -irreducible $G_{\mathbb Q_p}$-representation with distinct Hodge-Tate weights is uniquely determined by $a_p$. (This is -nowadays an exercise about weakly admissible filtered $\phi$-modules.) Finally, as $\rho$ is residually irreducible, it -follows that the lattice $T_p(E)$ is unique up to homothety (in particular, up to isomorphism). -Just as Tim notes, isogeny classes are countable. But there are uncountably many elliptic curves with good supersingular -reduction. (Fix a supersingular one over $\mathbb F_p$ and consider lifts of the $j$-invariant.) -I wonder how general a phenomenon this is that the Galois representation doesn't vary. -Kisin's result no longer seems to apply in this $l = p$ case (non-torsion coefficients). - -Now, here is why there is no contradiction with Serre-Tate theory. First, it is true that any two of the elliptic curves -above have isomorphic $p$-divisible groups over $\mathbb Z_p$: the reason is that $T_p(E)$ (with the Galois action) precisely -tells us what $E[p^\infty]$ is over $\mathbb Q_p$, and that Tate's full-faithfulness theorem says that this determines -$E[p^\infty]$ over $\mathbb Z_p$. Besides, the special fibre $E/\mathbb F_p$ is uniquely determined up to isogeny, -because we know $a_p$. [There's even an isogeny over $\mathbb F_p$, e.g., because the Dieudonne module is the same (see -below), but there must be an elementary way to see that.] -Serre-Tate theory says the following (specialised to our situation). The category of elliptic curves over $\mathbb Z_p$ is -equivalent to the following category $C$: the objects consist of triples $(\mathcal G, \overline E, f : \mathcal G \times \mathbb -F_p \to \overline E[p^\infty])$, where $\mathcal G$ is a $p$-divisible group over $\mathbb Z_p$, $\overline E$ is an elliptic curve -over $\mathbb F_p$ and $f$ is an isomorphism of $p$-divisible groups over $\mathbb F_p$. Morphisms consist of a map of -$p$-divisible groups over $\mathbb Z_p$ and a map of elliptic curves over $\mathbb F_p$ that are compatible with the -isomorphisms in the special fibre. In this equivalence, an elliptic curve $E/\mathbb Z_p$ maps to the triple $(E[p^\infty], -E \times \mathbb F_p, can)$, where $can$ is the canonical isomorphism. -Suppose we have two elliptic curves $E_1$, $E_2$ over $\mathbb Z_p$ as above (i.e., supersingular reduction and $p > -3$). Then we know that $E_1[p^\infty] \cong E_2[p^\infty]$ and that $E_1 \times \mathbb F_p$ is isogenous to $E_2 \times -\mathbb F_p$. But in order to apply the theorem of Serre-Tate and deduce the existence of an isogeny $E_1 \to E_2$ we need to -know that we can choose a map $\alpha : E_1[p^\infty] \to E_2[p^\infty]$ of $p$-divisible groups over $\mathbb Z_p$ and a map -$\beta : E_1 \times \mathbb F_p \to E_2 \times \mathbb F_p$ of elliptic curves such that -$\alpha \times \mathbb F_p = \beta[p^\infty]$ as map of $p$-divisible groups $E_1[p^\infty] \to E_2[p^\infty]$ over $\mathbb F_p$. -The problem is that we have very little flexibility: there are only countably many homomorphisms $E_1 \times \mathbb F_p -\to E_2 \times \mathbb F_p$. But it's hard to lift a map of $p$-divisible groups over $\mathbb F_p$ to a map -of $p$-divisible groups over $\mathbb Z_p$: -The category of $p$-divisible groups over $\mathbb Z_p$ is equivalent, by results of Fontaine-Laffaille, to the -category of quadruples $(M,M^1,\phi,\phi^1)$, where $M$ is a finite free $\mathbb Z_p$-module, $M^1$ a $\mathbb Z_p$-direct -summand ("Hodge filtration"), and $\phi : M \to M$, $\phi^1 : M^1 \to M$ are $\mathbb Z_p$-linear maps such that $p\phi^1 = \phi|_{M^1}$ and $\phi(M) + -\phi^1(M^1) = M$. [In our case $M$ is of rank 2 and $M^1$ of rank 1.] -The Dieudonne module of the special fibre of the given $p$-divisible group is recovered as follows -(reference?): we need to give a $\mathbb Z_p[F,V]/(FV-p)$-module $D$ that is finite free as $\mathbb Z_p$-module. We take $D -:= M$, $F := \phi$, and $V := pF^{-1}$ (this is defined on $M[1/p]$ initially, but the final condition on $M$ forces it to -preserve $M$). In other words, one forgets the Hodge filtration. So lifting $\beta[p^\infty]$ to $\mathbb Z_p$ is difficult because a given map of Dieudonne -modules doesn't need to preserve the Hodge filtration. (The countable flexibility we have with $\beta$ is not enough to -move a given line to any other. Note that $M^1 \otimes \mathbb F_p$ is uniquely determined as the kernel of $\phi \otimes \mathbb F_p$; but apart from that $M^1$ is free to vary.) -[Finally, a small comment/question: I think Tate's result over finite fields assumes that $l$ is prime to the characteristic. -Milne-Waterhouse (1971) have a version for $l = p$ where $T_l$ is replaced by the Dieudonné module. In this case -$T_p E$ is just the first crystalline cohomology group. Is there a general crystalline analogue of the Tate conjecture?] -Added: I think the isogeny theorem can also fail in the case of good ordinary reduction, using the same argument. -Consider again elliptic curves over $\mathbb Q_p$, now of good ordinary reduction. Let's restrict to those elliptic curves of -a fixed reduction $\overline E$ over $\mathbb F_p$. We have $a_p \ne 0$, and let's assume for simplicity that the two roots -of $x^2-a_p x + p$ are in $\mathbb Z_p$, so they are $u$, $pv$ with $u$, $v \in \mathbb Z_p^\times$. Then there's a basis of -$M$ (the Fontaine-Laffaille object associated to $E[p^\infty]$) such that $\phi = \begin{bmatrix}u & \\\ & pv\end{bmatrix}$. The -possible choices of $M^1$ are precisely the lines generated by the vector $\begin{bmatrix} px\\\ 1\end{bmatrix}$, where $x -\in \mathbb Z_p$. The isomorphism class of $M$ (and thus of the Galois representation $T_p E$) only depends on the valuation -of $x$, because a diagonal change of basis leaves $\phi$ invariant. But there are uncountably many $x$ of any fixed positive -valuation (and thus deformations of $\overline E$ to $E$). By the above argument, these cannot all be isogenous.<|endoftext|> -TITLE: are irreducible representations with large fixed subspaces trivial? -QUESTION [21 upvotes]: Say that $G$ is a finite group, and $V$ is an irreducible representation of $G$, over an algebraically closed field $k$. Suppose that for every $g \in G$, there is some subspace $W_g \subset V$ which is (pointwise) fixed by $g$, such that the dimension of $W_g$ is at least half the dimension of $V$. -If $k$ has characteristic zero, then a simple argument with character tables shows that $V$ must be the trivial representation. (When $k = \mathbb{C}$, then $\sum_{g \in G} tr_V(g)$ has positive real part; so appealing to the Lefschetz principle, $\sum_{g \in G} tr_V(g)$ is nonzero whenever $k$ has characteristic zero.) -Is this still true when $k$ has positive characteristic? - -REPLY [15 votes]: Unless I have misread either your question or their result, Corollary 1.2 in "Average dimension of fixed point spaces with applications" by Bob Guralnick and Attila Mar\'oti includes a positive answer to the question. You can find this on the arxiv at -http://arxiv.org/PS_cache/arxiv/pdf/1001/1001.3836v1.pdf<|endoftext|> -TITLE: smooth cohomology of Lie groups -QUESTION [5 upvotes]: Let $G$ be a Lie group and $A$ a smooth $G$-module. Define $C^n(G,A)=\{ f: G^n \to A|~f~\text{is smooth}\}$ and $\partial^n: C^n \to C^{n+1}$ by the standard formula as used in the cohomology of abstract groups. I think this cohomology must be well studied. Can somebody provide me some references for this cohomology. A similar cohomology for topological groups has been studied by Hu and Heller using continuous cochains. -Also, can somebody tell me what kind of Lie group extensions $H^2(G,A)$ correspoind to. - -REPLY [4 votes]: This is indeed very well studied. The standard references are Borel-Wallach, Continuous cohomology, discrete subgroups and representations of reductive groups, Annals of Math. Studies 94, Princeton University Press (1980) and the more gentle book by A. Guichardet, Cohomologie des groupes topologiques et des algèbres de Lie Textes Mathématiques 2 Fernand Nathan, Paris (1980). But there has been a lot of progress since. For instance, there is some recent work by Crainic extending the theory to Lie groupoids and algebroids. -The continuous cohomology is usually attributed to Hochschild-Mostow and coincides with the smooth cohomology under reasonably weak hypotheses. -The interesting fact is that there is an intimate interplay between the continuous cohomology of, say a semi-simple Lie group and the cohomology of its Lie algebra and the de Rham cohomology of the associated symmetric space. One of the most important results is the van Est-isomorphism: - -Let $G$ be a semi-simple Lie group with finite center and no compact factors. Let $M = G/K$ be the associated symmetric space. Then the de Rham complex - \[ - \mathbb{R} \to \Omega^{0}(M) \to \Omega^{1}(M) \to \cdots - \] - is an injective resolution of $\mathbb{R}$. In particular, since a $G$-invariant differential form on $M$ is automatically closed, there is a natural isomorphism - \[ - H_{c}^{\ast}(G,\mathbb{R}) \cong \Omega^{*}(M)^{G} - \] - where the right hand side are the $G$-invariant differential forms on $M$. If $A$ is a sufficiently nice smooth $G$-module then the $A$-valued differential forms need not be closed, but one still has - \[ - H_{c}^{\ast}(G,A) \cong H^{\ast}(\Omega^{\ast}(M,A)^{G}) - \] - Moreover, one may identify this with relative Lie algebra cohomology $H^{\ast}(\mathfrak{g},\mathfrak{k};A)$. - -This immediately gives us an interpretation of $H_{c}^{2}(G,\mathbb{R}) = \Omega^{2}(M)^{G}$. Namely, if $G$ is simple, then $\Omega^{2}(M)^{G}$ is one-dimensional if and only if $G$ is Hermitian (when it is generated by the Kähler form on $M$), and otherwise it is zero. So the dimension of $H_{c}^{2}(G,\mathbb{R})$ corresponds to the number of of Hermitian simple factors of $G$. -Of course, one may also bring discrete subgroups into play, together with all their rich and beautiful connections to geometry and number theory. There are far too many things to mention here, so I'd better stop now. -Before I forget: I don't know of a direct interpretation of $H_{c}^{2}(G,A)$ as equivalence classes of suitable extensions of $G$. One problem is that in the topological category, it is not clear at all that there should be a smooth (or continuous) section of non-trivial extension.<|endoftext|> -TITLE: Books you would like to read (if somebody would just write them...) -QUESTION [187 upvotes]: I think that the title is self-explanatory but I'm thinking about mathematical subjects that have not received a full treatment in book form or if they have, they could benefit from a different approach. -(I do hope this is not inappropriate for MO). -Let me start with some books I would like to read (again with self-explanatory titles) -1) The Weil conjectures for dummies -2) 2-categories for the working mathematician -3) Representations of groups: Linear and permutation representations made side by side -4) The Burnside ring -5) A functor of points approach to algebraic geometry -6) Profinite groups: An approach through examples -Any other suggestions ? - -REPLY [3 votes]: A Bourbaki book on Category Theory.<|endoftext|> -TITLE: when is the power of a nonnegative polynomial a sum of squares? -QUESTION [28 upvotes]: There are polynomials that are not sum of squares. For example Motzkin gave the example $x^4y^2+x^2y^4+z^6-3x^2y^2z^2$ in 1967. -Is there a real polynomial $f\in{\mathbb{R}}[x_1,\ldots,x_n]$ in several indeterminates that is not a sum of squares but $f^N$ is a sum of squares for some odd integer $N>0$? -This question is interesting in the following sense. The notion of writing nonnegative polynomials $f$ as a sum of squares is to give an algebraic proof of the inequality $f\ge 0$. As per Motzkin's example, we know that this is not always possible. One way to resolve this is to follow Artin and use denominators. Another way (which I learnt from D'Angelo) is to show that $f^{N}$ is a sum of squares for some odd $N$. -This question is me wondering whether such a technique of consider the radical of sum of squares is vacuous. - -REPLY [15 votes]: Here's an explicit example. The polynomial $f=x^{4} y^{2}+x^{2} y^{4}-x^{2} y^{2}+1$ is not a sum of squares (as one can check using Motzkin's original proof or by computer). On the other hand, the polynomial $f^3$ can be written as a sum of squares, -$$f^3=c_1F_1^2+c_2F_2^2+\ldots+c_{19}F_{19}^2$$ where the coefficients $c_i$ and polynomials $F_i$ are listed below. -I guess I should mention the software I used for computing this, namely the package "SOS.m2" for Macaulay2. This package has a function 'getSOS' which spits out a sum of squares representation of a given polynomial. See this link for details. The point is that the problem of finding such a representation can be viewed as a problem of semi-definite programming, and can be solved in reasonable time if the degree is small. In particular, this gives the algorithm you mention for checking whether a polynomial is non-negative. -EDIT: If anyone is interested, I have uploaded the Macaulay2 code here. -Now for the coefficients $c_i$: -(c1..c19)=(146/17,146/17,146/17,4036391/1186250,4036391/1186250,4036391/1186250, - 74/25,1847624417319/1971413728310,431999528319079/461906104329750, - 1847624417319/1971413728310,1847624417319/1971413728310,431999528319079/ - 461906104329750,431999528319079/461906104329750,8243/10693,1032024/ - 1393067,16675964223443/35265267617884,16675964223443/35265267617884, - 389070/559013,16675964223443/35265267617884) - -And the polynomials $F_i$: -(F_1,...,F_19)=(-459/3650 x^4 y^4-1071/3796 x^4 y^2-1071/3796 x^2 - y^4+x^2 y^2-17/73,-17/73 x^6 y^3-1071/3796 x^4 y^5+x^4 - y^3-459/3650 x^2 y^3-1071/3796 x^2 y,-1071/3796 x^5 y^4-17/73 - x^3 y^6+x^3 y^4-459/3650 x^3 y^2-1071/3796 x - y^2,-65670975/137237294 x^5 y^4+8569925/68618647 x^3 y^6+x^3 - y^2-65670975/137237294 x y^2,8569925/68618647 x^6 - y^3-65670975/137237294 x^4 y^5+x^2 y^3-65670975/137237294 x^2 - y,x^4 y^4-65670975/137237294 x^4 y^2-65670975/137237294 x^2 - y^4+8569925/68618647,-175/629 x^5 y^3-175/629 x^3 y^5+x^3 - y^3-175/629 x y,x^4 y^2-421805182124/9238122086595 x^2 - y^4-80070895463/1231749611546,x^2 - y^4-1201063431945/17632633808942,-80070895463/1231749611546 x^6 - y^3-421805182124/9238122086595 x^4 y^5+x^2 - y,-421805182124/9238122086595 x^5 y^4-80070895463/1231749611546 x^3 - y^6+x y^2,x^5 y^4-1201063431945/17632633808942 x^3 - y^6,-1201063431945/17632633808942 x^6 y^3+x^4 y^5,-21157/107159 - x^5 y^3-21157/107159 x^3 y^5+x y,-21157/86002 x^5 y^3+x^3 - y^5,x^6 y^3,1,x^5 y^3,x^3 y^6)<|endoftext|> -TITLE: cube + cube + cube = cube -QUESTION [55 upvotes]: The following identity is a bit isolated in the arithmetic of natural integers -$$3^3+4^3+5^3=6^3.$$ -Let $K_6$ be a cube whose side has length $6$. We view it as the union of $216$ elementary unit cubes. We wish to cut it into $N$ connected components, each one being a union of elementary unit cubes, such that these components can be assembled so as to form three cubes of sizes $3,4$ and $5$. Of course, the latter are made simultaneously: a component may not be used in two cubes. There is a solution with $9$ pieces. - -What is the minimal number $N$ of pieces into which to cut $K_6$ ? - -About connectedness: a piece is connected if it is a union of elementary cubes whose centers are the nodes of a connected graph with arrows of unit length parallel to the coordinate axes. - -Edit. Several comments ask for a reference for the $8$-pieces puzzle, mentioned at first in the question. Actually, $8$ was a mistake, as the solution I know consists of $9$ pieces. The only one that I have is the photograph in François's answer below. Yet it is not very informative, so let me give you additional information (I manipulated the puzzle a couple weeks ago). There is a $2$-cube (middle) and a $3$-cube (right). At left, the $4$-cube is not complete, as two elementary cubes are missing at the end of an edge. Of course, one could not have both a $3$-cube and a $4$-cube in a $6$-cube. So you can imagine how the $3$-cube and the imperfect $4$-cube match (two possibilities). Other rather symmetric pieces are a $1\times1\times2$ (it fills the imperfect $4$-cube when you build the $3$-, $4$- and $5$-cubes) and a $1\times2\times3$. Two other pieces have only a planar symmetry, whereas the last one has no symmetry at all. -Here is a photograph of the cut mentioned above. - (source) - -REPLY [53 votes]: JHI's elegant lower bound of $8$ on $N$ is achieved by an explicit dissection. -I show my construction below; you might want to try to find -a solution yourself before proceeding $-$ it makes for a neat puzzle. -There may well be other ways to do it. -If somebody can make a "$3$-dimensional" graphic or picture of the -$8$-piece dissection, you're welcome to add it by editing my answer. -My diagrams are two-dimensional, labeling each piece with its height. -Fortunately the dissection is simple enough for this to be possible; -in particular, the eight pieces comprise four boxes and four L-shaped prisms. -This also made it possible to find the solution using just pencil and paper -on an otherwise uneventful international flight. -Begin by cutting the $6 \times 6 \times 6$ cube top to bottom -into three pieces, as shown in top view in the first square diagram. -Then cut each piece horizontally in two, dividing AB into $3+3$,$\phantom.$ -C into $4+2$, and D into $5+1$. Each AB piece is then further subdivided -into a box B and an L-shaped prism A. The second diagram shows (say) the -bottom layer of four pieces, and the third diagram shows the top. -Note that the AB subdivisions are not quite the same. - (source) -Pieces with the same color will come together to form a smaller cube. -The larger C piece is a $4$-cube, -and the two A pieces form a $3$-cube as shown. -It remains to construct the $5$-cube from the remaining five pieces. -The last two diagrams show the bottom and top of the $5$-cube. - (source) -The two $5$'s are the larger B piece, rotated to span the entire -height of the cube, and the thick D piece. -The thin D piece completes the bottom, with width $1$. -The top is filled by the thinner C piece and the smaller B, -both rotated to height 4. QEF -I guess that a physical model won't make for a hard puzzle to reconstitute -into either one or three cubes (e.g. the AB, C, and D parts of the $6$-cube -are independent) but would still make a nice physical model of the identity -$3^3 + 4^3 + 5^3 = 6^3$. -This dissection is specific to the solution $(a,b,c;d)=(3,4,5;6)$ of the -Diophantine equation $a^3+b^3+c^3=d^3$; I don't know whether an $8$-piece -dissection is possible for any other solution. JHI's analysis shows that -one can never get below $8$, and in some cases even that's not possible: -if $a -TITLE: Quiver varieties and the affine Grassmannian -QUESTION [7 upvotes]: Recently I was watching a talk: http://media.cit.utexas.edu/math-grasp/Ben_Webster.html and at the end the lecturer gave a correspondence (I was having trouble with subscripts so changed the notation a bit): -$$ \mathfrak{Q} (\lambda, \mu) \Leftrightarrow \mathfrak{M} (\lambda, \mu)$$ -Between Quiver varieties and slices of affine Grassmannian. I would like to learn how this exactly works (like precise definition of both objects is a start!). If you have the time and the patience you can post a lengthy response here or you could just point me towards the right place to start reading on my own. Thank you! - -REPLY [9 votes]: Sadly, as the speaker, I would also like to know how that correspondence works. Long story short, some collaborators and I came up with a conjecture that certain pairs of symplectic varieties (algebraic varieties with an algebraic symplectic form) were in some kind of duality with each other. There is still not a good definition of this duality at the moment, but it seems to coincide with certain kinds of S-duality in physics. Unfortunately, you're going to have to wait a few months to read our paper on this stuff, and even then it will be a lot more conjecture than theorem, especially on this point. -So, how do we justify this conjecture? By certain remarkable coincidences. In the case of the quiver varieties and affine Grassmannians, these coincidences are manifestations of fact that both varieties are geometric avatars of the representations of semi-simple Lie algebras (read Nakajima for quiver varieties and Mirkovic-Vilonen for affine Grassmannians). -The most important bit of the conjecture is that these varieties are actually geometric avatars for the representations of categorifications of Lie algebras in the sense of Rouquier or Khovanov-Lauda, by which I mean that the simple module categories over these categorifications should occur as categories of modules over quantizations of either the quiver variety, or the slice. Actually many other module categories, like those corresponding to tensor products should show up as well. This is not written down anywhere in the literature, but the type A special case can be cobbled together from some different sources, and will be in a forthcoming paper of myself, Tom Braden, Tony Licata and Nick Proudfoot. Which I should now get back to writing.<|endoftext|> -TITLE: homotopy type of connected Lie groups -QUESTION [20 upvotes]: Is there a simple proof (short and low-tech) of the following fact: -(E. Cartan) A connected real Lie group $G$ is diffeomorphic (as a manifold) to -$K\times\mathbb{R}^n$ where $K$ is a maximal compact subgroup of $G$. Moreover, -all maximal compact subgroups of $G$ are conjugate to $K$. -One nice corollary of this is that the group $K$ is a deformation retract of $G$ so these -two groups have the same homotopy type. In the case where $G$ is semi-simple one may deduce the first part of the result above using the existence of the Iwasawa decomposition which is not so trivial to prove! But here since I'm only asking for a diffeomorphism between two manifolds, there might be a simpler argument! - -REPLY [2 votes]: Yet another traditional favorite argument, in reality a variant on the more geometric argument of Cartan's, that applies immediately to $G=GL(n,\mathbb R)$ and $K=O(n,\mathbb R)$ uses the model $G/K\approx C$, the cone of positive-definite symmetric $n$-by-$n$ matrices, via a Cartan decomposition $G=C\cdot K$, which itself follows from the spectral theorem for symmetric operators. -Then, given a compact $K'\subset G$, pick any $x_o\in C$, and consider the average $x=\int_{K'} k\cdot x_o\;dk$. Since the cone $C$ is non-degenerate, $x\not=0$, and, by design, it is $K'$-invariant. Since $G$ is transitive on $C$, all isotropy groups of points are conjugate to the isotropy group of $1_n$, namely, $O(n,\mathbb R)$. -This idea can be adapted to some other of the classical groups.<|endoftext|> -TITLE: Volumes of n-balls: what is so special about n=5? -QUESTION [42 upvotes]: I am reposting this question from math.stackexchange where it has not yet generated an answer I had been looking for. - -The volume of an $n$-dimensional ball of radius $R$ is given by the classical formula -$$V_n(R)=\frac{\pi^{n/2}R^n}{\Gamma(n/2+1)}.$$ -It is not difficult to see that the "dimensionless" ratio $V_n(R)/R^n$ attains its maximal value when $n=5$. -The "dimensionless" ratio $S_n(R)/R^n$ where $S_n(R)$ is the $n$-dimensional volume of an $n$-sphere attains its maximum when $n=7$. - -Question. Is there a purely geometric explanation of why the maximal values in each case are attained at these particular values of the dimension? -[EDIT. Thanks to all for the answers and comments.] - -REPLY [7 votes]: Brian Hayes has very nice article (Wayback Machine) about the volume of the n-ball in the current issue of American Scientist (Nov 2011). In particular, he discusses the surprising fact that the maximum volume occurs at $n=5$.<|endoftext|> -TITLE: Mathematical "urban legends" -QUESTION [134 upvotes]: When I was a young and impressionable graduate student at Princeton, we scared each other with the story of a Final Public Oral, where Jack Milnor was dragged in against his will to sit on a committee, and noted that the class of topological spaces discussed by the speaker consisted of finite spaces. I had assumed this was an "urban legend", but then at a cocktail party, I mentioned this to a faculty member, who turned crimson and said that this was one of his students, who never talked to him, and then had to write another thesis (in numerical analysis, which was not very highly regarded at Princeton at the time). But now, I have talked to a couple of topologists who should have been there at the time of the event, and they told me that this was an urban legend at their time as well, so maybe the faculty member was pulling my leg. -So, the questions are: (a) any direct evidence for or against this particular disaster? -(b) what stories kept you awake at night as a graduate student, and is there any evidence for or against their truth? -EDIT (this is unrelated, but I don't want to answer my own question too many times): At Princeton, there was supposedly an FPO in Physics, on some sort of statistical mechanics, and the constant $k$ appeared many times. The student was asked: -Examiner: What is $k?$ -Student: Boltzmann's constant. -Examiner: Yes, but what is the value? -Student: Gee, I don't know... -Examiner: OK, order of magnitude? -Student: Umm, don't know, I just know $k\dots$ -The student was failed, since he was obviously not a physicist. - -REPLY [14 votes]: When I took analysis from Paul Sally, he claimed that a student once asked him in class, "Professor Sally, why is it called the p-adic norm? If it's a norm, what does it measure?" Without thinking, Paul loudly replied, "Well, it measures the p-ness of a number." -I suspect that he just substituted himself into an existing urban legend, yet I would not be surprised if it were true.<|endoftext|> -TITLE: snake lemma in category of groups -QUESTION [12 upvotes]: In Wiki, under the item "category of groups", it states that the snake lemma fails in category of groups, however the nine lemma is valid. However, in the preface of the book " Mal'cev, protomodular, homological and semi-abelian categories ", it says "And category theory could not grasp either the conceptual foundations of the homological lemmas: the Nine Lemma, the Snake Lemma, which remain valid and strongly meaningful in the category Gp of groups, even if this category does not belong to the abelian setting in which these lemmas are generally proved in a significant categorical way." -Does the snake lemma fail or not in the category of groups? Any counter example or reference concerning that? - -REPLY [5 votes]: I think one can have the Snake Lemma in the category of (non abelian) groups by the following way. -First, I remind what is an exact sequence of pointed sets (see, for example, "Local fields" of Serre in the chapter about non abelian cohomology) : a pointed set is just a set $A$ with a based point $x_A$. A morphism of pointed sets is defined to be a map sending the based point to the based point. A sequence $A \xrightarrow{f} B \xrightarrow{g} C$ is said to be exact if $f(A) = g^{-1}(x_C)$. -Now let's go back to the snake Lemma. A cokernel is a pointed set with the class of $0$ as based point (I denote $Coker(f:A \rightarrow B)$ to be the left cosets $B/Im(f)$ but it also works for right cosets, and I call it cokernel even if it is NOT the cokernel in the theory of category). One can check that the Snake Lemma holds in the sense I have given above (with the same proof as Alex said).<|endoftext|> -TITLE: Finite groups in which every character has real values: grading the representations -QUESTION [14 upvotes]: Let $G$ be a finite group. Then the irreducible complex representations of $G$ come in three sorts: real, complex and symplectic=quaternionic. The type of an irreducible character $\chi$ can be read of from the Frobenius-Schur indicator -$$ s_2(\chi) = \frac{1}{|G|}\sum_{g\in G} \chi( g^2 ) \in \{ 1,0,-1 \}. $$ -Now the following seems to be true (and has been used by Noah Snyder in his interesting answer to another question) but I can't see why: -Suppose all characters of $G$ have real values. (Equivalently, every element of $G$ is conjugate to its inverse.) Then it seems that the Frobenius-Schur indicator defines a grading of the irreducible representations, and thus of the character ring. This means that if $\chi$ and $\psi$ are irreducible characters with $s_2 (\chi) = s_2(\psi)$, then all the irreducible constitutents of $\chi\psi$ have indicator $1$, and if $s_2(\chi) = -s_2(\psi)$, then all constituents of $\chi\psi$ have indicator $-1$. Why is this actually true? Of course, for example in the first case, $\chi\psi$ is afforded by a real representation, so the symplectic representations must occur with even multiplicity. But why can they not occur at all? -Moreover, I would like to know if this generalizes to other fields than $\mathbb{R}$, using elements of a Brauer group instead of the Frobenius-Schur indicator. -EDIT: The statement turned out to be wrong in general (see below), so the original question is in some sense obsolete. A more appropriate questions would have been why this Frobenius-Schur indicator grading is there in some (many?) cases. - -REPLY [3 votes]: Here is a variation of the question which is true and generalizes readily to the Brauer group. -If $G$ and $H$ are two groups with two complex irreducible representations $V$ and $W$, then $V \otimes_{\mathbb{C}} W$ is an irrep of $G \times H$, and the Schur-Frobenius indicators multiply. -For the proof, I'll switch to the generalization that works for any Brauer group. Let $F$ be a field of characteristic zero and let $V$ and $W$ be two irreps of $G$ and $H$ over the field $F$. Then $V$ and $W$ might each represent an element of the Brauer group of $F$, because by Schur's lemma each of $\mathrm{End}_G(V)$ and $\mathrm{End}_H(W)$ is a division ring. I say "might" because it is not necessarily true that the centers of these division rings are $F$. If so, then $V \otimes_F W$ does not quite have to be $(G \times H)$-irreducible, but its summands land in the expected place in the Brauer group, using the equation -$$\mathrm{End}_{G \times H}(V \otimes_F W) \cong \mathrm{End}_G(V) \otimes_F \mathrm{End}_H(V).$$ -Taking the case $F = \mathbb{R}$, you get a relation that is equivalent to the previous paragraph. - -Here also is a remark about Noah Snyder's answer to the other question. He points out that the representation ring of $G$ is graded by the character group $Z(G)^*$ of the center of $G$. It sometimes so happens, when the characters of $G$ are all real, that the Schur-Frobenius indicator is given by a group homomorphism from $Z(G)^*$ to $\{\pm 1\}$. If so, then of course the indicator gives you a reduced grading of the representation ring of $G$. So, it would be interesting to know which finite or compact groups have this property, or an analogous property for another Brauer group. I think (not quite sure) that all compact simple Lie groups with real characters are examples.<|endoftext|> -TITLE: What sums of equal powers of consecutive natural numbers are powers of the same order? -QUESTION [8 upvotes]: Trivially $n^1=n^1$, and everyone knows that $3^2+4^2=5^2$. Denis Serre quoted $3^3+4^3+5^3=6^3$ in a recent MathOverflow question (which prompted this one). Are any other examples known? - -REPLY [3 votes]: Well, if we consider n consecutive 4th powers with initial a, -F(a,n) = a^4 + (a+1)^4 + (a+2)^4 + ... + (a+n-1)^4 -or, equivalently, -F(a,n) = (n/30)(-1+30a^2-60a^3+30a^4+30a(1-3a+2a^2)n+10(1-6a+6a^2)n^2+(-15+30a)n^3+6n^4) -it is easy to check that F(a,n) = y^4 (or even just y^2) has NO solution in the positive integers with BOTH {a,*n*} < 1000, with the exception of the trivial n = 1. (I had checked this with Mathematica some time back.) -If we relax your question and allow n 4th powers in arithmetic progression d equal to some kth power, then the smallest I found was 64 4th powers with common difference d = 2 starting with, -29^4 + 31^4 + 33^4 + ... + 155^4 = 96104^2 -P.S. The closed-form formula for general d is available, but I find it too tedious to include in this post.<|endoftext|> -TITLE: Commutative Ring of Finite Global Dimension -QUESTION [5 upvotes]: The only examples of commutative rings of finite global dimension I know are either: - -Dedekind domains (and fields as a degenerate special case) -Regular local rings -Rings constructed from the previous examples by taking direct sums, or forming the rings of polynomials over a ring of finite global dimension. - -Are there other examples? In particular, are there other examples that are finite-dimensional over a field $k$? -(Examples of rings of finite global dimension are easier to come by in the noncommutative case, but I'm specifically curious about the commutative case.) - -REPLY [9 votes]: At arsmath's request, I'm making this official. -(This is pretty standard commutative algebra, but I realize not everyone has gone through it.) -A commutative ring $R$ is regular if it's noetherian and its local rings are regular. -Using Serre's theorem e.g. Matsumura Commutative Ring Theory p 156, and the fact -that $Ext$ commutes with localization, we can see that any regular ring with finite Krull -dimension has finite global dimension. -To an algebraic geometer regular = nonsingular. So in particular, so there is a large -supply of basic examples arising as coordinate rings of nonsingular affine varieties. -This is a bit circular the way I'm saying it, but of course, you can test the condition -using the Jacobian criterion...<|endoftext|> -TITLE: A general theory of multiplicity-free actions of $G\times H$? -QUESTION [5 upvotes]: There seem to be a bunch of different results of the form "This nice -representation $V$ of $G\times H$ breaks up as $\bigoplus_\lambda V_\lambda -\otimes W_\lambda$, where the $(V_\lambda)$ are distinct irreps of $G$ -and the $(W_\lambda)$ are distinct irreps of $H$." -Examples I know (described algebraically): - -$G = S_n$, $H = GL(k)$, $V = ({\mathbb C}^k)^{\otimes n}$. -$G = H$, $V = Fun(G)$ (the Peter-Weyl theorem) -$G = GL(m)$, $H = GL(n)$, $V = Sym({\mathbb C}^m \otimes {\mathbb C}^n)$. - -I know there should be some analogous one with $O(m)$ and $Sp(n)$... -My questions: - -What are other examples? -Is there a general theory of these, that in particular would -encompass Peter-Weyl? (So not just about actions on symmetric algebras.) - -REPLY [2 votes]: When $K$ is a complex reductive linear algebraic group, an affine variety $X$ equipped with an action of $K$ is called multiplicity-free if its ring of functions $\mathbb{C}[X]$ is multiplicity-free as a $K$-module. When $K$ is connected, this means that a Borel subgroup has a dense orbit on $X$ (if $X$ is also normal, such a variety is called spherical). -In the question's example (2), $K = G \times G$ and $X = G$ (with $G \times G$ acting by left and right multiplication) and $V= \mathbb{C}[G]$. In example (3), $K=GL(m)\times GL(n)$ and $X = (\mathbb{C}^m \otimes \mathbb{C}^n)^*$. -There is a general theory of spherical varieties, and there are classifications. For example, -Kac[MR575790], Leahy[MR1650378] and Benson-Ratcliff[MR1382030] did the case where $X$ is itself a representation of $K$ (see also [Howe and Umeda, MR1116239] and [Knop, MR1653036]). Homogeneous spherical varieties were classified in [Kramer, MR0528837] and [Brion, MR0822838]. -Edit: As pointed out by Allen in the comments, the condition that $V$ be multiplicity-free as a $(G \times H)$-module is weaker than the condition on $V$ in the original question. Let's call a $V$ as in the question strongly MF. As Allen points out in another comment, $\mathbb{C}[X]$ is strongly MF if and only if $B_G \times U_H$ and $U_G \times B_H$ have open orbits on $X$ (here $B_G$ is a Borel subgroup of $G$, and $U_G$ a maximal unipotent subgroup of $G$).<|endoftext|> -TITLE: Decent texts on categorical logic -QUESTION [18 upvotes]: Recently I read the chapter "Doctrines in Categorical Logic" by Kock, and Reyes in the Handbook of Mathematical Logic. And I was quite impressed with the entire chapter. However it is very short, and considering that this copy was published in 1977, possibly a bit out of date. -My curiosity has been sparked (especially given the possibilities described in the aforementioned chapter) and would like to know of some more modern texts, and articles written in this same style (that is to say coming from a logic point of view with a strong emphasis on analogies with normal mathematical logic.) -However, that being said, I am stubborn as hell, and game for anything. -So, Recommendations? -Side Note: More interested in the preservation of structure, and the production of models than with any sort of proposed foundational paradigm - -REPLY [13 votes]: Well, for the record, there is also my -Practical Foundations of Mathematics (CUP, 1999).<|endoftext|> -TITLE: Decomposing posets into countably many chains -QUESTION [8 upvotes]: A conjecture of Galvin's is that the following is possible (which I take to mean that the consistency of the following can be proven relative to the consistency of something like ZFC, or ZFC plus some large cardinal): - -A partially ordered set $P$ can be decomposed into countably many chains iff the same is true of every suborder $P_0 \subseteq P$ of size at most $\aleph _1$. - -You can see it stated as Conjecture 3.3 in Todorcevic's "Combinatorial Dichotomies in Set Theory". -I'm interested in this conjecture and determining its consistency strength, but in order to get a feel for it I want to first look at it in a special case in which it's supposed to be (according to Todorcevic, if I understood what he told me correctly) provable from ZFC alone. I'll actually list three special cases of increasing generality; an answer to the last case would be ideal but I'd be happy to see an answer for the first case. - -Galvin's conjecture, restricted to posets $P$ which are the Cartesian product of two linear orders (with the obvious product ordering). -GC restricted to posets which are the Cartesian product of countably many linear orders (countable = finite or denumerable). -GC restricted to posets which are subsets of some Cartesian product of countably many linear orders. - -REPLY [5 votes]: I've investigated Galvin's Conjecture quite a bit over the past few years. Here is a preprint of mine on the subject. In there, I show that Galvin's Conjecture restricted to finite-dimensional posets (posets which are subsets of some cartesian product of finitely many linear orderings) is equivalent to Rado's Conjecture (which was extensively studied by Todorcevic). Unfortunately, the ℵ0-dimensonal case has eluded me; perhaps you will have better luck...<|endoftext|> -TITLE: Is there more than 1 way to make a 17-node graph such that there are no 4-cycles and each node has at least four edges? -QUESTION [6 upvotes]: I'm working on the 17x17 challenge, and this sub-problem has come up. I have one solution to this problem that you can see here: - -For some complex reasons (that I can elaborate on if needed) I know that this graph is unique if I add the following constraint: There has to exist a subset of at least 6 5 nodes that are not connected to each other and are not individually part of a 3-cycle. In the graph above, any 5-set of pink nodes satisfies this constraint. What I do not know is if it is still unique when I remove this extra constraint. -I'm coming from a CS background so I may be missing something basic from a graph theory perspective. Any references that may help me either prove that this is the case, disprove it by producing one or more different graphs or (even better) by producing all possible alternative solutions would be deeply appreciated. -EDIT: On closer examination, it seems the graph contains a double Hamiltonian circuit that uses every edge. No idea if this is relevant. [not interesting] - -REPLY [6 votes]: Yes, here are some -Graph 1, order 17. - 0 : 5 8 12 13; - 1 : 6 9 13 16; - 2 : 7 10 13 15; - 3 : 8 9 14 15; - 4 : 9 10 11 12; - 5 : 0 11 15 16; - 6 : 1 10 14 16; - 7 : 2 11 13 14; - 8 : 0 3 12 14; - 9 : 1 3 4 15; - 10 : 2 4 6 12; - 11 : 4 5 7 16; - 12 : 0 4 8 10; - 13 : 0 1 2 7; - 14 : 3 6 7 8; - 15 : 2 3 5 9; - 16 : 1 5 6 11; -Graph 2, order 17. - 0 : 5 8 14 15; - 1 : 6 9 15 16; - 2 : 7 11 13 16; - 3 : 8 9 10 13; - 4 : 9 11 12 14; - 5 : 0 10 11 15; - 6 : 1 10 12 16; - 7 : 2 12 13 15; - 8 : 0 3 14 16; - 9 : 1 3 4 15; - 10 : 3 5 6 13; - 11 : 2 4 5 16; - 12 : 4 6 7 14; - 13 : 2 3 7 10; - 14 : 0 4 8 12; - 15 : 0 1 5 7 9; - 16 : 1 2 6 8 11; -Graph 3, order 17. - 0 : 5 7 11 13; - 1 : 6 8 13 15; - 2 : 7 9 10 15; - 3 : 8 10 14 16; - 4 : 9 11 12 14; - 5 : 0 11 15 16; - 6 : 1 12 13 16; - 7 : 0 2 10 13; - 8 : 1 3 14 15; - 9 : 2 4 12 15; - 10 : 2 3 7 16; - 11 : 0 4 5 14; - 12 : 4 6 9 16; - 13 : 0 1 6 7; - 14 : 3 4 8 11; - 15 : 1 2 5 8 9; - 16 : 3 5 6 10 12; -Graph 4, order 17. - 0 : 5 7 14 16; - 1 : 6 8 12 13; - 2 : 7 9 10 12; - 3 : 8 10 11 15; - 4 : 9 11 13 14; - 5 : 0 13 15 16; - 6 : 1 11 12 16; - 7 : 0 2 10 14; - 8 : 1 3 14 15; - 9 : 2 4 12 15; - 10 : 2 3 7 13; - 11 : 3 4 6 16; - 12 : 1 2 6 9; - 13 : 1 4 5 10; - 14 : 0 4 7 8; - 15 : 3 5 8 9; - 16 : 0 5 6 11; -Graph 5, order 17. - 0 : 5 7 11 13; - 1 : 6 8 12 15; - 2 : 7 9 10 15; - 3 : 8 10 14 16; - 4 : 9 12 13 16; - 5 : 0 11 15 16; - 6 : 1 11 12 14; - 7 : 0 2 10 13; - 8 : 1 3 13 15; - 9 : 2 4 14 15; - 10 : 2 3 7 16; - 11 : 0 5 6 14; - 12 : 1 4 6 16; - 13 : 0 4 7 8; - 14 : 3 6 9 11; - 15 : 1 2 5 8 9; - 16 : 3 4 5 10 12; -Graph 6, order 17. - 0 : 5 7 12 13; - 1 : 6 10 13 16; - 2 : 7 8 14 15; - 3 : 8 9 10 12; - 4 : 9 11 13 14; - 5 : 0 10 11 15; - 6 : 1 12 15 16; - 7 : 0 2 14 16; - 8 : 2 3 11 12; - 9 : 3 4 13 15; - 10 : 1 3 5 14; - 11 : 4 5 8 16; - 12 : 0 3 6 8; - 13 : 0 1 4 9; - 14 : 2 4 7 10; - 15 : 2 5 6 9; - 16 : 1 6 7 11; -Graph 7, order 17. - 0 : 5 7 11 14; - 1 : 6 11 13 15; - 2 : 7 8 12 13; - 3 : 8 9 14 15; - 4 : 9 10 11 12; - 5 : 0 10 13 16; - 6 : 1 12 14 16; - 7 : 0 2 12 15; - 8 : 2 3 11 16; - 9 : 3 4 13 14; - 10 : 4 5 15 16; - 11 : 0 1 4 8; - 12 : 2 4 6 7; - 13 : 1 2 5 9; - 14 : 0 3 6 9; - 15 : 1 3 7 10; - 16 : 5 6 8 10; -Graph 8, order 17. - 0 : 5 7 13 14; - 1 : 6 11 13 15; - 2 : 7 8 10 12; - 3 : 8 9 15 16; - 4 : 9 10 11 14; - 5 : 0 11 12 16; - 6 : 1 12 14 15; - 7 : 0 2 10 13; - 8 : 2 3 14 16; - 9 : 3 4 12 13; - 10 : 2 4 7 15; - 11 : 1 4 5 16; - 12 : 2 5 6 9; - 13 : 0 1 7 9; - 14 : 0 4 6 8; - 15 : 1 3 6 10; - 16 : 3 5 8 11;<|endoftext|> -TITLE: Presentation of GL(n,p)? -QUESTION [6 upvotes]: When finding representations of finite groups over $\mathbb{F}_p$, (i.e. homomorphism from $G$ to $GL(n,\mathbb{F}_p$), it requires many times presentation of $GL(n,p)$. What is presentation of GL(n,p)? - -REPLY [2 votes]: See -MR1871620 (2002i:20068) -Chiaselotti, Giampiero(I-CLBR) -Some presentations for the special linear groups on finite fields. (English summary) -Ann. Mat. Pura Appl. (4) 180 (2001), no. 3, 359–372. -http://dl.dropbox.com/u/5188175/slnpresentations.pdf<|endoftext|> -TITLE: Chebotarëv data over number fields -QUESTION [9 upvotes]: Define a Chebotarëv datum over a number field $K$ to be a finite group $G$ together with a map $\mathfrak{p}\mapsto\gamma_{\mathfrak{p}}$ from a cofinite set of primes of $K$ into the set of conjugacy classes of $G$ such that for every conjugacy class $c\subset G$, the proportion of $\mathfrak{p}$ with $\gamma_{\mathfrak{p}}=c$ is $\operatorname{Card}c/\operatorname{Card}G$. -Two Chebotarëv data $(G,\gamma)$, $(G',\gamma')$ over the same number field are said to be equivalent if there is an isomorphism $\varphi:G\to G'$ such that $\varphi(\gamma_{\mathfrak{p}})=\gamma'_{\mathfrak{p}}$ for almost all $\mathfrak{p}$. If so, we identify the two. -Every finite galoisian extension $L$ of $K$ gives rise to a Chebotarëv datum $(\operatorname{Gal}(L|K),\gamma_{L|K})$ (Chebotarëv's density theorem). -Moreover, if $L_1$ and $L_2$ are two finite galoisian extensions of $K$ for which the associated Chebotarëv data $(\operatorname{Gal}(L_i|K),\gamma_{L_i|K})$ are equivalent, then $L_1=L_2$ (see Lemma 1, p. 174, of Mazur's recent article). -Question. Does every Chebotarëv datum over a given number field $K$ arise from some finite galoisian extension $L$ of $K$ ? - -REPLY [8 votes]: In addition to David's answer, one can provide explicit counterexamples, at least in specific cases. -For simplicity, take $K=\mathbf{Q}$ and $G=\{\pm 1\}$. Define the Chebotarëv datum $\gamma$ by $\gamma(p) = -1$ if $p \equiv 1 \pmod{4}$ and $\gamma(p)=1$ if $p \equiv 3 \pmod{4}$. If $\gamma$ comes from a quadratic field $L$ of discriminant $D$, then there exists a quadratic character $\chi : (\mathbf{Z}/D\mathbf{Z})^* \to \{\pm 1\}$ such that $\gamma(p) = \chi(p)$ for almost all primes $p$. We now have a contradiction by looking at primes $p \equiv 1 \pmod{4D}$ (there are infinitely many such primes). -In fact, by Dirichlet's theorem on primes in arithmetic progressions, the Chebotarëv datum $\gamma$ cannot be modified on a subset of natural density zero in order to come from a quadratic field.<|endoftext|> -TITLE: Why is it so hard to implement Haken's Algorithm for knot theory? -QUESTION [22 upvotes]: Why is it so hard to implement Haken's Algorithm for recognizing whether a knot is unknotted? (Is there a computer implementation of this algorithm?) - -REPLY [29 votes]: Regarding Haken's algorithm: It's not so hard to implement (it's essentially implemented in Regina, though at present you need to type a few lines of python to glue the bits together; a single "big red button" is on its way). However, it's hard to run, since the algorithm has exponential running time (and, depending on how you implement it, exponential memory use). -There are two facts that make Haken's algorithm easier to implement than many other normal surface decision algorithms: - -You only need to search through vertex normal surfaces, not fundamental normal surfaces (Jaco & Tollefson, 1995). Vertex normal surfaces are much easier (and much faster) to enumerate. -The test that you apply to each vertex normal surface is relatively simple (see if it describes a disk with non-trivial boundary). For other problems (notably Hakenness testing), the test that you apply to each vertex normal surface can be far more difficult than the original vertex enumeration. - -The reason Haken's algorithm is slow is that vertex enumeration is NP-hard in general. There are some tempting short-cuts: one is to run $3^n$ polynomial-time linear programs that maximise Euler characteristic over the $3^n$ possible combinations of quad types. However, experimental experience suggests that this short-cut makes things worse: solving $3^n$ linear programs guarantees $\Omega(3^n)$ running time even in the best case for a non-trivial knot. On the other hand, if you perform a full vertex enumeration (and you structure your vertex enumeration code well [1]) then you often see much faster running times in practice, even though the theoretical worst case is slower. -An aside (which has already been noted above): there are much faster heuristic tests for unknot recognition, though these are not always guaranteed to give a definitive answer. SnapPea has some, as does Regina. There are many fast ways of proving you have a non-trivial knot (e.g., invariants or geometric structures). One fast way of proving you have a trivial knot is to triangulate the complement and "greedily simplify" this triangulation. If you're lucky, you get an easily-recognised 1-tetrahedron solid torus. If you're unlucky, you go back and run Haken's algorithm. The interesting observation here is that, if your greedy simplification is sophisticated enough, you almost always get lucky. (This is still being written up, but see arXiv:1011.4169 for related experiments with 3-sphere recognition.) -Btw, thanks Ryan for dragging me online. :) -[1] arXiv:0808.4050, arXiv:1010.6200<|endoftext|> -TITLE: Why are they called Spherical Varieties? -QUESTION [22 upvotes]: My understanding is if you have a homogeneous space $X = G/H$ for a reductive group $G$ and $X$ is normal and has an open $B$-orbit, for a Borel subgroup $B$ then you call $X$ spherical. -Someone asked me where 'spherical' came from and I had no idea. I asked a few more knowledgeable people and they also didn't know. So now I ask the same question here. - -REPLY [34 votes]: Since I am being asked the same question repeatedly and since the given answers are not quite correct, I post another answer despite the thread being so old. -According to a talk by Domingo Luna around 1985, the term spherical variety is not derived from spheres, at least not directly. Firstly, spheres are way too atypical, e.g., their compactification theory is pretty pointless. Secondly, in invariant theory circles spheres are called quadrics anyway. -The true origin is a paper by Manfred Krämer from 1979 called "Sphärische Untergruppen in kompakten zusammenhängenden Liegruppen". Krämer had observed that one of the standard constructions for spherical functions generalizes from symmetric spaces to arbitrary homogeneous spaces $G/H$ with $G$ a compact Lie group. If $G/H$ contains not too many spherical functions (i.e., they commute under convolution) then he called $H$ a spherical subgroup and proceeded to classify them for simple $G$ in the paper mentioned above. -Around the same time it was realized (Vinberg-Kimelfeld) that Krämer's condition is equivalent to a Borel subgroup having an open orbit on the complexification of $G/H$, i.e., $G_{\mathbb C}/H_{\mathbb C}$ being spherical. Since Krämer's list does provide lots of non-trivial examples, Brion-Luna-Vust came up with their term. -So the implications are -$$ -\text{sphere}\Longrightarrow\text{spherical function}\Longrightarrow\text{spherical variety} -$$ -The first arrow this time makes sense since spherical functions were first seriously considered on spheres (Legendre and Gegenbauer polynomials).<|endoftext|> -TITLE: Resource for Measured Foliations and Whitehead Equivalence -QUESTION [6 upvotes]: I am having trouble making the so-called "Whitehead equivalence" explicit. -It is quite easy to draw a picture of what a Whitehead move is, given a foliation, as a kind of limit of isotopies of the foliation. What I would like to more carefully understand is precisely what is happening to the transverse measure. The goal is to understand why the map that takes a measured foliation to a map from the space of isotopy classes of closed curves to the real numbers is constant under Whitehead equivalence (which is why, I suppose, measured foliations are considered as Whitehead equivalence classes). -Can anyone point out a resource? The more explicit the better. With respect, I have been reading the book by A. Fathi, F. Laudenbach, and V. Po´enaru ("Thurston's Work on Surfaces"), and while it is quite useful, it does not make this point clear. -Thanks! - -REPLY [5 votes]: I don't know of a source other than FLP for understanding the Thurston compactification via measured foliations, but you might find it easier to understand this compactification using measured geodesic laminations. The nice thing about measured geodesic laminations is that you don't need clumsy operations like Whitehead moves to make sense of them (and there is a bijective correspondence between measured foliations and measured geodesic laminations, so they contain the same information). Two nice sources for this theory are Casson-Bleiler's book and Bonahon's paper -MR0931208 (90a:32025) -Bonahon, Francis(1-SCA) -The geometry of Teichmüller space via geodesic currents. -Invent. Math. 92 (1988), no. 1, 139–162. -Probably Bonahon's paper is what you want -- it explicitly constructs the Thurston compactification using measured geodesic laminations and has references to places that give full details about the correspondence between measured foliations and measured geodesic laminations. -Another nice source for laminations is Bonahon's unfinished manuscript "Simple Closed Curves on Surfaces", available here.<|endoftext|> -TITLE: Prime counting - any fast alternatives to the Lagarias-Miller-Odlyzko combinatorial method or the Lagarias-Odlyzko analytical methods? -QUESTION [7 upvotes]: I guess the question says it all - I'm trying to track down fast algorithms for prime counting to know what's out there. -I'm already familiar with the two algorithms mentioned in the title (essentially sections 3.6.1 and 3.6.2 of my 2001 copy of Crandall & Pomerance's Prime Numbers: A Computational Perspective, although from looking online, it looks like they might be section 3.7.1 and 3.7.2 of a more recent addition). -Are there any other families of algorithms in the ballpark of, say, O(n^2/3 + epsilon) time and O(n^1/2+epsilon) space or better for prime counting? And if so, where could I find them? Is there any good repository of various prime counting algorithms anywhere on the web? - -REPLY [5 votes]: I don't know if this might interest you, but there is a nice formula due to Lifchitz which allows to compute the parity of $\pi(x)$ in $O(\sqrt{x}\log x)$: if $\Pi(x)$ represent the number of prime powers up to $x$ then -$$ \Pi(x) \equiv \sum_{m\leq \sqrt{x}} \mu(m) \sum_{n\leq \sqrt(x/m^2)} \left\lfloor\frac{x}{nm^2}\right\rfloor \pmod{2} $$ -This is more like a comment, but I can't comment as I have not enough reputation.<|endoftext|> -TITLE: Extremal Obstructions to Gowers Uniformity -QUESTION [10 upvotes]: Recall the definition of the Gowers uniformity norm $\|f\|_{U^{k}(G)}$, -$$ - \|f\|_{U^{k}(G)} := \left( \mathbb{E}_{x,h_1,\ldots,h_k \in G} \Delta_{h_1} \ldots \Delta_{h_{k}} f(x) \right)^{2^{-k}} \, -$$ -where the operator $\Delta_h$ is a multiplicative analog of a derivative given by -$$ - \Delta_h f(x) := f(x+h) \overline{f(x)} \,, -$$ -and $G$ is a finite abelian group. I'm specifically interested in the case $G=\mathbb{Z}_d$ of integers modulo $d$, and $k=3$. Therefore, I'll just use the shorthand notation -$\|f\|_{U^{k}(\mathbb{Z}_d)} = \|f\|_{U^{k}}$. -I'm interested in functions $f:\mathbb{Z}_d \to \mathbb{C}$ which have some fixed value of $\|f\|_2$, say 1, meaning that -$$ - \|f\|_2^2 = \sum_{h \in \mathbb{Z}_d} f(h) \overline{f(h)} = 1\,. -$$ -Then my question is, - -What are the functions having unit 2-norm which minimize $\|f\|_{U^3}$? - -I can prove a lower bound of $\|f\|^8_{U^3} \ge \frac{2}{d^{4} (d+1)}$, so such functions cannot have arbitrarily small Gowers norm. This bound seems to be tight for all values of $d$ (via numerics) but there is no obvious function which provably saturates the bound for all $d$. -From what I can tell, it appears that such obstructions to Gowers uniformity, like the 2-norm constraint above, have been studied before. But I cannot tell if such extremal problems have been studied, or even if they are thought to be tractable. - -REPLY [5 votes]: I'm not quite sure where your lower bound comes from, but something close comes from functions such as -$$ f(x)=e(x^3/d)$$ -This (after rescaling by $d^{-1/2}$ to match your definition) has $L^2$ norm 1, and $U^3$ norm -$$ \|f\|_{U^3}^8\leq\frac{2}{d^5}$$ -In general, these `phase functions' of degree $n$ will give functions with small $U^n$ norm (because 'differentiating' such a phase function $n-1$ times gives a sum over linear phase functions and hence a lot of a cancellation), and I suspect the extremal example will be of this sort.<|endoftext|> -TITLE: Which Riemann surfaces arise from the Riemann existence theorem? -QUESTION [12 upvotes]: The following was already known to Riemann. Suppose that one is given a connected Riemann surface $X$, a finite set $\Delta \subset X$ and a homomorphism $\phi: \pi_1(X \backslash \Delta) \to S_d$ where $S_d$ is the permutation group on $d$ symbols. If this homomorphism is transitive, i.e. the image of $\phi$ acts transitively on $\{1, \ldots ,d\}$, then this data allows one to construct a unique connected Riemann surface $Y$ with a map $f: Y \to X$ which is a branched cover: when we restrict $f$ to $f^{-1}(X \backslash \Delta)$ it is a $d$-fold cover and around the branch points the monodromy is given by $\phi$. This is known as the Riemann existence theorem and it is proven by constructing the cover corresponding to kernel of $\phi$ and filling in the missing points with disks using $\phi$. If $X$ is compact, $Y$ will be as well. -My question is: given the Riemann sphere (or any other Riemann surface), which compact connected Riemann surfaces can one get if one is allowed to pick $\Delta$, $d$ and $\phi$ as above? This may seem like a trivial question: topologically any Riemann surface arises this way. But it is not clear to me what complex structures can arise. Alternatively, the question may be phrased as: is the map from such data to the disjoint union of moduli spaces of Riemann spaces of different genus surjective? -I would of course also be interested in literature discussing this or related questions. -Edit: the answers are correct that this question was easy. I was actually interested in the situation where we demand that $\phi$ assigns to a loop around a point in $\Delta$ permutation of consisting of disjoint cycles of length two, such that we get only branching of degree 2. -So I guess the updated question is: what restrictions can we place on $\phi$ such that we still get all compact connected Riemann surfaces this way? - -REPLY [13 votes]: The answer is simple: any complex structure arises in this way. -Indeed, any compact Riemann surface $X$ admits a holomorphic cover $\phi \colon X \to \mathbb{P}^1$. -This is straightforward in genus $0$ and $1$. -If the genus is at least $2$, then the linear system $|3K_X|$ is very ample, so it gives an embedding $\gamma \colon X \to \mathbb{P}^N$ and the map $\phi$ is obtained by composing $\gamma$ with a suitable projection. -We are using here the fact that every compact Riemann surface is a smooth complex projective curve; this follows from the existence of a meromorphic function on it, which is a non-trivial fact (see Andy Putman's comment). -EDIT. We can always chose $\phi$ so that the local monodromy is general, that is it is given by a single transposition around each branch point. In fact, projecting $X \subset \mathbb{P}^N$ from a general subspace of the right codimension, we obtain a plane curve $X' \subset \mathbb{P}^2$ whose singularities are at worst ordinary double points. Moreover, $X'$ has at most a finite number of bitangent lines; taking the projection $\pi_p \colon X' \to \mathbb{P}^1$ from a point $p$ not contained in any of the bitangent lines and composing with the map $X \to X'$ we obtain a finite cover $\phi \colon X \to \mathbb{P}^1$ whith the desired property.<|endoftext|> -TITLE: Average distance to a curve of fixed length -QUESTION [5 upvotes]: Let $C$ be a continuous curve in the unit square having length $L$. Is there a lower bound on the average distance between the points in the unit square and $C$, as a function of $L$? Is there an asymptotic behavior that's known as $L$ gets large? (other suggestions for tags are welcome) - -REPLY [5 votes]: Here is a rough answer. I think it has to give the right order of magnitude. -$1/L$. If you draw a zigzag curve that goes up and down $L$ times it has length approximately $L$. Each point is distance no more than $1/L$ from the curve. -On the other hand if you consider a neighbourhood of a curve of width $1/(4L)$ on each side, it has area bounded above by approximately $1/2$. This means that 50% of points are at a distance greater than $1/(4L)$ from the curve so the average distance is at least $1/(8L)$.<|endoftext|> -TITLE: Locally compact Hausdorff space that is not normal -QUESTION [21 upvotes]: What is a good example of a locally compact Hausdorff space that is not normal? It seems to be well-known that not all locally compact Hausdorff spaces are normal (and only a weaker version of Urysohn's Lemma holds in general in the locally compact Hausdorff case). However, I can't seem to think of any examples that demonstrate this, and I have tried all of the "standard" topological counterexamples such as the long line, etc. - -REPLY [5 votes]: Let me give a construction that gives many locally compact spaces that are not normal. If $X$ is a completely regular space and $C$ is a compactification of $X$, then $X$ is paracompact if and only if $X\times C$ is normal. Therefore if $X$ is locally compact but not paracompact, then $X\times C$ is locally compact but not normal. Furthermore, the paracompact locally compact spaces are precisely the spaces that can be partitioned into a collection of $\sigma$-compact clopen sets (recall that a space is $\sigma$-compact if and only if it can be written as a countable union of compact sets). In particular, if $X$ is a connected locally compact space that is not $\sigma$-compact and $C$ is a compactification of $X$, then $X\times C$ is not normal.<|endoftext|> -TITLE: Is there a Chern-Gauss-Bonnet theorem for orbifolds? -QUESTION [26 upvotes]: There's a Gauss-Bonnet theorem for compact 2-orbifolds(due to Satake, I think), which gives a relation between the curvature of a Riemannian orbifold and the orbifold topology(i.e. taking into account not just the structure of O as a Hausdorff topological space, but also the structure of the singular points.) The only proof that I've seen was very much like the classical one of the Gauss-Bonnet theorem, with geodesic triangulations, angle defects and so on. This approach doesn't generalise to higher dimensions to prove the Chern-Gauss-Bonnet theorem, and I haven't even found a conjectured Chern-Gauss-Bonnet formula. I'm especially interested in the four-dimensional case, where the integrand is amenable. One reason I'm interested is just to get a feel for how much more complicated orbifolds are than manifolds; on the one hand, many basic definitions seem to go through from the manifold world to the orbifold world, but the generalisation leads to significant complications in practice. On the other hand, Kleiner and Lott posted a paper on the arxiv in which they use Ricci flow to geometrise 3-orbifolds, so orbifolds certainly seem like a good arena in which to generalise differential geometry. Chern-Gauss-Bonnet is a bit of a benchmark for higher-dimensional Riemannian geometry, and I'd like to know the state of the art is in this case. - -REPLY [7 votes]: One could probably show that the Gauss-Bonnet integral is a topological invariant of the orbifold (i.e. independent of the Riemannian metric), by doing local deformations of the metric within orbifold atlases. If you take an atlas of the orbifold which is the quotient of a ball by a finite group, then for any two Riemannian metrics which agree in a neighborhood of the boundary, the Gauss-Bonnet integral should be the same (since one may work equivariantly in the ball cover). Any two Riemannian metrics may be connected by a sequence of such local deformations by using a partition of unity, and so the total Gauss-Bonnet integral is a topological invariant. The question is whether this invariant equals the Euler characteristic. -I think this may be shown by reduction to the good orbifold case by a cut and paste trick. If you have a codim-1 suborbifold, one may assume that the Riemannian metric is reflection symmetric in a neighborhood of the codim-1 suborbifold. Cut along the suborbifold, then mirror the resulting boundary to get a new orbifold. Clearly the Gauss-Bonnet integral and the Euler characteristic of the resulting orbifold is the same (the orbifold euler characteristic is additive upon gluing along mirrored suborbifolds). I think one should be able to iterate this to reduce to the case of a good orbifold, which would look like a ball quotient with a right-angled boundary pattern (like in Thurston and Davis' reflection trick). -Take a covering by ball-quotients, and assume that their boundaries intersect transversally. One may choose a metric on the orbifold so that the metric is locally a product metric near the boundary of each ball quotient. When several of these intersect, the metric will look locally like a product of a metric on a cube times the metric on the intersection (so that the boundaries meet orthogonally). Then cut along these codim-1 suborbifolds, and mirror the boundary. Davis' argument shows that these orbifolds are good, and thus the Gauss-Bonnet theorem applies. Since each region is contained in a ball quotient, it is good and you can take a manifold cover. Then reflect iteratively in the boundary by doubling along each subset of the boundary coming from a single boundary of a ball quotient. You end up with a closed manifold, so the orbifold is good.<|endoftext|> -TITLE: What can we learn from the tropicalization of an algebraic variety? -QUESTION [37 upvotes]: I often hear people speaking of the many connections between algebraic varieties and tropical geometry and how geometric information about a variety can be read off from the associated tropical variety. Although I have seen some concrete examples of this, I am curious about how much we can get out of this correspondence in general. More precisely, my question is the following: - -Which information of $X=V(I)$ can be - read off its tropicalization - $\mbox{Trop(X)}=\bigcap_{f\in I}\mbox{trop}(f)$? - -As a very basic example, it is known that $\dim(X)=\dim_{\mathbb{R}}\mbox{Trop}(X)$. - -REPLY [5 votes]: In arxiv.org/0805.1916 Sam Payne shows that one can reconstruct the analytification of a quasiprojective variety over a nonarchimedean field as the inverse limit of its tropicalisations.<|endoftext|> -TITLE: Why must a reducible flat SU(2)-connection over a homology sphere be trivial? -QUESTION [6 upvotes]: Let $M$ be a homology sphere. Suppose $P=M\times SU(2)$ is the trivial $SU(2)$ principal bundle. Let $R$ be all reducible connections on $P$. Here $A$ in $R$ is reducible if the gauge transformation group acting on $A$ has nontrivial stable subgroup. I want to see that the only flat connection in $R$ is the product connection. - -REPLY [5 votes]: I think the statement is that in a principal $SU(2)$-bundle $P$ over a connected manifold $M$ with $H_1(M;\mathbb{Z})=0$, every reducible flat connection has trivial holonomy and therefore trivializes $P$. -For a Lie group $G$, gauge transformations $u$ that stabilize a $G$-connection $A$ are covariant-constant, and therefore determined by their value $u(x)$ at a point $x\in M$. Since $u$ commutes with the parallel transport defined by $A$, it centralizes the holonomy group $H_A\subset G$ (the image of the holonomy representation $\rho\colon \pi_1(M,x)\to G$). This sets up an isomorphism between the stabilizer of $A$ and $C_G(H_A))/Z(G)$, where $C_G(H_A)$ is the centralizer of $H_A$ in $G$ and $Z(G)$ the center ($\pm I$ for $G=SU(2)$). See Donaldson-Kronheimer's book. -The centralizer of a non-central element in $SU(2)$ is a circle $\mathbb{T}$ (to see this, observe that any element is conjugate to a diagonal element, or else consider rotations in $SO(3)$). The centralizer of any non-abelian subgroup of $SU(2)$ lies in the intersection of two different circle-subgroups, and is therefore just the center. -When $H_1(M)=0$, every non-trivial homomorphism $\pi_1(M)\to SU(2)$ has non-abelian image, since the abelian ones factor through $H_1$. For any flat connection $A$ with non-trivial holonomy, the centralizer therefore equals the center, which makes $A$ irreducible.<|endoftext|> -TITLE: References for sign conventions in homological algebra -QUESTION [30 upvotes]: There is no shortage of sign conventions in homological algebra. And once these conventions are set out, there is no shortage of diagrams where an obvious commutative diagram on the underlying abelian groups has to be corrected by an appropriate sign. Many people take a fairly relaxed attitude towards these. -Unfortunately, the signs don't always "just work themselves out". Conrad's book "Grothendieck duality and base change" (see his webpage) gives an extensive list of diagrams, and several signs that are likely opaque for most students of the subject initially. (For example, I find it unlikely that most graduate students would successfully identify the two signs going with canonical isomorphisms $(A[p]) \otimes (B[q]) \cong (A \otimes B)[p+q]$, and the reason they're different.) Poincare duality and the cap product give another offender. -A number of the signs going with these diagrams have conceptual explanations going with them, but most of the references I know generally present sign conventions (and the signs going into the appropriate diagrams) as dropped from the sky. I'd like to know if there is a place in where I can direct students who might want to understand this, rather than having to explain myself. -Do there exist references in the mathematical literature giving a conceptual basis from which these signs might be derived? -(By no means am I claiming that I know most, or all, of the relevant literature.) - -REPLY [14 votes]: There is a short classical paper that has not been mentioned: -Boardman, J. M. -The principle of signs. -Enseignement Math. (2) 12 1966 191–194. -For products in homology and cohomology theories, there is a -systematic explanation of the products on the level of the -symmetric monoidal (homotopy) category of spectra that dictates -the signs. See Section 9 of the following (also on my web page: -http://www.math.uchicago.edu/~may/) -May, J. P. -The additivity of traces in triangulated categories. -Adv. Math. 163 (2001), no. 1, 34–73. -The literature is strewn with mistakes of signs, which can have -serious calculational effects. I once spent months tracking down -a contradiction of signs involving the Steenrod and Dyer-Lashof -operations. The mistake occurred in the unpublished errata of -Steenrod and Epstein, which was meant to correct signs in the -original classic text.<|endoftext|> -TITLE: Double Orthogonal Complement -QUESTION [11 upvotes]: Let $V$ be a complex inner product space. If $W$ is a closed subspace of $V$, we may define $W^\perp$ to be the subspace of all vectors $v \in V$ such that $\langle v | w\rangle =0$ for all $w \in W$. Repeating this construction a second time, we obtain the subspace $W^{\perp \perp}$, the double orthogonal complement of $W$. A moment's thought shows that $W \subseteq W^{\perp\perp}$, and $(W^{\perp\perp})^{\perp\perp}= W^{\perp\perp}$. If $V$ is complete, i.e., a Hilbert space, then $W = W^{\perp\perp}$. Presumably, this equality is not true in general, though I don't know of a counterexample. Thus, the equation $W = W^{\perp\perp}$ defines a class of closed subspaces of $V$. -If $X$ is a set, let $\ell^1(X)$ denote the complex inner product space of absolutely summable functions $X \to \mathbb{C}$ with the usual inner product $\langle g |f\rangle= \sum_{x \in X}\overline{g(x)}f(x)$. If $\mathcal H$ is a Hilbert space, let $\mathcal B^1(\mathcal H )$ denote the complex inner product space of trace class operators on $\mathcal H$ with the usual inner product $\langle b | a \rangle = \mathrm{Tr}(b^*a)$. -Which closed subspaces $W \subseteq \ell^1(X)$ satisfy $W = W^{\perp\perp}$? Which closed subspaces $W \subseteq \mathcal B ^1 (\mathcal H)$ satisfy $W = W^{\perp\perp}$? More formally, is it possible in each case to characterize the class of subspaces $W$ satisfying $W = W^{\perp\perp}$ in another natural or revealing way? -Edit. Here is a counterexample that shows that in general $W \neq W^{\perp\perp}$. Let $\ell^2$ be the usual Hilbert space of square summable sequences, and let $\ell^2_f$ be the subspace of sequences that are eventually zero. Let $s\in \ell^2$ be the sequence defined by $(s)_n = \frac 1 n$. Finally, let $W = \ell^2_f \cap \{s\}^\perp$, a closed subspace of $\ell^2_f$. -For all natural numbers $n$, $W$ contains a sequence whose $n^{th}$ element is $-n$, whose $(n+1)^{st}$ element is $n+1$, and whose other elements are $0$. We deduce that $W^\perp \cap \ell^2_f = \{ 0 \}$, so the double orthogonal complement of $W$ in $\ell^2_f$ is all of $\ell^2_f$. However, $W \neq \ell^2_f$ because $W$ does not contain the sequence $(1,0,0, \ldots)$. Thus $W$ is not equal to its double orthogonal complement in the complex inner product space $\ell^2_f$. - -REPLY [6 votes]: In relation to the question: is it possible to characterize the class of subspaces $W$ satisfying $W=W^{\perp\perp}$ in another natural or revealing way? A theorem of Amemiya and Araki shows that the partially ordered set of such subspaces forms an orthomodular lattice if and only if the surrounding inner product space is a Hilbert space.<|endoftext|> -TITLE: When are unions of isomorphic groups isomorphic? -QUESTION [7 upvotes]: I was thinking about how to prove $\operatorname{Br}(K)\cong H^2(\operatorname{Gal}(\bar{K}/K),\bar{K}^*)$ without having to introduce inductive limits and all the profinite stuff. So, I started wondering if the conditions of a direct system could be weakened for the category of abelian groups in a way that isomorphisms would be still preserved. This brought me to the following general question: -Let $G$ and $H$ be two abelian groups, not necessarily finite, $I$ an index set and $(G_i) _{i\in I}$ and $(H_i)_{i\in I}$ families of subgroups respectevely of $G$ and $H$ such that -(1) $\forall i\in I: G_i \cong H_i$ and -(2) $\bigcup_{i\in I}G_i=G$ and $\bigcup_{i\in I}H_i=H$. -Question 1: Can we conclude that $G\cong H$? -Question 2: If yes, can we drop "abelian"? -EDIT: I forgot to mention that the $G_i$ (and $H_i$) are also assumed to be distinct subgroups. - -REPLY [14 votes]: The answer is no. -For a counterexample, let $G_i=\mathbb{Z}$ be the integers and let $H_i=\frac1i\mathbb{Z}$, for positive natural numbers $i$. The union $\bigcup_i G_i=\mathbb{Z}$, but $\bigcup_i H_i=\mathbb{Q}$. -For the revised question, where you want $G_i$ and $H_i$ distinct, there are still counterexamples, such as $G_i=i\mathbb{Z}$ and $H_i=\frac1i\mathbb{Z}$. - -On a positive note, if you have a bit more coherence in your isomorphisms, then you can make the affirmative conclusion. That is, if we can find particular isomorphisms $\pi_i:G_i\cong H_i$ which agree on their common domains, then they will build together into an isomorphism of $G$ and $H$. That is, what you want is not merely that $G_i\cong H_i$, but rather that the way that $G_i$ sits inside $G$ is the same as the way $H_i$ sits inside $H$. More generally, if $I$ is not just a naked index set, but is a directed set, such that when $i\lt j$ in this order then we have maps $G_i\to G_j$ and $H_i\to G_j$ and the isomorphisms $G_i\cong H_i$ make a commutative system, then the direct limit $G$ of the $G_i$'s will be isomorphic to the direct limit $H$ of the $H_i$'s by universal property arguments.<|endoftext|> -TITLE: Commuting unitaries -QUESTION [17 upvotes]: Is the following true: -For every unit vectors $x_1,..., x_n$, $y_1,..., y_n$ in $\mathbb{C}^k$ -there exist a Hilbert space $H$, unitary operators $U_1,...,U_n$ and $V_1,...,V_n$ in $B(H)$ and unit vectors $x,y \in H$, -such that $[U_i,V_j]=0$ for every $1\leq i,j\leq n$ and -$$\langle x_i, y_j \rangle = \langle U_i V_j x,y\rangle \qquad\text{for every $1\leq i,j\leq n.$}$$ -EDIT: What about the same question, but with $H$ - finite dimensional? - -REPLY [5 votes]: This is not an answer to Kate's question, but a remark that is too long for a comment. I answers (negatively) Tracy's comment. I am sure all this is clear to Kate. -The content of my comment is that you cannot take all the $U_i$'s and $V_j$'s commuting, and this is related to Grothendieck's inequality (see Pisier's very nice recent survey). -Indeed, then the $C^*$-algebra generated by the $U_i$'s and the $V_j$'s would be commutative and thus isomorphic to some $C(X)$ for a compact space $X$, and the linear functional $A\mapsto \langle Ax,y\rangle$ would correspond to a (signed) measure $\mu$ of total mass at most $1$ on $X$. -This would mean that you could write $\langle x_i,y_j\rangle$ as $\int f_i g_j d\mu$ where $f_i$ and $g_j \in C(X)$ have modulus one. Multiplying $g_j$ by the Radon-Nikodym derivative of $\mu$ with respect to $|\mu|$ we could as well assume that $\mu$ be positive. But Grothendieck's inequality says that such a decomposition is in general not possible unless $\sup_i \|f_i\|_2 \sup_j \|g_j\|_2\geq K_G$ where $K_G>1$ is Grothendieck's constant (complex case) and $\|\cdot\|_2$ is the $L^2$ norm with respect to $\mu$. -In fact (see remark 2.12 here), it is even known that, in general, one cannot take diagonal $D_i$'s as in Tracy Hall's comment unless $\sup_i \|D_i\|_2 \geq c n \log n$ for some $c$. Here for a diagonal matrix $D$, by $\|D\|_2$ I mean $(\sum_k |D_k|^2)^{1/2}$.<|endoftext|> -TITLE: Is there a manifold structure on a space of conformal maps? -QUESTION [8 upvotes]: I would be very grateful for any information or pointers for the following: -1) Fix an open subset $U$ of $\mathbb{CP}^1$. a) Does the set of all holomorphic maps from $U$ to $\mathbb{C}$ (with the compact-open topology) have the structure of a manifold in any sense? b) Is there even a notion of a differentiable structure, and what is the tangent space at a typical point (e.g. at the identity)? Does the subset of maps that are conformal on $U$ (i.e. have non-vanishing derivative there) inherit any sensible structure? -2) Is it possible to allow the domain $U$ to vary, e.g. is it possible to consider a collection of all maps from all possible domains (say simply connected ones)? -(I am coming across these maps in the context of conformal loop ensembles (CLEs), which are random families of (countably many, a.s.) loops in $U$, and in order to express certain constructions on these CLEs it appears that one should consider "differentiating" in the space of conformal maps.) -Many thanks! -Update. Maybe some further thoughts: If I fix $U$ to be, say, the open unit disk, then the space of holomorphic maps on $U$ certainly forms a topological vector space. Let's call it $H$. Is this a manifold in any sense (Frechet, I suppose)? Is it smooth (under which notion of differentiability)? -Next, if I restrict to those maps which are conformal on $U$, let's call this $A$, I don't seem to get a vector space; though I think $A$ is a closed subset of $H$ (in the compact-open topology), not being conformal at a point in $U$ is an open condition(?). But what can be said about the topology of $A$? Does $A$ contain a subspace which is an affine space modeled on some space of holomorphic functions? (I.e. "conformal + holomorphic = conformal"?) - -REPLY [2 votes]: A quick comment: I assume you want $U$ to be "non-trivial" i.e. not equal to $\mathbb{C}$ itself; if it were, then the collection of such maps should be infinite dimensional (in particular, it would contain every polynomial). -So assume that $U$ is non-trivial. I'll also assume that $U$ is simply connected, though I'm pretty sure that you can do away with this assumption. Thus $U$ is biholomorphic to the unit disc in $\mathbb{C}$, so we will assume it is the unit disc. -The holomorphic self-maps of the unit disc contain the group $G = PSL_2(\mathbb{R})$ (this is its group of automorphisms, actually). This is a real 3-manifold, so if you restrict yourself to biholomorphisms, you're good. -However, it also contains the maps $z \mapsto z^k$, and so all conjugates of these maps by $G$. There might be something more you can say about this, but I'm not at the moment sure what.<|endoftext|> -TITLE: The different Branches of Arithmetics -QUESTION [8 upvotes]: ... "and then the -different branches of Arithmetic-- -Ambition, Distraction, Uglification, -and Derision." -(Alice in Wonderland, chapter IX: the Mock Turtle's story) - -As a child I wondered for a long time what was the exact meaning of the above partition of Arithmetics quoted in the strange Mock turtle's speech. Today, I like to think it refers to that unpleasant circumstance many of us experienced sometimes: I want to get a certain difficult result (Ambition) but there is a fatal error somewhere in my argument, or computation (Distraction) and when I realize it, the construction miserably falls (Uglification) with a inner feeling of scorn (Derision). A sort of micro - Greek tragedy. Is that what Lewis Carrol really had in mind? It is possible, as he was a mathematician himself, but then did he refer to a precise incident? Did he left any comment on that passage? Has anybody quoted that passage in this meaning later? - -REPLY [2 votes]: "Has anybody quoted that passage in this meaning later?" Does this qualify? -http://jennifergeldard.wordpress.com/2010/03/01/ambition-distraction-uglification-and-derision/<|endoftext|> -TITLE: Is there a "Riemann mapping theorem" for a circle in C^2 ? -QUESTION [5 upvotes]: The Riemann mapping theorem says that if you have a simple closed curve in $\mathbb{C}$, then there is an essentially unique way to map a holomorphic disc to the interior. Is there any reasonable statement to make about holomorphic discs bounding a simple closed curve in $\mathbb{C}^2$? -(At first I thought you should just project to some appropriate choice of coordinate axes and reduce to the one dimensional case, but it is now not at all clear to me why some such choice should exist.) - -REPLY [8 votes]: If you want to generalise the Reimannian mapping theorem to higher dimensions the boundary condition should be different. Namely, it is more reasonable to ask that the boundary of the disk is mapped to a submanifold $M$ of $\mathbb C^n$ or real dimension $n$ such that $TM\oplus J(TM)$ span $\mathbb C^n$ at each point of $M$ (such submanifolds are called totally real). In other words, you can not force the boundary of a holomorphic disk in $\mathbb C^2$ to be an arbitrary curve. The best that you can do is to chose a two-dimensional surface in $\mathbb C^2$ and then the disk will chose a curve on the surface that can be its boundary. Under this condition you can expect to get a finite dimension set of solutions. This is what Gromov is doing in his seminal paper -Pseudo-holomorphic curves in symplectic manifolds -http://www.ihes.fr/~gromov/PDF/9[45].pdf -Example. Let us consider the torus $\mathbb T^2$ in $\mathbb C^2$ given by $|z|=|w|$. Then obviously the line $z=w$ intersects $\mathbb T^2$ in a circle and we get a holomorphic disk with boundary on $\mathbb T^2$. Let us prove that the curves on $\mathbb T^2$ isotopic this circle and bounding a holomorphic disk in $\mathbb C^2$ form a family of real dimension $3$. Indded let $D^2$ be a holomorphic disk in $\mathbb C^2$ with boundary on $\mathbb T^2$ isotopic to the above circle. Consider the projections of $D^2$ to the lines $z=0$ and $w=0$. Since these projections are holomorphic it is not hard to see that they are one to one to maps to disks -$(|w|\le 1, z=0)$ and $(|z|\le 1, w=0)$. So $D^2$ is a graph of a holomorphic one to one map from one unit disk to the other. These maps (as we know well) form $SL(2,\mathbb R)$.<|endoftext|> -TITLE: Are two conjectural descriptions of the motivic t-structure (via cohomology and via affine varieties) known to be equivalent? -QUESTION [12 upvotes]: I know of two conjectural descriptions of the motivic $t$-structure for Voevodsky's motives. -The first one is based on the conjecture that Weil cohomology theories should yield exact and conservative functors on the category of mixed motives. In the paper -Hanamura M. Mixed motives and algebraic cycles, III// Math. Res. Letters 6, 1999, 61--82 -it was proved that this approach yields a t-structure indeed if several (more or less) 'standard' motivic conjectures are fulfilled. Actually, Hanamura proves this for his own triangulated category of motives, but his category is isomorphic to the Voevodsky's one (by a result of my own:)). -The second approach was proposed by Voevodsky himeslf (in his well-known letter to Beilinson); it is based on the idea that the 'mixed motivic cohomology' of a (smooth) affine variety should satisfy the Artin's vanishing theorem. -My question is: are these two approaches known to be equivalent (if certain 'standard' conjectures are fulfilled)? Or (equivalently) did anybody prove that Voevodsky's approach 'should work'? -Also, does there exist a reasonable extension of Voevodsky's approach to relative motives (over a base scheme $S$)? -P.S. After asking this, I recollected the paper -Beilinson A., Remarks on $n$-motives and -correspondences at the generic point, in: Motives, polylogarithms -and Hodge theory, part I, Irvine, CA, 1998, Int. Press Lect. Ser., -3, I, Int. Press, Somerville, MA, 2002, 35-46. -Yet I am not sure that it answers my question completely. - -REPLY [7 votes]: Okay, I'm not familiar with Beilinson's paper, but here's my take on this. First let's recall the two definitions. I will denote the triangulated category of motives over a field $k$ by $DM(k)$ (for any of the equivalent definitions that are available); I am taking $\mathbb{Q}$-coefficients and looking only at compact objects, although I'm not sure the last is necessary. Also, I am not good with homological notation, so I will use cohomological notation all along, beware ! For example, for me $X[1]$ will mean "$X$ put in degree $-1$" and the motive of $\mathbb{G}_m$ will be $\mathbb{Q}\oplus\mathbb{Q}(-1)[-1]$ (where $\mathbb{Q}$ is the unit for the tensor product, i.e., the motive of $Spec(k)$). Sorry, but I'm too afraid to make a mistake if I try to translate. -So, first here is Hanamura's definition of the $t$-structure. He assumes that all the Grothendieck standard conjectures, Murre's conjecture and the vanishing conjectures are true, and this implies in particular that any realization functor $H:DM(k)\longrightarrow D^b(F)$ (where $F$ is an appropriate field of coefficients) is faithful. He defines a $t$-structure, say $({}^H D^{\leq 0},{}^H D^{\geq 0})$ on $DM(k)$ by taking the inverse image by $H$ of the usual $t$-structure on $D^b(F)$. Of course, you have to prove that it is indeed a $t$-structure (and independent of the realization functor), and he does this. He calls the heart the category of mixed motives over $k$, say $MM(k)$. If $X$ is a variety over $k$, we can associate to it a (cohomological) motive $M(X)$ in $DM(k)$, and I will denote by $H^k_M(X)\in MM(X)$ the cohomology objects of $M(X)$ for Hanamura's $t$-structure. Hanamura also proves that every mixed motive has a weight filtration, that is a filtration whose graded parts are pure motives, and he proves that pure motives are semi-simple and that all irreducible pure motives are direct factors of motives of the form $H_M^k(X)(a)$, where $X$ is a smooth projective variety. -Now to Voedvodsky's definition. I have tried to understand it, then rewrite it in cohomological notation, so directions of maps and shifts may have changed, but I think I still got the spirit of it. What he does is something like this : Define a full subcategory ${}^VD^{\geq 0}$ of $DM(X)$ by the condition that an object $M$ is in it if and only if, for every affine scheme $f:U\rightarrow Spec(k)$ that is purely of dimension $n$, for every $m>n$ and every $a\geq 0$, $Hom_{DM(k)}(f_*\mathbb{Q}_U(a)[m],M)=0$. I would like to make a few remarks. First, what I denoted by $f_*\mathbb{Q}_U$ is just my $M(U)$ of the preceding paragraph, but I wrote it like this because it will make the generalization to a general base scheme $S$ more clear; my notation makes sense if I allow myself to remember that we now have categories of motives over a very general base and the 6 operations on them (and if I say that $\mathbb{Q}_U$ is the unit motive in the category of motives over $U$). Second remark, I added twists whereas Voedvodsky's definition doesn't have any. The reason I did this is because Voedvodsky makes a definition only for effective motives, and I didn't see how to make ${}^VD^{\geq 0}$ stable by $(1)$ unless I added it in the definition (but maybe it is not necessary). Third remark, remember, cohomological notation, and for me passing from effective to general motives means inverting $\mathbb{Q}(-1)$, not $\mathbb{Q}(1)$ (in my world, $\mathbb{Q}(-1)$ is effective). -Ah, yes, and then Voedvodsky defines ${}^VD^{\leq 0}$ as the left orthogonal of ${}^V D^{\geq 1}:={}^VD^{\geq 0}[-1]$. -Anyway, what is the motivation for Voedvodsky's definition ? Here are a few principles. First, motivic $t$-structure on motives over a base $S$ should correspond (by the realization functors) to the (selfdual) perverse $t$-structure on complexes of sheaves over $S$. Second, if $f$ is an affine map of schemes, then ${f_*}$ is right $t$-exact for the perverse $t$-structures. Third, for any scheme $U$, the constant sheaves over $U$ are concentrated in perverse cohomology degree $\leq dim(U)$. So, if I come back to my situation above : $f:U\longrightarrow Spec(k)$ is an affine variety over $k$, purely of dimension $n$, $a\geq 0$, $m>n$, then $\mathbb{Q}_U(a)[m]$ should be concentrated in degree $<0$ for the motivic $t$-structure on the category of motives over $U$, and so $f_*\mathbb{Q}_U(a)[m]$ should be concentrated in degree $<0$ for the motivic $t$-structure on $DM(k)$, and it should be left orthogonal to elements that are concentrated in degree $\geq 0$. What Voedvodsky says is that this is enough to characterize the elements concentrated in degree $\geq 0$. -From this, the natural generalization of Voedvodsky's definition to a general base scheme $S$ is obvious : replace affine schemes $U\longrightarrow Spec(k)$ by affine maps $U\longrightarrow S$ (or maps $U\longrightarrow S$ such that $U$ is affine, I don't think it will make a difference). -So, are the two $t$-structures the same ? I think so. A first obvious observation is that ${}^HD^{\geq 0}\subset{}^VD^{\geq 0}$, that is, every object of ${}^HD^{\geq 0}$ is right orthogonal to motives ${{f_*}\mathbb{Q}(a)[m]}$ as above. This follows from the faithfulness of the realization functor and the fact that this would be true in the usual categories of sheaves (see the remarks above). We also know that ${}^HD^{\geq 0}$ is the right orthogonal of ${}^HD^{\leq -1}$, by the definition of a $t$-structure. So, what we have to see is that ${}^VD^{\geq 0}$ is right orthogonal to ${}^HD^{\leq -1}$, that is, that a motive that is right orthogonal to every $f_*\mathbb{Q}_U(a)[m]$ as above is right orthogonal to the whole ${}^HD^{\leq -1}$. -Here is one way to do this : Let $C$ be the smallest full additive subcategory of ${}^HD^{\leq -1}$ that is stable by isomorphism, direct summand, extension and contains all the objects of the form $f_*\mathbb{Q}(a)[m]$ as above. It is enough to show that $C={}^HD^{\leq -1}$. Noting that ${}^HD^{\leq -1}$ is generated (in the same way : direct sumands, isomorphisms, extensions) by objects of the form $H^k_M(X)(b)[l]$, for $X$ a smooth projective variety, $l>k$ and $b\in\mathbb{Z}$, I think that this is an easy exercise, playing with hyperplane sections of smooth projective varieties. (I had a bit a trouble with the fact that $C$ is stable by Tate twists. We know that $M(U)(-1)$ is a direct factor of $M(U\times\mathbb{G}_m)[1]$, so stability by $(-1)$ is not a problem. But I couldn't show stability by $(1)$ unless I put it in the definition.) -edit: corrected dollar sign<|endoftext|> -TITLE: The continuity of Injectivity radius -QUESTION [16 upvotes]: Dear all, -when reading a book of M. Berger, I learned that the injectivity radius Inj(x) on a compact Riemannian manifold depends continuously on the point x. -When the manifold is complete and non-compact, Inj may not be continuous. -For example, Inj(x) decreases to zero when x moves to the most curved point on a paraboloid. However, it could be infinity at that point. -My question is, can we prove the continuity of Inj on a non-compact manifold under some conditions? -(I think that the weakest condition is to assume the finiteness of Inj.) -ps. I must admit that I don't know how to prove the continuity of Inj even on a compact manifold. I think that the argument should involve the stability of ODEs (the geodesic equation and Jacobi equation). If one of you have a reference about this, could you please tell me? thanks a lot! - -REPLY [12 votes]: The compactness is irrelevant; i.e if it is true for compact manifolds then the same is true for complete ones. (The same proof as in comact case works, but it is easier to do this way.) -If $R<\mathrm{InjRad}_p$ then one can construct a smooth metric on a sphere with an isometric copy of $B_R(p)$ inside. -If there is a sequence of points $x_n\to x$ such that -$$\lim\ \mathrm{InjRad}_{x_n}< \mathrm{InjRad}_x,$$ -apply above consruction for $R$ slightly smaller than $\mathrm{InjRad}_x$. -You get a compact manifold with non-continuous InjRad. -If there is a sequence of points $x_n\to x$ such that -$$\lim\ \mathrm{InjRad}_{x_n}> \mathrm{InjRad}_x$$ then apply above construction for $p=x_n$ for large enough $n$ and $R>\mathrm{InjRad}_x$. -That leads to a contradiction again.<|endoftext|> -TITLE: power series of the reciprocal... does a recursive formula exist for the coefficients -QUESTION [26 upvotes]: Let $f(x)=\sum _{n=0}^{\infty } b_nx^n$ and $\frac{1}{f(x)}=\sum _{n=0}^{\infty } d_nx^n$. Then the coefficients of the reciprocal of $f(x)$ can be written down. The first few terms are: -$d_0 = \frac{1}{b_0}$, -$d_1 = -\frac{b_1}{b_0^2}$, -$d_2 = \frac{b_1^2-b_0 b_2}{b_0^3}$ -$d_3 = -\frac{b_1^3-2 b_0 b_1 b_2+b_0^2 b_3}{b_0^4}$ -... -I was wondering if there is a general recursive (preferably not of course) formula for the coefficients of the reciprocal. If an arbitrary $n$ is given, can I write down a formula for $d_n$ (recursive or not)? -Regards -//edit: as the comments below suggest I think people are misinterpretating the question. I am not looking for someone to show me how to solve a system of linear equations by substitution... I want a formula for d_n, Since posting the question, I found such a formula for $d_n$ at http://functions.wolfram.com/GeneralIdentities/7/, see the section on Ratios of the direct function ... if anyone knows of how this formula is derived or any other references to it or similar formulas please let me know... thanks - -REPLY [20 votes]: Without loss of generality we can take $b_0$ to be 1, since -\begin{equation*}\sum_{n=0}^\infty b_n x^n = b_0\biggl( 1+\sum_{n=1}^\infty (b_n/b_0)x^n\biggr). -\end{equation*} -Then for $b_0=1$ we have -\begin{equation*} -\frac1{f(x)} = \biggl( 1+\sum_{n=1}^\infty b_n x^n\biggr)^{-1}\\ -=\sum_{m=0}^\infty (-1)^m\biggl( \sum_{n=1}^\infty b_n x^n\biggr)^m. -\end{equation*} -Expanding by the multinomial theorem and extracting the coefficient of $x^n$ gives -\begin{equation*} -\frac1{f(x)} = \sum_{n=0}^\infty \kern 3pt x^n \kern -5pt -\sum_{m_1+2m_2+3m_3+\cdots = n} (-1)^{m_1+m_2+\cdots} \binom{m_1+m_2+\cdots}{m_1, m_2, \ldots} b_1^{m_1} -b_2^{m_2}\cdots.\end{equation*}<|endoftext|> -TITLE: Groups with large negative deficiency -QUESTION [6 upvotes]: Hello, -I've just learnt the notion of deficiency of a group but I don't know how work with it. -I want to construct a group with large negative deficiency ; naively I think that $(Z_2)^n$ will work because we need $n$ generators and $n$ relations for the square of the elements being $1$, plus $n(n-1)/2$ relations of commutation ; but maybe we can do better ... -Another problem is to control the deficiency of a direct product. Still naively we take generators and relations in both groups and add relations of commutation but it certainly does not work. -So my two questions : - -Does there exist finite groups with arbitrarily large negative deficiency ? -If I fix a finitely presentable group G and consider the product of G with groups of arbitrarily large negative deficiency, can I obtain products with arbitraly large negative deficiency ? - -Thank you, -mister_jones - -REPLY [4 votes]: There even exist 2-generator finite groups with arbitrarily large negative deficiency. -Let $G$ be the standard wreath product of a cyclic group of order 2 with a cyclic group of order $n$. I don't know the exact negative deficiency of $G$, but I can show that it is at -least $\lfloor n/2 \rfloor$. -It is a standard result that the negative deficiency of a finite group is at least equal to the rank of its Schur Multiplier $M(G) = H_2(G)$. I can show that this is at least $\lfloor n/2 \rfloor$ by constructing a central extension $E$ of $G$ by an elementary abelian 2-group $Z$ of rank $r := \lfloor n/2 \rfloor$ with $Z \le E'$. -In outline, to define $E$, let $y_1,\ldots, y_n$ be generators of the base group $B$ of $G$, let $Z$ have generators $z_1,\ldots, z_r$, and in $E$ we put $y_i^2=1$ for all $i$ and $[y_i,y_j] = z_r$, where $r$ is the distance of $i$ from $j$ when the integers $1,\ldots,n$ are arranged around a circle each at distance 1 from the next. Then the action of the cycle in $G$ of order $n$ on $B$ induces the trivial action on $Z$, so we can define the required extension $E$.<|endoftext|> -TITLE: Rolling without slipping interpretation of torsion -QUESTION [26 upvotes]: This is, in a sense, a follow up to this question. -Hehl and Obukhov try to give an intuitive description of torsion. I am confused about their description. I am looking at the following paragraph on page 5: - -How can a local observer at a point $p$ with coordinates $x^i$ tell whether his or her space carries torsion and/or curvature? The local observer defines a small loop (or a circuit) originating from $p$ and leading back to $p$. Then he/she rolls the local reference space without sliding ... As a computation shows, the added up translation is a measure for the torsion and the rotation for the curvature. - -So, my input data is a manifold $M$ with a connection $\nabla$ on $T_* M$, and a path $\gamma : [0,1] \to M$. From this data, I am supposed to obtain an affine linear map $v \mapsto Av+b$ from $T_{\gamma(0)} M$ to $T_{\gamma(1)}(M)$. In particular, if $\gamma(0)=\gamma(1)=p$, I obtain an endomorphism of $T_p M$. Then $b$ is a measure of the torsion, and $A$ is a measure of the curvature. -I am happy with the curvature part. This is parallel transport: Given a tangent vector $u \in T_p M$, I am to find the unique local section $\sigma$ of $\gamma^* T_p(M)$ such that $\sigma(0) = u$ and $\nabla \sigma=0$. Then $Au = \sigma(1)$. -I have two confusions about the torsion part: -(1) Hehl and Obukhov cite their definition of rolling without slipping to five sources (hidden by the ellipses above). The most readable of the ones I have access to is Differential Geometry, by Sharpe. But Sharpe (at least in Appendix B) only gives a definition for the Levi-Civita connection, not for an arbitrary connection. I think I have guessed what the definition should be, but could someone please write it down so I can be sure? -(2) In any case, it is my understanding that the Levi-Civita connection should be torsion-free and that, for the Levi-Civita connection, rolling without slipping corresponds to physically rolling my manifold along a plane. So the statement should be, if I take a surface $S$ in $\mathbb{R}^3$, and physically roll it along a table top, when I get back to the same tangency point on $S$, I should be at the same point on the table. Physical experimentation has not made it clear to me whether or not this is true. However, if I roll my surface along a non-small path, this is definitely false: otherwise, ball bearings could not roll! -Technically, this is not a contradiction, because Hehl and Obukhov only speak of a small circuit. But the usual situation in differential geometry is that when some quantity vanishes everywhere on a manifold, then "small" can be replaced by "contractible". And the example of a rolling ball bearing definitely shows that rolling-without-slipping a plane along a contractible loop according to the LC connection can produce a nontrivial translation. -What's going on? - -REPLY [2 votes]: This was meant to be a comment over at Secret Blogging but was probably classified as spam (not that I disagree with that), so I stick it here instead. -An example to consider is $S^3$. The metric from the ordinary embedding into $E^3$ gives a constant curvature connection with no torsion. The geodesics are all great circles and splits into equivalent classes of parallel geodesics (the Hopf fibration). An observer travelling along a geodesic will observe how nearby geodesics twist around him. This is a higher dimensional analogue of how nearby geodesics in two dimensions are observed to do sinusidal oscillations when the curvature is positive. -The curvature form is an so(3)-valued two-form which integrated around (the interior of) a closed loop gives the rotation of a frame transported around the loop. Now, given an element of so(3) at a point of the manifold, it can be reinterpreted as an vector in the tangent space. (This is the usual so(3) <-> angular velocity vector isomorphism.) This turns the curvature form into a torsion form, giving a new connection with no curvature but with "constant" (homogeneous) torsion. Integrating the torsion form gives a tangent vector which is the translation of the tangent space when translated around a (not necessarily small) loop. -This absolute parallelism connection has the same geodesics as the constant curvature connection. But in this case, an observer travelling along a geodesic, I believe, does not observe any twisting of nearby geodesics. Instead he sees the nearby geodesics to be completely straight lines, but they lag behind (or run ahead?). In some way a torsion connection introduces an ambiguity in the velocity concept which would be interpreted by an observer that the immediate neighbourhood does not stay in place, it slips? -Now, bundles. Take the tangent bundle $TM$ of some manifold and release each tangent space from its point of contact with the manifold. This turns tangent spaces into affine spaces and the tangent bundle into an affine bundle $AM$. This bundle does not have a distinguished zero section, instead each and every point of an affine fiber have equally right to be considered a point of the underlying manifold. Then, giving a connection on $M$, it should the case that the connection has vanishing torsion if and only if every contractible closed loop in $M$ lifts to a closed loop in $AM$? -The connection in $R^3$ with straight lines as geodesics and when a frame is transported along a line, it spins, is given by $\nabla_XY=\nabla_YX=Z$, etc. This is again a constant curvature connection with no torsion.<|endoftext|> -TITLE: Recent results on the Gauss circle problem? -QUESTION [7 upvotes]: Possible Duplicate: -What is the status of the Gauss Circle Problem? - -The Gauss circle problem is the following: Let $N(r)$ denote the number of solutions in integer pairs $(i,j)$ to the inequality $i^2 + j^2 \leq r^2$ (namely, the number of lattice points on or inside the disc centered at 0 with radius $r$). Then it is easy to see we should have $N(r) = \pi r^2 + E(r)$ for some error term $E(r)$, where the key is to estimate $E(r)$. Gauss proved that $E(r) \leq 2\sqrt{2}\pi r$, and Landau showed that $E(r) \ne o(r^{1/2}\log^{1/4}(r))$. The conjecture is that $E(r) = O(r^{1/2 + \epsilon})$ for any $\epsilon > 0$. If the conjecture is true, then the squares will provide an explicit example of a subset of positive integers $A$ such that that the representation function $r_A(n)$ defined to be the number of ways of writing $n$ as a sum of two elements of $A$ satisfies $\displaystyle \sum_{j \leq n} r_A(j) = cn + O(n^{1/4}\log(n))$. Such sets $A$ exist by a result of I. Ruzsa in 1999, but no known examples exist. -So my question is, what is the best known result on the Gauss circle problem? Or in lieu of that, a good explanation on why this problem is so difficult? - -REPLY [8 votes]: See A. Ivic, E. Kratzel, M. Kuhleitner, W. G. Nowak, Lattice points in large regions and related arithmetic functions: recent developments in a very classic topic, Elementare und analytische Zahlentheorie, 89–128, Schr. Wiss. Ges. Johann Wolfgang Goethe Univ. Frankfurt am Main, 20, Franz Steiner Verlag Stuttgart, Stuttgart, 2006, MR2310176 (2008b:11105). -The review, written by Wolfgang Schwarz, begins, "This paper is an up-to-date (2004) and most interesting survey paper (with an emphasis on results established during the last few years) on estimates of the number of lattice points in large regions, with nearly 200 references. -"Section 1 deals with the Gauss' circle problem (Hardy's identity, O-estimates, lower bounds, mean-square estimates)." -Some other recent papers that might be of interest: -Martine Babillot, Points entiers et groupes discrets: de l'analyse aux systemes dynamiques, Panor. Syntheses, 13, Rigidite, groupe fondamental et dynamique, 1–119, Soc. Math. France, Paris, 2002. MR1993148 (2004i:37057) -Fernando Chamizo, Antonio Cordoba, Lattice points, Margarita mathematica, 59–76, Univ. La Rioja, Logroño, 2001. MR1882616 (2003j:11118) -M. N. Huxley, Integer points in plane regions and exponential sums, Number theory, 157–166, -Trends Math., Birkhäuser, Basel, 2000. MR1764801 (2002i:11101) -Wenguang Zhai, On higher-power moments of Δ(x). III, Acta Arith. 118 (2005), no. 3, 263–281. MR2168766 (2006f:11121)<|endoftext|> -TITLE: Spaces with same homotopy and homology groups that are not homotopy equivalent? -QUESTION [48 upvotes]: A common caution about Whitehead's theorem is that you need the map between the spaces; it's easy to give examples of spaces with isomorphic homotopy groups that are not homotopy equivalent. (See Are there two non-homotopy equivalent spaces with equal homotopy groups?). It's surely also true that the pair (homotopy groups, homology groups) is not a complete invariant, but can anyone give examples? That is, I'm looking for spaces $X$ and $Y$ so that $\pi_n(X) \simeq \pi_n(Y)$ and $H_n(X;\mathbb{Z}) \simeq H_n(Y; \mathbb{Z})$ but $X$ and $Y$ are still not (weakly) homotopy equivalent. -(Easier examples are preferred, of course.) - -REPLY [37 votes]: Following up on John's comment, one can consider $S^2$-fibrations over $S^2$. There are two of them since such fibrations are classified by $\pi_1(\textrm{Diff}^{+}(S^2))=\mathbb{Z}_2$. One of them is $S^2\times S^2$ while the other can be shown to be the connected sum of $\mathbb{CP}^2$ and $\overline{\mathbb{CP}}^2$. These two spaces have the same homology. They have the same homotopy groups since they both form the base of a $S^1$-fibration with total space $S^2 \times S^3$. However, the intersection forms are not equivalent and hence they are not homotopy equivalent. - -REPLY [14 votes]: Take a finite group $G$, a finite-dimensional $\mathbb{Q}$-vector space $V$ and two representations actions of $G$ that are inequivalent but the spaces of coinvariants both have the same dimension, say zero. -The first concrete example that comes to my mind is $G=Z/4$, $V=Q[i]$, with the two actions where the generator acts by $-1$ or $i$. These two actions are not even equivalent under outer automorphisms of $G$. -Let $n \geq 3$ be an odd integer and consider the Eilenberg-Mac-Lane space $K(V,n)$, which inherits two $G$-actions. The two Borel-constructions $EG \times_G K(V,n)$ have the same homotopy groups. But the $n$th homotopy groups are not isomorphic when considered as a $\pi_1$-module; so these two spaces are not homotopy equivalent. -The homology can be computed from the Leray-Serre spectral sequence of the fibration $EG \times_G K(V,n) \to BG$. Recall $E^{2}_{pq}= H_p (G; H_p (K(V;n)))$. -To begin with, $\tilde{H}_* (K(V,n), \mathbb{Z}) =V$ if $*=n$ and $0$ otherwise. Thus $E^{2}_{pq}=0$ unless $q=0$ (then it is the group homology $H_p(G;Z)$ or $(p,q)=(0,n)$, in which case it is $V_G=0$. Thus the projections $EG \times K(V,n) \to BG$ are homology equivalences. -Finally note that the Eilenberg Mac Lane spaces can be realized as abelian topological groups and $G$ acts fixing the basepoint. Thus the maps $EG \times_G K(V;n)) \to BG$ have sections, which are homology equivalences as well. So the construction even produces a homology equivalence between two spaces with abstractly isomorphic homotopy groups. -EDIT: since there is a homology equivalence between these two spaces, it follows that the homology with coefficients, the cohomology rings and even the actions of the Steenrod algebra for all primes are isomorphic. - -REPLY [9 votes]: A fairly easy obstruction is the action of the fundamental group on the homotopy groups. Here is an example where the homotopy groups are isomorphic as abstract groups, but not preserving the action of the fundamental group. -Form a fibration over a circle with fiber a wedge of spheres $S^2\vee S^2$. That is, take the wedge, cross with an interval, and identify the ends by a homotopy equivalence of the wedge. Homotopy equivalences are parameterized by $GL_2(\mathbb Z)$, the action on homology. The homotopy type of the space remembers the monodromy as the action of the fundamental group on the homotopy groups. The homotopy groups are that of the universal cover, which does not depend on the choice of monodromy. Compute the homology by the Serre spectral sequence. This involves the homology of $\mathbb Z$ acting on $\mathbb Z^2$ by the monodromy. If the monodromy is hyperbolic, the homology vanishes and the space has the homology of the circle. Thus two different hyperbolic matrices give spaces with isomorphic homotopy and homology groups. -In a completely different direction, there are examples of pairs of simply connected spaces such that the Postnikov truncations are equivalent, but the inverse limits of their Postnikov towers are not, but I think such examples have to be pretty large.<|endoftext|> -TITLE: Is there a connection between exceptional Galois groups and Ramanujan's partition congruences -QUESTION [5 upvotes]: There are three exceptional Galois groups $L_2(5)$, $L_2(7)$ and $L_2(11)$ . These are cited as one of Arnold's "trinities" and are connected with other trinities and the McKay Correspondence. -Ramanujan studied partition numbers and found congruence relations modulo powers of 5, 7 and 11. the recent dramatic breakthrough by Ken Ono throws some light on the reasons behind these congruences. -My question is whether there is any known connection between these two instances of the three primes 5, 7 and 11 appearing in these two places. I realise that these are just small numbers so it is not a great coincidence, but partitions are connected to other areas of mathematics so I wondered if some correspondence was known. - -REPLY [15 votes]: No. -(Moderator's note: I am adding the explanation of the above answer, as provided in comments given by the poster whose account was deleted long ago, to make this a proper answer.) -Philosophically, it's a little hard to prove that something is not a coincidence, so you should take my answer to mean that I understand both and that there is no relation. Perhaps one way to think of it is as follows. The congruences proved by Ramanujan for $5$, $7$, and $11$ are very similar to each other, more so than the exceptional isomorphisms $A_5 \simeq L_2(5)$ and $L_3(2) \simeq L_2(7)$. -More generally, there is a unified reason why congruences for the partition function exist, they arise by projection onto some finite dimensional space of ordinary $p$-adic modular forms of weight $−1/2$. This space has dimension which grows roughly linear in $p$. For $p < 12$ this space is zero dimensional, hence Ramanujan's congruences. For $p$ in some further range, this space is one dimensional, explaining the congruences of Atkin. For larger $p$, this space is still finite dimensional but consists of eigenforms for the Hecke operators with different eigenvalues, so the implied congruences are not so transparent. (Nick Ramsey is writing an appendix to Folsom-Kent-Ono which will explain some of this.) -On the other hand, the exceptional isomorphisms are, well, exceptional. They differ in flavor from each other, and there is nothing that persists of this nature for larger $p$. As for what happens to $p=2$ and $p=3$, they are absent from the "exceptional" list because the groups are solvable, whereas in the case of partitions they are absent because the generating function for the partitions is $q^{1/24} \eta^{−1}$, where $\eta = q^{1/24}\prod_{n=1}^\infty (1−q^n)$ is modular form. That means one is really studying the function given by -$$1/\eta(24\tau) = \sum a_m q^m,$$ -where $a_m = 0$ unless $m = −1+24n$, in which case $a_m$ is the number of partitions $p(n)$ of $n$. The methods above for $p=2,3$ give congruences for the coefficients of this series when $m$ is divisible by $2$ or $3$. Since $a_m=0$ in these cases, one does not deduce anything about congruences for the partition function $p(n)$. Putting this together, it seems that the connection between the fact that Ramanujan's congruences hold for the same set of primes as those for which $L_2(p)$ is simple is the law of small numbers.<|endoftext|> -TITLE: Are there oriented $4k+2$ manifolds such that $im(H_{2k+1}(M; Z/2)\to H_{2k+1}(M, \partial M; Z/2))$ has odd dimension? -QUESTION [7 upvotes]: The following fairly specific question comes up in a bordism computation I'm trying to do: -Are there compact $\mathbb Z$-oriented $4k+2$ dimensional manifolds with boundary $M$ such that $im(H_{2k+1}(M; \mathbb Z/2)\to H_{2k+1}(M, \partial M; \mathbb Z/2))$ has odd dimension as a $\mathbb Z/2$ vector space? -Clearly the answer is no if $k=0$. Also, I can show that this can't happen if $\partial M=\emptyset$ using a combination of Poincare duality and the universal coefficient theorem. But I haven't been able to rule out the possibility if the boundary is non-empty, or to construct examples. -Thanks. - -REPLY [7 votes]: Martin O's answer is very nice. So in an oriented $2n$-manifold with $n$ odd the mod $2$ self-intersection of any $n$-dimensional mod $2$ homology class is $0$. -Looking for a more geometric explanation of that, or anyway an explanation with no Steenrod operations in sight, I came up with the following (which is also related to John Klein's comment): -Let's assume that the given class is represented by an immersed $n$-manifold $M$. The mod $2$ self-intersection number is then the evaluation on the mod $2$ fundamental class of $M$ of the mod $2$ Euler class of the normal bundle of the immersion. So it comes down to the following: -Claim: Let $n$ be odd and suppose that $M$ is a closed $n$-dimensional manifold and $E$ is a rank $n$ vector bundle such that the total space of $E$, considered as a $2n$-manifold, is orientable. Then the mod $2$ Euler class of $E$ is $0\in H^n(M;\mathbb Z/2)$. -Proof: A rank $n$ vector bundle has a twisted integral Euler class, which belongs to $H^n(M;\Gamma)$, where $\Gamma$ is the coefficient system (locally isomorphic to $\mathbb Z$) associated with $w_1(E)$, the obstruction to orientability of $E$. The mod $2$ Euler class is the mod $2$ reduction of this, so it suffices if this twisted integral class is $0$. The (twisted) integral Euler class of a vector bundle of odd rank is always killed by $2$ (this is a standard fact in the oriented case, and it seems clear in the twisted case, too), so it suffices if the group $H^n(M;\Gamma)$ is torsion-free. But by Poincare duality it is isomorphic to $H_0(M;\mathbb Z)$, since $E$ and the tangent bundle of $M$ have the same orientability obstruction by hypothesis.<|endoftext|> -TITLE: Magic square on an infinite lattice -QUESTION [9 upvotes]: This question came to me while reading the discussion of magic square in the complex plane with equal integrals along every horizontal, vertical and diagonal "magic square in the complex plane with equal integrals along every horizontal, vertical and diagonal" -That question ran into trouble when people pointed out the difficulties involved in just making sense of the question, e.g., what kind of function, what kind of integral, etc. So I made up something halfway between the finite-discrete magic square (that is, the usual kind) and the continuous analogue raised in Question 53352: -What can you say about complex numbers $a_{i,j}$ such that for all $i$ and $j$ the sums $$\sum_{i=-\infty}^{\infty}a_{i,j}=\sum_{j=-\infty}^{\infty}a_{i,j}\lt\infty$$ That is, the sums along vertical and horizontal lines all converge and are all equal. -If the doubly-infinite sums are a worry, just insist that all the numbers be non-negative reals (alternatively, replace the lower limits of summation with zero). -Splitting into real and imaginary parts, we see that we may assume the numbers are real. Other than that, I haven't done much thinking about it. -I realize the number theory tag is not quite appropriate; I'm trying to comply with the direction, "Please try use at least one tag corresponding to an arXiv subject area." - -REPLY [9 votes]: Here are two elementary results about those magic square arrays. -Claim: The vector subspace of collections with finitely many nonzero entries has abasis consisting of the magic squares with $4$ nonzero entries arranged as -$$\begin{array}{cc}-1 & 1 \\\1 & -1\end{array}$$. -Proof: You can always add a multiple of such a magic square to eliminate the rightmost nonzero entry in the bottom row without extending the nonzero entries to the top or left. -Claim: Consider arrays with all indices positive. Any arbitrarily chosen convergent first row and first column with the same sums can be extended to a magic square. -Proof: This can be done so that the rest of the array is $0$ except for the diagonal and one off-diagonal. For $i=2,3,4, ...$ choose $a_{i,i}$ to make the $i$th column have the required sum. Then choose $a_{i+1,i}$ to make the $i$th row have the required sum. -I think infinite arrays are a little too flexible. You can specify large portions of the array arbitrarily, such as the diagonal and everything below it (subject to convergence), and you can complete the array to a magic square.<|endoftext|> -TITLE: Does any research mathematics involve solving functional equations? -QUESTION [53 upvotes]: This is a somewhat frivolous question, so I won't mind if it gets closed. One of the categories of Olympiad-style problems (e.g. at the IMO) is solving various functional equations, such as those given in this handout. While I can see the pedagogical value in doing a few of these problems, I never saw the point in practicing this particular type of problem much, and now that I'm a little older and wiser I still don't see anywhere that problems of this type appear in a major way in modern mathematics. -(There are a few notable exceptions, such as the functional equation defining modular forms, but the generic functional equation problem has much less structure than a group acting via a cocycle. I am talking about a contrived problem like finding all functions $f : \mathbb{R} \to \mathbb{R}$ satisfying -$$f(x f(x) + f(y)) = y + f(x)^2.$$ -When would this condition ever appear in "real life"?!) -Is this impression accurate, or are there branches of mathematics where these kinds of problems actually appear? (I would be particularly interested if the condition, like the one above, involves function composition in a nontrivial way.) - -Edit: Thank you everyone for all of your answers. As darij correctly points out in the comments, I haven't phrased the question specifically enough. I am aware that there is a lot of interesting mathematics that can be phrased as solving certain nice functional equations; the functional equations I wanted to ask about are specifically the really contrived ones like the one above. The implicit question being: "relative to other types of Olympiad problems, would it have been worth it to spend a lot of time solving functional equations?" - -REPLY [42 votes]: In additive combinatorics, one often seeks to count patterns such as an arithmetic progression $a, a+r, \ldots, a+(k-1)r$. When doing so, one is naturally led to expressions such as -$$ {\bf E}_{a,r \in G} f_0(a) f_1(a+r) \ldots f_{k-1}(a+(k-1)r)$$ -for some finite abelian group $G$ and some complex-valued functions $f_0,\ldots,f_{k-1}$. If these functions are bounded in magnitude by $1$, then the above expression is also bounded in magnitude by one. When does equality hold? Precisely when one has a functional equation -$$ f_0(a) f_1(a+r) \ldots f_{k-1}(a+(k-1)r) = c$$ -for some constant $c$ of magnitude $1$. One can solve this functional equation, and discover that each $f_j$ must take the form $f_j(a) = e^{2\pi i P_j(a)}$ for some polynomial $P_j: G \to {\bf R}/{\bf Z}$ of degree at most $k-2$. This observation can be viewed as the starting point for the study of Gowers uniformity norms, and one can perform a similar analysis to start understanding many other patterns in additive combinatorics. -In ergodic theory, cocycle equations, of which the coboundary equation -$$ \rho(x) = F(T(x)) - F(x)$$ -is the simplest example, play an important role in the study of extensions of dynamical systems and their cohomology. Despite the apparently algebraic nature of such equations, though, one often solves these equations instead by analytic means (and in particular, not by IMO techniques), for instance by using the spectral theory or mixing properties of the shift $T$, and exploiting the measurability or regularity properties of $\rho$ or $F$. (The solving of such equations, incidentally, is a crucial aspect of the ergodic theory analogue of the study of the Gowers uniformity norms mentioned earlier, as developed by Host-Kra and Ziegler.) -Returning to the more "contrived" functional equations of Olympiad type, note that such equations usually use (a) the additive structure of the domain and range, (b) the multiplicative structure of the domain and range, and (c) the fact that the domain and range are identical (so that one can perform compositions such as $f(f(x))$). In most mathematical subjects, at least one of these features is absent or irrelevant, which helps explain why such equations are relatively rare in research mathematics. For instance, in many branches of analysis, the range of functions (typically ${\bf R}$ or ${\bf C}$) usually has no natural reason to be identified with the domain of functions (which may ``accidentally'' be ${\bf R}$ or ${\bf C}$, but is often more naturally viewed in a more general category, such as that of measure spaces, topological spaces, or manifolds), so (c) is usually absent. Conversely, in dynamics, (c) is prominent, but (a) and (b) are not. The only fields that come to my mind that naturally exhibit all three of (a), (b), (c) (without also automatically exhibiting much richer algebraic structure, such as ring homomorphism structure) are complex dynamics, universal algebra, and certain types of cryptography, but I don't have enough experience in these fields to actually provide some interesting examples.<|endoftext|> -TITLE: Differential Geometry/General Relativity Computer Algebra -QUESTION [6 upvotes]: Hi, -could anybody recommend a CAS suited to DG/GR applications such as computation of connection coefficients or generating (and possibly solving) PDEs for, for example, an unknown metric of given curvature. Oh, and compatible Linux (I'm using Maple through wine but am having myriad problems. Also tried Maxima but I don't think it has a PDE solving tool.) -Thanks, -Mat - -REPLY [2 votes]: I recommend for Mathematica : -http://www.xact.es/ - it seems to be the most advanced package for General Relativity.<|endoftext|> -TITLE: sum of the reciprocal of the primes squared -QUESTION [21 upvotes]: Does anyone know of any information/work on this sum? I found absolutely nothing on the web about it. - -REPLY [36 votes]: This would be $P(2)$, where $P$ is the "prime zeta function," q.v. -A couple of very old references are C. W. Merrifield, The Sums of the Series of Reciprocals of the Prime Numbers and of Their Powers, Proc. Roy. Soc. London 33 (1881) 4–10. doi:10.1098/rspl.1881.0063. JSTOR 113877 -J. W. L. Glaisher, On the Sums of Inverse Powers of the Prime Numbers, Quart. J. Math. 25 (1891) 347–362. -EDIT: A more recent source is Steven R Finch, Mathematical Constants, page 95: The sum of the squared reciprocals of primes is $$N=\sum_p{1\over p^2}=\sum_{k=1}^{\infty}{\mu(k)\over k}\log(\zeta(2k))=0.4522474200\dots$$<|endoftext|> -TITLE: Iterated forcing and CH -QUESTION [8 upvotes]: I need some help with this theorem: if $P_\beta=\langle P_\alpha,\dot{Q}_\alpha:\alpha\leq\beta\rangle$, $\beta<\omega_2$, is a CSI of proper forcings, $P_\alpha\Vdash \lvert \dot{Q}_\alpha\rvert\leq\aleph_1$, and CH holds in the ground model, then $P_\beta$ forces the CH in the generic extension. -A proof of this fact appears in the Handbook, but I need a diferent one. Just for simplicity, suppose $\beta=\omega$, then the problem is how to prove that $P_\omega$ forces the CH. Can someone give a clue or an idea of how to proceed? -Thanks. - -REPLY [3 votes]: For any $\alpha < \omega_2$, the forcing notion $P_\alpha$ has a dense subset $P'_\alpha$ of cardinality $\aleph_1$. -This is proved in Shelah's "Proper Forcing" book. Jakob Kellner and I also give a sketch of this proof in our review of Uri Abraham's survey article on proper forcing (MR2768684). -As a consequence, each $P_\alpha$ will satisfy the $\aleph_2$-cc. The rest of the proof is as in the handbook (I do not know any essentially different proof): Counting "nice names" (i.e., names of reals that are defined from countably many antichains) shows that CH is preserved.<|endoftext|> -TITLE: What is the consistency strength of the failure of square, in terms of large cardinals -QUESTION [9 upvotes]: In Jech one can find a lower bound for the consistency strength of PFA in terms of large cardinals. I don't have my copy of Jech in front of me at the moment, but as I recall the presentation of this fact goes something like this: - -It's stated and proven that PFA implies the failure of $\square _{\kappa}$ for all $\kappa > \omega$. -It's stated that a result of Magidor's shows that PFA implies the failure of a weaker version of square, but the proof of this is not given in the chapter (it might appear later in the book, I can't remember). -It's stated that a result of Schimmerling's proves that the failure of this weak version of square implies the existence of a model with countably many Woodin cardinals, and this is also not proved in the chapter. - -My question is whether this weaker version of square is necessary to get a lower bound in terms of large cardinals, or whether there is some large cardinal lower bound on the consistency strength of "square fails everywhere" itself? - -REPLY [7 votes]: My understanding is that the weaker form of square was initially needed to get a better lower bound on the consistency strength of PFA but that, with improvements in our understanding of the fine structure of the current inner models, the original form of square now suffices to give the best lower bounds. -There are two versions of square (each with their associated ``weak square'' heierarchy): $\square_\kappa$ (formulated by Jensen) and $\square (\kappa)$ (formulated by Todorcevic). It is immediate from their definitions that $\square_\kappa$ implies $\square (\kappa^+)$ ($\square_\kappa$ and $\square (\kappa^+)$ are both statements about sequences of length $\kappa^+$ --- the apparent shift in indexing is purely notational). Todorcevic's theorem is that PFA implies the failure of $\square (\kappa)$ for every regular $\kappa > \omega_1$. It is often cited in the weaker form stated in item 1 of the question. -Traditionally, lower bounds on PFA have been obtained through the failure of $\square_\kappa$ at -a single singular strong limit cardinal. To my knowledge, there is currently no method for obtaining better lower bounds on PFA. -It was somewhat of a breakthrough when it was realized that the failure of $\square (\kappa)$ at successive values of $\kappa$ was apparently more powerful than the failure of successive instances of $\square_\kappa$. That it took so long for this to be noticed may be due to the fact that Todorcevic's result was commonly cited in its weaker form. I don't believe, however, that this gives an improved lower bound on the consistency strength of PFA. It does, however, yield more strength from the failure of any form of square at cardinals below $\aleph_\omega$. -It is at least my impression that it should be the case that the failure of square at all cardinals has a much higher strength than and individual failure of $\square (\kappa)$ or -$\square_\kappa$. Whether the global failure of $\square_\kappa$, $\square (\kappa)$ or the weak forms of square mentioned above yield different consistency strengths is a matter which is completely open and for which there is only wild speculation.<|endoftext|> -TITLE: Enriched locally presentable categories -QUESTION [7 upvotes]: Is there a standard reference for the theory (if it exists) of $\mathcal{V}$-enriched locally presentable categories? Here $\mathcal{V}$ is a cosmos. Does anything unexpected happens here in contrast to the case $\mathcal{V}=\text{Set}$ treated in "Locally Presentable And Accessible Categories" by Adamek & Rosicky? In particular I'm interested in the case $\mathcal{V} = \text{Cat}$. - -REPLY [3 votes]: One remark about the standard theory of enriched accessible categories is that the enriching category $\mathcal{V}$ is typically assumed to itself be locally presentable. This ensures that for sufficiently large $\kappa$, there is a good notion of $\kappa$-small limits and $\kappa$-filtered colimits -- for instance, they should commute in $\mathcal{V}$ -- which makes for a good notion of "smallness" for $\mathcal{V}$-enriched categories. -In principle, though, one should be able to develop a theory of $(\mathcal{V},\Phi)$-accessible categories based on any good duality between a class $\Phi$ of $\mathcal{V}$-limits and a class $\Phi^\vee$ of $\mathcal{V}$-colimits that commute in $\mathcal{V}$. (By "good", I mean that a condition called soundness should be satisfied.) Conceptually, it's not necessary for $\mathcal{V}$ to be locally presentable or for $\Phi$ to be the class of $\kappa$-small $\mathcal{V}$-limits. For example, in the unenriched case, the duality between finite products and sifted colimits can be developed in perfect analogy to the theory of $\kappa$-accessible categories for fixed $\kappa$, and one recovers the usual theory of Lawvere theories. Some aspects of such a general theory of locally $(\mathcal{V},\Phi)$-presentable categories are touched on by Lack and Rosický. -What's missing from this story is an interesting notion of "raising the index of accessibility". In the standard enriched setup, where $\mathcal{V}$ is locally presentable, we have (just like the unenriched case) a hierarchy of $\Phi$'s to use -- $\kappa$-small $\mathcal{V}$-limits for arbitrarily large $\kappa$ -- and interesting theorems to state that rely on letting $\kappa$ vary. I simply don't know any good examples of such a hierarchy of $\Phi$'s besides the standard ones to motivate the development of the sort of general theory I'm driving at.<|endoftext|> -TITLE: Are there any very hard unknots? -QUESTION [124 upvotes]: Some years ago I took a long piece of string, tied it into a loop, and tried to twist it up into a tangle that I would find hard to untangle. No matter what I did, I could never cause the later me any difficulty. Ever since, I have wondered whether there is some reasonably simple algorithm for detecting the unknot. I should be more precise about what I mean by "reasonably simple": I mean that at every stage of the untangling, it would be clear that you were making the knot simpler. -I am provoked to ask this question by reading a closely related one: can you fool SnapPea? . That question led me to a paper by Kaufmann and Lambropoulou, which appears to address exactly my question: http://www.math.uic.edu/~kauffman/IntellUnKnot.pdf , since they define a diagram of the unknot to be hard if you cannot unknot it with Reidemeister moves without making it more complicated. For the precise definition, see page 3, Definition 1. -A good way to understand why their paper does not address my question (by the way, when I say "my" question, I am not claiming priority -- it's clear that many people have thought about this basic question, undoubtedly including Kaufmann and Lambropoulou themselves) is to look at their figure 2, an example of an unknot that is hard in their sense. But it just ain't hard if you think of it as a three-dimensional object, since the bit of string round the back can be pulled round until it no longer crosses the rest of the knot. The fact that you are looking at the knot from one particular direction, and the string as it is pulled round happens to go behind a complicated part of the tangle is completely uninteresting from a 3D perspective. - -So here's a first attempt at formulating what I'm actually asking: is there a generalization of the notion of Reidemeister moves that allows you to pull a piece of string past a whole chunk of knot, provided only that that chunk is all on one side, so to speak, with the property that with these generalized Reidemeister moves there is an unknotting algorithm that reduces the complexity at every stage? I'm fully expecting the answer to be no, so what I'm really asking for is a more convincing unknot than the ones provided by Kaufmann and Lambropoulou. (There's another one on the Wikipedia knot theory page, which is also easily unknotted if you allow slightly more general moves.) -I wondered about the beautiful Figure 5 in the Kaufmann-Lambropoulou paper, but then saw that one could reduce the complexity as follows. (This will be quite hard to say in words.) In that diagram there are two roughly parallel strands in the middle going from bottom left to top right. If you move the top one of these strands over the bottom one, you can reduce the number of crossings. So if this knot were given to me as a physical object, I would have no trouble in unknotting it. - -With a bit of effort, I might be able to define what I mean by a generalized Reidemeister move, but I'm worried that then my response to an example might be, "Oh, but it's clear that with that example we can reduce the number of crossings by a move of the following slightly more general type," so that the example would merely be showing that my definition was defective. So instead I prefer to keep the question a little bit vaguer: is there a known unknot diagram for which it is truly the case that to disentangle it you have to make it much more complicated? A real test of success would be if one could be presented with it as a 3D object and it would be impossible to unknot it without considerable ingenuity. (It would make a great puzzle ...) -I should stress that this question is all about combinatorial algorithms: if a knot is hard to simplify but easily recognised as the unknot by Snappea, it counts as hard in my book. -Update. Very many thanks for the extremely high-quality answers and comments below: what an advertisement for MathOverflow. By following the link provided by Agol, I arrived at Haken's "Gordian knot," which seems to be a pretty convincing counterexample to any simple proposition to the effect that a smallish class of generalized moves can undo a knot monotonically with respect to some polynomially bounded parameter. Let me see if I can insert it: - - - -(J.O'Rourke substituted a hopefully roughly equivalent image for Timothy's -now-inaccessible image link.) -I have stared at this unknot diagram for some time, and eventually I think I understood the technique used to produce it. It is clear that Haken started by taking a loop, pulling it until it formed something close to two parallel strands, twisting those strands several times, and then threading the ends in and out of the resulting twists. The thing that is slightly mysterious is that both ends are "locked". It is easy to see how to lock up one end, but less easy to see how to do both. In the end I think I worked out a way of doing that: basically, you lock one end first, then having done so you sort of ignore the structure of that end and do the same thing to the other end with a twisted bunch of string rather than a nice tidy end of string. I don't know how much sense that makes, but anyway I tried it. The result was disappointing at first, as the tangle I created was quite easy to simplify. But towards the end, to my pleasure, it became more difficult, and as a result I have a rather small unknot diagram that looks pretty knotted. There is a simplifying move if one looks hard enough for it, but the move is very "global" in character -- that is, it involves moving several strands at once -- which suggests that searching for it algorithmically could be quite hard. I'd love to put a picture of it up here: if anyone has any suggestions about how I could do this I would be very grateful. - -REPLY [22 votes]: I did not see any mention of this paper by Marc Lackenby, probably because the question is quite old : -Lackenby, Marc, A polynomial upper bound on Reidemeister moves, Ann. Math. (2) 182, No. 2, 491-564 (2015). ZBL1329.57010. -Building on the techniques introduced by Dynnikov and adding some normal surface theory, he shows that any unknot can be unknotted using only a polynomial number of Reidemeister moves. -So if we interpret the "much more complicated" by "exponentially more complicated", it gives quite a strong "no" answer to - -is there a known unknot diagram for which it is truly the case that to disentangle it you have to make it much more complicated?<|endoftext|> -TITLE: Nontrivial circular arguments? -QUESTION [26 upvotes]: There is a famous circular argument for the Prime Number Theorem (PNT). It turns -out that there exists an infinite sequence of elementary-to-prove Chebyshev-type estimates -that taken together imply PNT. Unfortunately, the collective existence of all these proofs seems to require the PNT, so one must work hard a la Selberg and Erdos for an elementary proof. See Harold Diamond, Elementary methods in the study of the distribution of prime numbers, Bull. Amer. Math. Soc. N. S. 7 (1982), 553-589. -Now on the traditional view, circular proofs simply have no value at all. Yet one feels perhaps that the example in the previous paragraph has something to say. Just an illusion? Or does there exist a foundational framework where circular proofs of this special sort enjoy bona fide status? -Just to riff a little more, imagine that the Goldbach conjecture turns out independent of PA (or some other, perhaps weaker, axiom system for arithmetic). The truth of the Goldbach conjecture (required for independence!) would then imply the existence of a proof (trivial!) for any given even number that that one even number equals the sum of two primes. Now that obviously isn't very interesting compared to the example in the first paragraph, and perhaps the difference has something to do with the greater quantifier complexity of PNT? -As a side question, are there any similar stories in the lore of the Riemann Hypothesis? -For example, does RH imply the existence of an infinite sequence of zero-free region proofs of a known form that collectively amount to RH? - -EDIT: Lest this come up repeatedly, let me expand upon a remark I made in a comment below: -I would be interested to know if RH implies the existence of proofs of a known form in a known system within which one does not assume RH, such that the conclusions of all these proofs conjunctively yield RH. -Actually the answer to my question is "yes" though I find my own example unsatisfying -(rather like the Goldbach example): -We can check zeros up to a given magnitude rigorously by known (non-trivial) numerical techniques. RH predicts these tests will come out positive, but of course the tests -don't rely on RH. If we know they all come out positive, that's RH. I find this unsatisfying because the little proofs approximate the whole of RH so badly (compared to -how the Chebyshev estimates really do make one feel one has PNT for all practical purposes). Now a family of zero-free regions that union up to $\sigma >1/2$ where each new zero-free region had strong arithmetic consequences, that would seem interesting. - -REPLY [34 votes]: Perhaps an example of the kind of circularity you mention -arises with the self-reference phenomenon that arises in -connection with the incompleteness -theorems -and related applications. Specifically, Gödel proved -the fixed-point lemma that for any assertion $\varphi(x)$ -in arithmetic, there is a sentence $\psi$ such that PA, or -any sufficiently powerful and expressible theory, proves -that $\psi$ is equivalent to -$\varphi(\ulcorner\psi\urcorner)$. In other words, $\psi$ -is equivalent to the statement "$\psi$ has property -$\varphi$." Thus, statements in the language of arithmetic -can refer to themselves, and so self-reference, the stuff -of paradox and nonsense, enters our beautiful number -theory. -One famous example, used by Gödel to prove the first -incompleteness theorem, occurs when $\varphi(x)$ asserts, -"$x$ is the code of a statement having no proof in PA", for -then the resulting fixed point $\psi$ effectively asserts -"this statement is not provable". It follows now that it -cannot be provable, for then it would be a false provable -statement, and so it is true. Thus, it is a true unprovable -statement, establishing the first incompleteness theorem. -A dual version of this, however, exhibits your circularity -property in a stronger way. Namely, let us apply the fixed -point lemma to the formula $p(x)$ asserting "$x$ is the -code of a statement provable in PA." In this case, the -resulting fixed point $\psi$ asserts "this statement IS -provable." Consider now the following theorem of Löb: -Theorem.(Löb) If the Peano Axioms (PA) prove the implication (PA proves -$\varphi$)$\to\varphi$, then PA proves $\varphi$ -directly. -(And the converse is immediate, so PA proves that (PA -proves $\varphi$)$\to \varphi$ if and only if PA proves -$\varphi$.) -In the case of $\psi$ asserting "$\psi$ is provable," we -have that the hypothesis of Löb's theorem holds, and -so we may make the conclusion that yes, indeed, $\psi$ -really is provable! In other words, the statement "this -statement is provable" really is provable, although no -naive argument will establish this. - -The proof of Löb's theorem is itself a surprising -exercise in circularity, something like the following: -Theorem. Santa exists. -Proof. Let $S$ be the statement, "If S holds, then Santa -exists." Now, we claim that $S$ is true. Since it is an -implication, we assume the hypothesis, and argue for the -conclusion. So assume that the hypothesis of S is true; -that is, assume $S$ holds. But then the implication -expressed by $S$ is true. So the conclusion that Santa -exists is also true. So we have shown under the assumption -of the hypothesis of $S$ that the conclusion is true. So we -have established that $S$ holds. Now, by $S$, it follows -that Santa exists. QED -(Those who know the proof of Löb's theorem will agree -that the proof is fundamentally the same as the above, -except that it is fully rigorous nonsense instead of silly -nonsense!)<|endoftext|> -TITLE: How does Schlessinger's criterion sit with Grothendieck Existence (aka GFGA)? -QUESTION [5 upvotes]: Grothendieck Existence, which I imagine is the less well known result among the two, states the following: -Let $A$ be a noetherian ring that is complete w.r.t. a proper ideal $I$. Let $V$ be a proper $A$-scheme. Let $W$ be the inverse image of the locus of $I$ (as a subscheme of $V$). Let $\mathfrak{V}=(W,\mathcal{O}_{\mathfrak{V}})$ be the formal completion of $V$ along $W$. Then the functor $\mathcal{F} \mapsto \hat{\mathcal{F}}$ from the category of coherent $\mathcal{O} _ V$ -modules to the category of coherent $\mathcal{O}_{\mathfrak{V}}$-modules is an equivalence of categories. -This results leaves a Schlessinger taste in my mouth. It seems like we're showing a certain deformation is effective. Is this a consequence of Schlessinger's criterion? Does it go the other way around? Are they just completely unrelated and I'm just seeing patterns where there are none? - -REPLY [10 votes]: Schlessinger's criterion is a criterion for the pro-representability of a functor. This is the same thing is getting something like an object on the formal scheme $\mathfrak V$. Grothendieck's result tells us that this object comes from an actual object on $V$. Usually this is the step after pro-representability to get actual representability (and is usually formulated as saying that the formal deformation is "effective", a property that goes beyond pro-representability). This is precisely the setup of Artin's criteria for representability by an algebraic space.<|endoftext|> -TITLE: Does BQP^P = BQP ? ... and what proof machinery is available? -QUESTION [5 upvotes]: Update #3: -Over on TCS StackExchange, I have rated as "accepted" an ingenious construction by Luca Trevisan, which answers a two-part question (as reframed by Tsuyoshi Ito) that is in essence "Do runtimes for P require EXP resources to upper-bound? … are concrete examples known?" -Hopefully I have grasped correctly that, in brief, Luca's construction yields the answers "yes" and "yes for all practical purposes" (FAPP). -It will take awhile (for me anyway) to appreciate whether Luca's $M$-machines obstruct the $P$-time uniform reduction of ${BQP}^{P}\,\to\,{BQP}$ that is at the heart of the original question posed here on on MathOverflow, that question being, "Does BQP^P = BQP? ... and what proof machinery is available?", which in turn generalized a question that was posed by Dick Lipton and Ken Regan on their weblog Gödel's Lost Letter and P=NP, the question "Is Factoring Really In BQP? Really?" -After some further reflection (which may take a few days) I will attempt a summary back-trace of this chain of questions, which so enjoyably unites elements of mathematics, science and practical engineering, and will post that summary both here and on TCS StackExchange. -In the meantime, my thanks and appreciation are extended to everyone ... and further comments are very welcome, of course! - -Update #2: -I've plowed through a pretty considerable portion of the (excellent) references that Aram provided, and hope to write a summary soon. One topic that I have not found addressed in these references (or it may be that I am too inexpert to perceive it) has been asked as a separate MathOverflow question "Do runtimes for P require EXP resources to upper-bound? … are concrete examples known?". - -Update #1: -So far, this topic has a ratio of views-to-answers (presently 285-to-1) that is large relative to other MathOverflow questions ... greater diversity in the answers would be good ... perhaps this draft summary posted on Shtetl Optimized will stimulate more of the "flaming arrow" responses that were hoped-for ... although the references that Aram's answer provides are terrific, needless to say. - -Does $BQP^P = BQP$ ? That is, is $P$ low for $BQP$? -Here P is the standard complexity class that is associated to polynomial-time algorithms implemented on (classical) Turing machines, and BQP is the standard (quantum) complexity class Bounded Error Quantum Polynomial Time. -We have specifically in mind a logic-gate instantiation of BQP, that is, a polynomial-time uniform family of quantum circuits (that is, a standard gate-based quantum computer), and for P we have in mind a single-tape Turing machine. So in practical terms, the proposition $BQP^P = BQP$ is about the feasiblity of compiling algorithms in P into polynomially-many reversible logic gates. -This question arose on Dick Lipton and Ken Regan's weblog Gödel's Lost Letter and P=NP as a natural generalization of the question: "Is Shor's algorithm for factoring in BQP?," and it is intimately linked to what Nielsen and Chuang's textbook calls "the Principles of Deferred and Implicit Measurement." -For engineers, proof machinery associated to this question is at least as interesting as the answer itself, because this class of problems arises commonly in engineering practice, in the guise (for example) of compiling a procedural algorithm (written in C and instantiated in a thread, for example) into a circuit diagram (written in Verilog and instantiated on an FPGA, for example). In practice, these algorithm-to-circuit translations are not so easy to solve ... particularly when there is no formal proof that the procedural algorithm is in $P$, but rather, only the empirical (ie, oracular) observation that it so behaves. -One point is that in both practical engineering and in computational complexity theory, procedural algorithms are in general not accompanied by documentation of how they work—perhaps it is infeasible even in principle to document how all the algorithms in the complexity class P work? (an answer to this question too would be very welcome). -Hence the proof machinery for dealing with undocumented (and even undocumentable?) algorithms, both classical and quantum, has both fundamental and practical significance, beyond the significance of the answer to the question asked. This practical/physical context is discussed more fully on Dick and Ken's weblog. -It wasn't so easy to decide whether to post this question of here on MathOverflow versus TCS StackExchange ... in the end, MathOverflow was chosen in hope that the question will attract a mathematically broad range of answers. -Suggestions are especially welcomed, as to how this question might be amended, to make it better-posed and/or more broadly interesting. Thanks! :) - -REPLY [7 votes]: Yes, P is contained in BQP (Benioff, 1982; http://prl.aps.org/abstract/PRL/v48/i23/p1581_1 ) and $BQP^{BQP}=BQP$, for pretty much the same reason $BPP^{BPP}=BPP$. This second point first appeared (that I know of) as Cor 4.15 of BBBV'97: http://www.cs.berkeley.edu/~vazirani/pubs/bbbv.ps .<|endoftext|> -TITLE: Regularity for transport equation? -QUESTION [5 upvotes]: In the book of Evans the transport equation, -$$\frac{d}{dt} u + b\cdot \nabla u = 0, \quad u(t=0)=u_0,$$ -is solved by the method of charateristics for $b$ and $u_0$ smooth enogh (in terms of $\mathcal{C}^k$ for some $k$). -I am not an expert for hyperbolic equations and I see this equation as a prototyp of hyperbolic equations of first order and I thought that there are some results of this equation in Sobolev spaces. But I haven't found some (expect for strong solutions or the result of DiPerna and Lions; see below). -To be more precise: I wonder if there is any result of the type: If $b$ and $u_0$ are contained in some Sobolev spaces, then $u\in L^p(W^{k,q})=L^p(0,T;W^{k,p})$ for $k\geq 1$ and some $p,q$? I know the paper of DiPerna and Lions from 1989, where develop a theory of this equation in Sobolev spaces, but they don't study if the solution $u$ has a (weak) derivative (or even more). -Does anyone knows an article for this equation or (if not) a good reference (book or survey article) for beginners in hyperbolic equations? -Thanks a lot! -Edit: 1) $b$ can be assumed as divergence free, $\nabla \cdot b=0$. -2) Higher regularity (in terms of derivates for $u$) can be established by the method, Willi Wong mentioned in his comment. This question is okay now. Does anyone has a reference for a beginng guide for hyperbolic equations or a survey article with further archiements than DiPerna and Lions? - -REPLY [2 votes]: You can start with an interesteing extension - Ambrosio 2004 showed existence, uniqueness and stability results for $b\in BV$ and ${\rm div}_x b$ absolutely continuous with respect to Lebesgue measure.<|endoftext|> -TITLE: Geometric meaning of small extensions ? -QUESTION [12 upvotes]: Let $(A,\mathfrak{m}_A)$ be a local Artinian $k$-algebra with residue field $k$. Then the scheme $\mathrm{Spec}(A)$ can be loosely seen as a "fat point", or an "infinitesimal neighbourhood" of a point $\mathrm{Spec}(k)$. If the maximal ideal satisfies the condition $\mathfrak{m}^k=0$, then I think the spec of $A$ can be seen as a "$(k-1)$-th order infinitesimal neighbourhood" of the closed point. -Recall: a small extension of $A$ is an extension of local Artinian $k$-algebras $0\to I \to B\to A \to 0$ such that $\mathfrak{m}_B\cdot I =0$. - -Does the concept of a small extension have any geometric interpretation? - -Does it mean that the closed embedding $\mathrm{Spec}(A)\hookrightarrow \mathrm{Spec}(B)$ is such that it only "fattens" already existent "directions" of $\mathrm{Spec}(A)$, making it into a "higher order" fattening of the closed point but without adding new "directions" of fattening? Is my suggestion totally misleading? -Edit: as comments point out, my interpretation was not correct. - -REPLY [6 votes]: If you think of elements in a local Artin $k$-algebra $R$ as, say functions on the origin of $k^n$ which remember some (finite amount of) higher order information in the various $n$ directions, then a small extension $R'$ of $R$ is just another such ring with functions that remember "at most one order higher". -For example, let $R = k[x,y]/((x,y)^2)$ and consider the small extension $R' = k[x,y]/((x,y)^3)$. The elements of $R$ are functions which remember up to 1st order in the directions $x,y$. The elements of $R'$ are what we get if we take functions from $R$ and stick on some 2nd order terms in the $x,y$. -You can also just extend only in one direction. For example, $R'' = k[x,y]/(x^3,xy,y^2)$ is also a small extension of $R$, but the only 2nd order term we added was $x^2$, not $xy$ nor $y^2$. -However, if you take $A = k[x]/(x^2)$ and $A'' = k[x]/(x^4)$, then this is not a small extension because we went two orders up, from 1st order to 3rd order. -Generally, I interpret a small extension as one that only thickens our fat point by an order of at most 1 in each direction. -Things are more complicated if you add a new direction. If you take $R = k[x,y]/((x,y)^2)$ as before, and consider $R[z]/(z^2)$, then this is not a small extension because of the cross terms $xz$ and $yz$. However, if you get rid of them, then $R[z]/(z^2,xz,yz)$ is a small extension of $R$. -You could say that adding a new direction involves making two small extensions. The first one adds the new variable with no cross terms, and the second adds the cross terms. For instance, let $R = k[x]/(x^2)$, $R' = R[y]/(xy,y^2)$, and $R'' = R[y]/(y^2)$.<|endoftext|> -TITLE: Decidability of tiling R^2 -QUESTION [15 upvotes]: Does there exist a closed curve, with finite area and finite circumference, of which it is undecidable (in an axiomatic system where it is constructable) whether it can tile the plane? -I know the general problem of a set of polygons is undecidable, but I haven't found any information on the single tile case. - -REPLY [30 votes]: The answer by Nick S is essentially correct, but I can provide a little more detail. -It is not hard to show that the tiling problem (if a set of shapes admits a tiling of the plane) is decidable if every set of shapes that tile admit a periodic tiling. The algorithm is to put shapes together, look at the way two shapes can combine, 3 shapes and so on. If the shapes cannot tile you will eventually find a ball that cannot be covered, if they can tile periodically you will eventually find the patch that can go periodic. -The discovery of aperiodic tilings and the result that the tiling problem was undecidable were proved at the same time by Berger. In fact Berger used an aperiodic tiling to show that the tiling problem was undecidable (originally in his thesis). Note that he (and others) call it the "domino problem", as the question was originally posed for squares with coloured edges, called Wang tiles. -R Berger The undecidability of the domino problem -Mem. Amer. Math. Soc. 66 1966 -http://books.google.com/books/about/Undecidability_of_the_Domino_Problem.html?id=8AmiHD0Lbu8C -Berger's proof was significantly simplified by Rafeal Robinson. An important part of the simplification was considering a smaller set of aperiodic tiles. Berger's example required 20,426 tiles, whereas Robinson's required just 6. -R M Robinson, Undecidability and Nonperiodicity for Tilings of the Plane -Inventiones math. 12 177-209, 1971 -http://www.lif.univ-mrs.fr/~fernique/qc/robinson.pdf -Although it proved useful in these proofs, the existence of aperiodic tilings is not equivalent to the undecidability of the tiling problem. For one thing, by definition, the tiling problem is decidable for all known aperiodic tilings. The best case for the tiling problem is a result of Nicolas Ollinger. He shows that the tiling problem is undecidable for sets of 5 tiles using a construction based on polyominoes. -N Ollinger, Tiling the Plane with a Fixed Number of Polyominoes. -Proceedings of LATA 2009, Lecture Notes in Computer Science 5457, Springer 2009, pp. 638-647. -http://www.springerlink.com/content/x5nkq3640w6784w4/ -For the question of aperiodic sets of shapes, we have had examples with two tiles (e.g. the Penrose tiling) for a long time, although there are still only a handful of examples with a small number of shapes (<10 for example). The question of a single tile was a longstanding open question, and became known as the einstein problem (a german pun). The answer now depends on your definition of "tile". For a permissive definition (a tile is a compact set that is the closure of its interior, for example) the problem was solved by Joan Taylor, who is not a professional mathematician but decided to take on this particular problem over several years. This is a significant development, but perhaps not the final answer. The tile that she found is not connected. With respect to the setting of this question, the boundary is not a closed Jordan curve, and, for me, this means that it fails "the laser cutter test" I cannot easily make the thing! This result is the preprint given in Richard Borcherds comment: -J E S Socolar, J M Taylor An aperiodic hexagonal tile -Journal of Combinatorial Theory, Series A 118 pp. 2207-2231 2011 -http://arxiv.org/abs/1003.4279 -I wrote a little bit about the story here: -http://maxwelldemon.com/2010/04/01/socolar_taylor_aperiodic_tile/ -and you can also find Joan Taylor's original paper, with beautiful hand-drawn images: -http://www.math.uni-bielefeld.de/sfb701/files/preprints/sfb10015.pdf -On the hyperbolic plane the first strongly aperiodic set (having no symmetries of infinite order) was discovered in 2005, by Chaim Goodman-Strauss: -C Goodman-Strauss A strongly aperiodic set of tiles in the hyperbolic plane -Inventiones Mathematicae, 159, Number 1, 119-132, 2005 -http://math.uark.edu/A_Strong_Aperiodic_Set_of_Tiles_in_the_Hyberbolic_Plane%281%29.pdf -The tiling problem took a little longer, and was only shown to be undecidable recently (in 2006-7), independently by Maurice Margenstern and Jarkko Kari, who give distinct proofs. -M Margenstern, The Domino Problem of the Hyperbolic Plane is Undecidable -The Bulletin of the EATCS 93 220--237 2007 -http://arxiv.org/abs/0706.4161 -J Kari, On the Undecidability of the Tiling Problem -SOFSEM 2008: Theory and Practice of Computer Science -Springer Lecture Notes in Computer Science, 4910, 74-82 2008 -http://www.springerlink.com/content/673165n258t18741/<|endoftext|> -TITLE: Maximal ideal and Zorn's lemma -QUESTION [12 upvotes]: It is known that any ring A (say commutative with 1) has a maximal ideal. The proof uses Zorn's lemma. -Now I heard some people saying that if we assume A to be noetherian, then we don't need to use Zorn's lemma. The argument would basically be as follows: -"Suppose it doesn't have a maximal ideal. Then we can build an ascending chain of distinct ideals." -But, as far as I know it, we have to use Zorn's lemma in order to construct such an ascending chain. Am I right? -If I am right, is it still true (via some other argument) that we don't need to use Zorn's lemma to prove the result? -(EDIT: My definition of noetherian ring is that any ascending chain of ideals stabilizes.) - -REPLY [6 votes]: This is an elaboration of Joel's answer. Suppose $R$ is a ring in which every ascending chain $I_0 \subseteq I_1 \subseteq I_2 \subseteq \cdots $ of ideals stabilizes. -We show that every (non-trivial) ideal can be extended to a maximal ideal. Given an ideal $J$, define a sequence of ideals $I_n$ as follows: - -Let $I_0 = J$. -For $n \geq 0$, if $I_n$ is a maximal ideal then let $I_{n+1} = I_n$, otherwise choose -an ideal $I_{n+1}$ which is strictly larger than $I_n$. - -We used Dependent Choice to construct the sequence of ideals $I_n$. Because $I_n$ is an ascending chain of ideals it stabilizes. When it does, it reaches a maximal ideal that extends $J$. -So we used Dependent Choice to construct the ascending chain of ideals and Excluded Middle to decide whether an ideal is maximal or not. -In a particular case it may be possible to extend a non-maximal ideal to a larger one in a canonical way (for example, if we know how to well-order $R$ then we generate $I_{n+1}$ by adjoining to $I_n$ the first element which is not in $I_n$ yet). In this case we do not even need Dependent Choice.<|endoftext|> -TITLE: If a field extension gives affine space, was it already affine space? -QUESTION [33 upvotes]: Let $R$ be a commutative Noetherian $F$-algebra, where $F$ is a -field (perfect, say). Assume that $R \otimes_F \overline F$ is a polynomial ring over the -algebraic closure $\overline F$. -Does it follow that $R$ was already a polynomial ring over $F$? -I doubt it, but haven't had any luck constructing a counterexample. -The question arises because I'm trying to understand Bruhat cells in -flag manifolds for non-split algebraic groups. In split ones, -Bruhat cells are affine spaces. - -REPLY [11 votes]: First, since the OP is interested in Bruhat cells in flag varieties over nonsplit groups (over perfect fields), I think the question is probably unnecessary. I'd recommend Borel and Tits IHES publication "Groupes Reductifs," where they describe the theory of relative root systems for nonsplit groups, and much more. In and around Theorem 5.15, you can find the basic results on parabolic subgroups and Bruhat decomposition, in this relative setting. From that section, one can deduce that the appropriate Bruhat cells are isomorphic to unipotent groups (as varieties) and in characteristic zero, they are just affine spaces, not twists thereof. -So, I doubt that twists of affine spaces have any place in studying Bruhat cells for nonsplit groups reductive groups over perfect fields. -Now I have a few remarks regarding the issue of existence of nontrivial twists of affine space (though this is outside my expertise). As the comments indicate, this dances around the famous Jacobian conjecture, and demonstrates how far we must go to understand the group of polynomial automorphisms of affine $n$-space (over $\bar Q$, for example). -To understand this group of automorphisms $G(\bar Q)$, it helps to filter out the easy part. One finds a translation subgroup $T(\bar Q)$ of $G(\bar Q)$, given by automorphisms which are translations of the affine space; the quotient of $G$ modulo this translation subgroup can be identified with the subgroup $G^0(\bar Q)$ of automorphisms preserving the origin. Among these, one can look at the induced automorphism on the tangent space at $0$, an element of $GL_n(\bar Q)$. The origin-preserving automorphisms which act as the identity on the tangent space at $0$, is called the group of principal automorphisms, or $G^1(\bar Q)$. So there are exact sequences of $Gal(\bar Q / Q)$-modules: -$$1 \rightarrow G^1(\bar Q) \rightarrow G^0(\bar Q) \rightarrow GL_n(\bar Q) \rightarrow 1.$$ -$$1 \rightarrow T(\bar Q) \rightarrow G(\bar Q) \rightarrow G^0(\bar Q) \rightarrow 1.$$ -(I'm working over $\bar Q$ for convenience). By standard Galois cohomology, one finds that the possibility of twisted forms of affine $n$ space boils down to the Galois cohomology with coefficients in $G^1(\bar Q)$, the principal automorphisms. -Now the group $G^1(\bar Q)$ is very complicated. I think Shafarevich was the first to study such automorphism groups, as points of Ind-varieties (or Ind-affine varieties), and there are a few papers of Kambayashi (see one in J. of Algebra, for example) describing such an algebraic structure on $G^1$ which I'm more familiar with. -Here, in my opinion, is why dimension $2$ is so much easier to deal with. There's a famous result of Gabber, which states that if $P: A^n \rightarrow A^n$ is a polynomial automorphism (a polynomial map with polynomial inverse) of degree $d$, then the inverse of $P$ has degree at most $d^{n-1}$. For this reason, in dimension $2$, invertible polynomial maps have inverses of the same (or less) degree. This makes it easy to "filter" the automorphism group $G^1$ by degree -- Galois cohomology will have image in a subgroup $G_d^1(\bar Q)$ of $G^1(\bar Q)$ consisting of automorphisms of degree $\leq d$ for some $d$. This filtration makes computation of Galois cohomology easier, I'd guess. -In dimension $\geq 2$, I don't think anyone knows enough about the structure of $G^1(\bar Q)$ to compute its Galois cohomology.<|endoftext|> -TITLE: Maximal (non-abelian) extensions of number fields unramified everywhere -QUESTION [8 upvotes]: Hello! -Let $K$ be a number field. All abelian unramified extensions are contained in the Hilbert class field which is a finite extension 'maximal' with respect to this property. For general unramified extensions, is there a bound (depending on $K$) on the degree of an unramified extension over $K$? If so, does the compositum of all unramified extensions also have finite degree over $K$ in general? -Thanks for your attention! -ADDENDUM: as Hunter noticed the answer can be no even just even for solvable groups, when the field admits an infinite class field tower. But perhaps it is still interesting to study the question for extension having simple Galois group, and possibly their compositum. Is there anything known about this case? - -REPLY [4 votes]: It is worth pointing out a recent paper by Manabu Ozaki in the Inventiones where he proves that - -Given any finite $p$-group $G$ ($p$ - being a prime number), there exists a - number field $F$ such that the group - of $F$-automorphisms of the maximal - unramified $p$-extension of $F$ is - isomorphic to $G$. - -See his Theorem I. The paper is also available on the arXiv.<|endoftext|> -TITLE: Motivation of filtered colimits -QUESTION [11 upvotes]: I am trying to move in categorical algebra beyond the basics. A Lawvere theory L is a small category with finite products. (I know that there also is a functor $(skeleton(FinSet))^{op}\to L$, which restricts a number of sorts in the algebraic theory to 1. Lets drop that requirement for now.) How to convert some variety to a Lawvere theory is pretty clear for me. The link (varieties ↦ Lawvere theories) is clear in some elementary operations, like - -mapping an algebra by some functor F -↦ postcomposing F; -underlying functor -↦ precomposition of a functor between -Lawvere theories. - -Then filtered colimits come. Lets take for reference “Adámek. a categorical introduction to general algebra.” Chapter 2 “Sifted and filtered colimits” and chapter 3 “Reflexive coequalizers” are devoid of mentioning varieties. Why the definition of a filtered colimit is such? I suppose there should be more concrete explanations involving algebraic operations, this is called “algebra” after all. Google suggests few texts on this subject, but they are abstract too. Any references? -The claim “an arbitrary algebra is a filtered colimit of finitely generated algebras” is needed to construct the left adjoint to an underlying functor. Can anyone refer me to its proof? (Update 2011-01-29. Also I want a precise proof constructing that left adjoint.) (Update 2011-01-29. Thank you all for insightful answers and comments. I suspect that there is no direct link between filtered colimits and traditional algebra, i.e. it is an abstract thing that is needed for another abstract thing… I need to think it through to formulate further questions.) - -REPLY [5 votes]: The reason that Grothendieck originally considered filtered colimits and what is now known as the theory of accessible and locally presentable categories (I think named by Makkai and Paré) is as follows: -Let $x$ be an object of a category $C$ such that $hom_C(x,-)$ preserves $\alpha$-filtered colimits, then given any morphism $x\to colim F$ where $F:D\to C$ is an $\alpha$-filtered diagram, the morphism $x\to colim F$ factors through at least one $F(d)$ for some $d$ in $d$, and given any two factorizations through $F(d)$ and $F(d')$, there exists a majorant factorization through $F(d'')$ where $d''\geq d'$ and $d''\geq d$ extending the other two factorizations. -I leave this as an exercise (it is, if you will, proof by introspection) (Hint: Use the corresponding statement for sets (which are valid because the statements hold for hom-sets) to perform the necessary manipulations). -This very powerful technique is used, for instance, in the modern generalizations of the small object argument and in situations regarding Bousfield localizations (the notion of accessibility is absolutely essential for results like Jeff Smith's theorem, for instance). -See Clark Barwick's paper for a fairly detailed treatment with regard to its use in homotopy theory. I would also suggest taking a look at Appendices 1 and 2 of Lurie's Higher topos theory as well as the book of Makkai-Paré, and also the standard modern reference on the subject by Adamek and Rosicky.<|endoftext|> -TITLE: Is there an Infinite increasing sequence of primes with bounded second or larger differences? -QUESTION [7 upvotes]: The opposite question is if for any infinite increasing sequence of primes and any $k$ the sequence of the $k$-th order differences of the elements of the sequence is unbounded. -But if the question is true then there are integers $k$ and $B$ and a sequence of primes $q_1 < q_2 < \dots$ such that - $$ \lvert \Delta^k q_n \rvert \le B \quad \quad \forall n=1,2,\dots$$ -and it is natural to ask what can be the least value of $k$ for which such a sequence exists, can we have $k=2$? -Does somebody know of any result or heuristic in one or the other sense? -EDIT: I expect now the question to have a negative answer. The reason is that if we start with the first $k$ primes a given bound $B$, and follow every possible chain of primes with $k$th order differences bounded by $B$. From an element $q$ of one of these chains we get at most $2B+1$ possible succesors and as the "probability" of one them being prime is about $(2B+1)/\log q$ the expected number of "active chains" decreases to zero as $q$ increases. -By the way I have computed the size of the longuest chain of primes with bounded second differences starting at 2,3 and for small values of $B=2,4,6,8,10$ it gives $57, 421, 1860, 24661, 380028$, it seems to increase very roughly as $e^{2B}$, is that a reasonable estimate?. - -REPLY [5 votes]: If this were known for $k=2$, it would mean that for sufficiently large $n$ and some $A$, there is always a prime between $n^2$ and $(n+A)^2$. I believe that this is still open; see Legendre's conjecture. For comparison, the Riemann Hypothesis would imply there is a prime between $n^2$ and $(n+A \log n)^2$. -It is known that for large enough $n,$ there are primes between $n^3$ and $(n+1)^3$ (and you can lower that exponent from $3$). Perhaps you can use this to construct sequences with some differences bounded, although this is not immediate to me.<|endoftext|> -TITLE: A question about the topological proofs of Bott periodicity -QUESTION [28 upvotes]: There are purely topological proof of Botts periodicity theorem, the first one given by Dyer and Lashof. I am heading to discuss the proof -in my lecture course on homotopy theory (as a final chord and as an application of the Leray-Serre spectral sequence). -Textbook references that I have consulted are: Dyer "Cohomology theories", Mimura-Toda: "Topology of Lie groups", Switzer "Algebraic topology". -I also want to advertise a nice paper by Kono and Tokunaga which uses methods are well beyond the scope of my course. -There are several versions of the argument, but the idea is always the same. Here is a sketch (in the complex case; a similar argument works in the real -case, but it is at least six times as complicated). - -Source and target of the Bott map $\beta: BU \to \Omega SU$ are homotopy commutative $H$-spaces and $\beta$ is an $H$-map. -Therefore, $H_* (BU)$ and $H_* (\Omega SU)$ are commutative rings under the Pontrjagin product and $\beta$ induces a ring homomorphism. -Consider the classifying map $\lambda :\mathbb{CP}^{\infty} \to BU$ of the Hopf -bundle. Tracing throgh the definition of the Bott map shows that $\beta \circ \lambda: \mathbb{CP}^{\infty} \to \Omega SU$ is the adjoint of -the following map $\Sigma \mathbb{CP}^{\infty} \to SU$: $(z,l)$ goes to the linear map that is multiplication by $z$ on $l$ and $1$ on the -complement (OK, you have to normalize to make the determinant $1$). -$\beta \circ \lambda: \mathbb{CP}^{\infty} \to \Omega SU$ is injective in homology and the induced map $P [H_* (\mathbb{CP}^{\infty})] \to H_{*}(\Omega SU)$ -($P$ means polynomial algebra) is a ring isomorphism. This is the part that makes heavy use of spectral sequences. -Anything proven so far implies that $\beta$ is surjective in homology. -Since the homology groups of $BU$ and $\Omega SU$ are free abelian of the same rank (here on needs to know the additive structure of $H_* (BU)$), -$\beta$ is also injective in homology. -By the Hurewicz theorem, $\beta$ is a weak homotopy equivalence. - -Version 1 of my question: Is it possible to modify the argument to use cohomology instead? Or is there a reason why the following sketch does -not work?: - -prove the cup product structure on $H^* (BU)$ (I wish to do this anyway) -compute the additive structure of $H^* (\Omega SU)$. -Identify some concrete cohomology classes of $\Omega SU$, for example given as transgressions of classes of $SU$ or -double transgressions of classes -of $BSU$. -Prove that the images of these classes under $\beta$ generate $H^* (BU)$. - -This would have, in my opinion, the advantage that it uses the much more familiar cup product structure on $H^* (BU)$ instead of the Pontrjagin product structure on $H_* (\Omega SU)$. Therefore the proof is at least psychologically simpler. -A more technical and more focussed version of my question is: -Version 2: what are the inverse images $(\beta^*)^{-1}(c_k) \in H^{2k} (\Omega SU)$. I am looking for a formula that is explicit enough to show -that $\beta^*$ is surjective. -EDIT: one obvious answer could be that my sketch does not need the H-space structures and will fail for that reason. - -REPLY [18 votes]: There is a beautiful observation of my advisor John Moore that to my mind ought to be part of the focus of any such argument: the Hopf algebra $H_{\ast}(BU;Z)$, or equivalently $H^*(BU;Z)$ is self-dual. This is pure algebra and streamlines the presentation. My preferred presentation is in Section 21.6 of May and Ponto, More Concise Algebraic Topology, available at Amazon or University of Chicago Press. Incidentally, there is a proof of real Bott periodicity along the same lines due to Moore (and written up by Cartan if memory serves) in one of the 1950's Cartan seminars and another such proof by Dyer and Lashof. The funny thing is that each of these proofs finds one of the six equivalences one has to prove more difficult than all of the others, but they find difficulty with different ones: combining the easier of the respective arguments gives a clean and instructive homological proof.<|endoftext|> -TITLE: metric on the space of real analytic functions -QUESTION [12 upvotes]: Hello, -this question may be simple but I couldn't find a reference. -Let $E$,$F$ be real Banach spaces and $\Omega\subset E$ be a bounded domain and let $C_b^{\omega}(\Omega,F)$ be the vector space of bounded real analytic functions from $\Omega$ to $F$. Now I would like to know if there is a natural way to define a metric on $C_b^\omega(\Omega,F)$, that makes the space complete. -Concretely, I have a series of real analytic functions that converge uniformly as well as their Frechet derivatives and now I would like to know if their limit is analytic again. -My first idea was to show that $C_b^\omega(\Omega,F)$ is a closed subspace of $C_b^\infty(\Omega,F)$. Here $C_{b}^{\infty}(\Omega,F)$ is the set of infinitely Frechet-differentiable functions equipped with the usual set of seminorms (i.e. $\Vert f\Vert_k:=\Vert D^k f\Vert_\infty$) defining a Frechet space and thus a metric $d(f,g):=\sum_{k=1}^\infty 2^{-k}\frac{\Vert f-g\Vert_k}{1+\Vert f-g\Vert_k}$ which makes $C_{b}^{\infty}(\Omega,F)$ a complete metric space. -Actually, I have no idea if this is true at all. Could someone please confirm if this is the right thing to prove or not? -Regards, -Mirko - -REPLY [15 votes]: The problem is nontrivial already in the finite dimensional case $E= \mathbb R^d$, $F=\mathbb R$. The space $C^{\omega}(\Omega)$ of real-valued real analytic functions on the open bounded set $\Omega\subset \mathbb R^d$ does not have any obvious or natural metric which would make it a Fréchet space. -The good news is that there is a "canonical" topology which renders $C^{\omega}(\Omega)$ as a complete (reflexive nuclear separable) space. In fact, it is natural to endow $C^{\omega}(\Omega)$ with either an inductive limit or a projective limit topology but these two are equivalent on $C^{\omega}(\Omega)$ as was shown by Martineau in 1966. -For practical purposes, the topology can be described following the suggestion of Piero D'Ancona in his comment above. Let $\{U_j\}_{j\in\mathbb N}$ be a monotonically decreasing sequence of open sets of $\mathbb C^d$ such that $\Omega=\bigcap U_j$. Let $\{h_j\}_{j\in\mathbb N}$ be a sequence of bounded holomorphic functions $h_j:U_j\to\mathbb C$ such that $h_j|_{U_k}=h_k$ for $k\geq j$. Then a subbase element of the topology on $C^{\omega}(\Omega)$ has the form -$$\mathcal V_{j, K}=\left\{f\mbox{ is real analytic on }\Omega:\ \sup_{x\in K} \left|\partial^{\alpha} f\right|\leq C_j[\delta_j(K)]^{-|\alpha|}\ \mbox{ for every }\alpha\in\mathbb N^{d}_{0}\right\},$$ -where the set $K\subset\Omega$ is compact, $\delta_j(K)=\mbox{dist}\{K,\partial U_{j+1}\}$ and $C_j$ is a constant which depends on the supremum of $h_j$ on $U_{j+1}$. -A sketch of the construction in the finite dimensional setting can be found, for instance, in A Primer of Real Analytic Functions by Krantz and Parks. Hopefully, it generalizes to the case of Banach spaces in a straightforward way. -[EDIT. Concerning your specific question whether the limit of a sequence of real analytic functions is itself an analytic function. Let $f\in C^\infty(\mathbb T)$ be a periodic smooth but non-analytic function. Then the partial Fourier sums $S_N f$ converge to $f$ in the uniform metric with all their derivatives.]<|endoftext|> -TITLE: Projective resolution of modules over rings which are regular in codimension n -QUESTION [9 upvotes]: All rings are Noetherian and commutative, modules are finitely generated. -It is a theorem of Serre that over a regular ring $R$, every module has a finite projective resolution. -More generally, if $R$ is regular in codimension n, what can we say about projective resolution of modules over $R$? For example, is it true that every ideal with height less than n has a finite projective resolution? -Similarly, over a Noetherian seperated regular scheme $X$, every coherent sheaf has a finite resolution by vector bundles. The same questions can be asked for schemes as for rings. -Examples are extremely appreciated. Thanks! -Edit:It is not true that every ideal with height less than n has a finite projective resolution. As inkspot pointed out, if $R$ is normal, excellent, local and all height 1 ideals have finite projective dimension, then $R$ is factorial. So the local ring of a cone at origin gives a counterexample. -Since factorial is equivalent to $Cl(R)=0$ for $R$ normal, this makes me wonder for a local ring $R$ whether every ideal with height less than n has a finite projective resolution is equivalent to: - - -$R$ is regular in codimension n plus some other condition on the ring such as normal and excellent. -Some kind of "generalized divisor class group"(may be Chow group) vanishes. - - -If $R$ is not local, I think condition 2 should be replaced by something like: - -2'. Some part of $K_0(R)$ and $G_0(R)$ are isomorphic. - -where $K_0$ is the Grothendieck group of the category of projective modules over $R$, $G_0$ is the Grothendieck group of the category of finite generated modules over $R$. -Could above be true? - -REPLY [6 votes]: Dear Liu, -I like your updated question a lot. To make things easier to discuss, let me define the following properties for a Noetherian local ring $R$ and $n>0$: -($A_n$) every ideal with height less than $n$ has a finite projective resolution. -($B_n$) $R_P$ is regular for each prime $P$ of height at most $n$. -($C_n$) The Chow groups $CH^i(R)=0$ for codimensions $1\leq i \leq n$. -Your question asked whether $A_n$ implies $B_n$ and $C_n$. -It is not hard to see that $A_n$ implies $B_n$: localizing a resolution of $R/P$ over $R$ shows that the residue field of $R_P$ has finite projective dimension, which forces $R_P$ to be regular. -I do not know whether or not $A_n$ implies $C_n$, and I am probably not alone! As far as I known, it is a conjecture that Chow groups of codimensions at least one in any regular local ring are $0$. It is known if $R$ is essentially of finite type over a field (and perhaps a bit more generally, see this paper). -However, it is relatively easy to show that if $A_n$ holds, then $CH^i(R)$ is torsion for codimensions $1\leq i \leq n$. The brief reason is if an ideal $I$ has positive height and finite projective dimension, the class of $R/I$ is $0$ in the Grothendieck group $G_0(R)$, and this group is equal to the total Chow group after tensoring with $\mathbb Q$. -I think $A_n$ for even $n=2$ is a pretty strong condition. I would not be too surprised if it implies regularity of $R$. For example, it is not hard to show that $A_n$ implies $B_{n+1}$ if $R$ is Cohen-Macaulay and $n\geq 2$ (so if $\dim R=3$, $A_2$ forces $R$ to be regular). -UPDATE per the comments: If $R$ is not local, $A_n \Rightarrow B_n$ still holds. However, $C_n$ is hopeless. Just take any non-singular affine curve with non-trivial Picard group. Then any module still has finite projective dimension since the ring is regular. The trouble is that $A_n$ is a local condition, but $C_n$ is not.<|endoftext|> -TITLE: Software for rational homotopy theory -QUESTION [29 upvotes]: Does anybody know a software manipulating commutative differential graded algebras, and providing a computation of the minimal model? I tried to use the package DGAlgebras of Macaulay2, but I got several mistakes. For example the rational numbers are not allowed as ground ring, and the differential on a generator of a free CDGA is not allowed to be zero. Thanks - -REPLY [6 votes]: SageMath has a package for cdga's written by Miguel Marco and John Palmieri, with none of the restrictions mentioned: http://doc.sagemath.org/html/en/reference/algebras/sage/algebras/commutative_dga.html. -It includes a method to produce the minimal model of a cdga up to a prescribed finite degree.<|endoftext|> -TITLE: Elementary proof of algebraicity of Hecke eigenvalues in weight 1 -QUESTION [21 upvotes]: It's "well known" that, for any weight $k$ and level $N$, the space $S_k(\Gamma_1(N))$ of cusp forms of that weight and level has a basis in which all the Hecke operators act by matrices with entries in $\mathbb{Z}$; consequently all the Hecke eigenvalues are algebraic numbers (indeed algebraic integers). -I was reflecting on how to prove this while teaching an undergraduate course on modular forms. For $k \ge 2$ it's not hard: there's the Eichler-Shimura machinery which relates it to a question about cohomology, and the cohomology with $\mathbb{Z}$ coefficients does the job. Alternatively, and more or less equivalently, you use the pairing with modular symbols. Both of these methods break down for $k = 1$; the only argument I know that works in this case is to use the fact that $X_1(N)$ has a model as an algebraic variety, and weight $k$ modular forms correspond to sections of the $k$-th power of a line bundle that has a purely algebraic definition. But that's not really something I can stand up and explain to a class of undergraduate students! - -For cusp forms of weight $k = 1$, can the algebraicity of the Hecke eigenvalues be proved without quoting heavy machinery from arithmetic geometry? - -REPLY [14 votes]: Let $S = S_{\mathbf{Q}} = M_{13}(\Gamma_1(N),\mathbf{Q})$, and $S_{\mathbf{C}} = S \otimes \mathbf{C}$ denote the corresponding space of modular forms over $\mathbf{C}$. -Let $V \subset S \times S$ be the subspace cut out by pairs of forms $(A,B)$ satisfying the following equation: -$$A \cdot E_{12} = B \cdot \Delta $$ -As equations in the Fourier coefficients of $A$ and $B$ these are linear equations with coefficients in $\mathbf{Q}$. Since, by the $q$-expansion principle, a modular form can be recovered from some finite number of Fourier coefficients, $V$ is determined by the null space of some finite matrix with coefficients in $\mathbf{Q}$. Since a linear system over $\mathbf{Q}$ has the same rank over $\mathbf{C}$, it follows that $V_{\mathbf{C}} = V \otimes \mathbf{C}$, where $V_{\mathbf{C}}$ is the set of solutions in $S_{\mathbf{C}} \times S_{\mathbf{C}}$ of the same equations. -On the other hand, there is an isomorphism: -$$V_{\mathbf{C}} \rightarrow M_{1}(\Gamma_1(N),\mathbf{C})$$ -given by -$$(A,B) \mapsto \frac{A}{\Delta} = \frac{B}{E_{12}}$$ -The point is that $E_{12}$ and $\Delta$ do not have any common zeros, so the image of this map clearly consists of holomorphic forms. Hence the map is well defined. However, if $F$ has weight one, then $(A,B) = (F \cdot \Delta, F \cdot E_{12})$ maps to $F$, so the map is surjective. It is clearly injective, so it is an isomorphism. -It follows that the image of $V$ under this map gives a rational basis for $M_1(\Gamma_1(N))$. Since $V$ is then preserved by Hecke operators (as is obvious on $q$-expansions), the result follows.<|endoftext|> -TITLE: Which platonic solids can form a topological torus? -QUESTION [12 upvotes]: 8 cubes can be joined face-to-face to form a closed ring with a hole in it, with each cube sharing a face with only two others. The same can be done with 8 dodecahedrons. -Is the same possible with the other platonic solids? -What is the minimum number of each solid needed to form such a loop? -Given a convex regular d-polytope, is there a general way to determine if it can form a loop, by gluing their d-1 dimensional faces together? (Assuming the loop has a hole and no two objects intersecting, ie no two objects share a d-volume). -And is there a way to compute the minimum number of equal polytopes needed for this? - -Edit: added image -I am also looking for software that can be used to check for atleast small N - -REPLY [4 votes]: A long time ago a similar question arose in the sci.math newsgroup. I worked it out that in order to make a loop out of dodecahedra, one has to use sets of three dodecahedra in a row, as shown (four times) in the picture above -- otherwise one introduces a set of rotations in R^3 which has no relations which we can use to get the first and last faces to line up. See http://www.math-atlas.org/98/dodec_prf -dave<|endoftext|> -TITLE: Real algebraic sets bounded away from integer points -QUESTION [16 upvotes]: A subset $S$ of $\mathbb{R}^n$ is "bounded away from integer points" if for some positive $\epsilon$ every point in $S$ lies at a distance of at least $\epsilon$ from $\mathbb{Z}^n$. For example the line $x+y=1/2$ in $\mathbb{R}^2$ is bounded away from integer points, but the curve $x^2+y=1/2$ is not, because the points $(n+\frac{1}{4n},-n^2-\frac{1}{16n^2})$ for $n=1,2,\ldots$ lie on this curve. -Question: Can anyone give an algorithm to determine whether a system of polynomial equations with real algebraic coefficients cuts out a subset $S$ of $\mathbb{R}^n$ that is bounded away from integer points? Is there a simple description of all such subsets $S$? -Remark: I have a vague notion that if $S$ is bounded away from integer points then this must be ``trivially'' verifiable, perhaps because $S$ projects on a linear affine subset of $\mathbb{R}^n$ that is obviously bounded away from integer points, but this is little more than guesswork. I don't actually know that the problem is decidable, but I would be surprised if it were not. -UPDATE: (I'll use this section to collect my latest thoughts on the problem.) -The situation is fairly transparent in $\mathbb{R}^2$, and the real problem is how things generalize to higher dimensions. Let $S$ be as above. Let $\lfloor\cdot\rfloor$ be the floor function, which will be applied to points coordinatewise. Then I propose the following conjecture: -There exists some translate of $S$ bounded away from integer points if and only if the set of all points $\lfloor p\rfloor$ for $p\in S$ is contained in a finite union of linear-affine subspaces of $\mathbb{R}^n$ (which will be defined over the rationals). - -REPLY [2 votes]: This is not a really an answer, but you might want to look at the papers -MR1852803 (2002e:11085) -Vâjâitu, Marian(R-AOS); Zaharescu, Alexandru(R-AOS) -Integer points near hyperelliptic curves. (French summary) -C. R. Math. Acad. Sci. Soc. R. Can. 23 (2001), no. 3, 84–90. -11J25 (11D75 11G30) -and -MR1689552 (2001a:11117) -Huxley, M. N.(4-WALC-SM) -The integer points close to a curve. III. Number theory in progress, Vol. 2 (Zakopane-Kościelisko, 1997), 911–940, de Gruyter, Berlin, 1999. -11J54 (11P21) -The latter seems particularly relevant to your specific question.<|endoftext|> -TITLE: Which polynomials arise as formulas for a conjugate -QUESTION [15 upvotes]: For any integer $r \geq 2$, et $V_r$ be the set of polynomials $Q \in {\mathbb Q}[X]$ of degree $r-1$ such that there is an algebraic number $\alpha$ of degree $r$ , such that -$Q(\alpha)$ is a conjugate of $\alpha$. - It is not hard to see that $V_2$ consists exactly of $X$ and all the polynomials $a_0-X$, for $a_0\in {\mathbb Q}$. Have -the $V_r(r \geq 3)$ been studied? Is anything known about $V_3$ ? - -REPLY [19 votes]: $\newcommand\Z{\mathbf{Z}}$ -$\newcommand\Q{\mathbf{Q}}$ -If we replace $\alpha$ by $\alpha + \lambda$ for $\lambda \in \Q$ (translation), we may -replace $Q(X)$ by $Q(X - \lambda) + \lambda$. -Similarly, if we replace $\alpha$ by $\mu \cdot \alpha$ (dilation), we may replace $Q(X)$ by -$\mu \cdot Q(X/\mu)$. -Let's discuss the case $[\Q(\alpha):\Q] = 3$. -By translation, we may assume that the -coefficient of $X$ is trivial. By dilation, we may assume that -the leading coefficient is $1$ (this specifically -uses the fact that we are in degree $3$). -Hence $Q(X) = X^2 + c$ for some $c \in \Q$. -If a degree $3$ field $K = \Q(\alpha)$ contains at least two conjugates of -$\alpha$ then it contains -all the conjugates, and is therefore Galois with cyclic Galois group. -Thus $Q$ induces an isomorphism of $K$ of order three, and -$\alpha$ must be a root of -$Q(Q(Q(X))) - X$ but not a root of $Q(X) -X$. -Hence it is a root of $F(X) = (Q(Q(Q(X))) - X)/(Q(X) - X)$, which -is an explicit degree $6$ polynomial. -Suppose that $\beta = u + v \sqrt{d}$ is a root of $F(X)$, where $u$ and $v$ are rational and $d$ is -not a square. We may write $F(u + v \sqrt{d}) = R(u,v) + S(u,v) \sqrt{d}$, where -$R$ and $S$ are polynomials with coefficients in $\Q[a,b]$. - Since $F(\beta) = 0$, then -$R(u,v) = S(u,v) = 0$. Computing the resultant -of these polynomials with respect to $c$, we obtain the equation: -$$(1 + 2 u^2 + 4 u^2)(-1 - 18 u + 4u^2 + 8 u^3) v = 0.$$ -Since $u$ is rational, it follows that $v = 0$. Thus -$F(X)$ does not have any genuine quadratic solutions, -and hence $F(X)$ has -no factors that are quadratic. -(If we assume that $F(X)$ has a cubic factor, we can prove this in a slightly cleaner way. -Since $F(X)$ has a cubic factor, it has at most one quadratic factor. Yet $Q(\beta)$ and $Q(Q(\beta))$ are also quadratic roots of $F(X)$, and hence either $Q(\beta) = \beta$ or $Q(\beta) = \sigma \beta$. Both possibilities are impossible.) -We are assuming that $F(X)$ has a cubic factor corresponding to $\alpha$. -Then either $F(X)$ factors as a product of two cubics, or as a cubic times three linear factors. In either case, the cubics correspond to fields -which are Galois and hence cyclic of degree three. Hence the discriminant of $F(X)$ is a square (easy exercise). We compute explicitly that -$$\Delta_F = - (7 + 4 c)^3 (7 + 4 c + 16 c^2)^2,$$ -and hence we deduce the necessary relation: -$$7 + 4 c = - t^2$$ -for some $t \in \Q$. -Making the substitution $c = (-7 - t^2)/4$, -the polynomial $2^{6} \cdot F(X)$ factors -as $A(X,t) A(X,-t)$, where -$$A(X,t) = -1 + 7 t - t^2 + t^3 + 18 x - 4 t x + 2 t^2 x - 4 x^2 - 4 t x^2 - 8 x^3.$$ -As long as either this polynomial or its cousin -$A(X,-t)$ are irreducible, we obtain a cubic -with a root $\alpha$ such that $Q(\alpha)$ is a conjugate of $\alpha$. -The polynomial $A(X,t)$ is reducible in $t$ if and only if there exists a -rational point $A(X,t) = 0$. This turns out to be a rational curve, and -we deduce that it is reducible if and only if there exists a $u \in \Q$ -such that -$$t = \frac{1 + 2 u - u^2 - u^3}{u(u+1)}.$$ -On the other hand, $A(X,-t)$ is reducible if and only if -there exists a $v \in \Q$ such that -$$-t = \frac{1 + 2 v - v^2 - v^3}{v(v+1)}.$$ -Hence any forbidden $t$ corresponds to a solution to -the equation -$$\frac{1 + 2 u - u^2 - u^3}{u(u+1)} = - \frac{1 + 2 v - v^2 - v^3}{v(v+1)},$$ -which correspond to points on the curve -$$C:u + u^2 + v + 4 u v + u^2 v - u^3 v + v^2 + u v^2 - 2 u^2 v^2 - u^3 v^2 - u v^3 - u^2 v^3 = 0.$$ -This curve is smooth in the affine locus. -The corresponding projective curve has three points at $\infty$, and -two of the corresponding points are singular. -The singularities at these points are nodes (I think), -and thus using Plücker's formula, we deduce that -$C$ has genus: -$$ g = (d-1)(d-2)/2 - n = (4 \cdot 3)/2 - 2 =4.$$ -Thus $C$ has finitely many rational points (Faltings). -(Note: this calculation may have been wrong, but we won't actually use it.) -The curve $C$ has some obvious points at $\infty$ and when -$u(u+1)v(v+1) = 0$. -Make the substitution $u = x+y$ and $v = x-y$. Then -the equation becomes: -$$-x - 3 x^2 - x^3 + 2 x^4 + x^5 + y^2 + x y^2 - 2 x^2 y^2 - -2 x^3 y^2 + x y^4 = 0.$$ -This is a quadratic in $y^2$. Hence we obtain a degree two covering -$C \rightarrow E$, where $E$ is the curve -$$E:-x - 3 x^2 - x^3 + 2 x^4 + x^5 + z + x z - 2 x^2 z - -2 x^3 z + x z^2 = 0.$$ -Given a rational point on this curve, the discriminant -$\Delta$ is rational, and hence $E$ is isomorphic to: -$$\Delta^2 = 4(4 x^4 + 3 x^3 + x^2 + 2 x + 1) -= 4(1+x)(1+2x)(1-x+2x^2)$$ -This is birational to an elliptic curve which -turns out to have conductor $112$. Cremona's program - mwrank tells me has Mordell-Weil group -$\Z/2\Z$. Hence -the only rational points correspond to $x = -1/2$ and $x = -1$, -which pull back to the "obvious" rational points on $C$. Hence -we have determined $C(\Q)$ completely, and we see that -one of $A(X,t)$ or $A(X,-t)$ is always irreducible. -Hence we deduce: -If $\alpha$ is a cubic irrationality with $Q(\alpha) = \sigma \alpha \ne \alpha$, -then, replacing $\alpha$ by $\lambda \alpha + \mu$, -$Q(X)$ is of the form: -$$Q(X) = X^2 - \frac{7+t^2}{4}$$ -with $t \in \Q$. -Conversely, for any such $Q(X)$, there exists - a cubic irrationality $\alpha$ -such that $\sigma \alpha = Q(\alpha)$. -Example: $t = 1$, and $Q(X) = X^2 - 2$. Then $\alpha = 2 \cos(2 \pi/7)$. -For higher degree polynomials, things will be even more of a mess, because -there will be various possibilities corresponding to what the Galois closure is, -&. &. The answer will have the following flavor: It will correspond -to the rational points on a bunch of varieties minus - rational points on subvarieties corresponding to -degeneracies (which in this case turn out to be empty). - -REPLY [3 votes]: If $f(x)$ is in $V_r$, and $g(x)$ is the minimal polynomial of a corresponding element $\alpha$, -then $f(\alpha)$ is also a root of $g(x)$, so $g(f(\alpha))=0$; hence $g(x)$ divides $g(f(x))$. -Thus, $V_r$ consists of the polynomials $f(x)\in\mathbb{Q}[x]$ of degree $r-1$ for which -there is an irreducible degree-$r$ polynomial $g(x)\in\mathbb{Q}[x]$ satisfying -$g(x)\mid g(f(x))$. This perspective explains the result for $r=2$: the map $\beta\mapsto f(\beta)$ -must permute the two roots of $g(x)$, and the only linear polynomials which permute two numbers are $x$ and $a_0-x$ (this doesn't quite show $a_0-x\in V_2$, one has to finish as Victor did).<|endoftext|> -TITLE: Alexander duality in terms of morse function? -QUESTION [7 upvotes]: As well known, we can prove Poincare duality in terms of morse theory. -(By comparing two chain complexes obtained from two morse functions, $f\colon M\to \mathbb{R}$,$-f\colon M\to \mathbb{R}$ for smooth closed manifold $M$) -Of course, I can prove Alexander duality theorem by using usual Lefschetz-Poincare duality theorem and some excision arguments. -But, I want to prove Alexander duality theorem by using a morse function defined on $S^{n}-X$, for a submanifold $X$ of $S^n$. -Are there any references or proofs about this? - -REPLY [11 votes]: There are various versions of stratified Morse theory out there, the granddaddy being the Goresky-Macpherson work. But for these purposes you're fine just using Morse theory on manifolds with boundary, which is a straightforward generalization of plain old Morse theory. In this setting you get Alexander duality "seen" at the chain-complex level in some sense, kind of analogous to Poincare's proof using dual cell decompositions to a triangulation, of the Morse theory/handle proof you refer to. -It's a little different though. The idea is to take the standard height function on $S^n$, and to perturb it so that it restricts to a Morse function on $M \subset S^n$, and on the boundary of a smooth tubular neighbourhood to $M$. Critical points to the height function, restricted to $M$ split into two kinds -- the kinds that contribute cells to the decomposition of the tubular neighbourhood, and the kinds that contribute to the decomposition of the complement. So a $k$-cell for $M$ corresponds to a $(n-k-1)$-cell for the complement. And so on. -I can say more later. -edit: The key idea is that if a function $f : S^n \to \mathbb R$ is morse on a submanifold $M \subset S^n$, then on the boundary of a small tubular neighbourhood of $M$, it's also Morse. Moreover, critical points of $f$ on $M$ correspond to opposite pairs of critical points on the boundary of the tubular neighbourhood (one over the other, in pairs). Think of Milnor's Morse Theory example of a torus in $\mathbb R^3$, but now think of the torus as the boundary of a tubular neighbourhood of a circle. In stratified Morse theory, the critical points of $f$ on the boundary of the tubular neighbourhood break up into two types of cell attachments: (1) cells for the tubular neighbourood, and (2) cells for the complement. The way you can tell (1) and (2) apart, is that the "down" direction pointing outwards for which ever object the cell is contributing to. This is because in stratified Morse theory you modify the flow so that it stays in your manifold with boundary. If the vector field is orthogonal to the boundary and pointing outwards, there's no natural way to modify the flow, so you get a cell. On the other hand if it points inwards, you certainly can. -So if $M \subset S^n$ is a closed submanifold, let $\nu M$ be a compact tubular neighbourhood of $M$. Let $p \in M$ be a critical point of $f$ that contributes a cell of dimension $i$ to $M$ using the downwards flowline. There are two critical points to $f$ above in $\pi^{-1}(p)$ where $\pi : \partial \nu M \to M$ is bundle projection. The "top" critical point contributes an $(n-m-1+i)$-cell to $\partial \nu M$, also to $\overline{S^n \setminus \nu M}$. The "bottom" point contributes an $i$-cell to $\partial \nu M$, also to $\nu M$. -So if for the purpose of generating $M$'s CW-decomposition we use the upward flowlines, then at a critical point corresponding to an $m-i$ cell on $M$, we have an $n-1-(m-i)$ cell attachment for $\overline{S^n \setminus \nu M}$ using the downward flowlines for $f$. These cells are not only complementary dimensional in the Alexander duality sense, but you can write out very explicitly the bounding cells that intersect in a single point. -I hope that helps. It's best to draw a picture and keep track of the dimensions in a few examples before trying the general picture.<|endoftext|> -TITLE: The identity $\sum_n \ln(n) x^n = \sum_p ln(p)(\sum_k\frac{x^{p^k}}{1-x^{p^k}})$ -QUESTION [6 upvotes]: As in the famous Euler product identity, the primes occur on -only one side of the following: -$\sum_n \ln(n) x^n = \sum_p ln(p)(\sum_k\frac{x^{p^k}}{1-x^{p^k}})\ .$ -My basic question: Does this identity appear in the literature? -If not, does the function $\sum_n \ln(n) x^n$? (It seems -distantly related to polylogarithms.) Does it extend beyond the unit disk? -(Computation suggests that behaves quit calmly up to the unit disk - I haven't -detected visible evidence of a natural boundary.) -Is my(?) identity somehow equivalent to the Euler product identity? -Is here some obvious reason why it shouldn't be useful for studying -the distribution of primes? (For a start it, does prove the infinitude of primes - -if there were only finitely many primes, the coefficients on the left would -have a bounded average.) - -REPLY [10 votes]: For every two series $\{a _i\} _{i=1}^{\infty}$ and $\{b _i\} _{i=1}^{\infty}$, which satisfy $a _{n}=\sum _{k|n}b _k$ one has -$$\sum _{n\geq 1}a _n x^n=\sum_ {n\geq 1} b _n \frac{x^n}{1-x^n}$$ -this is known as "Lambert transformation". Therefore your identity is equivalent to the familiar statement -$$\log(n)=\sum_{k|n}\Lambda(n)$$ -with the von Mangoldt function. This identity is indeed equivalent to unique factorization and therefore to the Euler product identity. As to the literature, this identity is used in the beginning of the proof of the prime number thorem by Delange (1955), for example.<|endoftext|> -TITLE: Counting isomorphism classes of elliptic curves with specific torsion -QUESTION [9 upvotes]: Generally speaking, I am interested in counting the number of $\mathbb{F}_p$- isomorphism classes of elliptic curves containing a specific torsion subgroup, and I was wondering if there were any simple formulas for particular torsion subgroups. Any information on this topic is appreciated. -More specifically, I would like to count the number of $\mathbb{F_p}$-isomorphism classes of elliptic curves whose torsion subgroup contains $\mathbb{Z}/N \times \mathbb{Z}/N$, where $N$ is a small fixed integer, for instance one for which the modular curve $X(N)$ has genus 0. -I know that if you fix a prime $p$, and fix an isogeny class of such curves over $\mathbb{F}_p$, then Schoof (in "Nonsingular plane cubic curves over finite fields") has formulas which express the number of isomorphism classes of these curves in the given isogeny class in terms of class numbers of certain quadratic extensions of the rationals. So, for example, summing these expressions over all possible isogeny classes is an answer to the question. However, I don't know of any easy way to compute this sum of class numbers, and it seems that such a sum could potentially be expressed more simply. -Also, Schoof's result holds for all $N$, and I thought it might be possible for smaller $N>1$'s that things simplified a bit. -Thanks for the help! - -REPLY [8 votes]: To add to Aşağı Güzdək's nice answer, there is something you can do also when the genus is greater than zero which is also in some sense an exact formula. Namely, if $\Gamma$ is one of $\Gamma_0(N), \Gamma_1(N)$ or $\Gamma(N)$ [or probably some other congruence subgroups as well, but I am not familiar with this story], and $Y_\Gamma$ (resp. $X_\Gamma$) is the corresponding open (resp. compact) modular curve, then one has -$$ |Y_\Gamma(\mathbf{F}_p)| = p+1-\mathrm{Tr}(\mathrm{Frob}_p|E_2(\Gamma))-\mathrm{Tr}(\mathrm{Frob}_p|S_2(\Gamma)),$$ -where $E_k(\Gamma)$ (resp. $S_k(\Gamma)$) is the Galois representation attached to Eisenstein series (resp. cusp forms) of weight $k$ for $\Gamma$. The Galois representation $E_2(\Gamma)$ is given by the $\ell$-adic cohomology of $X^\infty_\Gamma = X_\Gamma \setminus Y_\Gamma$, hence the appearance of the number of cusps in the previous answer. In general one can for a given $\Gamma$ and p figure out which of the cusps are defined over what fields by looking explicitly at the modular interpretation of $X_\Gamma^\infty$ described in the paper by Deligne & Rapoport. -The trace of Frobenius on the space of cusp forms is the really interesting one. In many cases this can be computed for a given $\Gamma$ and $p$ by looking in tables like the Modular Forms Database of William Stein, or asking SAGE to spit them out for you (see e.g. http://www.sagemath.org/doc/reference/sage/modular/modform/cuspidal_submodule.html) -- for a prime p, the trace of $\mathrm{Frob_p}$ coincides with the trace of the Hecke operator $T_p$, which can be read off from the pth Fourier coefficients of a basis of normalized eigenforms for the space of cusp forms you are looking at. -It seems the only built-in functionality in SAGE is for the groups $\Gamma_0(N)$ and $\Gamma_1(N)$. But perhaps one can still do something for small N, e.g. using that conjugation by the matrix with entries $(0,-1;N,0)$ takes $\Gamma(N)$ to $\Gamma_0(N^2) \cap \Gamma_1(N)$ -- this tells you how to get cusp forms for $\Gamma(N)$ both from $\Gamma_0(N^2)$ and $\Gamma_1(N)$, and by the genus formula for $\Gamma(N)$ you can tell if this has produced all of them.<|endoftext|> -TITLE: Dense orbits in billiards -QUESTION [5 upvotes]: This should be true in a more general setting, but for simplicity consider billiards that are connected, compact subsets of the plane with boundary $C^2$ except at finitely many points. A ball (or a ray of light) rolls inside, going in straight lines, and upon collision with the wall, the orbit is reflected. -It is intuitive that a statement like the following is true: - -For almost every billiard, there exists an orbit that is dense everywhere inside it. - -However, as far as I know this is still open. In fact, the last thing I heard was that it had just been proven for the case in which the billiard was an obtuse triangle with certain restrictions (but I have since forgotten the source, unfortunately). - -Question: What is the current status of the problem? - -Thank you! -Clarification The question is not about rectangular billiard tables, but in general about the balls rolling in more general shapes. 'Almost always' would then have to be given a meaning within the space of curves. (In fact, the problem is trivial in rectangles because an orbit with irrational slope will do.) Also, this is not about having dense families of orbits, but a single orbit that is dense in the billiard. -I think the way 'almost always' should be defined is by requiring some generic property to hold. Think, for example, of the definition R. Abraham give of bumpy metrics. - -REPLY [3 votes]: It is possible for a billiard path to be dense and yet not be uniformly distributed. This is true for the triangle with angles $0.4\pi,0.3\pi,0.3\pi$. (See, for example, theorem 1.3 of http://homepages.math.uic.edu/~demarco/billiards.pdf.)<|endoftext|> -TITLE: Questions about spectra of rings of continuous functions -QUESTION [16 upvotes]: I have been thinking a bit about rings of continuous functions of various kinds -- how they motivate the more modern notion of the Zariski topology on the prime spectrum as well as how they fit into a more general picture. (The commutative algebra course that I had mentioned in an earlier question has recently begun.) Here are two constructions which I have been thinking a lot about: -For $X$ an arbitrary topological space, let $C(X)$ be the ring of continuous $\mathbb{R}$-valued functions on $X$. One has a canonical map $\mathcal{M}: X \rightarrow \operatorname{MaxSpec} C(X)$ by taking $x$ to the maximal ideal $\mathfrak{m}_x$ of all functions vanishing at $x$. This map is continuous when the -codomain is given the Zariski topology (e.g. as a subspace of the Zariski topology on the prime spectrum: but really, just take the same definition and restrict to maximal ideals.) -When $X$ is compact (by which I mean quasi-compact and Hausdorff) I proved in class that $\mathcal{M}$ is a homeomorphism. This was not as graceful as I might have hoped: I found myself having to introduce an auxiliary topology on the maximal spectrum -- the "initial", "weak" or "Gelfand" topology -- to see that it was Hausdorff and then only after having shown that $\mathcal{M}$ is a homeomorphism with the Gelfand topology did I deduce that the Gelfand topology coincides with the Zariski topology (using the characteristic property of Tychonoff spaces that any closed set is an intersection of zero sets of continuous functions). -I mentioned that the following are true in the general case: -(i) $\operatorname{MaxSpec} C(X)$ is compact in the Zariski topology. Therefore $X_T := \mathcal{M}(X)$, endowed with the subspace topology is Tychonoff. -(ii) In fact $\mathcal{M}: X \rightarrow X_T$ is the universal Tychonoff space on $X$ and the induced map $C(X_T) \rightarrow C(X)$ is an isomorphism of rings. (So we may as well assume $X$ is Tychonoff.) -(iii) For any Tychonoff space $X$, $\mathcal{M}: X \rightarrow \operatorname{MaxSpec} C(X)$ is nothing else than the Stone-Cech compactification. -I am still looking for a nice, self-contained reference for these facts. Gillman and Jerison's classic text has most of them, but spread out over a fairly large number of pages. For instance, can it be so hard to see that $\operatorname{MaxSpec} C(X)$ is Hausdorff no matter what $X$ is? I am struggling even with that! -The upshot here is that taking the ring of $\mathbb{R}$-valued functions and then the maximal spectrum has the effect of passing from an arbitrary space $X$ to its universal compact space. I find this very interesting. -Now consider $C_2(X)$, the ring of continuous functions from $X$ to $\mathbb{F}_2$ (the latter endowed with the discrete topology: what else?). It now seems that the above discussion goes through with "universal compact space" replaced by "universal Boolean (= compact and totally disconnected) space", and that this is a slightly different (better?) take on Stone duality than the one I wrote up in my notes. -But actually in between $\mathbb{R}$ and $\mathbb{F}_2$ I did something that is maybe silly: I threw off a remark to my students that it seemed interesting to think about the case of $\mathbb{Q}_p$-valued functions. But (without having written anything down, so I could well be mistaken) I now think that although the rings one gets in this way are of course not Boolean, their spectra still comprise precisely the Boolean spaces...and that maybe the same holds for functions with values in any totally disconnected topological field. Is this really the case? -The above is pretty rambly, so let me end with one crisp question which is at least part of what's eating me: - -Can one characterize the commutative rings with Hausdorff maximal spectrum? Or with Boolean maximal spectrum? Or with maximal spectrum some other interesting class of compact spaces? - -REPLY [8 votes]: For a commutative ring $A$, $MaxSpec(A)$ is Hausdorff if and only if $A/JacobsonRadical(A)$ is a Gelfand ring (i.e. all equations of the form $(1-xb)\cdot (1-y(1-b))=0$, with $b$ from the ring, are solvable in that ring). $MaxSpec(A)$ is boolean if and only if every element of $A/JacobsonRadical(A)$ is a sum of a unit and an idempotent. -The Jacobson radical of $A$ is the intersection of all maximal ideals from $A$. -Both characterisations can be found in -N. Schwartz, M. Tressl; Elementary properties of minimal and maximal points in Zariski spectra. Journal of Algebra 323 (2010) 698-728. -This paper also characterises further properties of $MaxSpec(A)$ in terms of algebraic properties of the ring (see section 11 for an overview).<|endoftext|> -TITLE: Logical equivalences for FTA -QUESTION [5 upvotes]: I hope this isn't a stupid question... -It's well known that (in the presence of various other axioms), Euclid's Postulate 5 ('parallel axiom') is equivalent to the Pythagorean Theorem. That is, assume Pyth. Thm. is true without Postulate 5, and you get the 'parallel axiom' as a theorem. -My question: Are there well-known (or not-so-well-known) theorems/properties of the ring of integers which are equivalent to the Fundamental Theorem of Arithmetic in this way? That is, things which are not just consequences of it, but imply it. -I have had a lot of trouble finding anything about this on the Net, but of course the words involved are not exactly unique! Please be gentle if there is something obvious I'm missing - I've put "elementary" as a tag by way of anticipating there is a clear answer. At least the statement is elementary! -Edit: I like all three answers for different reasons, and have voted up accordingly. None really answers my question, but that's because, upon further review, I think it's not well posed. After all, FTA is not an axiom like Postulate 5 (though of course one needs various axioms to prove it). -So maybe the answer about $a|bc$ is closest to what I was looking for, though as it happens I like to prove this first as well. Probably the best question would be how much one can prove in number theory without using the FTA. But that would be a different question! Thanks. - -REPLY [7 votes]: Pete Clark has an expository paper, Factorization in integral domains, that is relevant to your question. In the particular case of FTA, I think it's more fruitful to ask for conditions on a commutative ring that imply (or are equivalent to) unique factorization, rather than to ask the reverse-mathematical question of the logical strength of FTA.<|endoftext|> -TITLE: Conformal-symplectic geometry ? -QUESTION [6 upvotes]: I think in priciple it's possible to consider a theory of "conformal-symplectic manifolds", in an analogous fashion as the usual conformal geometry. -To spell out the spontaneous definitions: say that two symplectic forms $\omega_1$, $\omega_2$ on a smooth manifold $M$ are conformal to each other if there is a smooth positive function $\lambda \in \mathcal{C}^{\infty}(M,\mathbb{R}^{+})$ such that $\omega_1=\lambda\cdot \omega_2$ on $M$. Call a pair $(M,[\omega])$, with $[\omega]$ a conformal class of symplectic structures, a conformal-symplectic manifold. A smooth map $\varphi : M \to N$ between conformal-symplectic manifolds $(M,[\omega_1])$ and $(N,[\omega_2])$ is conformal-symplectic if $\varphi^*(\omega_2)\in [\omega_1 ]$. -Just out of curiosity, I would like to ask: - -Has such a theory been considered or studied? What can be said about these structures (provided it doesn't turn out to be somehow a "trivial" subject)? - -REPLY [6 votes]: There is a notion of conformal symplectic structure related to what you are asking. I refer to locally conformally symplectic manifolds. These are manifolds $M$ equipped with a non-degenerate two-form $\omega$ and a good open cover $\left\{ U_{a}\right\}_{a\in I}$ such that for every $U_{a}$ there exists a function $e^{f_{a}}\in C^{\infty}(U_{a})$ satisfying -$d\left( e^{f_{a}}\omega|_{U_{a}}\right)=0$ -This is equivalent to the existence of a flat real line bundle $L\to M$ with connection $\nabla$ that descends to a well-defined closed one-form $\varphi$ in $M$ satisfying -$d\omega + \varphi\wedge\omega =0$ -One can define the coboundary operator $d_{\varphi} = d +\varphi$ on the complex of forms $\Omega^{\bullet}(M)$, whose cohomology is the so-called Lichnerowicz cohomology, which in general is not equivalent to the standard de Rahm cohomology. The two-form $\omega$ satisfies $d_{\varphi}\omega = 0$ and it is thus a cocycle. For further information you can check Izu Vaisman's papers from the 70's and 80's on locally conformally symplectic and K\"ahler manifolds.<|endoftext|> -TITLE: Anti-concentration of Bernoulli sums -QUESTION [19 upvotes]: Let $a_1,\ldots,a_n$ be real numbers such that $\sum_i a_i^2 =1$ and let $X_1,\ldots,X_n$ be independent, uniformly distributed, Bernoulli $\pm 1$ random variables. Define the random variable -$S:= \sum_i X_i a_i $ -Are there absolute constants $\epsilon >0$ and $\delta<1$ such that for every $a_1,\ldots,a_n$ -${\mathbb P} [|S| \leq \epsilon ] \leq \delta$ ? - -REPLY [3 votes]: Your question is part of what's called Littlewood-Offord theory, which has seen a lot of progress lately in work of Tao and Vu and of Rudelson and Vershynin. Take a look at Section 1.2 and especially Theorem 1.5 of this paper by Rudelson and Vershynin for more precise results than in George's answer. (Incidentally, that paper also contains arguments based on the central limit theorem, in the Berry-Esseen form, along the lines of what Igor suggests.)<|endoftext|> -TITLE: Why is it OK to rely on the Fundamental Theorem of Arithmetic when using Gödel numbering? -QUESTION [12 upvotes]: Gödel's original proof of the First Incompleteness theorem relies on Gödel numbering. -Now, the use of Gödel numbering relies on the fact that the Fundamental Theorem of Arithmetic is true and thus the prime factorization of a number is unique and thus we can encode and decode any expression in Peano Arithmetic using natural numbers. -My question is, how can we use a non-trivial result like the Fundamental Theorem of Arithmetic in the meta-language that describes Peano Arithmetic, when, the result actually requires a proof from within Peano Arithmetic itself, not like other trivial things we believe to be true (i.e. existence of natural numbers and the axioms for addition and multiplication which we want to interpret in the natural way - Platonism)? -I understand we could instead use a different way of enumeration, e.g. a pairing function and the Chinese Remainder Theorem or simply string concatenation, but, then the need for a proof of uniqueness when encoding and decoding remains and in general, I am interested in the structure of Gödel's original proof. -Basically, I have two ideas of how it might be possible to resolve this: - -Prove the Fundamental Theorem of -Arithmetic within a different sound -(?) system. -Maybe there is nothing needed to be -done to 'resolve' this, because I am -just misinterpreting something and -it is actually acceptable to use provable -sentences of PA in the -meta-language. - -Any ideas about this? -EDIT: -I have realized how to make my question less confusing: -Say PA proves FTA. Then if we only assume PA is consistent, that does not rule out the possibility of FTA being false. Now, if FTA is false, then PA and the meta-language too includes a false statement and thus the whole proof is useless. -How is this resolved? Is it maybe related to the fact the originally we actually assume $\omega$-consistency and obviously, for each natural number $n$ separately its unique prime factorization can simply be found algorithmically? - -REPLY [2 votes]: Here is a way to think about it. When you use something like Fund Theorem of Arithmetic to prove Godel numbering, you are referring to the abstract set $\mathbf N$, which "exists". The Fundamental Theorem of Arithmetic is simply true about this set, irrespective of any logical system which one is constructing. If you like, you can prove the Fundamental Theorem of Arithmetic, more or less rigorously, in a metalanguage. But the proof in this metalanguage is completely orthogonal to any proofs one might construct in the logical system, such as Peano axioms.<|endoftext|> -TITLE: explicit linear representations of fundamental groups of surfaces -QUESTION [5 upvotes]: I am looking for an explicit representation of the fundamental group of a closed orientable surface of genus >1. I guess they should be abundant in degree 2. Did anyone see the explicit matrix construction of such a representation? Are there any integral ones? Maybe in higher degrees? - -REPLY [2 votes]: Here are the examples from Narasimhan and Seshadri that I mentioned in my comment above. (I could have the reference wrong; I'm actually taking this out of notes from a talk I gave many years ago.) -These examples are really not that exciting, but at least they give irreducible unitary representations of each degree. Writing the generators of $\pi_1 M^g$ as $a_i$ and $b_i$ ($i=1, \ldots, g$) we define a representation $\rho$ by sending $a_1$ to the diagonal matrix -$$A_1 = \left[\begin{array}{rrr} z_1 \\ & \ddots\\ && z_n \end{array}\right],$$ -where the $z_i$ are distinct, and sending $a_2$ to the permutation matrix -$$A_2 = \left[\begin{array}{rrrrrr} 0 & 0 & \cdots & 0 &1 \\ 1 & 0 & \cdots && 0 \\ -0 & \ddots & 0 & \cdots & 0\\ - 0 & \cdots & 1& 0 & 0\\ - 0 & \cdots & 0 & 1 & 0\end{array}\right].$$ All the other generators are sent to the identity matrix. Since the $b_i$ all go to the identity, all the commutators $[\rho(a_i), \rho(b_i)]$ are trivial and we have a representation. It's irreducible because the invariant subspaces of $A_1$ are just sums of eigenspaces, and these eigenspaces are permuted transitively by $A_2$. -One can vary this example in a number of ways to get other interesting examples.<|endoftext|> -TITLE: Galois action on one-dimensional quotients of l-adic cohomology -QUESTION [12 upvotes]: Let $A$ be an abelian variety of dimension $g$ over a number field $K$, and $\ell$ be a rational prime. Suppose that the Galois action on the $\ell$-adic cohomology $H^k(A, \mathbb{Z}_\ell) \otimes_{\mathbb{Z}_\ell} \overline{\mathbb{Q}_\ell}$ has a one-dimensional Jordan-Holder quotient. Is there a (conjectural or known) classification of the possible characters $\psi : Gal(K^{ab} / K) \to \overline{\mathbb{Q}_\ell}^*$ which can give the action on this quotient (in terms of $K$, $g$, and $\ell$)? -(When $g = 1$, there is a simple explicit description: For $H^0$ and $H^2$, it must always be the trivial and cyclotomic character respectively. For $H^1$, there can only be a one-dimensional Jordan-Holder quotient if the elliptic curve has CM, in which case there is a well-known explicit description. This follows from Serre's open image theorem, since any reducible subgroup of $GL_2(\mathbb{Z}_\ell)$ has infinite index.) -Even if there is not a good description in full generality, are there any interesting classes of examples (besides those examples coming from abelian varieties of CM-type) which admit an explicit description? -EDIT: I would like to know not just whether a particular $\psi$ can occur for some abelian variety over some number field (which is answered below by Aşağı Güzdək), but rather for which pairs $(K, g)$ it can occur for an abelian variety of dimension $g$ defined over the number field $K$. - -REPLY [9 votes]: Any character $\psi$ arising in this way will be de Rham. The only such characters arise (up to twist: EDIT by finite order characters) from Grössencharacters from some CM field. By purity, one may determine the infinity type of $\psi$. By the Tate conjecture (proved by Faltings), any such quotient which arises in $H^1$ of an abelian variety $A$ is induced from a morphism $A \rightarrow X$ where $X$ is (the Weil restriction of) a CM abelian variety.<|endoftext|> -TITLE: Are there functions satisfying the following integral condition? -QUESTION [5 upvotes]: Can we find two functions $f$ and $g$ that are reasonably defined nontrivial(not everywhere zero, $f\neq g$, not linear polynomials) functions such that the following condition is satisfied? -$$ f( \left(\int_{0}^{t} g(x) \ \text{d}x\right)) = g( \left(\int_{0}^{t} f(x) \ \text{d}x\right)) $$ -P.S.: I migrated this question from here on Math.SE. I am sure this site hosts very knowledgeable mathematicians that keeping on migrating to another site is foolish. I felt a really good feeling for some time as nobody answered my question. But is usually the case that: "There is a general principle that a stupid man can ask such questions to which one hundred wise men would not be able to answer. In accordance with this principle I shall formulate some problems." Vladimir Arnold -Motivation: The equation that I wrote out was not random. At least, the symmetry I find in it and the absence of an iota of clue at proceeding with any method makes me fall in love with finding a solution. Part of the motivation was to find a function that in some way resembles the exponential function. The exponential map is invariant under differentiation. So, the natural curiosity to find a nontrivial map invariant under integration. For obvious reasons, such map does not exist because of the presence of the constant of integration in indefinite integrals. Hence, I added an extra condition that would make the would-be function more nontrivial and more appealing. - -REPLY [4 votes]: Following Igor's comment(or his hunch), I started out with finding polynomial counterexamples of the type: $f(x)= ax^n $ and $ g(x)= bx^m $ on $(0,\infty)$. It is easy to see that such polynomials satisfy the integral condition only if $m=n$, and $a=b$ if $n$ is even or $a=\pm b$ if $n$ is odd. Then a class of counterexamples on $(0,\infty)$ can be constructed by letting $a= -b$ with odd $n\geq3$ .<|endoftext|> -TITLE: Is there a way to graphically imagine smash product of two topological spaces? -QUESTION [16 upvotes]: Recently I've been reading "Topology" by Klaus Janich. I find this book very entertaining as it contains lots of graphical illustrations that appeal to my "geometrical" imagination. In paragraph 3.6 Janich gives nice illustrations of concepts such as "cone over a set" or "suspension". In the same paragraph he defines a smash product of two topological spaces. Yet, in the version of a book that I posses, there is no image that would present this concept. My question is therefore as in the title: - -Is there a way to graphically imagine smash product of two topological spaces? - -I'm not sure whether this question is suitable for MO. Perhaps I should put it into a comunity-wiki mode? - -REPLY [18 votes]: To eloborate on what Arnav Tripathy said in a comment: if $X$ and $Y$ are compact, then $X \wedge Y$ is the one-point compactification of $(X \setminus \{x_0\}) \times (Y \setminus \{y_0\})$.<|endoftext|> -TITLE: Axiom of Choice in a weaker system -QUESTION [10 upvotes]: Is it known whether or not there is a consistent system of logic where two or all of the axiom of choice, well-ordering principle, and Zorn's lemma have no (known) proof of equivalence? -I was thinking about the old adage "The Axiom of Choice is clearly true, the well-ordering principle is clearly false, and nobody really knows about Zorn's lemma" and the content of Proof that pi is transcendental that doesn't use the infinitude of primes, particularly François G. Dorais' beautiful answer, where he explains the existence of systems where it isn't known whether infinite primes is necessary for the transcendence of $\pi$, and it occurred to my rather logic-ignorant self that perhaps a sufficiently weak system might allow for this adage to be true, at least in the form $X$ is (clearly or otherwise) true and $Y$ is (again, clearly or otherwise) false for suitable $X$ and $Y$ from the above list. I think it would be particularly poetic to find one in which $X$ is the Axiom of Choice and $Y$ is the well-ordering principle. - -REPLY [7 votes]: A small quibble: I would view these principles as primarily set-theoretic principles, not logical principles per se. It's actually a little bit tricky, because even if we choose to weaken the metalogic (that is, the logical principles we allow ourselves to apply to a set theory, from the outside as it were), one can also speak of the internal logic of a category of sets, which can be stronger than the metalogic. This fact is definitely relevant to the question here. -More exactly, even if the external logic applied to ZF is intuitionistic first-order logic, assuming the axiom of choice will force the law of excluded middle to hold internally. What does this mean, exactly? It means that for every subset $A \subseteq X$, we can, with the help of the axiom of choice, construct a complement, i.e., a subset $B \subseteq X$ such that $A \cap B = \emptyset$ and $A \cup B = X$, so that the lattice of subobjects of a set is a Boolean algebra. This result is due to Diaconescu, who was an early pioneer in topos theory. See the nLab article on excluded middle for a discussion. -I bring this up because proofs of the the three-way equivalence between AC, Zorn, and well-ordering use excluded middle, but if we assume AC, this is no obstacle to being able to prove Zorn's lemma. But not the other way around. I believe the situation is this: - -The proof of "well-ordering" implies "axiom of choice" is perfectly constructive. If we formulate AC as saying that every surjection $p: A \to B$ has a section $s$, then we can construct a section if $A$ is well-ordered: for each fiber $b \in B$, define $s(b)$ to be the least element of $p^{-1}(b)$. -Under intuitionistic logic, the axiom of choice implies internal excluded middle, and the usual proof that AC implies Zorn will then carry over. -Under intuitionistic logic, Zorn's lemma plus internal excluded middle can be used to construct a well-ordering on any set $X$. (The usual proof carries over: if a maximal well-orderable subset is not the whole set, we could extend the well-ordering in contradiction to maximality.) It follows from the previous bullet point that AC proves the well-ordering principle in intuitionistic logic. But (I believe -- I should check this carefully): under intuitionistic logic, Zorn's lemma by itself does not prove the (internal) law of excluded middle, and does not imply the well-ordering principle, hence cannot imply AC either. - -There is some material on this in the nLab. There may be other answers to this question which devolve on what one takes the underlying set theory to be, but the vanilla approach taken here, which I guess is in line with the question, is to take ZF but weaken the metalogic to get some form of IZF (although what I really had in mind was just to use topos theory as the baseline "set theory").<|endoftext|> -TITLE: blow-up proper varieties to projective ones -QUESTION [11 upvotes]: Let $X$ be a proper smooth algebraic variety (or algebraic space) over an algebraically closed field of characteristic $p.$ By Chow's lemma, there exists a projective variety $Z$ and a projective birational morphism $Z\to X.$ Of course $Z$ may not be smooth anymore. -My question is, -Could we make $X$ projective by a finite number of blow-ups? -In the standard proof of Chow's lemma, no blow-up is used. Both Chow's lemma and resolution of singularities apply to much more general situation, and I wonder if blow-ups is enough in this special case (that $X$ proper smooth). Since it would be a little optimal to require that the blow-up $\widetilde{X}$ is both projective and smooth (which would give the resolution of singularities in char. $p$ in this special case), I don't care if the subvarieties in $X$ which we blow-up are smooth or not. And the answer is yes (in a stronger way that $\widetilde{X}$ is also smooth) in characteristic 0, due to Moishezon. - -REPLY [3 votes]: So here's what I think should work (it is based on my comments above). I'm going to assume for simplicity that the variety means irreducible (just to avoid some silly complications, it is not really needed). -Start with $f : Z \to X$ projective and birational (as discussed above, $Z$ is contructed via Chow's Lemma). We don't know that $X$ is quasi-projective, so we can't apply Hartshorne II, 7.11 and thus immediately argue that $Z$ is itself a blow-up. -Set $U_i$ to be an open affine (or even quasi-projective) cover of $X$ and fix $V_i = f^{-1}(U_i)$. On each chart $U_i$, set $J_i$ to be an ideal sheaf such that $f_i : V_i \to U_i$ is the blow-up of $J_i$. -For each $i$, fix $I_i$ to be a (generically non-zero) ideal sheaf on $X$ such that $I_i |_{U_i} = J_i$. -Now fix $I = \prod_i I_i$. This is an ideal sheaf on $X$. On each open chart $U_i$, it is equal to $J_i \cdot \prod_{j \neq i} I_j|_{U_i}$. -Set $\pi : Y \to X$ to be the blow-up of $I$. We wish to show that $\pi$ factors through $f$, and is in fact a blow-up of $Z$, and so $Y$ is indeed projective. Set $W_i = \pi^{-1}(U_i)$. -All schemes involved are separated, and so to verify that $Y \to X$ actually factors through $Z \to X$, it is sufficient to work locally, so work on a chart $U_i$. -There we are comparing the blow-up of $J_i$ with the blow-up of $J_i \cdot \prod_{j \neq i} I_j|_{U_i} = J_i \cdot K_i$. $V_i$ is the blow-up of $J_i$ and $W_i$ is the blow-up of $J_i \cdot K_i$. However, I claim it is a straightforward exercise to verify that $W_i$ is the same as the blow-up of the ideal sheaf $(J_i \cdot K_i) \cdot \mathcal{O}_{V_i}$. -Let me give a hint as to how to do this. -Set $R = \Gamma(U_i, \mathcal{O}_X)$, set $J_i = (x_1, \dots, x_n)$ and $K_i = (y_1, \dots, y_m)$. Then the blow-up of $J_i$ is covered by affine charts $U_{i,l} = \text{Spec} R[x_1/x_l, \dots, x_n/x_l]$ where the gluings are the obvious ones. Likewise the blowup of $J_i \cdot K_i$ is covered by charts $U_{i,s,t} = \text{Spec} R[(x_1y_1)/(x_s y_t), \dots, (x_1y_m)/(x_s y_t), \dots, (x_ny_m)/(x_s y_t) ]$. Which are easily checked to be the blow-ups of $U_{i,s}$ at $(J_i \cdot K_i) \cdot R[x_1/x_l, \dots, x_n/x_l]$ (unless I've completely forgotten how to do this). -This sort of computation should be viewed as a generalization of the fact that the blow-up of an ideal is the same as the blow-up of a power of an ideal. -But this proves everything I claimed, right? $Y$ is exactly the blow-up of $Z$ at $I \cdot \mathcal{O}_Z$, and so $Y$ is projective, and is the blow-up of some ideal sheaf. -Of course, writing this as a sequence of blow-ups at subvarieties (and not subschemes/ideals) is probably much much harder.<|endoftext|> -TITLE: $C^n$ And Forcing: Reading a Recent Paper By Kunen -QUESTION [5 upvotes]: While reading a recent paper by Kunen arxiv.org/abs/0912.3733, which deals with PFA and the existence of certain differentiable functions, (defined on all of $\mathbb{R}$) which map certain $\aleph_1$-dense subsets of $\mathbb{R}$ onto other $\aleph_1$-dense subsets of $\mathbb{R}$. The technical details aside, he was able to show: - -Theorem 1.6 Assume PFA, and let $D,E,\subset \mathbb{R}$ be $\aleph_1$-dense. Then there exist exists an order preserving bijection $f:\mathbb{R} \rightarrow \mathbb{R}$, and $D^\ast \subseteq D$ such that $D^\ast$ is $\aleph_1$-dense, $f(D^\ast)=E$, - and - -For all $x \in \mathbb{R}$, $f'(x)$ exists and $0\leq f'(x)\leq 2$ -$f'(d) = 0$ for all $d\in D^\ast$ - - -It occurred to me that with a few modifications his method/forcing notion might be used to add other differentiable, or Lipschitz functions to some ground model. It follows that these new functions would in turn produce new $C^1$ functions, and so on. And in the end could result in new systems of differential equations, with absolutely strange behavior. -So my questions are the following: - -What is known about "messing with" the class of $C^n$ functions, via forcing? -Are there other examples of more exotic forcing notions which add smoother functions? - -Edit: took out weakly bit.. - -REPLY [6 votes]: Michael, why do you say that Kunen's result is about weakly differentiable functions? -This seems to talk about functions that are plainly differentiable on all of $\mathbb R$. -Or do you mean they are differentiable on $D^*$? -Adding either $C^1$-functions or Lipschitz functions by forcing seems to be nontrivial. -I have been interested in the question of how many continuous, Lipschitz, or $C^n$-functions -are needed so that the functions together with their inverses (as relations) cover all of -$\mathbb R^2$. In the Sacks model there is a family of $\aleph_1$ $C^1$-functions so that -the functions and their inverses cover all of $\mathbb R^2$ (this is due to Steprans). -The same is true for Lipschitz functions. -You can use ccc forcing to add a "small" family of continuous functions that covers -$\mathbb R^2$ in the above sense. However, the argument fails for Lipschitz and for -$C^1$-functions. -Kubis and Vejnar showed that one can force a countable family of Lipschitz functions -on the Cantor set such that the functions and their inverses cover an uncountable square. -An absoluteness argument now shows that already in the ground model some uncountable square is covered by countably many Lipschitz functions. -So in some sense, the Kubis-Vejnar forcing does not give you anything new. -On the negative side, $\mathbb R^2$ cannot be covered by less than continuum many functions that are twice differentiable. This seems to indicate that adding twice differentiable functions with interesting properties by forcing is difficult.<|endoftext|> -TITLE: Infinite simple Galois groups -QUESTION [7 upvotes]: Conjecturally, every finite group is the Galois group of some extension of the rationals. -This question made me wonder what is known about infinite -simple groups occurring as Galois groups. -What are the infinite simple groups that are expected to be Galois groups, i.e., profinite? Are they classified? Are there any examples of such extensions? - -REPLY [13 votes]: Th notion of simple is not very interesting for profinite groups. The "right" concept is just infinite. We say that a profinite group is just infinite if all its non-trivial normal closed subgroups are of finite index. $SL_2(\mathbb{Z}_p)$ is an example of a just infinite profinite group. Another example coming from Galois theory is the Nottingham group which is an open subgroup of index p-1 in the automorphism group of the field $\mathbb{F}_p((t))$. -There are many examples of just infinite profinite groups. There isn't much general theory except Wilson's dichotomy that they are either Branch Groups (e.g. Grigorchuk group or the Gupta-Sidki group) or they contain an open subgroup which is direct sum of hereditarily just infinte profinite groups (i.e. every open subgroup is also just infinite) and some results of Colin Reid.<|endoftext|> -TITLE: Are some numbers more irrational than others? -QUESTION [42 upvotes]: Some irrational numbers are transcendental, which makes them in some sense "more irrational" than algebraic numbers. There are also numbers, such as the golden ratio $\varphi$, which are poorly approximable by rationals. But I wonder if there is another sense in which one number is more irrational than another. -Consider the following well known irrationals: -$\sqrt{2}$, -$\varphi$, -$\log_2{3}$, -$e$, -$\pi$, -$\zeta(3)$. -The proofs of irrationality of these numbers increase in difficulty from grade-school arguments, to calculus, to advanced methods. Other probable irrationals such as $\gamma$ most likely have very difficult proofs. -Can this notion be made precise? Is there a well defined way in which, for example, $\pi$ is more irrational than $e?$ - -REPLY [5 votes]: In response to a question about the comparative irrationality of real numbers, I wrote, "A theorem only has a difficult or long proof until one finds an easy or short proof." Mark Sapir replied, "There are only [a] finite number of proofs of length 10^10 so your last statement is wrong." I would argue, instead, that there are many contentious and, perhaps, wrong assumptions and philosophical misconceptions implicit in Sapir's sentence. For example, one could state and prove the theorem, "The sum of the first three odd numbers is nine." One could state and prove another theorem, "The sum of first five odd numbers is 25." One could state and prove another theorem, "The sum of first 100 odd numbers is 10,000." Continuing, one gets to statements of length greater than 10^10, and whose proofs would be even longer. Of course, someone might have the clever idea of proving inductively that, for all positive integers n, the sum of the first n odd numbers is n^2, and then one has a short proof of infinitely many theorems. It is exactly this kind of "exponential collapse," often called "progress in science," that I meant when I wrote a proof is long until one finds a short proof. I would be very interested to know if someone can really prove that there exist theorems that do not have short proofs. It is more likely that there are only finitely many theorems, and they all have short, simple, and elegant proofs. This, of course, is simply another description of Erdos' famous book.<|endoftext|> -TITLE: The fiber of a Serre fibration -QUESTION [6 upvotes]: If $p:E\to B$ is a Serre fibration (assume it is surjective), then for each -$b\in B$ we get a comparison map $p^{-1}(b) \to F_b$, where $F_b$ is the homotopy -fiber of $p$ over $b$. -It is easy to see that these maps induce isomorphisms on $\pi_n$ for $n\geq 1$, but I -wonder about $\pi_0$. -Question: Is it true that $p^{-1}(b) \to F_b$ is a weak homotopy -equivalence? - -REPLY [2 votes]: Dear Jeff, -The map you describe is a homotopy equivalence. This is proved as Proposition 1.1 in the paper -Varadarajan, K. On fibrations and category. Math. Z. 88 1965 267–273.<|endoftext|> -TITLE: transfer kernels and the Schur multiplier -QUESTION [9 upvotes]: Let $\Gamma$ be a finite $2$-group, and let $G$ be any subgroup -of index $2$. Moreover, let Ver$: \Gamma/\Gamma' \to G/G'$ -denote the group theoretical transfer, and let $M(\Gamma)$ be -the Schur multiplier of $\Gamma$. -Is it true that -$$ | \text{ker}\ {\rm Ver}_{\Gamma/\Gamma' \to G/G'} | \le 2 | M(\Gamma) | \quad ? $$ -The answer is positive for a couple of groups with small multiplier, -such as cyclic $2$-groups, which have trivial Schur multiplier, -the groups of order $8$, or dihedral and quaternion groups. Does -current computer technology allow finding a counterexample by going -through 2-groups of moderate size? -Edit. Here's a little bit of background. Let $K$ be a quadratic number field whose 2-class group has type (2,2). It is known that the Hilbert 2-class field $K^1$ has cyclic -2-class group, and that the Galois group $\Gamma$ of the second Hilbert 2-class field is either (2,2) itself, a dihedral, quaternion, or semi-dihedral 2-group. Let $K_j/K$ (j=1, 2, 3) denote the three unramified quadratic extensions inside $K^1$, and let $G_j$ be the -Galois groups of $K^2/K_j$. Then $\Gamma$ is quaternion or semi-dihedral if and only if -in each of the extensions $K_j/K$, exactly one ideal class of order 2 capitulates (i.e., becomes principal). Now these are exactly the groups among the possible Galois groups with trivial Schur multiplier. On the other hand, the transfer of ideal classes corresponds, via Artin's reciprocity law, to the transfer map (Verlagerung) from the abelianization of $\Gamma$ to that of the $G_j$. Thus in this case, we find that the order of the Schur multiplier is equal to one half of the maximal order of the capitulation kernels. - -REPLY [3 votes]: Take the short exact sequence of modules $\mathbb{Z}\hookrightarrow Ind^\Gamma_G\mathbb{Z}\twoheadrightarrow\mathbb{Z}$ and apply $H_i(\Gamma,-)$. Note that this sequence is exact because $|\Gamma:G|=2$. You obtain the long exact sequence, noting that $H_i(\Gamma,Ind^\Gamma_H\mathbb{Z})\cong H_iG$ by Shapiro's Lemma. As I remarked in the comment attached to this post, the coefficient module from the latter $\mathbb{Z}$ in the short exact sequence has nontrivial $\Gamma$-action (I will denote this coefficient by $\tilde{\mathbb{Z}}$, where the action is multiplication by $-1$ via elements of the nontrivial coset of $\Gamma/G$). Keeping that in mind, we have in particular: -$H_2(\Gamma,\tilde{\mathbb{Z}})\stackrel{\delta}{\rightarrow}H_1\Gamma\stackrel{tr}{\rightarrow}H_1G\rightarrow H_1(\Gamma,\tilde{\mathbb{Z}})$. -Exactness implies $Ker(tr)=Im(\delta)=H_2(\Gamma,\tilde{\mathbb{Z}})/Ker(\delta)$, so that $|Ker(tr)|\le |H_2(\Gamma,\tilde{\mathbb{Z}})|$. And we know that $tr=Ver$ and $H_2\Gamma=M(\Gamma)$. -I want to claim that $|H_2(\Gamma,\tilde{\mathbb{Z}})|\le 2\cdot|H_2\Gamma|$ (the latter homology has $\mathbb{Z}$-coefficient with trivial action), but at this moment I am unsure how to prove it. Hatcher's Algebraic Topology textbook gives a long exact sequence for general coefficient systems (pg330), with $H_3G\stackrel{res}{\rightarrow}H_3\Gamma\rightarrow H_2(\Gamma,\tilde{\mathbb{Z}})\rightarrow H_2G\stackrel{res}{\rightarrow}H_2\Gamma$, so this could be of use. - -Chris Gerig<|endoftext|> -TITLE: On a Conjecture of Schinzel and Sierpinski -QUESTION [19 upvotes]: Melvyn Nathanson, in his book Elementary Methods in Number Theory (Chapter 8: Prime Numbers) states the following: - -A conjecture of Schinzel and Sierpinski asserts that every positive rational number $x$ can be represented as a quotient of shifted primes, that $x=\frac{p+1}{q+1}$ for primes $p$ and $q$. It is known that the set of shifted primes, generates a subgroup of the multiplicative group of rational numbers of index at most $3$. - -I would like to know what progress has been made regarding this problem and why is this conjecture important. Since it generates a subgroup, does the subgroup which it generates have any special properties? -I had actually posed a problem which asks us to prove that given any interval $(a,b)$ there is a rational of the form $\frac{p}{q}$ ($p,q$ primes) which lies inside $(a,b)$. Does, this problem have any connections with the actual conjecture? -I had actually posed this question on MATH.SE (Link : https://math.stackexchange.com/questions/18352/a-conjecture-of-schinzel-and-sierpinski ). - I did get a decent answer from Andreis Caicedo, but i would like to have more opinions of Mathematicians from this community. - -REPLY [18 votes]: The question can be written as follows: Given two positive integers $a$ and $b$, do there exist primes $p$ and $q$ such that -$$aq-bp=b-a?$$ -You would expect there to be not just one such pair of primes, but infinitely many pairs. For instance, if $a=2$ and $b=1$, then $q$ is a Sophie Germain prime, and everyone expects there to be infinitely many of those. Moreover, you should be able to replace the right side of the equation with a constant $c$, i.e., -$$aq-bp=c.$$ -The twin prime conjecture says that there are infinitely many solutions when $a=b=1$ and $c=2$. Polignac's conjecture implies infinitely many solutions when $a=b=1$ and for every even value of $c$. In general you should expect infinitely many solutions when there isn't some obvious congruence that forces finiteness; for instance obviously $a=b=c=1$ only has one solution. Moreover, it's natural to expect a specific slowly decreasing density of solutions using a heuristic estimate derived from the prime number theorem. -This question for all suitable $a$, $b$, and $c$ is in turn a special case of yet more general questions about linear patterns in the prime numbers. For instance, the statement that there are infinitely many arithmetic progressions of length 3 in the primes is the statement that there are infinitely many solutions to -$$p-q = q - r > 0.$$ -Now, it's a famous theorem of Tao and Green that there are infinitely many arithmetic progressions of primes of arbitrary length. Later Tao and Green did a more systematic study that established the existence of all kinds of linear patterns in the prime numbers. However, the Sieprinski-Schnizel conjecture, and its generalization in the previous paragraph, are part of the "rank 1 case" that they did not solve. (These are just my mental notes from a survey talk by Terry Tao that I once attended.) If they could have done the rank 1 case, it would have included the twin prime conjecture and I think that it would have implied the asymptotic Goldbach conjecture too, so that would have been even more amazing than what they did accomplish. -I have no idea whether this remaining rank 1 case is the same class of question as the Tao-Green results, but just harder; or whether it is so much harder that it is in a different class. Let's optimistically say that it's the former. If so, then what makes the Schinzel-Sierpinski conjecture interesting is that you should always expect infinitely many solutions in prime numbers to linear equations, unless there are only finitely many solutions because of a simple congruence. And I might say that the Tao-Green results are the main recent progress, even though they answered different questions.<|endoftext|> -TITLE: Would it make sense to have someone write separate teaching and non-teaching letters of reference for one candidate? -QUESTION [10 upvotes]: There's a friend of mine (seriously, not me) who is facing a slight dilemma. There's a particular senior mathematician who he knows well (several co-authored papers, etc.) who also is probably the person best placed to comment on his teaching abilities. Now, of course, this sounds very convenient; after all you can roll lots of praise into one letter, right? -But there's an issue with this: I know for a fact that many people on search committees don't read the letters that are supposed to concern teaching. Certainly, many will not to read them until one has gotten down to a fairly short list. So you don't necessarily want lots of praise for your accomplishments in general to be in a letter which has "(teaching)" next to it on MathJobs. I'm trying to think of what to suggest to my friend as a way out of this dilemma, and one idea which floated to the top of my mind was this: maybe this senior person could write two separate letters, one for teaching and one for everything else. Is this an insane idea? Is there some better way out of this dilemma? - -REPLY [6 votes]: This is a reasonable thing to do, and is definitely possible in mathjobs--indeed, I believe that we have just offered a job to an individual whose mathjobs-based application had distinct research and teaching letters from the same person. He also had a teaching letter from someone else, as well as a surplus of research letters, consistently with Pete's advice. -I'm not sure how this worked on the letter-writer's end, but on our end the only difference is that there are two files next to the letter-writer's name rather than the usual one. The modifier "(teaching)" which usually appears next to the name of the authors of teaching letters, does not appear next to the name of the letter-writer in question--though maybe this is only because it is quite generally the case that you're only allowed to specify one person as your teaching reference,* and the applicant would have specified the author of the other teaching letter. -*I don't remember whether this is true, but looking at other applications with two teaching letters suggests that it is.<|endoftext|> -TITLE: Shadow boundary on convex body in $\mathbb{R}^3$ -QUESTION [7 upvotes]: Let $S$ be the surface of a compact, convex, smooth ($C^\infty$) body in $\mathbb{R}^3$, -with strictly positive Gaussian curvature at every point of $S$. -Fix a direction $z$ in a Cartesian coordinate system, -and consider all the lines parallel to $z$ and tangent to $S$, -which form a topological cylinder enclosing $S$, -touching $S$ on the shadow boundary resulting from a light source at $z=+\infty$ -(yellow in the figure below). -Parametrize these lines from $s=0$ to $s=1$ around the cylinder, -and let $h(s)$ be the height of the point of tangency -to $S$ above the $xy$-plane, orthogonal to $z$. -My question is: - -Can $h(s)$ have an arbitrarily large number of local maxima and minima? - - -I am interested to learn if this shadow-boundary curve is "well-behaved" in some sense, -for smooth convex bodies. -Thanks for pointers/suggestions/counterexmaples! - -REPLY [2 votes]: This is not an answer but a shadow tangent Mohammad Ghomi has a wonderful paper -concerning a converse question: what are neccessary and sufficient conditions -on the shadows which insure the surface to be convex. His answer: -for all projection directions, the shadow must be connected and simply connected. -See: -Solution to the shadow problem in 3-space, in Minimal Surfaces, Geometric Analysis and Symplectic Geometry, Adv. Stud. Pure Math, 34 (2002) 129-142. Which you can find on his web page.<|endoftext|> -TITLE: Transversality in the proof of the Blakers-Massey Theorem. Is it necessary? -QUESTION [17 upvotes]: Assume one is given a commutative square of spaces -$A \quad \to \quad C$ -$ -\downarrow \qquad \qquad \downarrow$ -$B\quad \to \quad X$ -which is a pushout and in which each map is a cofibration. -If $A \to B$ is $r$-connected and $A\to C$ is $s$-connected, -then the Blakers-Massey theorem says that the square is -$(r+s-1)$-cartesian (this means that the map from $A$ into the -homotopy pullback of the remaining terms is $(r+s-1)$-connected). -The only proofs of the statement that I know of (at this level of - generality) make use of transversality. However, if all spaces are simply connected, there are -proofs which avoid transversality (for example, when $B$ is a contractible, one can -deduce it using the Serre exact sequence). -Question: Is transversality intrinsic to a proof of the theorem in the general - case? - -REPLY [3 votes]: I refer here to my recent answer to -What is the intuition behind the Freudenthal suspension theorem -where the results do not require simple connectivity for descriptions of the critical group, basically because the proofs do not use homological, i.e. abelian, methods. -May 30: The original Blakers-Massey results were related to triad homotopy groups, since the exact sequences involving these and relative homotopy groups showed the triad groups as the obstruction to excision. So there was a question of calculating these groups, and homology groups were used for this in the simply connected case, see the book by J.F. Adams A student's guide to algebraic topology. However such calculations in the non simply connected case do follow from a Generalised Van Kampen Theorem proved with J.-L. Loday. I have revised and updated a paper of mine ``Triadic Van Kampen theorems and Hurewicz theorems'', -Algebraic Topology, Proc. Int. Conf. Evanston March 1988, Edited -M.Mahowald and S.Priddy, Cont. Math. 96 (1989) 39-57. -and made it available as -http://pages.bangor.ac.uk/~mas010/pdffiles/VKTEVAN2.pdf -June 9, 2017 There is a nice book on "Cubical Homotopy" by Munson and Volic (CUP, 2015) which deals with a lot of these connectivity arguments. I am unclear whether these argument cover the connectivity results of Theorem 6.1 of -Brown, R. and Loday, J.-L. ‘Homotopical excision, and Hurewicz theorems for n-cubes of spaces’. Proc. London Math. Soc. (3) 54 (1) (1987) 176–192, -which gives algebraic as well as connectivity results.<|endoftext|> -TITLE: Constructions unique up to non-unique isomorphism -QUESTION [24 upvotes]: 1) Fields have algebraic closures unique up to a non-unique isomorphism. -2) Nice spaces (without base point) have universal covering spaces unique up to a non-unique isomorphism. -3) Modules have injective hulls unique up to a non-unique isomorphism. -Such situations can lead to interesting groups - the absolute Galois group, the fundamental group, and the "Galois" groups of modules introduced by Sylvia Wiegand in Can. J. Math., Vol. XXIV, No. 4, 1972, pp. 573-579. -I'd appreciate any insight into the abstract features of situations which give rise to this type of phenomenon. And I'd appreciate as many examples from as many parts of mathematics as possible. - -REPLY [4 votes]: Let me mention Sullivan's minimal models. -Every commutative differential graded $\mathbb{Q}$-algebra (cdga) $A^*$ concentrated in non-negative degrees and such that $H^0(A^*)=\mathbb{Q}$ admits a minimal Sullivan model $i:M^*\to A^*$ where $M^*$ is a free commutative graded algebra obtained from $\mathbb{Q}$ by adding generators of non-negative degrees so that the differential of each generator is a $\mathbb{Q}$-linear combination of products of length $\geq 2$ of the previous generators, and $i$ is a map of cgda's that induces a cohomology isomorphism (i.e., a quasi-isomorphism). -The minimal model is unique up to a non-unique isomorphism. More generally, if $f:A^*\to B^*$ is a map of cdga's and $j:N^*\to B^*$ is a minimal model of $B^*$, then there is a cdga map $g:M^*\to N^*$, defined up to cdga homotopy, such that $fi=gj$ up to cdga homotopy; moreover, if $f$ is a quasi-isomorphism, then $g$ is an isomorphism. -This reduces the classification of non-negative cdga's up to quasi-isomorphism (and as a consequence, the classification of simply connected topological spaces up to rational homotopy) to the classification of algebras of a certain kind up to isomorphism. -Of course, this example is similar to some mentioned before (in a sense it is the commutative analog of the answer of John Palmieri).<|endoftext|> -TITLE: Quasifibrations and homotopy pullbacks -QUESTION [11 upvotes]: I'm wondering about the theoretical placement of quasifibrations. -One nice thing about "weak fibrations" (maps homotopy equivalent in the category of maps to Hurewicz fibrations) is that a pullback square involving (one) weak fibration is a homotopy pullback square. -Is the corresponding result - true for quasifibrations in the Serre-Quillen context? That is, suppose $E\to B$ -is a quasifibration, and the square -$$ -\begin{array}{ccc} -P & \to & E -\cr\downarrow&pb&\downarrow -\cr -X& \to &B -\end{array} -$$ -is a categorical pullback. Then is it a homotopy pullback in the Quillen-Serre model -structure? - -REPLY [18 votes]: The definition of quasifibration (according to Dold & Thom, 1958) is: a map $f:E\to B$ such that for all $b$ in $B$, the canonical map from the fiber to the homotopy fiber is a weak equivalence. Pullbacks with respect to such maps are not generally homotopy pullbacks; an example was given in that 1958 paper (Bermerkung 2.3), which goes something like this: -Let $\newcommand{\R}{\mathbb{R}}B=\R\times \R$. Then $E$ will have the same underlying set as $B$, and $f$ will be the identity map. But we topologize $E$ by "tearing" along the positive $y$-axis. For instance, let $E$ have the smallest topology such that $f$ is continuous and the set $[0,\infty)\times (0,\infty)$ is open. -The space $E$ is still contractible with this topology (it deformation retracts to $\R\times -1$). Therefore, the homotopy fiber over any point of $B$ must be weakly contractible, and thus weakly equivalent to the actual one-point fiber. So $f$ is a quasi-fibration. -Let $X= \mathbb{R}\times 1\subset B$, and let $P$ be the pullback of $E$ over $X$. Then $P$ has two path components, while $X$ is contractible; this is not a homotopy pullback!<|endoftext|> -TITLE: Open problems in Euclidean geometry? -QUESTION [54 upvotes]: What are some (research level) open problems in Euclidean geometry ? - -(Edit: I ask just out of curiosity, to understand how -and if- nowadays this is not a "dead" field yet) -I should clarify a bit what I mean by "Euclidean geometry". By this term I mean, loosely, the study of the geometry of certain subsets of Euclidean space $\mathbb{E}^n$ from a point of view which is either the "classical" one (i.e. axiomatic), or one that involves more modern tools, but the problem in question has not to be "clearly" a problem within some other branch of maths such as differential or algebraic geometry, algebraic or general topology, analysis, or measure theory. -Some examples to clarify: - -the study of configurations of lines or affine subspaces is EG; but the algebro-topological study of hyperplane arrangements is not. -plane conics as defined via their metric property are objects of EG; but "algebraic curves" are not, unless they're defined by some "elementary enough property" (intentionally vague) involving the Euclidean metric. -root systems of Lie algebras are EG. -polyhedral cones are EG. -polytopes are EG. -tessellations of space with polytopes or analogous objects are in EG. -minimal surfaces in $\mathbb{E}^3$ are not EG. -fractal geometry (Julia sets, self-affine fractals...) is not EG. -not sure about convex bodies. If they're polyhedral I'd say their study fits in EG. -packings of spheres are EG. - -REPLY [9 votes]: The classification of all pentagons tiling the plane is not completely finished, -see for example MR3037862. (Bagina, O. G. Convex pentagons that tile the plane (types: 11112, 11122). (Russian) Sib. Èlektron. Mat. Izv. 9 (2012), 478–530.) -Update: This problem is now probably solved: see e.g. https://en.wikipedia.org/wiki/Pentagonal_tiling and references therein.<|endoftext|> -TITLE: Is there any book explaining in detail the book "Basic Number Theory" by André Weil as Dirichlet did to "Disquisitiones Arithmeticae" by Gauss? -QUESTION [9 upvotes]: Is there any book explaining in detail the book "Basic Number Theory" by Andre Weil as Dirichlet did to "Disquisitiones Arithmeticae"? -This is because I have read the two books mentioned above and I hope there will be one. - -REPLY [7 votes]: Indeed Ramakrishnan and Valenza 's book is a pretty good reference. -Perhaps we could give more specific answers if you were more precise about exactly where your difficulties are? -EDIT: Since we've been given precisions in the comments below, I can confirm R&V's book will nicely do for the basics of the theory ; to get further, from the top of my head, you'll want to have a go with : - -Cassels and Froehlich (editors), "Algebraic number theory" -Serre, "Corps locaux" (translated) -Neukirch, "Class field theory" -J.S.Milne's notes on class field theory on his website<|endoftext|> -TITLE: Amenable groups with finite classifying space -QUESTION [14 upvotes]: A group $G$ is said to be elementary amenable if it can be obtained from finite and abelian groups by subgroups, quotients, extensions and increasing unions. It is well-known that all such groups are amenable, i.e. allow for a finitely additive and $G$-invariant probability measure on $G$. Grigorchuk's group is an example of a finitely generated group which is amenable but not elementary amenable. Grigorchuk has also found a finitely presented example of such a group in An example of a finitely presented amenable group not belonging to the class EG, 1998 Sb. Math. 189 75. -However, since Grigorchuk's example -$$\langle a,c,d,t \mid a^2 =c^2 =d^2 =(ad)^4 =(adacac)^4 =e, -a^t = aca, c^t = dc, d^t = c \rangle$$ -obviously contains torsion it cannot have a finite classifying space. - -Question: Is there an example of an amenable group which is not elementary amenable and whose $BG$ is homotopy equivalent to a retract of a finite $CW$-complex? - -Equivalently: - -Question: Is there an example of an amenable group $G$, such that $G$ is not elementary amenable and the trivial $G$-module $\mathbb Z$ has a finite resolution by finitely generated projective $\mathbb ZG$-modules. - -These kind of questions are sometimes called Day's problem for a certain class of groups. - -REPLY [8 votes]: There are very few examples of amenable but not elementary amenable finitely presented groups. There exists a torsion-free example, an ascending HNN extension of the basilica group -(Bartholdi, Laurent; Virág, Bálint, Amenability via random walks. -Duke Math. J. 130 (2005), no. 1, 39–56 - there it is proved that the basilica group is amenable but not elementary amenable; and R. I. Grigorchuk and A. Z˙ uk, On a torsion-free weakly branch group defined by a -three state automaton, Int. J. Algebra Comput. 12(1–2) (2002) 223–246, -R. I. Grigorchuk and A. ˙Zuk, Spectral properties of a torsion-free weakly branch -group defined by a three state automaton, in Computational and Statistical Group -Theory (Las Vegas, NV/Hoboken, NJ, 2001), Contemporary Mathematics, Vol. 298 -(American Mathematical Society, Providence, RI, 2002), pp. 57–82 - there it is proved that this group has a finitely presented ascending HNN extension). Essentially this is the only known example (there are several other similar examples, see references in Bartholdi, Laurent, Eick, Bettina, Hartung, René -A nilpotent quotient algorithm for certain infinitely presented groups and its applications. -Internat. J. Algebra Comput. 18 (2008), no. 8, 1321–1344. ). I doubt very much that it has a finite classifying space or even has a finite projective resolution with finitely generated modules, because the basilica group itself is infinitely presented.<|endoftext|> -TITLE: Ramified cover of 3-ball -QUESTION [9 upvotes]: I it true that any connected oriented 3-manifold with nonempty boundary can be obtained as a ramified cover of 3-ball with a ramification at a link? - -link = a 1-dimensional submanifold with possibly nonempty boundary. - -If answer is "YES", can we choose in addition the restriction of the covering at the boundary? - -REPLY [8 votes]: Berstein and Edmonds prove in Cor. 6.3 that for an orientable 3-manifold $W$ with connected boundary, with a branched cover $\varphi: \partial W\to S^2$ of degree $n>3$, then there is a branched cover $\Phi: W\to D^3$ such that $\Phi_{|\partial W}=\varphi$. In another paper, Edmonds claims in Theorem 2.1 that Cor. 6.3 extends to maps $f: W\to D^3$ such that the boundary map is a branched cover of the same degree as $f$ (allowable). One can easily construct an allowable map $f:W\to D^3$ by mapping $\partial W$ to $S^2$ by a branched cover so that each component of $\partial W$ has positive degree $>2$ (with respect to the orientation induced by $W$), and extend to all of $W$ by coning off. Theorem 2.1 implies that this map is homotopic to a branched cover. -The hypothesis of degree $>2$ is necessary, since for example if one has a knot $K\subset S^3$ which is not (strongly) invertible, then $M=S^3-\mathcal{N}(K)$ is a manifold with torus boundary such that there is a degree 2 map $T^2=\partial M\to S^2$ which is the quotient of the elliptic involution, but which doesn't extend over $M$.<|endoftext|> -TITLE: In the quaternions, "any imaginary unit may be called i" -QUESTION [6 upvotes]: Introduction -Suppose we are trying to prove that $\rm PSO_3\times PSO_3$ is isomorphic with $\rm PSO_4,$ and we catch on to the idea of using the quaternions to do so. We realize (as in Conway & Smith's On Quaternions and Octonions, whence the quotation) that we can encapsulate $\rm PSO_3$ as the set of all maps $x\mapsto\bar qxq$ for a unit quaternion $q$ operating on imaginary quaternions $x,$ and go on to trying to understand why $\rm PSO_4$ is the set of all maps $\pm(x\mapsto\bar lxr)$ for unit quaternions $l,r$ operating on quaternions $x.$ -To show that all maps $\pm(x\mapsto\bar lxr)$ really are elements of $\rm PSO_4,$ we begin by showing this in the case $\bar l=\cos\theta+isin\theta,r=1.$ This is simple, since this operation rotates the plane spanned by $1$ and $i$ through an angle of $\theta,$ and rotates the plane spanned by $j$ and $k$ through an angle of $\theta$ at the same time. -But at this point, we really are done proving that all maps $\pm(x\mapsto\bar lxr)$ are elements of $\rm PSO_4,$ since, as in the book, "any imaginary unit may be called $i,$ and perpendicular one $j,$ and their product $k$" (although this was said at a different point), right-multiplication has the same geometric properties as left-multiplication, and the composition of any two elements of $\rm PSO_4$ is an element of $\rm PSO_4.$ -Idea -It makes all the intuitive sense in the world to me that "any imaginary unit may be called $i,$" and I can really visualize this geometrically. Furthermore, I could go back through a proof, change all the $i$'s to $u$'s for an arbitrary imaginary unit quaternion $u,$ etc. But suppose I had a collection of proofs written by someone who wasn't very careful, and she/he had used $i, j,$ and $k$ for simplicity and computed examples, stating at the end of each one that it generalizes to all quaternions. Suppose I pored through these proofs and discovered that about a third of them were careless to the point of being false, because of some lack of care in going back/forth between general ($u,v,w$) and specific ($i,j,k$) contexts. To formalize this imaginary formalization attempt, suppose I had a computer that understood category theory really well and wanted to scan these proofs in for it and get it to check whether this person's proofs really proved whatever facts from geometry they purported to prove. -Question -In the specific example, the notion that any imaginary unit may be called $i$ and left-multiplication is like right-multiplication can be dealt with at a first approximation using the notion of automorphisms. There is a ring automorphism sending $u$ to $i$ and vice versa, and there is a group automorphism between the multiplicative group of the quaternions and its opposite group. But I wonder if it follows directly enough that geometric facts can be proved "by example." - -Is there a category-theoretic context in which certain ring/group automorphisms are natural and in which their being natural is biconditional with their preserving geometric properties? - -To explain why the notion of an automorphism by itself might not be enough, we can imagine $\mathbb{Q}[\sqrt{2}]$ acting on $\mathbb{R}$ by multiplication. Multiplication by $\sqrt{2}$ preserves order, but multiplication by $-\sqrt{2}$ reverses it, so in the context of orderings, it would not be accurate to say "any square root of 2 may be called $\sqrt{2}.$" -What I'm envisioning is a category $\mathcal{C}$ with an object $\mathbb{H},$ as well as a morphism for each automorphism of $\mathbb{H},$ an object $\mathbb{H}^\star,$ and a morphism for each group automorphism of $\mathbb{H}^\star,$ perhaps geometric objects or morphisms as well, and whenever we speak of "an instance of the quaternions" we are really speaking of a functor $\mathcal{C}\to\mathcal{C}$ at least one existing for each possibility of an imaginary unit being called $i,$ a perpendicular one $j,$ and their product $k.$ I know this doesn't work out as stated, because the identity morphism doesn't go to the identity morphism, but perhaps there's a way to fix that. Then an automorphism of the quaternions can be viewed as a natural transformation between functors $\mathcal{C}\to\mathcal{C},$ one preserving the hidden (say, setwise) structure of the quaternions, and another preserving their apparent ($i,j,k$) structure and for some reason, because it's natural, the geometry comes out alright. - -REPLY [20 votes]: Just to elaborate on what is already in the comments, the algebra automorphisms of $\mathbb H$ act transitively on the set of pairs $(u,v)$ where $u$ and $v$ are imaginary quaternions of unit length that are orthogonal to one another. -To see this, I will include here some remarks on $\mathbb H$ and its automorphisms. Part of the OP's concern seems to be that it is not a priori automatic that metric concepts in $\mathbb H$ such as unit length or orthogonality (and hence the notion of being imaginary, since the imaginary quaternions are the orthogonal complement to $\mathbb R$ in $\mathbb H$) -are preserved by Aut$(\mathbb H)$, and so one of my goals is to show that this concern is not necessary. Indeed, this geometry is intrinsic to the quaternions, as we will see. -(This is not coincidence: Hamilton was led to his discovery by trying to algebraize the geometry of $\mathbb R^3$.) -Note first that imaginary quaternions are characterized by the condition that -$\overline{u} = - u$, and thus for such quaternions, $|u|^2 = -u^2$. Thus if $u$ is imaginary, $u^2$ is a non-positive real number. Converesly, -if $u^2$ is a negative real number, -then one sees that $u$ is imaginary (exercise), and so the imaginary quaternions are also -characterized by having non-positive real squares. In particular, the set of imaginary -quaternions is preserved by Aut$(\mathbb H)$. -On imaginary quaternions, the inner product $u\overline{v} + v\overline{u}$ is simply -$u v + u v$, and so is also preserved by Aut$(\mathbb H)$. In particular, metric concepts like "length one" and "orthogonal" are preserved by Aut$(\mathbb H)$. -If $u$ and $v$ are unit length orthogonal imaginary quaternions, we then have that -$u^2 = v^2 = -1$ (unit length condition) and that $u v = - v u$ (orthogonality condition). -Thus, from the defining relations of $\mathbb H$, we obtain an algebra map -$\mathbb H \to \mathbb H$ that maps $i$ to $u$ and $j$ to $v$ (and then $k$ to $u v$). -This map is non-zero (since $u$ and $v$ are non-zero, having unit length), and hence -is necessarily injective ($\mathbb H$ is a division ring, hence has no non-trivial ideals), -and thus in fact bijective (source and target are of the same dimension). -An automorphism of $\mathbb H$ is determined by its values on $i$ and $j$ (since they generate -$\mathbb H$), and so the previous discussion shows that in fact Aut$(\mathbb H)$ is the -same as the group or permutations of pairs $(u,v)$ of orthogonal pairs of unit -vectors in the imaginary quaternions (also known as $\mathbb R^3$). -This group is well-known: it is precisely $SO(3)$. (If you like, $u$ and $v$ determine -uniquely a mutually orthogonal vector --- their quaternionic product $u v$ --- which can be characterized geometrically in terms of $u$ and $v$ via the right hand rule; thus pairs $(u,v)$ are the same as positively oriented orthonormal bases of $\mathbb R^3$, permutations of which are precisely the group $SO(3)$.) -Incidentally, it is not coincidence that Aut$(\mathbb H) = SO(3)$. -Namely, there is a natural map $\mathbb H^{\times} \to $ Aut$(\mathbb H)$ (where -$\mathbb H^{\times}$ means the non-zero --- equivalently invertible --- quaternions), -given by mapping $q$ to the automorphism $x \mapsto q x q^{-1}$. -The kernel of this map is precisely the centre, and so it induces an injection -$\mathbb H^{\times}/\mathbb R^{\times} \hookrightarrow $ Aut$(\mathbb H)$. -Now the source of this map can be identified with the quotient of the unit quaternions -(which form a copy of $SU(2)$) by $\pm 1$, and of course $SU(2)/\{\pm 1\} = -SO(3)$. On the other hand, this injection is in fact a bijection (i.e. any automorphism -of $\mathbb H$ is inner), by the Skolem--Noether theorem. This puts the description of Aut$(\mathbb H)$ obtained above into a more general perspective.<|endoftext|> -TITLE: Length of shortest possible knot -QUESTION [7 upvotes]: Consider a line L in R^3 in the shape of a trefoil knot. Consider the surface S that is the union of all unit circles that have centers on this line and whose tangent vectors are all perpendicular to the tangent vector of L at the cirle's center. S does not intersect itself. -What is the shortest possible length of L? - -REPLY [19 votes]: The invariant you are talking about is usually called the "ropelength" of the knot. You can find some basic stuff at the wikpedia page http://en.wikipedia.org/wiki/Ropelength which also gives some good references. (Note that some people use unit circles, while other people use circles of diameter 1, so the reported ropelength differs by a factor of 2.) -The exact value of the ropelength is not known for any nontrivial knot. However in the case of the trefoil, there are some pretty good bounds. It is between 15.66 and 16.372 if we define ropelength using circles of diameter 1. The upper bound is believed to be tighter.<|endoftext|> -TITLE: What is known about links with a countably-infinite number of tame components? -QUESTION [5 upvotes]: I want to be clear about my phrasing, but I'm not a topologist, so when I say "knot" or "link" I mean the equivalence class under ambient isotopy of an embedding of the circle into $\mathbb{R}^3$. -For a while I have been looking for references to what I've taken to calling the "Fraïssé link" - this thing would have (an instance of) every tame link occurring as a subset, and you would build it the same way that the model-theoretic Fraïssé construction builds the rational numbers from amalgamations of finite linear orders, or the random graph from finite graphs. However, when applying Fraïssé's construction, just ignore the model-theoretic question of what language you're using, or what a structure on a knot is. Just use "is a sublink" to replace the idea of a (logical) embedding, isotoping components where you need to to make the amalgamation property work. -I believe the chain-construction part of Fraïssé's work still succeeds as a plain-old union of sets this way. I also think the construction should commute with passing from a knot-instance to a knot, so realizations of this big snarly thing in $\mathbb{R}^3$ should all be ambient-isotopic to each other, too, but I'm not certain about that. If so, this would be an example of the kind of object I'm asking about in my question: links with a non-finite number of components, but the lowest-level components are ordinary knots. -This does not look like the usual sense of "wild knot" that people talk about, but if I'm wrong please let me know. Searches for "universal homogeneous link" or things like that have not led me to anything that looks similar to this object. Same with casual discussions with some model theorists. It's natural enough that I suspect I'm just missing the right terminology. -Ultimately, I'd like to find the right language to do some model theory with this object - in particular, a language that can express the Reidemeister moves as axioms, that works with the Fraïssé construction. But my specific request is whether knot theorists have noted any results about this particular object, or about any other links that have a countably-infinite number of tame components? -Update: Model Theory Background -Let me say a bit more about the model theory background to address Ryan's comment: take an at-most countably-infinite collection of finitely-generated structures (such as all finite graphs, all f.g. torsion-free Abelian groups, or all finite sets that are linearly ordered) up to the appropriate idea of isomorphism for that structure. Given three properties on that collection - "hereditarity", "joint embedding", and "amalgamation" - Fraïssé showed that there is a unique (up to isomorphism) countable structure (the Fraïssé limit) that -1) admits every member of the collection as a substructure and has no other non-isomorphic finitely-generated-substructures (the technical term is the original collection is the age of the Fraïssé limit.) -2) in which any isomorphism between two of its f.g. substructures extends to an automorphism of the entire structure -Hereditarity: the original collection is closed under taking finitely-generated substructures of its members. -Joint Embedding: for any $A, B$ in the collection, there is a common $C$ in the collection into which they both embed. -The property that avoids Ryan's trivial limit is the third: -Amalgamation: given $A_1, A_2$ in the collection, with a common substructure $B$ and embeddings $f_1:B \rightarrow A_1$ and $f_2:B \rightarrow A_2$, there is a $C$ in the collection and embeddings $g_1:A_1 \rightarrow C$, $g_2:A_2 \rightarrow C$ where $g_1 \circ f_1 = g_2 \circ f_2$. -So, for example, if $A_1$ is the Hopf link, and $A_2$ is the unknot linked in some way with a right trefoil, putting them side-by-side would be a joint embedding, but an amalgamation would have to treat one of the unknots in $A_1$ as the same unknot in $A_2$. So Joel's comment is right about what I want - this process builds a structure that has every finite link relating to every other finite link in all possible ways. -"The amalgamation property" is also the model-theorist's response to "why is the limit of finite linear orders $\mathbb{Q}$ rather than $\mathbb{Z}$?" -In the context of model theory, the definition of embedding and substructure depends critically on the choice of logical language for the structure, but the construction of a chain of embeddings using amalgamation leading to the final limit object does not depend on the underlying language. My thought was to drop the logical part (for now), and proceed analogously replacing "finitely generated substructure" with "sublink with finitely many components". - -REPLY [3 votes]: One of the best ways to understand knots and links geometrically is to create orbifolds with the knot or link as singular locus. When they are order 2 singular sets, there's a lot of -good theory involving the sphere/torus decomposition (or JSJ decomposition). But if you -make them into say order 7 orbifold loci, there are fewer decomposing surfaces, and less -to think about. The geometrization theorem for orbifolds says that each of these -in the ordinary sense decomposes into pieces modeled on one of the eight kinds of locally -homogeneous geometry. I'm not familiar with the Fraïssé theory, but we're used to -passing to geometric limits of these objects as they become more and more complicated, -and they can be quite interesting even in relatively simple situations (for instance, -the limits of $k$-bridge knots or links. A great deal is understood. -One can similarly -pass to limits with more and more components, which I guess probably includes the kind of -limit you're interested in. One could consider these orbifolds, together with erasing maps -that erase some of the components, mapping more complicated ones back to simpler ones; such -maps always decrease the Gromov norm, which is the sum of volumes of the hyperbolic pieces. - I think these limits of geometric structures are probably much more understandable than a particular -embedding in space. The geometric structures could be arranged so the links tend not -to crowd together, in this system of geometric structures. -Assuming from Joel David Hamkins' comment that it's not the trivial kind of example Ryan Budney mentioned -(with disjoint copies in disjoint balls), the link components would need -to have a limit set, and the way the link components converge to the limit set could take -many possible different inequivalent forms, without changing the topology and relationships of all the link components. There's probably no one best limit set, so I would be surprised -if such an object could be made canonical up to isotopy of $\mathbb R^3$. - -REPLY [2 votes]: I don't think that knot theorists are going to be very interested in such infinite links, but they do occur sometimes in the wider area of geometric topology, for instance in the proof of theorem 1.1 here. Still I doubt that they have been studied per se. -My understanding is that you are thinking about proper tame links with infinitely many components, i.e. such that every compact subset of $\Bbb R^3$ meets only finitely many components of the link (in general, a continuous map is called proper if the preimage of every compact set is compact). If so, then they reduce to "tame" embeddings of $X=(S^1\times\Bbb N)^+$ in $\Bbb R^3$, where $\Bbb N$ denotes the countable discrete space and $+$ stands for the one-point compactification. To see this, consider the one-point compactification of $\Bbb R^3$, which is $S^3$, and remove some distant point $pt$ (which won't make any difference since in moving the $1$-dimensional space $X$ in $S^3$ one can easily arrange to avoid $pt$). -In general, there is not so much literature about knotting of spaces more general than polyhedra; most of what is known is summarized in the recent book by Daverman and Venema. In this light the space $X$ and its "tame" embeddings don't look very attractive, to be honest. - -so realizations of this big snarly thing in $\Bbb R^3$ should all be ambient-isotopic to each other, too, but I'm not certain about that. - -Are you saying that if $L$ and $L'$ are proper tame links of countably many circles in $\Bbb R^3$ with the property that every finite link is a sublink of both $L$ and $L'$, then $L$ is ambient isotopic to $L'$? -This is not the case: take $L$ to be the union of links $L_n$, where each $L_n$ is a finite link in a ball $B_n$ of radius $1/3$ centered at $(0,0,n)$, and the ambient isotopy types of the $L_n$ are precisely those of all non-split finite links (each occurring once). On the other hand pick some infinite link $L''$ that is `non-split at infinity'; for instance such that its $i$th component has a nonzero linking number with the $(i+1)$st component for each $i$. As long as $L$ and $L''$ are disjoint, their union $L'$ is a again link that is non-split at infinity; by this I more specifically mean that it is not isotopic to any link whose $i$th component lies in the ball $B_i$ for each $i$. In particular, $L$ is not isotopic to $L'$ (even non-ambiently). -EDIT: One has of course to specify exactly what is the language and the first-order theory, but it seems that with a sufficiently reasonable setup, the model theoretic conditions given in Scott's edit should amount to the following. -Rather than unions of sequences $\mathcal{L}$ of finite tame links $L_1\subset L_2\subset\dots$ up to ambient isotopy, we consider the sequences themselves, up to a weaker equivalence relation of pro-isotopy (defined below). -A "Fraisse link" is then the pro-isotopy class of a sequence $L_1\subset L_2\subset\dots$ satisfying the property in my comment above: whenever some $L_n$ occurs as a sublink of a finite link $L$, then this $L$ must be equivalent to a sublink of some $L_m$ relative to $L_n$ (that is, the identity on $L_n$ must extend to a self-homeomorphism of $\Bbb R^3$ sending $L$ onto a sublink of $L_m$). -Two sequences $\mathcal{L}$ and $\mathcal{L}'$ as above are pro-isotopic if each $L_i$ is equivalent to a sublink of some $L_j'$, $j=j(i)$, via a self-homeomorhpism $h_i$ of $\Bbb R^3$; and each $L_j'$ is equivalent to a sublink of some $L_k$, $k=k(j)$, via a self-homeomorphism $h_j'$ of $\Bbb R^3$, so that for each $i$, the composition $h_{j(i)}'h_i$ sends $L_i$ onto itself; and for each $j$, the composition $h_{k(j)}h_j'$ sends $L_j$ onto itself. -Pro-isotopy is indeed very similar to pro-homotopy, which brings to attention an equivalent definition of pro-isotopy (a la Pontryagin's original definition of pro-isomorphism and Siebenmann's definition of shape): pro-isotopy is the equivalence relation generated by the following two relations: 1) the relation of being a subsequence, 2) sequences $\mathcal{L}$ and $\mathcal{L}'$ are related if there exists a sequence of self-homeomorphisms $H_i$ of $\Bbb R^3$ such that $H_i(L_i)=L_i'$ and $H_{i+1}(L_i)=H_i(L_i)$. -[The proof that the two definitions of pro-isotopy are equivalent is by a standard argument: given $\mathcal{L}$ and $\mathcal{L}'$ that are pro-isotopic in the original sense, the sequence $h_1(L_1)\subset L_{j(1)}'\subset h_{k(j(1))}(L_{k(j(1))})\subset\dots$ has a subsequence that is also a subsequence of $\mathcal{L}'$; and on the other hand is related in the sense of (2) to the sequence $L_1\subset h_{j(1)}'(L_{j(1)}')\subset L_{k(j(1))}\subset\dots$, which in turn has a common subsequence with $\mathcal{L}$.] -The point of these definitions is that a Fraisse link (as defined above) is, obviously, unique. -Now that a Fraisse link is defined, I understand the original question as follows: can a Fraisse link be identified with an (ambient isotopy class of) an infinite link? -The answer is no. It is clear that sequences $L_1\subset L_2\subset\dots$ and $L_1'\subset L_2'\subset\dots$ are pro-isotopic if their unions are ambient isotopic. -But the converse does not hold, by the reasons that Bill Thurston gave in the end of his answer. I'm not sure that the above definitions of pro-isotopy and a Fraisse link are exactly what he had in mind, but anyhow with these definitions his argument makes sense.<|endoftext|> -TITLE: Hidden convexity -QUESTION [7 upvotes]: Suppose you are given a domain $\Omega \subset \mathbb{R}^n,$ and a (Morse) function $f: \Omega \rightarrow \mathbb{R},$ all of whose critical points are positive-definite. -The question is: is there a diffeomorphism $\phi: \Omega \rightarrow \Omega$ such that $f \circ \phi$ is a convex function? -Another edit The above is a typo (or braino): I actually meant $\phi: \Omega \rightarrow \Omega_1,$ for some $\Omega_1 \subset \mathbb{R}^n.$ -EDIT A bit of motivation: this comes from (many years ago) constructing a counterexample to the statement that a convex function has a unique minimum (which is true on a convex set, but not otherwise). If a function has a unique critical point, the question is fairly easy (just by looking at level sets, and sending them to round circles). - -REPLY [5 votes]: It seems that the following interpretation is not covered yet: the domain is allowed to change but $f$ is not required to be proper. In this case the answer is no. Take any connected domain on the plane and any Morse function with at least two local maxima. Remove all critical points (except possibly local minima) from the domain, now all critical points are local minima. -Near the two local maxima, there are two closed level curves $\gamma_1$ and $\gamma_2$ such that $\gamma_2$ lies in the "lower" component of $\mathbb R^2\setminus\gamma_1$ and vice versa, where by "lower" component of $\mathbb R^2\setminus\gamma_i$ I mean the one where the values of $f$ are locally (near $\gamma_i$) smaller than on $\gamma_i$. This property of a pair of curves is preserved by diffeomorphisms. But it cannot hold for a convex function because, for a convex $f$, the "lower" component must be the bounded (convex) region enclosed by the level curve, but two curves cannot be both inside each other's bounded regions.<|endoftext|> -TITLE: An $L^0$ Khintchine inequality -QUESTION [24 upvotes]: Suppose that $\epsilon_1,\epsilon_2,\ldots$ are IID random variables with the Bernoulli distribution $\mathbb{P}(\epsilon_n=\pm1)=1/2$, and $a_1,a_2,\ldots$ is a real sequence with $\sum_na_n^2=1$. Letting $S=\sum_n\epsilon_na_n$, the question is whether there exists a constant $c > 0$, independent of the choice of $a$, with -$$ -\mathbb{P}(\vert S\vert\ge1)\ge c.\qquad\qquad{\rm(1)} -$$ -That is, I am interested in finding a bound on the probability of the sum being within one standard deviation of its mean. -If true, this represents a particularly sharp version of the $L^0$ Khintchine inequality. Considering the example with $a_1=1$ and all other $a_i$ set to zero, for which $\mathbb{P}(\vert S\vert > 1)=0$, it is necessary that the inequality inside the probability in (1) is not strict. Also, considering the example with $(a_1,a_2,a_3)=(1/\sqrt2,1/2,1/2)$, it can be seen that $c\le1/4$. I wonder if it is possible to construct further examples showing that $c$ must, in fact, be zero? -For any $0 < u < 1$, it is easy to find a bound -$$ -\mathbb{P}(\vert S\vert > u)\ge c_u -$$ -for $c_u > 0$ a constant independent of $a$. Considering the case with $a_1=a_2=1/\sqrt{2}$ and all other $a_i$ set to zero, it is clear that $c_u \le 1/2$. In fact, it can be shown that $c_u=(1-u^2)^2/3$ will suffice (see my answer to this other MO question), but $c_u$ decreases to zero as $u$ goes to $1$, so this does not help with (1). Combining the Paley-Zygmund inequality with the optimal constants in the $L^p$-versions of the Khintchine inequality for $p > 0$ (see ref. 1 or 2) it is possible to give improved values for $c_u$, but it still tends to zero as $u$ goes to 1. -My apologies if this is either obvious or some well-known fact that I have missed, but I could not find any reference for it. This question is something that I originally thought about while writing up some notes on stochastic integration (posted on my blog), as the $L^0$-version of the Khintchine inequality can be used to prove the existence of the stochastic integral. However, it is not necessary to have something as strong as (1) in that case. More recently, it came up again while answering this MO question. -[Update: Its been some time since this question was posted and answered. Many thanks to Anthony, Iosif and Ravi. There is ongoing research on this problem, and it seems likely that the optimal value of $c$ is 7/32 as conjectured by Oleszkiewicz in the paper linked in Ravi's answer. See Some explorations on two conjectures about Rademacher sequences by Hu, Lan and Sun, where the optimal value of 7/32 is shown for sequences of length at most 7, but it is still open in general. Also, the preprint Proof of Tomaszewski's Conjecture on Randomly Signed Sums by Keller and Klein also includes the claim that their methods improve the best known value for $c$ to 1/8.] -Refs: - -Haagerup, The best constants in the Khintchine inequality, Studia Math., 70 (3) (1982), 231-283. -Nazarov & Podkorytov, Ball, Haagerup, and distribution functions, Preprint (1997). Available from Fedja Nazarov's homepage. - -REPLY [8 votes]: In 1996, Krzysztof Oleszkiewicz proved that $c = 1/10$ works. Here is the reference: K. Oleszkiewicz (1996), "On the Stein property of Rademacher sequences", Probability and Mathematical Statistics 16:1, 127-130. The paper is currently available on this page: -http://www.math.uni.wroc.pl/~pms/publications.php?nr=16.1<|endoftext|> -TITLE: Angle of a regular simplex -QUESTION [22 upvotes]: I find the following question embarrassing, but I have not been able to either resolve it, or to find a reference. - -What is the vertex angle of a regular $n$-simplex? - -Background: For a vertex $v$ in a convex polyhedron $P$, the vertex angle at $v$ is the proportion of the volume that $P$ occupies in a small ball around $v$. In symbols, $$\angle v=\lim_{\varepsilon\to 0} \frac{|B(v,\varepsilon)\cap P|}{|B(v,\varepsilon)|}.$$ Up to normalization, this definition agrees with the familiar definition of the angles in the plane, or the solid angle in $3$-space. - -REPLY [3 votes]: Here is a link to the original paper by Rogers mentioned by Joseph O'Rourke in his answer, with a short proof of the asymptotic formula for $F_n(\alpha)$, which is reproduced in the book Sphere Packings by Chuanming Zong. -It is also shown there that the multiplicative error in the formula is at most $1+ \frac{31}{12n} + O\left(\frac{1}{n^2}\right)$ in the case of the regular simplex, where $\sec 2\alpha=n$ (and thus $c=1$, or, $b=1$ in the paper).<|endoftext|> -TITLE: Hilbert's 17th Problem for smooth functions -QUESTION [11 upvotes]: Consider an open subset $U \subseteq \mathbb{R}^n$ and a smooth function $f\colon U \longrightarrow \mathbb{R}$ with $f(x) \ge 0$ for all $x \in U$. -It is then known (if I remember correctly: by Michor?) that $f = g^2$ with a function $g$ which can be shown to be twice differentiable but not $C^2$ in general. In particular, a smooth square root does not exist in general. -My question is whether $f$ can be represented as a sum of squares of smooth functions, i.e. the smooth version of Hilbert's problem 17, and if so, what is the minimal number of squares needed? - -REPLY [4 votes]: Quoting from the introduction of Brumfiel's book Partially Ordered Rings and Semi-Algebraic Geometry (1979, CUP LMS Lecture Notes 37): - -As a final remark on the birational interpretation of Artin's solution of Hilbert's 17th problem, consider a different category, that of smooth manifolds and smooth real valued functions. Then Paul Cohen has shown me that (i) there exist nowhere negative smooth functions on any manifold which are not finite sums of squares of smooth functions (in fact, the zero set can be a single point) and (ii) given any nowhere negative smooth $f$, there are $h$, $g$, both smooth and $h$ not a zero divisor, such that $h^2f = g^2$. The zero divisors are, of course, the smooth functions which vanish on some open set. - -If the statement is correct ("on any manifold"), then there should be a counterexample on $\mathbb{R}$. I would love to see how it is constructed! But it does not seem to have been published, because in the 2005 paper ("Sommes de carrés de fonctions dérivables", Bull. Soc. Math. France, 133, 619–639) in which he proves (inter alia) that at least a $C^\infty$ function $f$ on $\mathbb{R}$ that is nonnegative can be written as $g^2 + h^2$ where $g$ and $h$ are $C^m$ for $m$ arbitrarily large but finite, Jean-Michel Bony mentions these counterexamples and asserts that, as far as he knows, they have not been published (he points to the aforementioned book, and also the book Real Algebraic Geometry by Bochnak, Coste and Roy, where a similar statement is attributed to P. Cohen and D. Epstein: see §6.6.4).<|endoftext|> -TITLE: Ping Pong and Free Group Factors -QUESTION [9 upvotes]: This question concerns alternative characterizations of free group factors. The ping pong lemma is a well-known criteria for the freeness of a group. I've often wondered if there is a ping pong like criterion that can be used to determine if a given type $II_{1}$ factor is a free group factor, e.g. a ping pong-like criterion for the action of the factor on some Hilbert space. - -Question: Is there a ping pong lemma analogue for group von Neumann algebras? - -REPLY [12 votes]: I am guessing that the answer is "yes" if you interpret the question in the following way. Let $A_i$ be some subalgebras of a von Neumann algebra $(M,\tau)$ and assume that there are mutually orthogonal Hilbert subspaces $H_i$ of $H=L^2(M)$ so that for all $i$, $x (H\ominus H_i) \subset H_i$ whenever $x\in A_i$ with $\tau(x)=0$. -Let us also assume that $1 \perp \oplus_i H_i$ (probably this is not necessary). -Then if $y = x_1 \dots x_n$ with $x_j \in A_{i(j)}$, $i(1)\neq i(2)$, $i(2)\neq i(3)$, etc. and $\tau (x_j) = 0$, we have: -$x_n 1 \in H_{i(n)}$ since $1\in H\ominus H_i$; -$x_{n-1} x_n 1 \in H_{i(n-1)}$ since $x_n 1 \in H_{i(n)} \subset H\ominus H_{i(n-1)}$ (because $i(n)\neq i(n-1)$ and so $H(i(n))\perp H(i(n-1))$; -$x_{n-2} x_{n-1} x_n 1 \in H_{i(n-2)}$ since $x_{n-1} x_n 1 \in H_{i(n-1)}\subset H\ominus H_{i(n-2)}$, etc. Thus We get that $x_1\dots x_n 1 \in H_{i(1)} \perp 1$, so that $\tau(y)=0$. It follows that $A_1,\dots,A_n$ are freely independent. -(Conversely, if $M$ is generated by $A_1,\dots,A_n$ and they are free inside of $M$, then $L^2(M) = \mathbb{C}1 \oplus \oplus_k \oplus_{j_1\neq j_2, j_2\neq j_3,\dots} L^2_0(A_{j_1})\otimes \cdots \otimes L^2_0(A_{j_k})$, where $L^2_0(A_j) = \{1\}^\perp \cap L^2(A_j)$. Then you can take $H_j = \oplus_k \oplus_{j_1\neq j_2, j_2\neq j_3,\dots; j_1= j} L^2_0(A_{j_1})\otimes \cdots \otimes L^2_0(A_{j_k})$ and then $H_j$ are orthogonal and -$H\ominus H_j$ is taken to $H_j$ by any $x\in A_j $ with $\tau(x)=0$). -If you now make some assumption (e.g. that $A_j$ are finite-dimensional, abelian or hyperfinite) then it follows from Ken Dykema's results (see e.g. his paper on Interpolated free group factors in Duke Math J.) that the von Neumann algebra they generate inside of $M$ is an interpolated free group factor. This is similar to the assumption you have put on the group (since the subgroup generated by a single element in the ping-pong lemma is necessarily abelian). -On the other hand, you raise the much bigger question of whether there exists some criterion that singles out free group factors -- just as the various functional-analytical criteria were shown by Connes to be equivalent to hyperfiniteness. Unfortunately, not much in known in this direction (note that a similar question exists on the ergodic equivalence side of things: is there a functional-analytic way of recognizing treeable actions? Or Bernoulli actions of free groups?)<|endoftext|> -TITLE: Existence of Solution for a Differential Matrix Equation -QUESTION [5 upvotes]: I have some problems proving that a Differential Matrix Equation has a solution. I apologize if the question is too elementary, but I've found this theorem stated everywhere on the web without any reference or clue about how to prove it. -What I exactly mean with a Differential Matrix Equation is: -$X'=AX+B$. Where $A$ is a matrix of size $n\times n$ and $B$ is a column vector. Both $A$ and $B$ have coefficients which are holomorfic functions in a convex open set $\Omega$ and continuous on the closure $\bar\Omega$. We also have an initial condition $X(z_0)=U_0$, where $U_0$ is a column vector of complex coefficients. -So far I know how to prove it when $A$ has constant coefficients. However the proof cannot be applied as it uses the trick of $(e^{Az})'=Ae^{Az}$ to find an explicit solution and in general $(e^A)'\neq Ae^A$ if $A$ is nonconstant. -I've also read about Magnus Series, but I don't fully understand them. Also I'm only interested in the existence, so I'd prefer an easier way to prove that there are solutions. - -REPLY [3 votes]: This really sounds like homework. Anyway, try this. We work under the assumption that $A,B$ are entire functions, otherwise everything works the same but the results are local. -If $X(z)=u+iv$ is holomorphic, its derivative can be written as $X'(z)=u_x+iv_x$. Now separate real part and imaginary part of your system, what you get is a nice linear system of $2n$ equations for $2n$ unknown functions, in the variable $x$, with $y$ as a parameter. Standard advanced calculus results give you the existence of a smooth global solution, smoothly dependent on the parameter $y$ (don't ask for references, just differentiate the system w.r.to $y$ and repeat). Now do the same trick writing $X'(z)=v_y-iu_y$: you get another set of $2n$ smooth functions. But of course $X'(z)=AX+B=u_x+iv_x=v_y-iu_y$ which means that $u,v$ satisfy the Cauchy-Riemann conditions and $X$ is holomorphic.<|endoftext|> -TITLE: Hochschild homology of Fukaya category in mirror symmetry -QUESTION [8 upvotes]: Hi -Can one explain to me what is the Hochschild homology of Fukaya category? -I mean the definition. -You can use the notations of FOOO (Fukaya-Oh-Ono-Ohta) if it helps you to explain easier. -I know what the Fukaya category is but I am very poor when it comes to algebra. -Also please explain what is the corresponding Hochschild homology in B-side?(please include the definition too) - -REPLY [14 votes]: As Kevin comments, Hochschild homology and cohomology are defined for any $A_\infty$-category $\mathcal{A}$. That includes Fukaya categories of symplectic manifolds and dg enhancements of the bounded derived category of varieties. -The most concrete definition of Hochschild homology $HH_\ast(\mathcal{A},\mathcal{A})$ is via the cyclic bar complex. One takes the direct sum over all $d\geq 0$ and all sequences of objects $X_0,\dots, X_d$ of the tensor product -$$ \hom(X_d,X_0) \otimes \hom (X_{d-1},X_d) \otimes \dots \otimes \hom(X_1,X_2) \otimes \hom(X_0,X_1). $$ -You should picture this tensor product not as a linear chain but as circular one; the term $\hom(X_d,X_0)$ is special. In the case where $\mathcal{A}$ is the Fukaya category $\mathcal{F}(M)$ of a symplectic manifold $M$, the $X_i$ are (decorated) Lagrangian submanifolds, and when these are transverse the elements of $\hom(X_i,X_j)$ are linear combinations of intersection points between $X_i$ and $X_j$. So the Hochschild chain complex has a basis given by cyclic sequences of intersection points, one of them marked as special. -The boundary operator is given by taking some sequence of $k\geq 1$ cyclically adjacent terms in the cyclic tensor product and composing them via one of the $A_\infty$-structure maps $\mu^k$ so as to shorten the cyclic sequence by $k-1$. In the Fukaya categorical case, the $\mu^k$ count pseudo-holomorphic $(k+1)$-gons. One does this in all possible ways and sums with hard-to-fathom signs as in Abouzaid's paper -1001.4593 -(it would be wonderful if someone can tell me how to make these signs transparent). There is also a chain-lengthening contribution to the complex from the obstruction cochain $\mu^0$. -This concrete description has some real advantages; for instance, as Seidel noticed, there is a geometric description of a homomorphism from Hochschild homology to quantum cohomology -$$ HH_{\ast}(\mathcal{F}(M),\mathcal{F}(M)) \to QH^{\ast}(M) $$ -(this for closed $M$) which is expected to be an isomorphism. -For computations, two facts are noteworthy. First, Hochschild homology has Morita-invariance properties. For example, it is unchanged under passing to the category of twisted complexes, which is useful because one can restrict attention to some collection of objects that generate the derived category. Second, it is the derived tensor product of graded bimodules (see Sasha's answer), which means in practical terms that you can compute it using much smaller complexes than the cyclic bar complex. -The conjecture that $HH_{\ast}(\mathcal{F}(M),\mathcal{F}(M))\cong QH^{\ast}(M)$ is consistent with mirror symmetry. In that case, the twisted complexes on $\mathcal{F}(M)$ (technically, the idempotent completion thereof - this doesn't affect $HH_\ast$ either) are quasi-equivalent to a dg-enhanced bounded derived category on the mirror manifold $W$, defined over some non-archimedean Novikov-type field. -Here my understanding is rather feeble, but I think the story is that $HH_\ast$ for this dg category is isomorphic to Hochschild homology of the non-singular variety $W$ (for various equivalent definitions, see Swan's article). This is known to be isomorphic to sheaf cohomology $H^\ast(W, \Omega^\ast_W)$ of the algebraic differential forms, hence to ordinary cohomology of $W$, hence finally to cohomology of $M$. So, if you have an HMS theorem for $M$ and $W$, you at least know that $HH_\ast(\mathcal{F}(M),\mathcal{F}(M))$ is isomorphic to $QH^\ast(M)$. - -REPLY [2 votes]: To define Hochschild homology of a category $C$ one considers the category of its endofunctors $End(C)$. It has a natural tensor structure given by the composition of functors. Moreover, it has a distinguished object $1_C$ --- the identity functor of $C$. The Hochschild homology is defined as self-Tors of this object: -$$ -HH_\bullet(C) = Tor_\bullet^{End(C)}(1_C,1_C). -$$ -In some sense this is a meta-definition. To make it work you need to have some structure on $End(C)$ which would allow you to speak about Tors. Usually, one replaces $End(C)$ by an appropriate faithfully flat subcategory which has such a structure. For example, if $C$ is a dg-category, one considers the category of $C$-bimodules. It is a dg-category as well, each bimodule produces a functor (by tensoring), its natural tensor structure corresponds to the composition of functors, and it has a so-called "diagonal bimodule" corresponding to the identity functor. So, in the above formula you should consider $1_C$ as the diagonal bimodule and $End(C)$ as the category of bimodules over $C$. -The Fukaya category, is defined as derived of a certain $A_\infty$ category with which you can perform the same procedure as with dg-category (consider $A_\infty$-bimodules etc). This gives you the definition of Hochschild homology of a Fukaya category. -On B-side you consider the derived category of coherent sheaves. It doesn't have any specific dg-enhancement, but you can pick up any one you like and again apply the above definition.<|endoftext|> -TITLE: arXiv number and mathscinet -QUESTION [14 upvotes]: I recently submitted a paper to a journal with references included, most of them taken from mathscinet (for math papers) and from spires (for physics papers). Now mathscinet does not contain eprint numbers, whereas spires does. The referee requested me to include the arxiv numbers for all the math papers. I ended up doing this manually, but is there a more systematic way to do this? I raise this issue since arXiv numbers could be more important to readers who download papers online. - -REPLY [8 votes]: JabRef has support for updating arXiv, MathSciNet and a bunch of other identifiers (as well as revising existing bibliographic metadata with updates from web services). -There are two options. -First, select the entries and then use "Tools > Search document identifiers online > arXiv". This will automatically query the arXiv and puts the identifier in the eprint field. -Second, in the entry editor, choose "Update with bibliographic information from the web" (the circled arrow on the left toolbar in the entry editor). This will query the selected database and updates all bibliographic metadata. Usually, I use this feature to see if a preprint in my library is already published but it also works the other way around to discover the arXiv identifier for a published article. -Disclaimer: I'm one of the core developers of JabRef.<|endoftext|> -TITLE: Baum-Connes conjecture -QUESTION [14 upvotes]: There are many versions of the Baum-Connes conjecture (the original, coarse, with coefficients, etc.). I would like to know what group theory results are needed in order to prove or disprove one of these conjectures. - -REPLY [7 votes]: I want to second David Ben-Zvi's recommendation of Nigel Higson's ICM lecture, though there is a bit of progress that has been published since it was written. Fortunately, you needn't venture further than Nigel's webpage to learn more: his paper "Counterexamples to the Baum-Connes Conjecture" with Lafforgue and Skandalis gives some more recent counter-examples to BC with coefficients and coarse BC. -A few other comments... -As far as I know, the ordinary Baum-Connes conjecture is known to be true for more or less every group that anyone can name. The experts (of which I am not one) seem to suspect that the conjecture is not true in general but that finding a counter-example requires new techniques for constructing discrete groups. -There are a few fairly general techniques for proving the conjecture. The first and most conceptually satisfying is the "Dirac / dual Dirac" method which involves using KK-theory to imitate Atiyah's operator theoretic proof of Bott periodicity. As far as I know there is no evidence that this technique is not powerful enough to prove injectivity in general, but there are examples of groups for which BC is true but the Dirac / dual Dirac method cannot prove surjectivity. I suggest Higson and Guentner's "Group C* algebra's and K-theory" for a discussion of these issues - the paper starts more or less from scratch and reads like a short textbook. -The second method is more of a philosophy: BC tends true for groups which act on nice spaces in a minimally nice way. For example, it is known to be true for a-T-menable groups (see "E-theory and KK-theory for groups which act properly and isometrically on Hilbert space" by Higson and Kasparov). One of the frontiers of the conjecture is in the realm of p-adic groups - I think Paul Baum is still quite active here - and one of the basic techniques seems to be to get p-adic groups to act on buildings. -I hope that helps! For more advice, I would really suggest talking directly to Nigel - I don't think BC is quite as central to his interests these days, but I would guess he's as up on the subject as anyone.<|endoftext|> -TITLE: Is a biideal of a Noetherian Hopf algebra automatically Hopf? -QUESTION [9 upvotes]: Let $H$ be a Hopf algebra over a field $k$, and $I$ be a biideal of $H$. I am looking for conditions that guarantee that $I$ is a Hopf ideal (that means $S\left(I\right)\subseteq I$). -One condition that definitely works is that $\dim\left(H / I\right) < \infty$ (where $\dim$ means dimension as a $k$-vector space). This is a well-known consequence of the criterion that a bialgebra $A$ is a Hopf algebra if and only if the $k$-linear map $A\otimes A\to A\otimes A,\ x\otimes y\mapsto xy_{(1)}\otimes y_{(2)}$ is bijective. (This criterion must be applied to $H$ and $H / I$.) -I suspect that some kind of Noetherianness of $H$ or $H / I$ (note that I am not specifying left or right or bi, because I have no idea) would make another criterion. My suspicion is based on the commutative $H$ case. Does anyone see a proof or a quick counterexample? - -REPLY [6 votes]: The answer is yes. -In "Quotients of Hopf Algebras", Warren D. Nichols, Comm. Algebra 6(1978), 1789-1800, proves that if $I$ is a bi-ideal then it will be a Hopf ideal under any of the following conditions: -$H/I$ is finite dimensional -$H/I$ is commutative -$H$ is pointed -$H$ is cocomutative -What you want is a generalization of the above, see corollary 2.4 in "The largest Hopf subalgebra of a bialgebra" by M. S. Eryashkin and S. M. Skryabin. They prove that a bi-ideal $I$ is a Hopf ideal provided that $H/I$ is weakly finite. In particular every bi-ideal is a Hopf ideal when $H$ is left or right Noetherian, or when $H$ satisfies a polynomial identity.<|endoftext|> -TITLE: Direct sum of injective modules over non-Noetherian rings -QUESTION [8 upvotes]: By the Bass-Papp theorem, if every direct sum of injective $R$-modules is injective then $R$ is Noetherian. I would like to know if there exists an injective module over $R$ non-Noetherian, that splits as infinite direct sum of nonzero (injective) $R$-modules. - -REPLY [12 votes]: A module is called $\Sigma$-injective if a direct sum of arbitrarily (equivalently, countably infinitely) many copies of that module is injective.  So it suffices to find an example of a $\Sigma$-injective module over a non-noetherian ring.  Apart from silly examples such as a direct product of two rings one of which is one-sided noetherian and the other of which is not, the main theorem of C. Megibben, “Countable injective modules are $\Sigma$-injective,” Proc. Amer. Math. Soc. 84 (1982), no. 1, 8–10, says what the title indicates.  This gives all sorts of examples of $\Sigma$-injective modules over non-noetherian rings.<|endoftext|> -TITLE: Locales and Topology. -QUESTION [35 upvotes]: As someone more used to point-set topology, who is unfamiliar with the inner workings of lattice theory, I am looking to learn about the localic interpretation of topology, of which I only have a limited understanding. As such, I have some questions: - -What are some accessible texts or online references on the subject? -What are some recent results in point-free topology that are unique to the subject, i.e., not translations of results from general topology into localic language? - -REPLY [4 votes]: As for question 2.: it's hardly recent, but since I didn't see it mentioned in this thread, let me mention that the Tychonoff theorem for locales does not require the axiom of choice and is a piece of purely constructive mathematics. So this cannot be in any way a translation of any of the proofs familiar from general topology; proving it involves genuinely new ideas specific to locale theory. See also this MathOverflow answer to the question "What is your favorite proof of Tychonoff's theorem?", and particularly the comments appearing below.<|endoftext|> -TITLE: Generators of the Odd Dimensional Quantum Spheres -QUESTION [11 upvotes]: As is well-known, the $(2N-1)$-quantum sphere $S^{2N-1}_q$ is defined to be the invariant subalgebra of $SU_q(N)$ under the coaction $\Delta_R = (id \otimes \pi) \circ \Delta$, where $\Delta$ is the comultiplication of $SU_q(N)$, and $\pi: SU_q(N) \to U_q(N-1)$ is the Hopf algebra surjection defined by setting, for $i,j \neq 1$, $\pi(u^i_1)=\pi(u^1_j)=0$, $\pi(u^1_1)$ = det$_q^{-1}$, and $\pi (u^i_j) = u^{i-1}_{j-1}$. (Recall that the invariant subalgebra of a $H$-coaction $\Delta_R$ on vector space $V$ is the subspace of all elements $v$ for which $\Delta_R(v) = v \otimes 1$.) An oft quoted result is that $S^{2N-1}_q$ is generated, as an algebra, by the elements $u^i, S(u^1_i)$, for $i=1, \ldots N$. Now it is clear that these elements are invariant, but it is far from clear (at least to me) that generate all the invariant subspace. Can anyone see why? The usual references given are in Russian and, even at that, are unavailable on the web. - -REPLY [5 votes]: This is shown in the book Quantum Groups and Their Representations, by Klimyk and Schmudgen. The result you ask for is Proposition 63 in Chapter 11. I'd expand more upon this but I have to give a talk shortly, and anyway it's all there in the book. -Edit: I guess I should say that the point here is the representation theory. Classically, we can think of the algebra of functions on the sphere as the representation of $SU(N)$ induced by the trivial representation of the subgroup $U(N-1)$. Then Frobenius Reciprocity tells you which representations of $SU(N)$ occur as summands in the function algebra, and hence which matrix coefficients generate it. These things all transfer over to the quantum setting because they are phrased in terms of representations.<|endoftext|> -TITLE: Simple but serious problems for the edification of non-mathematicians -QUESTION [25 upvotes]: When people graduate with honors from prestigious universities thinking everything in math is already known and the field consists of memorizing algorithms, then the educational system has failed in one of its major endeavors. -If members of the next freshman class will take just one one-semester math course before becoming the aforementioned graduates, here's what I think I might do (and this posting is indeed a question, as you will see). I would not have a fixed syllabus of topics that the course must cover by the end of the semester. I would assign very simple but serious problems that I would not tell the students how to do. A few simple examples: - -$3 \times 5 = 5 + 5 + 5$ and $5 \times 3 = 3 + 3 + 3 + 3 + 3$. Why must this operation thus defined be commutative? -A water lily has a single leaf floating on the surface of a pond. The leaf doubles in size every day. After 16 days it covers the whole pond. How long will it take two such leaves to cover the whole pond. (Here lots of students say "8 days". I might warn them against that. This is the very hardest problem assigned in an algebra course that I taught, according to most of the students.) -Here is a square circumscribing a circle. [Illustration here.] Here is how you use this to see that $\pi<4$. [Explanation here.] Now figure out how to prove that $\pi > 3$ by a similarly simple argument. -Multiples of 12 are 12, 24, 36, 48, 60, 72, 84, ...... Multiples of 18 are 18, 36, 54, 72, 90, ..... The smallest one that they have in common is 36. Multiples of 63 are 63, 126, 189, 252, 315, 378,..... Multiples of 77 are 77, 154, 231, 308, 385,.... Could this sequence go on forever without any number appearing in both lists? (Usual answer: Yes. It will. Because 63 and 77 have nothing in common.) Is it the case that no matter which pair of numbers you start with, eventually some number will appear in both lists? - -I said simple but serious, the latter meaning they will actually learn something worth learning about mathematics or about how to think about mathematics. Not all need be as elementary as these. With some of the less elementary problems I might sketch a solution or write out a solution in detail and then ask questions about the solution. -I would not fix in advance the date at which problems were to be turned in, but would set deadlines after discussion reveals that serious difficulties are overcome. I might also do some "teasing" concerning various math topics not covered. -HERE'S THE QUESTION: Which published books of problems can participants in this forum recommend for this purpose? Why those ones? - -REPLY [3 votes]: How about The Theory of Remainders by Andrea Rothbart. -I remember back in the day I was struggling with the concept of modular arithmetic and randomly came across the book above. It's really well written in an unorthodox way as a dialogue between two people talking about modular arithmetic. The book introduces basic concepts of abstract algebra and has plenty of "simple, but serious" exercises. If I recall correctly, it did a really good job of motivating the concept of fields. Above anything, it was written with a high school audience in mind, so incoming freshmen should not be deterred by the level of difficulty. I also found the style of the book engaging. I dare say I was bitten by the number theory bug shortly after reading it.<|endoftext|> -TITLE: possible CM-types of abelian varieties -QUESTION [8 upvotes]: Fix a CM-field $K$ of degree $2g$. Let $A$ be a polarized abelian variety of dimension $n$ over $\mathbb{C}$, with an isomorphism $\theta : K \to End_{\mathbb{C}}(A) \otimes_{\mathbb{Z}} \mathbb{Q}$. (So $n$ is a multiple of $g$.) -Then, the tangent space to the identity of $A$ defines an $n$-dimensional complex representation $\Phi$ of $K$ (which Shimura calls the type of $(A, \theta)$). Write -$\tau_1, \tau_2, \ldots, \tau_g, \rho \tau_1, \rho \tau_2, \ldots, \rho \tau_g$ -for the different embeddings of $K$ into $\mathbb{C}$, where $\rho$ denotes complex complex conjugation. Then (as shown by Shimura), the representation $\Phi$ decomposes as a direct sum -$\Phi = \bigoplus_{\nu = 1}^g (r_\nu \cdot \tau_\nu \oplus s_\nu \cdot \rho \tau_\nu)$ where $r_1 + s_1 = r_2 + s_2 = \cdots = r_g + s_g = \frac{n}{g}$. -What are the possible values of $(r_1, s_1, r_2, s_2, \ldots, r_g, s_g)$ as $A$ ranges over all complex abelian varieties, while $K$ remains fixed? Can they be arbitrary nonnegative integers satisfying the above constraint, or does fixing $K$ impose further conditions? - -REPLY [5 votes]: There are certain additional conditions on ``multiplicities" that are spelled out in the original Shimura's paper. Let me discuss a couple of interesting cases. -The case of CM type, i.e. when $n=g$. Let $\Phi$ be the corresponding CM type, which is a $n$-element set of embeddings of $K$ into the field $C$ of complex numbers. For each pair of complex-conjugate embeddings $K \to C$ exactly one of them lies in $\Phi$. -If $K$ does not contain a proper CM subfield then everything is fine, i.e., the endomorphism algebra of $A$ is $K$. However, if $K$ contains a proper CM subfield $L$ then a ``bad" choice of $\Phi$ would imply that the endomorphism algebra of $A$ is a matrix algebra over $L$ rather than $K$. In order to guarantee that the endomorphism algebra is $K$, one should require that if $\tau_1, \tau_2 \in \Phi$ and their restrictions to the maximal totally real subfield $L_0$ of $L$ coincide then their restrictions to $L$ also coincide. (Of course, one should require it for all CM subfields.) -For example, if $n=g=2$ and a quartic CM field $K$ contains an imaginary quadratic subfield then $K$ must contain another imaginary quadratic subfield as well! In this case one may check that the endomorphism algebra of $A$ is always bigger than $K$ (and $A$ is isogenous to a square of a CM elliptic curve). -The case of an abelian surface $A$ and an imaginary quadratic field $K$, i.e., $g=1, n=2$. It is known that the endomorphism algebra of an abelian surface is never an imaginary quadratic field. (Actually, as was pointed out by Frans Oort, this assertion remains true in characteristic p.) More precisely, suppose we are given an embedding of $K$ in the endomorphism algebra of $A$. Then either $A$ is isogenous to a square of an elliptic curve with multiplication by $K$ or the endomorphism algebra of $A$ is an indefinite quaternion algebra over the rationals.<|endoftext|> -TITLE: Which compact groups have nonisomorphic irreducible representations of the same dimension? -QUESTION [10 upvotes]: If $\Gamma$ is a compact simply-connected semisimple Lie group, then the Weyl Dimension Formula tells us exactly which dimensions it can act irreducibly on. -For certain $\Gamma$, it is easy to find pairs of nonisomorphic representations of the same dimension: -1). $A_n (n\geq 2)$, $C_n$ ($n$ = A116940(k)), $D_n (n\geq 4)$, and $E_6$ each have pairs of fundamental irreducible representations of the same dimension. -2). Additionally $G_2$ has two irreducible representations of dimension 77. -Furthermore, given that $\Gamma$ has one pair of nonisomorphic representations of the same dimension, it is easy to prove (using the Weyl Dimension Formula) that it has infinitely many such pairs. -Question 1: Among the remaining groups not mentioned above, which are known to have pairs of nonisomorphic irreducible representations of the same dimension? -Question 2: For obvious reasons $A_1$ cannot have such pairs, but are there any other cases where one can rule out the existence of such pairs? -It would seem that since the Dimension Formula so greatly restricts the possible dimensions for the other groups that in the long run there must be pairs of irreducibles of the same dimension; if anyone knows of any results along the lines of such heuristic forcing arguments those would be useful as well. -Edit: Robert's answer below reminded me that such pairs also occur for: -3). $B_n$ when $n$ = A116940(k) just as in the $C_n$ case because of a result I proved awhile back. More generally, one can show that if $B_n$ has such a pair then so does $C_n$ and conversely if $C_n$ has such a pair then so does $B_n$, so Robert's comment also shows that $C_4$ and $C_5$ have such pairs (the pairs arising from Robert's examples are in dimensions 11354112 and 38928384 for $C_4$ and 24741150720 for $C_5$). -4). $F_4$ has two irreducible representations of dimension 1053 that I had completely forgotten about in my list. - -REPLY [7 votes]: We give a table for irreducible representations up to dimension $2^{15}$ in the supplement of our preprint http://arxiv.org/abs/1012.5256v1 -I immediately found the following examples with respect to the first question: -(1) $B_3$ occurs twice in dimension 112, 168, etc. -(2) $B_4$ occurs three times in dimension 2772, twice in dimension 9504, etc. -(3) $B_5$ occurs twice in dimension 23595. -(4) $F_4$ occurs twice in dimension 1053. - -REPLY [5 votes]: Without offering a complete answer to the stated question, I'd first ask what significance the answer would have (one way or the other) in terms of Lie theory? I'd also want to extract the essential numerical problem, which only concerns the numerator polynomial in Weyl's dimension formula. (The denominator is constant and doesn't affect the outcome.) -Given an irreducible root system belonging to a simple Lie type of rank $r$, the polynomial in the numerator is a product of $n$ factors each of which is a $\mathbb{Z}$-linear combination of $r$ variables $x_1, \dots, x_r$ with positive coefficients depending on the root system. Here $n$ is the number of positive roots. The question is when if ever this polynomial can take the same value at two different $r$-tuples of strictly positive integers. If it does take the same value twice, it will take the same value infinitely many times: multiply each $r$-tuple by the same positive integer. -From the viewpoint of root systems (compact groups being far in the background now), types $A_r$ with $r>1$, $D_r$ with $r>3$, and $E_6$ have graph automorphisms which guarantee a positive answer to the question. Beyond this I can only see a possibility of accidental positive (or negative) answers as in the case of $G_2$. By now the question is only remotely connected to Lie theory, via the fixed linear combinations of the variables occurring as factors in the given polynomial.<|endoftext|> -TITLE: Example of sequences with different limits for two norms -QUESTION [33 upvotes]: I was explaining to my students that if there is an inequality between two norms, then there is an inclusion between their spaces of convergent sequences, with matching limits. I then proceeded to show examples of such inequalities on the normed spaces they knew, and counterexamples of sequences which converge for a norm and not for another, stating the equivalence of norms in finite dimension, etc. -It is then that I wondered about the following : does there exist a vector space, two norms on that vector space and a single sequence which converges for both norms, but with different limits? -The first remark is that such a counter-example cannot exist in finite dimension ; and one first has to find "really inequivalent norms", which do exist : consider the space of polynomials in one variable, and define norms on it by summing the absolute values of the coefficients : - - first with a weight $1$ for every coefficient ; - second with $2^n$ or $2^{-n}$ depending on the parity of the degree $n$. - -It's now easy to find a sequence going to zero for the first and not for the second, and a sequence going to zero for the second and not for the first - so there can't be an inequality between those. -Notice this is all over the real or complex numbers, though the question could be amusing in a more general setting. - -REPLY [4 votes]: I'm not sure if you would consider this an "example", but it seems this is not as widely known as I would expect. -Theorem. Let $X$ be a Banach space and let $||\cdot||_1, ||\cdot||_2$ be non-equivalent norms on $X$. Then there exists a sequence $(x_n)$ in $X$ and $x\neq y \in X$ such that $x_n \to x$ with respect to $||\cdot||_1$ and $x_n \to y$ with respect to $||\cdot||_2$. -Proof. By the bounded inverse theorem it must be that the identity map $\iota: (X, ||\cdot||_1) \to (X, ||\cdot||_2)$ is discontinuous. Thus by the closed graph theorem it must be that the graph of $\iota$ is not closed in $(X, ||\cdot||_1) \times (X, ||\cdot||_2)$. Since the graph is not closed, we may choose a sequence $(x_n, \iota(x_n))$ in the graph converging in $(X, ||\cdot||_1) \times (X, ||\cdot||_2)$ to some $(x,y)$ such that $(x,y)$ is not in the graph. Converging in $(X, ||\cdot||_1) \times (X, ||\cdot||_2)$ means that $x_n \to x$ in $(X, ||\cdot||_1)$ and $\iota(x_n) = x_n \to y$ in $(X, ||\cdot||_2)$. But $(x,y)$ not being in the graph means that $y \neq x$.<|endoftext|> -TITLE: When is a extension of $\mathbb{Z}$ by a free group a CAT(0) group? -QUESTION [22 upvotes]: The question has an easy answer, if one replaces free by free abelian: Then the resulting group is always solvable and a solvable subgroup of a CAT(0) group is virtually abelian. -If the resulting was CAT(0), then the chosen automorphism $\varphi$ in $\mathbb{Z}^n\rtimes_\varphi \mathbb{Z}$ would have finite order - otherwise the group would not be virtually abelian. -Now one can ask the same question for the free group instead or the free abelian group. -I would like to know for which automorphisms $\varphi$ of the free group $F_n$ the group $F_n\rtimes_\varphi \mathbb{Z}$ is CAT(0). -I only know, that $F_n \times \mathbb{Z}$ is CAT(0). I think that if the chosen automorphism has finite order, then the result should be CAT(0) (although I don't have a proof). And I do not know automorphism, that gives a non-CAT(0) group. - -REPLY [10 votes]: Seems worth recording here: there are some recent new examples of CAT(0) free-by-cyclic groups due to Rylee Lyman, see https://arxiv.org/abs/1909.03097v3. As she explains, "These are the first families comprising infinitely many examples for each rank of the nonabelian free group." Also, they're not relatively hyperbolic, which contrasts with the hyperbolic Hagen-Wise examples.<|endoftext|> -TITLE: Convexity radius of a Lie Group -QUESTION [5 upvotes]: Is there a nice formula/method to find the convexity radius of a matrix Lie group (the manifold can be noncompact) ? -Edited based on comments: -Definition : Convexity Radius (Berger - Panoramic View of Riemannian Geometry) -The convexity radius of a Riemannian manifold $M$ is the -infimum of positive numbers $r$ such that the metric open ball $B(m,r)$ is -convex for every $m ∈ M$. - -REPLY [4 votes]: For simply connected Lie group with bi-invariant metric it is half of the distance to the cut locus.<|endoftext|> -TITLE: Exact simulation of SDE -QUESTION [6 upvotes]: Consider a one dimensional SDE of the form $dX_t = \mu(X_t) dt + \sigma dW_t$, where $\sigma>0$ is a constant. Under mild regularity assumptions on $\mu(\cdot)$, one can exactly simulate trajectories of this SDE: because $\sigma$ is constant, one can first exactly simulate a (scaled) Brownian motion $dY_t = \sigma dW_t$ and use the fact that (Girsanov) $\text{Law}(x)$ and $\text{Law}(Y)$ are equivalent to do some kind of rejection sampling on the Wiener space. -See here for more details. -If $\sigma(\cdot)$ is not constant, in the one dimensional case, one can always find a function $\Psi$ such that $Z_t = \Psi(W_t)$ is of the form $dZ_t = \hat{\mu}(Z_t) dt + \sigma(Z_t) dW_t$: this follows from the fact that any $1$-dimensional continuous function is a derivative. This shows that a large class of $1$-dimensional SDE can be exactly simulated. -Question: the situation is quite different in $\mathbb{R}^d$ for $d \geq 2$: is there any diffusion $dX_t = \mu(X_t)dt + \sigma(X_t) \cdot dW_t$ that can be exactly simulated and that cannot be obtained through rejection sampling based on the process $Z_t = \Psi(W_t)$ for a well chosen function $\Psi:\mathbb{R}^n \to \mathbb{R}^d$. - -REPLY [4 votes]: Beskos and Roberts' whole approach relies on being able to transform the SDE to one with unit diffusion coefficient. If you can't do that, then the bridged process law isn't equivalent to the law of a Brownian bridge. This means the Radon-Nikodym derivative isn't bounded and so you can't do rejection sampling. -Beskos et al. submitted a discussion paper on this topic to the journal of the royal statistical society (available here). Dan Crisan suggests that a similar approach might work if one were to use other tractable bridges - say, Bessel bridges. In that case, it looks like you don't have to transform your SDE to one with unit diffusion coefficient. However, the authors show that this approach is equivalent to one using Brownian bridges.<|endoftext|> -TITLE: What is the motivation for a vertex algebra? -QUESTION [78 upvotes]: The mathematical definition of a vertex algebra can be found here: -http://en.wikipedia.org/wiki/Vertex_operator_algebra -Historically, this object arose as an axiomatization of "vertex operators" in "conformal field theory" from physics; I don't know what these phrases mean. -To date, I haven't been able to gather together any kind of intuition for a vertex algebra, or even a precise justification as to why anyone should care about them a priori (i.e. not "they come from physics" nor "you can prove moonshine with them"). -As far as I am aware, theoretical physics is about finding mathematical models to explain observed physical phenomena. My questions therefore are: -What is the basic physical phenomenon/problem/question that vertex operators model? -What is the subsequent story about vertex operators and conformal field theory, and how can we see that this leads naturally to the axioms of a vertex algebra? -Are there accessible physical examples ("consider two particles colliding in an infinite vacuum...", etc.) that illustrate the key ideas? -Also, are there alternative, purely mathematical interpretations of vertex algebras which make them easier to think about intuitively? -Perhaps people who played a role in their discovery could say a bit about the thinking process that led them to define these objects? - -REPLY [50 votes]: The original motivation for vertex algebras is explained briefly in the original paper http://www.jstor.org/stable/27441 as follows. For any even lattice one can construct a space $V$ acted on by vertex operators corresponding to lattice vectors. More generally one can write down a vertex operator for every element of $V$. These vertex operators satisfy some complicated relations, which are then used as the definition of a vertex algebra. In other words, the original example of a vertex algebra was the vertex algebra of an even lattice, and the definition of a vertex algebra was an axiomatization of this example. -This was motivated by my attempt to understand Igor Frenkel's work on the Lie algebra with Dynkin diagram the Leech lattice. Frenkel had constructed a representation of this Lie algebra using vertex operators acting on the space $V$, and I was trying to use his work to understand its root multiplicities. I did not use any insights from conformal/quantum/topological field theory or operator product expansions when defining vertex algebras (as implied by some of the other answers), for the simple reason that I had barely heard of these concepts and had almost no idea what they were. -This is not all that helpful for understanding what a vertex algebra really is. A better view is to regard them as something like a commutative ring with an action of the (formal) 1-dimensional additive group. In particular any such ring is canonically a vertex algebra. The difference between rings with group actions and vertex algebras is that the "multiplication" of a vertex algebra from $V\times V$ to $V$ is not defined everywhere: it can be thought of as a "rational map" rather than a "regular map" in some sense. More precisely if we write $u^g$ for the action of a group element $g$ on $u$, then the product $u^gv^h$ is not defined for all group elements $g$ and $h$, but behaves formally like a meromorphic function of $g$ and $h$, which in particular may have poles when $g$ and $h$ are trivial. Making sense of this informal idea produces the definition of a vertex algebra. (For more details see the unpublished paper http://math.berkeley.edu/~reb/papers/bonn/bonn.pdf.) This means that vertex algebras behave like commutative rings: for example, one should define modules over them, tensor products of modules, and so on.<|endoftext|> -TITLE: Hodge decomposition in Betti cohomology -QUESTION [7 upvotes]: The broad, generic and badly posed question may be formulated in this way: - -Let $X$ be a compact Kälher manifold (or even a projective one). If one considers the Hodge decomposition $H^k(X, \mathbb{C}) = \bigoplus_{p+q=k} H^{p,q}(X)$ and interprets the left hand side as Betti cohomology $\text{Hom}(\pi_1(X), \mathbb{C})$, which homomorphisms fit into which direct summand? - -I'm sure this is well studied, but couldn't find any reference. Without a precise question there is probably not much to say, so let us look at a few motivation and give more precise statements below. -Motivations -I was trying to study the easy cases of Simpson's non-abelian cohomology (cfr. for example his paper "Moduli of representations of the fundamental group of a smooth projective variety"), and the first possible idea is to reduce oneself to the abelian case. So some questions which are of interest in the non-abelian case should become easy in the abelian context, and the above provides the basics for one of them: Simpson defines Betti and Dolbeaut non-abelian cohomology, then proves that the constructed spaces are in fact naturally homeomorphic, and studies interactions between objects defined in these "different worlds". In the abelian context this boils down to interpreting the cohomology as singular cohomology on one side and as the direct sum of the Dolbeaut cohomology on the other. -More specific questions -One can in particular consider any given subgroup $G \subset \mathbb{C}$, and seek only those homomorphisms which take values in this subgroup (if $G$ is a subfield this means considering $H^k(X, G)$, otherwise there may be some torsion in this cohomology group which is ignored if we just take the subgroup of $H^k(X, \mathbb{C})$ as above). Then, can one say whether this subgroup, call it $H_G \subset H^k(X, \mathbb{C})$, is contained in some pure part of the Hodge decomposition $H^{p,q}(X)$? I would have thought that this depended entirely on the nature of $X$, but on second thought it seems that something can always be said: for example, if $k = 1$, then $H_{\mathbb{R}}$ is never contained in any pure part of some weight (since they are interchanged by conjugation), while some work I did would lead one to think that if $G = \lbrace k \pi i \rbrace_{k \in \mathbb{Z}}$, then $H_G \subset H^{0,1}(X)$, at least for $X$ projective such that $H_1(X, \mathbb{Z})$ has no torsion (I'm absolutely not sure of the rightness of this proof, and if this fact was true there should really be a more straight idea then the one I thought of). Furthermore, for $k = 2$, then always $H_{\mathbb{R}} \cap H^{1,1}(X) \neq \emptyset $ since there is the Kähler form, while the question if $H_{\mathbb{Z}}$ is a subset of $H^{1,1}(X)$ is by Kodaira embedding theorem equivalent to $X$ being projective. -Hence a slightly more focused question could be: - -For which classes of compact Kähler manifolds and which (in case cyclic) subgroups $G \subset \mathbb{C}$ one knows the Dolbeaut type of the elements of $ H_G $ inside some $H^k(X, \mathbb{C})$? - -The question may become more interesting (hence, probably won't have any sensitive answer) if one allows (in case small) variations of the Hodge structure, by considering a family $X \to S$. I do not know much about variations of Hodge structures, so this could be well known. In this case, what is surely known is that the question for $k = 2$ and $G = \mathbb{Z}$ is false (small deformations of projective varieties need not be projective) and for $G = \mathbb{R}$ is true (small deformation of Kähler manifolds are Kähler). In this case one may want to drop the smoothness hypothesis, or, on the other hand, require the morphism $X \to S$ to be both projective and smooth. - -REPLY [3 votes]: This is really a comment on Donu Arapura's answer, but it seemed large enough to deserve it's own post. Working again in the case of $GL_1$, Simpson considers three spaces: -$M_{betti}$: The space of $\mathbb{C}^*$-local systems on $X$. If you like, you can think of this as smooth complex line bundles with an integrable connection. -$M_{DR}$: The space of holomorphic line bundles $L$ on $X$, equipped with an integrable holomorphic connection. (Being compatible with a holomorphic connection forces $c_1(L)$ to be $0$ in $H^2(X)$.) -$M_{Dol}$: The space of holomorphic line bundles $L$ on $X$, with $c_1(L)= 0$ in $H^2(X)$, and equipped with an $\mathrm{End}(L)$-valued $1$-form. $\mathrm{End}(L)$ is naturally isomorphic to $\mathcal{O}$, so this is just a $1$-form, but it is the $\mathrm{End}(L)$ version which generalizes to higher $GL_n$. -The first space is $\mathrm{Hom}(\pi_1(X), \mathbb{C}^*) = H^1(X, \mathbb{Z}) \otimes \mathbb{C}^*$. In this latter form, it has a natural algebraic structure, as a multiplicative algebraic group. -The second space is an affine bundle over $\mathrm{Pic}^0(X)$. Each fiber is a torsor for $H^0(X, \Omega^1)$, so we can describe this space by giving a class in $H^1(\mathrm{Pic}^0(X), \mathcal{O}) \otimes H^0(X, \Omega^1)$. By GAGA, this cohomology group on $\mathrm{Pic}$ is the same algebraically or analytically; viewing it algebraically, we get an algebraic structure on $M_{DR}$. -The third space is simply $\mathrm{Pic}^0(X) \times H^0(X, \Omega^1)$ (for larger $n$, this vector bundle can be nontrivial). For obvious reasons, this has an algebraic structure. -The relations between these spaces are the following: All three are diffeomorphic. $M_{betti}$ and $M_{DR}$ are isomorphic as complex analytic varieties, but have different algebraic structure. $M_{Dol}$ and $M_{DR}$ are not isomorphic as complex analytic varieties, rather, $M_{Dol}$ is the vector bundle for which the affine bundle $M_{DR}$ is a torsor. -You might enjoy writing this all down in coordinates for $X$ an elliptic curve. As smooth manifolds, all three spaces should be $(\mathbb{C}^*)^2$.<|endoftext|> -TITLE: Frobenius elements from the point of view of étale fundamental groups -QUESTION [12 upvotes]: The goal of this question is to find a "geometric" definition of Frobenius element in $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. -Here are two definitions that don't work, but that should help explain what I mean. Fix an algebraic closure of $\mathbb{Q}$ and, for a prime $p$, fix an algebraic closure of $\mathbb{F}_p$. If there were a canonical morphism $f : \mathbb{Q} \to \mathbb{F}\_p$, then the induced map $f\_{\ast} : \pi_1(\text{Spec } \mathbb{F}_p) \to \pi_1(\text{Spec } \mathbb{Q})$ of étale fundamental groups (basepoints the algebraic closures above, suppressed) would be an obvious way to define Frobenius elements: we would just take the image of a generator of $\pi_1(\text{Spec } \mathbb{F}_p) \simeq \hat{\mathbb{Z}}$. (The general motivation being: "gee, if I had a category $C$ and a functor, let's call it $\pi_1 : C \to \text{Grp}$, wouldn't it be nice if I could find an object, let's call it $S^1$, with $\pi_1(S^1) \simeq \mathbb{Z}$, so I could probe elements of $\pi_1(X)$ using morphisms $S^1 \to X$," and it seems for schemes that $\pi_1(S^1) \simeq \hat{\mathbb{Z}}$ is the next best thing.) -But of course this is nonsense since no such $f$ exists. The next obvious thing (from my perspective, being someone who knows nothing about all this étale stuff) is to look at the canonical morphism $f : \mathbb{Z} \to \mathbb{F}\_p$, which induces a map $f\_{\ast} : \pi_1(\text{Spec } \mathbb{F}_p) \to \pi_1(\text{Spec } \mathbb{Z})$, and then to look at the canonical morphism $g : \mathbb{Z} \to \mathbb{Q}$ and the induced map $g\_{\ast} : \pi_1(\text{Spec } \mathbb{Q}) \to \pi_1(\text{Spec } \mathbb{Z})$. Perhaps the Frobenius elements at $p$ are nothing more than the inverse image under $g\_{\ast}$ of the image of a generator of $\pi_1(\text{Spec } \mathbb{F}_p)$ under $f\_{\ast}$ (up to conjugacy to account for changes in basepoint). -But this is also nonsense since $\pi_1(\text{Spec } \mathbb{Z})$ is trivial. - -So what is the correct version of this construction? - -Here's my guess: instead of using $\mathbb{Z}$, we have to use the localization $R = \mathbb{Z}_{(p)}$. As before there are morphisms $f : R \to \mathbb{F}_p, g : R \to \mathbb{Q}$ inducing maps $f\_{\ast} : \pi_1(\text{Spec } \mathbb{F}_p) \to \pi_1(\text{Spec } R)$ and $g\_{\ast} : \pi_1(\text{Spec } \mathbb{Q}) \to \pi_1(\text{Spec } R)$, and maybe now something like the statement "the Frobenius elements in $\pi_1(\text{Spec } \mathbb{Q})$ at $p$ are the inverse image under $g\_{\ast}$ of the image of a generator of $\hat{\mathbb{Z}}$ under $f\_{\ast}$ (up to conjugacy)" is finally true. Is it? And what does $g\_{\ast}$ look like? - -REPLY [7 votes]: Your picture is right. The one thing I'll say (and this is somewhat tangential to your question) is that, for me, the analogue of $\pi_1(S^1)$ viewed as $\pi_1(\mathbb{C}\setminus\{0\})$ is not quite $\pi_1(\operatorname{Spec}\mathbb{F}_p)$, but rather the tame fundamental group of $\operatorname{Spec}\mathbb{Z}_p^{nr}$, where $\mathbb{Z}_p^{nr}$ is the maximal unramified extension of $\mathbb{Z}_p$ (i.e. its universal cover: it's simply connected and therefore analogous to a disk around the origin in $\mathbb{C}$). In other words, $\pi_1(S^1)$ classifies $n$-fold covers of the circle (or of a punctured disk around the origin). Similarly, the tame fundamental group classifies $n$-fold covers of the 'punctured disk' $\operatorname{Spec}\mathbb{Z}_p^{nr}\setminus{p}$. The only hitch is that $n$ has to be prime to $p$ in this setting. This part of the Galois group is in the `geometric' direction (nothing's happening to the residue field of the point), while the part where the Frobenii live is in the arithmetic direction (all about extensions of $\mathbb{F}_p$).<|endoftext|> -TITLE: Conformal embedding of Riemann surfaces into 3-space -QUESTION [23 upvotes]: I can't seem to find any work on the following question: Can every (closed, of finite type) Riemann surface $S$ be realized as an embedded (or even immersed) smooth surface in Euclidean $3$-space, where by realized, I mean that the induced conformal structure on $T \subset \mathbb{E}^3$ is the conformal structure on $S?$ The answer is obviously yes when $S$ is of genus $0,$ but that's where the obvious statements end -- I don't know the answer for $g=1.$ - -REPLY [8 votes]: I have thought about this question before, but at the moment I can't remember links or references. Nonetheless, many years ago I thought of a sketch of an argument that should eventually work to prove that it's always possible. Namely, start with any smooth embedding of $S$ into $\mathbb{R}^3$. This surface has some conformal structure, and any other conformal structure is described by an ellipse field well-defined up to a scale. Then it is intuitive that that you can match the ellipse field by "wrinkling" $S$, in other words by making $S$ locally wavy in the long direction of the ellipses. More rigorously, $S$ is replaced by a graph of a function that is locally sinusoidal. You don't even have to match it exactly; approximations are good enough if you can encircle the desired conformal structure in Teichmuller space. There is a similar trick for a different purpose in the video "Outside In". -Of course it's messy to make all of the required transitions between the patches of wrinkles on the surface. Moreover the surface is not flat, only flat to first order, and a function cannot be graphed rectilinearly, only approximately so. I think that Garsia's paper uses the same idea or a similar idea, but he slogs through all of the approximations to make it work.<|endoftext|> -TITLE: Does there exist a measurable function which is not a.e. "strongly" measurable? -QUESTION [24 upvotes]: More specifically, letting $I=[0,1]$, do there exist $f,E$ with $E$ a (necessarily nonseparable) Banach space and $f$ a bounded Lebesgue measurable function $I\to E$ such that $f$ is not equal almost everywhere to a pointwise limit of a sequence of simple Lebesgue measurable functions? Here "simple" means having finite range, and Lebesgue measurability of $f$ means that $f^{-1}[A]$ is a Lebesgue measurable set in $I$ for every Borel set $A$ in $E$ . -If such $f,E$ exist, and if one constructs $F=L^p(I,E)$ along the lines of Problem 4 on page 120 in Richard M. Dudley's book Real Analysis and Probability, Wadsworth 1989, one obtains $F$ where the set of (equivalence classes of) simple functions is not dense. If instead of Lebesgue measurable functions one here considers the Borel measurable ones, this pathology is not possible by Corollary 4.2.7 on page 97 and Problem 10 on page 99 loc. cit. -I guess that an explicit construction might not be possible, and one should (somehow) use something based on the axiom of choice. - -REPLY [36 votes]: No. In fact, every Lebesgue measurable function $f\colon I\to E$ is equal almost everywhere to a limit of simple Lebesgue measurable functions. As you hint at in the question, this is easy to show in the case where $E$ is separable. The general situation reduces to the separable case due to the following result. For a full proof, see Fremlin, Measure Theory, Volume 4 Part I, Lemma 451Q. - -Theorem 1: Let $(X,\Sigma,\mu)$ be a finite compact measure space, $Y$ a metrizable space, and $f\colon X\to Y$ a measurable function. Then, there is a closed separable subspace $Y_0$ of $Y$ such that $f^{-1}(Y\setminus Y_0)$ is negligible. - -That is, $f$ has essentially separable image. Restricting $f$ to the complement of a negligible set reduces the problem to the situation where the codomain is separable, in which case it is a limit of simple functions. Compactness of the space $(X,\Sigma,\mu)$ means that there is a family $\mathcal{K}\subseteq\Sigma$ such that any subset of $\mathcal{K}$ with the finite intersection property has nonempty intersection, and such that $\mu$ is inner-regular with respect to $\mathcal{K}$. That is, $\mu(E)=\sup\{\mu(K)\colon K\in\mathcal{K},K\subseteq E\}$ for every $E\in\Sigma$. In particular, the Lebesgue measure is compact by taking $\mathcal{K}$ to be the collection of compact sets under the standard topology. -The proof of Theorem 1 is rather tricky, involving what Fremlin describes as "non-trivial set theory". It rests on the following two results. - -Theorem 2: Any metrizable space has a $\sigma$-disjoint base $\mathcal{U}$. That is, $\mathcal{U}$ is a base for the topology, and can be written as $\bigcup_{n=1}^\infty\mathcal{U}_n$ where each $\mathcal{U}_n$ is a disjoint collection of sets. - -(Fremlin, Measure Theory, Volume 4 II, 4A2L (g-ii)) - -Theorem 3: Let $(X,\Sigma,\mu)$ be a finite compact measure space and $\{E_i\}_{i\in I}$ be a disjoint family of subsets of $X$ such that $\bigcup_{i\in J}E_i\in\Sigma$ for every $J\subseteq I$. Then, $\mu\left(\bigcup_{i\in I}E_i\right)=\sum_{i\in I}\mu(E_i)$. - -(Fremlin, Measure Theory, Volume 4 I, 451P). -Theorem 3 is particularly remarkable, as it extends the countable additivity of the measure to arbitrarily large unions of sets. -Once these two results are known, the proof that $f$ has essentially separable image in Theorem 1 is straightforward. Let $\mathcal{U}=\bigcup_{n=1}^\infty\mathcal{U}_n$ be a $\sigma$-disjoint base for $Y$. Let $\mathcal{V}_n$ be the collection of $U\in\mathcal{U}_n$ such that $\mu(f^{-1}(U)) = 0$. By countable additivity, $\mathcal{U}_n\setminus\mathcal{V}_n$ is countable. Also, $\{f^{-1}(U)\colon U\in\mathcal{V}_n\}$ is a disjoint collection of negligible subsets of $X$ and, by measurability of $f$, any union of a subcollection of these is measurable. It follows from Theorem 3 that its union is negligible. That is, $f^{-1}\left(\bigcup\mathcal{V}_n\right)$ is negligible. Setting $Y_0=Y\setminus\bigcup_n\bigcup\mathcal{V}_n$ then, by countable additivity, $f^{-1}(Y\setminus Y_0)$ is negligible. Also, $\bigcup_n(\mathcal{U}_n\setminus\mathcal{V}_n)$ restricts to a countable base for the topology on $Y_0$, so it is separable (in fact, it is second-countable). -Finding a $\sigma$-disjoint base for the topology on $Y$ is easy enough. Following Fremlin, you can do this by well-ordering $Y$ and letting $(q_n,q^\prime_n)$ be a sequence running through the pairs $(q,q^\prime)$ of rationals with $0 < q < q^\prime$. Letting $\mathcal{U}_n$ be the collection of sets of the form -$$ -G_{ny}=\left\{x\in Y\colon d(x,y) < q_n, \inf_{z < y}\,d(x,z) > q_n^\prime\right\} -$$ -(over $y\in Y$) gives a $\sigma$-disjoint base. -The really involved part of the proof is in establishing Theorem 3. I suggest you look in Fremlin for the details, but the idea is as follows. By countable additivity, only countably many $E_i$ can have positive measure so, removing these, we can suppose that every $E_i$ is negligible. Also, restricting $X$ to the union of the $E_i$ if necessary, we can suppose that $X=\bigcup_iE_i$. Then define the function $f\colon X\to I$ by $f(x)=i$ for $x\in E_i$. Using the power set $\mathcal{P}I$ for the sigma-algebra on $I$, $f$ will be measurable. Then let $\nu=\mu\circ f^{-1}$ be the image measure on $(I,\mathcal{P}I)$. Fremlin breaks this down into two cases. -a) $\nu$ is atomless. As with any finite atomless measure space, there will be a measure preserving map $g\colon I\to[0,\gamma]$ for some $\gamma\ge0$, with respect to the Lebesgue measure $\lambda$ on $[0,\gamma]$. Using compactness, it can be shown that the sets on which $\lambda$ and $\nu\circ g^{-1}$ are well-defined coincide (precisely, $\mu$ is compact, so it is perfect, so $\nu\circ g^{-1}$ is perfect and therefore is Radon). The existence of non-Lebesgue sets will then give a contradiction unless $\gamma=0$, so $\mu(X)=0$. -b) $\nu$ has an atom $M\subseteq I$: In this case, $\mathcal{F}=\{F\subseteq M\colon\nu(M\setminus F)=0\}$ is a non-principal ultrafilter on $M$ which is closed under countable intersections. Again making use of compactness of $\mu$, this can be used to derive a contradition, but it requires some tricky set theory. I refer you to Fremlin (451P) for the full details of this argument. -Update: I will, however, give a brief overview of the ideas involved in (b). It is possible to reduce the problem to the case where $M$ is a regular uncountable ordinal and $\mathcal{F}$ is a normal ultrafilter. Using $[S]^n$ to denote the collection of size-$n$ subsets of a set $S$ and $[S]^{ < \omega}=\bigcup_{n=0}^\infty[S]^n$ for the collection of finite subsets, normal ultrafilters have the following property. - -If $\mathcal{S}\subseteq[M]^{ < \omega}$ then there exists an $F\in\mathcal{F}$ such that, for each $n\ge0$, $[F]^n$ is either a subset of $\mathcal{S}$ or disjoint from $\mathcal{S}$. - -See, Frelim (4A1L). This contradicts compactness as follows. Set $G_i=\bigcup\{E_j\colon j\in M, j\ge i\}$. Then choose $K_i\in\mathcal{K}$ with $K_i\subseteq G_i$ and $\mu(K_i) > 0$. Let $\mathcal{S}$ consist of the finite subsets $S\subset M$ such that $\bigcap_{i\in S}K_i=\emptyset$. Choose $F\in\mathcal{F}$ as above. It is not possible for $[F]^n$ to be a subset of $\mathcal{S}$. Otherwise, every $x\in X$ would be contained in no more than $n$ of the sets $\mathcal{K}^\prime=\{K_i\colon i\in F\}$. So, $\sum_{i\in F}\mu(K_i)\le n\nu(M)$. But, as this sum is over an uncountably infinite set of positive numbers, it should be infinite. Therefore, $[F]^{ < \omega}\cap\mathcal{S}=\emptyset$, and $\mathcal{K}^\prime$ satisfies the finite intersection property. So, by compactness, $\bigcap_{i\in F}G_i\supseteq\bigcap\mathcal{K}^\prime\not=\emptyset$. This contradicts the fact that, as $F\in\mathcal{F}$ is an unbounded subset of $M$, this intersection is empty.<|endoftext|> -TITLE: Is there a name for groups with presentations of the following type: $\langle a, b \mid [b, b^{a^i}] \; (i\in \mathbb{Z}), b^{c_0}(b^a)^{c_1}\ldots (b^{a^s})^{c_s}\rangle$, (where $c_0,\ldots,c_s$ are integers)? -QUESTION [6 upvotes]: The presentation is a bit prettier if we set $b_i := b^{a^i}$ for $i\in \mathbb{Z}$: - - - -$\langle a, b \mid [b, b_i] \; (i\in \mathbb{Z}), b_0^{c_0}b_1^{c_1}\ldots b_s^{c_s}\rangle$, - - - -Since the subgroup generated by the $b_i$ is abelian, we can even write the last relator additively as $\sum_{i=0}^s c_ib_i$. -These groups occur in a paper Derek Holt and I are writing on subgroups of finitely generated soluble groups. They are all metabelian, and if the $c_i$ are relatively prime, they are also torsion-free. -I would like to call these groups by their proper name if they have one, especially as we have not so far managed to come up with a good name. -Since they are fairly straightforward groups, I expect it is possible they have already come up somewhere else and been named, but unfortunately presentations are difficult to search for on MathSciNet. -EDIT: I might as well mention our current name for these groups. We use the notation $G(\mathbb{c})$, where $\mathbb{c} = (c_0,\ldots,c_s)$, for the group with the above presentation. This I am happy with. But our current name for the groups in general is `Gc-groups', which I don't like very much, and so I was hoping to find out that they already have a better name. - -REPLY [2 votes]: These groups are virtually wreath products of finitely generated Abelian groups with $\mathbb Z$. More precisely the subgroup $G_s=\langle a^s, b\rangle$ is a wreath product of a 1-related Abelian group (the relation is your product of powers) and $\langle a^s\rangle$, and your group is an extension of $G_s$ by a cyclic group of order $s$. I do not think there exists a special name for these groups. As with every metabelian group which is Abelian-by-cyclic, you can get a lot of properties of it by looking at the corresponding $\mathbb{Z}[b,b^{-1}]$-module. - Update. I just realized that this group is not a virtual wreath product. One can construct the group as follows. Consider the ring $R=\mathbb{Z}[t,t^{-1}]$ and the ideal $I$. there generated by the polynomial $f=c_0+c_1t+\ldots+c_st^s$. The group $\mathbb{Z}$ acts on the Abelian group $R/I$ by left multiplication (which are automorphisms of the additive group of $R/I$). Your group is the corresponding semidirect product $R/I\rtimes \mathbb{Z}$. One can also represent this group by $2\times 2$ matrices using the Magnus embedding. Anyway this does not answer your question about the name of the group. I still do not think it has a name.<|endoftext|> -TITLE: A particular combinatorial proof of Wilson's theorem -QUESTION [20 upvotes]: I (probably re)invented a very short combinatorial proof of Wilson's theorem that I perennially teach my students. If it occurs in the literature and someone can tell me where first (or even at all), I would like to attribute credit properly. -I actually prove $p! - p(p-1) \equiv 0 \mod p^2$. -$p!$ counts bijective function from ${\Bbb Z}/p$ to ${\Bbb Z}/p$ and ${\Bbb Z}/p \times {\Bbb Z}/p$ acts on these by $f(x)\stackrel{(a,b)}{\rightarrow} f(x-a)+b$. Excluding functions -of the form $cx+d, c\not=0$, all orbits have size $p^2$. - -REPLY [18 votes]: According to Dickson's History of the Theory of Numbers, this proof was first found by J. Petersen, Tidsskrift for Mathematik (3), 2, 1872, 64-5. (Petersen divides everything by 2, but the idea is the same.) Dickson gives a number of references to rediscoveries of Petersen's proof in the 19th century. Petersen was also apparently the first to discover the combinatorial proof of Fermat's theorem.<|endoftext|> -TITLE: Mass of spinor genus, positive integral quadratic forms -QUESTION [7 upvotes]: There seems to be general opinion that, for positive integral quadratic forms in at least three variables, spinor genera in the same genus all have the same mass (not representation measures of some number, that is different, indeed some recent authors write of representation mass of numbers and it throws me off). Authors R. Scharlau and R. Schulze-Pillot have been loosely mentioned in this regard. -Does anyone know for sure, and in particular know a simple (published) reference for positive forms? Evidently there is an analogous formulation where it is true for indefinite forms, see The Hirzebruch-Mumford volume for the orthogonal group and applications, by -Valery Gritsenko, Klaus Hulek, G.K. Sankaran, -http://arxiv.org/abs/math/0512595 -I should add that Kneser (1961) is sometimes mentioned in this regard, but my take is that, while his methods can be used to reconstruct the result, he is not explicit about the mass. -Here is computer output on an example that appears in Benham, Earnest, Hsia, Hung (1990), formula (3.8), positive ternary forms, including one of the 29 known spinor regular forms that are not regular. The sextuple a b c d e f refers to the form -$T(x,y,z) = a x^2 + b y^2 + c z^2 + d y z + e z x + f x y,$ with discriminant -$\Delta = 4 a b c + d e f - a d^2 - b e^2 - c f^2.$ -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= - - ===Discriminant 343 ==Genus Size== 3 - 343 = 7^3 - All forms in regular spinor genus represent 1 ------------------------------ - Spinor genus misses 1 4 16 64 121 256 - 484 529 841 - 343: 2 7 8 7 1 0 auto 4 Level 196 irreg spin candidate ---------------------------size 1 - Spinor genus misses no exceptions - 343: 1 2 49 0 0 1 auto 8 Level 196 - 343: 1 7 14 7 0 0 auto 8 Level 196 ---------------------------size 2 -Disc 343 -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= - -REPLY [5 votes]: Better late than never: -In the adelic setup, the mass or measure of a lattice $L$ (the german word Maß translates to measure in English) is given as -$\mu(O(V)\backslash O(V)O_{\mathbb A}(L))$, where $\mu$ is the (Tamagawa-normalized) Haar measure on the adelic orthogonal group $O_{\mathbb A}(V)$ of the underlying quadratic space $V$ and $O_{\mathbb A}(L)$ is the adelic isometry group of $L$; it is proportional to $\frac{1}{\vert O(L)\vert}$ in the definite case (here $O(L)$ is the global group of isometries of $L$, which is finite in the definite case). -If we sum that up over a set of representatives of the classes in the spinor genus of $L$ we obtain $\mu(O(V)\backslash O(V)O'_{\mathbb A}(V)O_{\mathbb A}(L))$, where $O'_{\mathbb A}(V)$ is the group of adelic orthogonal transformations of $V$ which have spinor norm $1$ at each place. -If we replace here the lattice $L$ by another lattice $\sigma L$ in the genus of $L$ (with $\sigma \in O_{\mathbb A}(V)$), we have -$$O(V)O'_{\mathbb A}(V)O_{\mathbb A}(\sigma L)=O(V)O'_{\mathbb A}(V)\sigma O_{\mathbb A}(L)\sigma^{-1}=O(V)O'_{\mathbb A}(V)O_{\mathbb A}(L),$$ -since $O'_{\mathbb A}(V)$ contains all commutators. -So, the measures for the spinor genus of $L$ and for that of $\sigma L$ are the same. -The argument is implicit in Kneser's 1961 article and also in Weil's 1962 article, but unfortunately neither of these authors bothered to write down the statement explicitly.<|endoftext|> -TITLE: Are affine groups over rings of integers finitely generated? -QUESTION [8 upvotes]: I'll begin by saying that I'm not sure what I want to ask specifically, but pretty sure what in general, so please don't hold my misunderstandings against me, but do comment on them. -I know that the unit group of a number field is finitely generated, and so is $SL_2(\mathbb{Z})$. I understand that so are $SL_n(\mathbb{Z})$ (or was it $GL$?). -1) What is a known positive generalisation? -I also know that the subgroup of an abelian variety of points over a number field is finitely generated. I noticed that this is relevant after reading Franz Lemmermeyer's "Higher Descent on Pell Conics III. The First 2-Descent" (arxiv). The paper contains a proof that the unit group of a quadratic number field is finitely generated - using heights. -The way I think about it is this: the norm equation isn't a projective variety, so we make up for that by considering it over the integers. So we have heights and parallelogram laws and a proof of finitely generated. -2) Is there a single proof for Mordell-Weil, Dirichlet's Unit Theorem, and any to answer to (1), at the same time, that uses some kind of underlying concept to projective-ness and integral-ness? -I think (2) is more far fetched than (1), so feel free to ignore it :) - -REPLY [9 votes]: It seems to me that if you are thinking of affine groups, then the appropriate result is that $S$-arithmetic subgroups of reductive linear algebraic groups over number fields are finitely generated. Over function fields there are exceptions (which I think are known explicitly). -This includes examples such as $SL_2(\mathbb{Z})$, as well as $S$-units in number fields. -For number fields the proof depends on the existence of compact (equivariant) retracts of fundamental domains for the action of the group on suitable spaces -- reduction theory, combined with the finite generation of the group of units. -However the Mordell-Weil theorem (finite generation of rational (or equivalently integral) points on an abelian variety, is a rather deeper result.<|endoftext|> -TITLE: Does there exist an event independent of a given sigma-algebra? -QUESTION [12 upvotes]: The following question came up in a discussion with my advisor: - -Let $(\Omega, \mathcal F, \mathbb P)$ be a non-trivial probability space, and suppose that $\mathcal G$ is a proper sub-$\sigma$-algebra of $\mathcal F$. Does there exist an event $U \in \mathcal F$ such that - -$U$ is independent of $\mathcal G$, and -$0 < \mathbb P(U) < 1$ ? - - -The answer is surely yes, but we can't seem to prove it. Our initial approach was to consider the space $L^2(\Omega, \mathcal G)^\perp$ of finite-variance random variables orthogonal to $\mathcal G$ (i.e. $\mathbb E(X|\mathcal G) = 0$ a.s.). Since $\mathcal G$ is a proper sub-$\sigma$-algebra of $\mathcal F$, this space is non-trivial. Choose a non-constant $X \in L^2(\mathcal G)^\perp$ and let $U = \{X < c\}$. For some $c$ the event $U$ has non-trivial probability. -However, orthogonality does not imply independence. For a simple example, take independent random variables $Y \sim \operatorname{Bernoulli}(1,p)$ and $Z \sim N(0,1)$, then set $\mathcal G = \sigma(Y)$ and $X = YZ$. Certainly $X$ is not independent of $\mathcal G$ though one easily can check that $\mathbb E(X|\mathcal G) = 0$ a.s. Our strategy above might find $X$; could we modify our strategy to find the independent $Z$ instead? - -REPLY [9 votes]: No. If the probability space contains atoms then you can easily construct sub-$\sigma$-algebras which are not independent to any nontrivial events. -However, we can still construct examples, even for measures without atoms. Suppose that $\Omega=[0,1]$ is the unit interval, $\mathcal{F}=\mathcal{B}([0,1])$ is the Borel sigma-algebra and $\mathbb{P}=\lambda$ is the Lebesgue measure. Let $\mathcal{G}$ be the sub-$\sigma$-algebra generated by sets of the form $[0,x)\cup[\frac23,\frac23+\frac34x^2)$ for $0\le x\le\frac23$. I claim that all events in $\mathcal{F}$ independent to $\mathcal{G}$ have probability zero or one. -In fact, if $f\in L^1(\Omega,\mathcal{F},\mathbb{P})$ then we can calculate the conditional expectation $g=\mathbb{E}[f\mid\mathcal{G}]$ directly. Writing $y=\frac23+\frac34x^2$ for $0\le x\le\frac23$, we can use $\frac{dy}{dx}=\frac32x$ to deduce that -$$ -g(x)=g(y)=\frac{f(x)+3xf(y)/2}{1+3x/2} -$$ -for almost every $x\in[0,\frac23)$. If $A\in\mathcal{F}$ is independent to $\mathcal{G}$, so that $\mathbb{P}(A\mid\mathcal{G})=c$ is constant, then consider taking $f$ to be the indicator function of $A$. Then, $f(x)+3xf(y)/2=(1+3x/2)c$ almost everywhere. However, the left hand side can only take the values $0,1,3x/2,1+3x/2$, from which we can deduce that $c=0$ or $c=1$. -This example only works because the conditional measure (or disintegration) $\mathbb{P}(\cdot\mid\mathcal{G})$ is atomic, even though $\mathbb{P}$ is not.<|endoftext|> -TITLE: Naive question about constructing constructible sheaves. -QUESTION [21 upvotes]: In algebraic geometry, an etale sheaf on a Noetherian scheme is called constructible if the scheme has a finite stratification by locally closed subschemes such that the pullback of the sheaf to each of the subschemes is lcc (locally constant constructible). This definition seems to make sense both for sheaves of abelian groups and sheaves of sets. There is a similar definition in algebraic topology and I have no doubt that analogues of my questions can be formulated in the algebraic topology world. -I don't really have such a good feeling for these constructible sheaves, and, when re-reading the definition recently, I realised that I could easily formulate simple questions about building such things, that I could not answer. Let me ask the most general question first, which might be too general to have a reasonable answer. -General question. Say I have a Noetherian scheme $X$ and I am given a stratification by finitely many locally closed subschemes $U_i$. Say I have lcc sheaves on each of the $U_i$. What additional data do I need to glue these sheaves together to get a constructible sheaf on $X$? -If I am lucky then there is some slick cocycle-related answer and we can all go home happy. But I am not a big fan of these general types of question, so here are a few completely specific special cases---the simplest non-trivial (as far as I am concerned) ones, which I am already a bit hesitant on. -Q2 Let $X$ be the affine line over the complexes. Let $P$ denote the origin and let $U$ be $X\backslash P$. Up to isomorphism, how many constructible sheaves of sets are there on $X$, whose pullback to $U$ is the constant sheaf of sets of size 1 (represented by the identity $U\to U$) and whose pullback to $P$ is the constant sheaf of sets of size 2 (represented by $P\coprod P\to P$)? -There are, I guess, two notions of isomorphism: one can imagine that the stalk above $P$ is a given set of size 2 and demand that isomorphisms must induce the identity on this set, or one can just demand that the isomorphism induces a bijection on the stalk at $P$. I've noticed that different people interpret the question in different ways so let me just ask that the isomorphism induces an arbitrary bijection, to fix ideas. -I am a little scared of sheaves of sets because the initial and the terminal objects in the category of sets are not isomorphic, and this tripped me up a bit when I was trying to answer Q2. We don't have such a problem in the category of abelian groups so perhaps that's a safer place to be: -Q3 $X$ the affine line, $P$ the origin and $U$ the rest, as before. How many isomorphism classes of constructible sheaves of abelian groups on $X$ are there, whose stalk at $P$ is cyclic of order 4 and whose pullback to $U$ is constant and cyclic of order 2? -What is the motivation for Q2 and Q3? There are several answers. The motivation is certainly not "I need to know the answers to these questions!". One possible answer is "I feel like I cannot say I understand constructible sheaves unless I can answer these basic questions". Another is "what I really want to know is the answer to the general question, but I am worried the general question is too vague so I am asking special cases of it in order to get a feeling for the general question". - -REPLY [15 votes]: I wonder if this may suffice for your purposes: suppose you have a scheme $X$, cut up into an open subscheme $U$ and a closed subscheme $Z$ supported on the complement of $U$ in $X$. Let $j:U\to X$ and $i:Z\to X$ be in the inclusions. Let $F$ be an \'etale sheaf of sets on $X$. One derives from $F$ a triple -$$ -(j^*F, i^*F, \phi: i^*F\to i^*j_*j^*F). -$$ -The $\phi$ piece comes from restricting the adjunction morphism $F\to j_*j^*F$ to $Z$. The formation of the triple is functorial in $F$. The resulting functor provides an equivalence of categories from the category of \'etale sheaves on $X$ to the category of triples -$$ -(A,B,\phi:B\to i^*j_*A), -$$ -where $\phi$ is any sheaf morphism. -To the extent that you can compute $j_*$ and $i^*$, you can thus describe sheaves on $X$ in terms of sheaves on $U$ and $Z$. Building up from the bottom of a stratification of $X$ by locally closed subschemes, you could thus (granting an ability to compute $i^*$ and $j_*$) describe sheaves on $X$ constructible with respect to your stratification. (Performing the computations is on par with proving that $j_*$ preserves constructibility.) -For example, in your Q2, you are extending $A = 1_U$ across $Z = P$, and you want $B = 1_Z + 1_Z$. The sheaf $i_*j^*A$ is $1_P$. Since this sheaf is terminal, there is a unique constructible $F$ on $X$ that restricts to $A$ on $U$ and $B$ on $Z$; it corresponds to the triple $(A,B,\phi)$, where $\phi:B\to i^*j_*A$ is the unique sheaf morphism. -In your Q3, since you are working with abelian sheaves, you should consider $A=(\mathbf{Z}/2\mathbf{Z})_U$ and $B=(\mathbf{Z}/4\mathbf{Z})_Z$ as abelian sheaves, and you should require $\phi$ to be a homomorphism of abelian sheaves. Since -$$ -i^*j_*A = (\mathbf{Z}/2\mathbf{Z})_Z, -$$ -there are two choices for $\phi$ in a triple $(A,B,\phi)$, corresponding to $1\mapsto 1$ and $1\mapsto 0$.<|endoftext|> -TITLE: Graphs with many triangles but few complete graphs on 4 vertices -QUESTION [7 upvotes]: Let $G$ be a graph on $n$ vertices with $an^2$ edges containing at most $an^2/2$ copies of $K_4$. If there are cubically many triangles, say $cn^3$, then there is at least one edge that is not contained in any $K_4$. -Note that it is necessary to require that the number of $K_4$'s is significantly smaller than the number of edges. Otherwise we could take a complete tripartite graph and add an edge in each of the three colour classes. -So far my proof attempts using averaging arguments failed. Maybe there are counterexamples? If so, does it help to assume in addition that $a$ is bigger than $1/4$? -Background/Motivation -If the statement is true it implies that asymptotically $5n^2/16$ is the smallest size of an antichain in $2^{[n]}$ that is maximal among the antichains containing only 2-sets and 4-sets: basically, the $K_4$'s and the nonedges in the graph are the 4-sets and the 2-sets in the antichain. - -REPLY [11 votes]: Your statement may be true for large enough values of $c$, but it is not true for all constants $c>0$. -Specifically, for small enough values of $c$, form a counterexample $G$ consisting of the disjoint union of two subgraphs: - -$K_{n\sqrt 2,n\sqrt 2,n\sqrt 2}$ together with one extra edge in each component of the tripartition as in your example -$K_{n\sqrt{12},n\sqrt{12}}$ together with a perfect matching in each component of its bipartition. - -(Obviously these numbers of vertices are not all going to be integers, so round them.) -Then there are approximately $18n^2$ edges ($6n^2$ in the first subgraph and $12n^2$ in the second), approximately $9n^2$ $K_4$'s ($6n^2$ in the first subgraph and $3n^2$ in the second), approximately $(3\sqrt 2 + 4\sqrt 3)n$ vertices, and approximately $(2\sqrt 2)n^3$ triangles (mostly in the first subgraph). So this example has only half as many $K_4$'s as edges, as you ask, and every edge belongs to at least one $K_4$, with $c\approx\frac{2\sqrt 2}{(3\sqrt 2 + 4\sqrt 3)^3}\approx 0.00203$.<|endoftext|> -TITLE: What are an adviser's responsibilities? -QUESTION [11 upvotes]: There seems to be a wide variance in what faculty members view as their responsibilities as a student’s adviser. I understand that some advisers go above and beyond what's expected and we can't hold everyone to their standards. I also understand that there is no one right method and two advisers with very different methods can be equally successful. However, it does seem useful to understand what we view as the baseline. -My question simply is: what are an adviser's role and responsibilities? -This question refers to both research as well as professional development (support, networking, job search, etc). - -REPLY [7 votes]: Apart from all the straightforward things that too many people don't do (meet regularly, help find a problem and suggest a plan of attack, help package partial progress into complete if modest theorems, etc) here's the big one: -Be honest if things are going badly! -Ask your student what their goals are. This basically means ask when do they expect to graduate and what kind of job do they want. If they are not on track, say so, because I bet you they didn't know that. Please do not tell them they're not ready at the last second, or write them a lukewarm letter while smiling to their face.<|endoftext|> -TITLE: Random permutations of Z_n -QUESTION [23 upvotes]: In "The maximum number of Hamiltonian paths in tournaments" by Noga Alon, the author states the following without proof (equation 3.1): -"Consider a random permutation $\pi$ of $\mathbb{Z}_n$. What is the probability that $\pi(i+1)−\pi(i) \mod{n} < n/2$ for all $i$?" -The claim is that this is $(2+o(1))^{−n}$, which makes sense and seems like it should be a standard argument. However, I have not been able to come up with a short proof, nor have I been able to find a proof in the literature. -Does anyone know of a complete proof? - -REPLY [5 votes]: For odd $n$ the answer to your question (as stated!) can be found in Noga Alon's paper. Namely, the number of permutations in your question equals the permanent of an $n\times n$ matrix $A$ in which each row and each column has $(n-1)/2$ ones and $(n+1)/2$ zeros. Therefore $2/(n-1)*A$ is doubly stochastic, so by van der Waerden's conjecture (proved by Egorichev and Falikman in 1981) the requested probability is $\geq n!^{-1}((n-1)/2)^n n!/n^n=(1/e+o(1))2^{-n}$. -For even $n$ the answer is similar. Then the number of permutations in your question equals the permanent of an $n\times n$ matrix $A$ in which each row and each column has $n/2$ ones and $n/2$ zeros. Therefore $2/n*A$ is doubly stochastic, so similarly as before the requested probability is $\geq n!^{-1}(n/2)^n n!/n^n=2^{-n}$. -On the other hand, for all $n$ the considered permanent is $\ll \sqrt{n} n!/2^n$ by Noga Alon's main theorem in the paper, hence the requested probability is $\ll \sqrt{n}2^{-n}$. -Of course this does not explain why (3.1) in Noga Alon's paper holds. This is a statement about random cyclic permutations $\pi$ satisfying $\pi(i+1)−\pi(i) \mod{n} < n/2$ for all $i$.<|endoftext|> -TITLE: Notation for a representable functor -QUESTION [8 upvotes]: For an object $X$ of a category, $h_X$ is the contravariant functor represented by $X$, i.e. $h_X = Hom(-,X)$. -Question a) Who invented this notation? (My guess: Grothendieck) -b) Is there a special reason why the letter $h$ was chosen? Is it in an abbreviation for "homomorphism"? - -REPLY [12 votes]: Let me answer the questions in order. -a) It was invented by Grothendieck, see EGA I, Springer edition, especially chapter 0, discussion of representable functors. -b) Quite possibly is a shortcut for $Hom$. Sometimes the letter $y$ is used (for Yoneda). The trouble is when you are considering the representable functor defined over several categories, e.g. a category and a subcategory. -Further evidence on a) The notation is already on SGA 3 and 4. There are several exposés by Grothendieck in Henri Cartan's seminar from 1960/61 in which he explains his point of view of Teichmüller's space through representable functors in the analytical category and he uses the notation $h_X$. -I an not aware of anyone else using these ideas at that time. Cartan's seminar is available at numdam: -http://www.numdam.org/article/SHC_1960-1961__13_1_A7_0.pdf -See also Bourbaki seminar, exposé 195 (February 1960) -http://www.numdam.org/article/SB_1958-1960__5__369_0.pdf -Bonus: If you, instead of considering contravariant functors $\mathrm{Sch}^{o} \to \mathrm{Set}$, use covariant functors $\mathrm{Aff} \to \mathrm{Set}$ the notation used in EGA is $h_X^{o}$. Perhaps the reason is that Yoneda's map is contravariant in this case.<|endoftext|> -TITLE: Modeling question: how often does "the world's oldest person" die? -QUESTION [38 upvotes]: This story yesterday (no need to follow the link to understand the question!) -http://www.cnn.com/2011/US/02/01/texas.oldest.person.dies/index.html?hpt=T2 -reminds me that I've often wondered about the following: -Suppose you have a fixed mortality curve $f(x)$ expressing the probability of a person's remaining alive at age $x$. Suppose also, for simplicity, you have a steady state population, so a given fixed birth rate. How would one compute from $f(x)$, and the birth rate, the probability distribution $\rho$ governing the time intervals from one "world's oldest person dies" event to the next? -Actually though I think I could probably write down explicit integrals to answer my own question literally, but I don't expect they would say much as such. So I'm really looking for a softer answer that would explain what features of $f(x)$, what measures of the heaviness of its tail I guess, dominate the behavior of $\rho$ and how? - -REPLY [4 votes]: There was a discussion on this quite a long time ago at sci.mat. See: -http://groups.google.com/group/sci.math/browse_thread/thread/531c39a2e5723df8/528137c752566266?lnk=gst&q=oldest+person#528137c752566266<|endoftext|> -TITLE: complex algebraic morphisms as topological maps: every morphism is a topological fibration on a Zariski dense open subset? -QUESTION [7 upvotes]: In writing up a paper, we need references (and help) for the following -facts which are probably well-known. They concern morphisms of complex algebraic -varieties as continious maps in complex topology. (Here by 'complex topology' I mean -the topology induced by the metric on $\Bbb C$). -(i) Is every algebraic morphism of complex algebraic varieties necessarily -a fibration in the (non-noetherian) complex topology on a -Zariski-open Zariski-dense subset ? - That is, does there exists a Zariski open subset $Y^0$ of $Y$ such that - $f_{\Bbb C}: f_{\Bbb C}^{-1}(Y^0(\Bbb C)) \longrightarrow Y^0(\Bbb C)$ -is a fibration in the (non-noetherian) complex topology ? -(ii) If $f:W\rightarrow Y$ is a dominant rational map of irreducible complex varieties, with Y normal, then the index of the image of $\pi_1(W) \rightarrow \pi_1(Y)$ -divides the number of irreducible components of a generic fibre. -Does something like this holds in prime characteristic ? - (In char 0 this appears in [Janos Kollar, - "Shaferevich maps and automorphic forms", Lemma 2.10.2]; - is there a more standard reference as I find it hard to follow - the proof there, not being an algebraic geometer.) -(iii) Stein factorisation. Every proper morphism $f:X\rightarrow Y$ -of algebraic varieties decomposes as $X\rightarrow^{f_1} X' \rightarrow^{f_2} Y$ -such that $f_1$ has connected fibres and is a fibration in complex topology over a Zariski open dense subset, and $f_2$ is finite and etale on an Zariski open dense subset ? -The last question only makes sense if (i) is not always true; without the bit about complex topology Stein factorisation appears in EGA. These questions came up when trying to define a noetherian "Zariski-type" topology on the universal covering space of a complex algebraic variety that is weaker than the complex analytic topology, sort of a model of etale topology... - -REPLY [5 votes]: I have also wondered about question (i) in the past, and fortunately the answer is yes. -Here is a reference: -Verdier, Stratification de Whitney et theoreme de Bertini-Sard, Invent 1976, Cor 5.1 -The result is probably also contained in Thom, Bull AMS 75 (1969), but it may -be harder to extract (at least it was for me). -Wouldn't (iii) follow from (i) + Stein factorization, or is there something -that I'm missing? [In rereading your question, I realized you posed this only in the event -that (i) failed.]<|endoftext|> -TITLE: When does f-nef imply nef (after twisting?) -QUESTION [7 upvotes]: For a surjective morphism $f \colon X \to Z$ of smooth projective varieties, a line bundle $L$ on $X$ is $f$-nef if is nef on the fibers of $f$. More precisely, for every curve $C$ that is mapped to a point, the degree of $L$ restricted to $C$ should be larger than zero. -Now it is not true that an $f$-nef bundle becomes nef if I tensor it with a sufficiently large power of an ample line bundle on $Z$. There is a counterexample in the Positivity book of Lazarsfeld, Example 1.7.12. -Are there any conditions I can put on $L$ such that this statement becomes true? I'm interested in the case where $L = \mathcal{O}_X (-D)$ for an effective divisor D. - -REPLY [3 votes]: The following is maybe obvious, and isn't exactly what you asked, but perhaps is still worth spelling out. -As Sándor's answer shows, if $L=O_X(-D)$ is $f$-nef (where $D$ is effective) then in fact $D$ must be pulled back from $Z$, so $L$ belongs to the subspace $f^\ast(N^1(Z))$ of $N^1(X)$. In general, for any class $f^\ast L$ in $f^\ast(N^1(Z))$ we can twist with a line bundle $f^\ast(A)$ (where $A$ is a sufficiently ample bundle on $Z$) to get something nef, because $A+L$ is ample on $Z$ for sufficiently ample, and then $f^\ast(A+L)$ is nef --- indeed, it's even semi-ample. -So we only have a problem when there are f-nef classes which are not pulled back from $Z$. In particular, we can have a problem when the subspace $K$ of f-numerically trivial classes is strictly bigger than $f^\ast(N^1(Z))$. -This is what happens in the example you cite from Lazarsfeld: there $K$ has dimension 2, whereas $f^\ast(N^1(Z))$ has dimension 1 (since $Z$ is a curve). Drawing a picture of what happens there is instructive, and explains why the f-ample and f-nef cases behave differently. The nef cone of $X$ is a 3-dimensional round cone, and $K$ is a 2-dimensional subspace of $N^1(X)$ which lies tangent to the cone. They meet in a single ray, which is exactly $f^\ast(Nef(Z))$. It's geometrically clear that starting from a point on $K$ which is not on the line $f^\ast(N^1(Z))$ and adding elements of $f^\ast(Nef(Z))$, we can never get into the nef cone; on the other hand, starting with an f-ample class, which means exactly a point in the open half-space on the same side of $K$ as the nef cone, and moving in the direction of $f^\ast(Nef(Z))$, we eventually end up in the interior of $Nef(X)$, i.e. in the ample cone. -Here is my amateurish attempt to illustrate that: -Nef http://www.freeimagehosting.net/uploads/f7858c4e0d.jpg<|endoftext|> -TITLE: Differential Hochschild Cohomology, general tools? -QUESTION [5 upvotes]: Background: in deformation quantization one wants to construct formal associative deformations (the star products $\star$) of the algebra of smooth complex-valued functions on a Poisson manifold $M$ in direction of the Poisson bracket. The cochains in order $\hbar^r$, where $\hbar$ is the deformation parameter, are not just any $2$-cochains of $C^\infty(M)$ but typically one requires them to be either continuous with respect to the usual $C^\infty$-topology, or even bi-differential. Beyond the deformation of $C^\infty(M)$ one also is interested in deforming all kinds of modules over $C^\infty(M)$ once a star product $\star$ is fixed in such a way that one still obtains modules. Also here the "classical" modules usually come from geometry (sections of bundles etc) and hence allow for the notion of differential cochains. -The deformation problem is mainly governed by the Hochschild cohomology of $C^\infty(M)$ with values in the corresponding endomorphisms of the "classical" module one wants to deform (or the algebra itself). However, the additional requirement of being differential brings us to a sub-complex. Thus the computation of its cohomology is a new problem which in the cases I know has to be done more or less by hand... -For a general commutative algebra one still can define mutlidifferential operators (with values in modules) a la Grothendieck in a completely algebraic way (reproducing the above differential operators in the case of $C^\infty(M)$). This gives a sub-complex of the Hochschild complex which I would like to understand: for the algebraic situation and also for continuous cochains (under some conditions) one has the usual tools of homological algebra to identify the Hochschild cohomology as certain Ext groups etc. -However, the additional requirement of being differential does not seem to fit into this nice algebraic abstract nonsense theory. I guess it would be very nice to have these sort of tools also available in the differential setting, which, after all, is entirely algebraic in its nature. -So my question is: are there any possibilities to transfer the usual notions of Ext etc to the differential case? - -REPLY [7 votes]: If I understand the question correctly Stefan is asking for an Ext interpretation of the polydifferential Hochschild cochain complex. Elements of this are not just continuous linear maps $C^\infty(M)^{\otimes n} \to C^\infty(M)$, but they have to be polydifferential operators. (This version of Hochschild cohomology is used in Kontsevich's formality theorem). -Anyway, one can understand the polydifferential condition as follows. Consider the jet bundle $J$ on $M$; this is an infinite-rank vector bundle whose fibre at a point $p \in M$ is the algebra of formal power series at $p$. If we choose coordinates $x_1,\dots, x_n$ at $p$, then we can identify the fibre $J_p$ as $\mathbb{R}[[x_1,\dots,x_n]]$. -It's standard that $J$ is a left $D$-module. Further, the obvious product on the fibre of $J$ makes $J$ into a commutative algebra in the symmetric monoidal category of left $D$-modules. -Then, one can take Hochschild cochains of $J$ in the symmetric monoidal category of left $D$-modules. -This is the same as the complex of poly-differential Hochschild cochains. The key point is that $D$-module maps $J^{\otimes n} \to J$ are the same as polydifferential operators. -Of course, this means that you can apply any of the standard interpretations of Hochschild cohomology in this context (e.g. $\operatorname{Ext}_{J \otimes J}(J,J)$). -One needs a little care with these definitions, because $J$ is a topological $D$-module. However, if you take continuous $D$-module maps and appropriately completed tensor products you get the right answer.<|endoftext|> -TITLE: Drawing planar graphs with integer edge lengths -QUESTION [19 upvotes]: It is well known that every planar graph has an embedding such that every edge is drawn as a straight line segment (Fáry's Theorem). Kemnitz and Harborth made the following stronger conjecture -Conjecture 1. Every planar graph has a straight line embedding with integer edge lengths. -I was wondering if it is possible to attack this problem with the following approach. -Conjecture 2. Let $X:=\{ x_1, \dots, x_n \}$ be a finite set of points in the plane such that no three points of $X$ are colinear. For any $\epsilon >0$, there exists $X':=\{x_1', \dots, x_n'\}$ such that for all $i, j \in [n]$ - -$d(x_i, x_i') < \epsilon$, -$d(x_i', x_j') \in \mathbb{Q}$, and -no three points of $X'$ are colinear. - -To prove Conjecture 2, it suffices to prove the following conjecture. -Conjecture 3. Let $X:=\{ x_1, \dots, x_n \}$ be a finite set of points in the plane such that no three points of $X$ are colinear and all pairwise distances are rational. Then the set of points which are at rational distance from all points in $X$ is a dense subset of the plane. -Conjecture 3 is trivial for $n=1$ and easy for $n=2$. Almering proved it for $n=3$, and I think it is open for $n>3$. Note that Conjecture 3 is (essentially) a weakening of: -Conjecture 4. There exists a dense subset of the plane with all pairwise distances rational. -This question was posed by Ulam in 1945 (see this mathoverflow question for more background). So, the reason I like Conjecture 3 is that it is still strong enough to prove Conjecture 1, but appears much weaker than Conjecture 4. -Unfortunately, Conjecture 3 is beyond my limited area of expertise. Hence: - -Question. What are the prospects for proving Conjecture 3? A proof or disproof would be fantastic. However, even arguments suggesting that it is true/false but say beyond current technology would be most welcome. - -REPLY [3 votes]: I think conjecture 3 is actually stronger then conjecture 4. -I prove $C_3\implies C_4$: -Pick any sequence of integers $a_n$, which contains all integers infinite times. -Pick any enumeration of all squares $s_n$ in the plane with corners at rational coordinates. -Then assuming conjecture 3, at step $n$ we can find a point with rational distances in $s_{a_n}$ which is not collinear to any previously used rationals, since there are only finitely many straight lines in our set so far. -At step $\omega$, we have $\omega$ many rationals in every square, so a dense set of all rational distances.<|endoftext|> -TITLE: Orientation and generalized cohomology theories -QUESTION [12 upvotes]: Let h* be a multiplicative generalized cohomology theory and $E \rightarrow X$ a real vector bundle. - -Is it true that, if $E$ is orientable with respect to h*, then it is also orientable with respect to the singular cohomology with coefficients in $h^{0}(pt)$, for $(pt)$ a space with one point? -Is it true that, if $E$ is orientable with respect to the singular cohomology with coefficeints in $\mathbb{Z}_{p}$, for $p > 2$, then it is also orientable with respect to the singular cohomology with coefficients in $\mathbb{Z}$? - -Added later: I thought to a possible simple answer using weak orientability, actually proving what stated in the last comment. I wrote this as a comment to answer 3. - -REPLY [3 votes]: At the risk of being obvious, let me make some general comments: -For a vector bundle $E\to X$ of rank $n$ let $E_x$ be the fiber over $x\in X$. For a multiplicative cohomology theory $h$, the cohomology $h^*(E_x,E_x-0)$ is a free module of rank $1$ over the graded ring $h^*$, with generator in $h^n$. Such a generator is called an orientation at the point $x$. A Thom class (also called an orientation) is an element of $h^n(E,E-0)$ that restricts to such a generator for every $x$. -There is also the other sense of orientation: a continuous choice of pointwise orientation for all points. In the special case of ordinary cohomology, in other words when the graded ring $h^*(point)$ is all in degree zero, an orientation in this weak sense implies an orientation in the Thom-class sense; but not in general.<|endoftext|> -TITLE: Rational points and finite étale covers -QUESTION [13 upvotes]: This question is broadly about non-trivial examples where a map $f:Y\to X$ of smooth projective $k$-varieties ($k$ not algebraically closed) is such that existence of $k$-points (or in fact $k'$-points for $k'/k$ a finite extension) on $X$ ensures their existence on $Y$. For example, if $f$ is a birational $k$-map, the Lang-Nishimura Theorem guarantees an element of $Y(k)$ if $X(k)$ is not empty. -An interesting example is the case of $f:Y\to X$ being a finite étale cover. In general it will not be the case that a point in $X(k)$ lifts to one in $Y(k)$ (for example elliptic curves and an element in the Tate-Shafarevich group). Are there however examples or conditions that guarantee this happens? -We say that a variety $X/k$ is potentially dense if $X(k')$ is dense in $\bar{X}$ for some finite extension $k'/k$. Campana's classification program conjectures that $X$ is potentially dense if and only $X$ is special (i.e. has trivial core), so in this case, since it seems that a finite étale cover remains special if the base is special, we'd have that the arithmetic of special varieties is preserved under étale covers. But does potential density pull back under finite étale covers? -Related bibliography and any comments on the aforementioned or related topics mentioned are welcome! - -REPLY [4 votes]: For finite étale morphisms I found the following result in a survey article of Graber, Harris, Mazur and Starr: -Call a flat morphism $\pi:X\to B$ of schemes over a number field $F$ with $B$ smooth and irreducible arithmetically surjective if for every finite degree extension $L/F$, the induced mapping of rational points $X(L)\to B(L)$ is surjective. -$\textbf{Proposition:}$ Let $B$ be an integral, integrally closed scheme over a number field $F$. A finite flat morphism $\pi:X\to B$ defined over $F$ is arithmetically surjective if and only if it admits a section over $F$.<|endoftext|> -TITLE: Does this formula have a rigorous meaning, or is it merely formal? -QUESTION [51 upvotes]: I hope this problem is not considered too "elementary" for MO. It concerns a formula that I have always found fascinating. For, at first glance, it appears completely "obvious", while on closer examination it does not even seem well-defined. The formula is the one that I was given as the definition of the cross-product in $\mathbb R^3 $ when I was first introduced to that concept: -$$ - B \times C := - \det -\begin{vmatrix} - {\mathbf i } & {\mathbf j } & {\mathbf k } \\\\ - B_1 & B_2 & B_3 \\\\ - C_1 & C_2 & C_3\\\\ -\end{vmatrix} -$$ -On the one hand, if one expands this by minors of the first row, the result is clearly correct---and to this day this is the only way I can recall the formula for the components of the cross-product when I need it. But, on the other hand, the determinant of an $n \times n$ matrix whose elements are a mixture of scalars and vectors is undefined. Just think what happens if you interchange one element of the first row with the element just below it. In fact, as usually understood, for a determinant of a matrix to be well-defined, its elements should all belong to a commutative ring. But then again (on the third hand :-) if we take the dot product of both sides of the formula with a third vector, $A$, we seem to get: -$$ - A \cdot B \times C = - A \cdot - \det -\begin{vmatrix} - {\mathbf i } & {\mathbf j } & {\mathbf k } \\\\ - B_1 & B_2 & B_3 \\\\ - C_1 & C_2 & C_3\\\\ -\end{vmatrix} - = - \det -\begin{vmatrix} - A_1 & A_2 & A_3 \\\\ - B_1 & B_2 & B_3 \\\\ - C_1 & C_2 & C_3\\\\ -\end{vmatrix} -$$ -and of course the left and right hand sides are well-known formulas for the (signed) volume of the parallelepiped spanned by the three vectors, $A, B, C$. Moreover, the validity of the latter formula for all choices of $A$ indicates that the original formula is "correct". -So, my question is this: Is there a rigorous way of defining the original determinant so that all of the above becomes meaningful and correct? - -REPLY [17 votes]: Back in the 19th century, when people had been experimenting with determinants a lot, they might have interpreted the above definition of $B\times C$ in terms of quaternions. If $i$, $j$, and $k$ denote basis elements of $\mathbb H$ and -$${\mathbf x}=x_1i+x_2j+x_3k,$$ -$${\mathbf y}=y_1i+y_2j+y_3k\quad$$ -are pure imaginary elements of $\mathbb H$, then the vector part $\Im(\mathbf{xy})$ of the Hamilton product -$\mathbf{xy}$ is equal to the determinant -$$\Im(\mathbf{xy})=\Im(\mathbf{x})\times \Im(\mathbf{y})=\det -\begin{vmatrix} - i & j & k \\\\ - x_1 & x_2 & x_3 \\\\ - y_1 & y_2 & y_3\\\\ -\end{vmatrix}.$$ -There is a note by Sir Arthur Cayley where he introduces the notion of a quaternion determinant. He mentions several identities of the form -$$ \det -\begin{vmatrix} - {\mathbf x} & {\mathbf x} \\\\ - {\mathbf y} & {\mathbf y} - \\\\ -\end{vmatrix} - = - -2\det -\begin{vmatrix} - i & j & k \\\\ - x_1 & x_2 & x_3 \\\\ - y_1 & y_2 & y_3\\\\ -\end{vmatrix} -$$ -and -$$ \det -\begin{vmatrix} - {\mathbf x } & {\mathbf x } & {\mathbf x } \\\\ - {\mathbf y } & {\mathbf y } & {\mathbf y } \\\\ - {\mathbf z } & {\mathbf z } & {\mathbf z } \\\\ -\end{vmatrix} - = - -2\det -\begin{vmatrix} -{3} & i & j & k \\\\ - x_0 & x_1 & x_2 & x_3 \\\\ - y_0 & y_1 & y_2 & y_3\\\\ - z_0 & z_1 & z_2 & z_3\\\\ - \end{vmatrix} -$$ -where $\mathbf x$, $\mathbf y$, $\mathbf z$ are arbitrary quaternions -$${\mathbf x}=x_0+x_1i+x_2j+x_3k, \mbox{ etc.}$$<|endoftext|> -TITLE: singly-generated monoids in mathematics -QUESTION [13 upvotes]: There are many, many examples in mathematics of operations $s$ satisfying $ss = s$ (i.e., idempotent operations). -Not quite as common, but still numerous, are operations $s$ satisfying $sss = s$, specifically, Galois connections from a poset to itself; see my recent post Abstract nonsense attribution and the articles referred to in replies to that post. -My (casual) question is, are there examples of important operations satisfying other relations of this kind, such as $sss = ss$? (This particular relation reminds me of the story of an article by X that ends with three footnotes: "The author thanks Y for translating the preceding article", "The author thanks Y for translating the preceding footnote", and "The author thanks Y for translating the preceding footnote"; infinite regress is avoided because, as X explained, "While I may not be able to translate a sentence I can certainly copy one!" Can anyone provide a web-reference for this? I think it's in Littlewood's Miscellany.) - -REPLY [5 votes]: While reading your question I was reminded of Kuratowski's closure-complement problem. Here we start with an arbitrary subset $X$ of a topological space, and are allowed to apply the two operations of closure and complement. It turns out that for any $X$, we get at most 14 distinct sets by applying these two operations. If we let $k$ denote the closure operator and $c$ denote the complement operator, then the following relations imply the result - -$kk=k$ -$cc=id$, and -$kckckck=kck$. - -So, to answer the question, if we set $s=ck$ (take the closure of a set and then take the complement of the result), we get the relation $ssss=ss$. -I'll end by mentioning that there do exist $X$ where all 14 sets are possible. For example, under the usual topology of the reals, -$(0,1) \cup (1,2) \cup \{3\} \cup ([4,5] \cap \mathbb{Q})$ -is one such set.<|endoftext|> -TITLE: What's the Kirby Diagram of a universal $\mathbb{R}^4$? -QUESTION [18 upvotes]: What's the Kirby diagram of a universal $\mathbb{R}^4$? - -Background -Define $\mathcal{R}$ as the set of smoothings of $\mathbb{R}^4$. For two oriented elements $R_1$, $R_2$ in $\mathcal{R}$ we can define their end sum $R_1 \natural R_2$ if we are given two proper embeddings $\gamma_i : [0, \infty) \rightarrow R_i$. -We remove a tubular neighborhood of $\gamma_i((0, \infty))$ from each $R_i$ and glue the resulting $\mathbb{R}^3$ boundaries together respecting orientations. The result is the end sum $R_1 \natural R_2$ of $R_1$ and $R_2$. As $\gamma_i$ is unique up to ambient isotopy, $R_1 \natural R_2$ is well defined up to diffeomorphism. -In "A universal smoothing of four-space" Freedman and Taylor proved the existence of an element $U \in \mathcal{R}$ such that for any $R \in \mathcal{R}$ the end sum $U \natural R$ is diffeomorphic to $U$. This $U$ is the universal $\mathbb{R}^4$. -Foreground -In "An invariant of smooth 4-manifolds" Taylor defines an invariant $\gamma(R) \in \{0,1,2,\ldots,\infty \}$ for $R \in \mathcal{R}$. Taylor defines $\gamma(R)$ to be $sup_K \{ min_X\{ \frac{1}{2} b_2(X) \} \}$, where $K$ ranges over compact $4$-manifolds smoothly embedding in $R$ and $X$ ranges over closed, spin $4$-manifolds with signature $0$ in which $K$ smoothly embeds. (Actually, Taylor defines $\gamma$ for all smooth $4$-manifolds, but we don't need this detail here.) -Taylor goes on to prove that if $R \in \mathcal{R}$ and $\gamma(R) > 0$, then any handle decomposition of $R$ has infinitely many three handles. -In "4-Manifolds and Kirby Calculus" Stipsicz and Gompf prove, see page 376, that $\gamma(U) = \infty$. Thus, any Kirby Diagram of $U$ must have infinitely many three handles. -Question -What's the Kirby diagram of a such a $U$? - -REPLY [25 votes]: I would also like to know the answer to that. As far as I know, it is still a difficult, unsolved problem. The bit about 3-handles is a clue, but I haven't found any way to make use of it.<|endoftext|> -TITLE: Analogy of Parseval identity for Legendre Transform ? -QUESTION [24 upvotes]: Parseval's identity states that the sum of squares of coefficients of the Fourier transform of a function equals the integral of the square of the function, or -$$ \sum_{-\infty}^{\infty} |c_n|^2 = (1/2\pi)\int^\pi_{-\pi} |f(x)|^2 dx $$ -where the $c_i$ are the Fourier coefficients. -The Legendre-Fenchel transform can be viewed as a generalization of the Fourier transform. For a given function $f : X \rightarrow R$ over a vector space $X$ which has dual $X^{*}$, the transform $f^* : X^* \rightarrow R$ is defined as: -$$ f^*(p) = \sup_{x \in X}\ \langle x, p\rangle - f(x) $$ -where further $p = f'(x)$ is denoted as $x^*$. -So my question is: -Is there any natural generalization of Parseval's identity to relate $f^*$ and $f$ ? To be specific, I'm trying to relate quantities like $\|x-y\|$ to $\|p - q\|$ where $p = x^*, q = y^*$ - -REPLY [15 votes]: I think the identity you want is -$$2\inf_x f(x)=\inf_x(f^\ast(x)+f^\ast(-x))\mbox{.}$$ -(I'm skipping a bunch of conditions required of $f$ to make this hold. We'll need convexity at least.) -Let's use $\oplus$ for infimal convolution and let $g(x)=f(-x)$. -By definition $(f\oplus g)(x)=\inf_y(f(x-y)+g(y))$. -Infimal convolution gives us $(f\oplus g)^\ast=f^\ast+g^\ast$. -So -$$\begin{align} -\inf_x(f^\ast(x)+f^\ast(-x)) &=& \inf_x (f^\ast(x)+g^\ast(x))\\ -&=& \inf_x (f\oplus g)^\ast(x)\\ -&=& \inf_x\inf_y(f(x-y)+f(-y))\\ -&=& 2\inf_y f(-y)\\ -&=& 2\inf_x f(x)\mbox{.} -\end{align} -$$ -That second last equality is because if $f$ takes a minimum value at $-y$ then clearly $f(x-y)\ge f(-y)$ for any $x$ so the double infimum is attained when $f(x-y)$ and $f(-y)$ are "aligned" at $x=0$. -It looks less symmetrical than Parseval's theorem. But if $f$ is real, then Parseval's theorem gives -$$ -\int f(x)^2dx=\int\hat f(\omega)\hat f(-\omega)d\omega -$$ -Addendum: here's a detailed breakdown of the analogy: -I assumed you wanted Parseval's theorem on the Fourier transform, not the identity for Fourier series. This replaces the infinite sum with an integral. -According to the analogy we replace all integrals with infima. All multiplications with additions. All Fourier transforms with Legendre transforms. Squaring $x$ is multiplication of $x$ by itself, so that becomes adding $x$ to itself resulting in $2x$. Apply all of these replacements to -$$ -\int f(x)^2dx=\int\hat f(\omega)\hat f(-\omega)d\omega -$$ -and we get -$$2\inf_x f(x)=\inf_x(f^\ast(x)+f^\ast(-x))\mbox{.}$$ -Compare with Fenchel duality when $A=1$ (in the notation of that wikipedia page).<|endoftext|> -TITLE: Is there a canonical height on the Weil-Chatelet group? -QUESTION [12 upvotes]: Jon Hanke and I were just chatting and realized we didn't know the answer to the following question. If E is an elliptic curve over a number field, is there in any sense a "canonical height" on the Weil-Chatelet group H^1(G_Q,E)? (Note that we have no clearly articulated criteria for what "canoncial" means or what kind of compatibility with the group law one should demand -- since the group is torsion, any exact imitation of the usual definition will give everything height 0.) - -REPLY [11 votes]: In my opinion, instead of a "height" on the Weil-Chatelet group, one should consider a "depth", using the local duality between the points on an elliptic curve and the elements of the Weil-Chatelet group. Working over $Q$ for simplicity, there is a Pontrjagin duality between locally compact abelian groups: -$$WC_p \times E_p \rightarrow U(1),$$ -where $WC_p = H^1(G_p, E)$ and $G_p$ is the absolute Galois group of $Q_p$ ($WC_p$ is the local Weil-Chatelet group at $p$) and $E_p = E(Q_p)$ is the group of points on $E$ over $Q_p$. -At the real place, the duality is between $WC_R$ and $\pi_0(E_R)$. But one might as well define an "extended Weil-Chatelet group" (whose interpretation is not clear to me) by letting $WC_R'$ be the Pontrjagin dual of $E_R$. There is an exact sequence of locally compact abelian groups: -$$1 \rightarrow WC_R \rightarrow WC_R' \rightarrow Z \rightarrow 1,$$ -to explain why I'd call $WC_R'$ an extended Weil-Chatelet group. -Now, there is a filtration on $E_p$, which can be used to define the local height. Namely, choosing a Neron model, and letting $E_p^0$ denote the preimage of the nonsingular points in the special fibre, and $E_p^1$ the preimage of the identity element in the special fibre, there is a further filtration: -$$E_p^0 \supset E_p^1 \supset E_p^2 \supset \cdots.$$ -When $E$ has nonsingular reduction at $p$, the local height function can be defined by -$\lambda_p(e) = n \log(p)$, if $e \in E_p^n$ and $e \not \in E_p^{n+1}$. -So, the natural thing on Weil-Chatelet groups is the dual increasing filtration. Define $WC_p^n$ to be the annihilator of $E_p^n$ with respect to local duality. This increasing filtration on $WC_p^n$ has been studied by McCallum ("Tate duality and wild ramification", Math. Ann. 288 (1990) 553-558) and Yamazaki ("On Tate duality for Jacobian varieties", J. of Number Theory 99 (2003) 298-306); it has interpretations in terms of splitting fields and Brauer groups, I recall. -Now, if $w \in WC_p$, define the depth of $w$, written $d_p(w)$ to be the smallest $n$ for which $w \in WC_p^n$. If $w \in WC_Q$ (the global Weil-Chatelet group), then $d_p(w) = 0$ for almost all $p$, and one could define a global depth of $w$ by $d(w) = \prod_p p^{d_p(w)}$ (mapping $w$ to appropriate local Weil-Chatelet groups). For example, elements of Sha will have global depth zero (not accounting for archimedean places). -Now I haven't worked out the archimedean details, or primes of bad reduction, but using Pontrjagin duality and the Fourier transform it shouldn't be so terrible to write down the "local depths" in these cases. The resulting global depth function will have the nice properties that elements of Sha will have depth zero, and for any $N$, there will only be finitely many (assuming finiteness of Sha, and using Poitou-Tate duality) elements of $WC_Q$ of depth bounded by $N$. -Now, if any period-index specialists (you know who you are) out there would like to describe in more detail the elements of $WC_Q$ of depth bounded by $N$, I'd love to know more!<|endoftext|> -TITLE: Computational cost of converting between 3-manifold presentations -QUESTION [7 upvotes]: Given a 3-manifold presented as a triangulation, a Heegaard splitting, or a Dehn surgery, what is the computational cost of converting to the other two presentations? I would like Heegaard splittings to be given as a word in the standard Dehn twist generators, and Dehn surgery as a word in the standard braid group generators. -For instance, suppose we want to triangulate a Heegaard splitting. Although I can't find a good reference for this, I think it can be done in time which is linear in both the genus and the length of the Dehn word. Start with a triangulation of a handlebody such that there's a strip of triangles along each of the canonical curves. To implement a Dehn twist, glue tetrahedra along the strip so that the result is a sequence of 2-2 Pachner moves on the surface triangulation, like in this picture: -http://www.iqc.ca/~galagic/dehntriangles.jpg -What about the other directions (e.g., converting a triangulation into a Dehn surgery)? Can they also be done efficiently? -edit: to whatever extent it may matter, I am primarily interested in simplicial triangulations, i.e., a single edge may not form a loop. - -REPLY [8 votes]: I believe all these translations are in principle easy. The challenge is in implementing them cleanly and efficiently; the translations can be annoying and confusing. -As you describe, to go from a Heegaard splitting -to a triangulation, it's just a matter of a sequence of Pachner moves. If you allow (as is -usually sensible to do) non simplicial complex triangulations, where edges are allowed to form a loop in the manifold, then you can use one-vertex triangulations of the surfaces. There is only a finite set of these up to isomorphism, and the Pachner moves correspond to the one-skeleton for an equivariant cell division of Teichmuller space of the surface with a distinguished point (the vertex). For a finitely generated group, the translation from one set of generators to another has linear cost. The same principle works here, for any -fixed genus: it's a translation from generators for the mapping class group to -a set of generators to a mapping class groupoid generated by the Pachner moves. (Lee Mosher -in particular has studied this correspondence in detail). The linearity still holds, -or at least nearly holds (this depends on the details of definitions) - when you consider surfaces of every genus together, if you use Dehn twists -around a system of curves where each curve only meets a bounded number of other -curves (as is the usual convention). -If you allow ideal triangulations for the manifold minus some finite collection of curves, you can do even better: the number of simplices needed is linear in the number of powers -of Dehn twists using standard generators. -To go in the other direction, a triangulation is practically a special case of a Heegaard splitting: a regular neighborhood -of the 1-skeleton union its complement. If you want the handlebodies described in -standard form, it's essentially just a matter of choosing a spanning tree for the 1-skeleton and dual 1-skeleton, plus some method to give a homeomorphism from the regular neighborhood -of the spanning tree to a sphere with a set of distinugished points. -If a Heegaard diagram is described as a nonseparating system of g simple curves on the boundary of a genus g handlebody to which disks are attached in the complementary handlebody, this can be translated into a gluing map expressed as a word in Dehn twists in a reasonably straightforward way; this also gives a Dehn surgery description. In fact, Lickorish described a method in his paper showing that all 3-manifolds are obtained by Dehn surgery on links. I believe the number -of powers of Dehn twists needed should be a linear function of the number of bits used to describe the -$g$ curves using either traintracks or normal curve coordinates.<|endoftext|> -TITLE: Reference request: representations of unipotent groups have a fixed point. -QUESTION [6 upvotes]: I'm looking for a reference for the following standard result: -Let $U$ be a unipotent algebraic group over an algebraically closed field $k$ (of any characteristic); then any algebraic representation of $U$ has a fixed point. -Statements of Engel's theorem for the analogous statement about Lie algebras seem to be ubiquitous. I can also find the statement that connected solvable groups always preserve a line in many places (for example, Borel, Theorem III.10.4). Combining this with the fact that unipotent groups have no non-trivial characters gives me the result I need. But it would be nice to have a place to which I could refer for the precise statement about unipotent groups. - -REPLY [5 votes]: Theorem 17.5 in Humphreys's Linear Algebraic Groups seems to be the result you want. (Also, doesn't the result for solvable groups only imply the corresponding result for connected unipotent groups? The proof for unipotent groups doesn't require connectedness.) - -REPLY [5 votes]: This is Kolchin's theorem, first proved in Kolchin's paper http://www.jstor.org/stable/1969399<|endoftext|> -TITLE: Which vector spaces are duals ? -QUESTION [23 upvotes]: Every finite-dimensional vector space is isomorphic to its dual. -However for an infinite-dimensional vector space $E$ over a field $K$ this is always false since its dual $E^\ast$ is a vector space of strictly larger dimension: $dim_KE \lt dim_K E^\ast $ (dimensions are cardinals of course). This is a non-trivial statement for which our friend Andrea Ferretti has given an astonishingly unexpected proof here. This implies for example that a vector space of countably infinite dimension over a field $K$, like the polynomial ring $K[X]$, cannot be the dual of any $K$-vector space whatsoever. -So an infinite-dimensional vector space is not isomorphic to its dual but it could be isomorphic to the dual of another vector space and my question is: which vector spaces are isomorphic to the dual of some other vector space and which are not? -In order to make the question a little more precise, let me remind you of an amazing theorem, ascribed by Jacobson (page 246) to Kaplanski and Erdős: -The Kaplanski-Erdős theorem : Let $K$ be a field and $E$ an infinite-dimensional $K$-vector space . Then for the dual $E^\ast$ of $E$ the formula $dim_K (E^\ast) = card (E^\ast)$ obtains. -So now I can ask -A precise question : Is there a converse to the Kaplanski-Erdős condition i.e. if a $K$- vector space $V$ (automatically infinite dimensional !) satisfies $dim_K (V) = card (V)$ , is it the dual of some other vector space : $V \simeq E^\ast$? For example, is - $\mathbb R ^{(\mathbb R)}$ - which satisfies the Kaplanski-Erdős condition ( cf. the "useful formula" below) - a dual? -A vague request : Could you please give "concrete" examples of duals and non-duals among infinite-dimensional vector spaces? -A useful formula : In this context we have the pleasant formula for the cardinality of an infinite-dimensional $K$-vector space $V$ ( for which you can find a proof by another of our friends, Todd Trimble, here ) -$$ card \: V= (card \: K) \; . (dim_K V) $$ - -REPLY [15 votes]: The $\mathbf{R}$-vector space $\mathbf{R}^{(\mathbf{R})}$ has dimension $\operatorname{card} \mathbf{R}$ by definition, so it is isomorphic to $\mathbf{R}^{\mathbf{N}}$ (by the Erdős-Kaplansky theorem and because $\operatorname{card}(\mathbf{R}^{\mathbf{N}}) = \operatorname{card} \mathbf{R}$). So $\mathbf{R}^{(\mathbf{R})}$ is isomorphic to the dual of $\mathbf{R}[X]$. -In general, your precise question is equivalent to the following purely set-theoretical question (which seems difficult). By the useful formula, the identity $\operatorname{card} V = \operatorname{dim}_K V$ is equivalent to $\operatorname{card} K \leq \operatorname{dim}_K V$. Let $\kappa = \operatorname{card} K$ and $\lambda = \dim_K V$, and assume $\kappa \leq \lambda$. Does there always exist an infinite cardinal $\alpha$ such that $\kappa^\alpha = \lambda$? (here $\alpha$ is meant to be the dimension of the vector space whose dual is $V$). In general, Stephen's answer shows that there are counterexamples. -EDIT : in order to explain why the question is difficult, consider a field $K$ such that $\operatorname{card} K = \aleph_1$ (for example, one could take $K=\mathbf{Q}((T_i)_{i \in I})$ with $\operatorname{card} I =\aleph_1$). Take a $K$-vector space $V$ of dimension $\aleph_1$. Then $V$ is a dual if and only if there exists $\alpha \geq \aleph_0$ such that $\aleph_1 = (\aleph_1)^\alpha$, which amounts to say that $\aleph_1 = 2^{\aleph_0}$. In other words, $V$ is a dual if and only if the continuum hypothesis holds.<|endoftext|> -TITLE: Reference: Learning noncommutative geometry and C^* algebras -QUESTION [21 upvotes]: I am starting to study noncommutative geometry and ${\rm C}^*$ algebras so my question is: -Does anyone know a good reference on this subject? -I would like a basic book with intuitions for definitions and this kind of things. I come from algebraic geometry, so if the book talks a bit about the relation with algebraic geometry, that would be highly appreciated. -I have already taken a look at Connes's book but I find it too hard, and I'm currently studying Landi's but it lacks lots of proofs (he refers to several papers). -Thanks in advance. - -REPLY [4 votes]: Let me add some new-ish books to the mix that I liked and deal with some topics in $C^*$-algebras that have picked up some steam in recent years: -If you want an introduction to the state of the art in the theory of $C^*$-algebras then the book $C^*$-algebras and Finite-Dimensional Approximations by Brown and Ozawa is what you want to read. If you always wanted to know what completely positive maps are and how they are used to define nuclearity or if you want to remind yourself about the intricacies of tensor products of $C^*$-algebras, if you want to know everything about group $C^*$-algebras and exactness or if you work in von Neumann algebras and want to see if any of the things you do still work in the $C^*$-algebra case, it is all in this book! Apart from that it contains a lot of humorous footnotes and is a great read. Gosh, I sound like a car salesman, but I am a huge fan of this book. -Next up is Classification of Nuclear $C^*$-algebras. Entropy in Operator Algebras by Rørdam and Størmer. This is a book about more advanced topics, but if you are looking for everything related to the classification program of nuclear, simple $C^*$-algebras, which has had some recent breakthroughs, then this is your book. It gives a good description of the purely infinite case for example, so if you always wanted to read up on what makes the Cuntz algebras $\mathcal{O}_2$ and $\mathcal{O}_{\infty}$ so special, then give it a shot. The reason why I am neglecting the entropy part in this description is entirely due to my own laziness: It is still on my reading list. -If you want to know how algebraic topology, in particular sheaf cohomology, can be used to classify continuous trace $C^*$-algebras or if you want to learn about Morita equivalences and Brauer groups, then Morita Equivalence and Continuous-Trace $C^*$-Algebras by Raeburn and Williams is the book of your choice. It contains a proof of the Imprimitivity Theorem and that stuff you always wanted to know about the whole induction and restriction business.<|endoftext|> -TITLE: Why is the Hodge Conjecture so important? -QUESTION [113 upvotes]: The Hodge Conjecture states that every Hodge class of a non singular projective variety over $\mathbf{C}$ is a rational linear combination of cohomology classes of algebraic cycles: Even though I'm able to understand what it says, and at first glance I do find it a very nice assertion, I cannot grasp yet why it is so relevant as to be considered one of the biggest open problems in algebraic geometry. Which are its implications? -Going a bit further, what about the Tate conjecture? - -REPLY [178 votes]: Let $K$ be one of the following fields: the complex numbers, a finite field, a number field (and we could amalgamate the last two into the more general case of a field finitely generated overs its prime subfield). -In each case we can consider the category of smooth projective varieties over $K$, with morphisms being correspondences [added: modulo the relation of cohomological equivalence; -see David Speyer's comment below for an elaboration on this point]. (I am really thinking of the category of pure motives, but there is no particular need to invoke that word.) -In each case we also have a natural abelian (in fact Tannakian) category in play: in the complex case, the category of pure Hodge structures, and in the other cases, the category of $\ell$-adic representations of $G_K$ (the absolute Galois group of $K$) (for some prime $\ell$, prime to the characteristic of $K$ in the case when $K$ is a finite field). -Now taking cohomology gives a functor from the category of smooth projective varieties to this latter category (via Hodge theory in the complex case, and the theory of etale cohomology in the other cases). The Hodge conjecture (in the complex case) and the Tate conjecture (in the other cases) then says that this functor is fully faithful. -The consequence (if the conjecture is true) is that we can construct correspondences between varieties simply by making computations in some much more linear category. -Let me give an example of how this can be applied in number theory: -Fix two primes $p$ and $q$, and let $\mathcal O_D$ be a maximal order in the quaternion algebra $D$ over $\mathbb Q$ ramified at $p$ and $q$, and split everywhere else (including at infinity). -Let $\mathcal O_D^1$ denote the multiplicative group of norm one elements in $\mathcal O_D$. -Since $D \otimes_{\mathbb Q} \mathbb R \cong M_2(\mathbb R)$, we may regard $\mathcal O_D^1$ as a discrete subgroup of $SL_2(\mathbb R)$, and form the quotient $X := \mathcal O_D^1\backslash \mathcal H$ (where $\mathcal H$ is the upper half-plane). -We may also consider the usual congruence subgroup $\Gamma_0(pq)$ consisting of matrices in $SL_2(\mathbb Z)$ that are upper triangular modulo $pq$, and form $X_0(pq)$, the compacitifcation of $\Gamma_0(pq)\backslash \mathcal H$. -Now the theory of modular and automorphic forms and their associated Galois representations show that $X$ and $X_0(pq)$ are both naturally curves over $\mathbb Q$, and that there -is an embedding of Galois representations $H^1(X) \to H^1(X_0(pq))$. Thus the Tate conjecture predicts that there is a correspondence between $X$ and $X_0(pq)$ inducing -this embedding. Passing to Hodge structure, we would then find that the periods of holomorphic one-forms on $X$ should be among the periods of holomorphic one-forms on $X_0(pq)$. -Now the theory of $L$-functions shows that the periods of holomorphic one-forms on $X_0(pq)$ in certain cases compute special values of $L$-functions attached to modular forms on $\Gamma_0(pq)$. So putting this altogether, we would find that (given the Tate conjecture) we could compute special values of $L$-functions for certain modular forms by taking period integrals on the curve $X$. In certain respects $X$ is better behaved than $X_0(pq)$, and so this is an important technique in investigating the arithmetic of $L$-functions. -Now it happens that in this case the Tate conjectures is a theorem (of Faltings), and so -the above argument is actually correct and complete. -But there are infinitely many other analogous situations in the theory of Shimura varieties where the Tate conjecture is not yet known, but where one would like to know it. -There are also similar arguments where one uses the Hodge conjecture to move from information about periods to information about Galois representations and arithmetic. -I should also say that there are lots of partial results along the above lines in these cases; there are many inventive techniques that people have introduced for getting around the Hodge and Tate conjectures. (Deligne's use of absolute Hodge cycles, applications of theta corresondences by Shimura, Harris, Kudla, and others, ... .) But the Hodge and Tate -conjectures stand as fundamental guiding principles telling us what should be true, and which we would dearly like to see proved. -Summary: algebraic cycles are very rich objects, which straddle two worlds, the world of period integrals and the world of Galois representations. Thus, if the Hodge and Tate conjectures are true, we know that there are profound connections between those two worlds: we can pass information from one to the other through the medium of algebraic cycles. If we had these conjectures available, it would be an incredible enrichment of our understanding of these worlds; as it is, people expend a lot of effort to find ways to pass between the two worlds in the way the Hodge and Tate conjectures predict should be possible. -Another example, added following the request of the OP to provide an example just involving complex geometry: If $X$ is a K3 surface, then from -the Hodge structure on the primitive part of $H^2(X,\mathbb C)$, one can construct an -abelian variety, the so-called Kuga--Satake abelian variety associated to $X$. The -construction is made in terms of Hodge structures. One expects that in fact there should -be a link between $X$ and its associated Kuga--Satake abelian variety provided by some correspondence, but this is not known in general. It would be implied by the Hodge conjecture. (This constructions is discussed in this answer and in the accompanying comments.)<|endoftext|> -TITLE: What is the different in the cyclotomic tower over a finite ramified extension of Qp? -QUESTION [6 upvotes]: If $K_n$ is the field $\mathbb{Q}_p(\mu_{p^n})$, then it's easy to see that the relative different $\mathcal{D}(K_n / K_{n-1})$ is $(p)$ for all $n \ge 2$. -What happens if I take an arbitrary, probably totally ramified, finite extension $L/\mathbb{Q}_p$ and look at the tower $L_n = LK_n$? It's clear that $\mathcal{D}(L_n / L_{n-1})$ divides $(p)$, and one can show (using a general result of Tate on $\mathbb{Z}_p$-extensions) that its valuation tends to 1 as $n \to \infty$; but is it true that it's equal to $(p)$ for all sufficiently large $n$? - -REPLY [5 votes]: Here's a relatively easy counterexample. Take $p=2$, $L=\mathbb{Q}_2(2^{1/3})$. I did some direct computation and saw that $v_2(\mathfrak{D}^{L_2}_L)=2/3$, $v_2(\mathfrak{D}^{L_3}_{L_2})=5/6$. Looks like there's a pattern. But there's a better way. -We're in a situation where not only $L/\mathbb{Q}_2$ is tamely ramified of degree $3$, but also every $L_n/K_n$. Now we use the functoriality of the Hasse-Herbrand transition function: if $k\subset F\subset K$, then $\varphi^K_k=\varphi^F_k\circ \varphi^K_F$. Use the relation $\varphi^{L_n}_{\mathbb{Q}_2}=\varphi^{K_n}_{\mathbb{Q}_2}\circ \varphi^{L_n}_{K_n}= \varphi^L_{\mathbb{Q}_2}\circ \varphi^{L_n}_ L$ and the fact that a tamely ramified extension has all the transition-function action at the origin. That is, the function is $y=x$ for $x\le 0$, but $y=x/e$ for $x\ge 0$, where $e$ is the ramification index. So as real functions, $\varphi^L_{\mathbb{Q}_2}=\varphi^{L_n}_{K_n}$, namely this is just $y=x/3$. Consequently, the transition function of $L_n/L$ is gotten by conjugating that of $K_n/{\mathbb{Q}_2}$ with the tame transition function. The effect is to multiply all coordinates of vertex points by $3$. -But we also know the transition function of $K_n$ over ${\mathbb{Q}_2}$: its vertices are at all $(2^{i-1}-1,i-1)$ for $2\le i\le n$.The new vertices are at $(3,3)$, $(9,6)$, $(21,9)$, etc. This means that the lower breaks of $L_n/L$ are at $3(2^{i-1}-1)$ for $2\le i\le n$, and in particular the unique break of $L_n/L_{n-1}$ is at $3(2^{n-1}-1)$. -Now use the formula -\begin{align}{ -v_F(\mathfrak{D}^F_k)=\sum_{j\ge 0}\bigl(|G_j|-1\bigr) -}\end{align} -where the $G$'s are the lower ramification groups, and where in this case all the numbers being added up are $1$ or $0$, to see that $v_{L_n}\bigl(\mathfrak{D}^{L_n} _ {L_{n-1}}\bigr)=3(2^{n-1}-1)+1=3\cdot2^{n-1}-2$. Divide by the ramification number of $L_n$ over $\mathbb{Q}_2$ to get $1-1/(3\cdot 2^{n-2})$, agreeing with my computations for $n=2$ and $n=3$. -It's not an issue of tame versus wild ramification in the extension $L/\mathbb{Q}_2$, either. I used $L=\mathbb{Q}_2(2^{1/4})$ to find that the numbers are $1-3/2^m$; the argument is similar but a bit more delicate, since you have no a priori idea of what the transition function of $L_n/K_n$ might be.<|endoftext|> -TITLE: does the j-invariant satisfy a rational differential equation? -QUESTION [14 upvotes]: Let $j$ be the Klein $j$-invariant (from the theory of modular functions). -Does $j$ satisfy a differential equation of the form $j^\prime (z) = f(j(z),z)$ for -any rational function $f$? - -REPLY [2 votes]: A third-order differential equation for $j(\tau)$ is gotten via the Schwarzian derivative. The result is (1.13) of the paper by Harnad: http://arxiv.org/abs/solv-int/9902013<|endoftext|> -TITLE: How to compute cohomology groups of a closed subscheme Z of projective space, defined by a homogeneous polynomial of degree d? -QUESTION [5 upvotes]: Let $Z = \mathrm{Proj}\,k[x_{0},x_{1},\ldots,x_{r}]/f$ be a closed subscheme of degree $d$, i.e., $f$ is a homogeneous polynomial of degree $d$, and $\mathcal{O}_{Z}(1)=i^{*}\mathcal{O}_\mathbb{P}(1)$. Could someone please help me compute all the cohomology groups? I took the short exact sequencence given by the surjection from the closed immersion, but im not sure how twisting and taking long exact sequence will solve it! - -REPLY [6 votes]: If you are just looking for cohomology of the structure sheaf (or of the twists $\mathcal{O}_X(n)$), then it should be pretty easy. -You have the following short exact sequence. -$$0 \to O_{\mathbb P} (-d + n) \to O_{\mathbb P}(n) \to O_Z(n) \to 0 $$ -Take cohomology. Because $\mathbb{P}$ is projective space (of whatever dimension) almost all of the cohomology groups $H^i(P, \mathcal{O}_{\mathbb{P}}(k))$ vanish. Whatever doesn't vanish is relevant to computing the cohomology of $\mathcal{O}_Z(n)$. -For example, the only two relevant parts of the sequence are: -$$0 \to H^0(\mathbb{P}, O_{\mathbb P} (-d + n)) \to H^0(\mathbb{P}, O_{\mathbb P}(n)) \to H^0(Z, O_Z(n)) \to 0.$$ -and -$$0 \to H^{\dim Z}(Z, O_Z(n)) \to H^{\dim Z + 1}(\mathbb{P}, O_{\mathbb P} (-d + n)) \to H^{\dim Z + 1}(\mathbb{P}, O_{\mathbb P}(n)) \to 0.$$ -So for example, the dimension of $H^{\dim Z}(Z, O_Z(n))$ is the same as the dimension of $H^{\dim Z + 1}(\mathbb{P}, O_{\mathbb P} (-d + n))$ minus the dimension of $H^{\dim Z + 1}(\mathbb{P}, O_{\mathbb P} (-n))$. -See Chapter III, Section 5 of Hartshorne for example. -For a generalization to complete intersections, see Exercise 5.5 in that same section of Hartshorne. -Of course, if you trying to compute the cohomology of some other sheaves, then it will be much much harder and it will depend on the sheaves as Emerton above suggests.<|endoftext|> -TITLE: A book you would like to write -QUESTION [67 upvotes]: Writing a book from the beginning to the end is (so I heard) a very hard process. Planning a book is easier. This question is dual in a sense to the question "Books you would like to read (if somebody would just write them)". It is about a book that you feel you would like to write (if you just have the time). A book that will describe a topic not yet properly discribed or give a new angle to a subject that you can contribute. -The question is meant to refer to realistic or semi-realistic projects (related to mathematics). Answers about book projects based on existing survey articles or lecture notes can be especially useful. -Of course, If you had some progress in writing a book mentioned here please please update your answer! - -REPLY [3 votes]: I would like to write 'the best book on differential geometry', a genuinely pedagogical, useful and visual introduction to differential geometry for students who are coming to the subject for the first time with no prior knowledge. -None of the existing textbooks really meet these criteria in my opinion. -Edit: There is a new book out by Tristham Needham called ''Visual Differential Geometry'' which claims to meet my criteria above, hopefully this lives up to expectations.<|endoftext|> -TITLE: Special values of L-functions as periods -QUESTION [35 upvotes]: If $M$ is a pure motive over $\mathbb{Q}$, one cas define its $L$-function $L(M,s)$ which conjecturaly is a meromorphic function over $\mathbb{C}$ with finitely many poles. -For example, when $M=\mathbb{Q}$ is the trivial motive, $L(\mathbb{Q},s)$ is the Riemann Zeta function $\zeta(s)$. -There is a famous concjecture saying that all values of $L(M,s)$ at integers $n$ (which are not zero or poles, say; otherwise replace value by principal value) are periods -- that is -to say have real and imaginary parts that can be expressed as (according to Wikipedia's definition) differences of volumes of region in of Euclideans pace given by polynomial -inequalities with rational coefficients. -This conjecture is due (if I am not mistaken) to Deligne in the case where $n$ is a so-called critical value of $L(M,s)$ and to Beilinson in general. I won't recall the definition of critical here, but I can say that for $\zeta(s)$ the critical values are $n=2,4,6,8,\dots$ and $n=-1,-3,-5,\dots$. Of course, the conjecture of Deligne in this case was known to Euler who proved that $\zeta(2n)$ is a rational times the volume of the unit ball in $\mathbb{R}^{4n}$, and $\zeta(1-2n)$ is rational, for $n \geq 1$ (the later made completely rigourous by Riemann). -Now my question: - -In which case (if any) where $n$ is not a critical value is it known that L(M,n) is a period? - -Here is a second question, related to the first: - -Do you know a good survey on the progresses oon Beilinson's conjecture? - -REPLY [33 votes]: In the case $M$ is the spectrum of a number field (so that $L(M,s)$ is the Dedekind zeta function associated to the number field), it is known thanks to Borel's theorem that all non-critical values $L(M,n)$ are indeed periods. EDIT : I should add that it is very easy to prove that $\zeta(n)$ is a period for every $n \geq 2$, by the following computation : -\begin{equation*} -\zeta(n)=\sum_{k=1}^{\infty} \frac{1}{k^n} = \sum_{k=1}^{\infty} \int_0^1 \cdots \int_0^1 (x_1 \cdots x_n)^{k-1} dx_1 \cdots dx_n = \int_0^1 \cdots \int_0^1 \frac{dx_1 \cdots dx_n}{1-x_1 \cdots x_n}. -\end{equation*} -There is a pole at $(1,\ldots,1)$ in the last expression (one can further regularize), but since the integral is absolutely convergent, this is sufficient to prove that $\zeta(n)$ is a period. -In the case $M$ is the motive associated to a (classical) newform $f$ of weight $k \geq 2$, the non critical values are $L(f,m)$ with $m \geq k$. Beilinson's theorem states that each of these values is given, up to a standard factor, by the determinant of a regulator matrix (you can think about it as an analogue of the class number formula if you wish). In fact, in this case, the regulator matrix has size 1, so we just have a number. Unravelling the definition of Beilinson's regulator map, this implies that $L(f,m)$ is indeed a period (if one considers $L'(f,k-m)$ instead, maybe one has to invert a power of $\pi$). Note however that in general this expression as a period is far from explicit. -In more complicated cases like the symmetric powers of the motive associated to a modular form, then (to my knowledge) almost nothing is known, except in the case of CM elliptic curves, for which there is a general theorem by Deninger. -It is quite difficult to give an exhaustive list of all results in this area, and I may have forgotten to mention important results. In any case, it would be indeed nice to have such a list. So please don't hesitate to complete this answer. -For a good survey on Beilinson's conjectures, you may want to look at Nekovar's article "Beilinson conjectures". The link with periods is well-explained in Kontsevich-Zagier's article "Periods". Another reference I have in mind is Flach's article "The Equivariant Tamagawa Number Conjecture : A survey".<|endoftext|> -TITLE: Why is "P vs. NP" necessarily relevant? -QUESTION [41 upvotes]: I want to start out by giving two examples: -1) Graham's problem is to decide whether a given edge-coloring (with two colors) of the complete graph on vertices $\lbrace-1,+1\rbrace^n$ contains a planar $K_4$ colored with just one color. Graham's result is that such a $K_4$ exists provided $n$ is large enough, larger than some integer $N$. -Recently, I learned that the integer $N$ determined by Graham's problem is known to lie between $13$ and $F^{7}(12)$, which is called Graham's number, a number which beats any imagination and according to Wikipedia is practically incomprehensible. Some consider it to be the largest number which was ever used in a serious mathematical argument. -One way of looking at this is the following: Graham's problem of deciding whether for given $n$ and given coloring such a planar $K_4$ exists takes constant time. Indeed, if $n$ is small you have to look, if $n$ is large you are done. However, this takes polynomial time (check all four-tuples of vertices) for all practical purposes (assuming that $N$ is close to its upper bound). -I am sure someone can now cook up an algorithm which needs exponential time for all practical purposes but constant or polynomial time in general. -2) I also learned that the best proof of Szemerédi's regularity lemma yields a bound on a certain integer $n(\varepsilon)$ which is the $\log(1/\varepsilon^5)$́-iteration of the exponential function applied to $1$. This bound seems ridiculous in the sense that it does not even allow for interesting applications of this result (say with $\varepsilon=10^{-6}$) to networks like the internet, neural networks or even anything practically thinkable. At this point, this is only an upper bound, but Timothy Gowers showed that $\log(1/\varepsilon)$-iterations are necessary. -Again, it seems that one could cook up reasonable algorithmic problems which have solutions which are polynomial time but practically useless. Maybe one can do better in concrete cases, but this then needs additional input. - -Coming closer to the question, what if finally $P=NP$ holds, but the proof involves something like the existence of a solution to Graham's problem or an application of Szemerédi's regularity lemma, so that finally the bounds of the polynomial time algorithm are for some reason so poor that nobody even wants to construct it explicitly. Maybe the bounds are exponential for all practical purposes, but still polynomial. -I often heard the argument that once a polynomial solution for a reasonable problem is found, further research has also produced practicable polynomial time algorithms. At least for Graham's problem this seems to fail miserably so far. - -Question: Is there any theoretical evidence for this? - -Now, maybe a bit provocative: - -Question: Why do we think that $P\neq NP$ is necessarily important? - -I know that $P$ vs. $NP$ is important for theoretical and conceptional reasons, but what if finally $P=NP$ holds but no effective proof can be found. I guess this wouldn't change much. -EDIT: Just to be clear: I do not dismiss complexity theory at all and I can appreciate theoretical results, even if they are of no practical use. - -REPLY [26 votes]: The $P \ne NP$ problem is the best way we know to formulate the belief (which was expressed even before the problem was formally stated) that certain specific algorithmic problems (such as finding a Hamiltonian cycle in a graph) requires exponential number of steps as a function of their description. The formulation is based on the important notion of a nondeterministic algorithm. The conjecture that P is not equal to NP is the basis of a mathematical theory of complexity which is remarkably beautiful. It is likely that a proof for the conjecture will lead to better understanding of the limitations of computation which will go beyond the conjecture itself. (Unfortunately we are very far from such a proof.) A counterexample (which is not expected) may lead to a major change of our reality and not just our understanding of it. (I suppose this is one of the main reasons in believing the conjecture.) Your concern is that maybe, because of the asymptotic nature of the conjecture, we can question its real relevance. Namely problems we regard as intractable may still be even if P=NP because of huge constants in the polynomial involved, or perhaps even if NP not equal P there can be efficient algorithm to relevant cases of intractable problems. These are serious concerns that are often raised and sometimes there are efforts to study them scientifically. -Overall, it looks that these asymptotic concerns are secondary compared to the problem itself. Moreover, there are many examples that the asymptotic behavior and practical behavior are similar beyond what the theory dictates (but a few examples also in the other direction). There are other related concerns regarding the relevance for the $P \ne NP$ problem. Is the worst case analysis relevant to practical problems? Is the hardness apply when we are interested in approximate solution rather than in exact solution? Both these questions (like the asymptotic issue) can be regarded as secondary in importance compared to the $NP \ne P$ problem itself but otherwise very important. Indeed both hardness of approximation and average case analysis are very central research topics. -(A side remark, there is something unnatural about the way Graham's numbers are used in the question. You are dealing with a decision problem where the answer is always yes when n is sufficiently large. And there is a simple polynomial time algorithm as a function of n to decide the answer for a every given n. So the relevance of this particular example (which is interesting) to the question asked about NP complete problems (which is also interesting) is not so strong.) -Finally, regarding relevance. Andreas raised the interesting possibility that NP=P but the constant involved are so huge that the polynomial algorithm for solving NP complete problems is utterly not practical. A related interesting possibility is that there even exists a practical polynomial time algorithm for NP complete problems but that finding this algorithm itself is computationally intractable. As I said, both these possibilities are considered unplausible. Even if true they will not harm the relevance of the NP=P problem in making the central distinction between intractable and tractable problems. In order to make these concerns interesting one should come up with a theoretical framework to study them, and then study them fruitfully. (As I mentioned above, this was done for related concerns regarding approximation and regarding average case behavior.) - -REPLY [13 votes]: It is clearly relevant whether there exist feasible algorithms for NP-complete problems, where I am using the term "feasible algorithm" in the informal sense of an algorithm that we could actually run quickly in practice on problems of practical interest. Call this the Main Question. -The relevance of the $P = NP$ question lies in its perceived relevance to the Main Question. Clearly, to prove that $P = NP$ is a relevant question, it would suffice to prove the lemma that $P$ is precisely the class of problems with feasible algorithms. -Of course, the lemma is obviously false, as you point out and as others have pointed out (e.g., Yuri Gurevich). Still, there is enough overlap between $P$ and the class of problems with feasible solutions that settling the $P=NP$ question would be a major partial step towards answering the Main Question. It is in this sense that $P=NP$ is a relevant question.<|endoftext|> -TITLE: Are there smooth bodies of constant width? -QUESTION [21 upvotes]: The standard Reuleaux triangle is not smooth, but the three -points of tangential discontinuity can be smoothed as -in the figure below (left), from the Wikipedia article. -However, it is unclear (to me) from this diagram -whether the curve is $C^2$ or $C^\infty$. -Meissner’s tetrahedron is a 3D body of constant width, -but it is not smooth, as is evident in the right figure -below. -    -My question is: - -Are there $C^\infty$ constant-width bodies in $\mathbb{R}^d$ (other than the spheres)? - -The image of Meissner’s tetrahedron above is taken -from the impressive work of -Thomas Lachand–Robert and Edouard Oudet, -"Bodies of constant width in arbitrary dimension" -(Math. Nachr. 280, No. 7, 740-750 (2007); pre-publication PDF here). Here is a link to Wayback Machine.) -I suspect the answer to my question is known, in which case a reference would suffice. Thanks! -Addendum. -Thanks to the knowledgeable (and rapid!) answers by Gerry, Anton, and Andrey, -my question is completely answered—I am grateful!! - -REPLY [7 votes]: This relevant paper was just released: - -Howard Resnikoff. "On Curves and Surfaces of Constant Width." Apr. 2015. -(arXiv abstract.) - - -"...we provide a Fourier series-based construction that produces arbitrarily many new surfaces of constant width." - - -      -![ConstWidth][1]<|endoftext|> -TITLE: Upper bounds for ranks of modular jacobians -QUESTION [8 upvotes]: The following question came to me earlier as a "side question"; something I'd like to know, but which is not totally necessary for what I'm thinking about or doing: -Consider the genus 32 curve $X_0(389)$, and denote its Jacobian variety as $J_0(389)$. -I am interested in finding an upper bound for the Mordell-Weil rank of $J_0(389)(\mathbb{Q}(i))$. -After thinking about this for some time, I turned to Google, which threw up [1]. Apparently, assuming the Birch-Swinnerton-Dyer conjecture, there is an absolute constant $C > 0$ such that for all primes $q$ sufficiently large, we have -$ \mbox{rank } J_0(q)(\mathbb{Q}) \leq C \mbox{ dim} J_0(q) $. -[Ideally this equation would be in the center] -The point of that paper is to show that $C = 6.5$ will do (existence of $C$ having been proved in an earlier paper by the same authors), but, "assuming [also] the Riemann Hypothesis for automorphic $L$-functions, Iwaniec, Luo and Sarnak have recently proved that one could take $C = \frac{99}{100}$". -Now I reckon 389 is "sufficiently large", which means (if you believe those conjectures) an upper bound for the rank over $\mathbb{Q}$ is 31. But this feels like it is way too big, maybe because there are generally so few rational points on modular curves. - -Does anyone know how to get a better upper bound? Or am I wrong in hoping for a smaller upper bound? - -Furthermore, since I'm interested in $\mathbb{Q}(i)$-rank, there is the following: - -What is the biggest the rank can jump by when going from $J_0(389)(\mathbb{Q})$ to $J_0(389)(\mathbb{Q}(i))$? - -I guess this last question can be asked in greater generality, replacing the Js with any abelian variety $A/\mathbb{Q}$. Can the rank jump be arbitrarily large when passing from $\mathbb{Q}$ to $\mathbb{Q}(i)$? Or is there a bound in terms of the dimension of $A$, say? -It's my bedtime now, so I'll pick this thread up in 9 or so hours. -[1]: "Explicit Upper Bound for the (Analytic) rank of $J_0(q)$". E. Kowalski, P. Michel. Israel J. Math, 2000. Preprint Available here - -REPLY [7 votes]: To partly answer your last two questions: It's not too hard to write down a sequence of abelian varieties $A_i/\mathbf{Q}$ such that $\mathrm{rank}A_i(\mathbf{Q})=0$ but $\mathrm{rank}A_i(\mathbf{Q}(\sqrt{-1}))\to \infty$ as $i\to\infty$. More precisely, if $p\equiv 1\ \mathrm{mod}\ 4$ is large enough, there is some newform $f$ of weight $2$ on $\Gamma_0(p^3)$ such that -a. $L(s,f)$ has root number $+1$, and $L(1/2,f)\neq0$, which in fact implies $L(1/2,f^{\sigma})\neq0$ for all Galois conjugates of $f$ by some work of Shimura; -b. $L(s,f\otimes \chi_{-4})$ has root number $-1$ (which is automatic from a. and my choice of $p$) and $L'(1/2,f \otimes \chi_{-4})\neq 0$, which again implies $L'(1/2,f^{\sigma} \otimes \chi_{-4})\neq 0$ for all Galois conjugates of $f$ by the Gross-Zagier formula and its generalizations. -The existence of such a newform follows from the asymptotic evaluation -$\sum_{f\in S_2^{new}(\Gamma_0(p^3)),\ \varepsilon(f)=+1}L(1/2,f)L'(1/2,f\otimes \chi_{-4}) \sim cp^3$ -for some real $c>0$, which nowadays is a standard application of the Petersson formula (see e.g. the final chapter of Iwaniec and Kowalski's book). Given such an $f$, let $A_f$ be the corresponding optimal quotient of $J_0(p^3)$. Then a. guarantees that $\mathrm{rank}A_f(\mathbf{Q})=0$ by the work of Kolyvagin-Logachev/Zhang/Longo/Tian-Zhang, while b. guarantees that $\mathrm{rank}A_f(\mathbf{Q}(\sqrt{-1}))=\mathrm{dim}A_f$ by the same group of people, and $\mathrm{dim}A_f \geq \frac{p-1}{2}$ (see e.g. this question), so we're done. -This whole argument works with $\mathbf{Q}(\sqrt{-1})$ replaced by any fixed imaginary quadratic field. If you demand the same dichotomy for a sequence of $A$'s of bounded dimension, I imagine it's not known.<|endoftext|> -TITLE: varieties with points in number fields -QUESTION [37 upvotes]: Let $V$ be a projective variety, defined over $\mathbb{Q}$. Suppose that for every number field $K \neq \mathbb{Q}$, there is a $K$-point of $V$. Does it follow that $V$ has a $\mathbb{Q}$-point? -More generally, what are the restrictions on the possible sets $S$ of number fields $K \supset k$ such that for some projective variety $V$ defined over $k$, we have that $V$ has a $K$-point if and only if $K \in S$? Note that there has to be some nontrivial restriction here, since there are only countably many projective varieties defined over number fields, but uncountably many sets of number fields closed under extensions. - -REPLY [8 votes]: A simple case where the answer to your question is positive is provided by Heegner's Lemma, an observation made by Heegner in his solution of the class number 1 problem and extracted from his proof by Birch: if $y^2 = f_4(x)$, where $f_4$ is a quartic polynomial whose leading coefficient is not a square, has a solution in a number field of odd degree, then it has a rational point. This lemma has a cohomological interpretation, where the claim essentially boils down to the observation that if $a^n$ is a square for some odd $n$, then so is $a$. -I expect that there are other instances of Heegner's lemma for other types of curves, but I don't know anything specific.<|endoftext|> -TITLE: Do plain functors out of monoidal categories factor into a nontrivial monoidal part and a plain part? -QUESTION [5 upvotes]: Given symmetric monoidal closed categories $C, D$, a symmetric monoidal (not closed!) functor $F:C\to D$ factors into two parts: - -the first is a symmetric monoidal closed functor from $C$ to a "halfway house" $C'$, followed by -a symmetric monoidal functor from $C'$ to $D$. - -To get $C'$, you do two changes of base: - -since $C$ is closed, consider it a $C$-enriched category and then apply $F$ to its hom objects to get a $D$-enriched category, and then -apply the "points" functor $\mbox{hom}(1, -):D \to \mbox{Set}$ to get a plain category. - -Does something similar always happen when a functor fails to preserve the structure of the source and target categories? In particular, does a plain functor between monoidal categories factor into a nontrivial monoidal functor followed by another plain functor? - -REPLY [5 votes]: Regarding the specific example, the construction of $C'$ can be tightened up as follows. The -first functor, from $C$ to $C'$ is not just hom-preserving (closed) but bijective on objects. The second functor, from $C'$ to $D$, has the property that it is fully faithful on maps out of the monoidal unit. -(This is enough to determine $C'$ uniquely. As is implicit in Scott's comment, you need more than just the fact that the first leg is closed, otherwise you could take the first map to be the identity.) -So you could consider the (2-)category of symmetric monoidal closed categories, and (strong) symmetric monoidal functors. Any such morphism $F:C\to D$ comes with a canonical comparison -$F[c,d]\to[Fc,Fd]$, and we can call it (strongly) closed if these comparisons are invertible. There's a class E of morphisms consisting of those which are bijective on objects and closed, and a class M of morphisms consisting of those for which $F$ induces a bijection between maps $i\to c$ and maps $Fi\to Fc$, for all $c$, where $i$ is the monoidal unit. (You might call these maps pre-fully-faithful, or something like that.) These classes E and M are closed under -composition. I haven't checked in detail, but it looks like they have a reasonable chance of -being orthogonal to each other, and so you would have a factorization system. -I don't see a way of doing something similar with mere functors between monoidal categories. But perhaps this is too much to expect. In the other example, although the internal hom is not preserved, there is a canonical comparison. So perhaps rather than looking at plain functors you should look at the lax monoidal ones (often just called monoidal, with the word strong being added to mean preservation up to isomorphism). Now it is true that every lax monoidal functor $C\to D$ factorizes as a lax monoidal $C\to C'$ followed by a strict monoidal $C'\to D$, moreover in a universal (initial) way. For any monoidal category $C$ there is a lax monoidal $p:C\to C'$ with the property that composition with $p$ induces a bijection between lax monoidal $C\to D$ and strict monoidal $C'\to D$, for any $D$. (Notice -that the order is opposite to that in your example: the strict map comes second.) -This situation is quite common, and holds for many different types of structure. It is much -less common, but does sometimes happen that there's a universal factorization $C\to D'\to D$ with the map $C\to D'$ strict and the map $D'\to D$ non-strict. -Your example involves an extra ingredient involving the bijective-on-objects and pre-fully-faithful conditions.<|endoftext|> -TITLE: Given a family of curves, when does there exist a fibered surface over Spec Z parametrizing them? -QUESTION [12 upvotes]: Let $X_p$ be a projective curve over the finite field $\mathbf{F}_p$ (i.e. a projective $\mathbf{F}_p$-scheme pure of dimension 1) for every prime number $p$. Let $X_\mathbf{Q}$ be a projective curve over $\mathbf{Q}$. When does there exist a flat projective integral normal scheme $X$ over Spec $\mathbf{Z}$ such that the fibre above $p$ is $X_p$ for every prime p and the generic fibre is $X_\mathbf{Q}$? -Suppose that $X$ exists. Then the generic fibre is a smooth projective connected curve over $\mathbf{Q}$. Furthermore, $X_p$ is almost always smooth over $\mathbf{F}_p$. Also, the arithmetic genus is constant in the fibres of $X$. Moreover, the Hilbert polynomial of $X_p$ is independent of $p$. -Example. Suppose that $X_p$ is a supersingular elliptic curve for all $p$. Then there does not exist such an $X$. - -REPLY [16 votes]: I suspect that even if you had a single curve over $\mathbb{F}_p$, you might not find a lift -to $\mathbb{Q}$. Below I sketch an argument that works under the assumption that $\mathcal{M}_g$ does not have Zariski dense set of points. If you believe the conjectures of Lang on rational points, this assumption should be satisfied as soon $\mathcal{M_g}$ is of general type, e.g. if $g \geq 24$. -Lemma. -Suppose that $g$ is an integer such that $\mathcal{M}_g(\mathbb{Q})$ is not Zariski dense in $\mathcal{M}_g$. Then for all sufficiently large primes $p$, there are smooth curves $C_p$ of genus $g$ defined over $\mathbb{F}_p$ that are not reductions of curves defined over $\mathbb{Q}$. -Proof. -By assumption, the set $Z:=\overline{\mathcal{M}_g(\mathbb{Q})}$ is a proper closed subset of $\mathcal{M}_g$. In particular, the dimension of $Z$ is smaller than the dimension of $\mathcal{M}_g$. By the Lang-Weil estimates, the number of points of $\mathcal{M}_g$ modulo $p$ grows as a polynomial in $p$ of degree $\dim \mathcal{M}_g$, since $\mathcal{M}_g$ is irreducible modulo $p$. Similarly, the number of points of $Z$ modulo $p$ grows as a polynomial in $p$ of degree at most $\dim Z$ (the "at most" comes from the fact that $Z$ need not be geometrically integral). Thus, for sufficiently large $p$, there will be points of $\mathcal{M}_g(\mathbb{F}_p)$ that are not contained in $Z(\mathbb{F}_p)$. These points correspond to smooth curves of genus $g$ defined over $\mathbb{F}_p$ that are not reductions of curves defined over $\mathbb{Q}$, as required.<|endoftext|> -TITLE: Can an integer or rational sequence satisfy some bounded order recurrence $\mod \ $ almost all primes but doesn't satisfy such in $\mathbb{Q}$? -QUESTION [5 upvotes]: Can an integer or rational sequence satisfy some bounded order recurrence $\mod \ $ almost all primes but doesn't satisfy such in $\mathbb{Q}$? -The recurrences $\mod p$ can be different, possibly depending on $p$ and the recurrence need not be linear, any recurrence will do. - -REPLY [4 votes]: YES, define $a_k$ by $a_0=0$, $a_k=a_{-k}$ for $k<0$ and for $k>0$ $a_k=m!$ where $m\ge 1$ is as large as possible subject to $k$ being a multiple of $m!$. Then $$a_{k+m!} \equiv a_k \mod m.$$ However we have $a_j=a_{m\cdot m!+j}$ for $1 \le j\le m!-1$ but not for $j=m!$ when $a_{m!} \ne a_{(m+1)!}$ so the sequence can't satisfy a recurrence of finite order. -By request: Here it is from $a_{-9}$ to $a_{33}:$ -$${\small \cdots 1,2,1,6,1,2,1,2,1,}\mathbf{0}\small{,\overline{1,2,1,2,1,6,1,2,1,2,1,6,1,2,1,2,1,6,1,2,1,2,1,24},1,2,1,2,1,6,1,\cdots}$$ -$a_0=\mathbf{0}$ and all other terms are positive. The length 24 sequence with the overline keeps repeating except the $4!=24$ is $5!=120$ in positions 120,240,360,480,600 (but not 720) 840,960 -Three notes: - -One could use the least common multiple of $\lbrace1,2,3\cdots m\rbrace$ in place of $m!$ -Since the question asked only about primes one could make it have period #$p$ (p primorial) $\mod p$ -If $a_k$ satisfies a recurrence of order $n$ mod $m$ then it is periodic $\mod m$ with a period $P=P_m$ which is no greater than $m^n$. Hence it is enough to ask: " If $$ is an integer sequence which is periodic mod $m$ (with a period $P_m$ depending on $m$) for every $m$, must it satisfy a finite recurrence?<|endoftext|> -TITLE: Is this 1974 claim still valid? -QUESTION [21 upvotes]: In G. F. Simmons' Differential Equations book (p.141), the following claim is made: -“... As a matter of fact, there is no known type of second order linear differential equation- apart from those with constant coefficients, and those reducible to these by changes of the independent variable--which can bee solved in terms of elementary functions.” -About 36 years have now passed since this statement made its published appearance. Is the remark still true or was it false even before? -My motivation is the analogous theory of integrals of finite combinations of elementary functions where you know certain large and important classes of functions whose integral is expressible as an elementary function( or as a finite combination of such). As we know, there is a somewhat complete classification as to which integral of finite combination of elementary function of one variable is expressible as a finite combination of elementary functions.(References are Piskunov's book and this ). So it seemed to me that it is natural to care whether such theorems existed for differential equations. -Thanks. - -REPLY [12 votes]: I think that the questionner could be addressed to the mathematical subject named Differential Galois Theory. -A first accessible reading could be J.F. Ritt "Integration in finite terms. Liouville's theory of elementary methods", Columbia University Press, 1948. -The corresponding wikipedia page could be another starting point. http://en.wikipedia.org/wiki/Differential_Galois_theory<|endoftext|> -TITLE: showing the subgroup membership problem is undecidable for $F_2 \times F_2$ -QUESTION [6 upvotes]: Let $F_2$ denote the free group of rank 2. Does anybody have a fast proof that the subgroup membership problem is undecidable for $F_2 \times F_2$? I saw a really fast proof last semester that started with a group with undecidable word problem and used that group to construct subgroups of $F_2 \times F_2$, but I've lost my notes from the talk. I tried looking at the original proof from the 1960's but I couldn't get an English translation. Thanks! - -REPLY [8 votes]: Here is a completion of the proof in Andy's answer. One needs to show that the subgroup is finitely generated (which is not true if $G$ is not finitely presented). The subgroup (also called the Mihailova subgroup) is generated by the elements $(a,a)$ and $(1,r)$, where $a$ runs over the generators of $F(S)$ and $r$ runs over the finite set of defining relations of $G$. The fact that the Mihailova subgroup (it consists of all pairs $(x,y)$ with $f(x)=f(y)$) is generated by these elements is easy. Clearly for every generator $(u,v)$ ($u=v=a$ or $u=1, v=r$) we have $f(u)=f(v)$. Conversely, if $f(u)=f(v)$, that is $f(uv^{-1})=1$ in $G$, the word $uv^{-1}$ can be represented as a product of conjugates of defining relators and their inverses. From this product it is easy to represent the pair $(u,v)$ as a product of generators (see, for example, this paper and the references there). For example, if $uv^{-1}=ara^{-1}$, then $(v,u)=(a,a)(1,r)(a,a)^{-1}(v,v)$ where $(v,v)$ is an obvious products of pairs of the form $(a,a)$.<|endoftext|> -TITLE: Is there a preferable convention for defining the wedge product? -QUESTION [50 upvotes]: There are different conventions for defininig the wedge product $\wedge$. -In Kobayashi-Nomizu, there is $\alpha\wedge\beta:=Alt(\alpha\otimes\beta)$, -in Spivak, we find $\alpha\wedge\beta:=\frac{(k+l)!}{k!l!}Alt(\alpha\otimes\beta)$, -where $\alpha$ and $\beta$ are any forms of degree $k$ and $l$ respectively, and $Alt(\cdot)$ take the alternating part of the tensor. -But, is there a rationale to prefer one of them among the others? -If not, what do you prefer? and for what reason? - -REPLY [2 votes]: I prefer the convention as in Kobayashi-Nomizu, as this is the natural one when using the modern approach to algebraic structures. In this set-up, the exterior algebra of a vector space $V$ is the quotient $\bigwedge V = TV/I$, where $TV$ is the tensor algebra of $V$, and I is the bilateral ideal generated by elements of the form $x\otimes x$. -As discussed in the Appendix of the reference "Euclidean Clifford Algebra", in this situation, one should define -$ -\alpha \wedge \beta := Alt (\alpha\otimes\beta) -$<|endoftext|> -TITLE: Nonfree projective module over a regular UFD? -QUESTION [19 upvotes]: What is the simplest example of a domain $R$ which is regular (in particular Noetherian) and factorial which admits a finitely generated projective module that is not free? -In fact I'll be at least somewhat happy with any example, since I can't think of one at the moment. -Some brief comments: $R$ needs to have Krull dimension greater than one or else it is a PID. The module in question needs to have rank greater than one, because the hypotheses force -the Picard group to be equal to the divisor class group and the divisor class group to be trivial. And famously, by work of Quillen and Suslin, one cannot take $R$ to be a polynomial ring over a field. Oh yes, and of course $R$ can't be local (or even semilocal, I suppose). I'm already out of ideas... -P.S.: If you can get an easier example by removing the hypothesis of finite generation, I'd be interested in that as well. - -REPLY [2 votes]: Let $A=\mathbb R[x_0,\ldots,x_n]/(x_0^2+\ldots+x_n^2-1)$ be the coordinate ring of $S^n$. Let $P$ be the kernel of the surjection $A^{n+1}\rightarrow A$ defined by $(x_0,\ldots,x_n)$. If $n\not=1,3,7$, then $P$ is Not free. Clearly $P\oplus A \sim A^{n+1}$. For $P$ not free, note that $P\otimes C(S^n,\mathbb R)$ corresponds to tangent bundle of $S^n$, which is not free, unless $n=1,3,7$. This gives examples of projective $A$-modules of rank $=$ dimension $A$ which is stably free but not free. Note that when rank $P>$ dimension $A$, then $P$ is cancellative (Bass' 1964), i.e. $P\oplus A^t\sim Q\oplus A^t$ implies $P\sim Q$. The above example says that Bass result is best possible. Note that Suslin (~1977) has proved that if $A$ is affine algebra over algebraically closed field, then projective $A$ modules of rank = dimension $A$ are also cancellative. Hence, if we replace $\mathbb R$ by $\mathbb C$ in $A$, then $P$ is free.<|endoftext|> -TITLE: Who is this guy : Z.A. Melzak (wrote Companion to Concrete Mathematics) ? -QUESTION [6 upvotes]: Author : Z.A. Melzak -Book Title : Companion to Concrete Mathematics. -Publication : Dover renewed 2004 2 volumes in one. Copyright 1972/1976. -I found this book extremely nice. -To whet your appetite he talks about reification as well as some plain and less plain way to accelerate series convergence. -In a few words it is a rare blend of concreteness and conceptualization. It smells a bit like Concrete Mathematics (by Knuth and co..) but the -only information I found was an an depth review by Klamkin (not entirely positive though). -QUESTION : I would like to find connections : information about the author, and for those who appreciate this book or heard loadably about it: what others books/works are in the same vein ? - -REPLY [17 votes]: According to the Mathematics Genealogy Project, Zdzislaw Alexander Melzak did his graduate studies at MIT and became a professor at UBC. I'm not sure whether his name might have actually been spelled Zozislaw, though (I guess someone Polish would know which is correct). The UBC Library has online a photo of him from 1966.<|endoftext|> -TITLE: Does this “flipping lexicographic” ordering have a standard name? -QUESTION [8 upvotes]: I’ve run into the following straightforward variant of lexicographic ordering, and am wondering if it has a standard name. I’ve been calling it the flipping lexicographic ordering, for evident reasons. I could also imagine it getting called the parity lexicographic ordering, but a brief search suggests that that’s used for some slightly different orderings. -$\newcommand{\x}{\mathbf{x}} \newcommand{\y}{\mathbf{y}} \newcommand{\N}{\mathbb{N}} \newcommand{\fl}{\mathrm{fl}} \newcommand{\lfl}{\;\sqsubset^\fl\;}$ -For sets $\x, \y \in \binom{\N}{m+1}$, write $\x = \{x_0 < \ldots < x_m\}$, $\y = \{y_0 < \ldots < y_m\}$. -Definition. $\x \lfl \y$ if $\x$ and $\y$ differ first in the $i$th place, and - -$i$ is even, and $x_i < y_i$; or -$i$ is odd, and $y_i < x_i$. (This is the flip!) - -As for ordinary lex, there’s also a nice inductive characterisation: Write $\x = \{x_0\} \cup \x^{\geq 1}$, and $\y = \{y_0\} \cup \y^{\geq 1}$, similarly to above. Then $\x \lfl \y$ if and only if either $x_0 < y_0$, or $x_0 = y_0$ and $\y^{\geq 1} \lfl \x^{\geq 1}$. (Again, note the flip.) -Does this ring any bells with anybody? -(Of course, $\lfl$ has obvious generalisations beyond $\binom{\N}{m+1}$; I’m sticking to that case here partly for definiteness, mainly since that’s the specific case I’m interested in.) - -Background: I’ve been playing around with implementing the algorithms from Ross Street’s “The Algebra of Oriented Simplices” (and related papers) in Haskell/Agda, and this ordering turns out to make a computationally convenient stand-in for his $\lhd$ order, in places. - -REPLY [6 votes]: I found an example in the mathematical literature where the same ordering on words, and more specifically continued fractions, is called "alternating lexicographic" order. I guess that there are other examples too, and that this is name can be considered standard. The term "boustrophedonic order" also appears in the mathematical literature, but it seems to mean something different. The boustrophedonic order on the English alphabet is AZBYCXDW... . In my opinion, calling your ordering boustrophedonic is clever, but I think that "alternating lexicographic" is more consistent as well as more standard, since it is an alternating combination of the lexicographic and colexicographic (or lex and colex) orderings.<|endoftext|> -TITLE: Isomorphism of semidirect products -QUESTION [15 upvotes]: Let $N,H$ be groups, $\phi \colon H\rightarrow Aut(N)$ be a homomorphism, $f\in Aut(N)$ and $\hat{f}$ be an inner automorphism of $Aut(N)$ induced by $f$. Then $$N\rtimes_{\phi} H \cong N\rtimes_{\hat{f}\circ\phi}H.$$ Also, if $\psi \in Aut(H)$, then $$N\rtimes_{\phi}H\cong N\rtimes_{\psi\circ\phi}H.$$ - -What are sufficient conditions for isomorphism of semidirect products? Are there any other criteria for isomorphism of semidirect products? - -REPLY [8 votes]: This is not an answer to your question, but there is another way that an isomorphism can arise. It is possible for two semidirect products $N_1 \rtimes H_1$ and $N_2 \rtimes H_2$ (with $N_1 \cong N_2$, $H_1 \cong H_2$) to be isomorphic as groups, but for there to be no isomorphism that maps $N_1$ to $N_2$. An example of this is the group -$G = \langle x,y,z \mid x^{29}= y^{29}=z^7=1, xy=yx, x^z=x^7, y^z=y^{16} \rangle,$ -which is an extension of $C_{29} \times C_{29}$ by $C_7$. Let $N_1$ and $N_2$ be the normal subgroups of order 29 generated by $x$ and $y$. Then $G/N_1 \cong G/N_2$ is the unique nonabelian group of order $29 \times 7$, but there is no automorphism of $G$ that maps $N_1$ to $N_2$, so this group can be expressed as a semidirect product of $C_{29}$ by the nonabelian group of order $29 \times 7$ in two different ways. -You might prefer ot restrict your attention to isomorphisms between $N_1 \cong N_2$ and $H_1 \cong H_2$ that map $N_1$ to $N_2$. If you do that, then I don't know the answer to your question in general, but I believe that in the special case when $N$ is an elementary abelian $p$-group, all isomorphisms arise in the ways you have described in your post. I don't feel like trying to write down a proof right now! - -REPLY [3 votes]: Aside from the two ways you've mentioned, I think there are just two other ways of getting an isomorphism (i.e. any isomorphism of semidirect products with same $N$ and $H$ is some combination of all four): - -Change $H$ to a different complement. This corresponds to multiplying $\phi$ pointwise by inner automorphisms of $N$. These are classified by (non-abelian) 1-cocycles $\mathrm{Z}^1(H,N)$, or since you've already incorporated global conjugations in $N$, you can use the non-abelian cohomology $\mathrm{H}^1(H,N)$. This coincides with the usual cohomology if $N$ is abelian. Note that this doesn't affect the induced map $\tilde{\phi}: H \to \mathrm{Out}(N)$. -Change $N$ to a different normal subgroup of the total group which happens to be isomorphic (and have isomorphic complement.) Edit: This case is hard to deal with in general, but one situation it can be avoided is if you only consider semidirect products in which $N$ plays some special role, like being the derived subgroup. Derek elaborates more on this case in his answer.<|endoftext|> -TITLE: "Must read" papers in algebraic K-theory? -QUESTION [19 upvotes]: I'm mainly interested (graduate student) in surgery theory and geometric topology. -If I have a chance to suggest "must read" papers in geometric topology for beginner, -I'm very glad to suggest "Topological Library" books volume 1,2,3 -(including monumental papers of Smale,Milnor,Kervaire-Milnor,Thom,Serre,Novikov...) -available in the following cite.(volume 3 is not available in English edition up to now) -http://www.amazon.com/Topological-Library-Characteristic-Structures-Everything/dp/9812836861/ref=sr_1_1?s=books&ie=UTF8&qid=1296894607&sr=1-1 -Question: What are "must read" papers in algebraic K-theory? -(I hope that most of them can be readable with basic understanding about classical K-theory such as Rosenberg's text or Milnor's ann. math. studies book) - -REPLY [3 votes]: On the Lichtenbaum-Quillen Conjectures from a Stable Homotopy-Theoretic Viewpoint by Stephen A. Mitchell (around 60 pages) contains unbelievable amount of the very key information on both algebraic K-theory, stable homotopy theory, and their most exciting interactions. When I started reading it, I thought I knew some of both, and was overwhelmed by the breathtaking panorama the author was skillfully presenting, showing only the very essence and at the same time managing to convey deepness and importance of several highly technical methods. Fantastic paper!<|endoftext|> -TITLE: Spectral techniques for genus of a graph -QUESTION [8 upvotes]: A generic question: -are there any spectral techniques to estimate the genus of a graph? I am interested in complete balance multipartite graph. - -REPLY [5 votes]: Yes, there are techniques. For graphs of fixed genus and $n$ vertices, the second lowest eigenvalue of the laplacian is of order $O(1/\sqrt{n}),$ where the hidden constant depends on the genus (in an explicit way -- this follows from the Cheeger inequality and the separator theorems of Lipton and Tarjan, see eg, the paper of Spielman and Teng called "Spectral partitioning works). The dependence on the genus can be made quite explicit, so if you do that, you will get a lower bound on the genus in terms of the size of the graph and $\lambda_2.$ - -REPLY [4 votes]: I believe that this is the subject of Jon Kelner's paper Spectral partitioning, eigenvalue bounds, and circle packings for graphs of bounded genus. In particular, he proves a lower bound of the form $O(g/n)$ for $\lambda_2$, resolving a conjecture of Spielman and Teng.<|endoftext|> -TITLE: Deligne's letter to Piatetskii-Shapiro from 1973 -QUESTION [14 upvotes]: Could anyone point me to a place where I could find Deligne's letter to Piatetskii-Shapiro from 1973? It is cited for example in Berkovich's "Vanishing cycles for formal schemes II". - -REPLY [23 votes]: I have typeset Deligne's letter, and placed the result here: -http://www.math.ias.edu/~jaredw/DeligneLetterToPiatetskiShapiro.pdf -I have made some minor edits so that the text reads more naturally to a native speaker of English. Also I made a few annotations where I truly believe there is an error in the original. Any other errors are mine. -The letter struck me with how much it accomplishes in such a short space. Actually, I would say that the meat of the argument is confined to the final three pages, with the rest there only to establish notation. This letter ought to be required reading for anyone studying automorphic forms in arithmetic!<|endoftext|> -TITLE: History of the Lagrange Inversion Theorem -QUESTION [10 upvotes]: I'm doing research on the history of the Lagrange inversion theorem. The earliest predecessor I've found is the one referenced by De Morgan; viz. Jo. H. Lambert's construction in Observationes Variae in Mathesin Puram, Acta Helvetica, Vol. 3, 1758, pp. 128-168. -If anyone knows of an earlier construction I'd greatly appreciate hearing about it. -Thanks in advance. -Cheers, Scott - -REPLY [8 votes]: If you count any inversion of a power series as a predecessor of Lagrange -inversion, then I believe the earliest examples are Newton's inversion of the -log series to obtain the exponential series, and inversion of the inverse sine -series to obtain the sine series. The exponential series is in his De methodis -(1671), p.61, and the sine series is in his De analysi (1669), p. 233, 237. -Also, de Moivre, Philosophical Transactions 20 (1698), pp. 190-193, gave a -more general formula in a paper entitled "A method of extracting the root of an -infinite equation."<|endoftext|> -TITLE: Complexity of random knot with vertices on sphere -QUESTION [26 upvotes]: Connect $n$ random points on a sphere in a cycle of -segments between succesive points: - -      - -I would like to know the growth rate, with respect to $n$, of the crossing number -(the minimal number of crossings of any diagram of the knot) -$c(n)$ of such a knot. -I only know that $c(n)$ is $O(n^2)$, -because it is known -that the crossing number is upper-bounded by -the stick number $s(n)$: -$$\frac{1}{2}(7+\sqrt{ 8 c(K) + 1}) \le s(K)$$ -for any knot $K$. -And $s(n) \le n$ is immediate. -I feel certain this has been explored but I am not finding it -in the literature. Thanks for pointers! - -REPLY [3 votes]: I did some experiments on the degrees of Alexander polynomials, as suggested by Bill Thurston, and here is a table of values: -{{10, 1}, {20, 3}, {30, 7}, {40, 7}, {50, 19}, {60, 35}, {70, - 43}, {80, 73}, {90, 73}, {100, 115}, {120, 123}, {140, 175}, {160, - 233}, {180, 259}} - -(the first number is the number of vertices, the second is the degree of the Alexander polynomial). -Or, for those of us more visually-minded, here is a plot: - -I would describe the growth rate as at most $N^{3/2},$ consistent with Bill's theory.<|endoftext|> -TITLE: Video lectures of mathematics courses available online for free -QUESTION [410 upvotes]: It can be difficult to learn mathematics on your own from textbooks, and I often wish universities videotaped their mathematics courses and distributed them for free online. Fortunately, some universities do that (albeit to a very limited extent), and I hope we can compile here a list of all the mathematics courses one can view in their entirety online. -Please only post videos of entire courses; that is, a speaker giving one lecture introducing a subject to the audience should be off-limits, but a sequence of, say, 30 hour-long videos, each of which is a lecture delivered in a class would be very much on-topic. - -REPLY [2 votes]: I recently found the YouTube channel of the university of Uppsala (Uppsala Algebra : https://www.youtube.com/channel/UCPWnhR29VHTAk7rZUEDQdDQ/playlists) -It contains mostly courses by Walter Mazorchuk on representation theory of finite groups, Lie algebras and the category O (and on linear algebra which if of less interest for me) and a course on commutative algebra and algebraic geometry by Seidon Alsaody. -I mostly watched the courses by W. Mazorchuk and they are very good.<|endoftext|> -TITLE: Clifford Action for Kahler Manifolds -QUESTION [6 upvotes]: I'm working with Kahler manifolds at the moment and looking at their spin$^c$ structure. I really don't know much about spin$^c$ structures in general and don't have enough time to learn it all at present, so I was hoping someone might be able to show me a shortcut in this case. -As far as I understand, for the usual spin$^c$ structure on an $N$-dimensional Kahler manifold $M$, the spinor bundle $S$ is given by the the direct sum of all the anti-holomorphic, that is $S = \bigoplus_{i=1}^N \Omega^{(0,i)}(M)$. Let -$$ -\nabla^s:S \to S \otimes \Omega^{(1,1)}, ~~~~~~~~~ s \mapsto \sum s_i \otimes \omega_i -$$ -be the spin$^c$ connection. -Is there a simple direct algebraic description of the Clifford action -$$ -c:S \otimes \Omega^{(1,1)} \to S -$$ -in this case? For example, an initial stupid guess might be, for $\omega_i = \omega_i^h + \omega_i^{ah}$, with $\omega_i^{h} \in \Omega^{(1,0)}$ and $\omega_i^{ah} \in \Omega^{(0,1)}$, we would have -$$ -c(\sum_i s_i \otimes \omega_i) = \sum_i s_i\omega^{ah}_i. -$$ -I am quite sure this is complete rubbish, but it illustrates the kind of result-from-heaven I'm hoping exists - -REPLY [4 votes]: You ask for a short-cut, and I'm going to interpret this as asking "How can I see that on a Kaehler manifold $X$ the forms $\Omega^{0,\bullet}_X$ are a complex Clifford module, without first going through all that stuff about spin groups and their representations?" -Warning: My scalar factors are probably wrong. -I'm going to start with the Hodge Laplacian -$$\Delta = 2(\overline{\partial}\;\overline{\partial}^\ast+\overline{\partial}^\ast\overline{\partial})$$ -acting on $S:=\Omega^{0,\bullet}_X$. -What makes $\Delta$ a Laplacian is that it's a second order differential operator whose symbol $\sigma \colon T^\ast X \to \mathrm{End}(S)$ is the quadratic map $\lambda \mapsto - \| \lambda\|^2 \mathrm{Id}_S$. (The symbol is just the coefficient matrix of the leading (i.e. second) derivative terms in $\Delta$, written invariantly.) The symbol of $\overline{\partial}$ is $\lambda\mapsto \lambda^{0,1}\wedge \cdot$, that of $\overline{\partial}^\ast$ is $\lambda\mapsto \iota(g(\lambda))$ where $g\colon T^\ast X \to TX$ is the metric tensor, and from this you can check the symbol of $\Delta$. -$\Delta$ has an evident square root, $D=\sqrt{2}(\overline{\partial}+\overline{\partial}^\ast)$, and this is a Dirac operator: a first order differential operator whose square is a Laplacian. The (complexified) symbol $\rho\colon T^*X\otimes \mathbb{C}\to \mathrm{End}(S)$ of a Dirac operator $D\colon \Gamma(S)\to \Gamma(S)$ makes $S$ into a Clifford module. The natural Clifford multiplication on $(0,\bullet)$-forms is the symbol of $D$, that is -$$ \rho(\lambda) = \sqrt{2}(\iota(g(\lambda))+\lambda^{0,1}\wedge \cdot ) \in \mathrm{End}(S). $$ -In fact, this Clifford module is a $Spin^c$-structure, which means that pointwise on $X$ it's a isomorphic to a standard Clifford module.<|endoftext|> -TITLE: Localic locales? Towards very pointless spaces by iterated internalization. -QUESTION [14 upvotes]: One can think of locales as (generalizations of) topological spaces which don't necessary have (enough) points. Of course when one studies locales, one "actually" studies frames, -certain sorts of lattices, but reverses the arrows so that story has a less algebraic and more geometrical/topological flavor. -So while "localic spaces" make lack for a "sufficient" set of point, by their very nature they do have a sufficient set of "open sets" (scare quotes because these open sets arise as a primitive notion and not as actual sets of anything). - -Curiosity drive us on: could one go a step further and make the open sets of locales as ghostly as locales make the points of spaces ghostly? - -Let me put the question more technically (but correct me if you think now that I'm asking the wrong question): One often interprets algebraic objects such as "group objects" in sufficiently nice categories. So locales with ghostly open sets might show up as localic locales, or what amounts to the same, "coframic frames" -- diagrams in the opposite category to the category of frames, specifically diagrams that define internal frames. Though such things must have at least two points (the empty set and the total space) corresponding to the nullary operations in the algebraic theory (just as a localic group must have at least one point, the identity, a priori it seems that they might have no more. -In this spirit - -How would one construct interesting examples of coframic frames = localic locales? - -Even if such things have few classical points, and few classical open sets, they will still have plenty of classical (open) sets of open sets. - -By what formalism might one keep on going, iterating the internalization? - -The previous question implicitly suggests levels of internalization indexed by natural numbers? - -Could someone propose a definition for a level of internalization indexed by a limit ordinal? - -REPLY [2 votes]: Let me give an example of a category whose objects should be thought of as being “localic locales.” -Consider the category $\mathcal{F}_{U}$ consisting of all pairs $(X,M)$ where $X$ is a set and $M$ is an ultrafilter on $X$. If $(X,M),(Y,N)$ are objects in $\mathcal{F}_{U}$, then a function $f:X\rightarrow Y$ is said to be a morphism if $f^{-1}[R]\in M$ whenever $R\in N$. Let $\mathcal{G}_{U}$ be the quotient category $\mathcal{F}_{U}/\simeq$ where if $f,g:(X,M)\rightarrow(Y,N)$ are morphisms, then $f\simeq g$ if and only if $\{x\in X\mid f(x)=g(x)\}\in U$. This category was originally formulated in Andreas Blass's dissertation. -One should consider the objects in $\mathcal{G}_{U}$ as being open set free locales. If $(X,M)$ is an object in $\mathcal{G}_{U}$, then the “points” in this “space” are the elements in the intersection $\bigcap M$. If $M$ is non-principal, then $\bigcap M=\emptyset$, so the “space” $M$ is completely point-free. The “open sets” in the “space” $(X,M)$ are the equivalence classes in $P(X)/M$. In this case, $(X,M)$ only has two points, namely $X/M$ and $\emptyset/M$. -The category $\mathcal{G}_{U}$ even has its own version of Stone-duality since the dual of $(X,M)$ will be an ultrapower $\mathcal{A}^{X}/M$ for a suitable structure $\mathcal{A}$.<|endoftext|> -TITLE: Fill in the blanks: "1Cob is the free ____ category on a ____" -QUESTION [6 upvotes]: This is probably straightforward, but I'm having trouble writing down a precise statement. "Everyone knows" that the cobordism category $\text{2Cob}$ (all manifolds compact and oriented) is the free symmetric monoidal category on a commutative Frobenius object. What is the analogous statement for $\text{1Cob}$? -It looks something like the free symmetric monoidal category on an object with a (left and right) dual, but I'm not sure if I'm interpreting the orientation on points correctly. - -REPLY [5 votes]: Qiaochu, yes, but you don't need to say "left and right dual" because a left dual is a right dual in a symmetric monoidal category. It would be enough to say "with a left dual", or say it as Theo did. -An equivalent way of describing it is "the free compact closed category generated by a single object". For some information on free compact closed categories, there is an old paper by Kelly and LaPlaza, Coherence for compact closed categories, Journal of Pure and Applied Algebra 19 (1980), pp. 193–213. The description in terms of 1-cobordisms is implicit in that paper.<|endoftext|> -TITLE: How canonical is cofibrant replacement? -QUESTION [24 upvotes]: Quillen's original definition of a model category included noncanonical factorization axioms, one being that any map can be factored into a cofibration followed by an acyclic fibration. More recent references have strengthened this axiom by assuming that this is actually a functorial factorization. -If you take a "classical" model category (not exactly in Quillen's sense - I would still like to assume that it has all small limits and colimits), it becomes natural to ask whether it has such a functorial factorization. More, one might wonder how canonical this factorization is, even just on the level of objects. -So in this situation, let's say that we have either: - -a very large category whose objects are "factorization functors", and whose morphisms are natural transformations between them (of necessity natural weak equivalences), or -a very large category whose objects are functorial cofibrant replacements for objects, and whose morphisms are again natural weak equivalences. - -Are there easy examples where these categories are empty? Are there examples where they are nonempty, but the functorial factorization is "noncanonical" in the sense that the category of factorizations is noncontractible? Does this category of factorizations become contractible under stronger assumptions (such as cofibrant generation)? - -Added later: It turns out that the question of homotopy type has a boring answer (and, embarassingly, one answerable by the standard techniques). Let's suppose we have one factorization functor $F$, so that any arrow $g:x \to y$ factors canonically as $$x \stackrel{c(g)}{\to} F(g) \stackrel{f(g)}{\to} y.$$ -Then, naturally associated to any other such functor $G$, we get a third replacement functor abusively written as $G \circ F$, obtained by applying $G$ to the map $c(g)$; this gives a factorization $$x \to (G\circ F)(g) \to F(g) \to y$$ and the first map, together with the composite of the latter two maps, gives a new factorization. -There are natural transformations of factorization functors $F \leftarrow G \circ F \rightarrow G$, and this is natural in $G$; this provides a two-step homotopy contracting the space of factorization functors down to the constant $F$. -(However, the responses to the first question have already informed me quite a bit!) - -REPLY [10 votes]: Dear Tyler, -The other replies here have summarised quite well what has been done by Emily and myself in this regard, but it might be useful to point something out concerning the later addition you have made to your question. You describe a construction that, from two functorial factorisations $F$ and $G$, produces a third, $G \circ F$. This, as you observe, comes equipped with a natural transformation $G \circ F \to F$. However, in trying to define similarly a natural transformation $G \circ F \to G$, one must use lifting properties; the component at $g$ being obtained by considering a square with $x \to G \circ F(g)$ on its left, and $G(g) \to y$ on its right. The problem is that these components are highly unlikely to constitute a natural transformation; the naturality squares most likely do not commute. Even if the functorial factorisations in question were derived from Quillen's small object argument, and the liftings chosen were the ones described in that argument, naturality is still unlikely to obtain: for the construction of those liftings requires the making of some non-canonical choices (more specifically, the choices made in Hovey's book, p.33, line 6, "there is a $\beta < \gamma$") which materially affect the resulting fillers, and are unlikely to cohere for different choices of $g$. -The notion of "natural" / "algebraic" weak factorisation system could be understood as an attempt to rectify precisely this problem. Here one has provided not only the functorial factorisation, but in addition, canonical choices of filler, which, in particular, will cohere sufficiently to allow the construction of the natural $G \circ F \to G$ sought above. That every (cofibrantly generated) weak factorisation system can be made into an instance of this notion relies on modifying the small object argument in such a way as the chosen fillers it provides no longer rely on the making of non-canonical choices (to be more precise: the non-canonical choices are still there, but the outcome is now independent of them). -Richard<|endoftext|> -TITLE: Riemann Zeta Function connection to Quantum Mechanics. -QUESTION [31 upvotes]: I feel like this question is probably wrong for MO, (too low level, perhaps unclear) but my curiosity has got the better of me: -I hear that the Riemann Zeta Function and its zeros have applications to quantum mechanics, as well as other fields. I do not understand these connections, and because of this the following question came up: -In theory, is it possible through physical experiments (particle experiments) to approximately calculate the first few zeros of the Riemann zeta function? -In other words, (using the explicit formula) could we write down the $n^{th}$ prime number (up to a given margin of error/probability of correctness) only from doing quantum mechanical experiments? -(If there are conjectures/facts that we cannot prove, but would answer the question, I would be happy to hear those too) -Thanks! - -REPLY [4 votes]: This may be buried in one of the references above, but for those don't wish to go through them all... -The zeta function can arise as the trace of Hamiltonians governing physical systems. For example in an experiment to measure the Casimir effect (two perfectly conducting plates placed very close to each other) the force they exert on each other has a formula that involved the the derivative of the Riemann zeta function evaluated at $-\frac{1}{2}$. This has been experimentally validated to a reasonable amount of precision. -This may not get you zeros, but it gets you certain values.<|endoftext|> -TITLE: Affine bundles over varieties -QUESTION [7 upvotes]: Let $\mathbb{A}^n_k$ be the Affine $n$-space over an algebraically closed field $k$. -Let $X$ be a variety over $k$. What would be the right definition of an "Affine bundle" i.e bundle of fiber type $\mathbb{A}^n_k$ over $X$ (I mean local triviality in zarisky topology,or -etale .. )?. When can one get a vector bundle from an "Affine bundle" , more -precisely (I think !) if I assume the structure group of the affine bundle to -be $Aut_{Var_k}(\mathbb{A}^n_k)$, when can one get a reduction of the structure group -to $GL_n(k)$? - -REPLY [3 votes]: The very interesting paper "Locally polynomial algebras are symmetric algebras, Invent. Math. 38 (1976/77), no. 3, 279–299. MR 0432626 (55 #5613), by H. Bass, E. H. Connell, and D. L. Wright, gives some answers to this question. It says that if everybody is affine and the bundle is locally trivial in the Zariski-topology, then it is a vector bundle.<|endoftext|> -TITLE: The story about Milnor proving the Fáry-Milnor theorem -QUESTION [68 upvotes]: This question is similar to a previous one about "urban legends", but not the same. It is established that Milnor proved the Fáry-Milnor theorem as an undergraduate at Princeton. For the record, Fáry was a professor in France and proved the result independently. Milnor has a solely authored paper in the Annals that acknowledges Fox. The problem had been posed by Borsuk in 1947. In Milnor's version of the theorem, the infimal total curvature of any smooth knot equals its bridge number. There are many stories to that Milnor thought it was a homework problem or a test question, that he came to class late and solved it on the spot, etc. My question is: Is there any good evidence that when Milnor solved this problem, that he thought it was anything other than an open problem? I am not so interested in answers with "names withheld to spare embarrassment", nor otherwise in digressions or mischievous answers. I am more interested in a convincing citation, either to print or to a named person. (Like Milnor himself, although he has the right not to discuss the matter.) Frankly I think that some versions of this Milnor story are a little bit tasteless. At least in my own mind, I'd like to have it in a more dignified form, if it's true at all. - -The answer by "none" is very good and at first I accepted it. But when I checked Nasar's book, to my surprise it does not completely put the issue to rest. Nasar says that Milnor, as a freshman, showed the proof to his differential geometry professor Albert Tucker with the request, "Would you be good enough to point out the flaw in this attempt. I'm sure there is one, but can't find it." (Tucker then passed around the proof to Fox and Chern.) It would make no sense for Milnor to say this if he truly thought that it was homework. But Nasar also contradicts this inference with the statement, "The story goes that Milnor mistook the conjecture for a homework assignment." For this statement she cryptically cites "Princeton University Archives". -Besides the conflicting evidence, the mistaken-for-homework story as attributed to Milnor is also suspiciously similar to the confirmed story about Dantzig. - -Update: I sent e-mail to Milnor and I got a short reply that begins, "It seems a pity to contradict such a pleasant tall tale; but for my version see..." As I see it, the story that Milnor mistook Borsuk's curvature conjecture for homework has now died three deaths: (1) Milnor says that it's not true. (2) It isn't consistent with either Milnor's or Tucker's account of what happened. (3) It is similar to a true story about Dantzig that has mutated and that has also been pasted onto other mathematicians, such as for instance Fefferman. -By 1991, Tucker said that "as a bad joke", he called Borsuk's conjecture "an assignment". But this is 40 years after the fact, when history was already muddied by the mutating story that started with Dantzig. Actually it's not necessarily so bad, if it happened, because it should have been clear from context that it was only a joke. -Milnor referred me to a short autobiographical account, "Growing up in the Old Fine Hall". This version of the story says that Tucker first discussed Fenchel's theorem that total curvature of any topological circle is at least $2\pi$, and then stated Borsuk's conjecture; then a few days later Milnor had a draft of a proof. This account emphasizes the mathematics more than the human drama. Even so, it's just not consistent with the "thought it was homework" story. -In various sources (many provided in the second answer by "none" below), Tucker gives a consistent account that Milnor first thought that his proof was wrong and asked Tucker to find the mistake. This detail is decidedly absent from Milnor's published accounts. Obviously he asked Tucker to check his proof; that could easily be confused with being sure that there is a mistake. Maybe only two people were there; I would leave it at saying that one of them says so and the other one doesn't say. -I re-accepted the first answer given by "none" since it is most of the whole story. - -Finally the true case of Dantzig. One reason that this story is so well-known, and a reason that it has mutated so much and gotten a little tacky, began with a coincidence on a plane trip to or from California. Dantzig happened to be seated next to mega-Reverend Robert Schuller and told him his account. Schuller relayed it to his flock with some silly exaggerations, and the story made its way into many other churches. Eventually Dantzig's name disappeared from this story spreading mostly among non-mathematicians. It would then be easy to fill in the name of some other mathematician. This history is told in the book "Curses! Broiled Again!" (which apparently got it from the College Mathematics Journal) For the record, in the true version of the story, Dantzig was in graduate school in statistics, it was 1939, he came in late to class, he saw two open problems on the board and thought they were homework. The professor was Neyman, who must have been a bit impenetrable to his students, because he didn't tell Dantzig for six weeks what had really happened. - -REPLY [8 votes]: Check this video of a 1965(!) lecture of Milnor (which incidentally is great). At minute 2:00 the presenter mentions the homework story. I thought this would be the ultimate document as Milnor would come up to the stage, thank him for his kind remarks, and say something like no, it wasn't really like that... But he comes up to the stage and his very first words are "In order to start talking about differential topology I have to first give a few definitions..." -Oh well...<|endoftext|> -TITLE: Fulfilling Pythagoras' Dream using Nonstandard Models of Arithmetic and/or Surreal Numbers -QUESTION [14 upvotes]: Pythagoras and his followers believed that the Universe was made of numbers. Specifically, they thought that if you compared any magnitudes of the same kind, say the lengths of two objects, you would always get a whole number ratio. Then someone came up with a proof that the side length of a square and its diagonal are not in a whole number ratio, thereby demonstrating that space could not be completely described by Number. This seemed catastrophic for the Pythagoreans, but then mathematicians eventually grew to accept "irrational" numbers, and the rest is history. -But I'm curious how things would work out if Pythagoras and his peers had chosen to take a different route, one in which they didn't have to give up their dream of directly characterizing all magnitudes using whole numbers. The standard proof of sqrt(2) being irrational goes like this: suppose that sqrt(2)=a/b, where a and b contain no common factors. Then we show that a and b must have a common factor of 2. But what if the fraction a/b could NOT be written in simplest terms, but instead you could keep dividing a and b by 2 and never get them to be coprime? Then you'd be able to construct an infinite descending sequence of natural numbers, which contradicts the well-ordering principle/mathematical induction. -So I'm trying to find a nonstandard model of arithmetic in which, among other things, sqrt(2) is rational. Countable nonstandard models of PA are not well-ordered, which is what I want, but first order induction is still sufficient to prove that sqrt(2) is irrational. So we would have to find nonstandard models of a weaker system, like Robinson's Q. Q also has the advantage of having computable nonstandard models. So I want to find a nonstandard model of Q, let's call it S, and then construct the real numbers using ordered pairs of elements of S. But the least upper bound axiom of the real numbers is actually sufficient to prove second-order induction for the natural numbers. So the system of real numbers I'm constructing should not satisfy LUB; it should only satisfy the first-order properties of the real numbers. -So to sum up: I want a nonstandard model S of Robinson Arithmetic which is not well-ordered, and such that we can define operations on S^2 which would make S^2 a nonstandard model of the first-order theory of real closed fields. (EDIT: As François said, a simpler way to say this is that I want the field of fractions of S to be real closed.) Is that even possible, and if so would S even be a computable nonstandard model? -I've been thinking about this problem a while, and I think Conway's Surreal Numbers might be promising. The analogue of the integers for Surreal numbers are called Omnific integers. Omnific integers are not well-ordered, and it turns out that you can find Omnific integers a and b such that a^2=2b^2, which are both hopeful signs. And even better, the set of ratios of Omnific integers actually satisfies the first-order theory of real closed fields! If Omnfic integers really do satisfy all the properties I want, then I have a few questions about them. Do they constitute a nonstandard model of Robinson arithmetic? Does there exist an axiomatization of them in first order logic? In other words, can we define them alone, without resorting to the full-blown definition (which involves set theory) of the Surreal numbers? On a related note, can they even be put in a set, or do they form a proper class? I think the latter may be true, in which case I may have to find some subclass of the Omnific integers which is an actual set (because I think it may not be legitimate to take the Cartesian product of proper classes). But in that case, will that subclass still satisfy all the properties I want? -Any help would be greatly appreciated. -Thank You in Advance. - -REPLY [10 votes]: First note that the axioms of Robinson Arithmetic are usually straightforward to verify in any structure. Surely the nonnegative omnific integers satisfy these basic axioms. Note that a nonstandard model of Robinson Arithmetic cannot be wellordered because of the axiom which says that every nonzero element has an immediate predecessor. -You can push a little farther than Robinson Arithmetic and still have recursive models. Open induction is the statement that induction holds for quantifier-free formulas (in the language of Robinson Arithmetic). Shepherdson (A non-standard model for a free variable fragment of number theory, MR161798) has shown that there are recursive models of open induction too. Furthermore, he showed that one can get such models where $\sqrt2$ is rational, for example. I described Shepherdson's construction in my previous answers here and here. In this construction, the field of fractions of the model is not necessarily real closed, but if my memory serves correctly there is a variant of the construction does do that.<|endoftext|> -TITLE: 2-morphisms in structured 2-categories -QUESTION [7 upvotes]: There are many $2$-categories, which are first specified by certain categories with extra structure; then the $1$- and $2$-morphisms are functors and natural transformations that preserve the extra structure. I want to understand the general procedure in finding the "correct" definitions of these $2$-morphisms, if there is any. -Example 1: Objects are tensor categories. Then $1$-morphisms should be tensor functors (some allow them to be lax) and $2$-morphisms are natural transformations $\eta$ which are compatible with the tensor structure. This means that $\eta(1)$ is an isomorphism and that for every pair of objects $x,y$ we have a commutative diagram which identifies $\eta_{x \otimes y}$ with $\eta_x \otimes \eta_y$. -Example 2: Take as objects cocomplete categories. Then $1$-morphisms are cocontinuous functors and $2$-morphisms are natural transformations $\eta$ which preserve colimits. The latter means that that for every colimit $\colim_i x_i$ the morphism $\eta(x)$ is the colimit of the morphisms $\eta(x_i)$. But wait, this is automatically true! This follows easily from the cocontinuity of the functors and the naturality of $\eta$. In how far is this "coincidence"? -So far I have never seen this definition of a "cocontinuous natural transformation", but actually this property is used very often when dealing with natural transformations in this situation. So perhaps it should be included in the definition? For example the "correct" definition of a homomorphism $f : G \to H$ of groups includes that $f$ preserves the unit, inversion and multiplication, although everyone knows that multiplication is enough and unfortunately some authors then take the "wrong" definition and get the correct one by a lemma. I hope it's clear that I don't want to offend anyone here and there is no "correct" definition, but perhaps the one which fits best into general patters of category theory. -Example 3: Objects are symmetric tensor categories. Then $1$-morphisms are tensor functors which preserve the symmetry (the functor $F$ maps the symmetry $x \otimes y \cong y \otimes x$ to the symmetry $F(x) \otimes F(y) \cong F(y) \otimes F(x)$; again this is a commutative diagram) and $2$-morphisms are natural transformations $\eta$ which are compatible with the tensor structure as in Example 1 and also are compatible with the symmetry. But what should this compatibility mean? Actually I have not been able to write down a diagram which connects $\eta$ with the symmetry and does not directly follow from the naturality. So perhaps we cannot even formulate a compatibiltiy condition here? Again I'm interested in how far this is "coincidence". - -REPLY [4 votes]: A. This is really just an aspect of Mike Shulman's answer, but could be of some use in particular cases. -There's a 2-categorical limit called the power (or cotensor) of an object $B$ by the arrow-category $2$. This is an object $B^2$ with the property that morphisms from $A$ to $B^2$ are in bijection with pairs of morphism from A to B with a 2-cell between them. For example -if B is a category then $B^2$ is the functor category $[2,B]$. If $B$ is a monoidal category then $B^2$ is $[2,B]$ with the evident (pointwise) monoidal structure. -In each of your examples, and more generally in Mike's setting, this limit exists in the -structured 2-category, and is preserved by the forgetful 2-functor into Cat. Normally you would prove this given the definition of 2-cell. But you can also turn this around. Given a -structure on B, if you know how to make $B^2$ into a structured object, then you can use this to define the structured 2-cells. -In examples where the structure is given by a 2-monad, and in particular in examples which -involve structure described by operations $B^n\to B$, natural transformations between these, and equations, then you can always do this in a "pointwise way". -(But if you choose a strange way to make $B^2$ into a structured object you will get a strange notion of 2-cell.) -Suppose, for example, that $B$ is a monoidal category. Once you agree to make $[2,B]$ monoidal in the pointwise way, then you can define a monoidal transformation to be a monoidal functor with codomain $[2,B]$, and this will agree with the standard definition which you referred to. -In the case of a cocomplete category $B$, you don't need to choose how to make $[2,B]$ -cocomplete, it just is. And then you can consider cocontinuous functors with codomain $[2,B]$; once again this will give no extra condition to be satisfied by a natural transformation between cocontinuous functors -The case of symmetric monoidal categories can be treated in the same way. -B. Regarding the case of symmetric monoidal categories, there is a general phenomenon here. As you add structure to your objects in the form of operations $B^n\to B$ (like a tensor product) you generally introduce preservation conditions on both morphisms and 2-cells (although there are special cases, as in your Example 2, where the 2-cell part is automatic). But if you introduce structure in the form of natural transformations between the operations $B^n\to B$ (such as a symmetry), this results in new preservation conditions for the morphisms but not for the 2-cells. -C. Despite all this, there can be more than one choice for the 2-cells. The general principles described by Mike (and by me) would suggest that if our structure is categories with pullback, so that our morphisms are pullback-preserving functors, the 2-cells should be all natural transformations between these. But sometimes it's good to consider only those natural transformations for which the naturality squares are pullbacks. (These are sometimes called cartesian natural transformations.) See this paper for example.<|endoftext|> -TITLE: Can a rational family of genus-g curves have generic gonality? Can it be Brill-Noether general? -QUESTION [11 upvotes]: We know that M_g is general type for g large enough. In particular, the generic genus-g curve is not contained in a (non-isotrivial) rational family parametrized by P^1. In fact, the high-genus curves I know how to build over C(t) all have low gonality; it's easy to make a hyperelliptic curve y^2 = f(t,x), and with a little more work you can make curves over k(t) which are 3-gonal, 4-gonal, or 5-gonal. But do we know whether there's a curve over C(t) whose gonality is close to the generic value, which is on order g/2? -A colleague suggested that the linear system of the generator of Pic on a K3 with Picard number 1 would be a good place to look for these. I don't immediately see the proof that these guys have big gonality, but it certainly seems reasonable. -So more generally, one might ask -- are there genus-g curves over C(t) which are "Brill-Noether generic," i.e. which are not distinguished from the generic genus-g curve by the presence of any g^r_d? -Note that I am not asking "what do we expect the union of all rational curves on M_g to look like, assuming we believe Lang's conjecture?" -- that question is too intimidating. Rather, I'm asking whether there is some Brill-Noether locus on M_g which MIGHT contain all the rational curves. Having phrased it this way, I imagine this must be a question with some literature attached, but I wasn't able to find it. - -REPLY [11 votes]: One can construct pencils of k-gonal curves of genus g by taking a K3 surface S with -Pic(S) generated by two classes: an ample class C with self-intersection 2g-2, and an elliptic curve E, so self-intersection 0, such that C.E=k. Every curve in the linear system -|C| has gonality k, and the pencil of minimal degree is computed by the restriction of E to C (use for this the work of Green-Lazarsfeld). Here k varies from 2 to (g+2)/2, so you cut through all the gonality strata in this way. -These rational curves will not be covering the k-gonal locus M^1_{g, k} inside M_g. As to which of these loci are uniruled, what is known is that when g is large, the k-gonal locus becomes of general type when k exceeds g/3. It is not clear how optimal this bound is. -The follow-up question of Felipe, as to whether anyone has written down a Brill-Noether general curve over Q (and unbounded genus), is much more difficult, and as far as I am aware nobody knows.<|endoftext|> -TITLE: Is every virtual knot group an HNN extension? -QUESTION [11 upvotes]: A basic fact in knot theory is that a knot group $\pi(K)$ is an HNN extension of $\pi(F)$, the fundamental group of a Seifert surface complement. A nice discussion of this may be found in Chapter 11 of An introduction to the theory of groups by Rotman. This property means that a knot group is completely determined by the fundamental group of a Seifert surface complement, plus a choice of meridian. I wonder whether this is somehow the real reason that Seifert surfaces are important in knot theory. Morally, the fact that a knot group is an HNN extension means that all the information about a knot which you might care about is contained in any of its Seifert surfaces. -To remind you of the explicit group-theoretic statement, there is an isomorphism -$\phi\colon\thinspace \frac{\pi(F)\ast \langle m\rangle}{\mathcal{N}}\longrightarrow \pi(K),$ -where $\langle m\rangle$ is the infinite cyclic group generated by the meridian, and $\mathcal{N}$ is the smallest normal subgroup of the free product $\pi(F)\ast \langle m\rangle$ containing the elements -$m^{-1}\mu^+(z)m(\mu^{-}(z))^{-1}\qquad z\in \pi(F),$ -with $\mu^{\pm}$ denoting the pushoff maps. -There is a natural notion of a virtual knot group, by assigning a formal generator to each arc of a virtual knot diagram, and a Wirtinger relation to each real crossing (virtual crossings are ignored). Any Wirtinger presentation of deficiency $0$ or $1$ can be realized as a virtual knot group by Theorem 3 of a paper by Se-Goo Kim. - -Question: Is every virtual knot group an HNN extension? (edit: over a finitely generated group?) Can the base group be described in terms of a group generated by a commutator at each real crossing? If not, is a virtual knot group "almost" an HNN extension in some useful sense? - -I'm interested in this question because I wonder whether invariants coming from Seifert surfaces can be read off Gauss diagrams in any systematic way. Are Seifert surfaces an essential feature of knots, as opposed to virtual knots; or are they a non-essential luxury? - -REPLY [4 votes]: According to a theorem of Kuperberg, a virtual knot corresponds -canonically to an embedding of a knot in a thickened surface $K\subset \Sigma_g\times [0,1]$ of minimal -genus $g$ (up to -homeomorphism). There is therefore another natural fundamental group -associated to the knot, namely the fundamental group of the knot complement -$\pi_1(\Sigma_g\times [0,1] - K)$. This group certainly splits as an HNN extension (in many ways). The fundamental group of the virtual knot is obtained -from this by killing the two peripheral subgroups corresponding to -$\Sigma_g \times \{0,1\}$. One may think of this as the fundamental -group $\pi_1( S\Sigma -K)$, where $S\Sigma$ is the suspension. -If $K$ is homologically trivial in $\Sigma\times [0,1]$, then -one could take an embedded minimal genus surface $F \subset \Sigma \times [0,1]$ spanning -$K$, so $\partial F=K$. Unfortunately, though, there is no canonical choice of homology -class for this surface. One has a geometric splitting of $S\Sigma-K$ along $F$, however -$F$ might not be $\pi_1$-injective in this space since Dehn's lemma is not -available. -If $K$ is not homologically trivial in $\Sigma \times [0,1]$, it is still homologically -trivial in $S\Sigma$, so one could take a surface bounding it (which must intersect -a singular point of $S\Sigma$). One could think of this as taking a minimal -genus surface giving a cobordism between $K$ and an embedded curve in $\Sigma \times \{0,1\}$. Again, there is not a canonical homology class ($H_2(S\Sigma)=\mathbb{Z}^{2g}$) and the surface may not be $\pi_1$-injective (in fact, there are virtual knots where the longitude is trivial in the virtual knot fundamental group). -Also, I don't think that linking numbers are well-defined (again, since $H_2(S\Sigma)$ is -large), so it's not clear how -to obtain an Alexander polynomial from such a surface.<|endoftext|> -TITLE: The third axiom in the definition of (infinite-dimensional) vector bundles: why? -QUESTION [24 upvotes]: Serge Lang's Differential and Riemannian Manifolds is a no doubt the best available reference for the theory of not-necessarily-finite-dimensional differential manifolds, but unfortunately it suffers the defect of containing no exercises and few examples. This makes it difficult to learn the subject from this book, especially if one is say a graduate student who is also still in the process of learning functional analysis. -One place where an example would really have been helpful is in the context of the definition of vector bundle (pp. 40-41), which involves three axioms that Lang labels VB1 - VB3. The third one, VB3, states that, in coordinate overlaps, the mapping of points of the base space into the automorphisms of the fibers induced by coordinate changes should be a morphism. As Lang notes, this axiom is redundant in the finite-dimensional case because of the following result (p.42): -Proposition 1.1. Let $\mathbf{E}$, $\mathbf{F}$ be finite-dimensional vector spaces. Let $U$ be open in some Banach space. Let $f: U \times \mathbf{E} \to \mathbf{F}$ be a morphism such that for each $x \in U$, the map $f_x : \mathbf{E} \to \mathbf{F}$ given by $f_x(v) = f(x,v)$ is a linear map. Then the map of $U$ into $L(\mathbf{E},\mathbf{F})$ given by $x \mapsto f_x$ is a morphism. -However, this result is apparently false in the infinite-dimensional case. The problem is that Lang does not provide an example showing this; and nor does he discuss why smoothness of the map from $U$ into $L(\mathbf{E},\mathbf{F})$ (or, in the specific case of interest, from $U_i \cap U_j$ into $Laut(\mathbf(E))$ is necessary or convenient for whatever purposes such infinite-dimensional bundles are used for. -So if I may ask: what would be a counterexample to the above proposition in the infinite-dimensional case? Even more to the point, where does one go looking for such a counterexample? Can I take $U = \mathbf{F} = \mathbb{R}$, and $\mathbf{E} = \ell_2$? Can we make even continuity fail, i.e. is VB3 necessary even for $C^0$ - manifolds? I know we can't make $f$ bilinear, since $L^2(\mathbf{E}, \mathbf{F}; \mathbf{G}) \cong L(\mathbf{E}, L(\mathbf{F},\mathbf{G}))$ -- but this is what makes the question mysterious to me, because my understanding is that "continuity failures" in infinite dimensions arise from non-convergence of sequences (so that you can't just "write everything in a matrix and see that the entries are continuous/smooth"), in which case you ought to be able to exhibit the phenomenon in the simplest case of (bi)linear maps; but the aforementioned isomorphism blocks this. So why does a fundamental difference between finite- and infinite- dimensional spaces suddenly appear when we switch from linear to nonlinear maps? Why doesn't the fact that $f$ is a two-argument morphism provide bounds that would force $x \mapsto f_x$ to be a morphism as well, just like in the bilinear case? -Also, why can't we just "do without" Lang's VB3 in the case of infinite-dimensional manifolds? - -REPLY [2 votes]: In fact, the underlying question is the one of the STRUCTURE (algebraic, differential) that you allow on the group of isomorphisms of the fibers. As you certainly know, for a vector bundle E of rank n, there is a (canonical) principal bundle GL(E) with fibers the linear group GL_n. This Lie group is analytic, and with very few possible topologies (up to equivalences), where as infinite dimensional Lie groups have many pathologies (this is an actual research subject). The bundles come AFTER Lie groups in the construction so... -So that, the book of Lang is perhaps very good (if you say so...) but not sufficiently up-to-date and I should say very formal. Try more recent references, e.g. Kriegl, Michor, "the convenient setting for global analysis" (1997) to be convinced of that. I do not say that it is THE reference, but this is a serious one to have a rigorous viewpoint of ONE actual approach of the subject.<|endoftext|> -TITLE: Computing H_2 from pi_1=Z and pi_2 -QUESTION [7 upvotes]: (Related question: What part of the fundamental group is captured by the second homology group?) -Suppose I have a path-connected space $X$ for which $\pi_1(X)=\mathbb{Z}$. Suppose I know $\pi_2(X)$ as a $\pi_1(X)$-module, and I want to compute $H_2(X)=H_2(X;\mathbb{Z})$. -Claim: $H_2(X)$ is isomorphic to the largest quotient of $\pi_2(X)$ on which $\pi_1(X)$ acts trivially (ie the group of coinvariants). -One can prove this by passing to universal covers (assuming $X$ is locally path-connected, blah, blah, blah). The Cartan-Leray spectral sequence of the regular cover $\tilde{X}\to X$ degenerates at the $E_2$ page (since $K(\mathbb{Z},1)$ is a circle) and it follows that $$H_2(X)\cong H_0(\mathbb{Z};H_2(\tilde{X}))\cong H_0(\mathbb{Z};\pi_2(\tilde{X}))\cong H_0(\mathbb{Z};\pi_2(X)).$$ -This tells me the answer, but teaches me nothing. -Question: Is there a more elementary proof of this fact? (I am prepared to accept that "elementary" might still involve Postnikov towers, obstruction theory, etc) -Edit: Many thanks for all the helpful answers, I wasn't expecting four alternatives! A remark and a question: - -This is true with an arbitrary free group replacing $\mathbb{Z}$, and -it seems to be well known. Has anyone seen it written down anywhere? - -REPLY [3 votes]: There is a cofibrant way of showing this, i.e. forgetting about coverings, Postnikov towers, fibrations, spectral sequences, etc. There are very nice and simple algebraic models for low-dimensional homotopy types. The simplest are crossed modules, which are group homomorphisms $$\partial\colon C_2\longrightarrow C_1$$ such that $C_1$ acts on the right of $C_2$ and the following two equations are satisfied: -$$\partial(x_2^{x_1})=x_1^{-1}x_2x_2, \qquad x_2^{\partial(y_2)}=y_2^{-1}x_2y_2.$$ -Crossed modules can be regarded as non-abelian chain complexes $C_*$ concentrated in degrees $1$ and $2$. The subscript indicates the degree of each element. Notice that the first equation says that $\partial$ is $C_1$-equivariant if we let $C_1$ act on itself by conjugation. -The homology of $C_*$ is usually regarded as homotopy groups: -$$\pi_1C_*=C_1/\partial(C_2),\qquad \pi_2C_*=\ker\partial.$$ -Notice that $\pi_1C_*$ acts on the right of $\pi_2C_*$. -The canonical example of a crossed module is the homomorphism -$$\partial \pi_2(X,Y)\longrightarrow \pi_1Y$$ -associated to any pair of spaces $(X,Y)$. The fundamental crossed module of a connected CW-complex $X$ with $1$-skeleton $X^1$ is -$$\partial\colon\pi_2(X,X^1)\longrightarrow \pi_1X^1.$$ -To any crossed module $C_*$ we can associate a two-step chain complex -$$\cdots\rightarrow 0\rightarrow C_2^{ab}\otimes_{\mathbb{Z}[C_1]}\mathbb{Z}\stackrel{\bar{\partial}}\longrightarrow C_1^{ab}\rightarrow 0\rightarrow \cdots$$ -by abelianizing $C_1$ and $C_2$ and killing the action of $C_1$ on $C_2^{ab}$. If $C_*$ is the fundamental crossed module of $X$ then the homology of this chain complex is $H_1(X)$ and $H_2(X)$ in the corresponding degrees. -Now assume $\pi_1(X)\cong\mathbb{Z}$. Then the natural projection $C_1=\pi_1X^1\twoheadrightarrow \pi_1X\cong\mathbb{Z}$ has a section $i\colon \pi_1X\rightarrow \pi_1X^1$. This section gives rise to a homotopy equivalence of crossed modules: -$$\begin{array}{rcccl} -&\pi_2X&\stackrel{0}\longrightarrow&\pi_1X&\\\ -{\text{inclusion}}&\downarrow&&\downarrow&\scriptstyle i\\\ -&\pi_2(X,X^1)&\longrightarrow&\pi_1X^1& -\end{array}$$ -In particular, the chain complexes associated to these two crossed modules are quasi-isomorphic. The chain complex of the upper crossed module, given by the trivial homomorphism $0\colon \pi_2X\rightarrow \pi_1X$, is simply -$$\cdots\rightarrow 0\rightarrow \pi_2X\otimes_{\mathbb{Z}[\pi_1X]}\mathbb{Z}\stackrel{0}\longrightarrow (\pi_1X)^{ab}\rightarrow 0\rightarrow \cdots,$$ -hence we recover the well-known isomorphism $(\pi_1X)^{ab}=H_1X$ and what we wanted to obtain $\pi_2X\otimes_{\mathbb{Z}[\pi_1X]}\mathbb{Z}=H_2X$.<|endoftext|> -TITLE: Gap in an argument in Fulton & Harris? -QUESTION [16 upvotes]: I'm reading through the two chapters in Fulton and Harris on the representation theory of $\mathfrak{sl}(3,\mathbb{C})$, in preparation for lecturing on them this week. I'll use F&H's notation, so that $\Gamma_{a,b}$ is the irreducible representation of highest weight $a e_1 - b e_3$ (with $e_i$ the character taking a diagonal matrix to its $(i,i)$ entry). -In the discussion of the multiplicities of the weights occurring in $\Gamma_{a,b}$, it seems to me that there's a gap -- so I am wondering what I'm missing here (or if perhaps there really is a gap). The issue arises at the sentence "To begin with..." on page 184, so let me give a quick summary of the argument to that point. Let $V$ be the standard representation of $\mathfrak{sl}(3,\mathbb{C})$ on $\mathbb{C}^3$, and $V^*$ its dual. The weights of ` $W :=\mathrm{Sym}^a V \otimes \mathrm{Sym}^b V^*$ lie on shrinking concentric hexagons $H_0,H_1,\ldots$ (with $H_0$ the outermost hexagon, and with $H_i$ degenerating to a triangle for $i$ sufficiently large). The multiplicities of $W$ are constant along each hexagon. For the argument to go through it suffices to show that the only highest vectors of $W$ are the unique ones (up to scaling) that occur at the unique dominant vertices of each $H_i$ up to $i = \textrm{min}(a,b)$, each of which contributes a $\Gamma_{a-i,b-i}$ to $W$. -The claim in the sentence "To begin with..." is that this follows for essentially combinatorial reasons: if there were a highest weight vector of weight $\alpha$, where $\alpha$ lies on $H_i$ but is not the dominant vertex, then "the multiplicity of $\alpha$ in $W$ would be strictly greater than [the multiplicity of the dominant vertex of $H_i$]." But why? Obviously this is correct if one considers only the contributions from highest weight vectors lying on $H_i$ or in its interior. But it seems a priori possible that some $\Gamma_{a-j,b-j}$ with $j < i$ (i.e., the irreducible constituent coming from one of the higher weight vectors on a hexagon lying outside $i$) could contribute a larger multiplicity to the dominant vertex of $H_i$ than to $\alpha$. If you haven't yet proved that the multiplicities of $\Gamma_{a,b}$ are also constant along the hexagons, I don't immediately see how this argument is going to give an inductive proof of that claim. Am I missing something elementary here, or does one have to work harder than F&H claim in order to finish the argument? - -REPLY [15 votes]: I'm still on the fence as to the pedagogical value of the "lecture" approach taken by F&H, but anyway it seems essential in a formal classroom presentation to articulate clearly which features of irreducible representations are or aren't known in general at each step of the exploration of examples in low ranks. F&H defer to later chapters the rigorous general treatment including the Weyl group and its action on weights. Initially they work out some details of examples starting with type A. These are illuminating and especially useful in concrete applications to algebraic geometry. But eventually the student has to confront more intricate behavior, especially in the case of tensor product decompositions. -Having said this, I suspect that the informal discussion in F&H may be somewhat oversimplified on page 184. Though I'd have to review the previous discussion more thoroughly to be sure about that. The structure of the weight diagram for -an arbitrary irreducible representation of $\mathfrak{sl}_3(\mathbb{C})$ does depend ultimately on the Weyl group symmetry involved, without which it's hard to develop from first principles. Along the way you have to get control over weight strings in terms of the restriction to rank one, etc. The particular type of tensor product studied here in rank two is very nice but also very special in having summands with multiplicity one. But how to make it all rigorous without relying on too much general theory is a balancing act. -My own solution, for what it's worth, is to have students take turns talking about this material and arguing about the details. They can learn a lot that way.<|endoftext|> -TITLE: Construction of abelian varieties from Hilbert modular forms? -QUESTION [15 upvotes]: Some experts tell me that the construction of abelian varieties from -Hilbert modular forms is an (apparently difficult) open problem. However, -in view of the construction of $l$-adic Galois representations due to Carayol for instance, -it is not clear what exactly the obstruction to the usual method of taking -the quotient of the jacobian (of the associated quaternionic Shimura curve) by the -`annihilator' of the associated quaternionic eigenform would be. -To be slightly more precise, consider the setting of Carayol "Sur les représentations galoisiennes modulo $l$ attachées aux formes modulaires" (Duke Math. Journal, 1986). That is, let $F$ be a totally real field of degree $d$, with set of real places $\lbrace \tau_1, \ldots, \tau_d \rbrace$. Fix integers $k \geq 2$ and $w$ having the same parity. Let $D_{k,w}$ denote the representation of $\operatorname{GL_2}({\bf{R}})$ that occurs via unitary induction as $\operatorname{Ind}(\mu, \nu)$, where $\mu$ and $\nu$ are the characters on ${\bf{R}}^{\times}$ given by -\begin{align*} -\mu(t) &= \vert t \vert ^{\frac{1}{2}(k-1-w)}\operatorname{sgn}(t)^k; ~~ -\nu(t) = \vert t \vert ^{\frac{1}{2}(-k+1-w)}. -\end{align*} Fix integers $k_1, \ldots k_d$ all having the same parity. Let -$\pi \cong \bigotimes_v \pi_v$ be a cuspidal automorphic representation of -$\operatorname{GL_2}({\bf{A}}_F)$ such that for each real place $\tau_i$ of $F$, there is an isomorphism $$\pi_{\tau_i} \cong D_{k_i, w}.$$ It is well know that such -representations correspond to holomorphic Hilbert modular -forms of weight ${\bf{k}}=(k_1, \ldots, k_d)$. If $d$ is even, then assume -additionally that there exists a finite prime $v \subset \mathcal{O}_F$ where -the local component $\pi_v$ is an "essentially square integrable" (i.e. special or -cuspidal) representation of $\operatorname{GL_2}(F_v)$. Let $B/F$ be a quaternion -algebra that is ramified at $\lbrace \tau_2, \ldots, \tau_d \rbrace$ if -$d$ is odd, and ramified at $\lbrace \tau_2, \ldots, \tau_d, v \rbrace$ -if $d$ is even. Let $$G = \operatorname{Res}_{F/{\bf{Q}}}(B^{\times})$$ be the associated algebraic group over ${\bf{Q}}$. Hence, we have an isomorphism -$$G({\bf{R}}) \cong \operatorname{GL}({\bf{R}}) \times \left( \mathbb{H}^{\times} \right)^{d-1},$$ where $\mathbb{H}$ denotes the Hamiltonian quaternions. Let $\overline{D}_{k,w}$ denote the representation of $\mathbb{H}^{\times}$ corresponding to $D_{k,w}$ via Jacquet-Langlands correspondence. We then consider cuspidal automorphic representations $\pi' = \bigotimes_v \pi_v'$ of $G({\bf{A}}_F)$ such that $\pi_{\tau_1}' \cong D_{k_1, w}$ and $\pi_{\tau_i} \cong \overline{D}_{k_i, w}$ for $i = 2, \ldots, d$. Such representations should (I believe) correspond to modular forms of weight ${\bf{k}} = (k_1, \ldots, k_d)$ on the indefinite quaternion algebra $B$. To be slightly more precise, let $S_{\bf{k}}(\mathfrak{m})$ denote the finite dimensional ${\bf{C}}$-vector space of quaternionic modular forms of weight ${\bf{k}}$ and level $\mathfrak{m} \subset \mathcal{O}_F$ on $B$. Write $\mathfrak{d} =\operatorname{disc}(B)$. The space $S_{\bf{k}}(\mathfrak{m})$ comes equipped with actions of the standard Hecke operators $T_v$ for all primes $v \nmid \mathfrak{m}\mathfrak{d}$, and with Atkin-Lehner involutions for all prime powers $v^e \mid \mathfrak{m}\mathfrak{d}$. The Jacquet-Langlands correspondence induces a "Hecke equivariant" isomorphism of spaces \begin{align*} S^B_{\bf{k}}(\mathfrak{m}) &\cong S_{\bf{k}}(\mathfrak{m}\mathfrak{d})^{\operatorname{\mathfrak{d}-new}}, \end{align*} where $S_{\bf{k}}(\mathfrak{m}\mathfrak{d})^{\operatorname{\mathfrak{d}-new}}$ denotes the space of cuspidal Hilbert modular forms of weight ${\bf{k}}$ that are new at primes dividing $\mathfrak{d}$. -Anyhow, at least when we assume ${\bf{k}} = (2, \ldots, 2)$, a standard argument shows that there is a $G({\bf{A}}_f)$-equivariant isomorphism $\Gamma(\omega) \cong S^B_{\bf{k}}(\mathfrak{m})$, where $\omega$ is the sheaf of homomorphic $1$-forms on the complex Shimura curve \begin{align*} M(\bf{C}) &= G({\bf{Q}}) \backslash G({\bf{A}}_f) \times X/H.\end{align*} Here, $X = {\bf{C}} - {\bf{R}}$, and $H \subset G({\bf{A}}_f)$ is a compact open subgroup of level $\mathfrak{m}$. Let $M$ denote Shimura's canonical model of this curve (defined over $F$). Let $J$ denote the Jacobian of $M$. Let ${\bf{T}}$ denote the subalgebra of $\operatorname{End}(J)$ generated by Hecke correspondences on $M$. My question is whether or not the following construction can or has been made rigorous. Namely, in the setup above, start with a Hilbert modular eigenform ${\bf{f}} \in \pi$, and consider an associated quaternionic eigenform $\Phi \in \pi'$. Viewing $\Phi$ as an eigenform for the Hecke algebra ${\bf{T}}$, consider the homomorphism $\theta_{\Phi}:{\bf{T}} \longrightarrow E$ that sends a Hecke operator acting on $\Phi$ to its corresponding eigenvalue. Here, $E = E_{\Phi}$ denotes the extension of ${\bf{Q}}$ generated by all of the eigenvalues of $\Phi$. Let $I_{\Phi} = \ker{\theta_{\Phi}}$. Consider the quotient \begin{align*} A &= J/I_{\Phi}J. \end{align*} Is $A$ not an abelian variety associated to the Hilbert modular eigenform ${\bf{f}}$? Or is this completely trivial, with the subtle part being the task of showing that $\dim(A) = [E: {\bf{Q}}]$? -A more naive question to ask is why Shimura's construction cannot be generalized -directly for a cuspidal Hilbert modular form ${\bf{f}} \in S_{\bf{k}}(\mathfrak{m})$. Also, how does taking weight ${\bf{k}} = {\bf{2}}$ make the problem simpler? Apologies if parts of this question were somewhat vague, I have sketched matters for simplicity/space. - -REPLY [15 votes]: There is no problem with constructing an abelian variety $A$ for most Hilbert modular forms of parallel weight $2$, the issue is finding such a variety for all $\pi$. In particular, when $d = [K:\mathbf{Q}]$ is even, there is a local obstruction to the existence of a corresponding Shimura curve which realizes the Galois representation associated to $\pi$. -In particular, if $\pi$ has "level one", then no such Shimura curve exists. - To construct the Galois representation in this case one has to use congruences; this was done by Taylor in the late 80's. -This issue is also discussed here: -Are there motives which do not, or should not, show up in the cohomology of any Shimura variety?<|endoftext|> -TITLE: Open map D⁴ → S² -QUESTION [52 upvotes]: Is it possible to construct an embedding $D^4\hookrightarrow S^2\times -\mathbb R^2$ -such that the projection $D^4\to S^2$ is an open map? -Here $D^n$ denotes closed $n$-ball. -An open map D⁴ → S². -It is easy to construct an embedding $D^3\hookrightarrow S^3$ such that -its composition with Hopf fibration $f_3:D^3\to S^2$ is open. -Composing $f_3$ with any open map $D^4\to D^3$, -one gets an open map $f_4:D^4\to S^2$. -The map $f_3$ is not a projection of embedding $D^3\hookrightarrow S^2\times\mathbb R$. -(We have $f_3^{-1}(p)=S^1$ for some $p\in S^2$ and $S^1$ can not be embedded in $\mathbb R$.) -I still do not understand if one can present $f_4$ as a projection of an embedding $D^4\hookrightarrow S^2\times\mathbb R^2$. - -REPLY [2 votes]: This is too long for a comment but maybe it'll help me clarify what you're looking for. Interpret $D^4$ as the unit compact ball in $\mathbb C^2$. -$$ D^4 = \{ (z_1,z_2) \in \mathbb C^2 : |z_1|^2+|z_2|^2 \leq 1 \} $$ -There is a function $f : D^4 \setminus \{(0,0)\} \to S^2$ given by $f(z_1,z_2) = z_2/z_1$, where we're thinking of $S^2$ as the Riemann sphere. $f$ is an open map. But $f$ is also a composite: -$$ D^4 \setminus \{(0,0)\} \to S^2 \times \mathbb C \to S^2 $$ -the 1st map $D^4 \setminus \{(0,0) \} \to S^2$ being $(z_1,z_2) \longmapsto (z_2/z_1, z_2)$ and the second map being $(z_1,z_2) \longmapsto z_1$. -The first map is an embedding, and the 2nd map is a projection. My original post had the domain as $D^4$ but that makes no sense. Okay, maybe I'm starting to wrap my head around the question you're asking. This map above has the property that the restriction $S^3 \to S^2$ is the Hopf fibration, which as a map $S^3 \to S^2$ is not null-homotopic. So it's impossible to extend the above construction to a map $D^4 \to S^2 \times \mathbb C$. If you don't leave the world of submersions this means your map $S^3 \to S^2$ has to be a null-homotopic submersion, but such things do not exist -- a submersion would have to be a circle bundle over $S^2$ and those are only Hopf fibrations. -So if there is a positive answer to your question, the map $S^3 \to S^2$ has to have have some degeneracies.<|endoftext|> -TITLE: Locally compact abelian groups -QUESTION [8 upvotes]: First, some preliminaries: - -Define an "LCA group" to be a locally compact Hausdorff abelian topological group. -Define "smooth manifold" in a way that requires Hausdorffness, but not connectedness or paracompactness. Define a "Lie group" to be a smooth manifold with smooth group operations. Note that with these definitions, any discrete topological space is a manifold, and any discrete topological group is a Lie group. - -Now: -I have been told that any LCA group A has a compact subgroup K such that A/K is a Lie group. -However, I have not been able to extract this result from the literature. For some attempts, see this post to the n-Category Cafe. -Can anyone find a proof of this result, or prove it? - -REPLY [8 votes]: Corollary 7.54 in Hoffman and Morris, The Structure of Compact Groups, seems to be what you want. - -REPLY [6 votes]: I am not too familiar to Lie groups, but I think that your claim can follow from the following Theorem which is proven in Rudin: -If A is locally compact abelian group, then there exists a loc.cpct.abelian group $H$ such that $A$ is isomorphic to some $R^k \times H$, and $H$ contains an open compact subgroup $K$. -Reference: This is on page 95 in Deitmar Echterhoff "Principles of Harmonic Analysis", see also When does a LCA group not contain a (closed) infinite cyclic subgroup?.<|endoftext|> -TITLE: Does derived algebraic geometry allow us to take quotients with reckless abandon? -QUESTION [21 upvotes]: So one of the major problems with the categories of schemes and algebraic spaces is that the "correct quotients" are oftentimes not schemes or algebraic spaces. The way I've seen this sort of thing rectified is either by moving a step up the categorical ladder or by defining some nonstandard quotient to "fix" things. That is, the (sheafy) quotient of a scheme by an étale equivalence relation is an algebraic space, and the (stacky) quotient of an algebraic space by a smooth groupoid action is an algebraic stack (and it is my understanding that these descriptions characterize alg. spaces and stacks up to equivalence). -Derived algebraic geometry gives us a number of powerful tools and has some very nice features: We have a whole array of new, higher dimensional, affine objects (coming from simplicial commutative rings), and a good supply of higher-categorical objects, which we get "all at once", as it were, rather than piecemeal one-level-at-a-time descriptions. -Does the theory of derived algebraic geometry give us enough "n-categorical headroom" (to quote a recent comment of Jim Borger) to take quotients of geometric objects "with reckless abandon" (not a quote of Jim Borger)? - -REPLY [13 votes]: There is more than one way that derived algebraic geometry generalizes ordinary algebraic geometry. The new affines don't help you much with quotients, which are (homotopy) colimits, but they give you well-behaved intersections, which are (homotopy) limits. On the other hand, you can consider functors from affines (new or old) to a category like simplicial sets that has better quotient behavior than plain sets. This gives you a notion of derived stacks, and I believe they behave well under many colimits. -I'm not sure what you mean by "reckless abandon". I tend to make mistakes when I'm not careful with my mathematics, even if I'm looking at derived algebraic geometry.<|endoftext|> -TITLE: Diagonal map and "infinitesimal points" -QUESTION [9 upvotes]: Let $f:X\to Y$ be a morphism between schemes. To construct the relative sheaf of differentials on $X$ (relative to $Y$), we first consider the diagonal map $\Delta: X \to X\times_Y X$ and then define $\Omega_{X/Y} = \Delta^{-1} \mathscr{I}/\mathscr{I}^2$ where $\mathscr{I}$ is the sheaf representing the immersion $X\to X\times_Y X$ (it's the kernel of $\mathscr{O}_{X\times_Y X} \to \Delta_* \mathscr{O}_X$. -Algebraically, this works out fine, due to the theory of abstract Kahler derivation defined on algebras. Is there a way to actually see the motivation behind this? -Moreover, what's the analog in higher infinitesimal approximation (instead of just 1st order one given by the differentials)? What's the (say, "analytic") insight behind the relationship between higher infinitesimal and higher diagonal? - -REPLY [6 votes]: I would like to add another answer to this old question. Consider -the case $X = Spec(A)$, $Y = Spec(R)$. Just to fix ideas, suppose that $A = R[T]$. If $f\in A$ and $a_0\in R$, one can consider the Taylor expansion of $f$ around $a_0$: -$$f(T) = \sum_i \frac{f^{(i)}(a_0)}{i!}\cdot (T-a_0)^i\in R[T].$$ -Now there is no reason why we should take a rational point $a_0 : R[T] \to R$ and in fact -we can consider the Taylor expansion around an arbitrary $S$-valued point $a_0 : R[T]\to S$. The Taylor expansion will then be naturally an element of $S\otimes_R R[T]$. Taking the universal point $S = R[T_0]$, $a_0 = T_0$, we see that the ``universal Taylor expansion'' of $f$ is -$$ -f(T) = \sum_i\frac{f^{(i)}(T_0)}{i!}\cdot (T-T_0)^i\in R[T_0,T]. -$$ -If we write $R[T_0,T] = R[T]\otimes_R R[T]$, then we rewrite the above as -$$ -1\otimes f(T) = \sum_i\left(\frac{f^{(i)}(T)}{i!}\otimes 1\right)\cdot(1\otimes T-T\otimes 1)^i -$$ -Looking mod $(1\otimes T-T\otimes 1)^2$ we get: -$$ -1\otimes f(T) \equiv f(T)\otimes 1 + (f'(T)\otimes 1)\cdot (1\otimes T-T\otimes 1)\pmod{(1\otimes T-T\otimes 1)^2} -$$ -Now, in this particular case, $I =\ker(A\otimes_R A\to A)$ is generated by -$1\otimes T-T\otimes 1$. Hence we see that $I/I^2$ is simply the space of linear terms -of Taylor expansions and the canonical map $d : A\to I/I^2$ is simply sending a function -$f\in A$ to the linear term in its Taylor series. Note that $1\otimes T-T\otimes 1$ is usually denoted by $dT$. -This also explains nicely what happens in higher degree. We can introduce the algebras -$P^n = (A\otimes_R A)/I^{n+1}=R[T_0,T]/(T-T_0)^{n+1}$, the ring of Taylor expansions of degree $\leq n$ where -the terms of degree at most $n$ of the Taylor expansion live. There is a natural map -$d^n : A\to P^n$, sending $a$ to $1\otimes a$ which is simply sending $a$ to its Taylor expansion. -This explanation works exactly the same if $A/R$ is smooth (instead of $A = R[T]$), because -locally on $A$ there is an etale map $F\to A$ where $F$ is a polynomial $R$-algebra and this -map induces an isomorphism on $I/I^2$ and $P^n$ more generally.<|endoftext|> -TITLE: Is volume--preserving an intrinsic property? -QUESTION [7 upvotes]: Let $M$ be a compact smooth manifold without boundary. A Riemannian metric $g$ on $M$ induces a volume measure (or Lebesgue measure) $m_g$ on $M$. -A diffeomorphism $f:M\to M$ is said to be {volume--preserving} if $f_*(m_g)=m_g$, that is, for each Borel subset $A\subset M$, $f_*(m_g)(A):=m_g(f^{-1}A)=m_g(A)$. Or equivalently the Jacobian (determinant of the tangent map $Df_x:T_xM\to T_{fx}M$) satisfies $J(f,m_g)(x)=1$ for every point $x\in M$. -If we change the Riemannian metric to $g'$ and the induced measure $m_{g'}$, the volume--preserving property with respect to $g$ is slightly distorted: there exists a uniform constant $C\ge1$ such that -$C^{-1}\le J(f^n,m_{g'})(x)\le C$, for every point $x\in M$ and every time $n\in\mathbb{Z}$.---- $(*)$ -My question is: is $(*)$ a characterization of volume--preserving? -That is, for a given $f\in\mathrm{Diff}(M)$, if $(*)$ holds for some arbitrarily chosen Riemannian metric $g$, does there exist a Riemannian metric $g'$ such that $J(f,m_g)=1$? -Thanks! - -REPLY [2 votes]: Just to add a positive result. For circle diffeomorphisms ($C^1$) one has that if the rotation number is irrational, a necessary and sufficient condition to be linearizable (by a $C^1$ map, which in particular implies you will preserve a $C^1$-length) is to have bounded derivatives (that $log |(f^n)'|$ is uniformly bounded). So, in this context you have your answer. This is a consequence of Gottshalk-Hedlund theorem, see for example the book by deMelo-VanStrien.<|endoftext|> -TITLE: What should I read before reading about Arakelov theory? -QUESTION [37 upvotes]: I tried reading about Arakelov theory before, but I could never get very far. It seems that this theory draws its motivation from geometric ideas that I'm not very familiar with. -What should I read to learn about the geometry that in turn inspired the ideas in Arakelov theory? - -REPLY [44 votes]: To get a nice overview of how and why Arakelov theory started you could read the introduction to R. de Jong's Ph.D. thesis on -http://www.math.leidenuniv.nl/~rdejong/publications/ -I remember that being very helpful to me. -To avoid too many complex analytic difficulties you should stick to the case of arithmetic surfaces (i.e. integral regular flat projective 2-dimensional $\mathbf{Z}$-schemes). The complex analysis involved is all "Riemann surfaces theory". An elementary and thorough treatment of this is given in P. Bruin's master's thesis -http://www.math.leidenuniv.nl/~pbruin/ -Arakelov theory provides an intersection pairing on an arithmetic surface $X$. The idea is to add vertical divisors on $X$ above the "points at infinity" on Spec $\mathbf{Z}$ (or Spec $O_K$ ). There will be two contributions: finite and infinite. To get a good understanding of the finite contributions I recommend reading Chapter 8.3 and 9.1 of Q. Liu's book. -I remember that after reading these texts the article by Faltings was much more readible to me. I also enjoyed the very nice asterisk by Szpiro on the subject, Séminaire sur les pinceaux de courbes de -genre au moins deux (all in French, last page has an English abstract). -Here's some advice on what you shouldn't read when you just start. I wouldn't start immediately reading the papers by Gillet and Soulé (unless you really want too). The complex analysis is very involved. The paper by Bost "Potential theory and Lefschetz theorems for arithmetic surfaces" introduces the most general intersection theory (based upon $L^2_1$ Green functions) and should also be left for later reading in my opinion. -To learn Arakelov theory the proofs don't really help me understand the statements for they are based upon moduli space arguments usually (e.g. the proof of the Noether formula). Therefore, I would also recommend you skip most of the proofs on a first reading. -What did help is seeing how Arakelov theory gets applied. I recommend the recent book by Couveignes, Edixhoven, et al. available here -http://arxiv.org/abs/math/0605244<|endoftext|> -TITLE: Minimal prerequisite to reading Wiles' proof of Fermat's Last Theorem -QUESTION [29 upvotes]: May I respectfully ask what the minimal background needed to read Wiles' proof of Fermat's Last Theorem is? -I'm not an expert on number theory, but out of curiosity I wanted to understand - at a cursory level if possible - the outline of the proof. -Thank you to all responders in advance. -My background: Junior-year undergraduate in Theoretical Physics. - -REPLY [8 votes]: Here is a good set of notes by Nigel Boston. I find them very readable and fairly self contained. -http://www.math.wisc.edu/~boston/869.pdf<|endoftext|> -TITLE: When does $P(a-b)=0$ for $a\ne b$ ensure $P(0)=0$? -QUESTION [9 upvotes]: Let $n$ be a positive integer. How large must be a set $A\subset F_2^n$ to ensure that if $P$ is a quadratic polynomial in $n$ variables, vanishing at all non-zero points of the sumset $2A:=\{a_1+a_2\colon a_1,a_2\in A\}$, then also $P(0)=0$? -Considering the situation where $P(x_1,\ldots,x_n)=\sum x_ix_j+\sum x_i+1$, and $A$ consists of $0$ and the elements of the standard basis, we see that having $|A|>n$ is not enough. On the other hand, if $|A|>C2^{3n/4}$ with an appropriate absolute constant $C$, then $A$ contains a $3$-dimensional affine subspace, whence $2A$ contains a $3$-dimensional linear subspace, and the assertion is easy to deduce. Where exactly between $n$ and $2^{3n/4}$ lies the truth? - -REPLY [5 votes]: Let $A$ be a maximal subset of $F^n_2$ with $P(A+A)\equiv 0$ but $P(0)\neq 0$. By finding a maximal linearly independent subset of $A$, and applying an appropriate linear transformation, we may assume $A$ contains the basis elements $e_1,\ldots,e_k$ and that all other elements of $A$ are linear combinations of these. We may further assume that $P$ is a polynomial in $x_1,\ldots,x_k,$ since we can throw out the other terms without changing our hypotheses. Then using that $P(e_i)=P(e_j)=P(e_i+e_j)=0$ and $P(0)=1$, one can deduce -$P=\sum_{\leq k} x_ix_j+\sum_{\leq k} x_i + 1$. -Finally, if $A$ contained anything besides the basis elements $e_i$ (and possibly zero*), $A+A$ would necessarily contain a non-zero element with either 3 or 0(mod 4) non-zero coordinates, for which $P$ does not vanish. So in fact, $|A|>n+1$ suffices. -*Actually, if $n\equiv 2 (\text{mod }4)$, we need $|A|>n+2$, since we can take the $e_i$'s, $0$, and $(1,1,\ldots,1)$.<|endoftext|> -TITLE: Hochschild H^1 (R,M) = 0 vs. H_1 (R,M) = 0 where R is a ring and M is an (R,R)-bimodule -QUESTION [7 upvotes]: Let $k$ be a commutative ring (with unity). Let $R$ be a $k$-algebra (with unity, but not of necessity commutative). -Let $M$ be an $\left(R,R\right)$-bimodule where $k$ acts in the same way from the left and from the right (I'd call this an $\left(R,R\right)_k$-bimodule, but I haven't seen this notation anywhere). There are two ways to define the Hochschild homology and cohomology of $R$ with coefficients in $M$: either as the homology of the standard complex tensored with $M$ rsp. the cohomology of Hom of the standard complex and $M$, or as $\mathrm{Tor}$ and $\mathrm{Ext}$. As far as I understand, these two definitions are equivalent only if $R$ is a projective $k$-module, which I don't want to require here. -Question 1: So let me define Hochschild cohomology and homology through the standard complex. Then, Löfwall's text seems to silently hint at the fact that if a $k$-algebra $R$ satisfies $\mathrm{H}^1\left(R,M\right)=0$ for all $\left(R,R\right)$-bimodules $M$, then it also satisfies $\mathrm{H}_1\left(R,M\right)=0$ for all $\left(R,R\right)$-bimodules $M$. While this is clear from homological algebra in the case when $R$ is a projective $k$-module, is this true otherwise? And how is it proven? -(Remark: A $k$-algebra $R$ such that $\mathrm{H}^1\left(R,M\right)=0$ for all $\left(R,R\right)$-bimodules $M$ is said to be zero-dimensional (in Löfwall's text) or separable (in the modern sense of this word).) -Question 2: The same text gives a counterexample for the opposite direction (if $\mathrm{H}_1\left(R,M\right)=0$ for all $\left(R,R\right)$-bimodules $M$, then $\mathrm{H}^1\left(R,M\right)=0$ for all $\left(R,R\right)$-bimodules $M$). In this counterexample, $k$ is a field and $R$ is commutative, but infinite-dimensional. I assume that counterexamples fade when we impose some more restrictive conditions on $k$ and $R$. What about finite-dimensional $R$? What about finitely-generated-as-algebras $R$? If $k$ is not a field anymore? If $R$ is not commutative anymore? - -REPLY [6 votes]: (Let me write $A$ for your $R$, because I will mix letters up if not...) -The standard complex for $A$ over $k$ is exact, independently of the projectiveness of $A$ over $k$; call $d$ its differential. Then we have a short exact sequence -$$ -0 -\to -\frac{A\otimes_kA\otimes_kA}{d(A\otimes_kA\otimes_kA\otimes_kA)} -\to A\otimes_k A -\to -A -\to -0 -$$ -Call $K$ the quotient appearing here, and call $\iota$ the map $K\to A\otimes_kA$ -Now suppose $H^1(A,K)=0$ (with cohomology defined using the standard complex, as you wanted; this hypothesis is implied by the hypothesis that $H^1(A,M)=0$ for all $M$ that you wanted to consider, of course!) This means that every bimodule map $f:K\to K$ can be factorized through $\iota$, so there exists a $\bar f:A\otimes_kA\to K$ such that $\bar f\circ\iota=f$. In particular, if we take $f=\mathrm{id}_K$ we see that the injective map $\iota$ splits. It follows that the short exact sequence of $A$-bimodules above itself splits, and that the map $A\otimes_kA\to A$ given by multiplication is split by a map $\phi:A\to A\otimes_kA$. Let $e=\phi(1)=A\otimes_kA$. -Consider now the standard complex, -$$ -\cdots -\to -A\otimes_kA\otimes_kA\otimes_kA -\to -A\otimes_kA\otimes_kA -\to -A\otimes_kA -\to -A -$$ -Using $\phi$ (or $e$), you can construct a retraction of this complex, so that when you tensor it with an arbitrary bimodule $M$ over $A^e$ you get an acyclic complex. In particular, $H_1(A,M)=0$, defined again as you wanted. -The first two components of that retraction are -$$ -a\in A\mapsto ae\in A\otimes_kA, -$$ -$$ -a\otimes b\in A\otimes_kA\mapsto ae\otimes b\in A\otimes_kA\otimes_kA, -$$ -and it keeps going like that. - -As for your second question (and I consider now the situation in which $A$ is $k$-projective): -When $H_1(A,M)=0$ for all $M$, one says that the the weak dimension of $A$ as an $A^e$-module is zero and writes $\operatorname{wdim}A=0$. When $H^1(A,M)=0$ for all $M$, then one says that the projective dimension of $A$ as a bimodule is zero, and writes $\operatorname{pdim}A=0$. -As you know, $$\operatorname{pdim}A=0\implies\operatorname{wdim}A=0$$ and the converse is false. If $A^e$ is noetherian, then the converse holds (see, for example, McConnell and Robson's bible on noetherian rings); this includes the case in which $A$ is finitely dimensional over $k$ or, more generally, if $A$ is commutative and finitely generated over $k$ as an algebra. If $A$ is not projective over $k$, I have no idea what happens (and you should probably ignore that situation if you are reading up on Hochschild cohomology for the first few times! :) )<|endoftext|> -TITLE: Minimal surface which divides a convex body into two regions of equal volume -QUESTION [6 upvotes]: Question. Given a convex body $\Omega$, what is the shape of a surface $\Gamma$ of minimal area which divides $\Omega$ into two regions of equal volume? - -Background/motivation. -A 2D version of the question was posed by Michael Goldberg in Monthly: find the shortest curve which divides a convex quadrilateral into two equal areas. In the latter case the solution is given by a circular arc perpendicular to two sides of the quadrilateral (or just a segment of a straight line in degenerate situations). This follows from the observation that a circular sector is the shortest curve which, together with two sides of an angle, encloses a fixed area. -Goldberg himself offered a physically intuitive solution to the problem: - -The curve may be considered as a restraining -member under tension produced by internal fluid pressure in the restricted area. -The ends of the curve are free to slide along the sides. Hence, the curve must be -normal to two sides of the quadrilateral. Furthermore, since the fluid pressure -is uniform, the curve must take the form of a circular arc. - -The same approach suggests that in the general multidimensional case the solution is given by a spherical "cap" that intersects $\partial \Omega$ orthogonally. Now, assuming that the intuition is correct, is there a simple formal proof of this conjecture? -Edit. As Sergei Ivanov points out the minimal surface in question need not be a spherical cap for dimensions $>2$. -A modified question: is there always at least one solution to the problem? Should one impose any smoothness conditions on $\partial\Omega$ to guarantee existence? - -REPLY [14 votes]: This is a classical problem on which we know the existence, thanks to the Geometric Measure Theory. In space dimension $n$, the solution is a hypersurface which is smooth of constant curvature away from a possible singular set of dimension at most $n-7$. For instance, if $n=3$, it is smooth up to the boundary, and it meets $\partial\Omega$ orthogonally. -If you relax the equal volume condition, and ask for a hypersurface of smaller area that divides $\Omega$ into two pieces of volumes $\epsilon\Omega$ and $(1-\epsilon)\Omega$, then the surface tends to a boundary point $p$ where the curvature has a critical point. See for instance the discussion in the recent PhD Thesis of Paul Laurain (ENS de Lyon).<|endoftext|> -TITLE: p-groups as Sylow subgroups -QUESTION [8 upvotes]: There are some nice families of groups as $S_n, A_n$, $GL(n,q)$, $SL(n,q)$, and they are useful; we know their elements, and we can get small groups as subgroups of these groups. Is it possible to get every $p$ group as a Sylow-p subgroup of some group in such families of groups? ( For example, the non-abelian groups of order 8 are Sylow-2 subgroups of SL(2,3) and S4; there are five non-abelian groups of order 16, having no element of order 8, and one of them, namely $D_8 \times C_2$, is Sylow-2 subgroup of $S_6$. Also $SD_{16}$ is Sylow-2 subgroup of GL(2,3).) - -REPLY [2 votes]: The answer to the original question is a definite no. There are many $p$-groups which do not occur -as Sylow $p$-subgroups of classical groups, symmetric groups, or close relatives. As noted in -various comments, embedding a finite $p$-group in a symmetric or classical group is an easy matter. -But it is very difficult for a finite $p$-group $P$ to be a Sylow $p$-subgroup of a group $G$ with no factor group of order $p$. The most general results in this direction are probably by George -Glauberman. For example, when $p \geq 5$ and $P$ is a finite $p$-group whose outer automorphism -group is a $p$-group, then any finite group $G$ with $P$ as a Sylow $p$-subgroup has a factor -group of order $p$. There are meaningful senses in which ``most" finite $p$-groups have outer -automorphism group a $p$-group.<|endoftext|> -TITLE: geodesic 2-dimensional submanifolds of a Riemannian manifold -QUESTION [8 upvotes]: Possible Duplicate: -Must a surface obtained by exponentiating a plane in a tangent space of a Riemannian manifold be geodesically convex? - -The one dimensional geodesic submanifolds of a given Riemannian manifold $(M,g)$ are just geodesics. So one can can wonder, how to construct 2-dimensional geodesic submanifolds. Lets first consider the following question: -Given any point $x\in M$ and a two dimensional subspace of $V\subset T_xM$. Then the exponential map restricted to a sufficiently small ball around $0\in V$ gives an embedding of the open disc into $M$. When is it a geodesic submanifold? -Note that there are many spaces, that have this property at every point and at every tangent plane, like $\mathbb{S}^n,\mathbb{H}^n,\mathbb{R}^n$ and (if I am not mistaken) products of those. So one can also ask: -What properties must the metric $g$ have to ensure, that at every point $x\in M$ and at every two dimensional subspace $V\subset T_xM$ the exponential map $B_\varepsilon(0)\subset V\rightarrow M$ gives locally geodesic surfaces? -A "general" manifold should not have this property I think. It would be nice to have a simple and short counterexample. - -REPLY [6 votes]: Your examples (i.e. space forms) are the only manifolds with the property that the exponential map sends 2-dimensional disks to totally geodesic surfaces. One way to see this is using Jacobi vector fields. -More precisely, let $X$ and $Y$ be two orthogonal vectors in some tangent space $T_xM$ and assume that the exponential map sends (a neighbourhood of 0 of) the plane spanned by $X$ and $Y$ into a totally geodesic surface $S$. If you denote $\gamma_s(t)=exp_x(t(X\cos(s)+Y\sin(s)))$, then -$$J:=\frac{\partial\gamma}{\partial s}$$ -is a Jacobi field along the geodesic $\gamma_0$ and thus satisfies (denoting $\gamma_0$ by $\gamma$): -$$\nabla^2_{\dot\gamma,\dot\gamma}J=-R_{J,\dot\gamma}\dot\gamma.$$ -Since $J$ and $\dot\gamma$ are tangent to $S$, which is totally geodesic, and moreover $R_{J,\dot\gamma}\dot\gamma$ is orthogonal to $\dot\gamma$, we see that $R_{J,\dot\gamma}\dot\gamma$ ahs to be proportional to $J$. A standard argument then shows that $M$ has constant sectional curvature (provided that $dim(M)>2$). I can give more details about this if you need.<|endoftext|> -TITLE: prime ideals in regular local rings -QUESTION [15 upvotes]: Suppose $R$ is a regular local ring. Let $m$ be the maximal ideal. Then, if the dimension of $R$ is $n$, there is a regular sequence of size $n$, say $x_1,x_2,...,x_n$ s.t. $m=(x_1,x_2,...,x_n)R$. Further, the ideals $(x_{i_1},...,x_{i_j})$ with $i_1,...,i_j\in {1,...,n}$, are prime. -Can we make similar statements about any other kind of prime ideals in a regular local ring $R$? Specifically, do any other prime ideals satisfy the condition: if the ideal is minimally generated by a certain set of generators, then every subset of the generators defines a prime ideal? One example in light of the first paragraph, are the prime ideals generated by a subset of the regular sequence that generates the maximal ideal. -Also, when does a regular sequence define a prime ideal in a regular local ring $R$, and when does a maximal regular sequence define a maximal ideal? - -REPLY [7 votes]: As Sandor pointed out, a necessary condition is that the prime ideal $P$ is a complete intersection. Here is a proof that it is also sufficient. It will suffice to prove the following: -Claim: Let $(R,m)$ be a Noetherian local ring and $x\in m$ a regular element on $R$. If $R/(x)$ is a domain, then so is $R$. -Proof: Suppose $ab=0$ in $R$. Then modulo $x$, one of them say $a$, must be $0$. So $a=xa_1$, thus $x(a_1b)=0$. As $x$ is regular, $a_1b=0$, and continuing in this fashion one of $a,b$ must be divisible by arbitrary high power of $x$, so it must be equal to $0$. -As for an example which is not a part of a regular s.o.p, take something like $P=(x^2+y^2+z^2, u^2+v^2+w^2)$ in $\mathbb C[[x,y,z,u,v,w]]$.<|endoftext|> -TITLE: Borel Lemma for vector-valued functions -QUESTION [18 upvotes]: The classical Borel Lemma states that for an arbitrary sequence $(v_n)_{n \in \mathbb{N}_0}$ of complex numbers there is a smooth function $f\colon \mathbb{R} \longrightarrow \mathbb{C}$ with Taylor coefficients at $0$ given by the $v_n$. Some generalizations work for functions of $d$ variables and also for values taken in an arbitrary Fréchet space $V$ instead of the complex numbers. -The proofs I know use essentially the fact that $V$ has a countable set of seminorms defining the topology. On the other hand, taking the space of compactly supported smooth functions with its usual LF topology as $V$ and $v_n$ with increasing support gives easily a counter-example that for this sequence we can not have a smooth $f\colon \mathbb{R} \longrightarrow C^\infty_0(\mathbb{R})$ with $v_n$ being the Taylor coefficients, unless the $v_n$ are all in same $C^\infty_0(K)$ for a fixed compact subset $K$. -So my question is to which lcs one actually can extend the Borel lemma? Are Fréchet spaces the end of the story? -EDIT: There is of course a stupid way to extend it beyond Fréchet: whenever you have a coarser lc topology on $V$ then every smooth function with respect to the orignal one is also smooth with respect to the coarser one. So if $V$ is Fréchet then every coarser topology on $V$ will also have a valid Borel Lemma. Examples are e.g. the operator topologies on the bounded operators on a Hilbert space (soooorry for overlooking this in the first try). -So the refined question is: are there other lcs with topologies not dominated by a Fréchet one for which the Borel Lemma holds? - -REPLY [6 votes]: Yes, there are other spaces in which Borel's Theorem holds (property (BT)). -Example 1. Note that every cartesian product of locally convex spaces with (BT) has (BT). In particular, any uncountable power ${\mathbb R}^I$ has (BT). Yet, ${\mathbb R}^I$ cannot be given a topology ${\mathcal O}$ which makes it a Fréchet space and is finer than the product topology ${\mathcal T}$. -In fact, for any such topology ${\mathcal O}$, we could choose a decreasing sequence of absolutely convex $0$-neighbourhoods $U_n$ in $({\mathbb R}^I,{\mathcal O})$ which form a basis of $0$-neighbourhoods. Let $C_n$ be the closure of $U_n$ in $({\mathbb R}^I,{\mathcal T})$. -Now $({\mathbb R}^I,{\mathcal T})$ is a Baire space (see Oxtoby, J.C., Cartesian products of Baire spaces, Fundam. Math. 49 (1961), 157-166). -As in the usual proof of the open mapping theorem, we see that each $C_n$ -is a $0$-neighbourhood in $({\mathbb R}^I,{\mathcal T})$. Since each $0$-neighbourhood in $({\mathbb R}^I,{\mathcal T})$ contains some $C_n$, we see that $({\mathbb R}^I,{\mathcal T})$ would be first countable, which is absurd. -This contradiction shows that ${\mathcal O}$ cannot exist. -Example 2. Recall that the usual proof of Borel's Theorem (for $V$ a Fréchet space) furnishes a smooth function $f\colon {\mathbb R}^d\to V$ of the form $f(x)=\sum_{\alpha \in{\mathbb N}_0^d} \;v_\alpha h(m_\alpha x)x^\alpha$, where $h\colon {\mathbb R}^d\to[0,1]$ is a cut-off function around $0$ and the $m_\alpha$ are positive constants. It can always be achieved that $m_\alpha\to\infty$ as $|\alpha|\to\infty$. As a consequence, the above sum is a finite sum (almost all summands vanish) for each fixed $x\in{\mathbb R}^d$, and likewise for all partial derivatives of the summands. -It is clear from this observation that the completeness of $V$ does not play a role: The construction works just as well if $V$ is any (not necessarily complete) metrizable locally convex space. -Let $V$ be a metrizable locally convex space whose dimension as an abstract vector space is infinite and countable. By the preceeding, $V$ has (BT). Yet, there is no finer vector topology on $V$ making it a Fréchet space. -Let me mention that the product topology on ${\mathbb R}^I$ from Example 1 cannot be refined to a non-complete metrizable locally convex vector topology ${\mathcal O}$ either (by the same proof as above).<|endoftext|> -TITLE: universal property of the determinant bundle -QUESTION [5 upvotes]: Let $X$ be a nontrivial ringed space (i.e. all stalks are nonzero). To every locally free module $M$ on $X$ of constant rank $n$ we can associated it's determinant $\det(M)$, which is a line bundle and is defined as the $n$th exterior power of $M$. We can also define $\det(M)$ if the rank is not assumed to be constant. It's surely locally constant, so $X$ is partitioned into the open subsets $(X_n)_{n \geq 0}$, where $M$ has constant rank $n$, and we can glue the $\det(M|_{X_n})$ to get the determinant $\det(M)$. There is no doubt that this is well-defined, but it seems to me a bit uncanonical. -For example look at the functoriality: If $f : M \to N$ is a homomorphism, then $\det(f) : \det(M) \to \det(N)$ is first defined on the intersections on the open subsets on which $\det(M)$ and $\det(N)$ are defined, and then glued. Then a small argument is needed to prove that this is, indeed, a functor. Isn't this ugly? Therefore: -Question Is there a characterization of the functor $\det(-)$ from locally finite free modules to line bundles which does not depend on partitions of $X$? -For example, for every $n \geq 0$, the functor $\det(-)$ on locally free modules of rank $n$ has a universal property, namely $\text{Hom}(\det(M),L)$ corresponds to alternating maps $M^n \to L$. You can write down a similar universal property when $n$ is a fixed locally constant function on $X$. But my question is if we can do it without fixing $n$. - -REPLY [5 votes]: There is a paper by Knudsen and Mumford that gives a thorough treatment of determinants of perfect complexes. Mumford has put a copy online here. -Edit: Knudsen's 2002 paper that Theo Buehler linked in the comment below appears to resolve the question in a more canonical way than the older paper. It also has a neat letter by Grothendieck at the end.<|endoftext|> -TITLE: Do the tails of the decimal expansion of pi form a dense set in [0,1]? -QUESTION [6 upvotes]: Let $a_n=10^n \cdot \pi$. Is the set of numbers $\{a_n-\lfloor a_n \rfloor : n \in \mathbb{N}\}$ dense in [0,1]? -What is the best known result near this question? -Apparently John Nash asked this on an undergraduate analysis exam (according to an anecdote told by Seymour Haber, recounted in Sylvia Nassar's biography of Nash). - -REPLY [4 votes]: The comments have pretty much said all there is to be said about $\pi$. I'll just note that the fractional part of $10^n\alpha$ is known to be not just dense but uniformly distributed in $[0,1)$ for all real $\alpha$, except for a set of measure zero. There is no reason to think $\pi$ is in this exceptional set, and no expectation of a proof anytime soon that it isn't.<|endoftext|> -TITLE: Nice applications of the spectral theorem? -QUESTION [20 upvotes]: Most books and courses on linear algebra or functional analysis present at least one version of the spectral theorem (either in finite or infinite dimension) and emphasize its importance to many mathematical disciplines in which linear operators to which the spectral theorem applies arise. One finds quite quickly that the theorem is a powerful tool in the study of normal (and other) operators and many properties of such operators are almost trivial to prove once one has the spectral theorem at hand (e.g. the fact that a positive operator has a unique positive square root). However, as hard as it is to admit, I barely know of any application of the spectral theorem to areas which a-priori have nothing to do with linear algebra, functional analysis or operator theory. -One nice application that I do know of is the following proof of Von Neumann's mean ergodic theorem: if $T$ is an invertible, measure-preserving transformation on a probability space $(X,\mathcal{B},\mu)$, then $T$ naturally induces a unitary operator $T: L^2(\mu) \to L^2(\mu)$ (composition with $T$) and the sequence of operators $\frac{1}{N}\sum_{n=1}^{N}T^n$ converges to the orthogonal projection on the subspace of $T$-invariant functions, in the strong operator topology. The spectral theorem allows one to reduce to the case where $X$ is the unit circle $\mathbb{S}^1$, $\mu$ is Lebesgue measure and $T$ is multiplication by some number of modulus 1. This simple case is of course very easy to prove, so one can get the general theorem this way. Some people might find this application a bit disappointing, though, since the mean ergodic theorem also has an elementary proof (credited to Riesz, I believe) which uses nothing but elementary Hilbert space theory. -Also, I guess that Fourier theory and harmonic analysis are intimately connected to the spectral theory of certain (translation, convolution or differentiation) operators, and who can deny the usefulness of harmonic analysis in number theory, dynamics and many other areas? However, I'm looking for more straight-forward applications of the spectral theorem, ones that can be presented in an undergraduate or graduate course without digressing too much from the course's main path. Thus, for instance, I am not interested in the use of the spectral theorem in proving Schur's lemma in representation theory, since it can't (or shouldn't) be presented without some prior treatment of representation theory, which is a topic in itself. -This book by Matousek is pretty close to what I'm looking for. It presents simple and short (but nevertheless impressive and nontrivial) applications of linear algebra to other areas. I'm interested in the more specific question of applications of the spectral theorem, in one of its versions (in finite or infinite dimension, for compact or bounded or unbounded operators, etc.), to areas which are not directly related to the theory of linear operators. Any suggestions will be appreciated. - -REPLY [3 votes]: You mention the proof of the mean ergodic theorem by Von Neumann. Here is another result from ergodic theory that makes use of the spectral theorem. -Theorem -Let $(X, {\cal T}, \mu)$ be a probability space, $T: X \rightarrow X$ a measure preserving transformation. The following properties are equivalent. -$\bullet$ For all $f,g \in L^2$, -$$ -{1\over N}\sum_{k=1}^N \ \Bigl| \int f\circ T^n g \,d\mu -\int f d\mu \int g d\mu \Bigr| \rightarrow 0. -$$ -$\bullet$ There is no function $f \in L^2$ and $\theta \in {\bf R}$ such that -$$ -f\circ T = e^{i\theta} f. -$$ -If any of these properties is satisfied, the transformation $T$ is said to be weak-mixing. -I don't know of any proof of (2) => (1) that does not use some version of the spectral theorem.<|endoftext|> -TITLE: How to place k bishops on an nxn chessboard -QUESTION [14 upvotes]: In how many different ways can k bishops be placed on an nxn chessboard such that no two bishops attack each other? Please try to respond with a formula and explanation. - -REPLY [9 votes]: See my book -http://problem64.beda.cz/silo/kotesovec_non_attacking_chess_pieces_2013_6ed.pdf -, page 234-236. Number of ways to place k non-attacking bishops on an n x n board, (n>1)<|endoftext|> -TITLE: A question about imaginary Turing machines. -QUESTION [5 upvotes]: Let T be a Turing machine which, when started on an infinite blank tape extending to the left and to the right, -is programmed to print out in some recursively enumerable order, all the theorems of ZFC (one after another) on -the squares of the tape. It is also programmed to halt when and only when (if ever) it prints a theorem that is -the negation of a theorem which it has already printed. Does anyone know of any estimates that have been made of -the number of states and the number of symbols which would be sufficient for the machine T to have available in -order to be able to accomplish this task-i.e. to test the consistency of ZFC? - -REPLY [3 votes]: The Turing Machine you describe here can actually be constructed (from a practical standpoint also), but it would be tedious and not of much practical use. -First note that the finite set of symbols $\{ \in, \forall, \exists, x, \prime, \land, \lor, \lnot, \rightarrow, \leftrightarrow \}$ would be more than sufficient to represent all statements expressible in the language of set theory (prime used to differentiate the countably infinite set of variables: $x, x^{\prime}, x^{\prime\prime},\ldots$). We can have the scrap work occur to the left of the place where we list out all of our ZFC theorems. The (highly inefficient) process can be described as follows: - -(1) Copy the tape counter, which can represent $n$ by $n$ successive primes written from right to left. -(2) Run a Turing Machine that on an input of $n$ primes, writes the first $n$ ZFC axioms (separated by spaces only working left on the tape) according to some recursive (computable) numbering. -(3) Run a Turing Machine that on an input of $n$ formulas, applies the first $n$ rules of inference in first-order logic (in the sense that there are an infinite number of substitution rules, each applying a specific change of variable) on each of the formulas already on the working part of the tape and writes them all to the left. -(4) Iterate through each formula on the left, determining whether it's the same as one already in your actual list or a negation of one. If not, add the formula to your list. If it's a negation of one, add the formula, reset your pointer to the beginning of the actual list, and HALT. -(5) Add a prime to the original counter, reset the pointer to the beginning of the prime list, and go to step (1). - -In a nutshell, this procedure at every stage $n$, lists all ZFC theorems provable in $n$ steps (not already listed) by limiting oneself to the first $n$ rules of inference and the first $n$ axioms according to some recursive (computable) enumeration. Note that the the number of states required for such a Turing Machine is sensitive to ZFC only in the listing of the axioms (i.e., the number of states for the Turing machine carrying out step (2)). -The problem with such a Turing machine is that it exhibits no effective intelligence in its procedure. Its running time for producing theorems would be exponential as a function of the length of their shortest proofs and emulating it in a programming language such as C++ would not change this fact. Even if there were an obvious contradiction in ZFC with a proof requiring $1000$ lines (considering all of the formal manipulations, this isn't very long), it would most likely take somewhere on the order of $2^{1000}$ years to find. Therefore, the issue is not with the size of the Turing machine, but its running time. Indeed, there is an ongoing research program in automated theorem proving to only choose "smart" paths of deduction rather than mindlessly trying them all. -Finally let me address an issue with the number of symbols and coding. Anything that can be done with a Turing machine having a countable alphabet can be carried out by a Turing Machine having a binary alphabet (with a likely introduction of new states) under suitable coding. This is because there are only finitely many states and hence finitely many possible transitions. Specifically, given a finite collection of subsets $\{S_1, \ldots, S_n\}$ of a countable alphabet of symbols representing the possible transitions from one state to another, we can assign each symbol a Natural number so that each of the $S_n$ are finite unions of recursive sets (assigning first the $n$-intersection to a computable infinite coinfinite set of Natural numbers; then doing the same for each of the $(n-1)$-intersections with the remaining unassigned symbols and Natural numbers, down to the individual $S_i$) and hence recursive. -Also, with coding, your question admits an annoying answer. This is similar to a question I asked in Asymptotic density of provable statements in ZFC. -Specifically, if ZFC is inconsistent, then a Turing machine printing two contradictory statements and then halting would get the job done. Otherwise, we can bijectively associate the set of Natural numbers with the set of all binary strings. By taking any computable infinite and coinfinite subset $A$ of the Natural numbers, we can code the statements in the language of set theory so that the ZFC theorems are precisely the Natural numbers (binary strings) from $A$. This means that a Turing Machine could require an arbitrarily large number of states if $A$ is suitably complex or very few if $A$ were something simple.<|endoftext|> -TITLE: Is there a notation for the symmetric / antisymmetric subspaces of a tensor power that distinguishes them from the symmetric / exterior power? -QUESTION [6 upvotes]: Let $V$ be a finite-dimensional vector space over a field $k$, say of characteristic $0$. The symmetric group $S_n$ acts on the tensor power $V^{\otimes n}$ in the obvious way, and this action defines two subspaces of $V^{\otimes n}$, the subspace on which $S_n$ acts via the trivial character and the subspace on which $S_n$ acts via the antisymmetric character. -Question 0: Is the construction of these subspaces functorial in $V$? If it is, are the corresponding functors naturally isomorphic to the symmetric and exterior powers, and if that's true, are the corresponding natural isomorphisms unique? -If the answers to Question 0 turn out more or less like I suspect, we should not regard these subspaces as completely synonymous with the symmetric power $S^n V$ and the exterior power $\Lambda^n V$, respectively, since these are naturally thought of as quotients of $V^{\otimes n}$. (This issue recently came up in another MO question.) -Question 1: Is there an established notation in the literature which respects this distinction? - -REPLY [2 votes]: The subspace of $\Sigma_n$-invariants of $V^{\otimes n}$ is called the $n$th divided power of $V$ (at least when $V$ is a free module).<|endoftext|> -TITLE: Period rings for Galois representations -QUESTION [15 upvotes]: I have some questions concerning period rings for Galois representations. -First, consider the case of $p$-adic representations of the absolute Galois group $G_K$, where $K$ denote a $p$-adic field. Among all these representations, we can distinguish some of them, namely those which are Hodge-Tate, de Rham, semistable or crystalline. This is due to Fontaine who constructed some period rings : $B_{HT}$, $B_{dR}$, $B_{st}$ and $B_{crys}$. -Constructing the ring $B_{HT}$ is not very difficult and it is quite natural. -Does someone have any idea where $B_{dR}$ comes from ? -For $B_{crys}$, I guess it was constructed to detect the good reduction of (proper, smooth ?) varieties. I don't know anything of crystalline cohomology but does someone have a simple explanation of the need to use the power divided enveloppe of the Witt vectors of the perfectisation (?) of $\mathcal{O}_{\mathbb{C}_p}$ ? -As for the ring $B_{st}$, once you have $B_{crys}$, I think the idea of Fontaine was to add a period from Tate's elliptic curve, which have bad semistable reduction. Does someone knows if Fontaine was aware that adding just this period will be sufficient or was it a good surprise ? -Finally, why there is no period rings for global $p$-adic Galois representations ? - -REPLY [9 votes]: Beilinson's results (two papers, one mentioned by Keerthi and the other here) have been generalised by Bhargav Bhatt; his paper also introduces a global period ring $A_{ddR}$ for global Galois representations!! The ring $A_{ddR}$ is a filtered $\hat{Z}$-algebra equipped with a Gal$(\bar{Q}/{Q})$-action. -A (one of the many) beautiful result in this paper is the following theorem: -Let $X$ be a semistable variety over $Q$. Then the log de Rham cohomology of a semistable model for $X$ is isomorphic to the $\hat{Z}$-etale cohomology of $X_{\bar{Q}}$, once both sides are base changed to a localization of $A_{ddR}$ (whle preserving all natural structures on either side).<|endoftext|> -TITLE: Verifying coefficients of modular forms -QUESTION [14 upvotes]: Hi, -I was wondering about good techniques that one can use to show that given certain coefficients, they are the Fourier coefficients of a cusp form, assuming we know the desired weight and level. I am aware of Weil's "converse theorem", but am not aware of any examples of it being used to prove something is a modular form. So, even a pointer to an example of that would be useful. I'm curious if there are other more direct methods as well. Thanks! - -REPLY [2 votes]: The search for a modular form of some weight and level, whose Fourier coefficients match a given list of initial coefficients, has been implemented in PARI/GP with the function mfsearch.<|endoftext|> -TITLE: Generalized Gauss map -QUESTION [5 upvotes]: Given a surface M in Euclidean space, we have the generalized-Gauss-map G, i.e. map the tangent spaces into the Grassmannian G(2,n). What is the relation between DG and the second fundamental form of M, and the Gauss curvature? - -REPLY [2 votes]: The differential of the Gauss map is the 2nd fundamental form.<|endoftext|> -TITLE: Commutator formulas in a universal enveloping algebra -QUESTION [5 upvotes]: I am interested in finding formulas for commutators of symmetrized monomials in a universal enveloping algebra. Let $C(x_1,\ldots, x_n)= (1/n!)\sum x_{\sigma(1)}\cdots x_{\sigma(n)}$ where the sum runs over all permutations, and $x_i \in L$ for some LIe algebra $L$. This is an element of $UL$. -Now, reasonable combinatorics shows that, in $UL$, we have, -$ -[C(x_1,\ldots, x_n), l] = \sum_{i=1}^n C(\ldots,[x_i,l],\ldots) -$ -for $l\in L$. -I am looking for formulas for $[C(x_1,\ldots, x_n), C(y_1,\ldots, y_m)]$ in terms of -symmetrized monomials and brackets. Even for $n=m=2$ the number of terms gets fairly large. If anyone knows where I can find such things I would be very grateful. - -REPLY [4 votes]: This is probably not yet a final answer but may shine some additional light on the problem: -For simplicity, I assume that $L$ is finite-dimensional and defined over the reals (for some other field of char $0$, the following should still work). -The symmetrization map can be viewed as a linear map -\begin{equation} -\sigma\colon \mathrm{S}(L) \longrightarrow \mathrm{U}(L) -\end{equation} -from the symmetric algebra over $L$ into the universal envelopping algebra. It is now possible (essentially via PBW) to show that this is a fitration compatible linear bijection. Thus it allows to pull-back the product of $\mathrm{U}(L)$ to $\mathrm{S}(L)$. The result is the star product of Gutt / Drinfel'd (both in 1983, I guess). A further canonical isomorphism yields that the symmetric algebra is nothing else than the poylnomials on the dual $L^*$ (suppose $L$ is finite-dimensional for convenience) Thus your question is equivalent to the following task: -What is the Gutt star product commutator of two (homogeneous) polynomials on $L^*$? -Gutt has computed many properties of this star product and fan almost explicit formula. However, it essentially involved the full BCH series of $L$, so my guess is that a complete answer might be as complicated as computing BCH. -The Gutt star product can be characterized nicely as follows: take $x, y \in L$ and view them as linear polynomials on $L^*$ as usual. Then form the formal exponential functions $e_{\hbar x} (\alpha) = \exp(\hbar \alpha(x))$ and similarly for $e_{\hbar y}$ where $\hbar$ is your formal parameter. Then $\star_{\mathrm{Gutt}}$ is uniquely determined by -\begin{equation} - e_{\hbar x} \star_{\mathrm{Gutt}} e_{\hbar y} - = - e_{\mathrm{BHC}(\hbar x, \hbar y)} -\end{equation} -I hope I got the signs right :) You can find this formulas also in Section 8 of q-alg/9707030 (published in Commun. Math. Phys.). There are many more papers on the Gutt star product, so a little MRsearch will probably give some addition info. -The solution of your problem is now obtained by differentiating the above equation with respect to $\hbar$ sufficiently often and use polarization afterwarts. But as I sad, you need to know BCH quite well to efficiently do that. In the end you take commutators\ldots<|endoftext|> -TITLE: A solvability theorem -QUESTION [5 upvotes]: There is a theorem which says: -Let $G$ be a finite group. Suppose that every maximal subgroup of $G$ has index equal to a prime or the square of a prime. Then $G$ is solvable. -Reading existing proofs and other results, I have cobbled together my own proof: -Proof. Suppose, to the contrary, that a counterexample exists. Let $G$ be a minimal counterexample. Since $G$ is nontrivial, let $p$ be the largest prime factor of $|G|$, and let $P$ be a Sylow $p$-subgroup of $G$. -If $P$ is normal in $G$, then the maximal subgroups of $G/P$ correspond to maximal subgroups of $G$ containing $P$. These have index equal to a prime or the square of a prime, so $G/P$ is a smaller group for which the condition holds and therefore $G/P$ is solvable. Then, since $P$ is also solvable, $G$ is solvable. Therefore we assume that $P$ is not normal in $G$. -Then the normalizer $N_{G}(P)$ is a proper subgroup of $G$. Then let $M$ be a maximal subgroup of $G$ containing $N_{G}(P)$. Then $[G:M] = \frac{[G:N_{G}(P)]}{[M:N_{G}(P)]} = \frac{[G:N_{G}(P)}{[M:N_{M}(P)]} \equiv \frac{1}{1} = 1 \mod{p}$. -Since $M$ is a proper subgroup of $G$, $[G:M] \geq p+1 > p$. Since $p$ is the largest prime factor of $|G|$, $[G:M]$ is not prime. Therefore there is a prime $q$ such that $q^{2} = [G:M]$. Since $q \neq p$ and $q | |G|$, $q < p$. Then $p \nmid q-1$ so $q^{2} \equiv 1 \mod{p}$ implies $q \equiv -1 \mod{p}$. Then, since $q < p$, we have $q = p-1$. Since $p$ and $q$ are both prime, we have $p=3$ and $q=2$. -Since $p=3$ and $p$ is the largest prime factor of $|G|$, $|G| = 2^{a}3^{b}$ for nonnegative integers $a,b$. This means all maximal subgroups of $G$ have index $2$, $3$, $4$, or $9$. The Frattini subgroup $\Phi(G)$ is nilpotent, so, since $G$ is a counterexample, $G/ \Phi(G)$ must be unsolvable. -$\Phi(G)$ is the intersection of all maximal subgroups of $G$. Since the conjugate of a maximal subgroup is also a maximal subgroup, $\Phi(G)$ is the intersection of the cores of the maximal subgroups of $G$. If $N$ and $M$ are normal subgroups of $G$ with $G/N$ and $G/M$ solvable, then $G/(N \cap M)$ is also solvable. Therefore, since $G/ \Phi(G)$ is unsolvable, there is a maximal subgroup $K$ of $G$ such that $G/Core_{G}(K)$ is unsolvable. -The quotient $G/Core_{G}(K)$ is the image, in the symmetric group $S_{[G:K]}$, of the action-on-cosets homomorphism based on the subgroup $K$. Therefore, if $G/Core_{G}(K)$ is unsolvable, the symmetric group $S_{[G:K]}$ must be unsolvable. This means that $[G:K] > 4$, so that $[G:K] = 9$. -Then $G/Core_{G}(K)$ is a transitive (in fact, primitive, since $K$ is maximal in $G$) unsolvable subgroup of the symmetric group $S_{9}$. Also, since $|G| = 2^{a}3^{b}$ for some nonnegative integers $a,b$, $|G/Core_{G}(K)| = 2^{\alpha}3^{\beta}$ for some nonnegative integers $\alpha , \beta$. From now on, we denote $G/Core_{G}(K)$ by $H$. -The contradiction is obtained by showing no such subgroup of $S_{9}$ exists: -First of all, since $H$ is a subgroup of $S_{9}$, we have $\alpha \leq 7$ and $\beta \leq 4$ (from Lagrange's Theorem and the factorization of $9!$). -If $\alpha = 7$, then $H$ is a primitive subgroup of $S_{9}$ containing a whole Sylow $2$-subgroup of $S_{9}$, and thus containing a transposition. Then $H = S_{9}$, contradicting $|H| = 2^{\alpha}3^{\beta}$. Therefore $\alpha \leq 6$. -If $\beta = 4$, then $H$ is a primitive subgroup of $S_{9}$ containing a whole Sylow $3$-subgroup of $S_{9}$, and thus containing a $3$-cycle. Then $H$ contains $A_{9}$, for a similar contradiction. Therefore $\beta \leq 3$. -Since $H$ is transitive on $9$ points, $9 | |H|$ so $\beta \geq 2$. -If $|H| | 864$ (equivalently, if $\alpha \leq 5$), then $H$ is solvable: -It suffices to prove this claim when $|H| = 864$. The number of Sylow $3$-subgroups of $H$ is $1$, $4$, or $16$. If this number is $1$ or $4$, then a Sylow $3$-subgroup normalizer is a solvable subgroup of $H$ having index at most $4$. Any group having a solvable subgroup of index at most $4$ is solvable, so $H$ is solvable. Therefore assume that the number of Sylow $3$-subgroups of $H$ is $16$. -$H$ acts transitively by conjugation on its Sylow $3$-subgroups. A Sylow $3$-subgroup $P$ fixes itself and acts without fixed points on the other Sylow 3-subgroups. Since $9 \nmid 16-1$, one of these suborbits must have exactly $3$ points in it. This gives us two Sylow $3$-subgroups $R, Q$ such that $[R: R \cap Q] = [Q: R \cap Q] = 3$. This means $R \cap Q$ is normalized by both $R$ and $Q$, so that $N_{H}(R \cap Q)$ has more than one Sylow $3$-subgroup of $H$. How many Sylow $3$-subgroups does it have? $4$ or $16$. -If $N_{H}(R \cap Q)$ has $16$ Sylow $3$-subgroups, then $|N_{H}(R \cap Q)|$ is a multiple of $16$ and $27$, so it is a multiple of $432$ and it is $432$ (in which case $N_{H}(R \cap Q)$ is normal in $H$ because its index is $2$) or $864$ (in which case $R \cap Q$ is normal in $H$, $R \cap Q$ is solvable because it is a $3$-group, and $H/(R \cap Q)$ is solvable because a Sylow $2$-subgroup of it is a solvable subgroup of index $3$, and a group with a solvable subgroup of index at most $4$ is solvable). -If $N_{H}(R \cap Q)$ has $4$ Sylow $3$-subgroups, then its order is a multiple of $27$ and $4$, so its order is a multiple of $108$. Then $[H : N_{H}(R \cap Q)] | 8$, and the solvability of $H$ follows (since a group with a solvable subgroup of index at most $4$ is solvable), except possibly when $N_{H}(R \cap Q)$ is a maximal subgroup of $H$. -If $N_{H}(R \cap Q)$ is maximal in $H$, let $H$ act on its cosets and let $L$ be the kernel of this homomorphism from $H$ to the symmetric group $S_{8}$. -Since $27 | |H|$ but $27$ does not divide $8!$, $3 | |L|$. -If $9 | |L|$, then $L$ is solvable because $L$ is a $3$-group and $H/L$ is solvable because a Sylow $2$-subgroup of $H/L$ is a solvable subgroup of index $1$ or $3$. -If $3$ exactly divides $|L|$, then $H/L$ is a primitive subgroup of $S_{8}$ whose order is a multiple of $9$. Then $H/L$ contains a whole Sylow $3$-subgroup of $S_{8}$ and thus $H/L$ contains a $3$-cycle. Then, since $H/L$ is primitive on $8$ points, $H/L$ contains $A_{8}$, contradicting $|H| = 864$. -Now it only remains to handle the cases when $\alpha = 6$, or, equivalently, when $|H| = 576$ or $1728$. -If $|H| = 576$, then $H$ is solvable: -The number of Sylow $3$-subgroups of $H$ is $1$, $4$, $16$, or $64$. -If the number is $1$ or $4$, then $H$ has a Sylow $3$-subgroup normalizer (itself obviously solvable) which has index $1$ or $4$ in $H$. Therefore $H$ is solvable. -If $H$ has $64$ Sylow $3$-subgroups, each is self-normalizing. Since they are abelian (for they have order $9$), the Burnside $p$-complement Theorem applies to show that $H$ has a normal Sylow $2$-subgroup and so is solvable. -Therefore assume that $H$ has exactly $16$ Sylow $3$-subgroups. -As before, we may obtain two Sylow $3$-subgroups $Q, R$ such that $[Q: Q \cap R] = [R: Q \cap R] = 3$. Then let $x$ be chosen so that $ < x > = Q \cap R$. Then the centralizer $C_{H}(x)$ contains two Sylow $3$-subgroups of $H$, so it contains at least $4$ Sylow $3$-subgroups of $H$. In fact, the number of Sylow $3$-subgroups of $H$ it contains is $4$ or $16$. -If it is $16$, then $C_{H}(x)$ is a subgroup of index $4$ in $H$. $C_{H}(x)$ is solvable because $C_{H}(x)/ < x >$ has a Sylow $2$-subgroup of index $3$, so $H$ is solvable. -So assume that the number of Sylow $3$-subgroups of $H$ in $C_{H}(x)$ is $4$. In this case, $C_{H}(x)/ < x >$ is a group of order $12$ which has $4$ Sylow $3$-subgroups, so $C_{H}(x)/ < x > \cong A_{4}$. Then the subgroup $V$ of order $4$ in $A_{4}$ lifts to a subgroup $N$ of order $12$ in $C_{H}(x)$. Since $V$ is normal in $A_{4}$, $N$ is normal in $C_{H}(x)$. Moreover, since $N$ centralizes $x$, $N$ is abelian. Therefore $N$ has a characteristic subgroup $W$ of order $4$ which is normal in $C_{H}(x)$. -But also $W$ is contained in a Sylow $2$-subgroup $P$ of $H$, and $N_{P}(W) > W$. Therefore $N_{H}(W)$ contains $C_{H}(x)$ which has order $36$, and $N_{P}(W)$, whose order is a multiple of $8$. Therefore $|N_{H}(W)|$ is a multiple of $72$, so it is $72$, $144$, $288$, or $576$. -$N_{H}(W)/W$ has order dividing $144$, so it is solvable. Therefore $N_{H}(W)$ is solvable. If $|N_{H}(W)|$ is $144$, $288$, or $576$, then $H$ has a solvable subgroup of index at most $4$ and is therefore solvable. Therefore, assume $|N_{H}(W)| = 72$. -Since the solvability of $H$ follows if $N_{H}(W)$ is not maximal in $H$, assume $N_{H}(W)$ is maximal in $H$. -Then let $H$ act on the cosets of $N_{H}(W)$, and let $L$ be the kernel of the homomorphism obtained from $H$ to $S_{8}$. If $|L|$ is a nonmultiple of $3$, $H/L$ contains a whole Sylow $3$-subgroup of $S_{8}$ and so contains a $3$-cycle. Then $H/L$ is primitive on $8$ points and contains a $3$-cycle, so $H/L$ contains $A_{8}$, contradicting $|H| = 576$. -So assume that $|L|$ is a multiple of $3$. Then $H/L$ is a transitive subgroup of $S_{8}$, so $8 | |H/L|$. This means that $|L| | 72$, so $|L| | 864$ and $L$ is solvable. Then, since $3 | |L|$, $H/L$ has a Sylow $2$-subgroup of index at most $3$ and so is solvable. Therefore $H$ is solvable, as claimed. -We now come to the final case, in which it will be shown that $S_{9}$ has no primitive subgroup $H$ with $|H| = 1728$: -If $H$ has a subgroup of index $2$, then it is solvable and therefore $H$ is solvable. Therefore we assume that $H$ has no subgroup of index $2$, so that $H < S_{9}$ implies $H < A_{9}$. Since $H$ is a transitive group on $9$ points, a point stabilizer in $H$ has order $192$. Since all of the subgroups of order $192$ in $A_{8}$ are conjugate, in $S_{8}$, to the stabilizer, in $A_{8}$, of the partition $12|34|56|78$ of the $8$ indices, we conclude that $H$ is doubly transitive on $9$ points and that any $2$-point stabilizer has a third fixed point. -How many sets of $3$ points arise as the fixed point sets of $2$-point stabilizers in $H$? This number is $\frac{\binom{9}{2}}{\binom{3}{2}} = 12$. Since $H$ is doubly transitive, all the $2$-point stabilizers in $H$ are conjugate in $H$. Therefore $H$ also acts transitively on the $12$ sets of $3$ that arise as fixed point sets of $2$-point stabilizers. -This means that the stabilizer $M$, in $H$, of one of these sets of $3$ has order $144$. -How does $M$ act on the other $6$ points? It can't act faithfully, since $S_{6}$ has no subgroup of order $144$ (for $S_{n}$ has no subgroup of index strictly between $2$ and $n$, except when $n = 4$). Then a nontrivial element of $M$ fixing the $6$ points can only be a $3$-cycle or a transposition, so that $H$ is a primitive permutation group containing a transposition or a 3-cycle and thus $H$ contains $A_{9}$, contradicting $|H| = 1728$. -My question is: what is the easiest way to prove this solvability theorem? (The Burnside $p^{a}q^{b}$ Theorem is too magical via character theory, and too hard without it, for my taste.) - -REPLY [12 votes]: Here is an easier proof. -Starting as you did, we get $|G| = 2^a3^b$. Now let $N$ be minimal normal in $G$. Since the hypothesis on maximal subgroups is inherited by $G/N$, it follows (working by induction on $|G|$) that $G/N$ is solvable, so it suffices to show that $N$ is solvable. We can thus assume that $N$ is neither a 2-group or a 3-group. Let $P$ be a Sylow 3-subgroup of $G$, so $S = P \cap N$ is a Sylow $3$-subgroup of $N$, and thus $1 < S < N$, and hence $S$ is not normal in $G$. Let $M$ be a maximal subgroup of $G$ containing ${\bf N}_G(S)$. Since $P \subseteq M$, we see that $|G:M|$ is a power of $2$, so it is $2$ or $4$. By the Frattini argument, $NM = G$, so $|N:N \cap M|$ is $2$ or $4$. Then $N$ has a proper normal subgroup of index dividing $4!$. Then $N' < N$, so $N' = 1$ and $N$ is abelian. This completes the proof.<|endoftext|> -TITLE: sums of fractional parts of linear functions of n -QUESTION [11 upvotes]: As $\alpha$ and $\gamma$ range uniformly over $[0,1]$, what is the typical (e.g. median or root-mean-square) order of magnitude of $C_m (\alpha,\gamma)$ := $\sum_{1 \leq k \leq m} \left( {\rm frac}(k\alpha+\gamma) - \frac12 \right)$ where frac($x$) denotes the fractional part of $x$? -I'd settle for an answer in the case where $\gamma = 0$. -I know there are articles that address the question where $\alpha$ is fixed (going back to Hardy), but they don't immediately answer my question. Perhaps one could cobble together an answer using results about how the magnitude of $C_m (\alpha,\gamma)$ is bounded in terms of the continued fraction convergents for $\alpha$, along with results about how the convergents grow for a generic real number. - -REPLY [5 votes]: I think I can show that -$$\sum_{1 \leq h,k \leq N} \frac{GCD(h,k)^2}{hk}$$ -grows linearly. But I get the constant is -$$\sum_{ GCD(i,j)=1} \frac{1}{\max(i,j) i j}$$ -This constant is incredibly close to $3$. (I am omitting the $1/12$, so my $3$ is your $0.25$.) My intuition is that they can't be equal, but they agree to a lot of digits, so I am not sure. See below for my computations. -UPDATE: This constant is $3$, due to an identity of Euler. See Marty's answer here. I'll leave the numeric work below for those might be curious how to approximate things like this. - -Let's group the sum according to $GCD(h,k)$. So we have -$$\sum_d \sum_{\substack{1 \leq h,k \leq N \\ GCD(h,k)=d}} \frac{d^2}{hk} = \sum_d \sum_{\substack{1 \leq i,j \leq N/d \\ GCD(i,j)=1}} \frac{d^2}{d^2 ij} = \sum_d \sum_{\substack{1 \leq i,j \leq N/d \\ GCD(i,j)=1}} \frac{1}{ij}$$ -where $h=di$ and $k=dj$. Grouping on $(i,j)$, we have -$$\sum_{\substack{1 \leq i,j \leq N \\ GCD(i,j)=1}} \frac{\lfloor N/\max(i,j) \rfloor}{ij} = N \sum_{\substack{1 \leq i,j \leq N \\ GCD(i,j)=1}} \frac{1}{\max(i,j) i j} + O \left( \sum_{\substack{1 \leq i,j \leq N \\ GCD(i,j)=1}} \frac{1}{i j} \right)$$ -The error term is $O(\log N)^2$, so that's not the dominant term. -Once we check that the sum converges, this will show that your rate of growth is linear with that coefficient. We'll drop the $GCD(i,j)=1$ condition, since that just makes the sum smaller. -$$\sum_{i,j} \frac{1}{\max(i,j) i j} = \sum_{n} \frac{1}{n^2} \left( 2 + \frac{2}{2} + \frac{2}{3} + \cdots + \frac{2}{n-1} + \frac{1}{n} \right) = \sum_{n} n^{-2} O(\log n).$$ -Here $n=\max(i,j)$. The final sum converges by the integral test, so the original one does as well. - -Now, what is the value of this sum? Notice that, if $(i,j) = (g i', g j')$ with $GCD(i', j')=1$, then $\max(i,j)i j = g^3 \max(i',j') i' j'$. So, if we sum over all pairs, instead of just the relatively prime ones, then we multiply by a factor of $\sum g^{-3} = \zeta(3)$. So we want to compute -$$\sum_{1 \leq i,j} \frac{1}{\max(i,j) i j}$$ -and, in particular, we want to know how it compares to $3 \zeta(3)$. As we showed above, we can simplify this sum to -$$\sum_{n} \frac{1}{n^2} \left( 2 + \frac{2}{2} + \frac{2}{3} + \cdots + \frac{2}{n-1} + \frac{1}{n} \right).$$ -Now, is this actually the same as $3 \zeta(3)$? I had Mathematica compute the sum of the first $10,000$ terms, using $20$ digit precision for all intermediate computations. If Mathematica can be trusted, the result is $3.6040133$. Now, $2+2/2+2/3+\cdots +2/(n-1) + 1/n = 2 \log n + 2 \gamma + O(1/n)$. So I approximated the rest of the sum by the integral $\int_{10000}^{\infty} 2 (\log t + \gamma) dt/t^2$. (Here $\gamma$ is the Euler gamma constant.) According to Mathematica, this integral is $0.0021575$. The error in approximating a decreasing sum by an integral is bounded by the first term, which is $1.8 \times 10^{-7}$. The error in approximating the harmonic number by a $\log$ should be something like $\int_{10000}^{\infty} dt/t^3 = 5 \times 10^{-9}$; I don't have the energy to turn this into a rigorous bound. So the sum should be $3.6040133 + 0.0021575 \pm 2 \times 10^{-7} = 3.6061708 \pm 2 \times 10^{-7}$. (That error lines up with the last digit given.) -And what is $3 \zeta(3)$? I kid you not, it is $3.6061707$, right in range! So they might be equal, but, if so, I can't see why. -UPDATE: OK, I went back and improved my approximations in two ways: (1) I replaced the Harmonic number $H_n$ by $\log n + \gamma + (1/2) n^{-1} - (1/12) n^{-2} + (1/120) n^{-4}$ and (2) I approximated the sum of the terms past $10000$ by the first few terms of the Euler-Macluarin approximation, up to the $B_4$ term. The result, doing my internal computations with $20$ digits of accuracy: 3.6061707094787828562. The numerical value of $3 \zeta(3)$: Exactly the same! -Something is going on here. If you don't mind, I'll ask a separate question about what.<|endoftext|> -TITLE: Collecting proofs that finite multiplicative subgroups of fields are cyclic -QUESTION [87 upvotes]: I teach elementary number theory and discrete mathematics to students who come with no abstract algebra. I have found proving the key theorem that finite multiplicative subgroups of fields are cyclic a pedagogical speedbump. For example, Serre's proof in A Course in Arithmetic runs a full page, requires introducing Euler's $\phi$-function, and depends on a counting argument that might seem to beginners too clever or magical for a cornerstone result. -I'd like to have a collection of proofs of this fact, to compare their advantages, -to match their viewpoints to my various audiences, to contrast for my students, etc. -To get the ball rolling, here's the shortest argument I can think of (and if it's in the literature somewhere I'd love a reference). -Induction on the order of the subgroup. So suppose multiplicative -subgroup $G$ of field $F$ has order $n$. If $n=p^k$ with $p$ prime and $G$ isn't -cyclic, all $p^k$ elements of $G$ satisfy $x^{p^{k-1}}-1=0$, impossible. -If $n=ab$, $gcd(a,b)=1$, then $(\cdot)^a:G\rightarrow G$ has a kernel $A$ of size at most $a$ and a range $B$ of size at most $b$ (since the $y\in B$ satisfy $y^b=1$), so $|A|=a$, $|B|=b$, and a product $xy$ of cyclic generators $x,y$ for $A,B$ respectively generates $G$. - -If you know published proofs distinctly different from either of these, please cite a source. No need to spell out the details, but please mention a key feature to help avoid duplicates. If you have your own favorite approach, please share it. - -REPLY [2 votes]: Here's my version of a proof, which seems to me to be very elementary. I'll prove the following assertion, which clearly applies to finite subgroups of the multiplicative group of a field: -Let $G$ be a finite abelian group such that, for every prime $p$, there are at most $p$ elements $g\in G$ satisfying $g^p=1$. Then $G$ is cyclic. -I argue by induction on $n=|G|$. So let $G$ be a non-trivial group satisfying the hypothesis. Then there exists an element of $G$ with order $>1$, and hence (by taking an appropriate power) an element of $G$ with order $p$ for some prime $p$. The main observation I will use is that the $p$th power map on $G$ is a homomorphism, and therefore defines a $p$-to-$1$ surjective function $\phi\colon G\to G^p$ (as the hypothesis on $G$ implies that the kernel of $\phi$ must have order $p$.) -The subgroup $G^p$ has order $n/p -TITLE: Is the Baily--Borel compactification functorial? -QUESTION [5 upvotes]: The following question seems pretty natural, but searching online -and looking in some obvious places didn't turn up much, so maybe -I can ask it here. (Disclaimer: I'm a newcomer to this topic, so -apologies if the question is obviously misguided.) -Suppose that $V_1 = X_1/\Gamma_1$ and $V_2= X_2/\Gamma_2$ are -arithmetic quotients of Hermitian symmetric domains. Let $V_1^\ast$ -and $V_2^\ast$ be their respective Baily--Borel compactifications. Now suppose -we have an analytic map $f: V_1 \rightarrow V_2$. Does it extend -to a morphism $f^\ast: V_1^\ast \rightarrow V_2^\ast$? -If the answer in general is no, are there any nontrivial cases in which it is yes? (For instance, the simplest example that comes to mind is the case where $X_1=X_2$, $\Gamma_1 \subsetneq \Gamma_2$, and $f: V_1 \rightarrow V_2$ is the quotient.) - -REPLY [3 votes]: A useful reference might be the article "Satake Compactification and extension of Holomorphic Mappings", Inv.Math. 16, 237-248, 1972, by Kiernan and Kobayashi. They show that if the map $V_1 \to V_2$ is induced from a map $X_1 \to X_2$ then it extends. In particular, the answer is positive for your "simplest example", though that presumably follows from the construction of the compactification.<|endoftext|> -TITLE: Interpolation of Sobolev spaces -QUESTION [9 upvotes]: I know quite a bit about the abstract theory of Interpolation of Banach spaces. Today I had a colleague from Environmental sciences (who used to be in our Applied Maths department) come and ask me about (complex) interpolation of Sobolev spaces. I was, in the end, able to explain enough to give him a "black box" which seemed to do enough (recover norm estimates for odd cases from formula established for even cases by partial integration techniques). -Now, the book which he'd been pointed to (by another book) was "Interpolation spaces" by Bergh and Lofstrom. I've read this, of course, but it takes a (almost caricatured) pure maths approach: you have to read it cover to cover to catch all the definition etc. So my question is: - -Does anyone know of a "friendly" (applied maths style) approach to complex interpolation of Sobolev (and related function) spaces? - -I'm guessing that this must exist, as it's only a small step from the classical Riesz-Thorin interpolation, which must be used by lots of people who don't particularly care about abstract Banach spaces. -Edit: Perhaps I'm being dismissive or confusing or something about "applied maths style". I don't wish to be! The book my colleague showed me said something like: "The odd case follows by interpolation. (This is not an easy argument, and we do not give it. See, for example, the book of Bergh+Lofstrom.)" I'm sort of taking this as a baseline. Many thanks for the suggestions so far-- I'll leave this open a bit longer, and then accept an answer. - -REPLY [7 votes]: What exactly does your colleague need interpolation for? I guess he just needs to extend some inequalities to intermediate values of the parameters. Then he can use the black box approach and his problem is reduced to computing interpolation spaces between given couples of Banach spaces. Then, two possibilities: -1) The interpolation spaces have already been computed. There is a vast literature on this, and he would not need to really study it but just check the statements. Besides the books already mentioned I would add Bennett and Sharpley, Interpolation of Operators, and a few books by H.Triebel with a similar name (Interpolation is the keyword). -2) In the unlucky case his spaces have not been considered, then he has to delve into the theory a little, and try to compute the spaces himself. Bergh and Lofstrom is a strange book, full of results, but with several imprecisions which can cause the beginner a few nightmares. Better start with Bennett and Sharpley which is crystal clear and reliable, keeping BL on the side for a comparison. Lunardi's book is also quite good but less comprehensive (at least from the early version I have). -3) If all else fails, try real interpolation, Peetre style. The theory is much easier to grasp, and closer to approximation and convexity methods he might be familiar with.<|endoftext|> -TITLE: Does the following series converge? -QUESTION [6 upvotes]: Does the following series converge? $\sum_{n=1}^{\infty} \vert \sin n \vert ^{n}$ - -REPLY [6 votes]: The question has basically been answered in the comments by David Speyer and SJR. It is a theorem of Chebyshev that that for any irrational $\alpha$ and any real $\beta$, the inequality -$$|\alpha n - k - \beta| < 3/n$$ -has infinitely many solutions. In particular, take $\alpha = 1/(2\pi)$ and $\beta = \frac12$. Then one gets that $n$ is so close to an odd multiple of $\pi$ that $|\sin n|^n$ converges to 1 for these values. Even if you took $|\sin n|^{n^2}$, these values would be bounded away from 0. Certainly if the terms of a series do not converge to 0, then the series does not converge.<|endoftext|> -TITLE: What does Rng^{op} look like? -QUESTION [9 upvotes]: There are several well-known dualization results in category theory, i.e. that such-and-such a well-known category D is isomorphic to the opposite C^{op}. Does anyone know of such a result concerning what the opposite category to Rng, rings (*-monoid on +-group) with their homomorphisms, looks like? -I ask (naively I should add) because I'm curious if there's a natural "algebraic" structure on homology dual to that of the ring-structure of cohomology. - -REPLY [33 votes]: Here is a too-serious answer to your question, along with answers to a couple questions I think you should be asking: -The category you're interested in, as noted by others, is the category of coalgebras / corings, which is emphatically not the opposite category of rings --- but we're going to see exactly what's different between the two in the nicest case. To start things off, here's a definition of a coalgebra so that we're all on the same page: an $R$-coalgebra is an $R$-module $A$ together with morphisms $\Delta: A \to A \otimes_R A$ and $\epsilon: A \to R$ satisfying - -Coassociativity: $(\Delta \otimes \operatorname{id}) \circ \Delta = (\operatorname{id} \otimes \Delta) \circ \Delta$, -Counitality: $(\epsilon \otimes \operatorname{id}) \circ \Delta = (\operatorname{id} \otimes \epsilon) \circ \Delta = \operatorname{id}$ after identifying $R \otimes_R A$ and $A \otimes_R R$ with $A$. - -Coalgebras are familiar objects in algebraic topology, and you've already found the biggest source of them. Suppose you have a cohomology theory $E$ whose coefficient ring is a (graded) field. For any space $X$, the constant map $X \to \mathrm{pt}$ induces a map in homology $E_* X \to E_*$ which serves as the counit for $E_* X$. Toward a comultiplication, we have maps $E_* X \xrightarrow{\Delta} E_* (X \times X) \leftarrow E_* X \otimes_{E_*} E_* X$, but asking for the right-hand map to have an inverse is the same as asking for a Kunneth isomorphism. This only happens under restrictions on $X$ or restrictions on $E$ --- such as when $E_*$ is a field, which is one reason we requested that. -Now, let's lay our cards on the table and just announce some dualities we see in front of us: - -There's your opposite category $\mathsf{Rings}^{op}$, which is dual to rings in the sense that $(\mathsf{Rings}^{op})^{op}$ is equal to $\mathsf{Rings}$. -In the language of the comments below, coalgebras are Eckmann-Hilton duals of algebras. This is really a statement about how we produced the definition above: we took the definition for an algebra, and we flipped all the arrows around. -There's also the notion of linear algebraic duals: $V^\vee = \operatorname{Hom}_k(V, k)$. To avoid some very serious technicalities, we'll want to work in the nicest, most familiar setting possible: modules of finite rank over a ground ring that's a field. - -Now, we would like to compare these three ideas. There is another category of interest floating around: the category of $k$-algebras has an associated category of presheaves $\widehat{\mathsf{Algebras}_k}$ $=$ $\operatorname{Functors}(\mathsf{Algebras}_k, \mathsf{Sets})$ which receives a map $\mathsf{Algebras}_k^{op} \to \widehat{\mathsf{Algebras}_k}$ described by the left side of the $\operatorname{Hom}$-functor: $X \mapsto \operatorname{Hom}(X, -)$. This assignment, called the Yoneda embedding, is a functor into a cocomplete category which is an equivalence onto its image and whose image is codense --- these are consequences of the Yoneda lemma. That the target of the Yoneda embedding is cocomplete makes it a much nicer category to play around in, and so it's worth considering what this embedding's use is. -I claim there's a relation between the category of $k$-coalgebras and $\widehat{\mathsf{Algebras}_k}$. Again, to make linear algebra behave nicely, we need to encode finiteness restrictions into our setup, and to make that happen we'll turn to "$k$-formal schemes". The classical $\operatorname{Spec}$ construction in algebraic geometry also gives a contravariant functor off the category of $k$-algebras which is an equivalence onto its image. Rather than fussing with what a Zariski spectrum is, since we're just playing around with categories, I will instead take the Yoneda embedding to be my definition of $\operatorname{Spec}$ and the presheaf category to be something dimly, vaguely, sorta like the category of schemes. Representable presheaves (i.e., those in the image of $\operatorname{Spec}$) are called affine schemes. Plenty of constructions from algebraic geometry transfer almost without comment; for instance, defining $\mathbb{A}^1 = \operatorname{Spec} \mathbb{Z}[x]$, we recover the functor $\mathcal{O}(X) = \operatorname{Hom}_{\widehat{\mathsf{Algebras}_k}}(X, \mathbb{A}^1)$, which in the case of an affine $X = \operatorname{Spec}(A)$ gives $\mathcal{O}(\operatorname{Spec} A) \cong A$. -A scheme $X$ will be called finite if it is $\operatorname{Spec}$ of an algebra of finite dimension as a $k$-module. These, too, are in ample supply in algebraic topology. If $X$ is a compact pointed space, then the algebra $H^* X$ will be finite in the sense we need. Of course, algebraic topology gets done on more than compact spaces, so we need to broaden our perspective a little bit: we can ask instead that $X$ be compactly generated, so that if $X_\alpha$ denotes the collection of compact subsets of $X$ directed by inclusion, we then have $X = \operatorname{colim} X_\alpha$. In the case that $X$ is a CW-complex, it is sufficient to take $X_\alpha$ to be just the finite subcomplexes of $X$. We might then be interested in the scheme $X_E := \operatorname{colim} \operatorname{Spec} E^* X_\alpha$. Such a scheme which occurs as the colimit of a directed system of finite $k$-schemes is called a $k$-formal scheme. -So now we have a suitable notion of a 'finite' scheme that still captures all our interesting (and frequently large) cohomology rings. Here's the comparison I promised early on: the functor taking a formal scheme $\operatorname{colim} \operatorname{Spec} A_i$ to the $k$-coalgebra $\operatorname{colim} A_i^\vee$ is an equivalence of categories. There's almost nothing to say because the sea of definitions we've made take care of most everything, but there is one key lemma: given a $k$-coalgebra, you need to know that you can write it as the colimit of finite $k$-coalgebras. The exact lemma that gets used is: if $E$ is a finite dimensional subspace of a $k$-coalgebra $A$, then there exists a subcoalgebra $F \subseteq A$ which is finite as a $k$-vector space and which contains $E$ (i.e., $E$ can be finitely enlarged so that it becomes closed under comultiplication). If you push around elements a bit you'll see that that's the case (and that this requires working over a ground field). Then, straight after, here's the second big assertion: the assignments $X \mapsto E_* X$ and $X \mapsto X_E$ are equivalent under the above equivalence of categories. -Now, this construction is delicate, and the limitations are not meant to be taken lightly! You need a ring structure on the underlying (co)homology theory to even dream of having products, you need Kunneth isomorphisms to make sense of the coalgebra structure, you need a coefficient field to have good linear-algebraic duality, and there are still potential problems with supercommutativity that we haven't addressed. But, when all the stars align and God smiles on us, this is what the coalgebra structure on homology is supposed to mean: it's another presentation of the formal scheme associated to the (perhaps more familiar) ring structure on cohomology. -To counterbalance that caveat, that's not to say that this point of view is not immensely useful. Here are some sample applications: - -All of chromatic homotopy: $\mathbb{C}\mathrm{P}^\infty$ ($ = B\mathrm{U}(1) = \mathrm{Gr}_1$) carries the structure of an $H$-group, and there are a whole sea of cohomology theories $E$, called complex-orientable, for which $\mathbb{C}\mathrm{P}^\infty_E$ is (noncanonically) isomorphic to $\hat{\mathbb{A}}^1 = \operatorname{colim} \operatorname{Spec} E^*[x] / \langle x^n \rangle$, an object which behaves a lot like an infinitesimal one-dimensional Lie group. This "formal Lie group" carries an immense amount of information about the cohomology theory $E$, and the "space" of available formal Lie groups carries an immense amount of information about stable homotopy theory as a whole. -There are a variety of partial theorems in the following spirit: if $E$ and $F$ are complex oriented cohomology theories and $F$ is represented by the spaces $F_k$ in the sense that $F^k X = [X, F_k]$, then $\bigoplus_k E_* F_k$ behaves like $\operatorname{Hom}(\mathbb{C}\mathrm{P}^\infty_E, \mathbb{C}\mathrm{P}^\infty_F)$. Goerss has shown this when $E = H\mathbb{F}_p$ and $F$ satisfies a certain condition on $F_* \Omega^2 S^3$ (which is also satisfied for a class of complex-oriented spectra called Landweber exact). (Addendum: Goerss spends a lot of time setting up a "super" version of Dieudonne modules, which is meant to address in part issues with supercommutativity ignored/avoided here.) -Armed with an ample supply of Kunneth isomorphisms, Ravenel and Wilson, the progenitors of the idea above, computed these coalgebras in the cases where $E$ and $F$ ranged in $K(n)$ (Morava $K$-theory), $E(n)$ (Johnson-Wilson theories), $BP$ (Brown-Peterson theory), and $H\mathbb{Z}/p^j$ (singular theories / Eilenberg-Mac Lane spaces). For instance, one can define the free (supercommutative) algebra on a group object in the category of $k$-formal schemes, and it turns out that $H_*(K(\mathbb{Z}/p^j, q); \mathbb{F}_p)$ is the free alternating algebra in this sense on the formal group scheme $B\mathbb{Z}/p^j_{H\mathbb{F}_p}$. A similar statement can be made for $K(n)_* K(\mathbb{Z}/p^j, q)$. -The above ideas have stable versions too: the homology of spectra $E_* F$ is to be thought of as the scheme of isomorphisms $\operatorname{Iso}(\mathbb{C}\mathrm{P}^\infty_E, \mathbb{C}\mathrm{P}^\infty_F)$, and so, for instance, the dual of the Steenrod algebra, an object of classical interest, can be thought of as $\mathcal{O}$ of the automorphisms of a particular formal Lie group $\hat{\mathbb{G}}_a$. Relatedly, the statement that the dual Steenrod algebra coacts on the homology coalgebras $H_* X$ is straightened out in the category of formal schemes by saying that $\operatorname{Aut} \hat{\mathbb{G}}_a$ acts on the formal scheme $X_{H\mathbb{F}_2}$. This has also received classical interest, though not in this language: the final part of Thom's thesis on calculations in the real bordism ring amount to showing that this $\operatorname{Aut} \hat{\mathbb{G}}_a$ action on the homology of the real bordism spectrum is free. -Finally, this is an example of a broader phenomenon: often the schemes associated to rings have enlightening interpretations in moduli theoretic terms. A great many classical objects in stable homotopy theory lead double lives in this framework as moduli spaces, and the geometry of the moduli space frequently informs us on how the original topological objects behave. - -Anyway, this is all to say that you should definitely care deeply about the coalgebra structure on homology, as it's one way to get into formal schemes, where everything is interesting and magical and great.<|endoftext|> -TITLE: Riemannian metrics on non-paracompact manifolds -QUESTION [12 upvotes]: After proving the existence of Riemannian metrics on manifolds, one of the students asked if the "paracompactness" is necessary. Of course the standard proof with the partition of unity -uses this assumption, and for the only non-paracompact manifold I know, namely the long line, a similar argument seems to show the existence of a metric. So, just out of curiosity, does there exist a "manifold" (of course, non-paracomapact) which does not admit any Riemannian metric? - -REPLY [21 votes]: On the contrary, the long line does not have a Riemannian metric. Every countable subset of the long line has a least upper bound, so if it were Riemannian then a geodesic ray in the long direction would have to stop short of the end. But then the Riemannian metric would break down at the "endpoint" of this geodesic. -More generally, it is a theorem of Stone that every metric space is paracompact, hence every Riemannian metric space is too.<|endoftext|> -TITLE: Is there evidence whether undergraduate math courses improve problem-solving? -QUESTION [12 upvotes]: The most commonly stated reason for why mathematics should be a required condition for graduating is }to improve problem-solving skills". Usually it's taken for granted that taking a mathematics course does improve one's ability to solve problems. Does anyone know of any studies that either back that up or contradict it? -Edit: I would also be interested in studies backing up claims that taking a math course improves logical reasoning, especially for mathematics courses for non-majors. - -REPLY [5 votes]: (I don't think my answer directly answers the question, but I'm hoping it would be useful.) -I assume that when you say "problem solving" you mean mathematical "problem-solving as a skill" ("being able to obtain solutions to the problems other people give you to solve," Schoenfeld, 1992). -I was unable to find any studies that answer the question "Does taking an ordinary undergraduate mathematics course improve one's ability to solve (mathematical) problems?" (where ordinary means the instruction is not explicitly targeted at improving problem solving skills). -But there have been studies that show that undergraduates taking certain "problem-solving courses" experienced "marked shifts in [their] problem solving behavior" (e.g., Schoenfeld, 1987, p. 207). -As I understand it, researchers in mathematics education usually don't consider questions of the type "does the ordinary way of teaching improve this skill/understanding?" important (where "ordinary" is usually referred to as "traditional"). They usually consider it more valuable to ask questions of the type "what way of teaching will improve this skill/understanding?" -A good reference is -Schoenfeld, A. H. (1992). Learning to think mathematically: Problem solving, metacognition, and sense-making in mathematics. In D. Grouws (Ed.), Handbook for Research on Mathematics Teaching and Learning (pp. 334-370). New York: MacMillan. -which uses some material from -Schoenfeld, A. H. (Ed.). (1987). Cognitive Science and Mathematics Education. New Jersey: Erlbaum. -Chapter 2 (Foundations of cognitive theory and research for mathematics problem-solving, by E. A. Silver) and Chapter 8 (What's all the fuss about metacognition? by A. H. Schoenfeld) of the 1987 Schoenfeld book are particularly useful.<|endoftext|> -TITLE: What is the shortest Ph.D. thesis? -QUESTION [48 upvotes]: The question is self-explanatory, but I want to make some remarks in order to prevent the responses from going off into undesirable directions. -It seems that every few years I hear someone ask this question; it seems to hold a perennial fascination for research mathematicians, just as quests for short proofs do. The trouble is that it has strong urban-legend tendencies: someone will say, "So-and-so's thesis was only $\epsilon$ pages long!" where $\epsilon \ll 1$. It will often be very difficult to confirm or disconfirm such claims, since Ph.D. theses are often not even published, let alone readily available online. If you Google around for a while, as I did, you will find many dubious leads and can easily waste a lot of time on wild goose chases. Frankly, I'm a bit fed up with this state of affairs. I am therefore asking this question on MO in the hope that doing so will put this old question to rest, or at least establish provable upper bounds. -I would therefore request that you set yourself a high standard before replying. Don't post a candidate unless you're sure your facts are correct, and please give some indication why you're so sure. Read the meta discussion before posting. (Note that the meta discussion illustrates that even a MathSciNet citation isn't always totally definitive.) Include information about the content and circumstances of the thesis if you know it, but resist the temptation to gossip or speculate. -I'm not making this question community wiki or big-list because it should ideally have a definite answer, though I grant that it's possible that there are some borderline cases out there (perhaps there are theses that were not written in scholarly good faith, or documents that some people would regard as equivalent to a Ph.D. thesis but that others would not, or theses in subjects that are strictly speaking distinct from mathematics but that are arguably indistinguishable from mathematics dissertations). -Finally, to anticipate a possible follow-up question, there is a list of short published papers here (search for "Nelson"). Note that the question of the shortest published paper is not as urban-legendy because the facts are easier to verify. I looked up the short papers listed there myself and found them to be quite interesting. So in addition to trying to settle an urban legend, I am hoping that this question will bring to light some interesting and lesser known mathematics. - -REPLY [44 votes]: David Rector's thesis ("An Unstable Adams Spectral Sequence", MIT 1966) is 9 pages, according to the record at the MIT library. I haven't seen the actual thesis for many years, but I'm pretty the actual mathematical content takes about 3 pages total, and is largely identical to the published version in Topology (1966, same title), which is 3 pages plus bibliography. (Dan Kan, his advisor, likes short papers.) - -REPLY [9 votes]: I already posted this on meta where there was some discussion of whether the page count was correct. My guess is that it is, so I will post it here too: -MR2615548 -Martens, Henrik Herman Buvik -A NEW PROOF OF TORELLI'S THEOREM. -Thesis (Ph.D.)–New York University. 1962. 12 pp.<|endoftext|> -TITLE: Decomposition of induced representations in S_n -QUESTION [7 upvotes]: Let C be a cyclic subgroup of S_n. -How does the representation $Ind_C^{S_n}\rho$, where $\rho$ is some representation of $C$, decompose into irreducible components? -Is there are a way to know which components appear with multiplicity 1? - -REPLY [8 votes]: There is a combinatorial way to decompose $Res_C^{S_n}S^\lambda$ for an irreducible $S_n$-module $S^\lambda$. We use the notion of the "major index" for a standard tabuleau of shape $\lambda$. If $C=C_n$, the result is obtained by Kraśkiewiz-Weyman. See also Garsia's paper "Combinatorics of the free Lie algebra and the symmetric group"(Theorem 8.4) and Reutenauer's book "Free Lie algebras"(Theorem 8.8 and 8.9). This result can be generalized to any cyclic subgroup in Jöllenbeck-Schocker's paper "Cyclic characters of symmetric groups".<|endoftext|> -TITLE: On a proof of the existence of tubular neighborhoods. -QUESTION [5 upvotes]: Studying analysis on manifolds, I have found, in the proof of the existence of tubular neighborhoods, a reference to theorem 3.1.2 in "Topologie algebrique et theorie des faisceaux" of Godement. -Without going through the machinery of the sheaves, at least now, is it possible to bypass the Godement's result? -And, if yes, what is an accessible (not sheaf-theoretic) route? -This is the initial setting: -$J:N\rightarrow M$ is a smooth embedding. -$\pi:E\rightarrow N$ is a vector bundle, and $s_0:N\rightarrow E$ is the zero section of $\pi$. -$\psi:U\rightarrow M$ is a smooth map from a neighborhood $U$ of $s_0(N)$ in $E.$ -$\psi$ is a local diffeomorphism in each point of $s_0(N),$ and $\psi\circ s_0=j$. -At this point there is a reference to the argument of Godement in order to prove that: -(*)There exists a neighborhood $V$ of $s_0(N)$ in $U$ such that $\psi|_V$ is a diffeomorphism. -What is the argument (differential geometric, not sheaf-theoretic) in order to conclude (*)? -Added by Mariano: I now have a copy in my hands. Theorem 3.1.2 reads (my translation): - -Let $$0\to\mathscr L'\to \mathscr L\to\mathscr L''\to0$$ be a short exact sequence of sheaves of abelian groups. If $\mathscr L'$ is flasque, then for all open sets $U$ there is a short exact sequence $$0\to\mathscr L'(U)\to \mathscr L(U)\to\mathscr L''(U)\to0$$ - -He remarks that we therefore have a short exact sequence of pre-sheaves. - -REPLY [19 votes]: In the finite-dimensional setting, it's possible to construct tubular neighborhoods without anything like Godement's lemma. Many sources simply rely on a point-set topology argument that's based on the same idea as Godement's lemma (to be precise, I'm talking about the argument on p. 109 of Lang's book Differential and Riemannian Manifolds, which he says follows Godement). I'll explain another approach. -The idea is to use a Riemannian metric on the manifold $M$, which also induces a Riemannian metric on $TM$ (viewed as a manifold in its own right). The geodesic distance then gives a (topological) metric on $TM$. If $Y\subset M$ is a (not necessarily closed) submanifold, then a simple metric geometry argument can then be used to find a neighborhood of the zero section of $N(Y)$ (thought of as the perpendicular complement of $TY$ inside $TM$) on which the exponential map is injective. The key fact about the exponential map $f$ is that every point in the zero section of $N(Y)$ has a neighborhood on which $f$ is a diffeomorphism onto an open subset of $M$. (Edit: Note that in the finite dimensional setting, $N(Y)$ is automatically a locally trivial vector bundle. This does not seem to be the case for arbitrary infinite dimensional Riemannian manifolds, as discussed here: Orthogonal complements in Hilbert bundles. Hence the discussion that follows does not work in as great generality as arguments based on Godement's lemma.) -The general metric geometry fact is this: - -Consider a metric space $T$ and subspaces $X, Y$, and $D$ such that $Y \subset X$ and $Y\subset D$. (Think: $T = TM$, $X$ is the zero section of $TM$, $Y$ is a submanifold of $M$, and $D$ is the domain of the exponential map, lying inside $NY$.) Let $f: D\to X$ be a continuous map that restricts to the identity on $Y$ (think: $f$ is the exponential map). Assume further that for each $y\in Y$ there exists $\epsilon_y >0$ such that $f$ restricted to $B_{\epsilon (y)} (y, D) = \{z\in D \,:\, d(z,y) < \epsilon(y)\}$ is a homeomorphism onto an open subset of $X$. Then there exists a subspace $D'$, open in $D$, on which $f$ is injective. - -Proof. For each $y\in Y$, $f(B_{\epsilon (y)/2} (y, D))$ is open in $X$, hence contains $B_{\epsilon'(y)} (y, X)$, for some $\epsilon'_y < \epsilon_y/4$ (remember that $f(y) = y$). Now consider the inverse image $Z_y$ of $B_{\epsilon'_y} (y, X)$ under the restriction of $f$ to -$B_{\epsilon (y)/2} (y, D)$. Since $f$ is a homeomorphism when restricted to this ball, $Z_y$ is open as a subset of $D$. Now I claim that $f$ is injective on $D' = \bigcup_{y\in Y} Z_y$. Say $f(z_1) = f(z_2) = y_0$ with $z_1 \in Z_{y_1}$ and $z_2\in Z_{y_2}$, and assume $\epsilon_{y_1} \geq \epsilon_{y_2}$. Then we have -$$d(z_2, y_1) \leq d(z_2, y_2) + d(y_2, y_0) + d(y_0, y_1) < \epsilon_{y_2}/2 + \epsilon'_{y_2} + \epsilon'_{y_1} $$ -$$< \epsilon_{y_1}/2 + \epsilon_{y_2}/4 + \epsilon_{y_1}/4 \leq \epsilon_{y_1}$$ -(for the second inequality, note that by definition, $y_0 = f(z_i) \in f(Z_{y_i}) \subset B_{\epsilon'_{y_i}} (y_i, X)$ for $i=1, 2$). -So $z_2$ and $z_1$ both lie in $B_{\epsilon_{y_1}} (y_1, D)$, and since $f$ is injective on this ball we have $z_1 = z_2$. -This argument is useful for the construction of equivariant tubular neighborhoods in certain infinite-dimensional settings. See http://arxiv.org/abs/1006.0063; I just updated it so I guess the new version will show up tomorrow. The equivariant version of the above argument is in Proposition 2.3 or 2.4, depending on the version.<|endoftext|> -TITLE: Are there any notion of 'almost primes' known to have small gaps? -QUESTION [12 upvotes]: A notorious question with prime numbers is estimating the gaps between consecutive primes. That is, if $(p_n)_{n \geq 1}$ is the canonical enumeration of the primes, then set $g_n = p_{n+1} - p_n$. It is shown that $g_n > \frac{c \log(n) \log \log(n) \log \log \log \log(n)}{(\log \log \log(n))^2}$ infinitely often, but a precise estimate is not known. -My question is, is there a 'natural' superset of the primes that are of interest (say, the set of numbers that are either primes or product of two primes) such that the gap between consecutive members is well known or well estimated? - -REPLY [5 votes]: How about practical numbers? -They aren't a superset of primes, but they are a "notion of 'almost primes'" as the title requests. -Hausman and Shapiro showed in 1984 that practical number gaps satisfy $\frak{g}$$_n < 2 \cdot \frak{p}$$_n^\frac{1}{2} + 1$. On the other hand, for primes, the best known bound is the Baker-Harman-Pintz or BHP bound published in 2000: $g_n \in p_n^{0.525} + O(1)$. Conditional on the Riemann hypothesis we have $g_n \in p_n^{\frac{1}{2}+o(1)}$, but not $g_n \in O(p_n^\frac{1}{2})$.<|endoftext|> -TITLE: Consistency strength needed for applied mathematics -QUESTION [21 upvotes]: Given that we can never proof the consistency of a theory for the foundations of mathematics in a weaker system, one could seriously doubt whether any of the commonly used foundational frameworks (ZFC or other axiomatisations of set theory, second-order PA, type theory) is actually consistent (and hence true of some domain of objects). One of the ways to justify a certain framework for the foundations of mathematics is by adopting an empiricist stance in the philosophy of mathematics and argue that mathematics must be right because it correctly explains natural phenomena that we observe (i.e. is needed in empirical sciences), and that hence some foundational framework unifying our mathematical knowledge is justified. -Now different foundational frameworks have different consistency strengths. For example, ZFC with some large cardinal axiom (which one might want to accept in order to do category theory more comfortably) has a greater consistency strength than just ZFC. The above justification would only justify a foundational framework of a given consistency strength if that consistency strength is needed for some application of mathematics to empirical sciences. -Have there been any investigations into the question which consistency strength in the foundational framework is needed for applied mathematics? Is there any application of mathematics to empirical sciences which requires a large cardinal? Is maybe something of consistency strength weaker than ZFC enough for applied mathematics? Have any philosophers of mathematics asked questions like these before? - -REPLY [8 votes]: Edited 3/10/2017 -Very large cardinals around the rank-into-rank area potentially have applications in cryptography. Rank-into-rank cardinals produce self-distributive algebras which may be used as platforms or to produce platforms for authentication schemes and key exchange protocols. Therefore, if these self-distributive algebras are seriously considered as platforms for new cryptosystems, then one will need all of the large cardinal hierarchy in order to research applied mathematics. -Recall that a rank-into-rank embedding is an elementary embedding $j:V_{\lambda}\rightarrow V_{\lambda}$. Let $\mathcal{E}_{\lambda}$ denote the collection of all rank-into-rank embeddings $j:V_{\lambda}\rightarrow V_{\lambda}$. Define a binary operation $*$ on $\mathcal{E}_{\lambda}$ by letting $j*k=\bigcup_{\alpha<\lambda}j(k|_{V_{\alpha}})$. Then $(\mathcal{E}_{\lambda},*)$ satisfies the self-distributivity identity $j*(k*l)=(j*k)*(j*l)$. If $\gamma$ is a limit ordinal with $\gamma<\lambda$, then define a congruence $\equiv^{\gamma}$ on $(\mathcal{E}_{\lambda},*)$ by letting -$j\equiv^{\gamma}k$ if and only if $j(x)\cap V_{\gamma}=k(x)\cap V_{\gamma}$ whenever $x\in V_{\gamma}$. Then my question and answer shows that the algebra $(\mathcal{E}_{\lambda}/\equiv^{\gamma},*)$ is locally finite and hence locally computable. -The non-abelian group based cryptosystems can generally be modified to produce cryptosystems that could use any left-distributive algebra as a platform. For example, in this paper, the authors have modified the Anshel-Anshel-Goldfeld key exchange to produce a cryptosystem that could use any left-distributive algebra as a platform. Furthermore, the Ko-Lee key exchange also could be modified to produce a cryptosystem for left-distributive algebras. In this paper, Using shifted conjugacy in braid-based cryptography, Dehornoy proposes an authentication scheme that can use any self-distributive algebra as a platform (though you need to tweak this cryptosystem if the self-distributive algebra is not left-cancellative) rather than just conjugacy on groups (Dehornoy had the platform shifted conjugacy on braid groups in mind, but this platform was later shown to be insecure). -Dehornoy in 4 remarks that the combinatorial complexity of the Laver tables and the fact that the classical Laver tables produce extremely fast growing functions suggests that the classical Laver tables or similar structures may be a good platform for his authentication scheme or some other cryptosystem. However, the classical Laver tables are currently a very insecure platform for all cryptosystems. First of all, $A_{48}$ is the largest classical Laver table which has even been computed. Therefore, the classical Laver table based cryptosystems only provide at most 48 bits of security. Furthermore, the homomorphisms between the classical Laver tables allow one to easily write computer programs that break all cryptosystems based on the classical Laver tables. My generalized Laver tables which you can compute online here are not a secure platform for self-distributive algebra based cryptosystems either since it is easy to factorize elements in generalized Laver tables. Fortunately, unless I have simply overlooked something, the ternary Laver tables so far appear to be plausible platforms for these self-distributive algebra based cryptosystems. Click here for a ternary Laver table calculator on my website. -Of course, it is way to soon to comment on the security or the insecurity of these ternary Laver table based cryptosystems, and much more research needs to be done on cryptography based on Laver-like algebra. These line of investigation have barely been researched, but I have recently proposed a polymath project to promote an investigation into these directions of inquiry.<|endoftext|> -TITLE: Amenability of groups II -QUESTION [14 upvotes]: Are there any non-amenable group $G$ with the property: -There exists $C<1$ such that for every finite set $S\subset G$ there exists a set $F\subseteq S$ such that $|F|\geq C |S|$ and $F$ generates an amenable group. -UPDATE: -Here is some motivation for the question: https://mathoverflow.net/questions/55075/amenability-of-groups-iii - -REPLY [9 votes]: A few thoughts. First you can assume that your group is finitely generated. Second, it does not contain free non-amenable subgroups, free Burnside groups, etc. Moreover, every subgroup of your group also has the same property. Now, you can look at the (not very large) list of known counterexamples to von Neumann's problem (non-amenable groups without free subgroups) to convince yourself that an example of such a group is not known. I strongly suspect that examples do not exist, that is every group satisfying your property must be amenable. I think that the following property may hold for all finitely generated groups: for every infinite finitely generated group $G$ and every $\epsilon\gt 0$ there exists a finite subset $S\subset G$ such that every subset of $S$ containing at least $\epsilon|S|$ elements generates $G$. (By Gromov's thesis this statement, if true, must be trivial.) For example, in the case of $\mathbb Z$, any sufficiently large (of cardinality bigger than $\frac{2}{\epsilon}$) set of primes is such. If it is true, then every group with your property is amenable. -For the free group $F(a,b)$, the set can be constructed as $S=\{a^pb^q, p,q\in I\}$, where $I$ is any set of natural numbers such that for every $x_1\lt x_2 \lt...\lt x_n\in I$ the numbers $x_2-x_1, x_3-x_1,...,x_n-x_1$ are relatively prime (here $n$ depends on $\epsilon$, $|I|$ also depends on $\epsilon$). Indeed, if $I$ is large enough, a subset with $\ge \epsilon|S|$ elements will contain $n$ elements of the form $a^pb^{q_1}, a^pb^{q_2},..., a^pb^{q_n}$ where $q_1\lt q_2\lt...\lt q_n$ are from $I$. Then the subgroup generated by these elements will contain $b$. Similarly it will contain $a$. Instead of the free group, one can take any group $G$ generated by $a,b$ where the set $S$ is infinite. It might be enough to prove my conjecture in general. - Edit. One needs to be more careful in choosing the set $S$ for the free group, but the general idea seems to be correct.<|endoftext|> -TITLE: Is there an odd-order group whose order is the sum of the orders of the proper normal subgroups? -QUESTION [53 upvotes]: For a finite group G, let |G| denote the order of G and write $D(G) = \sum_{N \triangleleft G} |N|$, the sum of the orders of the normal subgroups. I would like to call G "perfect" if D(G) = 2|G|, since then the cyclic group of order n is perfect if and only if the number n is perfect. But the term "perfect group" is taken, so I'll call such a group immaculate. -My question is: - -Does there exist an immaculate group of odd order? - -Since the cyclic immaculate groups correspond one-to-one with the perfect numbers, a "no" answer would immediately prove the famous conjecture that there are no odd perfect numbers. However, perhaps someone can easily see that there is a non-cyclic immaculate group of odd order, proving that the answer is "yes". -Here's what I know. There are no abelian immaculate groups except for the cyclic ones. (Edit: more generally, if $D(G) \leq 2|G|$ then every abelian quotient of $G$ is cyclic. Proof: not hard, and given here.) However, there do exist nonabelian immaculate groups, e.g. $S_3 \times C_5$ (of order 30). Derek Holt has computed all the immaculate groups of order less than or equal to 500. Their orders are -$$ -6, 12, 28, 30, 56, 360, 364, 380, 496 -$$ -(Integer Sequence A086792). Of these, only 6, 28 and 496 are perfect numbers; the rest correspond to nonabelian immaculate groups. Some nonabelian immaculate groups of larger order are also known, e.g. $A_5 \times C_{15128}$, $A_6 \times C_{366776}$, and, for each even perfect number n, a certain group of order 2n. But these, too, all have even order. -Edit: Steve D points out that p-groups can never be immaculate. This also appears as Example 2.3 here; it follows immediately from Lagrange's Theorem. I should have mentioned this, as it rules out an easy route to a "yes" answer. - -REPLY [14 votes]: Almost a decade late to the party, but here is another example: -$$(C_7 \rtimes C_{3^2}) \times (C_{19^2} \rtimes C_5) \times C_{11^2 \cdot 197} = \text{SmallGroup}(63, 1) \times \text{SmallGroup}(1805, 2) \times C_{23837}.$$ -where $\text{SmallGroup}(a,b)$ denotes the $b$th group of order $a$ according to the database of GAP. Then: -$$\frac{D(C_7 \rtimes C_{3^2})}{|C_7 \rtimes C_{3^2}|}\frac{D(C_{19^2}\rtimes C_5)}{|C_{19^2}\rtimes C_5|}\frac{D(C_{11^2})}{|C_{11^2}|}\frac{D(C_{197})}{|C_{197}|} = -\frac{95}{63}\cdot \frac{2167}{1805}\cdot \frac{133}{121}\cdot \frac{198}{197} $$ -$$=\frac{5 \cdot 19}{3^2\cdot 7} \cdot\frac{11 \cdot 197}{5\cdot19^2}\cdot \frac{7 \cdot 19}{11^2}\cdot \frac{2 \cdot3^2\cdot 11}{197}=2.$$ -This group is of order $2710624455$ and is thus about 100 times smaller than François Brunault's example. -I found this group for a project for one of my courses. Immaculate groups seem to be very rare in general, examples of order are even harder to find. For further results, see http://www.math.ru.nl/~bosma/Students/JorisNieuwveld/A_note_on_Leinster_groups.pdf<|endoftext|> -TITLE: Can Eisenstein's lattice point proof of quadratic reciprocity be generalized? -QUESTION [22 upvotes]: I'm referring to this proof. The key formula ("Eisenstein's Lemma") is -$$\left(\frac{q}{p}\right)=(-1)^{\sum_{u}\lfloor\frac{qu}{p}\rfloor},\text{ where $u=2,4,\ldots,p-1$}$$ -The sum in the exponent is easily seen to count the number of lattice points in this rectangle that are below the diagonal and having even $x$-coordinate, - -where $p$ and $q$ are our primes in question, and via some clever flipping, quadratic reciprocity pops out. -I seem to recall in my first summer at PROMYS, some counselors tried to work out a version for quartic reciprocity, using (if I remember correctly) a similarly-defined lattice in $\mathbb{Z}[i]\times\mathbb{Z}[i]$. However, I don't remember the details, or if they were successful. I'd be interested to see if a "low-tech" proof using lattice point counting can work for cubic or quartic reciprocity laws (or even more generally, but I suspect that would be overly optimistic). - -REPLY [20 votes]: Gauss's (unpublished and largely unknown) proof of the quartic reciprocity law -probably used lattice point arguments. The details were supplied by several authors -at the end of the 19th century (for references, see e.g. Hill's article below). -A modern approach using geometric ideas similar to those above was provided in several articles by R. Hill, such as this one. -Edit (2015). For reconstructions of Gauss's ideas see the recently published book - Gauss's reciprocity laws -in number theory (in German).<|endoftext|> -TITLE: Most wanted Diophantine equations -QUESTION [15 upvotes]: For most of my life, one single (family of) Diophantine equation(s) dominated the list of the world's most celebrated unsolved mathematical problems. Perhaps the world we live in now has grown too sophisticated for such heavy focus on such a narrow question (visions of Fermat's MO postings getting closed as "Too localized"), but if not: - -What specific unsolved Diophantine equations would today's number theorists most like to crack? (And why - historical provenance, application to another part of mathematics, "test" question for a major arithmetical theory, etc.) - -"Specific" means, for example, "solve BSD, etc. to find an algorithm to decide the solvability of all elliptic curves" doesn't count, nor is this is the place to talk about -anything like the $abc$ conjecture. -Answers should not depend on any sort of coding, Matiyasevich-style. - -REPLY [6 votes]: You may be interested in Some open problems about diophantine equations; it says, "We have collected some open problems which were posed by participants of an instructional conference (May 7-11, 2007) and a subsequent more advanced workshop (May 14-16, 2007) on solvability of Diophantine equations, both held at the Lorentz Center of Leiden University, The Netherlands. Among the 22 problems posed: -Find all integer solutions to the equation $x^2-x=y^5-y$. -Do the same for the equation ${x\choose2}={y\choose5}$. -Determine all arithmetic progressions of the form $a^2,b^2,c^2,d^5$ where $a,b,c,d$ are integers with $\gcd(a,b)=1$. -Are there infinitely many positive integer solutions to ${x^3-1\over y^3-1}=z^2$?<|endoftext|> -TITLE: Extending arithmetic functions to groups -QUESTION [20 upvotes]: Thinking along the lines of Tom Leinster's fascinating recent question, I'm wondering more generally about how to extend questions about natural numbers to groups, with the cyclic groups representing the natural numbers, and normal subgroups representing divisors. For example, let $\mathbb{G}$ be the set of all isomorphism classes of finite groups. Then an "arithmetic function" could be defined to be simply a function $f:\mathbb{G}\rightarrow\mathbb{C}$. Here are a few analogs I wrote down rather quickly (so perhaps there are better proposals for generalizations than these): -$$\text{id}(G)=|G|\quad\quad\quad \epsilon(G)=\begin{cases}1\text{ if $G$ is trivial}\\ 0\text{ otherwise}\end{cases}\quad\quad\quad z(G)=0$$ -$$\sigma_k(G)=\sum_{N\,\triangleleft \,G} |N|^k \quad\quad\quad\quad\phi(G)=|G|\prod_{\substack{N\,\triangleleft \,G \\ |N|\text{ prime}}}\left(1-\frac{1}{|N|}\right)$$ -Tom Leinster's question is whether there is a solution to $\sigma_1(G)=2|G|$. - -Question 1: What is known, if anything, about these functions? What is a good proposal for an analog of the Mobius function (I couldn't think of one offhand)? Can anyone demonstrate that they satisfy formulas that are analogs of their natural-number counterparts, or perhaps instead give examples that would indicate that these functions act weird and aren't nice generalizations to make? - -I suspect that if these functions act badly, the most likely fix would be to redefine $\mathbb{G}$ to be isomorphism classes of finite abelian groups. All subgroups are normal, and the structure theorem makes things much more controlled. I would guess that defining (for example) what it means for two groups to be "coprime", and hence what it means for an "arithmetic function" to be "multiplicative", would go over much more smoothly with abelian groups. -Going further: given two "arithmetic functions" $f,g:\mathbb{G}\rightarrow\mathbb{C}$, we can define a "Dirichlet convolution" by -$$(f\ast g)(G)=\sum_{N\,\triangleleft \,G}f(N)g(G/N)$$ -I've got to say, my jaw dropped a bit when I wrote that down. But one immediate difference I can see, somewhat discouraging, is that $\ast$ would not be abelian, since we aren't guaranteed that there are any normal subgroups $M\triangleleft G$ isomorphic to $G/N$ (see this MO question), much less that if $M\cong G/N$ then $G/M\cong N$. However, the function $\epsilon$ is still a left and right identity for $\ast$, and $z$ is still a zero. - -Question 2: Can anyone prove or disprove that $\ast$ is associative? If it is, we at least get a non-commutative ring under $\ast$ and pointwise addition (that $\ast$ distributes over pointwise sums is obvious). - -Now I suspect I am really getting "greedy" with my generalizing. We could further define "Dirichlet series", such as the zeta function (note that this is not the same thing as the zeta function of a group): -$$\zeta_{\mathbb{G}}(s)=\sum_{G\in\mathbb{G}}\frac{1}{|G|^s}=\lim_{n\rightarrow\infty}\sum_{\substack{G\in\mathbb{G}\\ |G|\leq n}}\frac{1}{|G|^s}$$ -I seriously doubt there is any hope for an Euler product-like expression. But perhaps, if we restricted ourselves to finite abelian groups... -Also, I'm familiar with the result that the number of isomorphism classes of groups of order $p^n$ grows as $p^{\frac{2}{27}n^3+O(n^{8/3})}$, and I imagine the growth rate is at least as bad for not-necessarily $p$-groups. - -Question 3: Is there any $s>0$ for which $\zeta_{\mathbb{G}}(s)$ converges? - -I wrote this question rather fast, and I welcome any feedback on how to improve it, or make it more appropriate for MO. Should I break this up into multiple questions? Is it too open-ended? - -REPLY [7 votes]: It looks like some of these ideas appear in Philip Hall's paper, "The Eulerian functions of a group," Quart. J. Math. Oxford Ser. 134-151, 1936. He discusses both an Euler $\phi$ function and a Möbius function defined on groups. -The Möbius function is as Tom Leinster describes in his answer to Q1. -The function $\phi_n(G)$ is the total number of distinct $n$-bases of the group $G$, where an $n$-basis is any ordered set of $n$ elements of $G$ which generate $G$. Thus, if $G$ is cyclic of order $m$, $\phi_1(G)$ is the usual Euler function $\phi(m)$.<|endoftext|> -TITLE: Is a complex manifold projective just because its blow-up at a point is ? -QUESTION [21 upvotes]: Consider a compact connected complex manifold $X$ of dimension $n$. Siegel proved in 1955 that its field of meromorphic functions $\mathcal M (X)$ has transcendence degree over $\mathbb C$ at most $n$. Moishezon studied those complex manifolds for which the degree is $n$, and consequently these manifolds are now called Moishezon manifolds. -In dimension 2, every Moishezon surface is projective algebraic according to a theorem of Chow-Kodaira proved in 1952, long before Moishezon formally introduced his concept. However in dimensions 3 and more, there exist nonprojective Moishezon manifolds (you can see an example in Shafarevich's book Basic Algebraic geometry). -Nevertheless a Moishezon manifold $X$ is close to projective: Moishezon's main result is that after a finite number of blow-ups with smooth centers, $X$ becomes algebraic projective. So if a blow-up of $X$ is projective, you can't deduce that $X$ was projective. However this main result says nothing about the dimensions of the manifolds you blow up. I've heard it claimed that if only one point is blown-up, you can't get from a non-projective to a projective manifold, but I could obtain neither precise reference nor proof. -Hence my question : -If a compact complex manifold becomes projective algebraic after blowing-up a point, was it already projective algebraic? - -REPLY [5 votes]: There is a sledgehammer available for this particular nut: Mori's results on extremal rays, in his paper (Annals 1979?) on smooth projective varieties $Y$ where $K_Y$ is not nef: if $Y=Bl_PX$, with exceptional divisor $E$, then any line in $E$ will span an extremal ray and then can be contracted in the category of projective varieties. Since all curves in $E$ lie in the same ray, $E$ must be contracted to a point, so the result of the contraction must be $Y$.<|endoftext|> -TITLE: Geometric meaning of Cohen-Macaulay schemes -QUESTION [54 upvotes]: What is the geometric meaning of Cohen-Macaulay schemes? -Of course they are important in duality theory for coherent sheaves, behave in many ways like regular schemes, and are closed under various nice operations. But whereas complete intersections have an obvious geometric meaning, I don't know if this is true for CM schemes. Perhaps we can make somethinkg out of the following theorem: A noetherian ring $R$ is CM iff every ideal $I$ which can be generated by $ht(I)$ many elements is unmixed, i.e. has no embedded associated primes. Also, Eisenbud suggests that Cor. 18.17 in his book "Commutative algebra with a view toward algebraic geometry" reveals some kind of geometric meaning, but perhaps someone can explain this in detail? -Every integral curve is CM. Now assume that you are given a surface, given by some homogeneous equations, how do you "see" if it is CM or not? - -Hailong's answer contains the link to How to think about CM rings?, which is pretty the same question as mine and has already some very good answers. So I appologize for this duplicate. But the answers here reveal some more insights, thanks! - -REPLY [52 votes]: [EDIT: I rewrote the first couple of paragraphs, because I realized a better way to say what I had in mind.] -There are many ways to define dimension and some of them give the same answer some of them don't. -Depth is a sort of dimension. Perhaps not the most obvious, but one that works well in many situation. -In general we count dimension by chains and the main difference between Krull dimension and depth is about the same as the difference between Weil divisors and Cartier divisors. -For simplicity assume that we are talking about finite dimensional spaces. Infinite dimension can be dealt with by saying that it contains arbitrary dimensional finite dimensional spaces where we may substitute "Krull dimension" or "depth" in place of "dimension". -I usually think of Krull dimension as going from small to large: We start with a (closed) point, embed it into a curve, then to a surface until we get to the maximal dimension. However, for comparing to depth it is probably better to think of it as going from large to small: Take a(n irreducible) Weil divisor, then a(n irreducible) Weil divisor in that and so on until you get to a point. -In contrast, when we deal with depth we take Cartier divisors: We start with the space itself (or an irreducible component), then take a(n irreducible) hypersurface, then the intersection of two hypersurfaces (such that it is a "true" hypersurface in each irreducible component this condition corresponds to the "non-zero divisor" provision)=a codimension $2$ complete intersection, and so on until we reach a zero dimensional set. -So, I would say that the geometric meaning of Cohen-Macaulay is that it is a space where our intuition about these two notions giving the same number is correct. I would also point out that this does not mean that necessarily all Weil divisors are Cartier, just that one cannot get a longer sequence of subsequent Weil divisors than Cartier divisors. - -Another, less philosophical explanation is the following: -Cohen-Macaulay means that depth = dimension. $S_n$ means that this is true up to codimension $n$. Then one may try to give geometric meaning to the $S_n$ property and say that CM means that all of those properties hold. -So, -$S_1$ --- means the existence of non-zero divisors, i.e., that there exists hypersurfaces that are like the ones we imagine. -$S_2$ --- is perhaps the most interesting one, or the one that is the easiest to explain. See this answer to another question where it is explained how it corresponds to the Hartogs property, that is, to the condition that functions defined outside a codimension $2$ set can be extended to the entire space. -$S_3$ --- I don't have a similarly nice description of this, but I am sure something could be made up, or some people might even know something nice. One thing is sure. This means that every ("true") hypersurface has the $S_2$ property, which has a geometric meaning as above. -[...] -So, one could say that -$S_n$ means that every ("true") hypersurface has the $S_{n-1}$ property, which we already described. -I realize that this description of $S_n$ may not seem satisfactory, but in practice, this is very useful. I would also add that in moduli theory it is actually important to know that some properties are inherited by hypersurfaces (=fibers of morphisms), so saying that hypersurfaces are $S_2$ is actually a good property. -More specifically, for example, the total space of a family of stable (resp. normal, $S_2$) varieties is $S_3$ ("is" as in "has to be"). Then one might (as in Shafarevich's conjecture, see Parshin's Theorem, Arakelov's Theorem, Manin's Theorem, Faltings' Theorem) study deformations of these families (say the embedded deformations of a curve in the moduli stack of the corresponding moduli problem). Then the total space of these deformations ought to be $S_4$ on account of their fibers having to be $S_3$ since their fibers have to be $S_2$. This actually explains why it is not entirely bogus to say that $S_4$ means that codimension $2$ complete intersections satisfy the Hartogs property. -This actually reminds me another thing that is important about CM. A lot of properties are inherited by general hypersurfaces. The CM property is inherited by all of them. This makes them perfect for inductive proofs. - -One way to see that a surface is CM is that normal $\Rightarrow$ CM. Of course, the point is that normal is equivalent to $R_1$ and $S_2$, so it always implies $S_2$ and if the dimension is at most $2$, then $S_2$ is the same as CM. If you have a non-normal surface, but it is non-normal only because it has normal crossings in codimension one, then it is CM. You may also try to test directly for the Hartogs property mentioned above: -A reduced surface $S$ is Cohen-Macaulay if and only if for any $P\in S$ and any regular function $f$ defined on $U\setminus \{P\}$ for an open set $P\in U\subseteq S$ there exists a regular function $g$ on $U$ such that $g_{|U\setminus \{P\}}=f$. - -REPLY [7 votes]: I would argue that the reason why Cohen-Macaulayness is interesting is that any non-algebraic connection is quite subtle, yet they exist. Some of them are already mentioned in this question (nice intersection property, good duality, combinatorial meaning etc). -In my opinion, the reason is the presence in the Cohen-Macaulay definition of depth, a homological condition. Any time you can connect a homological to a geometric property, one has some interesting and non-trivial result. -To your last question, how to see that a (I assume projective, the answer would be different in the affine case) surface is Cohen-Macaulay. Suppose your surface $S$ is given as $Proj(R/I)$, here $R=k[x_0,\cdots, x_n]$ and $I$ is some homogenous ideal. For simplicity I will assume that $I$ is a prime ideal. Then Cohen-Macaulayness for $S$ is the same thing as $Ext_R^{n-1}(R/I,R)$ is only supported at the irrelevant ideal $(x_0,\cdots, x_n)$ (use the duality between $Ext$ and local cohomology). So you can only "see" it after some significant work. -This may sound overly complicated, but that is the nature of what we are dealing with. A closely related analogue (hinted at in Karl's answer) is the notion of normality. It is a very useful notion, but can be very tricky to detect without prior knowledge.<|endoftext|> -TITLE: Geometric abelian class field theory -QUESTION [8 upvotes]: There is a very nice geometric proof of Deligne for the Artin Reciprocity in the geometric setting, namely for a smooth, projective, geometrically irreducible curve $C$ over a finite field $\mathbb{F}_{q}$, with function field $K=k(C)$, and idele group $\mathbb{I}_{K}:=\prod^{'}_{p\in|C|}K^{*}_{p}$ there is a one-to-one correspondence between the finite quotients of the double quotient space $k(C)^{*}\backslash\mathbb{I}_{K}/\prod_{p \in |C|}\widehat{\mathcal{O}_{p}^{*}}$ (which is isomorphic to $Pic_{C}(\mathbb{F}_{q})$) and the finite quotients of $\pi^{ab}_{1}(C)$. -Now on the other hand the Artin Reciprocity Law for function fields states (e.g. in Artin-Tate: Class field theory) that the group $\mathbb{I}^{0}_{K}/K^{*}$ of norm 1 idele classes is isomorphic via the Reciprocity map to $Gal(\bar{K}^{ab}/K\bar{k})$. -My questions would be: - -These two statements seem to me first as different statements, don´t they? -If we put aside Deligne´s geometric proof for the geometric statement (not seriously and not so for long :-)) then how could one prove the geometric statement using the "number theoretic" Reciprocity Law for function fields? - -REPLY [5 votes]: They are different statements. What Deligne proves is the unramified case, i.e. the description of abelian extensions of $K$ unramified everywhere. If you could extend his argument to affine curves then you could possibly prove Artin reciprocity by his method. Going the other way should not be difficult. Have you looked at Serre's book "Groupes algebriques et corps de classes"?<|endoftext|> -TITLE: A non-trivial property of all groups -QUESTION [24 upvotes]: This question appeared in my answer to this question, but it seems to be interesting in itself. Let $G$ be an infinite finitely generated group, $\epsilon\gt 0$. Is there a finite subset $S\subset G$ such that every subset of $S$ with at least $\epsilon|S|$ elements generates $G$? If the answer is "yes", it should have a trivial proof by Gromov's thesis (every property of all finitely generated groups is either false or trivial). - Update. In view of Stephen's answer and Kevin's comment below, perhaps a more correct question is this: - -Is it true that if we represent an infinite group $G$ as a union of a finite number of subsets, then one of these subsets generates a finite index subgroup of $G$? - -Compare with Adreas Thom's question. - -REPLY [13 votes]: This concerns the updated question. The answer is yes. -If $G$ is represented as the union of a finite number of subsets $A_1, \dots,A_n$, then one of the subsets generates a finite index subgroup of $G$. Indeed, let $G_i$ be the subgroup which is generated by $A_i$. Then, $G$ is the union of the $G_i$. -B.H. Neumann proved that if $G$ is the union of a finite number of left cosets of subgroups, then one of the subgroups is of finite index (see this post). In particular, there exists some $i$ such that $G_i$ is of finite index.<|endoftext|> -TITLE: Random knot on six vertices -QUESTION [30 upvotes]: This question is inspired by Joseph O'Rourke's beautiful question on random knots. Choose an random ordered 6-tuple of points on the unit sphere in $\mathbf{R}^3$, and form a knot by connecting successive pairs of points in the 6-tuple by sticks (see the picture at Joseph's question). By known results on stick numbers, the resulting knot will either be the unknot or the trefoil knot. What is the probability of producing one or the other? - -REPLY [27 votes]: I wrote a program in Mathematica to sample knots from this distribution and test what proportion are the trefoil knot. -In order to tell if a given knot is the unknot or the trefoil, the program first checks the total curvature of the knot and applies the Fary-Milnor theorem: if the curvature is less than $4 \pi$, then it's the unknot. Half the time, this test identifies the unknot. I think it should be possible to compute the exact probability of the curvature being too small. -Next, the program projects the knot onto 100 random planes. If any of these projections has less than 3 crossings, then we are again considering the unknot. This test eliminates all but ~1% of cases. -Finally, if we're still not done, the program takes the projection with the least number of crossings and checks if the resulting knot diagram is tricolorable. Usually this diagram has three crossings and this test might be a bit of a sledgehammer, but this test completely distinguishes the unknot from the trefoil. (I don't use this test first because my implementation is very slow.) -In a test run of 10,000 random knots, 68 knots were determined to be the trefoil. The computation took about 12 minutes. Here's one of the trefoils it found: - -The code follows. As usual, beware of bugs. -(* Random points, projections, those sorts of things *) -randsph[] := Normalize@Table[RandomVariate@NormalDistribution[], {3}] -randknot[] := Table[randsph[], {6}] -close[x_] := Join[x, {First[x]}] -project[ x_, frame_ ] := Flatten[frame[[2 ;; 3]] . Transpose[ {x} ]] -framify[x_] := Orthogonalize@{x, randsph[], randsph[]} -rotate[{x_, y_}] := {-y, x} -halfintersecthelper[a_, b_, c_, - d_] := (a - c) . rotate[b - a] / ((d - c) . rotate[b - a]) -halfintersect[a_, b_, c_, d_] := - 0 <= halfintersecthelper[a, b, c, d] <= 1 -intersect[a_, b_, c_, d_] := - halfintersect[a, b, c, d] && halfintersect[c, d, a, b] -nintshelper[cknot3_, frame_] := - Module[{cknot2 = (project[#1, frame] &) /@ cknot3}, - Table[If[Abs[i - j] > 1 && Abs[i - j] != 5 && - intersect[cknot2[[i]], cknot2[[i + 1]], cknot2[[j]], - cknot2[[j + 1]]], {i, - halfintersecthelper[cknot2[[j]], cknot2[[j + 1]], cknot2[[i]], - cknot2[[i + 1]]], - If[over[cknot3[[i]], cknot3[[i + 1]], cknot3[[j]], - cknot3[[j + 1]], frame], +1, -1], {Min[i, j], Max[i, j]}}, {0, - 0, 0, 0}], {i, 1, 6}, {j, 1, 6}]] -nints[cknot3_, frame_] := (#1[[3 ;; 4]] &) /@ - Union[Select[Flatten[nintshelper[cknot3, frame], 1], #1[[3]] != 0 &]] -curvature[cknot3_] := - Total@Table[ - VectorAngle[cknot3[[i + 1]] - cknot3[[i]], - cknot3[[1 + Mod[i + 1, 6]]] - cknot3[[i + 1]]], {i, 1, 6}] -overhelper[a_, b_, c_, d_] := (b - a)\[Cross](d - c) -over[a_, b_, c_, d_, frame_] := - overhelper[a, b, c, d].(c - a) overhelper[a, b, c, d].frame[[1]] > 0 - -(* Can this knot be tricolored? *) -vars[seq_] := x /@ Range@Length@seq -domains[xs_] := And @@ (#1 == 0 || #1 == 1 || #1 == 2 &) /@ xs -nonconstant[seq_] := ! - And @@ Table[x[i] == x[i + 1], {i, 1, Length[seq] - 1}] -overs[seq_] := - And @@ Module[{n = Length[seq]}, - Table[If[seq[[i, 1]] == +1, x[i] == x[1 + Mod[i, n]], True], {i, 1, - n}]] -names[seq_] := Union[(#1[[2]] &) /@ seq] -overname[seq_, n_] := - x@First@Flatten[Position[seq, {+1, n}, {1}, Heads -> False]] -undername1[seq_, n_] := - x@First@Flatten[Position[seq, {-1, n}, {1}, Heads -> False]] -undername2[seq_, n_] := - x[1 + Mod[First@Flatten[Position[seq, {-1, n}, {1}, Heads -> False]], - Length[seq]]] -overunder[seq_, n_] := - Mod[overname[seq, n] + undername1[seq, n] + undername2[seq, n], - 3] == 0 -overunders[seq_] := And @@ (overunder[seq, #1] &) /@ names@seq -conditions[seq_] := - domains[vars@seq] && overs@seq && overunders@seq && nonconstant@seq -tricolor[seq_] := FindInstance[conditions@seq, vars@seq] - -(* Init *) -overalltrials = 0; -overallcount = 0; - -(* Random trials! *) -First@ - Timing@Module[{trials = 10000, nframes = 100, count = 0, frames, i, - j, k, crossings, ncrossings, pgood, projn, projj}, - frames = framify /@ Table[randsph[], {nframes}]; - For[i = 1, i <= trials, i++, - k = close[randknot[]]; - (* Angles *) - pgood = If[curvature[k] >= 4 Pi, 0, -1]; - (* Projections *) - projn = 20; - projj = 0; - For[j = 1, j <= nframes && pgood == 0, j++, - crossings = nints[k, frames[[j]] ]; - ncrossings = Length@crossings/2; - If[ncrossings < 3, pgood = -1]; - If[ncrossings < projn, projn = ncrossings; projj = j]; - ]; - If[pgood == 0, crossings = nints[k, frames[[projj]]]; - pgood = If[tricolor@crossings != {}, +1, -1];]; - (* Record *) - If[pgood == +1 && count == 0, testk = k; - testf = frames[[projj]]]; - If[pgood == +1, count++]; - ]; - overalltrials += trials; - overallcount += count; - ] -overallcount -overalltrials -overallcount / overalltrials * 100. - -(* Draw a trefoil knot found by the random trials *) -nints[testk, testf] // MatrixForm -pk = Map[project[#, testf] &, testk ] -Graphics3D[{Thickness[0.02], Opacity[1], Specularity[White, 50], - Line[testk, VertexColors -> {Red, Yellow, Green, Cyan, Blue, Purple, - Red}]}, Axes -> False, PlotRange -> All, Boxed -> False]<|endoftext|> -TITLE: Generation of finite index subgroups -QUESTION [21 upvotes]: Related to a question by Mark Sapir (see here) and a question by Kate Juschenko (see here), let me ask the following: - -Question: Let $G$ be a finitely generated group and let $\varepsilon>0$. Is there a finite subset $S$ such every subgroup $H \subset G$ with $|H \cap S| \geq \varepsilon \cdot |S|$ is of finite index? - -This would answer Kate's question and be in the spirit of Mark's suggestion. -EDIT: Bill Thurston's argument below gives for each $\varepsilon>0$ the existence of a compactly supported function $f_{\varepsilon} \colon G \to [0,1]$ with $\sum_{g \in G} f_{\varepsilon}(g)=1$ such that the following holds: - -If $\sum_{h \in H} f_{\varepsilon}(h) \geq \varepsilon$ for some subgroup $H$, then $H$ is of finite index. - -The remaining question is, whether $f_{\varepsilon}$ can be taken to be $|S|^{-1} \chi_S$ for some finite subset of $G$. - -REPLY [15 votes]: (Edited to add more detail about random walks on infinite graphs) -Let $G$ be a group with generators $g_1, \dots, g_n$, and let $\epsilon > 0$ be given. -Let $K$ be a cell complex with $n$ edges corresponding to the generators with -fundamental group $G$ (this can be obtained by sewing on 2-disks for each of a possibly infinite set of relators). -We can visualize a subgroup $H$ in terms of the covering space $K_H$ of $K$ -having fundamental group $H$. If $H$ has infinite index, then this covering space is infinite. Consider a random walks on the vertices of $K_H$ (that is, the coset space $G/H$) where at each time, there is equal probability of staying fixed or going to one of the $2n$ neighbors. The probability of being on any particular vertex at time $t$ evolves by a discrete form of the Laplacian: it is a convex combination of the probability of being at the $2n+1$ -elements in the neighborhood of radius 1 (the graph might have loops or multiple edges -between vertices, so it is a weighted avereage). -Claim: the sequence of probability distributions obtained by random walks of length $k$ on the 1-skeleton of $K_H$ tends toward $0$ at every node of the graph. Note -that for subgroups of finite index, this is process is a positive linear transformation -of a finite-dimensional vector space, some power of it is strictly positive, so it -converges to the unique positive eigenvector of the Perron-Frobenius theorem: the -uniform distribution. One way to understand the situation for an infinite graph -is to consider the list of probabilities arranged in -decreasing order. The maximum value always decreases, unless all its neighbors -have the same value. Therefore, the sequence of maximum values converges. The rate of decrease -is at least as great as the difference of the maximum from the next highest value, so -for the maximum to converge, it is necessary that the next highest value converge to the same value. Similarly, the sum of the first two values decreases at least in proportion -to the difference of the minimum from the third highest value (since at least one of the -two points with the two highest values has a neighbor with value not greater than -the third highest value). In general, the cumulative sum of the first $n$ highest -values always decreases, and at a rate at least as great as the step to the $n+1$st highest values. It follows that all probabilities converge pointwise to a constant function. -For an infinite graph, this implies they all converge to 0. -This argument is uniform over all subgroups of infinite index $H$. -Therefore, for any -$\epsilon$, after some time $T(\epsilon)$, the random walk will have probability less than $\epsilon$ -of being at the base point of $K_H$. In other words, the probability distribution -of the random walk in $G$ has sum over $H$ less than $\epsilon$. -To summarize, for some time $T(n, \epsilon)$ -depending only on $\epsilon$ and the number $n$ of generators for $G$, for any -subgroup $H$ of infinite index (or even index less than say $2/\epsilon$) the time -$T(n,\epsilon)$ probability -of the random walk being in $H$ is less than $\epsilon$. -I think this already answers the question in spirit: instead of a test set, this gives -a test function of compact support. I believe an extension of this argument should answer -the problem as posed, but at the moment I don't have the free energy to try to work out the details.<|endoftext|> -TITLE: Is there a "disjoint union" sigma algebra? -QUESTION [16 upvotes]: I'm looking for a measure-theoretic analogue to the disjoint union topology, or for work on the $\sigma$-algebra generated by canonical injections. More formally: -For an indexed family of sets $\{A_i\}_{i \in I}$, define $\psi_i : A_i \to A$, $a \mapsto (a,i)$ (the canonical injections), where $A = \bigcup_{i \in I} (A_i \times \{i\})$. Then the disjoint union $\sigma$-algebra $\mathcal{A}$ is the finest $\sigma$-algebra on $A$ such that for every $i \in I$, $\psi_i$ is $\mathcal{A}_i$-$\mathcal{A}$-measurable. -There are so many analogies between measure theory and topology that I've been surprised at how difficult it's been for me to find anything on this. -I'd appreciate references to any related ideas as well. - -REPLY [10 votes]: David H. Fremlin's "Measure Theory", vol 2, 214K, gives this construction explicitly. He also proves some elementary properties, but unfortunately stops short of universal properties such as in Peter's insightful answer.<|endoftext|> -TITLE: Longest element of Weyl groups -QUESTION [18 upvotes]: What is a reduced expression of the longest element of each type of Weyl group. For type $A_n$ it is just $s_n(s_ns_{n-1})...(s_n...s_1)$. I know for type $B_n,C_n,E_7,E_8$,$G_2$ and $D_n$ (n even) it is just $-id$, although I don't have an explicit reduced expression for them. For type $D_n$ (n odd) and type $E_6$ I don't know what are the longest elements. Any reference where it is written explicitly? - -REPLY [20 votes]: 2-color your Dynkin diagram, black and white. Let $w$ be the product of the white simple reflections, $b$ the product of the black. Note that $w$ and $b$ are well-defined, as the reflections you're multiplying to make them, commute. You'll have to pick the order if you want an actual word, in what follows. -If $G$ is not $A_{even}$: the affinization of the diagram is also 2-colorable, so you can choose the affine vertex to be white. Let $\chi = w b$, a Coxeter element. The Coxeter number $h$ is even, and $\chi^{h/2} = w_0$. So you get a reduced word $wbwbwb\ldots wb$, where the total number of letters is $h$ (and each letter is a product of commuting reflections). -If $G$ is, unfortunately, $A_{even}$: you have to pick $w$ vs. $b$, and the diagram automorphism shows that the choice is unavoidable. The Coxeter number is odd. But you still get a reduced word, $wbwb\ldots bw$, again with $h$ letters.<|endoftext|> -TITLE: How were moduli spaces defined before functors? -QUESTION [28 upvotes]: People today in algebraic geometry will typically define a moduli space to be the space which represents the functor of families of whatever object they are interested in studying. -However, I am fairly certain that moduli spaces in algebraic geometry have been around for much longer than functors have. So my questions are: -How did algebraic geometers define moduli spaces before functors? Was there a precise definition they employed? Or was it more informal i.e. "Difficult to define, but we'll know one when we see one"? If there wasn't a precise definition, then how could they employ moduli spaces in actual proofs? - -REPLY [20 votes]: I will give an answer, but first I would like to clarify the question. It seems to me that most commenters have misinterpreted the question. The question is not how people managed to construct different examples of moduli spaces before they had the tool of the language of functors. The question is the following: A moduli space is supposed to be a space whose points are in bijection with the isomorphism classes of some type of object. But it's not just that; for, if we only needed to find a bijection, then the sets would merely need to have the same cardinality. But we need more: namely, we need the geometry of the space to somehow reflect the nature of the objects in a "natural way." Now, we may have an intuitive idea of what this means, and in many cases we might be able to recognize when some space is not just in bijection with a class of objects but actually has geometry that reflects those objects. But the question is: how can we precisely state what this means? -Nowadays, we have the language of functors and functor of points, and we look not just at a single set, but at isomorphism classes of the given type of object over arbitrary bases, given a functor. We then say that a space is a moduli space for those objects if it represents that functor (or represents it up to isomorphisms - the coarse vs fine distinction is not too relevant for this discussion); note that now, this determines not only the set of points of our space, but it actually (by Yoneda's lemma) determines the geometry of our space. -So the question is the following: before the notion of functors and functor of points, how did people rigorously define what it meant for the geometry of the moduli space to reflect the geometry nature of the set of isomorphism classes of objects to be parametrized. I should add that this is a question that I have been curious about myself for a long time. -Now, according to Newstead's text Introduction to Moduli Problems and Orbit Spaces, - -The word "moduli" is due to Riemann, who showed in his celebrated paper of 1857 on abelian functions that an isomorphism class of Riemann surfaces of genus $p$... However, it is only very recently that one has been able to formulate moduli problems in precise terms and in some cases to obtain solutions to them. - -This book (at least the edition I'm looking at) was written in 1977, which gives some perspective on this statement. -In the ensuing chapter, Newstead goes on to define a family of objects parametrized by a variety. The definition is quite simple: it is a morphism of varieties $X \to S$ such that the fiber of any point $s$ of $S$ (i.e. its pre-image in $X$) is an object of the type in question. Even if one does not have the language of functors, my guess is that this idea could motivate a more precise notion of what a moduli space is. -János Kollár's draft book on moduli spaces also gives some hints: - -The classical literature never differentiates between the linear system as a set and the linear system as a projective space. There are, indeed, few reasons to distinguish them as long as we work over a fixed base field $k$. If, however, we pass to a field extension $K/k$, the advantages of viewing $|L|$ as a $k$-variety appear. - -The first sentence suggests that there was not a precise definition of moduli space in classical literature. It also suggests a natural idea leading to the functor of points, i.e. that over a field k, we might want to look and objects parametrized over different field extensions of k. -In general, as you can see from the history mentioned in his text, people didn't necessarily have precise definitions of moduli spaces, but they did understand that the geometry (well, the parameters) of the moduli space should correspond to the coefficients of the defining equations of the objects in question to be parametrized. -Finally, in Dieudonné's Historical Development of Algebraic Geometry, the author states on p.837 - -the precise meaning of this result that Riemann surfaces of genus $g$ are parametrized by $3g−3$ complex parametrized was to remain until very recently among the least clarified concepts of the theory. - -While this doesn't answer the question, it might be of interest to note that Dieudonné later notes in his section on Grothendieck's functor of points - -in particular, one transfers in that way to the theory of schemes many classical constructions such as projective spaces..., and one is able to give a general meaning to the concept of "moduli" introduced by Riemann for curves - -suggesting slightly that there was no general meaning before this point. -That's the best I can do for now. An expert in the history of algebraic geometry might have more to say, but the sources I have seem to point in the direction of saying that there was not a precise definition until much later<|endoftext|> -TITLE: Superfluous definitions -QUESTION [17 upvotes]: It is well known that the axioms of a ring R with unity 1 imply that the underlying group must be commutative. -For if a and b are elements of R, and writing + for the group operation then applying the distributive property one has -$$ -\begin{align} -a+a+b+b&=a*(1+1)+b*(1+1)\\\\ -&=(a+b)*(1+1)\\\\ -&=(a+b)*1+(a+b)*1\\\\ -&=a+b+a+b, -\end{align} -$$ -whence $a+b=b+a$. -For educational purposes, are there more (not only algebraic) examples of such superfluous definitions? - -REPLY [4 votes]: A complete lattice is a poset in which every subset has both an infimum and a supremum. Existence of the infimum for every subsets is already enough. -P: Every element is a lower bound for the empty set, so the infimum is the largest element and a largest element exists, so every set has an upper bound. The infimum of all upper bounds is easily seen to be an upper bound- and a smallest one at that. -Nonnegativity in the definitions of metrics and norms is redundant but often included. -P: $0=d(x,x)\leq d(x,y)+d(y,x)=2d(x,y)$ for all x,y. The case of norms is similar. -A regular Hasudorff space is just a regular space that is $T_0$. -P: Take two points, one is not in the closure of the other. Separate that point and the closure of the other by open sets. This also separates the points.<|endoftext|> -TITLE: Why is the dimension of Gaussian variables is bounded by the dimension of the space? -QUESTION [6 upvotes]: I'm looking at a probabilistic proof of a local version of Dvoretzky's theorem in Pisier's manuscript "Probabilistic Methods in the Geometry of Banach Spaces." - -For each $\epsilon >0$ there is a number $\eta^\prime(\epsilon) > 0$ with the following property. Let $X$ be a Gaussian r.v. with values in a Banach space $B$ of dimension $N.$ Then $B$ contains a subspace $F$ of dimension $n = [\eta^\prime(\epsilon) d(X)]$ which is $(1+\epsilon)$-isomorphic to $\ell^n_2.$ Conversely, if $B$ contains a subspace $F$ with $F \stackrel{1+\epsilon}{\sim} \ell^n_2,$ then there is a $B$-valued Gaussian r.v. $X$ such that $d(X) \geq (1+\epsilon)^{-2}n.$ - -This statement uses the "dimension" $d(X)$ of a Gaussian variable, -$$ d(X) = \mathbb{E}\|X\|^2/\sigma(X)^2, $$ -where -$$ \sigma(X)^2 = \sup \{ \mathbb{E} \xi(X)^2 \mid \|\xi\|_{B^\star} \leq 1 \} $$ -is the weak variance of $X.$ -For this to match the usual $n = O(\log N)$ statement of the theorem, you'd need a lower bound on $d(X)$ of order $\log N,$ and as a sanity check an upper bound of $O(N).$ -Any hints or references on how to show these two bounds on $d(X)$? Pisier states that the upper bound $d(X) \leq N$ is easy to show, but I've not been able to prove even that. - -REPLY [5 votes]: None of these estimates is trivial. The upper bound follows from John's theorem that the Banach-Mazur distance between an N dimensional normed space and an N dimensional Euclidean space is at most $\sqrt N$. The lower bound is more involved and uses the Dvoretzky-Rogers lemma. A good reference for the Gaussian approach to Dvoretzky's theorem is Pisier's book: The volume of convex bodies and Banach space geometry. For a more geometrical approach see: Milman and Schechtman: Asymptotic theory of finite-dimensional normed spaces. -The theorem and it's proof is presented in other books as well.<|endoftext|> -TITLE: Webpages for specialized communities -QUESTION [30 upvotes]: First, my apology for this soft question. My excuse was that I really think this may be of interest to the community. I would understand if it gets closed (I have advocated closing some soft questions myself, so it's fair). -Background/Motivation: Recently at an AMS meeting I decided to attend a special session outside my interests and had a lot of fun. Also, to learn about what is going on in a different but related area, I found it very useful to look at some recent conferences and start by skimming the list of speakers and abstracts/notes. So a website with recent and upcoming conferences would be useful. They are not always easy to find with Google. -Requirements: Since this is soft, let's take some care to make sure it will not be too easy to answer. So let's define "specialized community" to be "there is at least one AMS special session devoted to (part of) it. Also, the more useful the website is the better (for example, one should preferably contains list of conferences, and perhaps more) -Examples: here is one I visit frequently: for commutative algebra -Another one I found recently and like a lot: Finite dimensional algebras - -Surely there must be others? - -REPLY [2 votes]: A few more for Logic: -Mathematical Logic Around the World contains a lot of useful information about Logic departments, Logic journals, LaTeX for Logicians etc. The events section is a little outdated though. -The European Set Theory Society has conference announcements as well as slides from the conferences. -There is also the Young Set Theory Network, which has conference announcements, and hopefully will have more things in the future.<|endoftext|> -TITLE: How many elements does it take to normally generate a group? -QUESTION [19 upvotes]: This is a terminology question (I should probably know this, but I don't). Given a group $G$, consider the minimal cardinality $nr(G)$ of a set $S \subset G$ such that $G$ is the normal closure of $S$: $G = \langle\langle S \rangle \rangle$ (nr is short for normal rank). In other words, how many elements in $G$ do we need to kill to produce the trivial group? What is this invariant called? "Corank" and "normal rank" seem to mean other things, and I'm not sure what other terms to search for. -Also, what methods are there to get a lower bound on $nr(G)$, say when $G$ is finitely generated? A trivial lower bound in this case is $rk(H_1(G))$, since clearly $nr(A)=rank(A)$ for $A$ a finitely generated abelian group. One has $nr(G)\leq rank(G)$, since it suffices to kill a generating set, and if $G\to H$ is a surjection, then $nr(G)\geq nr(H)$. - -REPLY [6 votes]: What you define as $nr(G)$ is indeed commonly referred to as the weight of a group, often written $w(G)$. In general, computing the weight of a finitely generated group, or even any sensible lower bounds on the weight, is very difficult. Wiegold posed the following problem in 1976: -Is it true that every finitely generated perfect group is the normal closure of one element? (i.e., has weight 1). -This can be found as problem 5.52 in the Kourovka notebook: http://arxiv.org/abs/1401.0300 -Wiegold's question is `true' in the case of finite groups, and also solvable groups. See M. Chiodo, Finitely annihilated groups, Bull. Austral. Math. Soc. 90, No. 3, 404-417 (2014). In particular, Corollary 5.7 states: -Let $n > 1$ and let $G$ be a finite or solvable group. Then $w(G) = n$ if -and only if $w(G^{ab}) = n$, and $w(G)=1$ if and only if $w(G^{ab})= 1$. --Maurice<|endoftext|> -TITLE: Arithmetic progressions modulo $p$ under the squaring map -QUESTION [12 upvotes]: I feel that the following problem should be known, but I'm not sure where to look for it. -Fix a real constant $\frac{1}{2} \ge \epsilon > 0$. For varying primes $p$, Let $A_p$ denote the set of residue classes coming from the first $\lfloor p \epsilon \rfloor$ integers. Let $B_p$ denote the squares (modulo $p$) of the elements of $A_p$. Then one might ask whether -$$\lim_{p \rightarrow \infty} \frac{|A_p \cap B_p|}{|A_p|} =^{?} \epsilon.$$ -It's true for $\epsilon = \frac{1}{2}$, but that's a degenerate case where $B_p$ can essentially be replaced by $\mathbf{F}^{\times 2}_p$, in which case the answer follows from any non-trivial upper bound on character sums (say the Polya-Vinagradov inequality). Is it true more generally? - -REPLY [7 votes]: This is a variant of a common theme. It should follow from more or less standard exponential sums estimates. The general buzzword is Erdos-Turan inequality. The answer should be yes and it might follow from the results of: -A. Granville, I. E. Shparlinski and A. Zaharescu, On the -distribution of rational functions along a curve over $\mathbb{F}_p$ and residue races, J. Number Theory, 112 (2005), 216--237. -or C. Cobeli and A. Zaharescu, On the distribution of the $\mathbb{F}_p$-points on an affine curve in $r$ dimensions, Acta Arithmetica 99 (2001), 321--329. -Or perhaps even earlier papers.<|endoftext|> -TITLE: Is the "closedness of the image of operator" needed in the defintion of Fredholm operators? -QUESTION [8 upvotes]: in the Higson and Roe's book "analytic K-homology" just after the definition of the Fredholm operator there is a remark (2.1.3 you can see at it onlin at Google books (click here)) which claims that in the definition of Fredholm operators the condition of being closed for image of operator is superfluous. -It seems that in the proof $cokernel(T)$ is supposed to be $\frac{H_2}{Image(T)}$ instead of $\frac{H_2}{\overline{Image(T)}}$ which cause the newly defined operator $\tilde{T}$ to be immediately surjective. (which is not obviouse for me when I take $coker(T)= \frac{H_2}{\overline{Image(T)}}$ (as it is expected when we consider coker at the category of Hilbert spaces with bounded operators)). -I was wondering if there is anybody help me to find out what is goning on in this proof or prove or disprove the statement in a clear way? -Edit: -So You agree with me that this proof is true just if we take $cokernel_1(T)=\frac{H_2}{Image(T)}$. -Now here is the question: what if we put $cokernel_2(T)=\frac{H_2}{\overline{Image(T)}}$? I mean if $cokernel_2(T)$ is finite dimension can we say that $cokernel_1(T)$ is finite dimensional also? -In other word: Is there any bounded operator $T:H_1\to H_2$ which has finite dimensional $coker_2(T)$ (please note the index 2) and $ker(T)$ but it is not a Fredholm operator? - -REPLY [4 votes]: If $X$ and $Y$ are Banach spaces, $T:D(T)\subset X\to Y$ is a closed operator and there exists a closed subspace $N$ of $Y$ such that $\operatorname{Im}(T)\cap N=\{0\}$ and $\operatorname{Im}(T)\oplus N$ is closed, then $\operatorname{Im}(T)$ is closed. -See Theorem IV.5.10 in A.E. Taylor and D.C. Lay. Introduction to functional analysis, 2nd ed. Krieger 1980.<|endoftext|> -TITLE: Why is the Laplacian ubiquitous? -QUESTION [35 upvotes]: The title says it all. -I'm wondering why the Laplacian appears everywhere, e.g. number theory, Riemannian geometry, quantum mechanics, and representation theory. And people seems to care about their eigenvalues. Is there any deep reason behind this? - -REPLY [3 votes]: The Laplacian is a second order differential operator which is both linear -and invariant under any rigid body motion (rotations and translations). -There is no simpler non-trivial operator with these properties. In modeling many -physical phenomena, one expects these two properties to arise, so sometimes -the simplest approximation based on physical principles leads to the Laplacian.<|endoftext|> -TITLE: Whitehead doubles of any knots -QUESTION [8 upvotes]: I was curious about the fact that the Whitehead doubles of all knots have Alexander polynomial equal to 1, which is the same as a unknot. How to prove this? - -REPLY [10 votes]: This is slightly more mechanistic variant of Ian's response. When a knot is a Whitehead double, that means you've obtained it by doing a splicing construction (Larry Siebenmann's terminology, originally -- this is a particular formalism for describing satellite operations on knots, one that's friendly with the JSJ-decomposition for 3-manifolds). Alexander polynomials behave very nicely with respect to splicing, I'll describe it a bit below. - (source) -The input in a splicing construction is an $(n+1)$-component link $L=(L_0,L_1,\cdots,L_n)$ such that the sublink $(L_1,L_2,\cdots,L_n)$ is the trivial $n$-component link. The other input is $n$ knots $K_1,\cdots,K_n$. The splice knot I like to denote $L \bowtie K$. In the above Whithead double case, the input link $L$ is the Whitehead link, $n=1$ and the input knot $K$ is the figure-8 knot. - (source) -For the knot above, the input link $L$ has the same complement as the Borromean rings (but it isn't quite the Borromean rings). $n=2$ and the two knots are both trefoils (with the same handedness). -Alexander polynomials behave well under splicing, in particular: -$$\Delta_{L\bowtie K}(t) = \Delta_{L_0}(t) \Delta_{K_1}(t^{l_1}) \Delta_{K_2}(t^{l_2}) \cdots \Delta_{K_n}(t^{l_n})$$ -where $l_i$ is the linking number between $L_i$ and $L_0$ in the link $L$. -So the "reason" the Alexander polynomial of a Whitehead double is trivial (from this perspective) is that (1) the linking numbers of the components of the Whitehead link are zero, and (2) the "base" component of the Whitehead link ($L_0$) has trivial Alexander polynomial (both components are unknots so they have to have trivial Alexander polynomials). -Anyhow, the proof of this is just the generalization of Agol's argument to this setting, by lifting everything to the Abelian cover, Mayer-Vietoris type arguments. -But just staring at the above formula you see there are all kinds of other constructions on knots that produce new knots with trivial Alexander polynomials. More precisely, the example below is of a spliced knot which is the splice of the Borromean rings with two figure-8 knots, so its Alexander polynomial is trivial. There's no requirement to use figure-8 knots, any knots work but the way splicing works is you have to be quite careful about your framing conventions. - (source: Wayback Machine) -A diagram that indicates the framing conventions: -alt text http://s2.postimage.org/7h4kvwycx/blah3.jpg -So this knot also has trivial Alexander polynomial because $L_0$ is the unknot and all the respective linking numbers are zero.<|endoftext|> -TITLE: Why is a general curve automorphism-free? -QUESTION [15 upvotes]: Fix an algebraically closed field $k$. Why is the general curve over $k$ of genus $g \ge 3$ automorphism-free? -I am particularly interested in seeing an argument that does not go by induction and specialization to a singular genus $g$ curve. -Let's say a curve is a smooth, projective, connected $1$-dimensional $k$-scheme. - -REPLY [14 votes]: One way to do it is through deformation theory, provided we only consider -automorphism groups $G$ of order not divisible by the characteristic (one may of -course assume that it is cyclic of prime order). Then the -the moduli space (or just a miniversal deformation) of all curves of genus $g>1$ -is smooth with tangent space at the curve $C$ equal to $H^1(C,T^1_C)$. The -tangent space of the sublocus where the action of $G$ extends is equal to -$H^1(C,T^1_C)^G$ and hence all curves in a neigbourhood of $C$ has an action of -$G$ only when $G$ acts trivially on $H^1(C,T^1_C)$. The (Brauer) character of -the action can be computed by the holomorphic Lefschetz trace formula (resp. of -a lifting of $(C,G))$ and is seen to be non-trivial. (To this I guess one has to add -that there is a finite stratification of the moduli space where the automorphism -group is fixed on each stratum.) -Addendum: In principle this method could also handle automorphisms of order equal to the characteristic (say), what needs to be shown is that they act non-trivially on $H^1(C,T^1_C)$. I haven't thought about that though. -As suggested by Mariano one can also count parameters: Assume $G$ is cyclic of -prime order. If the order is not equal to the characteristic one can use the -Hurwitz formula to get bounds on the genus for the quotient curve and the number -of critical values of the quotient map. Counting parameters for the quotient -curve and the critical values always gives a value which is smaller than $3g-3$ -(not a difficult calculation but also not altogether pleasant). -Addendum: The calculation turns out not to be that difficult. Indeed, if we are dealing with an automorphism of prime order $\ell$, if $C\to C'$ is the quotient map and we have $r$ critical values, then the Hurwitz formula gives -$$ -3(g-1) = 3\ell(g'-1) + \frac32(\ell-1)r, -$$ -where $g'=g(C')$, and on the other hand we want to show (when $g'>1$) that $3(g-1)>3(g'-1)+r$ which follows immediately as $\frac32(\ell-1)>1$. When $g'=1$ we want to show that $3(g-1)>1+r-1=r$ ($1$ parameter for varying the elliptic curve and by automorphisms we may fix one critical value), i.e., $\frac32(\ell-1)r>r$, i.e., $(3\ell-5)r>0$ which is always OK. Finally, with $g'=0$ we are OK if $3(g-1)>\max(r-3,0)$. We may assume $r>3$ and then $r\geq 6$ for reasons of divisibility which easily gives that we are OK unless $r=6$ and $\ell=2$ which gives $g=2$. -The case when -the order is equal to the characteristic is even messier, one has to look at the -local contribution at a critical value to the genus of $C$ which is of -Artin-Schreier type and then bound the number of local parameters in such covers -(we here have parameters even when the critical values and the quotient curve -stay constant). One could also however use the result of Oort saying that a -curve with such an automorphism lifts equivariantly.<|endoftext|> -TITLE: Prime factorization of n+1 -QUESTION [12 upvotes]: If $n=\prod_{i=1}^{k} p_i^{e_i}$ is a prime factorization of integer $n$. -Is there a quick way to find the prime factorization of $n+1$? -Or the only way to do it is recalculating the whole factorization? -Any references and/or articles on this problem? - -REPLY [4 votes]: There are two questions asked: -Q1. Does the prime factorization of $n$ give a quick way to find the prime factorization of $n+1$? -A1.No it does not (as noted in other answers). Knowing $n$ is prime is of no use. -Q2. Is it ever of any use or do you always just have to start from scratch? (as I will choose the rephrase the question.) -A2. It is sometimes of use but not usually. When $n$ is a power of a smaller number there may be some help. Since the case of Fermat numbers was raised I will comment that factoring numbers of the form $2^e+1$ is somewhat easier (relative to the size) than numbers of the form $2^e+3$. Given $n=2^e$ we know that $2^{e/f}+1$ is a factor of $n+1$ where $f$ is any odd divisor of $e$. So that is a start to factoring $n+1$. In the case $n=2^{2^e},$ $n+1$ might be prime. To test primality one has Pépin's test. The numbers involved are so huge that this is not practical very far out. ALSO it is known that any candidate factor must be of the form $k2^e+1$ so in attacking $2^{2^{20}}+1$ (known composite, no factors known) we have already ruled out 99.9999% of the possible factors.<|endoftext|> -TITLE: Does Hölder continuity imply smoothness for the CMC equation: $u:D^2\rightarrow\mathbb{R}^n$, $\Delta u = 2H\partial_xu\times\partial_yu$, $H$ constant? -QUESTION [6 upvotes]: Context: I am currently reading through the freely available lecture notes from Tristan Riviere (here) on the applicability of integration by compensation in the analysis of various geometrically motivated PDEs. -I have attempted to find something in the vast literature to the following effect: suppose $u:D^2\rightarrow\mathbb{R}^n$, $u\in W^{1,2}$, $\Delta u = f(u,\nabla u) = 2H\partial_xu\times\partial_yu$. Then -$$u\in C^{0,\alpha}(D') \rightarrow u\in C^\infty(D''),$$ -where $D'' \subset\subset D' \subset\subset D$. -In other words, an interior estimate. With the little regularity on hand, it appears to be very difficult. I find this surprising because most of the time, proving the Hölder regularity of the solution is the 'most difficult' part. I have the feeling that I am missing an obvious reference or a well-known folklore argument. -Showing the Hölder continuity of the solution relies on deriving a Morrey-type estimate with the help of the Wente lemma. In the process of doing this, one also shows that -$$\sup_{\rho < 1/2, p\in B_{1/2}(0)} \rho^{-\alpha} \int_{B_\rho(p)} |\Delta u|$$ -is bounded. This implies that $f\in\mathcal{H}^1$. I just include this extra detail in case this question fits into a general framework of optimal interior regularity for Poisson's equation on a disk when the right hand side is Hardy. (This is the reason for the earlier form of my question.) -Can anybody help? - -REPLY [4 votes]: I'm too lazy to type-up the proof myself, so I'll send you to a reference. -Chang, S.-Y. A., Wang, L. and Yang, P. C. (1999), "Regularity of harmonic maps". CPAM has the proof in Section 3. Once you get $C^{1,\gamma}$ you immediately get RHS is in $C^\gamma$ and the rest follow by standard elliptic regularity. -Note that the structure of the equation (RHS being of the form $d(u\cdot du)$) is only used for Wente's lemma. For the upgrade of regularity one uses a Caccioppoli type inequality. -(BTW, the Chang-Wang-Yang result bypasses the Hardy space estimates. For that the result can be found in the original paper of Helein, though I'd guess the material is also in his book if you don't read French.)<|endoftext|> -TITLE: Tropical homological algebra -QUESTION [18 upvotes]: Has anyone out there thought about homological algebra over the tropical semifield $\mathbb{T}$? For example, I'm interested in the Hochschild homology and cyclic homology of tropical algebras, if this can be made to make sense. -[Edit] Perhaps I should add some motivation. Topologists and geometric group theorists have been interested in the moduli space of metric graphs for at least 25 years now, mainly because of its appearance as a classifying space for automorphism groups of free groups. This space turns out to have a second identity as the moduli space of tropical curves, and people in tropical geometry tell me that it should probably in fact carry the structure of a tropical orbifold. -This means that, in addition to homological invariants built from (sheaves of) continuous of PL functions on the space (containing essentially the information of rational homotopy), one can try to build and study homological invariants made from the tropical structure sheaf. I'm interested in what sort of geometric information these invariants might carry. Is there perhaps some new information hiding in here? -It is always exciting when one finds that an object one has known for many years turns out to have a hidden new structure. -Following Zoran's comment, it looks like Durov has constructed, among many other things, a model category structure on complexes of modules over $\mathbb{T}$. This means that, in principle, something like homological algebra can be done. But it's a different matter to explicitly develop homological algebra. -So, to expand on my original question, here are some explicit questions. - -What is a tropical chain complex? -Is there an explicit tropical -analogue of the usual Hochschild -chain complex and does it compute -the correct derived functor? -Ordinary Hochschild homology carries -a Gerstenhaber algebra structure. Is -there an analogous structure on a -tropical Hochschild homology? -Same questions for cyclic homology. - -REPLY [4 votes]: Your question has two parts - the main question, existence of tropical homotopical algebra, seems to be nicely resolved by Zoran's answer. The questions about Hochschild homology though are rather context independently true - I am not familiar with the Durov story, but assuming he provides a symmetric monoidal model category (and hence symmetric monoidal $\infty$-category) that should suffice. Given an algebra object $A$ in any symmetric monoidal $\infty$-category, we can define its Hochschild homology and cohomology, as self-tor and ext as bimodules (see e.g. here - note that what I'm referring to as the homology and cohomology corresponds to the chain level versions of these notions, not the cohomology groups). The former has an $S^1$ action, with invariants giving cyclic homology, and can be calculated by a version of the cyclic bar construction. Regarding the Gerstenhaber structure on Hochschild COhomology, again the answer is yes by a general form of the Deligne conjecture (see Lurie's Higher Algebra - the part that until last week was called DAG VI..) - there's an $E_2$ structure on the Hochschild cohomology. In fact if your algebra $A$ is separable (dualizable as $A$-bimodule) one obtains the full structure of a framed two-dimensional topological field theory (via Lurie's Cobordism Hypothesis theorem), which includes all these operations and many more. (The Hochschild homology and cohomology are assigned to the circle with two different framings in this story.)<|endoftext|> -TITLE: Triangulating hypercubes -QUESTION [8 upvotes]: Motivation: I'm working on a computational problem at the moment, and have some very good routines for natively working with simplicial complexes and calculating homology, but the structures I'm dealing with arise naturally as cubical complexes. -Problem: Is there an efficient way to triangulate the n-cube, i.e. calculate a (relatively) small list of n-simplices on the same vertices as the cube, and which define a simplicial complex spanning the cube? -I've done some reference-chasing and there seems to be no decently-sharp estimate (as an upper or lower bound) for the asymptotic complexity of the problem, although the best upper-bounds I'm aware of (for the size of the smallest solution-set) seem to indicate something exponentially smaller than factorial (see Haiman, 91). This paper also exhibits a lower bound, given below -$\frac{2^n\,n!}{(n+1)^{{}^{(n+1)/2}}}$ -Orden and Santos improved the upper bound somewhat, by reducing the base of the exponential. - -REPLY [6 votes]: A slight improvement on the lower bound was recently obtained by Alexey Glazyrin (http://arxiv.org/abs/0910.4200, http://www.sciencedirect.com/science/article/pii/S0012365X12003974). Basically, it changes the $2^n$ factor in Nick's initial post to an $e^n$. Smith's hyperbolic volume method gave an $\sqrt{6}^n$. -But, of course, as Greg points out, the real open question is whether the simplicity of the cube grows as the $c^n n!/n^{n/2}$ of the lower bounds or as the $c^n n!$ that can be achieved by actual constructions (or something in between).<|endoftext|> -TITLE: Counterexamples to Kollár's conjecture -QUESTION [6 upvotes]: In "Singularities of theta divisors, and birational geometry -of irregular varieties", Jour. AMS 10, 1 (1997), 243–258, L. Ein and R. Lazarsfeld provided a counterexample to Kollár's conjecture, i.e. an irregular 3-fold of general type of maximal Albanese dimension with Euler characteristic = 0. Their example can be trivially generalized to higher dimension. -My question is: are other counterexamples to Kollár's conjecture known? Where can I find them? -Thank you very much for the time dedicated to me -Best regards - -REPLY [6 votes]: I just saw this question (better late than never). The following references come to mind (the first one has an example which somewhat improves on the example of Ein-Lazarsfeld, the others have a variety of related results): -The Example at the end of Chen and Hacon "On the irregularity of the image of the Iitaka fibration. Comm. in Algebra Vol. 32, No. 1, pp. 203-215 (2004), doi: 10.1081/AGB-120027861 . -Example 5.6 in arXiv:1111.6279 On the Iitaka fibration of varieties of maximal Albanese dimension. Zhi Jiang, Martí Lahoz, Sofia Tirabassi. -arXiv:1105.3418 Varieties with vanishing holomorphic Euler characteristic. J. A. Chen, O. Debarre, Z. Jiang. -Cai, Jin-Xing; Chen, Jungkai Alfred: A note on characterizations of Abelian varieties by topological invariants. Manuscripta Math. 112 (2003), no. 1, 15–19, doi: 10.1007/s00229-003-0384-2.<|endoftext|> -TITLE: Automorphism group of a scheme -QUESTION [20 upvotes]: I have a probably stupid question on schemes ... - Let $S$ be a scheme, and let $A = \mathsf{Aut}(S)$ be its automorphism group. Does $A$ carry -a scheme structure itself, that is, can one see $A$ as a group scheme ? - Thanks ! - -REPLY [7 votes]: This is a partial answer to THC's question about a "direct" way to do this. -Let $X$ be a smooth projective variety such that $\omega_X$ is ample. Then there is some $m\in\mathbb N$ such that $\mathscr L=\omega_X^{\otimes m}$ is very ample, so $X$ has an embedding into $\mathbb P^N=\mathbb P(H^0(X,\mathscr L))$. -Obviously $\mathscr L$ is invariant under any automorphism, and hence so is $H^0(X,\mathscr L)$. This implies that the automorphisms of $X$ extend to automorphisms of the ambient $\mathbb P^N$. In other words ${\rm Aut}(X)$ can be identified with a quotient of a subgroup of ${\rm Aut}(\mathbb P^N)$ and hence it inherits a natural scheme structure. -On the other hand it actually also follows that in this case ${\rm Aut}(X)$ is finite (see below why) and in fact it seems to me that this argument could only possibly work if something close to that is true. -Assume that $X$ is embedded into a projective space and all automorphisms of $X$ are induced by automorphisms of the ambient projective space. Then ${\rm Aut}(X)$ has a scheme structure (as above) and let ${\rm Aut}^\circ (X)$ denote the connected component of ${\rm Aut}(X)$ containing the identity. Then the quotient ${\rm Aut}(X)/{\rm Aut}^\circ (X)$ is a discrete group which is the quotient group of a linear algebraic group and hence it is finite. -The action of ${\rm Aut}^\circ (X)$ (in this case) exhibits $X$ as a uniruled variety and hence if $X$ is not uniruled (e.g., when $\omega_X$ is ample) it implies that ${\rm Aut}(X)$ is finite. -In other words if the scheme structure on ${\rm Aut}(X)$ given by the Hilbert scheme argument has infinitely many components, then it cannot be the quotient of a subgroup of ${\rm Aut}(\mathbb P^N)$. This happens for instance for K3 surfaces with an infinite automorphism group, in particular for any supersingular K3.<|endoftext|> -TITLE: Complicating an Example by Toen (motivations for DAG) -QUESTION [14 upvotes]: I'm trying to read (the introduction of) a survey by Toen on Derived Algebraic Geometry, specifically the "Simplicial Presheaves and Derived Algebraic Geometry" one. -He motivates the introduction of DAG as a means to construct moduli spaces. His example is the moduli of linear representations of a group admitting a finite presentation. -Now, DAG (AFAIU) enlarges the theory of stacks in two directions: a derived bit and a stacky bit. The derived bit concerns replacing rings with more general ring-like objects. The stacky bit comes from using stacks of oo-categories. -His motivation for the `derived' direction comes from the fact that in constructing a moduli space he gets a tensor product, which isn't considering higher Tor's. (I guess I'm sort of OK with that, but for no concrete reason) -His motivation for the `stacky' direction comes from taking quotients of a 'rigidified' moduli problem, which admits a moduli space (the aforementioned tensor product). The problem here is the usual fact that you want to remember the isomorphism groups of objects. -Unfortunately this only motivates us to introduce stacks, not oo-stacks. - -So my question is: Can we complicate this example a bit more (but hopefully not too much) so that we need to use higher stacks? - -I understand that if we have already enhanced our geometry along the derived direction then we actually need oo-stacks to keep track of notions of equivalence which are weaker than isomorphisms. -So a parallel question might be: - -if we don't derive our geometry first, do we need to introduce higher stacks? - -Comments on the derived are also very much appreciated, thanks! - -REPLY [2 votes]: The moduli of linear categories (or abelian categories) in naturally a $2$-stack. You can take a look at the PhD thesis (in french) of Mathieu Anel (a former student of Bertrand Toën). -When you compute the tangent complex of this $2$-stack then you get the $2$-truncation of the Hochschild complex. If you consider the corresponding derived stack then you get the full Hochschild complex. -By the way, higher stacks are introduced to allow quotients, while derived schemes are introduced to allow fiber products.<|endoftext|> -TITLE: Prime avoidance in adjacent degrees -QUESTION [8 upvotes]: Let $\mathfrak{p}_1, \dotsc, \mathfrak{p}_k$ be relevant homogeneous primes ideals in the graded ring $R := \Bbbk[x_0, \dotsc, x_n]$, where $\Bbbk$ is a field. Prime avoidance (in Eisenbud's terminology) tells us that there exists a nonconstant homogenous polynomial $f \not\in \cup_i \mathfrak{p}_i$. Is there an easy way to see that for some $d \geq 1$, there exist homogeneous $f \in R_d$, $g \in R_{d+1}$, neither of which is contained in $\cup_i \mathfrak{p}_i$? - -Note: If $\Bbbk$ is infinite, this is easy: we can use the fact that no vector space is a finite union of proper subspaces to find $\alpha \in R_1 \smallsetminus \cup_i \mathfrak{p}_i$, and then take $f=\alpha$, $g = \alpha^2$. -Motivation: If the $\mathfrak{p}_i$ are the associated primes of a closed subscheme $X$ of $\mathbb{P}^n_{\Bbbk}$ then we can pull back $g/f$ to a nice rational section of $\mathcal{O}_X(1)$. Consequently, we know that $\mathcal{O}_X(1)$ comes from a Cartier divisor. Using the fact that any line bundle can be twisted to a very ample line bundle, this shows that the Cartier divisor group surjects onto $Pic(X)$. - -REPLY [6 votes]: The answer is yes. The following is extracted from a preprint of Gabber-Liu-Lorenzini. - -Let $B=\oplus_{n\ge 0}B(n)$ be a graded ring. Let $I=\oplus_{n\ge 0}I(n)$ be a homogeneous ideal of $B$. Let $\mathfrak p_1,\dots,\mathfrak p_r$ be homogeneous prime ideals of $B$ not containing $B(1)$ and not containing $I$. Then there exists an integer $n_0 \geq 1$ such - that for all $n\ge n_0$, $I(n)\not\subseteq \cup_{1\le i\le r} \mathfrak p_i$. - -Proof. We proceed by induction on $r$. If $r=1$, choose $t\in B(1)\setminus \mathfrak p_1$ and a homogeneous element $\alpha\in I\setminus \mathfrak p_1$, say of degree $n_0$. Then -$t^{n-n_0}\alpha\in I(n) \setminus \mathfrak p_1$ for all $n\ge n_0$, as desired. -Let $r\ge 2$ and suppose that the lemma is true for $r-1$. We can suppose -that $\mathfrak p_i$ is not contained in $\mathfrak p_r$ for all $i\ne r$, so that -$I\mathfrak p_1\cdots\mathfrak p_{r-1} \not\subseteq \mathfrak p_r$. Similarly, -we can suppose that $\mathfrak p_r$ is not contained in $\mathfrak p_i$ for all -$i\ne r$, so that -$I\mathfrak p_r\not\subseteq (\mathfrak p_1\cup \ldots\cup\mathfrak p_{r-1})$. -Hence, we can apply the case $r=1$ and the induction hypothesis -to obtain that there exists $n_0$ such that for all $n\ge n_0$, -there are homogeneous elements -$f_n\in I\mathfrak p_1\cdots\mathfrak p_{r-1} \setminus \mathfrak p_r$ and -$g_n\in I\mathfrak p_r\setminus (\mathfrak p_1\cup \ldots\cup\mathfrak p_{r-1})$ -of degree $n$. It is easy to check that -$f_n+g_n\in I(n)\setminus \cup_{1\le i\le r}\mathfrak p_i$, as desired.<|endoftext|> -TITLE: Maximizing the Smallest Eigenvalue of a Diagonally Dominant Matrix -QUESTION [10 upvotes]: Assume that we have a full-rank diagonally dominant matrix $A$, all the diagonal elements of which are positive, all the non-diagonal elements are negative, and the sum of the absolute values of the non-diagonal elements is equal to the diagonal element. More precisely: -\begin{equation} -A=[a_{i,j}] \qquad a_{i,i}>0 \qquad a_{i,j} \leq 0 \textrm{ for $i \neq j$} \quad \textrm{and } \quad a_{i,i}=\sum _{j\neq i}|a_i,_j | -\end{equation} -We also have a positive diagonal matrix $D$ whose trace is constant and equal to $K$: -\begin{equation} -d_{ij}=0 \textrm{ for } i \neq j \qquad d_{ii} \geq 0 \quad \textrm{and } \quad K=\sum_{i}d_{i,i} -\end{equation} -What is the matrix $D$ such that the smallest eigenvalue of the matrix sum $T=A+D$ is as large as possible? -In other words, how can we find the $d_{ii}$ 's such that we maximize the smallest eigenvalue of $T=A+D$? -Thank you all in advance! :-) - -REPLY [3 votes]: If the A matrix were symmetric (so the eigenvalues are real), then you could just solve a semidefinite programming problem (SDP) to find the matrix D (and 'lambda'). In particular, maximizing the smallest eigenvalue ('lambda') of the matrix A + D in your case would be equivalent to the SDP (over variables D and lambda): -max lambda -such that: A + D >= lambda I - Trace(D) = K - D_{ii} >= 0 -where the first '>=' denotes 'greater-or-equal in the cone of positive semidefinite matrices', and I is the identity matrix. -There are several MATLAB-based packages in which you can formulate and solve problems like this (for instance, Yalmip and CVX). You'll probably also need a solver for SDPs (e.g., sedumi or sdpt3). -Good luck, - -Dan<|endoftext|> -TITLE: Models for P map in EHP sequence -QUESTION [9 upvotes]: The E and H maps in the EHP sequence have models that aren't too hard to define. To review, the E map is induced on homotopy by $E: \Omega^n S^n \to \Omega^{n+1} S^{n+1}$ sending a map to its suspension. The H map has a nice model once one knows about James's theorem that $\Sigma \Omega \Sigma X$ splits as $\bigvee_i \Sigma X^{\wedge i},$ where $X^{\wedge i}$ is $X$ smashed with itself $i$ times. When $X$ is $S^n$ one has $\Sigma \Omega S^{n+1}$ then projects -onto the "quadratic" wedge factor $\Sigma (S^n)^{\wedge 2} \cong S^{2n + 1}$. The adjoint of this map is $H : \Omega S^{n+1} \to \Omega S^{2n + 1}$. -My question is "what about $P$"? That is, what is a model for this map? As I understand it, the usual approach is to show that $E$ and $H$ fit into a fiber sequence (when localized at two, or integrally through a range(?)) and then $P$ is the connecting map. I was disappointed to learn that despite its name $P$ only coincides with a Whitehead product in some cases. Of course, there are general nonsense constructions of connecting maps for fibrations in terms of the maps one already has in hand, but I want to know: -Is there an "intrinsic" model for $P$, one which doesn't rely on the $E$ and $H$ maps in its definition? -I would also be interested in an intrinsic model for the map on homotopy, say in terms of framed bordism. - -REPLY [5 votes]: Dear Dev, -You probably know this, but the map $P$ is the Whitehead product map in the metastable range in the following sense: Suppose $X = \Sigma Y$ is a suspension. Then the generalized Whitehead product map $$P:\Sigma Y \wedge Y\to \Sigma Y$$ coincides with map -from the homotopy fiber of $$E: \Sigma Y \to \Omega \Sigma (\Sigma Y)$$ into the domain of $E$ in the metastable range (roughly thrice the connectivity of $Y$). -By "coincide," I mean that since $E\circ P$ has a preferred null-homotopy, one has a preferred map -$$ -\Sigma Y \wedge Y \to \text{hofiber}(E) -$$ -which is a metstable equivalence. It is in this sense that we write "$P$ " for the -connecting map in the EHP sequence (which is a metastable homotopy fiber sequence). -However, I do not believe this map will integrally factor up to homotopy through $\Omega^2\Sigma (Y\wedge Y)$ even when $Y$ is a sphere (except in the cases when $Y = S^{n-1}$ is an $H$-space: $n=2,4,8$). -So your question seems to live in the world of 2-localized spheres, and not the spheres themselves.<|endoftext|> -TITLE: Nilpotency of a group by looking at orders of elements -QUESTION [24 upvotes]: For any finite group $G$, let -$$\theta(G) := \sum_{g \in G} \frac{o(g)}{\phi(o(g))},$$ -where $o(g)$ denotes the order of the element $g$ in $G$, and where $\phi$ is the Euler totient function. -It is not too hard to see that if $G$ is nilpotent, then $\theta(G)$ is in fact equal to $\sigma(|G|)$, i.e. the sum of the divisors of $|G|$. However, it seems that $\theta(G)$ is always less than or equal to $\sigma(|G|)$, and that equality holds if and only if $G$ is nilpotent. -My question is twofold: (1) Is this claim true? (2) What kind of "natural" properties of groups (such as nilpotency) are there that can be checked by only looking at the orders of the elements of the group? - -REPLY [6 votes]: (I post this as an answer, since I don't want to enter the below math into the comment box. It answers Tom de Medt's question on my comment to Robert Guralnick's answer.) -Nilpotency can be checked by looking only at the orders of elements: Write $e_G(k)$ for the number of elements of order $k$ in $G$. As I remarked in that comment, nilpotency of $G$ implies multiplicity of $e_G$ on coprime elements, since $G$ is the direct product of the Sylow subgroups. -Indeed, one can make somewhat more precise statements: Let $p$ be a prime and $|G|_p$ the $p$-part of $|G|$, and note that -$$ \sum_{i\geq 0} e_G(p^i) \geq | G |_p ,$$ -with equality if and only if there is only one Sylow $p$-subgroup of $G$. So one can tell from the orders whether $G$ has a normal Sylow $p$-subgroup. $G$ is nilpotent iff all Sylow subgroups are normal. -Looking at the multiplicity of $e_G$ was perhaps more complicated than necessary, but here is a justification of the assertion from my comment: If $e_G$ is multiplicative, then -$$ |G| = \prod_{p } |G|_p - \leq \prod_{p } \sum_{i\geq 0} e_G(p^i) - = \sum_{k\geq 0} e_G(k) = |G|, $$ -and thus the "$\leq$" must be a "$=$", which again implies all Sylow $p$-subgroups are normal.<|endoftext|> -TITLE: Fragmenting a homeomorphism of a compact manifold -QUESTION [7 upvotes]: Let $M$ be a compact manifold and let $f : M \rightarrow M$ be a homeomorphism which is isotopic to the identity. We will say that $f$ can be fragmented if it satisfies the following property. Let $\mathcal{U}$ be any open cover of $M$. There then exists homeomorphisms $f_1,\ldots,f_n$ from $M$ to itself which are isotopic to the identity and open sets $U_1,\ldots,U_n \in \mathcal{U}$ such that the support of $f_i$ is in $U_i$ and $f = f_1 \cdots f_n$. -It is known that if $M$ is a smooth manifold and $f$ is a diffeomorphism that is smoothly isotopic to the identity, then $f$ may be fragmented (and the resulting $f_i$ are also diffeomorphisms smoothly isotopic to the identity). Indeed, this is is one of the ingredients of Thurston's proof that the identity component of the group of diffeomorphisms of a compact manifold is simple. A proof can be found in chapter 2 of Banyaga's book "The structure of classical diffeomorphism groups". -However, the proof in the smooth case appears to use smoothness in a strong way. -Question : Can an arbitrary homeomorphism $f : M \rightarrow M$ which is isotopic to the identity be fragmented? - -REPLY [9 votes]: If I understand you correctly, this is Corollary 1.3. of Kirby and Edwards, Deformations of spaces of imbeddings, Ann. Math. (2) 93 1971 63–88. (I couldn't find an online version of the journal article.) -For $k$-parameter families of maps, but in the smooth or PL case, see Lemma B.0.4 of my recent paper with Scott Morrison (http://arxiv.org/pdf/1009.5025). - -REPLY [6 votes]: I am reasonably confident that this result is contained in the paper of Edwards and Kirby (Annals, 1971)<|endoftext|> -TITLE: Stratification of smooth maps from R^n to R? -QUESTION [8 upvotes]: I'm interested in stratifications of smooth maps $\mathbb{R}^n\to\mathbb{R}$ (or more generally of any $n$-manifold $M^n\to\mathbb{R}$). The codimension 0 stratum should be Morse functions, and the codimension 1 stratum should be Morse cancellations, e.g. the $t=0$ value of the following 1-parameter family of maps -$$ - (x_1,\ldots,x_n) \mapsto tx_1 + x_1^3 \pm x_2^2 \pm\cdots\pm x_n^2 . -$$ -Is there a good reference for the general codimension $k$ case? -Another way of phrasing the question: given a $k$-parameter family of smooth maps $F: P^k\times \mathbb{R}^n\to\mathbb{R}$, is there a known list of specific singularities such that we may assume that $F(p, \cdot)$ has only these singularities after a small perturbation? I suppose the way to start is to make $F$ Morse as a map from an $(n+k)$-manifold to $\mathbb{R}$, then look at the ways the coordinate axes of $P\times \mathbb{R}$ line up with gradients and the eigenspaces of the hessian of the Morse singularities of $F$. But I would rather cite the details than work them out for myself. -If the general case is messy (instability, cross-ratios, etc.), I would also be interested in an answer for $n=2$. - -REPLY [6 votes]: It looks to me that what you are really interested in is the Thom-Boardman stratification of the function space. For that I would recommend the well-written, Stable Mappings and Their -Singularities by Guillemin and Golubitsky (in the Springer GTM series).<|endoftext|> -TITLE: Martingales in both discrete and continuous setting -QUESTION [12 upvotes]: I am wondering, polynomials like -$S_n^4-6n S_n^2+3n^2+2n$ for $$S_n=\sum_{i=1}^n{X_i}$$ where $$\mathbb{P}(X_i=1)=\mathbb{P}(X_i=-1)=\frac{1}{2}$$ is a martingale (under the conventional filtration). While $$B_t^4-6t B_t^2+3t^2$$ for Brownian motion $B_t$ is also a martingale. -Note the difference between the two, and the similarity! -What's the general conclusion about the polynomials of $S_n$ and $n$, also about $B_t$ and $t$, to make them into martingales? -Thanks. - -REPLY [21 votes]: One knows that $P(S_n,n)$ is a martingale if and only if $P(s+1,n+1)+P(s-1,n+1)=2P(s,n)$ and -that $Q(B_t,t)$ is a martingale if and only if $2\partial_tQ(x,t)+\partial^2_{xx}Q(x,t)=0$. -Assume that $P(S_n,n)$ is a martingale and, for a given $d$ and for every $h>0$, let -$$ -Q_h(x,t)=h^{d}P(x/\sqrt{h},t/h), -$$ -in the sense that one evaluates $P(s,n)$ at the integer parts $s$ and $n$ of $x/\sqrt{h}$ and $t/h$. -If $Q_h\to Q$ when $h\to0$, writing $\partial_t$ and $\partial^2_{xx}$ as limits of finite differences of orders $1$ and $2$, one sees that $2\partial_tQ+\partial^2_{xx}Q=0$, hence $Q(B_t,t)$ is a martingale. -Example: $P(s,n)=s^2-n$. For $d=1$, $Q_h(x,t)=x^2-t$ hence $Q(x,t)=x^2-t$. -Other example: -$P(s,n)=s^4-6ns^2+3n^2+2n$. For $d=2$, $Q_h(x,t)=x^4-6tx^2+3t^2+2ht$ hence $Q(x,t)=x^4-6tx^2+3t^2$. -In the other direction, to deduce a martingale in $S_n$ and $n$ from a martingale in $B_t$ and $t$, one should probably replace each monomial by a sum of its first derivative. This means something like replacing $q(t)=3t^2$ by $\displaystyle\sum_{k=1}^n(\partial_tq)(k)=3n^2+3n$ but I did not look into the details. - -Edit (Thanks to The Bridge for a comment on the part of this answer above this line) -Recall that a natural way to build in one strike a full family of martingales that are polynomial functions of $(B_t,t)$ is to consider so-called exponential martingales. For every parameter $u$, -$$ -M^u_t=\exp(uB_t-u^2t/2) -$$ -is a martingale hence every "coefficient" of its expansion as a series of multiples of $u^i$ for nonnegative integers $i$ is also a martingale. This yields the well known fact that -$$1,\ B_t,\ B^2_t-t,\ B^3_t-3tB_t,\ B^4_t-6tB_t^2+3t^2, -$$ -etc., are all martingales. One recognizes the sequence of Hermite polynomials $H_n(B_t,t)$, a fact which is not very surprising since these polynomials may be defined precisely through the expansion of $\exp(ux-u^2t/2)$. -So far, so good. But what could be an analogue of this for standard random walks? The exponential martingale becomes -$$ -D^u_n=\exp(uS_n-(\ln\cosh(u))n) -$$ -and the rest is simultaneously straightforward (in theory) and somewhat messy (in practice): one should expand $\ln\cosh(u)$ along increasing powers of $u$ (warning, here comes the family of Bernoulli numbers), then deduce from this the expansion of $D^u_n$ along increasing powers of $u$, and finally collect the resulting sequence of martingales polynomial in $(S_n,n)$. -Let us see what happens in practice. Keeping only two terms in the expansion of $\ln\cosh(u)$ yields $\ln\cosh(u)=\frac12u^2-\frac1{12}u^4+O(u^6)$ hence -$$ -\exp(-(\ln\cosh(u))n)=1-\frac12u^2n+\frac1{24}u^4(2n+3n^2)+O(u^6). -$$ -Multiplying this by -$$ -\exp(uS_n)=1+uS_n+\frac12u^2S_n^2+\frac16u^3S_n^3+\frac1{24}u^4S_n^4+\frac1{120}u^5S_n^5+O(u^6), -$$ -and looking for the coefficients of the terms $u^i$ in this expansion yields the martingales -$$ -1,\ S_n,\ S_n^2-n,\ S_n^3-3nS_n, -$$ -and -$$ -S_n^4-6nS_n^2+2n+3n^2,\ -S_n^5-10nS_n^3+5(2n+3n^2)S_n. -$$ -Thus, in $M_t^u$, $B_t$ scales like $1/u$ and $t$ like $1/u^2$ hence Hermite polynomials are homogeneous when one replaces $t$ by $B_t^2$. The analogues of Hermite polynomials for $(S_n,n)$, from degree $4$ on, are not homogeneous in the sense of this dimensional analysis where $n$ is like $S_n^2$. Ultimately, this is simply because in $D_n^u$ one has to compensate $uS_n$ by $(\ln\cosh(u))n$, which is not homogeneous in $u^2n$. -Note that this argument of non homogeneity carries through to continuous time processes. For instance, the exponential martingales for the standard Poisson process $(N_t)_t$ are -$$ -\exp(uN_t-(\mathrm{e}^u-1)t), -$$ -and the rest of the argument is valid once one has noted that $\mathrm{e}^u-1$ is not a power of $u$.<|endoftext|> -TITLE: Can homologous submanifolds be connected by an immersed manifold with boundary? -QUESTION [9 upvotes]: Supposed I have an n-dimensional manifold M with a k-dimensional submanifold that is homologous to zero (or, equivalently, two homologous submanifolds). Can I always construct a k+1-dimensional manifold N and a smooth map $N\to M$ so that the boundary maps diffeomorphically to my submanifold? Can I just take abstract k+1-simplecies and glue them along boundaries to make N, and then somehow smooth it out? If not, is there some understandable obstruction? -I'm most interested in the smooth category, but if it makes more sense in some other category (or there is otherwise a better question I should've asked), do tell me. -Update: As I first asked it, the question was a bit stupid because I forgot about cobordisms. However, in the case I care about, this does not seem to be a problem, since I want the boundary of N to be a union of two submanifolds which are diffeomorphic to each other. - -REPLY [4 votes]: First of all, it doesn't matter whether or not the map is smooth. If you find any continuous map, then it will have a smooth approximation. -The other answers so far explain that the cobordism group gives you an obstruction to improving a singular-simplicial chain into a mapped-in manifold. In fact, it is easy to see that this is basically the only obstruction, and that null corbodism directly gives you a way to improve the simplicial chain. I'll work with integer coefficients rather than over $\mathbb{Z}/2$ so that things survive a little longer. Say that you have this $(k+1)$-dimensional cobounding chain. You can manifold-ize a $k$-dimensional face of the chain because various $k$-dimensional sheets meet the face with opposite sign and you can pair them. This is basically using the fact that the reduced 0-corbodism group is trivial. Then turn to the $(k-1)$-faces. Because of what you did to the $k$-faces, the sheets meet the $(k-1)$-faces in a collection of circles. But circles are null cobordant, so you can smooth them. You can continue in this way using the fact that oriented surfaces and 3-manifold are all null-cobordant. But when you try to improve a $(k-3)$-face, the link of the an incoming sheet can be a 4-manifold that is not null-cobordant, like $\mathbb{C}P^2$. Then you're stuck. -The obstruction is fundamental because the original null-homologous $k$-cycle could have been an embedded $\mathbb{C}P^2$, and the original cobounding chain could have been a cone over it.<|endoftext|> -TITLE: Exact factorization of finite groups -QUESTION [5 upvotes]: Suppose that $G=MN$ and $G=MP$ are two exact factorization of a finite group $G$. What is the relation between $M$ and $P$? Clearly if $G=MN$ then $G=M(mNm^{-1})$ is another factorization of $G$. Is this the only possibility to change $N$ into $P$? - -REPLY [2 votes]: The answer of the problem is the following: -A group G has two exact factorizations $G = M N = M P$ if and only if there exists a unitary bijective map (just a map, not a morphism of groups) $v : N \to P$ such that $n v(n)^{-1} \in M$, for all $n \in N$ and $v$ satisfy four natural compatibility conditions (not very transparent to write down here -- they are similar, mutatis mutandis, to the one given in Propositions 2.1 in http://front.math.ucdavis.edu/0903.5060 -- for example one of them is that v is a morphism of $M$-sets).<|endoftext|> -TITLE: Is there an elementary way to show the triangular inequality for this expression ? -QUESTION [8 upvotes]: Consider the space $X$ of all scalar products on $\mathbb{R}^n$. For a scalar product $s$ and a base $B:=b_1\ldots,b_n$ let $M_{s,B}$ denote the matrix, whose $(i,j)$-th entry is $(s(b_i,b_j))$ . Given two scalar products $s,s'$ one can find by PCA a orthonormal basis $B$ of $s$ such that $M_{s',B}$ is a diagonal matrix. By positive definiteness, all diagonal entries $\lambda_,\ldots,\lambda_n$ are positive. Let $d(s,s'):=\sqrt{\sum_{i=1}^n\log(\lambda_i)^2}$. I want to show, that this defines a metric on the set of all scalar products. Symmetry and Definiteness are clear. But why does it satisfy the triangular inequality ? -To be honest I already know, that it is a metric. This distance function comes from a Riemannian metric on the set of all scalar products. But I am looking for a simpler way (without computing the Levi-Civita connection and showing, that the geodesics satisfy the ODE and so on). -Furthermore the resulting space should be a $CAT(0)$-space. It would be nice if one could show the CAT(0) inequality directly. The geodesic from $s$ to $s'$ is given by -$$[0;1]\rightarrow X\qquad t\mapsto s_t,\mbox{ where } s_t(b_i,b_j)=\begin{cases}\lambda_i^t&i=j \\\ 0& i\neq j\end{cases}$$ - -REPLY [3 votes]: Does the following proof for triangle inequality for the Riemannian metric for posdef matrices help? -(I am currently travelling, so the proof is missing exact references; will add them later when I am near my books.) -The proof that recall below is one of my favorites, and I first saw it in a paper by R. Bhatia. I think this proof must be also available in his book: Positive Definite Matrices. -Let $A$ and $B$ be strictly posdef matrices. We wish to prove that the function -$$ -d(A,B) = \left(\sum\nolimits_i (\log\lambda_i(AB^{-1}))^2\right)^{1/2}, -$$ -is a metric; here $\lambda(X)$ is the vector of eigenvalues of $X$ sorted in -decreasing order. Clearly, the only non-trivial part is the -triangle-inequality. -Definition: Let $x$ and $y$ be vectors with entries sorted in -decreasing order. We say $x$ is majorized by $y$, written $x -\prec y$, if -$$ - \sum\nolimits_i^k x_i \le \sum\nolimits_i^k y_i,\quad 1 \le k \le - n,\quad \sum\nolimits_i^n x_i = \sum\nolimits_i^n y_i. -$$ -Definition: Let $\varphi : R^n \to R$. We say $\varphi$ is -Schur-convex, if $x \prec y$ implies that $\varphi(x) \le -\varphi(y)$. -Lemma 1 (Lidskii): Let $A$ and $B$ be posdef matrices. The following -majorization holds -$$ -\log \lambda(AB) \prec \log\lambda(A) + \log\lambda(B). -$$ -I think this Lemma is proved in Chapter 3 of R. Bhatia's Matrix Analysis. -Triangle-Inequality: Using Lemma 1 we have -\begin{align*} - \log \lambda(AB^{-1}) &= \log\lambda(C^{-1/2}AC^{-1/2}C^{1/2}B^{-1}C^{1/2} \\\\ - &\prec \log\lambda(C^{-1/2}AC^{-1/2}) + - \log\lambda(C^{1/2}B^{-1}C^{1/2})\\\\ - &= \log\lambda(AC^{-1}) + \log\lambda(CB^{-1}). -\end{align*} -Since $\|x\|_2$ is Schur-convex, and satisfies the -triangle-inequality, on applying it to both sides of the above -majorization we obtain -\begin{equation} - d(A,B) \le d(A,C) + d(C,B), -\end{equation} -proving the desired triangle inequality. In general, we could use any -symmetric gauge function to obtain a corresponding triangle-inequality.<|endoftext|> -TITLE: Examples of results which were surprising but later shown to be natural. -QUESTION [5 upvotes]: After Ramanujan formulated his conjectures on the Tau-function, and after the importance of the function was realized, it took the development of the theory of Modular forms for the complete resolution and understanding of the conjectures and the function itself.(For example, it was only later that people could explain the appearance of the mysterious index 24 in its definition.) -Another example is the problem of constructibility of regular polygons. The Ancient Greeks must have pondered the reason for their inability to construct certain polygons. But after Gauss, it now seems natural why one cannot construct a 11-gon using only a compass and straightedge. -In the above two cases, there is a common feature. There is a discovery which at first seems surprising or baffling. Only later, after sufficient developments in theory, was the mystery lifted. Are there any other such examples? - -REPLY [3 votes]: Gödel's incompleteess theorems -- knocked over the Hilbert program to prove mathematics was consistent and complete. -Smale's sphere eversion theorem - originally thought to be a counterexample showing an error in the proof, but the eversion is really there. -Butterfly effect (Edward Lorenz) -- looked like a numerical artifact in an ODE integrator, the idea of nonperiodic solutions hadn't been forseen. -Banach Tarski paradox -Monty Hall problem (ducks head) -Barrington's theorem for branching programs -PCP theorem -Rationality of Legendre's constant ;-)<|endoftext|> -TITLE: Nice application of generalized smooth spaces -QUESTION [22 upvotes]: I am a fan of category theory in general, and I appreciate that various brands of generalized smooth spaces (Diffeological spaces, Chen spaces, Frolicher spaces ...) form much nicer categories of spaces at the expense of having somewhat more convoluted objects. I might be interested in taking up the study of one form of generalized smooth space or another, but my conscience will not let me unless I see that they can actually buy me a more conceptual understanding of regular old manifolds. -So I would like a Big List of theorems about manifolds whose proof can be made significantly shorter or more conceptual by making use of generalized smooth spaces and maps between them. Something like a standard construction in the manifold setting becoming representable in the new setting, and this makes short work of some (previously) complicated theorem. - -REPLY [10 votes]: I think there are some theorems which are easier to prove in the diffeological framework, or as you say: for which the proof reveals more conceptual reasons. For example this one ? -Proposition Let $X$ be a connected diffeological space, let $\omega$ be a closed 2-form on $X$. Let $P_\omega \subset {\bf R}$ be its group of periods. If the group of periods $P_\omega$ is (diffeologically) discrete (that is, is a strict subgroup of $\bf R$) then there exists a family of non-equivalent principal fiber bundles $\pi : Y \to X$, with structure group the torus of periods $T_\omega = {\rm R} / P_\omega$, equipped with a connexion form $\lambda$ of curvature $\omega$. This family is indexed by the extension group ${\rm Ext}({\rm Ab}(\pi_1(X)), P_\omega)$. -This theorem is a generalization of the classical construction of the prequantization bundle of an integral symplectic (or pre-symplectic) manifold, that is the ones for which $P_\omega = a {\bf Z}$. Why such a generalization is interesting? Well, here are some comments: -1) The only condition for the existence of such "integration structures" is that the group of periods is diffeologically discrete, which is hidden in the classical construction by some technical hypothesis (countable at infinity or analog statements). -2) The space $Y$ is a quotient of the space ${\rm Paths}(X)$ on which the form $\omega$ is lifted modulo the action of a "Chain-Homotopy" operator (actually what is built by quotient is a groupoid and the bundle $Y$ is just the "half-groupoid"). So, the diffeological space ${\rm Paths}(X)$ is a master piece of this construction (but almost everywhere in diffeology), and the fact that diffeological spaces support differential forms (in particular ${\rm Paths}(X)$) with the whole tools of Cartan calculus is fundamental. -3) The generality of this theorem involve essentially "irrational tori", since in general the quotient $T_\omega$ is of course not a Lie group. -The last point illustrates why irrational tori are important in diffeology: or you accept these objects or you give up this (kind of) theorems. You may note that such a theorem doesn't exists in the restricted category of Frölicher spaces since irrational tori are trivial there. You may be happy with just the integral case, but in my opinion you miss a lot by not taking the whole generality of the construction, and putting fences where they do not exist. -I may give some other examples where diffeology give a shortcut for known classical theorems, and by the way extend them to objects which do not belong to the category of manifolds. - -Here is an example of a more conventional theorem: the homotopic invariance of De Rham cohomology. Differential forms and De Rham cohomology are well defined concept in diffeology, they apply in particular on space of paths of diffeological spaces, spaces of smooth maps, quotients etc. -We use here the Chain-Homotopy operator -$$ -K : \Omega^p(X) \to \Omega^{p-1}({\rm Paths}(X)) \quad \mbox{which satisfies} \quad K \circ d + d \circ K = \hat 1^* - \hat 0^*, -$$ - where $\hat 1$ and $\hat 0$ are the maps defined from ${\rm Paths}(X)$ to $X$ by $\hat 1(\gamma) = \gamma(1)$ and $\hat 0(\gamma) = \gamma(0)$. -Proposition Let $X$ and $X'$ be two diffeological spaces, let $f_0$ and $f_1$ be two homotopic smooth maps from $X$ to $X'$, let $\alpha$ be a closed $p$-form on $X'$. The pullbacks $f_0^*(\alpha)$ and $f_1^*(\alpha)$ are cohomologous. -Proof Let $\varphi : X \to {\rm Paths}(X')$ be the map defined by $\varphi(x) = [t \mapsto f_t(x)]$. The pullback by $\varphi$ of the identity $K(d\alpha) + d(K\alpha) = {\hat 1^*}(\alpha) - {\hat 0^*}(\alpha)$ gives $d(\varphi^*(K\alpha)) = f_1^*(\alpha) - f_0^*(\alpha)$. $\square$ -This is an example of simplification/generalization of a classical theorem by short-cuting the proof through diffeology. Here also the space of paths of a diffeological space, and the Chain-Homotopy operator, are crucial. May be something more fundamental is hidden behind that. Enxin Wu a Dan Christensen student is working on a possible Quillen model based on diffeology, it will give maybe some lighting on this question? - -BTW Frölicher spaces is equivalent to the full subcategory of what we call "reflexive diffeological spaces" (a work in progress with Y. Karshon and al), the ones whose diffeology is completely defined by the real smooth maps. They are the "intersection" of the category {Diffeology} and the category {Sikorski}. There are nice examples and counter-examples which illustrate the difference between these categories.<|endoftext|> -TITLE: Transversality in Morse theory for the (perturbed) geodesic action functional -QUESTION [6 upvotes]: I am interested in Morse homology on the loop space of a given compact (Riemannian) manifold. A small perturbation renders the geodesic action ("energy") functional Morse. Now I am interested in the Morse-Smale property, i.e. for any critical points x and y the unstable manifold of x intersects the stable manifold of y transversally. -Could anyone please provide a reference that a generic choice of metric on the loop space yields the Morse-Smale property? (Notice that the correct choice of perturbations of the metric is part of the problem.) I have difficulties finding an appropriate reference for this. -There seem to be two obvious ways to realize Morse-Smale transversality in this setting: -1. The abstract way: Here one considers a given Hilbert manifold with a metric. The space of perturbations consist of (some class of) metrics which are uniformly equivalent to the given one. This is for instance the approach followed by Abbondandolo/Majer: "Lectures on Morse homology for infinite-dimensional manifolds". The problem with this reference is that their space of perturbations is too big - the space in question is not separable. In particular the Sard-Smale theorem, which is crucial in this setting, cannot be applied. I have difficulties in writing down a separable Banach space of perturbations which is still enough to provide surjectivity of the linearized "master section". -2. The concrete setting: Obviously, it is not enough to consider metrics on the loop space which come from metrics on the base manifold. I do not know whether it suffices to consider metrics on the loop space which come from metrics on the base times $S^1.$ My problem is that the "master section" involves the gradient (w.r.t. the induced metric on the loop space) of the perturbed energy functional in question. I have no clue how to obtain a useful formula for its linearization. -So, could anyone please give me a hint about solving 1. or 2.? It is also possible that pursuing the paths 1. or 2. might not be a clever idea, in which case I would appreciate any advice. - -REPLY [5 votes]: Instead of considering the Morse homology of the energy functional my making it Morse-Smale, it might be easier for you (and geometrically more natural) to view the geodesic energy as a Morse-Bott functional, whose critical points appear in $S^1$-families. -The appropriate setting of Morse homology for Morse-Bott functions was worked out by Urs Frauenfelder in the Appendix of "The Arnold-Givental conjecture and moment Floer homology". -The Morse(-Bott) homology of the geodesic functional was considered by Abbondandolo and Schwarz in section 4 of "Estimates and Computations in Rabinowitz-Floer homology", see here: -Abbdondandolo-Schwarz<|endoftext|> -TITLE: When is an affine part of an elliptic curve isomorphic to an affine part of a norm equation? -QUESTION [6 upvotes]: Given a cubic number field and a basis $\{\gamma_1,\gamma_2,\gamma_3\}$ for it over the rationals, we can write down the norm equation $N(x_1\gamma_1+x_2\gamma_2+x_3\gamma_3)=1$. For almost all substitutions, say $x_1=c$, the resulting affine cubic curve is an affine part of an elliptic curve. -I was wandering what can be said of the converse. If we are given an elliptic curve over the rationals, is there a cubic number field such that for some substitution (any kind) in the norm equation, we get an affine curve isomorphic to an affine part of an elliptic curve? -I've started reading Serre's Algebraic Groups and Class Fields, which seems relevant, since its main results concern rational maps $C\rightarrow G$ from a curve to a commutative algebraic group, which is the case above. - -REPLY [3 votes]: I'm adding a second answer here because it doesn't appear to agree with my first answer. This second answer came from me trying to understand Felipe's arguments; it is possible that I am just rewriting what he said in more words. -Second answer: Let $X$ be a smooth cubic curve. Over $\mathbb{C}$, the ways to express $X$ as a linear combination of a product of three lines and a triple line are in bijection with triples $(P_1, P_2, P_3)$ of colinear cusps of $X$. Proof: Let $f = a L_1 L_2 L_3 + b L_4^3$. Then $L_4$ intersects $L_1$, $L_2$ and $L_3$ at one point each. Then $f$ restricted to $L_i$ vanishes to order $3$ at $P_i$. So $P_i$ is a cusp of $f$ and $L_i$ is the tangent line there. And $(P_1, P_2, P_3)$ all lie on $L_4$, so they are colinear. So, given an expression of $f$ as above, we find a triple of colinear cusps. -Conversely, given three colinear cusps $(P_1, P_2, P_3)$, let $L_4$ be the line through them and let $L_i$ be the tangent line to $P_i$. So $f$ restricted to $L_i$ is the cubic with an order $3$ root at $P_i$. So, choosing an appropriate scalar $b$, we have $f|_{L_1} = b L_4^3|_{L_1}$. Let $g = f - b L_4^3$. So $g$ vanishes on the line $L_1$; let $g = L_1 Q$, where $Q$ is a conic. Then $g$, restricted to $L_2$, vanishes to order $3$ at $P_2$. Since $L_1$ does not pass through $P_2$, this shows that $Q$ vanishes to order $3$ at $P_2$. But $Q$ is a conic, so this implies that $Q$ contains $L_2$. Similar, $Q$ contains $L_3$. So $Q=a L_2 L_3$ for some scalar $a$, and $f=a L_1 L_2 L_3 + b L_4^3$. So, given a triple of colinear cusps, we get such a linear representation. - -My confusion: If I count correctly, there are $12$ such triples of colinear cusps. Namely, given any two of the $9$ cusps, there is a unique way to complete it to such a triple. There are $\binom{9}{2} = 36$ pairs of cusps, and we get each such triple $3$ ways. How do I square this with the $810$ count I got earlier? -An additional note: I'd been working over $\mathbb{C}$. For the original question, we want to impose the additional conditions that $L_4$ has coordinates in $\mathbb{Q}$, and the group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ acts transitively on the $P_i$. In other words, that the cubic $f \cap L_4$ has no rational root. Someone better at computer algebra than I am should be able to turn that into the condition that a certain degree $12$ polynomial has a rational root, and that a certain degree $3$ polynomial does not.<|endoftext|> -TITLE: Does this sum equal zeta(3)? -QUESTION [14 upvotes]: Does: -$$\sum_{1 \leq ij$. -So $T=3$ iff $S=\zeta(3)$. -As I describe in the above linked thread, numerical computations suggest that the sums agree to $20$ digits of accuracy. What is going on? - -REPLY [26 votes]: Hi David, -This is the first example of a multiple zeta identity. Your sum S is just $\zeta(1,2)$, where the multiple zeta value is defined by: -$$\zeta(s_1, s_2, \ldots, s_k) = \sum_{0 < n_1 < n_2 < \cdots n_k} \left( \prod_{i=1}^k n_i^{-s_i} \right).$$ -Your identity $\zeta(1,2) = \zeta(3)$ was discovered by Euler according to Wikipedia.<|endoftext|> -TITLE: Poincaré duality with boundary conditions -QUESTION [12 upvotes]: If $M$ is a compact oriented manifold with boundary then by Poincaré duality the cohomology of $\Omega(M)$ (de Rham cohomology of $M$) is dual to the cohomology of $\Omega_0(M)$, where $\Omega_0(M)$ denotes differential forms vanishing on $\partial M$. This question is about a generalization of this fact to more complicated boundary conditions. -Suppose $M$ is a compact oriented $C^\infty$ manifold with corners. Its boundary is decomposed (by the corners) to faces (of dimension $\dim M -1$). -Let $V$ be a finite-dimensional vector space, and let us choose for every face $F\subset\partial M$ a subspace $V_F\subset V$. Let us consider the complex (of $V$-valued differential forms with boundary conditions given by $V_F$'s) -$$\Omega(M)_{V,\{V_F\}}= -\{\alpha\in\Omega(M)\otimes V;\quad \alpha|_F\in\Omega(F)\otimes V_F\text{ for all faces }F\}\subset\Omega(M)\otimes V.$$ -The "naive dual" of $\Omega(M)_{V,\{V_F\}}$ is $\Omega(M)_{V^*,\{\text{ann} V_F\}}$ with the pairing given by integration over $M$ ($\text{ann} V_F\subset V^*$ denotes the annihilator of $V_F$). -Is there a condition under which the pairing between the cohomologies of $\Omega(M)_{V,\{V_F\}}$ and of $\Omega(M)_{V^*,\{\text{ann} V_F\}}$ is perfect? What is the reason for the fact that the pairing is not always perfect? -Some remarks: - -The "dual complex" $\Omega(M)_{V^*,\{\text{ann} V_F\}}$ can be described as -$$\{\alpha\in\Omega(M)\otimes V^*;\quad \int_M\langle\alpha\wedge d\beta\rangle= - (-1)^{\deg\alpha +1}\int_M\langle d\alpha\wedge \beta\rangle\quad \forall\beta\in \Omega(M)_{V,\{V_F\}} \}.$$ -If $V_F=0$ for all $F$'s then we get the standard Poincaré duality for manifolds with boundary. -It is possible that manifolds with corners is not the right picture; the question might be about any "reasonable" division of $\partial M$ into "faces" (of dimension $\dim\partial M$). - -REPLY [5 votes]: An example where the duality fails is when $M^n$ is the closed unit ball $B^3 \subset \mathbb{R}^3$, and its boundary $S^2$ is divided into four quarters by 2 great circles. If $V = \mathbb{R}$, $V_F = V$ for 2 opposite quarters $F$ and $V_F = 0$ for the other two, then $H^1_{V, \{ V_F \}}(M) = 0$ while $H^2_{V^*, \{\text{ann} V_F\}} \cong \mathbb{R}$ (essentially, they are $H^1_c$ and $H^2_c$, respectively, of the product of an open 2-disc and a closed interval). -In a sense, the reason that the duality fails is that near the intersection of the two great circles, the set of boundary points where the forms are allowed to be non-zero is disconnected, and that no matter how small a neighbourhood we choose in $B^3$ for the intersection point, its cohomology will therefore not be entirely elementary. This can be prevented by demanding that every point in $\partial M$ has an "elementary" neighbourhood $U \cong \mathbb{H}^{n}$ such that - -the subdivision of $\partial U$ into faces is diffeomorphic to a complete fan (a subdivision of $\mathbb{R}^{n-1}$ into simplicial cones), -$V$ has a basis $\{e_i\}$ such that for each face $F$ meeting $U$, -$V_F$ is spanned by a subset, -for each $e_i$, the interior in $U$ of the union of the faces $F$ such that $e_i \not\in V_F$ is connected. - -Essentially, 1. says that the subdivision of $\partial M$ is sensible, 3. prevents situations like in the example above, and 2. makes sure we can state 3. sensibly when $\dim V > 1$ (see example in Trial's comment below). -I think that if $M^n$ is oriented with boundary and possesses such "elementary" neighbourhoods, then -$$H^k_{V, \{V_F\}}(M) \cong H^{n-k}_{c, V^*, \{ \text{ann} V_F\}}(M)^*$$ -where the subscript $c$ indicates the cohomology of a complex with compact supports. It should be possible to prove this using induction on a good cover (and the duality between the Mayer-Vietoris sequences for normal and compactly supported de Rham cohomology) like for standard Poincaré duality, provided that the statement is true for open subsets $U \subset M$ diffeomorphic to $\mathbb{R}^n$ and for the "elementary" neighbourhoods. -For $U \cong \mathbb{R}^n$ this is just usual Poincaré duality tensored with $V$. For an "elementary" neighbourhood $U$, -$$H^k_{V, \{V_F\}}(U) = \bigoplus_i H^{k}_{V_i, \{V_F \cap V_i\}}(U) $$ -$$H^k_{c, V^*, \{\text{ann} V_F\}}(U) = \bigoplus_i H^{k}_{c, V_i^*, \{\text{ann} (V_F \cap V_i) \}}(U), $$ -where $V_i$ is the span of the element $e_i$ of the basis from condition 2. -The terms on the right hand side all vanish, except that if $e_i \in V_F$ for all $F$ meeting $\partial U$ then $H^0_{V_i, \{V_F \cap V_i\}} \cong V_i$ and $H^{n}_{c, V_i^*, \{\text{ann} (V_F \cap V_i)\}}(U) \cong V_i^*$ -(3. is used to show that $H^{n-1}_{c, V_i^*, \{\text{ann} (V_F \cap V_i)\}}(U) = 0$). So the duality holds for the "elementary" neighbourhoods.<|endoftext|> -TITLE: Positivity of a rational function -QUESTION [8 upvotes]: A rational function is called positive if all its Taylor coefficients are positive. - Friedrichs-Lewy conjecture states the positivity of the rational function - \begin{eqnarray*}\frac{1}{ -(1-x)(1- y)+(1- y)(1-z)+(1-z)(1-x)} -= \sum\limits_{ k,m,n\ge0} -a_{k,m, n }x^k y^mz^n. \end{eqnarray*} -The conjecture was first proved by G. Szego. -Let $P_n=\prod\limits_{i=1}^n(1-x_i)$, is it true that the following rational function is positive \begin{eqnarray*}\frac{1}{ -\sum\limits_{i=1}^n\frac{P_n}{1-x_i}}=\sum\limits_{i_1,i_2\cdots, i_n\ge 0}a_{i_1,i_2\cdots, i_n }x_1^{i_1}x_2^{i_2}\cdots x_n^{i_n}\quad? \end{eqnarray*} -A reference: - -Armin Straub, Positivity of Szegö's rational function, Advances in Applied Mathematics - 41 Issue 2 (2008) pp 255–264, doi:10.1016/j.aam.2007.10.001, (Wayback Machine pdf) - -REPLY [9 votes]: The answer is yes. -This was already proved in Gabor Szegö's original paper from 1933: -G. Szegö, Über gewisse Potenzreihen mit lauter positiven Koeffizienten, Mathematische Zeitschrift, Volume 37, Number 1, 674-688, DOI: 10.1007/BF01474608 -The result can be found in Paragraph 3 "Verallgemeinerungen". However, apparently the simplified proof method due to Armin Straub which you mention applies also to this generalization.<|endoftext|> -TITLE: What is known about the category of monads on Set? -QUESTION [24 upvotes]: Monads on the category Set of sets and functions are somehow fundamental objects of category theory, and moreover they have important applications to computer science. We know of a good number of monads on Set, but they all appear (at least to me) as isolated examples (other than three big classes of them I'll discuss below). I'm wondering what is known about the category of monads on Set; call it $Mnd$. Below, I'll refer to a monad $(M,\eta,\mu)$ in $Mnd$ simply by its functor, $M$. -Clearly, the category $Mnd$ is not small because for any set $E$, there is a monad $X\mapsto X\amalg E$ and another monad $X\mapsto X^E$. It would be nice to "classify" monads so as to see these as two, rather than as a large number, of examples. Are there ways to "classify" objects in $Mnd$ to notice more broad patterns? Here's another class of them for example: those coming from algebraic theories (e.g. the free group monad, the free ring monad, etc.) -For what types of diagrams does $Mnd$ have limits or colimits? Clearly $Mnd$ has an initial object (the identity monad) and a final object ($X\mapsto\{\star\}$). -Suppose given a monad $M\in Mnd$, and suppose we know (1) the set $M(\emptyset),$ (2) the set $M(\{\star\})$, and (3) the function $M(\emptyset\to\{\star\})$. Can we say anything else about $M$? Can we limit the three possible answers to (1),(2),(3) above for monads? -Are any other interesting facts known about $Mnd$? - -REPLY [23 votes]: The category of all monads on Set (often this is called $Mnd$, and $Mon$ means the category of monoids) is not just large, but has large hom-sets. -But if you restrict it to the full subcategory $Mnd_f$ of all finitary monads, then you get a beautiful category: it is not just complete and cocomplete but locally finitely presentable. -It is even monadic over the category $Set^N$ of families of sets indexed by the natural numbers. -The finitary monads are the ones that correspond to finitary (one-sorted) algebraic theories, and so include the monads for monoids, commutative monoids, groups, rings, Lie algebras (over a given field), and so on. More formally, a monad is finitary when its endofunctor part $T:Set\to Set$ preserves filtered colimits, or equivalently when $T$ is the left Kan extension of its restriction to finite sets. -An example of a monad which is not finitary is the ultrafilter monad, whose algberas are compact Hausdorff spaces. -Instead of finitary monads, you can take monads of rank $\alpha$, for some regular cardinal $\alpha$ - these are the ones whose endofunctors preserve $\alpha$-filtered colimits, and which can be described in terms of $\alpha$-small operations and equations. The full subcategory $Mnd_\alpha$ of $Mnd$ consisting of the monads of rank $\alpha$ is also locally finitely presentable. For example $Mnd_f$ is just the case where $\alpha=\aleph_0$. -The inclusion $Mnd_\alpha\to Mnd$ does preserve colimits (in fact it has a right adjoint), and so $Mnd$ does have colimits of diagrams of monads with bounded rank. -An example of a monad which does not have a rank is the powerset monad, whose algebras are the complete semilattices.<|endoftext|> -TITLE: Good programs for drawing (weighted directed) graphs -QUESTION [45 upvotes]: Does anyone know of a good program for drawing directed weighted graphs? - -REPLY [2 votes]: Gephi is pretty user-friendly and has good data-visualization features, they describe it as "like Photoshop but for graph data".<|endoftext|> -TITLE: Presentations of simple groups -QUESTION [8 upvotes]: Finite simple groups (non-abelian) can generated by two elements. -Let $G=\langle x,y|x^l=y^m=(xy)^n=1,...\rangle$ be a finite simple group (non-abelian), and $\langle x,y|x^p=y^q=(xy)^r=1,...\rangle$ be another presentation of $G$. (Here, "..." means possibly more relations). -1) If $(1/l)+(1/m)+(1/n)<1$, then does it imply that $(1/p)+(1/q)+(1/r)<1$? -2) If $(1/l)+(1/m)+(1/n)=1$, then does it imply that $(1/p)+(1/q)+(1/r)=1$? -3) If $(1/l)+(1/m)+(1/n)>1$, then does it imply that $(1/p)+(1/q)+(1/r)>1$? -(For a group $G=\langle x,y|x^l=y^m=(xy)^n=1,...\rangle$ to be a group of symmetries of a compact orientable surface, there are some restrictions on the integers $l,m,n$ and genus of surface, due to Riemann-Hurwitz relation. If the integers (l,m,n) satisfy some conditions as above, then it will allow us to consider less number of presentations of finite simple groups to check for possibility of action of the group on the surface). - -REPLY [2 votes]: The only groups where 1/l+1/m+1/n is greater than one are the cyclic groups, the dihedral groups, and the groups A(4), S(4), and A(5). Therefore, A(5) is the only simple groups where 1/l+1/m+1/n>1. As mentioned above, it is also a group such that 1/p+1/q+1/r<1, so it is a counterexample to (1) and (3). -All groups such that 1/l+1/m+1/n=1 are solvable, so therefore (2) is vacuously true. -Therefore your statement can be made into the stronger (and more accurate) statement: The only simple group with presentation $G=\langle x,y|x^l=y^m=(xy)^n=1,...\rangle$, when 1/l+1/m+1/n>1 is the group A(5)<|endoftext|> -TITLE: Does the amenability problem for Thompson's group $F$ predate 1980? -QUESTION [19 upvotes]: The first place where the amenability problem for Thompson's group $F$ appears in the literature is, I believe, 1980 in a problems article by Ross Geoghegan. I have heard, however, vague comments to the effect that the problem was considered by other people before this. Does anyone have any knowledge about the existence of this problem prior to 1980? -Edit: Following Mark's advice offline, I wrote Richard Thompson to verify the details of Mark's answer. He did confirm that he considered the problem. He first observed that his group $F$ did not contain $\mathbb{F}_2$. He then discovered the material on amenability in Hewitt and Ross's text on abstract harmonic analysis. He then observed that $F$ was not elementarily amenable. This occurred sometime prior to his 1973 visit to University of Illinois at Urbana-Champaign to visit Day. -He did not, however, attend Greenleaf's series of lectures in 1967. Instead he read Greenleaf's 1969 text "Invariant Means on Topological Groups and Their Applications." It seems that the best date to attach to Thompson's consideration of the question is 1973. Of course Thompson never published his observations and they were not widely circulated. His observations mentioned above were rediscovered by others in the 1980s (the question itself by Geoghegan). -I have invited Richard to post an answer, in which case I will delete this edit. - -REPLY [22 votes]: I think the story is this. Greenleaf gave a series of lectures on amenability in Berkeley in 1967. He mentioned the Day-von Neumann's problem, in particular. At that time Thompson was a student in Berkeley and has discovered his group already, he also proved that all free subgroups of his group are cyclic. Thus the problem about amenability of $F$ was formulated by Thompson in 1967. It was not published then, and the first time the problem was published was in 1980 by Ross Geoghegan.<|endoftext|> -TITLE: What is the Hochschild cohomology of the dg category of perfect complexes on a variety? -QUESTION [15 upvotes]: Let $X$ be a quasi-projective variety over a field $k$. Let $D_{qcoh}$ be a dg enhancement of the unbounded derived category of quasi-coherent sheaves over $X$, and $D_{perf}$ its full subcategory of perfect complexes. -This question is about Hochschild cohomology in the dg category sense. The question in the title gives the gist, but a more precise question is the following. -There is a restriction homomorphism -$$ r\colon HH^\ast(D_{qcoh},D_{qcoh}) \to HH^\ast(D_{perf},D_{perf}). $$ -This is clear if one computes Hochschild cohomology using the bar complex. -Question: Is $r$ an isomorphism? -The left-hand side makes me nervous because of the unbounded complexes, but I think this is true in the case of smooth $X$. I'm interested in singular varieties (or schemes). What about the affine case? To be concrete, what about the case where $X$ is a local complete intersection? -Comments: The category $D_{qcoh}$ looks much more ferocious than $D_{perf}$. The reason for bringing it into the picture is that its Hochschild cohomology seems to be better understood. One possible reference (which also discusses similar results for perfect complexes, but only in the smooth case) is -Toen's article on derived Morita equivalence. If I understood correctly (did I?), $HH^\ast(D_{qcoh},D_{qcoh})$ is isomorphic to -$$HH^\ast(X):= Ext^\ast_{\mathcal{O}_{X\times X}}(\delta_{\ast}\mathcal{O}_X,\delta_{\ast}\mathcal{O}_X)$$ -where $\delta\colon X\to X\times X$ is the diagonal. And this is the thing I really want to compare to $HH^\ast(D_{perf},D_{perf})$, for the reason that I know how to compute it in examples. There is a local-to-global Ext spectral sequence converging to $HH^\ast(X)$, and for local complete intersections one can use the Hochschild-Kostant-Rosenberg isomorphism to understand the Ext-sheaf. -Motivation: I've been looking at manifestations of homological mirror symmetry in which one has an embedding of the Fukaya category of a symplectic manifold into $D_{per}$ for a mirror variety. I'd like to compute Hochschild cohomology of the Fukaya category via algebraic geometry. - -REPLY [18 votes]: The answer is yes - at least if you take for the definition of $HH^*$ the self-ext of the identity functor. -For any quasicompact quasiseparated scheme we know (thanks to Thomason-Trobaugh) that $D_{qc}(X)$ is compactly generated by the perfect complexes. This means that $D_{qc}(X)=Ind(D_{perf}(X))$ -- the quasicompact [dg enhanced throughout!!] derived category is the ind category (in the $\infty$-categorical sense) of the perfect one. -Now it's an easy consequence of various results in Lurie's Higher Topos Theory [and maybe by now appears in there explicitly?] that -$\bf Proposition$: Passing to ind-categories induces an equivalence of [symmetric monoidal] $\infty$-categories between small, idempotent-complete stable $\infty$-categories (with exact functors) and compactly generated presentable stable $\infty$-categories (with proper functors -- continuous functors that preserve compact objects). -(This is spelled out in my paper with Francis and Nadler.) This is very useful -- it means you can go back and forth freely between small concrete categories that you like and big flexible objects which have all (small) limits and colimits, where the adjoint functor theorem and many other things apply. -So the passage from perfect to quasicoherent is part of an equivalence of categories. -It also means in particular that all exact functors on perfect complexes are representable by quasicoherent sheaves on the square --- though unless you're smooth and proper these will not be precisely the perfect complexes on the square.. -Since Hochschild cohomology is given by self-Ext of the identity functor (which is certainly a proper functor), we can choose to calculate it either with the small categories or with the large categories. (For a reference for the equivalence between dg and stable $\infty$-categories I suggest this paper by Blumberg, Gepner and Tabuada.) -So now you might want to check that the cyclic bar construction for a small dg category does indeed calculate self-Ext of the identity functor. (Maybe you want to take the dg category to be pretriangulated, or first prove that both sides are invariant under Morita equivalence.) This is pretty clear I think - at least notationally it's easier if we assume our category has one generator (equivalently finitely many -- and this is always the case for [q-c,q-s] schemes), hence is just modules over a dg algebra. Then we observe the standard bar construction is a free resolution of the algebra $A$ considered as a bimodule (i.e. of the identity functor), and the cyclic bar construction computes the Ext. For the multiobject version I think there's an MO discussion already somewhere..<|endoftext|> -TITLE: Uniform boundedness of an $L^2[0,1]$-ONB in $C[0,1]$ -QUESTION [11 upvotes]: Assume that we have an orthonormal basis of smooth functions in $L^2[0,1]$. Are there useful practical criteria to determine whether the sup-norm of the basis functions has a uniform bound? I am sure people investigated this but I do not manage to find a useful result. Can someone offer me references, where I can read about this or a connected problem? -I would like to investigate the properties of some fourth order differential operators (with the corresponding boundary conditions) in $C[0,1]$. Since these operators are nice selfadjoint operators with compact resolvent in $L^2[0,1]$, and the corresponding eigenfunctions are analytic, it would be straightforward to ask whether the properties I need can be carried over using the basis. -Edit on 02/18/2011: -Though completely unrelated to my question, googling brought me the following interesting paper: -Toth and Zelditch, Riemannian manifolds with uniformly bounded eigenfunctions, Duke Math. J. Volume 111, Number 1 (2002), 97-132. -This gives me hope, however, that my question might have interested somebody... - -REPLY [6 votes]: I might be wrong, but there answer to you general question is 'no'. Take an arbitrary function with $L^2$ norm equal to one, then start an orthogonalization procedure (which leaves your first function fixed) and produce an ONB. You have no way to bound the sup norm of the first function. -But your actual needs seem much more natural, and standard. For an ONB which arises as the eigenfunction set of a given selfadjoint, positive, elliptic operator $L$ of order $m$, you can resort to an elementary Sobolev embedding. Let $f_j$ be the eigenfunction corresponding to the eigenvalue $c_j$. Then the sup norm of $f_j$ is less than the $H^{1/2+\epsilon}$ norm of $f_j$, which is equivalent to the $L^2$ norm of $L^{\frac 1{2m}+\epsilon}f_j=c_j^{\frac 1{2m}+\epsilon}f_j$ which is exactly $c_j^{\frac 1{2m}+\epsilon}$. In general I think you can not do better than that. When $L$ is the Laplace-Beltrami operator on some compact manifold, or powers thereof, there exist much more precise results.<|endoftext|> -TITLE: Ordinal-indexed homology theory? -QUESTION [16 upvotes]: Back when I was a grad student sitting in on Mike Freedman's topology seminar at UCSD, he posed the following question. Does there exist a good homology theory $H_{\alpha}(X)$ where $\alpha$ is an ordinal? The obvious ways of trying to define an $H_\omega(X)$ don't work out very well essentially because $S^\infty$ is contractible. Still, I've always thought it was an interesting question, and I wonder whether anyone has thought of this or knows of any ideas along these lines. - -REPLY [4 votes]: I'm not sure whether this is what you are after, but this paper, Semi-infinite cycles in Floer theory: Viterbo's Theorem, by Max Lipyanskiy, develops a theory of ($\omega$/2+k)-dimensional cycles in an $\omega$-dimensional manifold with a choice of polarization of its tangent space. -($k\in\mathbb{Z}$ above.)<|endoftext|> -TITLE: analytic vs. algebraic Gauss-Manin connection -QUESTION [21 upvotes]: There are the following two notions of "Gauss-Manin connection": - -The complex-analytic one: let $f:X\to S$ be a smooth family of complex manifolds. Then we obtain a local system $R^nf_{\ast}\mathbb{C}$ of complex vector spaces on $S$, defining a holomorphic vector bundle $\mathcal{V}=R^nf_{\ast}\mathbb{C}\otimes\mathcal{O}_S$ on $S$ with an integrable connection $\nabla :\mathcal{V}\to\mathcal{V}\otimes\Omega_S^1$. Now the vector bundle $\mathcal{V}$ can be identified with the relative de Rham cohomology $\mathcal{H}_{dR}^n(X/S)$ of the family, so we get a connection on the latter. -The algebraic one: let $f:X\to S$ be a smooth morphism of smooth schemes over a field $k$. Now Katz and Oda in "On the differentiation of De Rham cohomology classes with respect to parameters" (J. Math. Kyoto Univ. 8 (1968), pp. 199-213) construct an integrable connection on $\mathcal{H}_{dR}^n(X/S)$ as some boundary map in a certain spectral sequence. - -It is implicit in the literature that these two constructions are compatible, i.e. for a smooth family of smooth varieties over the complex numbers, the connection described in 1. is just the analytification of the one in 2. This sounds pretty reasonable as well. But thinking a bit about it, I was unable to come up with an argument, so could perhaps someone give me a hint where to find this or how to do it? - -REPLY [12 votes]: The two constructions are compatible. -Your first definition of the Gauss-Manin connexion is $ DR^{-1} (R f_* \mathbb{C}_X ) $. Here $DR : D^b_{hr}(\mathcal{D}_X) \to D^b_c( \mathbb{C}_X )$ and $DR(\mathcal{M}) = \omega_X \otimes^L_{D_X} \mathcal{M}$ is the analytic de Rham complex. This is an equivalence by the Riemann-Hilbert correspondance. It sends an $\mathscr{O}$-coherent $\mathcal{D}$-module (i.e. a vector bundle with an integrable connexion) to a local system (i.e. a locally constant sheaf). The inverse functor sends a locally constant $V$ to the vector bundle $\mathscr O_X \otimes_{\mathbb{C}} V$ together with the only connexion so that $V$ is the local system of horizontal sections in $(\mathscr O_X \otimes_{\mathbb{C}} V,\nabla)$. -Your second definition is a special case of the direct image $\mathcal{H}^n(f_+\mathscr O_X)$ in the sense of D-modules for $f$ smooth. -The (algebraic or analytic) de Rham complex on $X$ is filtered by -$$ - L^r\Omega_X^\bullet = f^*\Omega_Y^r \otimes \Omega_X^{\bullet-r} -$$ -This induces a spectral sequence -$$ - R^pf_*(Gr_L^q \Omega_X^\bullet) \Rightarrow R^{p+q}f_*(\Omega_X^\bullet) -$$ -But $Gr_L^q \Omega_X^\bullet = \Omega^q_S \otimes \Omega_{X/S}^{\bullet-q}$ and the Gauss-Manin connexion can be interpreted as the differential -$$ - R^nf_*\Omega_{X/S} \to \Omega^1_S \otimes R^{n}f_*\Omega_{X/S} -$$ -in the spectral sequence. Now the analyfication functor is compatible with inverse and direct images of $\mathcal{O}$-modules and it sends $\Omega^i_X$ to $\Omega^i_{X^{an}}$ so it sends one spectral sequence to the other. This shows that the analytic and algebraic Gauss-Manin connexions are compatible. -It remains to prove is that $DR$ is compatible with direct images -$$ - DR f_+ \mathcal{M} \overset{\sim}{\to} Rf_* DR \mathcal{M} -$$ -(for $f$ a smooth morphism of complex analytic varieties and $\mathcal{M} =\mathscr O_X$ a regular connexion). -For $f:X\to S$ an open immersion this is a theorem of Deligne (cf. Borel IV.6.1) and it is actually equivalent to regularity (by a theorem of Mebkhout I think). -For $f:X\to S$ proper this is done in (Borel VIII.15): analyfication induces the natural transformation $DR f_+ \to Rf_* DR $ and it is an isomorphism because of the projection formula. In the case $\mathcal{M} =\mathscr O_X$ you can make this a little bit more concrete. It is enough to prove that the natural transformation is an isomorphism on the fibers. But by proper base change, this means you can suppose $S$ is a point so this is equivalent to $H^n_{dR}(X^{an}) \to H^{n}(X^{an};\mathbb{C})$ being an isomorphism which is the Poincaré Lemma.<|endoftext|> -TITLE: Reference request: equivalence of formal group laws and Lie algebras in characteristic zero -QUESTION [6 upvotes]: Let $k$ be a field of characteristic zero. Wikipedia states that the natural functor from finite-dimensional formal group laws over $k$ to finite-dimensional Lie algebras over $k$ is an equivalence of categories but does not provide a reference, and I'm not sure where to find one. Does anyone know of one? -(Not really sure how to tag this. I am not necessarily interested in any particular application.) - -REPLY [6 votes]: As indicated by Serre's reference list, a basic source of ideas about formal group laws would be the papers by Michel Lazard in the period 1955-1965. Just do a quick search in www.numdam.org using his last name. One of his shorter papers is Sur les groupes de Lie formels à un paramètre. Bulletin de la Société Mathématique de France, 83 (1955), p. 251-274. But I'm not sure these papers will be explicit enough to answer your question completely. Lazard was especially interested in dealing with $p$-adic fields, while Dieudonne dealt further with fields of prime characteristic. Certainly Lazard did more than anyone else to establish the abstract foundations of formal group laws. -Another possibly more readable source would be Bourbaki's Groupes et algebres de Lie, Chapters II-III (especially Chapter II) together with the historical notes at the end of that volume. Here in particular the connections with work of Dynkin and others, along with the Baker-Campbell-Hausdorff formula, are emphasized. -ADDED: I think Emerton's brief answer and the added comments essentially answer the original question; but I wanted to point to the broader background sources as well. Serre's 1965 lecture notes Lie Algebras and Lie Groups (Benjamin) end up in Chapter V of Part II at Lie Theory, with the initial convention: "Unless otherwise specified, $k$ will denote a field complete with respect to a non-trivial absolute value." In some places it is required that $k$ be of characteristic 0; then it may be allowed to be just a $\mathbb{Q}$-algebra. -For some classical Lie theory, $k$ is assumed to be the real or complex field. -But much is also done with analytic groups over ultrametric fields, etc. So it's important to keep track of which kind of base ring or field you are working over. -Theorem 3 on page 5.28 is the theorem in question here. Serre gives a complete proof but relies heavily on previous material. By now he is making a distinction between the "formal" case and the "analytic" case; it's essential to be in the "formal" case when you want to work over an arbitrary field of characteristic 0. With these qualifications, I'd agree that for Theorem 3 a relatively intuitive proof is possible avoiding much of the surrounding "analytic" material.<|endoftext|> -TITLE: Why must nilpotent elements be allowed in modern algebraic geometry? -QUESTION [40 upvotes]: On the Wikipedia page1 about algebraic varieties https://en.wikipedia.org/wiki/Algebraic_variety, a sentence reads as follows: -[[A more significant modification is to allow nilpotents in the sheaf of rings. A nilpotent in a field must be 0: these if allowed in coordinate rings aren't seen as coordinate functions. -From the categorical point of view, nilpotents must be allowed, in order to have finite limits of varieties (to get fiber products).]] -So I am wondering if there is an intuitive example to get non-reduced 'schemes' from reduced 'algebraic varieties' (probably by taking fiber product or alike)? -1 Link to a revision from February 2011. - -REPLY [2 votes]: Question 1: "So I am wondering if there is an intuitive example to get non-reduced 'schemes' from reduced 'algebraic varieties' (probably by taking fiber product or alike)?" -Nilpotent elements are useful in mathematics (algebra/geometry, complex analysis, differential geometry) when studying derivatives, differential operators and tangent spaces. -Let $k$ be a field and let $A:=k[x],B:=k[x,y]$ be polynomial rings with $I:=(y-x)\subseteq B$ the ideal defined by the element $y-x$. Let $J(l):=B/I^{l+1}$ and let -$T^l:A \rightarrow J(l)$ be defined by $T^l(f(x)):=f(y)$. The left $A$-module $J(l)$ -is free of rank $l+1$ on the elements $(dx)^i:=(y-x)^i$ for $i=0,..,l$. You can prove this using induction. -Example 0. Let $k$ have characteristic zero. It follows from the binomial theorem and an induction that -$T^l(f(x))=\sum_{i=0}^l \frac{f^{(i)}(x)}{i!}(dx)^i$. -Example 1. Let $f(x):=x^2$. We get -$T^l(f(x)):=f(y)=y^2=(x+dx)^2=x^2+2xdx+(dx)^2=$ -$f(x)+ f'(x)dx+\frac{f^{(2)}(x)}{2!}(dx)^2$. -Hence the map $T^l$ is the Taylor expansion of the polynomial $f(x)$. The map -$T^l$ is a "differential operator" of order $l$ from $A$ to $J(l)$. If $k$ is the field of real numbers and $\mathfrak{m}:=(x-a)$ for $a\in k$ we may pass to the fiber -$J(l)(a):= \kappa(\mathfrak{m})\otimes_A J(l) \cong k\{(dx)^i\}$ -and as an element of $J(l)(a)$ it follows -$T^l(f(x)) = \sum_{i=0}^l \frac{f^{(i)}(a)}{i!}(dx)^i \in J(l)(a)$. -Hence in this case we get the value of the Taylor series of $f(x)$ of order $l$ at the -real number $a\in k$. -If $z=a+ib$ with $a,b$ real numbers and $b \neq 0$ it follows -$p(x):=(x-z)(x-\overline{z})=x^2-2ax+a^2+b^2 \in k[x]$ -is an irreducible polynomial with $k[x]/(p(x))\cong \mathbb{C}$. Choosing an explicit isomorphism $\phi: k[x]/((p(x)) \cong \mathbb{C}$ it follows -$0=\phi(p(x))=p(\phi(x))$ -hence $\phi(x)$ equals $z$ or $\overline{z}$. Hence if $\mathfrak{p}=(p(x))$ and we pass to the fiber $\kappa(\mathfrak{p})\otimes J(l):=J(l)(\mathfrak{p})$ we get an isomorphism -$J(l)(\mathfrak{p})\cong \mathbb{C}\{(dx)^i\}$. -If we "Taylor expand" $f(x)$ and pass to the fiber $J(l)(\mathfrak{p})$ -we get an element $T^l(f(\phi(x))) \in \mathbb{C}\{(dx)^i\}$, and the value of the Taylor series $T^l(f(\phi(x)))$ -depends on the choice of an isomorphism $\kappa(\mathfrak{p})\cong \mathbb{C}$. and the choice of $\phi(x)\in \mathbb{C}$. What does not depend on this choice is if the higher order derivative $\frac{f^{(i)}(x)}{i!}$ is zero or not. -Note: If $f(x) \in \mathbb{R}[x]$ is any polynomial it follows $T^l(f(x))=0$ in the fiber at $x-a$ iff $f^{(i)}(a)=0$ for $i=0,..,l$. This holds iff $f(x)=f_l(x)(x-a)^{l+1}$. Hence $T^l(f(x))=0$ in the fiber at $x-a$ iff $a$ is a zero of $f(x)$ of order $l+1$. If $p(x):=(x-z)(x-\overline{z})$ it follows $T^l(f(x))=0$ in the fiber at $p(x)$ iff $f(x)=f_l(x)p(x)^{l+1}$. Hence $T^l(f(x))=0$ iff $z$ and $\overline{z}$ are zeros of $f(x)$ of multiplicity $l+1$. -If $\pi: Spec(J(l))\rightarrow Spec(A)$ is the canonial map, it follows $Spec(J(l))$ is a "geometric vector bundle" in the sense of Hartshorne, Ex II.5.18. The scheme $Spec(J(l))$ is non-reduced but $Spec(A)\cong \mathbb{A}^1_k$ is reduced. This gives an intuitive explanation of why non-reduced schemes are important in the study of reduced schemes, answering Question 1. The construction of $J(l)$ can be done in the "language of algebraic varieties" using coherent sheaves. -This construction globalize to any scheme/differentiable manifold/complex manifold/etc. -Hence nilpotent elements and non-reduced ideals are useful in algebraic geometry, differential geometry and complex analysis. In these fields we study tangent spaces and differential operators. -For any smooth projective curve $C \subseteq \mathbb{P}^n_k$ there is a similar construction of a finite rank vector bundle $\mathcal{J}_C^l$ and a universal differential operator -$T^l: \mathcal{O}_C \rightarrow \mathcal{J}_C^l$ which locally "Taylor expands" sections: For any open subset $U \subseteq C$ we get a differential operator -$T^l(U): \mathcal{O}_C(U) \rightarrow \mathcal{J}_C^l(U)$ -and the map $T^l(U)$ Taylor expands the section $s\in \mathcal{O}_C(U)$. If $k$ is the field of real or complex numbers, we may view $s$ as a real valued or complex valued function, and in this case $T^l(U)(s)$ is the Taylor expansion of $s$. -Example 2. If $C$ is a complex holomorphic curve and $\mathcal{O}_C$ is the sheaf of holomorphic functions on $C$, there is a similar construction. There is a finite rank holomorphic vector bundle $\mathcal{J}_C^l$ and a differential operator -$T^l: \mathcal{O}_C \rightarrow \mathcal{J}_C^l$ -with the same properties. -Example 3. Complex holomorphic curves and nilpotent ideals. The holomorphic vector bundle $\mathcal{J}_C^l$ may be constructed using the ideal of the diagonal $\mathcal{I} \subseteq \mathcal{O}_{C \times C}$. The ideal $\mathcal{I}$ is a coherent sheaf of ideals, and one may prove that the quotient $\mathcal{O}_{C \times C}/\mathcal{I}^{l+1}$ is (as a left $\mathcal{O}_C$-module) locally trivial of rank $l+1$. The corresponding holomorphic vector bundle is isomorphic to $\mathcal{J}_C^l$. The pair $(C\times C, \mathcal{O}_{C\times C}/\mathcal{I}^{l+1})$ may be viewed as a "locally ringed space" with nilpotent elements in the structure sheaf. The sheaf $\mathcal{J}_C^l$ is supported on the diagonal $\Delta(C) \subseteq C \times C$, and it follows $\mathcal{J}_C^l$ is a sheaf of left and right $\mathcal{O}_C$-modules. The sheaf $\mathcal{J}_C^l$ is locally trivial as left and right $\mathcal{O}_C$-module, but these two structures are not isomorphic in general. If you consider the ideal sheaf $\overline{\mathcal{I}} \subseteq \mathcal{J}_C^l$ and choose any non-zero section $s \in \overline{\mathcal{I}}(U)$ for $U \subseteq C$ an open set it follows $s^{l+1}=0$. Hence the sheaf of rings $\mathcal{J}_C^{l}$ has nilpotent elements. When we restrict $\mathcal{J}_C^{l}$ to the diagonal $\Delta(C)$ -we get a sheaf of left and right $\mathcal{O}_C$-modules, but for local sections $s \in \mathcal{O}_C(U)$ and $\omega \in \mathcal{J}_C^l(U)$ it follows $s\omega \neq \omega s$. -You may have seen in Hartshorne exercise II.5.18 that for a scheme $(X,\mathcal{O}_X)$ there is an "equivalence of categories" between the category of locally free finite rank $\mathcal{O}_X$-modules and the category of finite rank geometric vector bundles, and there is a similar result valid for complex holomorphic manifolds. Given a complex holomorphic manifold $(Y, \mathcal{O})$ where $\mathcal{O}$ is the sheaf of complex holomorphic functions -on $Y$, it follows there is an "equivalence of categories" between the category of locally trivial finite rank $\mathcal{O}$-modules and the category of finite rank holomorphic vector bundles on $Y$. -In a calculus course in one variable you define the derivative $f'(x)$ of a real valued smooth function $f(x)$ using limits. In algebraic geometry we work over arbitrary fields (or rings) and we cannot "take limits". Using non-reduced ideals we can still formally "take derivatives" of sections of sheaves as explained above. In characteristic zero we get as you can observe above the correct Taylor expansion of any polynomial (or rational) function. In characteristic $p>0$ the notion is useful as well, ref. Question 1 in your question. -See also the discussion at: -https://math.stackexchange.com/questions/3928532/about-the-reducedness-in-algebraic-geometry/3942701#3942701<|endoftext|> -TITLE: Finitely generated algebra in which every element is annihilated by a non-zero polynomial -QUESTION [5 upvotes]: Let $K$ be a field, and $A$ a finitely generated associative algebra over $K$. We suppose that $A$ has a unit and that every element $x$ of $A$ is annihilated by a non-zero polynomial $P_x$ depending on $x$. Is $A$ a finite-dimensional vector space over $K$ ? -Under the additional assumption that there is an integer $d$ such that, for all $x \in A$, the degree of $P_x$ is less than $d$, is the result true? - -REPLY [9 votes]: This is the Kurosh problem, which has a negative solution. If I recall correctly, one exhibits an example using the Golod-Shafarevich lemma. -Wikipedia has a page on this, in fact. The example was constructed by Golod.<|endoftext|> -TITLE: need references regarding the elementary theory of free semigroup and free abelian groups -QUESTION [6 upvotes]: Recently, I read that two free abelian groups $S$ and $T$ have the same elementary theory if and only if rank$S$=rank$T$. Does anyone have a reference with a proof of this? Also, what is known about the elementary theory of non-abelian free semigroups? I know that non-abelian free groups of finite rank have the same elementary theory; does this imply the analogous statement for free semigroups? Thanks! - -REPLY [6 votes]: Two finitely generated free semigroups of different ranks have different elementary theories. Indeed, the set of elements $x$ such that $\forall z,t \neg(x=zt)$ is exactly the set of generators of the free semigroup. So the rank of a finitely generated free semigroup is elementary definable. -For the free Abelian group of rank $n$ the formula distinguishing it from any free Abelian group of rank $\gt n$ is this: -$$\exists x_1,...,x_{2^n} \forall y \exists z: yx_1=z^2 \vee yx_2=z^2 \vee ... \vee yx_{2^n}=z^2$$ -(we enumerate the subsets of the set of generators and for the subset number $i$ we denote the product of generators from that subset by $x_i$). It is easy to see that this formula holds in every free Abelian group of rank $\le n$ and does not hold if the rank is $\gt n$. For $n=1$ this is the same formula as in the answer of Henry Wilton. -The standard reference for elementary classification of Abelian groups is Szmielew, W. -Elementary properties of Abelian groups. Fund. Math. 41 (1955), 203–271. A shorter proof can be found in Kargapolov, M. I. On the elementary theory of Abelian groups. -Algebra i Logika Sem. 1 1962/1963 no. 6, 26–36 and Eklof, Paul C.; Fischer, Edward R. -The elementary theory of abelian groups. Ann. Math. Logic 4 (1972), 115–171 and Zakon, Elias -Model-completeness and elementary properties of torsion free abelian groups. -Canad. J. Math. 26 (1974), 829–840 (for torsion-free groups). - -REPLY [2 votes]: It is easy to prove that non-isomorphic free abelian groups (of finite rank) have distinct elementary theories, by exhibiting specific sentences that hold in one but not the other. For instance, $\mathbb{Z}$ is distinguished by the property that for some element $y$ (eg $y=1$), either $x$ or $x+y$ is even. In other words, the sentence -$\exists y~\forall x~\exists z~(x=2z) \vee (x+y=2z) $ -holds in $\mathbb{Z}$ but not in $\mathbb{Z}^n$ for any $n>1$. The same idea can be used to distinguish $\mathbb{Z}^m$ and $\mathbb{Z}^n$ for any $m\neq n$.<|endoftext|> -TITLE: Reference needed: Isomorphism on pi_1 and homology gives weak equivalence -QUESTION [8 upvotes]: Let $f : X \to Y$ be a map between a connected space $X$ and a space $Y$. If $\pi(f) : \pi_1(X) \to \pi_1(Y)$ is an isomorphism, and $H_n(f) : H_n(X, G) \to H_n(Y, G)$ is an isomorphism for all $n \ge 1$ and for any local system of coefficients $G$, then $X$ is weakly equivalent to $Y$. Does anyone have a reference (or proof) for this? - -REPLY [8 votes]: The proof in the simply connected case is well-known and is a consequence of the relative Hurewicz theorem. See Corollary 1, page 79 of Mosher and Tangora's book, Cohomology Operations and Applications in Homotopy Theory for this case. -If $X$ and $Y$ aren't 1-connected, then $f$ lifts to a map of universal covers -$\tilde f: \tilde X \to \tilde Y$ and your assumption about local coefficients implies that $\tilde f$ is a homology isomorphism. We can therefore apply the previous paragraph to show that $\tilde f$ is a weak equivalence. This implies that $f$ is since $f$ is a $\pi_1$-isomorphism.<|endoftext|> -TITLE: Universal property of the tangent bundle -QUESTION [28 upvotes]: If $X$ is a scheme (over some base scheme, but which I will ignore) its tangent bundle $T(X)$ is defined as the relative spectrum of the symmetric algebra of its sheaf of differentials. Combining the universal properties of these three constructions, we get a universal property of $T(X)$, namely: Defining $U[\epsilon] := U \times \text{Spec}(\mathbb{Z}[\epsilon]/\epsilon^2)$, we have a canonical bijection -$Hom(U,T(X)) \cong Hom(U[\epsilon],X)$ -Thus $T(-)$ is right adjoint to $(-)[\epsilon]$. -Question Is there a similar universal property of the tangent bundle of a manifold? -Here a manifold is assumed to be smooth and paracompact. I doubt that we can literally translate it, because $U[\epsilon]$ is $U$ as a topological space, but the structure sheaf now has the nilpotent $\epsilon$. My motivation comes from the observation that in every construction of the tangent bundle of a manifold I know, some nasty calculations with charts have to be made. I want to avoid this with the help of a universal property, proving that $T(\mathbb{R}^n)$ exists and then formally deducing the existence of $T(M)$ for every manifold. I know this can be done without using a universal property and that this would possibly not be the best construction or characterization of the tangent bundle, but it hopefully avoids irrelevant choices of charts. A similar question was asked here and here, but this seems to go in another direction. - -REPLY [10 votes]: The algebro-geometric definition does translate, almost literally. John Klein's comment to Martin's question provides one way of doing the translation. -Let $X$ be a manifold. Define a contravariant functor $F$ on manifolds by taking $F(Y)$ to be the set of pairs $(f, \delta)$ where $f : Y \rightarrow X$ is a morphism of manifolds and $\delta : C^\infty(X) \rightarrow C^\infty(Y)$ is a derivation with respect to the module structure on $C^\infty(Y)$ induced from $f$. Then $F$ is representable by $TX$. -Alternately, you may think of an element of $F(Y)$ as an extension of $f : Y \rightarrow X$ to a map of ringed spaces $Y[\epsilon] \rightarrow X$ where $Y[\epsilon]$ is the space $Y$, ringed by $C^\infty(Y)[\epsilon] / (\epsilon^2)$.<|endoftext|> -TITLE: When does a cosimplicial object compute homotopy colimits? -QUESTION [7 upvotes]: Suppose I want to compute the homotopy colimit of a diagram of spaces. There is a simple way of getting a simplicial space from this diagram, and a theorem tells me that taking the geometric realisation of this gets me the correct answer. -Now suppose I'm trying to compute homotopy colimits in some other model category - the category of baubles, say. I can still get a simplicial bauble from a diagram of baubles, but there is no longer any such thing as geometric realisation. However, I can define a realisation functor relative to a fixed cosimplicial bauble; in my previous example, I was using the cosimplicial space given by the standard n-simplices, but I could have defined a formally identical geometric realisation functor given any cosimplicial space. -My guess is that if my cosimplicial bauble is nice, I get the true hocolim of baubles by doing this. (For instance, I suspect 'simplicial chains on the standard simplices' suffices for the category of complexes of abelian groups.) My question is: what does nice mean here? Is there an easy-to-read set of sufficient conditions? - -REPLY [5 votes]: Dear Saul, -The answer to your question is the subject of chapters 16-19 of Phil Hirschhorn's book Model Categories and their Localizations. -To write out the answer in the general case would be prohibitively time consuming, but I'll write a little bit out. -Definition 19.1.5 Let $M$ be a framed model category, and let $\mathcal{C}$ be a small category. If $X$ is a $\mathcal{C}$-diagram in $M$, then the homotopy limit $\operatorname{holim} X$ is defined to be the equalizer of the maps -$$\prod_{\alpha\in \mathcal{C}} (\widehat{X}_\alpha)^{\mathfrak{N}(\mathcal{C}\downarrow \alpha)}\rightrightarrows \prod_{\sigma:\alpha\to \alpha'\in \operatorname{Arr}(\mathcal{C})} (\widehat{X}_{\alpha'})^{\mathfrak{N}(\mathcal{C}\downarrow \alpha)}$$ -Where $\widehat{X}_\alpha$ is the natural simplicial frame on $X_\alpha$. -In particular, the key concept here is the concept of a framed model category. A framed model category is defined to be a model category equipped with the data of cosimplicial and simplicial framing functors $M\to M^\Delta$ and $M\to M^{\Delta^{op}}$, where these are defined as follows: -A cosimplicial frame on an object $X$ is an object $\widetilde{X}$ equipped with a reedy weak equivalence $\widetilde{X}\to cc_\ast X$ where $cc_\ast X$ is the constant cosimplicial object defined by $X$. -A simplicial frame on an object $X$ is an object $\widehat{X}$ equipped with a Reedy weak equivalence $cs_\ast X\to \widehat{X}$ where $cs_\ast$ is the constant simplicial object defined by $X$. -In a framed model category, we require that we have functorial frames. It is proven in Hirschhorn's book that if the original model category has functorial factorizations, then there exist essentially unique (up to a contractible space of choices) functorial simplicial and cosimplicial frames on $M$ (Theorem 16.6.9 and Theorem 16.6.10). -I'm sure you won't have a hard time finding a copy (isn't Hirschhorn at MIT?). -If your model category is either combinatorial or cofibrantly generated (don't remember which, but I think you only need cofibrant generation. I believe that the defect with non-combinatorial cofibrantly generated model categories is that the diagram categories are not necessarily cofibrantly generated again, while combinatorial model categories actually are stable under exponentiation by small categories.), you can also define the holim and hocolim to be derived functors of the ordinary ones, since lim and colim give a quillen adjunction between the model structure on $M$ and its projective and injective diagram model categories.<|endoftext|> -TITLE: Hilbert modular forms -QUESTION [11 upvotes]: What is a good reference to learn about Hilbert modular forms both in the classical and adelic settings? For example, why usually one considers only parallel weight forms? Thanks. -Karl - -REPLY [9 votes]: I know of two pretty good books on Hilbert modular forms: Garrett's "Holomorphic Hilbert Modular Forms" and Freitag's Hilbert Modular forms. The latter only covers classical Hilbert modular forms whereas the former introduces both classical and adelic Hilbert modular forms. -Shimura's article "The special values of the zeta function associated with Hilbert modular forms" (Duke Math. Journal) is something of a classic and carefully builds up the theory of Hilbert modular forms (beginning with the classical theory and building up to the adelic theory). The reason that one constructs adelic Hilbert modular forms is to gain invariance under the full Hecke algebra (which is not automatic in the case of classical Hilbert modular forms over a totally real field of strict ideal class number greater than $1$). The Hecke operators play an extremely important role in Shimura's paper, so he spends quite a bit of time developing them. He also proves a number of other basic properties of Hilbert modular forms. -As Emerton suggested in his response, it is important to have a good grip on the classical case before moving on to the adelic case. This is especially important in Shimura's article, as the space of adelic Hilbert modular forms has two well-known and important direct sum decompositions. One is a direct sum over representatives of the strict ideal classes and has as its summands spaces of classical Hilbert modular forms. The other is a direct sum over Hecke characters extending a character $\psi: (\mathcal{O}_K/\mathfrak{N})^\times\rightarrow \mathbb C^\times$. Moving between all of these spaces is a bit subtle. -If you are looking for the quickest way to learn the basics of classical / adelic Hilbert modular forms I would probably suggest starting by reading the preliminary sections of a few papers. I say this because Hilbert modular forms can get VERY technical very quickly, to the point that one has trouble seeing the forrest for the trees. There have been a number of papers written within the last year or two on the computation of systems of Hecke eigenvalues of spaces of Hilbert modular forms. Most of these papers contain very good expositions on the basics of classical / adelic / quaternionic Hilbert modular forms. John Voight has written a few of these papers and has them available on his website. Another person active in this area is Lassina Dembele. -I do not quite agree with your assertion that people only (or even usually) consider the case of parallel weight. In fact, one often has good reason to consider Hilbert modular forms of non-parallel weight (as Shimura notes in the introduction of the aforementioned article). Much of the theory of classical (elliptic) modular forms generalizes rather straight-forwardly to the Hilbert modular case without restrictions on the weight vector. For example, in their paper "Twists of Newforms", Shemanske and Walling study the newform theory of arbitrary weight Hilbert modular cusp forms and show that for the most part, it is exactly as one would hope. -That said, there is one very good reason for restricting to the case of parallel weight: this is the only case in which Eisenstein series exist. Said differently, if the weight is not parallel then all Hilbert modular forms are cuspidal.<|endoftext|> -TITLE: How to show modularity of an elliptic curve? -QUESTION [27 upvotes]: In the days before [W, TW, BCDT], how did people show that specific elliptic curves over $\mathbb{Q}$ were modular? For instance, I was reading through a paper of Buhler, Gross and Zagier from 1985 on the curve 5077a, and they say that modularity can be checked by a finite computation in the 422-dimensional space of cuspforms of weight 2 and level 5077 (and remark at the end that Serre and Mestre have checked it). A google search brought up the name "Faltings-Serre method": was this the technique of choice? Also, are there any good references for it? - -REPLY [5 votes]: The answer is outlined in Don Zagier's 1985 paper Modular points, modular curves, modular surfaces and modular forms.<|endoftext|> -TITLE: Existence/Uniqueness of solutions to quasi-Lipschitz ODEs -QUESTION [10 upvotes]: Would the Picard–Lindelöf theorem still be true if the requirement that f be Lipschitz continuous in y was replaced with the requirement that f be almost Lipschitz in y? -If not, are there any moduli of uniform continuity weaker than Lipschitz continuity that it is known suffice, or results indicating that there can't be any? - -REPLY [21 votes]: Yes. This follows from the classical uniqueness theorem due to Osgood (the original paper appeared in 1898). - -Osgood's Criterion. Let $\omega(t,u)=\phi(t)\psi(u)$ where $\phi(t)\geq 0$ is continuous - on the interval $(0,a)$ and $\psi(u)$ is continuous on $\mathbb R_{+}$, $\psi(0)=0$, $\psi(u)>0$ for $u>0$, and - $$\int_{0}^{\epsilon}\phi(t)dt<\infty,\qquad \int_{0}^{\epsilon}\frac{du}{\psi (u)}=\infty$$ - for some $\epsilon>0$. Suppose that the mapping $f:[0,a]\times B_R(x_0)\to \mathbb R^d$ satisfies the condition - $$||f(t,x_1)-f(t,x_2)||\leq\omega (t,||x_1-x_2||)$$ - for any $t\in(0,a]$ and any $x_1,x_2\in B_R(x_0)$. Then the initial value problem - $$\dot{x}=f(t,x),\qquad x(0)=x_0$$ - has at most one solution on the interval $[0,\delta]$ with some $\delta>0$. - -Osgood's theorem allows for the mappings $f(t,x)$ which are discontinuous at $t=0$. (Actually, the condition that $\phi(t)$ is continuous on $(0,a)$ can be replaced with an assumption -of mere integrability.) Of course, the existence of a local solution is implied by the Peano theorem under the additional assumption that $f$ is continuous in $(t,x)$. -Moreover, Wintner showed that Osgood's uniqueness condition implies the convergence of successive Picard iterations to a local solution on a sufficiently small interval (A. Wintner, "On the Convergence of Successive Approximations", Amer. Journal of Math. Vol. 68 (1946), pp. 13-19).<|endoftext|> -TITLE: Proof that the homotopy category of a stable $\infty$-category is triangulated -QUESTION [18 upvotes]: I've been looking at various general strategies for proving that some category is triangulated, and Lurie manages to prove that a huge class of interesting examples of categories that we know about are triangulated in his book Higher Algebra (formerly DAG I-IV and VI).(EDIT: here's a link to the book) The trouble is that I am very new to this language, and so what he calls $\infty$-categorical notions that are basic and easily motivated' I see as foreign and unfamiliar. -The part I'm really interested in is the proof of the octahedral axiom on page 24 of Higher Algebra. He builds a diagram using a proposition from Higher Topos Theory that seems completely out of context (to me!). The proposition says: -``Suppose we are given a diagram of $\infty$-categories $\mathcal{C} \rightarrow \mathcal{D}' \leftarrow \mathcal{D}:p$, where $p$ is a categorical fibration. Let $\mathcal{C}^0$ be a full subcategory of $\mathcal{C}$. Let $\mathcal{K} \subset Map_{\mathcal{D}'}(\mathcal{C}, \mathcal{D})$ be the full subcategory spanned by those functors $F: \mathcal{C} \rightarrow \mathcal{D}$ which are $p$-left Kan extensions of $F\vert\mathcal{C}^0$. Let $\mathcal{K}'\subset \text{Map}_{\mathcal{D}'}(\mathcal{C}^0, \mathcal{D})$ be the full subcategory spanned by those functors $F_0: \mathcal{C}^0 \rightarrow \mathcal{D}$ with the property that, for each object $C \in \mathcal{C}$, the induced diagram $\mathcal{C}^0_{/C} \rightarrow \mathcal{D}$ has a $p$-colimit. Then the restriction functor $\mathcal{K} \rightarrow \mathcal{K}'$ is a trivial fibration of simplicial sets.'' -And Lurie says that, in order to prove (TR4), we use this ``repeatedly to construct a map from the nerve of the appropriate partially ordered set into $\mathcal{C}$.'' (See Lurie's book available for download on his webpage.) -Now, obviously this must be some sort of standard use of the proposition, but I would very much like to understand this one proof without reading all of Higher Topos Theory, so we have my question: - -Is there possibly a more easy-going reference for this proof? -or -If it doesn't require too much effort, would someone be willing to explain how the cited proposition applies in this instance? -or -Do I really just have to read Higher Topos Theory up through Chapter 4? - -REPLY [5 votes]: This point of view is surely somewhat inelegant and rusty, but I think that once it has been polished it exploits only basic definitions: take it as a motivation more than as a proof if you want. -Allow me to start from the beginning recalling that the octahedral axiom (TR4 for short) says that given three distinguished triangles -\begin{gather*} -X\xrightarrow{f}Y\to Y/X\to X[1] \\ -Y\xrightarrow{g}Z\to Z/Y\to Y[1]\\ -X\xrightarrow{gf}Z\to Z/X\to X[1] -\end{gather*} -I can arrange them in the following "braid": - -given this, there is a (non-unique) way to complete it with the arrows $s,t$ indicated. This is standard in any book about triangulated categories. -Now, in the $\infty$-categorical setting "distinguished triangles" are replaced by "fiber sequences" like -$$ -\begin{array}{ccc} -A &\to& B &\to& 0 \\ -\downarrow && \downarrow && \downarrow\\ -0 &\to& C &\to& A[1] -\end{array} -$$ -Thus, once translated the braid diagram in a stable $\infty$-category $\bf C$ we are in the following situation: - -(originally there were colours to help identify different fiber sequences: unfortunately codecogs does not support it!). Axiom TR4 says that we can find arrows $Y/X\to Z/X\to Z/Y$ such that the triangle $Y/X\to Z/X\to Z/Y\to (Y/X)[1]$ is distinguished. This means, at the end of the day, that not only the composite arrow $Z/Y\to Y[1]\to (Y/X)[1]$ can be embedded in a triangle, but also that this embedding can be done "in a coherent way". -Now for the proof. -The completion axiom (which I suppose to hold) implies that the diagram - -can be completed with an arrow $Y/X \xrightarrow{\phi} Z/X$ completing the square - -Now consider the ojects $V={\rm hocolim}\Big( Y/X \leftarrow{} Y \xrightarrow{g} Z \Big)$ and $W={\rm hocolim}\Big( 0\xleftarrow{} Y/X \xrightarrow{\phi}Z/X \Big)$; arrange them in the diagram - -2-out-of-3 now implies that the outer rectangle is a pushout, hence $W\cong Z/Y$. It remains to prove that $V\cong Z/X$; this follows from the 2-out-of-3 property applied to the last diagram:<|endoftext|> -TITLE: Codimension of Measurable Sets -QUESTION [22 upvotes]: I am currently teaching an advanced undergraduate analysis class, and the following question came up. -Intuition suggests that "most" subsets of $[0,1]$ are not Lebesgue measurable. However, the power set $\mathcal{P}([0,1])$ has the same cardinality as the collection of measurable sets, so it is not clear how to make this statement precise. -One method is to view $\mathcal{P}([0,1])$ as a vector space over $\mathbb{Z}_2$, with addition corresponding to symmetric difference of sets. Then the measurable sets $\mathcal{M}$ form a subspace, and the quotient $\mathcal{P}([0,1])/\mathcal{M}$ is clearly uncountable. -So my question is: what is the cardinality of $\mathcal{P}([0,1])/\mathcal{M}$? It seems like it should be $2^c$, just like $\mathcal{P}([0,1])$, but I don't know how to prove it. -Also, in what other senses are "most" subsets of $[0,1]$ non-measurable? - -REPLY [14 votes]: In 1917, Lusin and Sierpinski showed that the unit interval $[0,1]$ can be partitioned into $2^{\aleph_{0}}$ many pairwise disjoint sets each having Lebesgue outer measure 1; say, $X_{i}$, $i \in I$. Fix $i_{0} \in I$. -For each proper subset $J$ with $i_{0} \notin J \subset I$, let $S_{J} = \bigcup_{j \in J} X_{j}$. Then the sets $S_{J}$, $i_{0} \notin J \subset I$, are distinct modulo $\mathcal{M}$. -Nikolai N. Lusin and Waclaw Sierpinski, “Sur une décomposition d’un intervalle en une infinité non dénombrable d’ensembles non mesurables” [On a decomposition of an interval into a nondenumerably many nonmeasurable sets], Comptes Rendus Académie des Sciences (Paris) 165 (1917), 422-424. - -REPLY [9 votes]: I think this works to show that $\mathcal{P}/\mathcal{M}$ has cardinality $2^c$. It is a small variation on one of the usual constructions of Lebesgue non measurable sets. -Let $A\subset [0,1]$ be a Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$ and partition $A$ into two subsets $B$ and $C$ so that the outer measure of both is one-half and $C$ has cardinality $c$. For every subset $D$ of $C$ (maybe you need $C\sim D$ infinite), $B+D$ mod $[0,1]$ is not Lebesgue measurable. Now let $D$ vary over an almost disjoint family of subsets of $C$ that has cardinality $2^c$.<|endoftext|> -TITLE: How to explicitly compute lifting of points from an elliptic curve to a modular curve? -QUESTION [25 upvotes]: Say $E$ is an elliptic curve over the rationals, of conductor $N$. There's a covering of $E$ by the modular curve $X_0(N)$, and if you rig it right then you can define this map over $\mathbf{Q}$: there's a map $\pi:X_0(N)\to E$ of algebraic varieties over $\mathbf{Q}$. -Now say I have an explicit $\mathbf{Q}$-point $P\in E(\mathbf{Q})$. Its pre-image in $X_0(N)$ will be a finite set of points, all defined over number fields. Perhaps a bit more conceptually, the pullback of $\pi$ via the map $Spec(\mathbf{Q})\to E$ induced by $P$ is a scheme $Spec(A)$ where $A$ is a finite $\mathbf{Q}$-algebra. -How would one go about actually computing these number fields in an explicit example? (or computing $A$, if you like). One can do computations in Jacobians of modular curves so easily these days using modular symbols, so I would imagine this is easy for the experts. -As an explicit example let's take a non-torsion point $P$ on an elliptic curve of rank two (so one can't "cheat" and do the calculation using Heegner points or cusps)---for example let $P$ be some random non-zero small height element of Mordell-Weil mod torsion in the rank two curve of conductor 389. What number fields do the points in the modular curve that map to $P$ cut out? - -REPLY [8 votes]: I recently implemented an algorithm to determine these number fields by computing -the $j$-polynomial: Let $\varphi: X_0(N) \to E$ be a fixed modular parametrization -and $P \in E(\mathbb{Q})$. By j-polynomial I mean the polynomial $F_P(x) = \prod_{z : \varphi(z) = P}(x - j(z))$. There's a Laurent -series $x(q)$ with integer coefficients, which is the modular function $\frac{1}{x \circ \varphi}$ on $X_0(N)$, -We compute $x(q)$ via the gp function ell.taniyama. Then set $u = \frac{1}{j(q)}$, which is also an element in $\mathbb{Z}[[q]]$. Then using Linear algebra, one can find an irreducible polynomial $F$ such that -$F(x,u) = 0$. -Setting $x_0 = \frac{1}{x(P)}$ and the polynomial $j^{2 \deg \varphi}F(x_0,1/j)$ is a constant multiple of $F_P(j)F_{-P}(j)$. -Then we could use complex analytic method to determine which factor corresponds to $P$ and which is $-P$. -As an example we take the elliptic curve 121b1 with rank 1 and trivial torsion. $E(\mathbb{Q})$ is generated by -$P = (4:5:1)$. Then we compute some j-polynomials: - -$F_{-P}(x) = x^4 + 1421551441067913615636000 x^3 + 910640170936002098476853963114167004130307406250 x^2 - 55869041153225091624766256009488963324954953937500000 x + 1513370207928838475604980619812428055721351700525634765625$ -$F_{4P}(x) = x^{4} - \frac{1131444376477476487694208}{43} x^{3} + \frac{11389877706351841520907948036498862509059802748293120}{1849} x^{2} + \frac{831545351967828972021160038394755202358001953700040409088}{1849} x + \frac{16095967144279358005293903881120972455827496828529236714192896}{1849}$ -$F_{P}(x) = x^4 + 98823634118413525094400000 x^3 + 45688143672322270430861721600000000 x^2 - 496864268553728774541064273920000000000 x + 1577314437358442913340940353536000000000000$ - -Since atkin-lehner acts as +1 in this case the number fields in question are splitting fields of these polynomials, respectively. -I plan to compute with the rank 2 curve 389a mentioned in the original post. -Doing so is hard with the current computing power I have, so I was thinking -about replacing $u$ by an $\eta$-product with smaller valence. -Please let me know if this helps. I'll keep polishing this algorithm for the goal of including it in my coming up thesis:).<|endoftext|> -TITLE: When does $P(a−b)=0$ for $a≠b$ ensure $P(0)=0$? (Continued.) -QUESTION [14 upvotes]: As a natural (and expectable) extension of my earlier question: - -How large must be a set $A\subset F_2^n$ to ensure that if $P$ is a cubic polynomial in $n$ variables over the field $F_2$, vanishing at every non-zero point of the sumset $2A:=\{a_1+a_2\colon a_1,a_2\in A\}$, then also $P(0)=0$? - -(Here $n$ is a given positive integer, to be thought of as a growing parameter.) - -Ultimately, I want to know how large must $A$ be for every given degree $\deg P$. Say, if $P$ has degree zero, then, trivially, $|A|\ge 2$ suffices. Furthermore, it is easy to see that for $P$ linear, one needs $|A|\ge 3$ (while $|A|\ge 2$ is insufficient). For $P$ quadratic, it suffices to have $|A|\ge n+3$. In the case where $P$ is cubic, at least $|A|>2n$ is needed: consider, for instance, the set - $$ A=\{0,e_1,...,e_n,e_1+e_2,...,e_1+e_n\}, $$ -where $e_i$ are the vectors of the standard basis, and the polynomial - $$ P=\sum_{12^{7n/8}$, then $A$ contains an affine $4$-dimensional subspace; hence $2A$ contains a linear $4$-dimensional subspace, and the rest follows with a minor effort.) - -REPLY [5 votes]: Here is a small comment to Seva's answer, which allows to get smaller bounds for some fields. -Assume that we have two functions $f,g\in\mathbb F^A$ such that -$$\sum_{a\in A} f(a)Q(a)=\sum_{a\in A} g(a)Q(a)=0 \tag{$\ast$} $$ -for any polynomial $Q\in\mathbb F[x_1,\ldots,x_n]$ of degree $\deg Q\leqslant d$. Then -$$ -\sum_{a\in A,b\in A} f(a)g(b)Q(a,b)=0 -$$ -for any polynomial $Q$ of degree at most $2d+1$. (If $Q$ is a monomial, then our sum factorizes and one of factors is 0.) -Applying this with the polynomial $Q(x,y):=P(x-y)$, where $\deg P\leqslant 2d+1$ and $P(a,b)=0$ for $a,b\in A$, $a\ne b$, we get -$$ -0=\sum_{a\in A,b\in A} f(a)g(b)P(a-b)=P(0)\cdot \sum_{a\in A} f(a)g(a). -$$ -Now if $P(0)\ne 0$, then $\sum_{a\in A} f(a)g(a)=0$ for any two functions $f,g\in\mathbb F^A$ . Then functions $f,g$ satisfying ($\ast$). This condition determines a linear subspace of $\mathbb F^A$ of dimension at least $|A|-\binom{n+d}d$ (or $|A|-\sum_{i\leqslant d} \binom{n}i$, if we consider only multilinear polynomials). We have thus shown that if $P(0,0)\ne 0$, then this is an isotropic subspace (any two vectors are mutually orthogonal.) The maximal dimension of an isotropic subspace is well-studied. It cannot exceed $|A|/2$ by obvious reasons (such a space is contained in its own orthogonal complement). For real field there is no isotropic subspace even of dimension $1$. -See the answer by Robin Chapman to my old question here for references on isotropic subspaces.<|endoftext|> -TITLE: How large can a non-sumset be? -QUESTION [5 upvotes]: The theory of sumsets $A+B$ where $A$ and $B$ are finite subsets of an additive group $Z$ is extensively studied in additive combinatorics: finding long arithmetic progressions inside them, finding lots of subsets of this form, bounding its size above and below, and so on. -A fairly natural inverse question is the following. - -Is there a function $f$ such that if $\lvert A\rvert\gg f(\lvert Z\rvert)$ then $A=B+C$ where $B$ and $C$ are both fairly large sets? - -Since results such as Szemeredi's theorem and Ramsey theory suggest that sets can have lots of structure from cardinality conditions alone, and sumsets are very structured, this seems like a plausible hope. -The case for general finite additive groups may be too hard/trivially false, so I am (as usual in these questions) mostly interested in the cases $Z=\mathbb{Z}/N\mathbb{Z}$ and $Z=\mathbb{F}_p^n$. -I suspect that this sort of result is already known, or follows easily from another well known theorem, and would appreciate any reference and/or proof. - -REPLY [7 votes]: As far as I know, the only result of this sort is due to Alon, found here.<|endoftext|> -TITLE: What a geometer should know ... -QUESTION [5 upvotes]: I am wondering what are the prerequisites for being a modern geometer? It seems that the amount you have to know is just huge: differential geometry, differential topology, algebraic topology, algebraic geometry, symplectic geometry, ... and then there is all kinds of overlap. So my question is the following: what should you know (from the topics I mentioned and the topics that I forgot) to call yourself a geometer (lets say by the time you get your Ph.D.)? And where are the best books/articles covering all that? - -REPLY [15 votes]: The question asked effectively is "What does one have to know ... to call oneself a geometer?" and as several comments note, that question is tough to answer, nonspecific, and the answers are scary ... and yet the question swiftly picked up two "favorite" votes too. -If the question were rephrased as "What ideas are geometers pursuing?" then the MathOverflow community might be able to supply answers that are more specific, useful, and inspiring to students beginning their research. -Notable mathematicians have written many fine essays on this topic. Commonly these essays are more-or-less centered around a guiding idea that was articulated by Mac Lane in his Mathematics, Form and Function (1986) as follows (and I'm posting this as an "answer" solely to be able to format this quotation properly): - -Analysis is full of ingenious changes of coordinates, clever substitutions, and astute manipulations. In some of these cases, one can find a conceptual background. When so, the ideas so revealed help us understand what's what. We submit that this aim of understanding is a vital aspect of mathematics. [...]Effective or tricky formal manipulations are introduced by Mathematicians who doubtless have a guiding idea---but it is easier to state the manipulations than to formulate the idea in words.Just as the same idea can be realized in different forms, so can the same formal success be understood by a variety of ideas. A perspicacious exposition of a piece of Mathematics would let the ideas shine through the display of manipulations. - -Nowadays the notion of "geometry" has become so broadly generalized, as to be effectively identical to Mac Lane's notion of "a piece of mathematics whose ideas shine through the display of manipulations." -For systems engineers nowadays (me in particular) geometry is largely about the dynamical flow of complex systems ... and of course we want our engineering understanding to "shine through the display of manipulations" .. but it would be a grave mistake to imagine that geometric understanding of dynamical systems is all that geometry is about ... because geometry has evolved to become a much broader notion than that. -Therefore, a reasonable piece of advice to young researchers nowadays—in math, science, engineering, and even medicine, it doesn't much matter which—is not to ask oneself "What articles and books should I read?" without first asking oneself the organizing question "What articles and books will I someday want to write? What will be the core ideas? How will I explain these core ideas clearly?" -As soon as you can write down those ideas in even a hazy and uncertain form, then your research career will have begun ... as you learn, your ideas will slowly take concrete form ... and almost certainly you will be led to the study of geometry in its many modern forms. -AMSlogo http://faculty.washington.edu/sidles/misc/AMS_logo.png -Perhaps this is why, 2400 years after Plato's Academy first affirmed it, the members emblem of the American Mathematical Society until recently (and maybe still?) said in greek: "Let none but geometers enter here".<|endoftext|> -TITLE: Problem in Rick Miranda: finding genus of a Projective curve -QUESTION [29 upvotes]: I asked the following question in stack exchange (https://math.stackexchange.com/questions/21164/problem-in-rick-miranda-finding-genus-of-a-projective-curve) a few days ago, but didn't get any solution. Somebody please help me with it. -I have just started learning Riemann Surfaces and I am using the book by Rick Miranda: Algebraic curves and Riemannn Surfaces. #F in section 1.3 asks to determine the genus of the curve in $\mathbb{P}^3$ defined by the two equations $x_0x_3=2x_1x_2$ and $x_0^2 + x_1^2 +x_2^2 +x_3^2 = 0$. #G also has a similar question in which he asks to determine the genus of the twisted cubic. Please explain how to approach this type of question. - -REPLY [5 votes]: This was meant to be a comment on the ending remark of Sándor Kovács' answer, but it got too long to fit: -In a student seminar today, some people had the old edition of Miranda, and some had the new edition, so we had both the original problem and your degenerate version. (The old edition has the degenerate version). -The way we ended up seeing that solution set $X$ of $x_0x_1 = x_2x_3$ is $\mathbb{P}^1\times\mathbb{P}^1$ was to observe that we could rewrite it as $det\begin{pmatrix} x_0 & x_2 \\\\ x_3 & x_1 \end{pmatrix} = 0$ or $det\begin{pmatrix} x_0 & x_3 \\\\ x_2 & x_1 \end{pmatrix} = 0$. So both matrices must be rank 1. So we have two maps from $X$ to $\mathbb{P}^1$, namely the maps which send a point of $X$ to the corresponding element of the nullspace of one or the other matrix. This gives a map $X \to \mathbb{P}^1\times\mathbb{P}^1$, which is not too hard to compute explicitly (in fact $(\langle a,b\rangle, \langle c,d\rangle) \mapsto \langle ac,ad, bc, bd \rangle$ just as you say). So we are really looking at the zero set of $a^2c^2+a^2d^2+b^2c^2+b^2d^2=0$ in $\mathbb{P}^1\times\mathbb{P}^1$. At this point we basically followed the rest of your post. I just thought someone might like the observation about determinants!<|endoftext|> -TITLE: Homotopy type of the plane minus a sequence with no limit points -QUESTION [10 upvotes]: It is well known that the plane minus $n$ points is homotopy equivalent to a wedge of circles and hence its fundamental group is free on $n$ letters. -Question: Is the plane minus an infinite sequence of points having no limit point -homotopy equivalent to an infinite wedge circles? -I'm pretty sure that this could fail if the sequence had a limit point, since then the space in question might be something more like the Hawaiian earing. But is the sequence having a limit point the only thing that could break down? - -REPLY [2 votes]: Expanding on John's answer a bit: If $X = \{x_1,x_2,\ldots\} \subset R^2$ is a countable discrete set, then there is a diffeomorphism $f:R^2 \to R^2$ with $f(x_i)=i$. Proof: by Sards theorem, we can assume that $X \cap N =0$ (translate a bit if necessary). -Claim 1: we can find paths $p_n: [0,1] \to R^2$ which are embeddings, $p_n (0)=x_n$, $p_n (1)= n$ and such that the images of all the $p_n$ are all disjoint. -Proof by induction on $n$. Suppose that $p_1,\ldots ,p_{n-1}$ have been constructed. By homological duality theory, the complememt $R^2 \setminus \cup_{i=1}^{n-1} p_i ([0,1])$ is path-connected. We can then find a path from $x-n$ to $n$, avoiding $\cup_{i=1}^{n-1} p_i ([0,1])$ and also $X \cup N$. -If we have found these paths, we can find neighborhoods $U_n$ of the images, $U_i \cap U_j=\emptyset$ for $i \neq j$. There exist diffeomorphisms $f_n:U_n \to U_n$ with compact support and $f_i(x_i)=i$. These glue together to a global diffeomorphism $f$ with the desired property.<|endoftext|> -TITLE: Surjective implies local affine surjective? -QUESTION [6 upvotes]: Take scheme morphism $f: X\to Y$ and suppose $f$ surjective. If $y \in Y$ can one find affine open $V \subset Y$ containing $y$ and affine open $U \subset X$ such $f(U) = V$ ? -Thank you. -Later: Very good answer of Kevin shows it is not true. Is there hypothese which make it true ? -For example $X$ irreducible and/or $f$ faithfuly flat ? - -REPLY [7 votes]: If $f$ is open (e.g. $f$ finite type and flat over noetherian $Y$), then your condition is trivially satisfied: let $V'$ be any affine open neighborhood of $y$ and let $U'$ be an affine open subset of $X$ such that $y\in f(U')\subseteq V'$. Take a principal open subset $V'_h$ such that $y\in V'_h\subseteq f(U')$, then $V:=V'_h$ and $U:=U'_h$ are what you want. -A counterexample with $X$ irreducible and $f$ projective : consider $Y$ the affine plan, $y$ the origin and $f : X\to Y$ the blowing-up of $y$. For any affine open subset $V$ containing $y$, $f^{-1}(V) \to V$ is the blowing-up of $y$. If $f(U)=V$, then the complement of $U$ in $f^{-1}(V)$ is finite because $f$ is an isomorphism out of $y$. By Zariski's extension theorem, $O_X(U)=O_X(f^{-1}(V))=O_Y(V)$. This is impossible as $U$ is affine, because $U$ would be the image of a section $V\to f^{-1}(V)$, hence closed in (and then equal to) $f^{-1}(V)$.<|endoftext|> -TITLE: What about the classification of big finite simple groups? -QUESTION [10 upvotes]: How hard is it to classify all big finite simple groups, i.e., all finite simple groups larger than some sufficiently large constant? Alternatively - how hard is it to classify all finite simple groups up to finitely many exceptions? -Is any known proof of such a classification significantly easier or shorter than the classification of (all) finite simple groups? - -REPLY [2 votes]: Hrushovski asks roughly this question in his paper On Pseudo-Finite Dimensions. In section 6 he discusses the "classification of large finite simple groups" (in analogy to some of the related results which can be proven by model theoretic means, and end up only giving information about sufficiently large finite objects), and notes that the only known proof is via the full classification of finite simple groups.<|endoftext|> -TITLE: If the tensor product of two $kG$-modules is projeсtive, does either of them have to be projective? -QUESTION [11 upvotes]: Let $G$ be a finite group and $k$ a field, let us assume that char($k$) divides the group order. Let $kG$-mod denote the category of fintely generated $kG$-modules. -This category has as a tensor product $\otimes_{k}$ with diagonal $G$-action. -Given now $M,N\in kG$-mod such that $M\otimes_{k}N$ is projective, can we then conclude that either $M$ or $N$ had to be projective? -If not, can we ask for certain conditions on the field $k$ or the group $G$ such that the statement holds? - -REPLY [6 votes]: If $G$ is a $p$-group, this is almost never true (and I believe but may be wrong that for general $G$, it's completely governed by the $p$-Sylow). In that case, we can understand modules in terms of the support variety $V={\rm Proj} H^{2*}(G,k)$, the projective variety associated to the (even-dimensional) cohomology ring of $G$. To a finite $kG$-module $M$ we can associate its support in $V$, namely the support of Ext(M,M) as a graded $H^{2*}(G,k)$-module. This support is a closed subset of $V$, and conversely every closed subset is the support of some finite $kG$-module. Finally, the support of $M\otimes N$ is the intersection of the support of $M$ and the support of $N$, and a module is projective iff its support is empty. Thus $M\otimes N$ is projective iff $M$ and $N$ have disjoint support. -Thus unless $V$ is just a single point, it is possible to have non-projective modules whose tensor product is projective. For $V$ to be a point, the cohomology of $G$ must be a polynomial ring in one variable, up to nilpotent elements. By Quillen's theorem, this is the case iff all elementary abelian subgroups of $G$ are conjugate and rank 1. -In particular, it is true for cyclic groups, but otherwise it is almost always false. There's a simple argument to see directly that it holds for cyclic groups of order $p$: in that case, $kG$ can be identified with $k[x]/x^p$, and every indecomposable module is of the form $M_i=k[x]/x^i$ for some $i\leq p$. Such a module $M_i$ is projective iff $i=p$. If $M_i$ and $M_j$ are not projective, then $M_i\otimes M_j$ has dimension $ij$, which is not divisible by $p$. Thus $M_i \otimes M_j$ cannot be a sum of copies of $M_p$ and is hence not projective.<|endoftext|> -TITLE: An explicit description of $\operatorname{gr}(k \cdot G)$ for the filtration induced by the augmentation ideal? -QUESTION [10 upvotes]: Let $A$ be any bialgebra (associative, unital, etc.) over a ring $k$. Then among other things it has a counit $\epsilon : A \to k$, and hence an augmentation ideal $I = \ker \epsilon$, which is a Hopf ideal. Any ideal determines a filtration -$$ A \supseteq I \supseteq I^2 \supseteq \dots$$ -and hence an associated graded vector space -$$ \operatorname{gr} A = \frac A I \oplus \frac I {I^2} \oplus \frac {I^2} {I^3} \oplus \dots $$ -Since $\epsilon: A \to A/I = k$ is a morphism of bialgebras, $I$ is a "Hopf ideal", and hence $\operatorname{gr} A$ is a bialgebra. But it is graded with zero part a Hopf algebra, hence $\operatorname{gr} A$ is Hopf. -Moreover, $\operatorname{gr} A$ is generated by $I/I^2$ as an algebra, and each element of $I/I^2$ is primitive. So in particular $\operatorname{gr} A$ is generated by its primitive part, and hence is a quotient of some universal enveloping algebra of some graded Lie algebra $\mathfrak a$; moreover, $\mathfrak a$ is generated as a Lie algebra by its degree-$1$ part, which is precisely $I/I^2$. (Generically the surjection is not an isomorphism, as $A$ might be finite-dimensional but ${\rm U}\mathfrak a$ never is.) -For now, I am interested in the following special case. The ground ring $k$ is a field of characteristic $0$. $G$ is a discrete group, and $A = k\cdot G$ is its group algebra. Then I can calculate $I / I^2$ has as its basis the non-identity elements of the abelianization $G_{\rm ab}$ of $G$. $I^2/I^3$ is spanned by pairs $(g,h) \in G\times G$, modulo $(g,1) = 0$ and $(g,hk) = (g,h) + (g,k)$, and the same relations on the other side (and maybe more relations?), and the multiplication $(I/I^2)^{\otimes 2} \to (I^2/I^3)$ is $gh = (g,h)$. -Anyway, $\operatorname{gr} (k\cdot G)$ feels a lot like some homological construction, but I don't know much homology theory. So: - -Question: What's a hands-on description of $\operatorname{gr} (k\cdot G)$? How does it relate to other constructions I might have met? - - -Update: I definitely made an error in the above, which just means that I understand less than I thought. I'd like to explain an example, and then ask a second, more precise version of the above question. -Suppose that $G$ is a abelian. Then $k\cdot G$ is a commutative cocommutative Hopf algebra, and so is the algebra of functions on some affine algebraic group, which for want of a better name I'll call $G^\vee$ --- it's the "dual group" to $G$. The augmentation ideal then corresponds to the identity element $e\in G^\vee$, and the filtration is the filtration of the algebra of functions on $G^\vee$ in Taylor series. Then $\operatorname{gr}(k G)$ is the symmetric algebra of ${\rm T}^*_e G^\vee$. (If you complete at the augmentation ideal, you're writing down "formal power series near $e$".) -For example, when $G$ is the group with two elements and $\operatorname{char}(k) = 0$, then $I^n = I$ for $n>0$, and so $\operatorname{gr}(kG) = k$ is one-dimensional, not two-dimensional like $kG$. I was confused in my original question, because there are two kinds of filtrations on a vector space --- going up and going down --- and in one of them $\operatorname{gr}$ preserves dimensions, and in the other it may not. -So this shows that I may have been wrong when I wrote "Generically the surjection is not an isomorphism". Note that in the abelian case, $\operatorname{gr}(kG)$ is a symmetric algebra, and in particular is a universal enveloping algebra (the Hopf structure is the right one). -In the nonabelian case, I can't use quite as much geometric language, but I expect something similar should still be true: - -Updated question: Is $\operatorname{gr}(kG)$ a universal enveloping algebra? If so, how is the corresponding Lie algebra related to $G$ (which remember is just a discrete group)? - -REPLY [6 votes]: Daniel Quillen has answered this in -Quillen, Daniel G., On the associated graded ring of a group ring. J. Algebra 10 1968 411–418. -From the Mathematical Reviews by J. Knopfmacher: -"Let $KG$ denote the group algebra of a group $G$ over a field $K$ of characteristic $p$, and let $KG$ be filtered by the powers of its augmentation ideal. Then the author's main theorem states that the associated graded algebra of $KG$ is isomorphic to the universal enveloping algebra of the $p$-Lie algebra $\text{gr}^pG\otimes_ZK$, where $\text{gr}^pG$ is the graded $p$-Lie algebra defined by the $p$-lower central series of $G$. The proof of this interesting result is not obvious, and involves theorems of M. Lazard [Ann. Sci. École Norm. Sup. (3) 71 (1954), 101--190; MR0088496 (19,529b)]. As one corollary, it follows that, if $G$ and $G'$ are two finite $p$-groups whose group algebras over $Z/pZ$ are isomorphic, then $\text{gr}^pG\cong\text{gr}^pG'$."<|endoftext|> -TITLE: non-asymptotic Bertrand-type theorems for arithmetic progression -QUESTION [5 upvotes]: It is well known that primes of form $4k+3$, call them $3=q_1 < q_2 < \dots$ satisfy $q_{n+1}/q_n\rightarrow 1$ (and even $q_n=\frac{n}{2\log n}(1+o(1))$). I would be glad to see results of Dusart-type (Strong Bertrand postulate) with concrete bounds for $N$ such that $q_{n+1} < 1.1 q_n$ provided $q_n > N$ or something like this (I am not sure that I need a multiple exactly $1.1$) - -REPLY [4 votes]: Have a look at (Satz 9 and the preceeding calculations): -1935-10 P. Erdõs: Über die Primzahlen gewisser arithmetischer Reihen (in German), Math. Z. 39 (1935), 473--491; Zentralblatt 10,293. -Available at: http://www.renyi.hu/~p_erdos/Erdos.html<|endoftext|> -TITLE: Known additive bases with irregular counting function -QUESTION [6 upvotes]: For $A \subset \mathbb{N}$ and positive integer $h > 0$, define $r_{A,h}(n)$ to be the number of ways to write $n$ as the sum of $h$ (not necessarily distinct) elements of $A$. We say $A$ is an additive basis of order $h$ if $r_{A,h}(n) > 0$ for all $n$ sufficiently large. Well-known additive bases of finite order include the Waring bases ($\mathbb{N}^k$ for some $k > 0$) and the primes (Goldbach-Shrinel'man Theorem, the $h = 4$ case known as Vinogradov's Theorem). In both cases, the COUNTING function for $A$, which is the function $f(n) = |A \cap [1,n]|$, is quite regular. In particular, for the Waring bases we have $f(n) \sim n^{1/k}$ and for the primes we have $f(n) = \pi(n) \sim \frac{n}{\log(n)}$ (Prime Number Theorem). My question is are there any well-known cases where $f(n)$ is very irregular? In other words there exists some function $g(n)$ and constants $0 < c_1 < c_2 < 1$ such that $\displaystyle \liminf_{n \rightarrow \infty} f(n)/g(n) < c_1, \limsup_{n \rightarrow \infty} f(n)/g(n) > c_2$. I am interested in naturally occurring examples, and less interested in a specific construction of a basis with this property (that shouldn't be too difficult, though I have not tried yet). In particular, whether this property is worthy of serious study. - -REPLY [7 votes]: I guess the very natural example of numbers whose base 3 expansions have only 0's and 1's (a basis of order 2) has the property you mention (taking $g(n)$ to be $n^{2/3}$).<|endoftext|> -TITLE: Counterexamples in algebraic topology? -QUESTION [108 upvotes]: In this thread -Books you would like to read (if somebody would just write them...), -I expressed my desire for a book with the title "(Counter)examples in Algebraic Topology". -My reason for doing so was that while the abstract formalism of algebraic topology is very well-explained in many textbooks and while most graduate students are fond of the general machinery, the study of examples is somehow neglected. I am looking for examples that explain why certain hypotheses are necessary for theorems to hold. The books by Hatcher and Bredon contain some interesting stuff in this direction, and there is Neil Strickland's bestiary, which is mainly focused on positive knowledge. -To convey an idea of what I am after, here are a few examples from my private ''counterexamples in algebraic topology'' list. Some are surprising, some less so. - -The abelianization of $SL_2 (Z)$ is $Z/12$, the map $BSL_2(Z) \to BZ/12$ is a homology equivalence to a simple space. But it is not a Quillen plus construction, since the the homotopy fibre is $BF_2$ (free group on $2$ generators), hence not acyclic. See The free group $F_2$ has index 12 in SL(2,$\mathbb{Z}$) . -Maps $f:X \to Y$ which are homology equivalences, the homotopy groups are abstractly isomorphic, but though, $f$ is not a homotopy equivalence (a number of examples has been given in the answers to these questions: -Spaces with same homotopy and homology groups that are not homotopy equivalent?, -Are there pairs of highly connected finite CW-complexes with the same homotopy groups?). -Self-maps of simply-connected spaces $X$ which are the identity on homotopy, but not on homology (let $X=K(Z;2) \times K(Z;4)$, $u:K(Z;2) \to K(Z;4)$ be the cup square, -and $f:X\to X$ is given by $f(x,y):= (x,y + u(x))$, using that EM-spaces are abelian groups). There are also self-maps of finite simply connected complexes that are the identity on homology, but not on homotopy, see Diarmuid Crowleys answer to Cohomology of fibrations over the circle: how to compute the ring structure? -The stabilization map $B \Sigma_{\infty} \to B \Sigma_{\infty}$ induces a bijection on free homotopy classes $[X, B \Sigma_{\infty}]$ for each finite CW space $X$. However, it is not a homotopy equivalence (not a $\pi_1$-isomorphism). -The fibration $S^1 \to B \mathbb{Q} \to B \mathbb{Q}/\mathbb{Z}$ is classified by a map $f:B \mathbb{Q}/\mathbb{Z} \to CP^{\infty}$, which can be assumed to be a fibration with fibre $B \mathbb{Q}$. Now let $X_n$ be the preimage of the n-skeleton of $CP^{\infty}$. Using the Leray-Serre spectral sequence, we can compute the integral -homology of $X_n$ and, by the universal coefficient theorem, the homology of field coefficients. It turns out that this is finitely generated for any field, -and so we can define the Euler characteristic in dependence of the field. It is not independent of the field in this case (the reason is of course that the integral homology of $X_n$ is not finitely generated). -The compact Lie groups $U(n)$ and $S^1 \times SU(n)$ are diffeomorphic, their classifying spaces have isomorphic cohomology rings and homotopy groups, but the classifying spaces are not homotopy equivalent (look at Steenrod operations). - -Question: Which examples of spaces and maps of a similar flavour do you know and want to share with the other MO users? -To focus this question, I suggest to stay in the realm of algebraic topology proper. In other words: - -The properties in question should be homotopy invariant properties of spaces/maps. This includes of course fibre bundles, viewed as maps to certain classifying spaces. -Let us talk about spaces of the homotopy type of CW complexes, to avoid that a certain property fails for point-set topological reasons. -This excludes the kind of examples from the famous book "Counterexamples in Topology". -The examples should not be "counterexamples in group theory" in disguise. Any ugly example of a discrete group $G$ gives an equally ugly example of a space $BG$. -Same applies to rings via Eilenberg-Mac Lane spectra. -I prefer examples from unstable homotopy theory. - -To get started, here are some questions whose answer I do not know: - -Construct two simply-connected CW complexes $X$ and $Y$ such that $H^* (X;F) \cong H^* (Y;F)$ for any field, as rings and modules over the Steenrod algebras, but which are not homotopy equivalent. EDIT: Appropriate Moore spaces do the job, see Eric Wofsey's answer. -Let $f: X \to Y$ be a map of CW-complexes. Assume that $[T,X] \to [T;Y]$ is a bijection for each finite CW complex $T$ ($[T,X]$ denotes free homotopy classes). What assumptions are sufficient to conclude that $f$ is a weak homotopy equivalence? EDIT: the answer has been given by Tyler Lawson, see below. -Do there exist spaces $X,Y,Z$ and a homotopy equivalence $X \times Y \to X \times Z$, without $Y$ and $Z$ being homotopy equivalent? Can I require these spaces to be finite CWs? EDIT: without the finiteness assumptions, this question was ridiculously simple. -Do you know examples of fibrations $F \to E \to X$, such that the integral homology of all three spaces is finitely generated (so that the Euler numbers are defined) and such that the Euler number is not multiplicative, i.e. $\chi(E) \neq \chi(F) \chi(X)$? Remark: is $X$ is assumed to be simply-connected, then the Euler number is multiplicative (absolutely standard). Likewise, if $X$ is a finite CW complex and $F$ is of finite homological type (less standard, but a not so hard exercise). So any counterexample would have to be of infinite type. The above fibration $BSL_2 (Z) \to BZ/12$ is a counterexample away from the primes $2,3$, but do you know one that does the job in all characteristics?. Of course, the ordinary Euler number is the wrong concept here. - -I am looking forward for your answers. -EDIT: so far, I have gotten great answers, but mostly for the specific questions I asked. My intention was to create a larger list of counterexamples. So, feel free to mention your favorite strange spaces and maps. - -REPLY [12 votes]: What follows is merely a reference to the excellent answer and comment by Karol Szumiło in this mathoverflow question asked by Mike Shulman. There, Karol provides arguments and bibliographic sources which prove the failure of Brown representability in the homotopy category of unpointed spaces, and in the homotopy category of not necessarily connected pointed spaces. Please upvote Karol's comment and answer, together with the question. For convenience, I review below the main points of that discussion.$\newcommand{\Ho}{\mathrm{Ho}} -\newcommand{\op}[1]{#1^{\mathrm{op}}} -\newcommand{\Set}{\mathrm{Set}}$ -In the article "Splitting homotopy idempotents II", Peter Freyd and Alex Heller construct a very special homotopy idempotent $Bf:BF\to BF$, i.e. $Bf$ is an idempotent in $\Ho$, the homotopy category of spaces with the homotopy type of CW-complexes. Here, $F$ denotes Thompson's group, and $BF$ is its classifying space. Importantly, the homotopy idempotent $Bf:BF\to BF$ does not split, i.e. does not admit a retract in the homotopy category. This example was also constructed independently by Dydak in his 1977 article "A simple proof that pointed connected FANR-spaces are regular fundamental retracts of ANR's". -The idempotent $Bf:BF\to BF$ provides a retract $R$ of the representable functor $[-,BF]:\op{\Ho}\to\Set$, since idempotents do split in the category $\Set$ of sets. Then $R$ is necessarily half-exact, i.e. preserves small products and weak pullbacks; these are the conditions in Brown's representability theorem. However, $R$ is not representable since a representing object for $R$ would be a retract for $Bf:BF\to BF$ in $\Ho$. We can also apply the same argument to the pointed map $(Bf)_+ : (BF)_+ \to (BF)_+$ obtained by adding disjoint basepoints, giving a half-exact, non-representable functor $\op{(\Ho_\ast)}\to\Set$ on the pointed homotopy category. -More examples of the failure of Brown representability are described by Alex Heller in the article "On the representability of homotopy functors", which provides some more fascinating insights into the phenomenon of Brown representability. In particular, that article gives a functor $N:\op{\Ho}\to\Set$ which is half-exact, but is not even a retract of any representable functor. The functor $N$ is defined for each space $X$ by -$$ N(X)=\prod_{[x]\in\pi_0(X)} S\bigl(\pi_1(X,x)\bigr) $$ -where $S(G)$ is the set of normal subgroups of $G$, for each group $G$. Observe that the choice of $x\in X$ representing $[x]\in\pi_0 X$ is inconsequential. For a homotopy class of maps $[f]:X\to Y$ in $\Ho$, the function $N([f])$ is given by taking preimages of normal subgroups by $f_\ast:\pi_1(X,x)\to\pi_1(Y,f(x))$. Then $N$ is not a retract of a representable functor because normal subgroups of $\pi_1(X,x)$ can have arbitrarily large index if one varies the space $X$.<|endoftext|> -TITLE: Are squarefree monomial ideals on a regular system of parameters in a regular local ring radical? -QUESTION [7 upvotes]: Suppose $(R,m)$ is a regular, local ring. Let $x_1,x_2,...,x_n$ be a regular system of parameters. Let $I$ be an ideal generated by squarefree monomials in the $x_i$'s. Is $I$ a radical ideal? The motivation for this is the polynomial ring in finitely many indeterminates over a field (which although not local is regular) and the indeterminates generate the homogeneous maximal ideal. Then every ideal generated by squarefree monomials in the indeterminates is radical. -I was trying to see whether the proof from monomial case in polynomial rings carries over. There we express an ideal generated by squarefree monomials as an intersection of ideals generated by subsets of the indeterminates. Each such ideal is prime. In the case of a regular system of parameters in a regular local ring, any subset of a rsop's also generates a prime ideal. So the only obstruction in the proof is whether the modularity law holds for such ideals. - -REPLY [4 votes]: ADDED: here is a proof of the statement you need (namely the square free monomial ideal $I$ is a intersection of primes generated by subsets of parameters) without using the modularity property. We will use induction on $N=$ the total numbers of times the parameters appear in the generators of $I$. For example if $I=(xy, xz)$ then $N=4$. The statement is obvious if $N=1$. -Suppose $I$ has a generator (say $f_1$) which involves at least $2$ parameters. Pick one of these parameters, say $x$ and WLOG, we can assume $I=(f_1,\cdots, f_n, g_1,\cdots,g_l) $ such that $x|f_i$ for each $i$ but $x$ does not divide any of the $g_j$s. Let $F_i=f_i/x$. We claim that: -$$ I = (I,x) \cap (I,F_1)$$ -If the claim is true, we are done by applying the induction hypothesis to $(I,x)$ and $(I,F_1)$. One containment is obvious, for the other one we need to show if $xu \in (I,F_1)$ then $xu\in I$. -Write $$xu = f_2x_2 + \cdots f_nx_n + \sum g_jy_j + F_1x_1$$ which implies -$$x(u- F_2x_2 +\cdots F_nx_n) \in (g_1,\cdots, g_l, F_1) = I' $$ -$I'$ has minimal generators which do not contain $x$. By induction, $I'$ is an intersection of primes generated by other parameters, so $x$ is a NZD on $R/I'$. So $(u- F_2x_2 +\cdots F_nx_n) \in I'$, and therefore $xu \in I$, as desired. -REMARK: note that for this proof to work, you only need that all subsets of the sequence (not necessarily parameters) generate prime ideals. I guess it fits with your other question. - -So from the comments I will take your question as proving $J\cap (K+L) = J\cap K + J \cap L$ for parameter ideals (by which I mean ideals generated by subsets of a fixed regular s.o.p). -It will suffice to understand $I\cap J$ for two such ideals. To be precise, let $g(I)$ be the set of s.o.p generators of $I$. Let $P$ be the ideal generated by the intersection of $g(I),g(J)$, and $I', J'$ generated by $g(I)-g(P), g(J)-g(P)$. Then we need to show: -$$I \cap J = P + I'J' $$ -Since $R/P$ is still regular we can kill $P$ and assume that $g(I), g(J)$ are disjoint, and we have to prove $I \cap J = IJ$. This should be an easy exercise, but a slick and very general way is invoking Tor (which shows that this is even true for $I,J$ generated by parts of a fixed regular sequence).<|endoftext|> -TITLE: Comparing colimits in schemes with colimits in sheaves of sets -QUESTION [8 upvotes]: Suppose I have a diagram of schemes, and I know that the colimit exists in the category of schemes. How does this colimit compare with the colimit of the corresponding sheaves (I'm being nonspecific about the topology on purpose)? We always have a map from the colimit of sheaves to the colimit of schemes. Are then any conditions I can impose on my diagram so that this map is an isomorphism? Is there any reference where this issue is discussed? - -REPLY [6 votes]: My understanding is that it typically requires hard projective geometry to show that the sheaf-theoretic quotient is a scheme. (See Grothendieck's Seminaire Bourbaki talks in FGA.) I also think that showing it is an algebraic space is much easier, or at least can be done under wide and reasonable generality, but still requires some work. (See the work of M Artin.) But I think it is usually not hard to check whether a given scheme or algebraic space is the sheaf-theoretic quotient. In fact, unless I'm mistaken, it's easy enough that no one ever bothered to write it down. -First consider the (main) case where the colimit diagram is of the form $E\rightrightarrows X$, where the two maps are the projections of an equivalence relation on $X$. Then the quotient $X/E$ in the category of sheaves has the properties that (i) the map $X\to X/E$ is an epimorphism and (ii) the induced map $E\to X\times_{X/E} X$ is an isomorphism. Further, these properties characterize the quotient up to unique isomorphism. This is just general sheaf theory. -So if $f\colon X\to Y$ is a map such that the two compositions $E\rightrightarrows X \to Y$ agree, the induced map $X/E\to Y$ is an isomorphism if and only if $Y$ satisfies the properties above. So the candidate quotient map $f\colon X\to Y$ mush be a sheaf epimorphism, and we must have $E=X\times_{Y} X$. These are easy to check in many cases. For (i), a map $f\colon X \to Y$ is an epimorphism if and only if it has a section locally on a cover of $Y$ with respect to the Grothendieck topology in question. This is again just general sheaf theory. If $X$ and $Y$ are schemes, this is often easy to check -- for instance any smooth surjection has a section etale locally. And for (ii), you just need to check that two subobjects of $X\times X$ agree. Often they'll be closed subschemes of $X\times X$, so you just need to show they have the same ideal sheaf. -For example, the map $\mathbf{A}^1\to\mathbf{A}^1$ given by $z\mapsto z^2$ is the sheaf-theoretic quotient of $\mathbf{A}^1$ by the involution $z\mapsto -z$ in the fppf or fpqc topology, but not in the etale or Zariski topology. -Now consider the case where you're forming the sheaf-theoretic coproduct of a family of schemes. This is easy. It's representable by the disjoint union of the schemes in the family. (Maybe I using some weak properties of the topology here...) -For a general diagram of schemes, I think we can just combine the two previous cases. Represent such a diagram by a functor $I\to \mathrm{Schemes}$, where $I$ is an "index" category (small). Let $X'$ be the sheaf colimit $\mathrm{colim}_{i\in I} X_i$, and let $D$ be the disjoint union $\coprod_{i\in I} X_i$. Then the induced map $D\to X'$ is an epimorphism, and so $X'$ is the quotient of $D$ by the equivalence relation $D\times_{X'} D$. So to check whether a candidate quotient $Y$ is the actual quotient, you need to show that (i) the map $D\to Y$ is a sheaf epimorphism and (ii) the corresponding equivalence relation $D\times_Y D$ agrees with $D\times_{X'} D$. As above, I'd say it's typically not hard to show (i). To show (ii), you'd need to have some hands-on understanding of $D\times_{X'}D$. But I'm going to guess that this is not hard. I bet you just take the usual definition of colimits in the category of sets (disjoint unions modulo and explicit equivalence relation given by the maps in the diagram) and then repeat the definition in the category of sheaves, making sure to replace existential quantifiers by their local versions with respect to the topology in question. (This is getting a bit more abstract than I enjoy, so I'm calling it quits. But if you have a specific diagram, I might be willing to give it a shot.)<|endoftext|> -TITLE: Impact of discrepancy between Kunen's and Jech's definition of iterated forcing on full support iterations -QUESTION [15 upvotes]: One of the first things we usually learn when we study iterated forcing is that we can force over a model of ZFC + GCH to make the continuum function ($\lambda \mapsto 2^{\lambda}$) restricted to some set $S$ of regular cardinals follow any nondecreasing pattern satisfying König's Theorem ($\text{cof}(2^{\lambda}) > \lambda$). This is accomplished by an Easton product of adding the desired number of Cohen subsets to each cardinal in $S$. For example, assuming that the GCH holds in $V$, we can force $2^{\aleph_n} = \aleph_{n+2}$ for all Natural numbers $n$ with the full support $\omega$-iteration where we force with the ground model poset adding $\aleph_{n+2}$ many Cohen subsets of $\aleph_n$ at every stage $n$. -However, if we instead force with the posets adding $\aleph_{n+2}$ Cohen subsets of $\aleph_n$ from the successive extensions, then the iteration will preserve the GCH. Such a phenomenon is easily explained away by the fact that after forcing to add Cohen subsets of $\aleph_n$, we have necessarily changed the definition of the forcing adding Cohen subsets of $\aleph_{n+1}$ in the extension. - -Now consider a full support $\omega$-iteration $\mathbb{P}$ where at every stage $n$, we force to add a Cohen real. Since the poset adding a Cohen real is the set of finite partial binary sequences with domain $\omega$, all transitive models of ZFC agree on its definition. Therefore, the aforementioned concern disappears in this context. Because the forcing to add a Cohen real satisfies the countable chain condition (c.c.c.), it is proper. Then since proper forcing is closed under countable support iterations using Jech's definition of a forcing iteration, $\mathbb{P}$ will be proper and hence $\omega_1$-preserving. However, using Kunen's definition, the iteration can collapse $\omega_1$. For example, from Chapter VIII, Exercise (E4): -Let $\mathbb{P}_{\omega}$ be defined by countable supports, and let each $\pi_n$ be $(\text{Fn}(\omega, 2)){}^{\check{}}$). Show that $\mathbb{P}_{\omega}$ collapses $\omega_1$. -Kunen then notes that this problem goes away when full names are used. On the other hand, Jech seems to circumvent the problem altogether by defining $\mathbb{P}_n * \dot{\mathbb{Q}}_n$ to be the set of pairs $\langle p, \dot{q}\rangle \in \mathbb{P} \times \text{dom}(\dot{\mathbb{Q}}_n)$ such that all conditions from $\mathbb{P}_n$ force $\dot{q}$ to be in $\dot{\mathbb{Q}}_n$ rather than just requiring that the condition $p$ forces this as Kunen does. -What I would like to see then is an elaboration of the explanation of why these forcing iterations are different including the intuition behind these differences. Specifically, my question is as follows: - -Why does the full support $\omega$-iteration forcing with $(\text{Fn}(\omega, 2)){}^{\check{}}$ (forcing to add a Cohen real) at every stage collapse $\omega_1$ under Kunen's definition but not under Jech's? For example, how is it that every binary $\omega$-sequence from the ground model is coded in the generic filter using Kunen's definition but not Jech's? - -As a side request related to how this question came up, can you provide an example of a full support $\omega$-iteration that must have size at least $\aleph_2$, but each individual stage of forcing is necessarily proper and of size at most $\aleph_1$? - -REPLY [12 votes]: I hope I can answer this question in a reasonable way. The natural way to define the iteration $\mathbb P*\dot{\mathbb Q}$ is to consider all pairs $(p,\dot q)$ with -$p\in\mathbb P$ and $p\Vdash\dot q\in\dot{\mathbb Q}$. -Unfortunately this is a proper class. (The definition in Jech's book also gives a proper class.) -The problem in Kunen's definition of the iteration $\mathbb P*\dot{\mathbb Q}$ is not so much that he defines -$(p,\dot q)\in\mathbb P*\dot{\mathbb Q}$ if $p$ forces $\dot q$ to be in $\dot{\mathbb Q}$, -but that he only considers names $\dot q$ in the domain of $\dot{\mathbb Q}$. -He does this to make sure that the iteration $\mathbb P*\dot{\mathbb Q}$ is a set rather than a proper class. -But there are other ways to make sure that the iteration is a set, for example by allowing -arbitrary names for $\dot q$ that are in some sufficiently large $H_\chi$ (sets with transitive closure of size $<\chi$). -"Sufficiently large" depends on $\mathbb P$ and $\dot{\mathbb Q}$. -Now Jech only considers pairs $(p,\dot q)$ such that $\dot q$ is forced by $1_{\mathbb P}$ to be in $\dot{\mathbb Q}$. But this doesn't really make a difference since by the existential completeness lemma, if $p\Vdash\dot q\in\dot{\mathbb Q}$, then there is a name $\dot r$ which is forced by $1_{\mathbb P}$ to be a condition in $\dot{\mathbb Q}$ and such that $p\Vdash\dot q=\dot r$. -Now in the case of a two step iteration, the definitions by Jech and Kunen are equivalent, -and both are equivalent to the one that I suggested (arbitrary names in the second coordinate, but cutting off somewhere). -The right way to see this is to first show that if you cut off Jech's iteration at a sufficiently large $H_\chi$, then you get a dense subset of Jech's iteration. -This cut off version of Jech's iteration is dense in my version (with the same sufficiently large $H_\chi$). -Now Kunen's iteration is equivalent to mine since given $(p,\dot q)$ in my iteration, -there are $s\leq p$ and $\dot t$ in the domain of $\dot Q$ such that -$s\Vdash\dot q=\dot t$. We have $(s,\dot t)\leq(p,\dot q)$, showing that Kunen's iteration is dense in mine. -Things start to fall apart if we go to longer iterations, and iterations with infinite support. My argument above still shows that Kunen's iteration of arbitrary length, with finite supports, is equivalent to my version for long iterations. -This is because given a finitely supported condition in the long iteration in my sense, -say with support of length $n$, you can decrease the first $n-1$ coordinates to force the last coordinate to be something given by a name in the domain of the respective name of a partial order. Then you decrease the first $n-2$ coordinates to take care of the $n-1$-th coordinate and so on. -This argument clearly does not go through with countable supports. Hence the countable support iteration in Kunen's sense is not necessarily equivalent to my version or Jech's version. It is equivalent, however, if all the iterands are given by full names. -Now, what is the problem in terms of properness? The proof of properness of countable support iterations of proper forcing (and I am referring to the proof in Goldstern's -"Tools for your forcing construction") makes frequent use of the existential completeness lemma to cook up names for conditions. Now in the Kunen version of the iteration -it may happen that these names cannot be used since they may not be in the domain of the respective name for a partial order. -And in fact, as the exercise shows, even the countable support iteration of Cohen forcing -of length $\omega$ in Kunen's sense depends on which names you choose for the iterands: -Full names yield a proper forcing (which would be equivalent to the iteration in my sense), -but taking canonical names yields a forcing notion that collapses $\omega_1$. -I hope this helps.<|endoftext|> -TITLE: Primes represented by two-variable quadratic polynomials -QUESTION [21 upvotes]: I'm looking over a paper, "Primes represented by quadratic polynomials in two variables" [1] which attempts to characterize the density of the primes in two-variable quadratic polynomials. Its apparent improvement over previous works is that it covers not just quadratic forms plus a constant but quadratic forms plus linear forms plus a constant. -I would like to know what the current state of knowledge is for this sort of problem. - -This paper covers only the case of polynomials that "depend essentially on two variables". This has surely not been improved, else the H-L Conjecture E, F, etc. would have been solved. -The paper allows only integers, not half-integers, as coefficients, and so does not give the number of primes, e.g., in A117112. -It does not give the constant (or prove that a constant exists!) for the densities it finds. - -The third is my main interest at this point. References would be appreciated, whether to research papers, survey papers, or standard texts. I also have some interest in the cubic case, if anything is known. It seems that even characterizing which two-variable cubics represent an infinite number of primes is open...? Certainly the conditions Iwaniec gives for two-variable quadratics do not suffice to give an infinity of primes. -For example, with the polynomial $P(x,y)=ax^2+bxy+cy^2+ex+fy+g$, setting $\Delta=b^2-4ac$ and $D=af^2-bef+ce^2+g\Delta$ and $C(x)=\sum_{p=P(x,y)\le x}$1, [1] shows that -$$\frac{x}{(\log x)^{1.5}}\ll C(x)\ll\frac{x}{(\log x)^{1.5}}$$ -for all $P$ with $D\neq0$ and $\Delta$ not a square, but is it known that $C(x)\sim kx/(\log x)^{1.5}$ for some $k$? Are values of $e$ or $E$ known such that -$$e<\liminf\frac{C(x)(\log x)^{1.5}}{x}$$ -or -$$\limsup\frac{C(x)(\log x)^{1.5}}{x}< E$$ -? -Similarly, if $D=0$ or $\Delta$ is a square do we know when -$$\ell=\lim\frac{C(x)\log x}{x}$$ -exists and what its value is? -Related question: Who should I cite for these results? I'm never sure of when there might have been simultaneous discoveries or other reasons for priority issues. -[1] Iwaniec, H. (1974). Primes represented by quadratic polynomials in two variables. Acta Arithmetica 24, pp. 435–459; DOI: 10.4064/aa-24-5-435-459. -[2] Hardy, G. H., & Littlewood, J. E. (1923). Some problems of ‘Partitio numerorum’; III: On the expression of a number as a sum of primes. Acta Mathematica 44:1, pp. 1-70. - -REPLY [7 votes]: The modern reference work on the subject seems to be [1], but it spends only a page and a half on the subject of primes in multivariate quadratic polynomials (pp. 396-397). More than half this space is devoted to Iwaniec's 1974 result. The balance mentions Sarnak's application to the Problem of Apollonius and a result of "J. Cho and H. Kim" on counting primes in $\mathbb{Q}[\sqrt{-2}].$ So nothing there. -Pleasants [2] shows that, subject to a Davenport-Lewis [3] condition on the $h^*$ (a complexity measure on the cubic form part), multivariate cubic polynomials have the expected number of primes. Unfortunately this condition requires (as a necessary but insufficient condition) that there be at least 8 variables. Further, it double-counts repeated primes. -Goldoni [4] recently wrote a thesis on this general topic. His new results (Chapter 5) on the $h$ and $h^*$ invariants make it easier to use the results of Pleasants but do not extend them to cubic polynomials with fewer than 8 variables. -Of course I would be remiss in failing to mention the groundbreaking work of Heath-Brown [5], building on Friedlander & Iwaniec [6]. These results will no doubt clear the way for broader research, but so far have not been generalized. -So in short it appears that: - -Nothing further is known about primes represented by quadratic polynomials. -Apart from $x^3+2y^3$, almost nothing is known about which primes are represented by cubic polynomials, though some results are known for how often such polynomials take on prime values provided $h^*$ and hence the number of variables is large enough. - - -On the historical side, of course Fermat is responsible for the proof of the case $x^2+y^2$. I have references that say that Weber [7] and Schering [8] handled the case of (primitive) binary quadratic forms with nonsquare discriminants, but I haven't read the papers. Motohashi [9] proved that there are $\gg n/\log^2 n$ primes of the form $x^2+y^2+1$ up to $n$, apparently (?) the first such result with a constant term. He conjectured that the true number was -$$\frac{n}{(\log n)^{3/2}}\cdot\frac32\prod_{p\equiv3(4)}\left(1-\frac{1}{p^2}\right)^{-1/2}\left(1-\frac{1}{p(p-1)}\right)$$ -but as far as I know the constant still has not been proved even for this special form. -Edit: Apparently Bredihin [10] proved the infinitude of primes of the form $x^2+y^2+1$ some years before Motohashi. He only gave a slight upper-bound on their density, though: $O(n/(\log n)^{1.042}).$ (Motohashi improved the exponent to 1.5 in a later paper.) - -[1] Friedlander, J. and Iwaniec, H. (2010). Opera de Cribro. AMS. -[2] Pleasants, P. (1966). The representation of primes by cubic polynomials, Acta Arithmetica 12, pp. 23-44. -[3] Davenport, H. and Lewis, D. J. (1964). "Non-homogeneous cubic equations". Journal of the London Mathematical Society 39, pp. 657-671. -[4] Goldoni, L. (2010). Prime Numbers and Polynomials. Doctoral thesis, Università degli Studi di Trento. -[5] Heath-Brown, D. R. (2001). Primes represented by $x^3 + 2y^3$. Acta Mathematica 186, pp. 1-84; Wayback Machine. -[6] Friedlander, J. and Iwaniec, H. (1997). Using a parity-sensitive sieve to count prime values of a polynomial. Proceedings of the National Academy of Sciences 94, pp. 1054-1058. -[7] Weber, H. (1882). "Beweis des Satzes, dass, usw". Mathematische Annalen 20, pp. 301-329. -[8] Schering, E. (1909). "Beweis des Dirichletschen Satzes". Gesammelte mathematische Werke, Bd. 2, pp. 357-365. -[9] Motohashi, Y. (1969). On the distribution of prime numbers which are of the form $x^2+y^2+1$. Acta Arithmetica 16, pp. 351-364. -[10] Bredihin, B. M. (1963). Binary additive problems of indeterminate type II. Analogue of the problem of Hardy and Littlewood (Russian). Izv. Akad. Nauk. SSSR. Ser. Mat. 27, pp. 577-612.<|endoftext|> -TITLE: Image of norm map for local field -QUESTION [6 upvotes]: Let $F$ be a finite extension of $Q_2$ (2-adic field) or $F_2((x))$ (function field). Let $E/F$ be a separable extension of degree $2$. - -What is the image of the norm map $N_{E/F}$? -In particular - is it true that the index $[F^{\star} : N_{E/F}(E^{\star})]$ depends only on the ramification $e(E|F)$? - -REPLY [12 votes]: It might be worth working the example of $\mathbb{Q}_2$. Recall that $u \in \mathbb{Q}_2$ is a square if and only if $u$ is of the form $4^k (1+8 \ell)$ for $\ell \in \mathbb{Z}_2$. So $\mathbb{Q}_2^{\times} / (\mathbb{Q}_2^{\times})^2$ has order $8$, with representative elements being $1$, $3$, $5$, $7$, $2$, $6$, $10$ and $14$. This is an important computation for two reasons: -(1) Quadratic extensions of a characteristic zero field are of the form $K(\sqrt{a})$ for some nonsquare $a$, and two different $a$'s give the same extension if there ratio is a square. So there are $7$ quadratic extensions of $\mathbb{Q}_2$. The unramified one is $\mathbb{Q}_2(\sqrt{5})$. -(2) If $L$ is any quadratic extension, then $(\mathbb{Q}_2^\times)^2 \subset N_{L/\mathbb{Q}_2} (L^\times)$ as, for $a \in \mathbb{Q}_2$, we have $N(a)=a^2$. So we can describe the norm group by giving its image in the $8$ element group $\mathbb{Q}^{\times}_2/(\mathbb{Q}_2^{\times})^2$. -So, for example, in $\mathbb{Q}_2(\sqrt{3})$, the norms are elements of the form $a^2 - 3 b^2$. A little checking shows that the image in $\mathbb{Q}^{\times}_2/(\mathbb{Q}_2^{\times})^2$ is represented by $\{ 1, 1-3 \cdot 2^2, 3^2 -3, 1 - 3 \} \equiv \{ 1, 5, 6, 10 \}$. You can enjoy writing down the $4$ element subgroup of $\{ 1,3,5,7,2,6,10,14 \}$ corresponding to each of the $7$ quadratic extensions.<|endoftext|> -TITLE: Intended interpretations of set theories -QUESTION [14 upvotes]: In his Set Theory. An Introduction to Indepencence Proofs, Kunen develops $ZFC$ from a platonistic point of view because he believes that this is pedagogically easier. When he talks about the intended interpretation of set theory he says such things as, for example, that the domain of discourse $V$ is the collection of all (well-founded, when foundation is introduced) hereditary sets. -This point of view has always made me feel a bit uncomfortable. How can a variable in a first-order language run over the elements of a collection that is not a set? Only recently I realized that one thing is to be a platonist, and another thing is to believe such an odd thing. -A first-order theory of sets with a countable language can only prove the existence of countably many sets. Let me call them provable sets for short. Platonistically, we wish our intended interpretation of that theory to be one in which every provable set is actually the set the theory says it is. So we don't need our interpretation to contain every set, we just need that it contains at least the true provable sets. This collection is, really, a set, although it doesn't know it. -To be a bit more concrete, if one is a platonist and the cumulative hierarchy is what one has in mind as the real universe of sets, one can think that the $V$ of one's theory actually refers to a an initial segment of that hierarchy, hence variables run no more over the real $V$ but only over the elements of some $V_\alpha$. -There's a parallel to these ideas. For example, when we want to prove consistency with $ZFC$ of a given sentence, we do not directly look for a model of $ZFC$ where that sentence is true, but instead we take advantage of knowing that every finite fragment of $ZFC$ is consistent and that every proof involves only finitely many axioms. -My question is: then, is this position tenable or am I going awfully wrong? I apologize that this seems a philosophical issue rather than a mathematical one. I also apologize for stating things so simply (out of laziness). - -REPLY [2 votes]: Not being able to post comments, this "answer" intends to be a comment to the question: you say "For example, when we want to prove consistency with ZFC of a given sentence, we do not directly look for a model of ZFC where that sentence is true, but instead we take advantage of knowing that every finite fragment of ZFC is consistent and that every proof involves only finitely many axioms." But as far as I know, if we knew that every finite fragment of ZFC is consistent, we would know that $all$ of ZFC is consistent, wouldn't we? I think this would follow by the compactness theorem, hence we cannot assure that every finite fragment of ZFC is consistent. Please let me know if there's something I'm not understanding well here.<|endoftext|> -TITLE: Is this number already known to be transcendental? Is there a survey about up-to-date trascendence results? -QUESTION [9 upvotes]: Is the number $\sum_{n=1}^\infty \frac{1}{2^{n^2}}$ known to be transcendental? -Is there a survey with up-to-date transcendence results? - -REPLY [24 votes]: I have checked with Introduction to Algebraic Independence Theory, where it is mentioned in the preface (p. V) that - -D. Bertrand and independently D. Duverney, Ke. Nishioka, Ku. Nishioka, I. Shiokawa (DNNS) deduced results on algebraic independence of the values of theta-functions at algebraic points and in particular derived the transcendence of the sums $\sum_{n=1}^\infty q^{n^2}$ for any algebraic $q$ satisfying $0 < |q| < 1$. - -The precise references are not given but a little googling turned up the paper by D. Bertrand, "Theta Functions and Transcendence", The Ramanujan Journal, Vol. 1 (1997), pp. 339-350, which seems to be relevant. The second reference is DNNS, "Transcendence of Jacobi's theta series", Proc. Japan Acad. Ser. A Math. Sci., Vol. 72 (1996), pp. 202-203.<|endoftext|> -TITLE: Stronger version of the isoperimetric inequality -QUESTION [15 upvotes]: I have been searching for a version of the isoperimetric inequality which is something like: -$P(\Omega) - 2\sqrt{\pi} Vol(\Omega)^{1/2} \geq \pi (r_{out}^2 - r_{in}^2)$ where $r_{out}$ and $r_{in}$ are the inner and outer radius of a given set. There are of course details which I am missing such as what kind of sets this applies to (clearly connected and possibly simply connected). I was hoping somebody may recognize this inequality and be able to direct me to a source for it along with a proof. -Update: I'm curious if anyone can direct me to a some papers which relate the isoperimetric deficit to Hausdorff distance. Such as: -$P(\Omega)^2 - 4\pi Vol(\Omega) \geq C d_H(\Omega,B)^2$ whre $B$ is a sphere in $\mathbb{R}^2$ -which may be the inner or outer circle. -Update April 12: I would like to know if the first Bonnesen inequality written below is strictly stronger than the one in higher dimensions? In particular, if one considers the Fraenkel assymetry $\alpha(\Omega) = \min_B |\Omega \Delta B|$ where $|B|=|\Omega|$, does it hold on a bounded domain that -$ r_{out}^2 - r_{in}^2 \leq C \alpha(\Omega)$, -for some constant $C>0$? This seems like it should be true but I can't seem to find a concise proof of it. - -REPLY [7 votes]: A very good source of Bonnesen type inequalties is the paper by Rovert Osserman entitled Bonnesen style isoprimetric inequalities, Americam Math Monthly 86(1979) 1-29. Here is another link for the same paper through this page. Osserman's 1978 Bulletin AMS paper on the ioperimetric inequality is also a good related source.<|endoftext|> -TITLE: Principal bundles over groups -QUESTION [7 upvotes]: If we have an extension of groups (say algebraic groups or group schemes) $1\to F\to P\to G\to 1$, then $P$ is a principal $F$-bundle over $G$ (is it locally trivial?). How about going in the opposite direction? - -Question. Let $F$ and $G$ be groups. $P$ be a principal $F$-bundle over $G$. When does $P$ carry a structure of a group such that $1\to F\to P\to G\to 1$ is an extension? How to classify all such structures? - -Sounds connected with (some kind of) group cohomology of $G$ with coefficients in $F$, but I can't figure out what exactly this group is. Does the obstruction lie in some $H^2$, which is only a pointed set, and if it vanishes then the structures are classified by the corresponding $H^1$? The fact that the principal bundle itself corresponds to an element of $H^1(G, F)$ confuses me a bit. -For example, if $G$ is an abelian variety and $L$ a line bundle on it, then we have the corresponding principal $F=\mathbb G_m$-bundle $P_L$. When does it give an extension of $G$ by $\mathbb G_m$? - -REPLY [2 votes]: The 'some $H^2$' which classifies group extensions you are thinking of is actually cohomology $H^1$ with values in the crossed module $F \to Aut(F)$ (or the 2-group corresponding to it). -See http://ncatlab.org/nlab/show/nonabelian+group+cohomology for more details.<|endoftext|> -TITLE: Syntactically capturing complexity classes -QUESTION [10 upvotes]: Primitive recursive functions are syntactically constructible in the sense that from a set of "axioms" we can build every function in the set $PR$. This basicly means that we can build a machine that prints the definition for every function in $PR$. -Now, we can build hierarchies in the set $PR$ by adding some semantic restrictions. For example Grzegorczyk created hierarchy $\{\mathcal{E}_i\}$ by restricting the rate of growth of the functions in each level. -I found papers mentioning the fact that if we take the second level of Grzegorczyk-hierarchy and define $E_2 = \{ f\in\mathcal{E}_2| ran(f)\in\{0,1\}\}$ (i.e. give yet another semantic restriction), then $E_2$ encapsulates LINSPACE (to my understanding its not actually this straightforward, but the idea should come clear). -In this construction we started defining functions syntactically and added some semantic constraints to come up with a class of functions computable in linear space. -This motivates to ask if there exists any constructions which provide ways to deploy purely syntactic machinery to produce, say, all the Turing-machines that run in polynomial space / time / whatever complexity class? Or functions instead of Turing-machines? -Is this even possible? - -REPLY [5 votes]: Bellantoni and Cook's syntactic characterization of P, and Bellantoni's 1992 thesis should probably be mentioned: - -http://www.cs.toronto.edu/~sacook/homepage/ptime.pdf -http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.53.8024<|endoftext|> -TITLE: need a good reference for introduction to elementary theory of groups -QUESTION [7 upvotes]: I've recently become interested in the elementary theory of groups due to Sela and Myasnikov-Kharlampovich's work with free groups. I'd like a good introduction to the field of the elementary theory of groups, and in particular I'd like a reference to contain examples of group properties that cannot be read from a group's elementary theory. For example, it seems that the statements "G is vritually abelian" or "G is hopfian" could not be expressed with first-order sentences, but I don't have enough knowledge yet to determine if such things are true. Does anyone know such a reference? Or, specifically, does someone know a proof of the fact that being hopfian (or some other group property) can't be read from a group's elementary theory? Thanks! - -REPLY [4 votes]: You might try Champetier and Guirardel's Limit groups as limits of free groups. -It has a short section (section 5) on elementary and universal theory, though perhaps none of the "non-examples" you're looking for. It is, however, a pleasure to read and if you're interested in limit groups, Makanin-Razborov digrams, etc. I highly recommend it.<|endoftext|> -TITLE: AX=XB and the Cecioni--Frobenius theorem -QUESTION [7 upvotes]: The Frobenius--Cecioni theorem states that if $A$ and $B$ are square matrices with entries in a field $k$ then the dimension of the $k$ vector space of solutions of -$$ -AX=XB -$$ -is given by the sum -$$ -\sum_{i,j} \deg( \gcd(d_i(A),d_j(B))) -$$ -Here $d_i(\cdot)$ denotes the $i$th invariant factor of its argument. My question: How well known is this theorem in any particular context? (E.g., "I use it all the time and I work in number theory ...", "I've never heard of it and I taught abstract algebra for years ...") Are there any well known applications of this result? - -REPLY [6 votes]: Great, so now I know how it is called. I came across a corollary some time ago (C. Koc, AMM 2002, solution of problem #10813) and posted it on the MathLinks forum. From this corollary either "loup blanc" (another poster in that topic) or me reconstructed the theorem, at least in the case of an algebraically closed field. -I still do not know an answer to the natural question: For a given $n$, what are the possible values of $m$ such that there exist two $n\times n$ matrices $A$ and $B$ with $\dim\left\lbrace X\in k^{n\times n}\mid AX=XB\right\rbrace = m$ ? This is equivalent to asking for the possible values of $\sum\limits_{i} a_ib_i$ where $\left(a_i\right)$ and $\left(b_i\right)$ are two finite sequences of nonnegative integers satisfying $\sum\limits_i a_i\leq n$ and $\sum\limits_i b_i\leq n$. This is a purely combinatorial question. Maybe you happen to know an answer to it, since I am hardly the first person to pose it?...<|endoftext|> -TITLE: Consequences of the geometric properties of the eigencurve -QUESTION [11 upvotes]: The eigencurve $\mathcal{E}$ is a rigid-analytic space parametrizing certain $p$-adic families of modular forms and associated Galois representations. By constructing an auxiliary reduced rigid curve locally isomorphic to $\mathcal{E}$ it was shown that $\mathcal{E}$ is a curve. There is a morphism of rigid spaces $f: \mathcal{E} \to \mathcal{W}$ assigning to a modular form its weight-character. In general this is not a proper morphism of rigid spaces, but a "valuative criterion" was proven in some cases. - -As it is a "moduli space," one should be able to deduce certain properties about $p$-adic modular forms from the geometric structure of $\mathcal{E}$. What are some potential number-theoretic consequences of the "properness" of $f: \mathcal{E} \to \mathcal{W}$? How about other unknown geometric properties? - -I hope you can explain what this object is about for non-experts like me. Thank you! - -REPLY [11 votes]: The eigencurve is an honest moduli space---it parametrises families of finite slope overconvergent modular eigenforms (or more precisely, of systems of overconvergent finite slope Hecke eigenvalues)---but I know of no "natural" properties of p-adic modular forms that one can deduce from any geometric structure, other than e.g. the consequences of the fact that the eigencurve is 1-dimensional. -Properness just means that you can't have a family of overconvergent modular forms of finite slope that degenerates to an infinite slope form, which somehow isn't too surprising if you think about what the associated $(\phi,\Gamma)$-modules might look like. I think Alex Paulin has been thinking about making this train of thought rigorous. The fact that the eigencurve is 1-dimensional can be regarded as a statement about deformations of Galois representations---an overconvergent finite slope eigenform gives rise to a Galois representation with a crystalline period and one can now consider deforming such things and I guess one will end up proving that a certain $H^1$ is 1-dimensional. -As for "unknown geometric properties", I don't know of any reasonable conjectural statements about the eigencurve that have not already been conjectured. I made the observation that near the boundary of weight space the eigencurve seemed to be a disjoint union of annuli; Kilford and I proved a special case of this and Roe proved another one---and this has classical consequences for slopes of the $U_p$ operator. For example the eigenvalues of $U_2$ on weight $k$ level $\Gamma_1(4)$ modular forms, when $k$ is odd, have $2$-adic valuations which form an arithmetic progression (whatever the value of $k$). But I think this is nothing more than a curiosity at the moment.<|endoftext|> -TITLE: "Twisted" universal enveloping algebra? -QUESTION [7 upvotes]: Let $\mathfrak{g}$ be a $k$-Lie algebra, and $Q: \bigwedge^2 \mathfrak{g}^* \rightarrow k$; define $U_Q(\mathfrak{g})$ to be the quotient of the full tensor algebra over $\mathfrak{g}$ by the ideal generated by elements of the form $x\otimes y - y \otimes x -[x,y] - Q(x,y)$. This definition does not depends properly on $Q$ but only in its cohomology class in the Chevallay cohomology. -Has anyone seen this kind of algebra appear somewhere and/or has a name for them? They appeared to me from a deformation... - -REPLY [10 votes]: These algebras were considered by Ramaiengar Sridharan a long time ago. See [Sridharan, R. Filtered algebras and representations of Lie algebras. Trans. Amer. Math. Soc. 100 1961 530--550. MR0130900 (24 #A754)] -If the map you are using to twist is not a Chevalley-Eilenberg cocycle, then things are ugly. In particular, you do not get a PBW-basis of the algebra (the cocycle condition is equivalent to the BPW property, in fact; this was in the general context of quadratic Koszul algebras a few years ago) -By the way: I call them Sridharan enveloping algebras, and I have heard others do the same.<|endoftext|> -TITLE: Why can we define the moment map in this way (i.e. why is this form exact)? -QUESTION [8 upvotes]: Given a symplectic manifold $(X, \omega)$ and a group $G$ acting on $X$ preserving the symplectic form, we define the moment map $\mu : X \to \mathfrak{g}^*$ so that -$$ -\langle d\mu(v), \xi\rangle = \omega\big(\xi^*(x), v\big) -$$ -where $\xi \in \mathfrak{g}$, $\xi^*$ is the vector field generated by $\xi$, and $v \in T_x(X)$. -Why can we do this? Why is $\mu$ defined? -I read the above definition as follows. Define the $\mathfrak{g}^*$-valued 1-form $\theta$ via -$$ -\theta(v) = \omega(\xi^*,v). -$$ -Then this form is exact, and so we define $\mu$ so that $d\mu = \theta$. -Why can we do this? Having done some simple computations, the fact that $G$ preserves the symplectic form seems to be imply that exactness holds (and from the above definition it is necessary), but I don't know why this should be the case. -My question is: - -Why does the fact that $G$ preserves the symplectic form imply that the 1-form $\theta$ defined above is exact? - -Also: - -Is the preservation of the symplectic form equivalent to the exactness of $\theta$? - -REPLY [6 votes]: There is yet another interpretation of the already mentioned obstructions for the existence and uniqueness of momentum maps (moment mappings, moment maps, momentum mappings....) in terms of Poisson geometry. This even generalizes then to Poisson manifolds and makes things perhaps a little bit more transparent: -For the existence of a (non-equivariant) momentum map you ask Poisson vector fields to be Hamiltonian, the obstruction therefore lies in the first Poisson cohomology, which is precisely the quotient of Poisson vector field modulo Hamiltonian vector fields. -In the symplectic case this quotient becomes canonically isomorphic to the first deRham cohomology as symplecticl vector fields correspond to closed one-forms while Hamiltonian ones correspond to exact one-forms via the musical isomorphism induced by $\omega$. -The uniqueness (and also the equivariance) is then controlled by the zeroth Poisson cohomology which are th functions with trivial Hamiltonian vector field, also called the Poisson center. In the symplectic and connected case, these are just the constant functions, but in the general Poisson case this might be a much more interesting cohomology. -The advantage of this point of view is, perhaps, the fact that now all obstructions are somehow arising from the same complex: the canonical complex associated to a Poisson structure $\pi$. As a vector space this is just the sections of the exterior algebra of the tangent bundle (i.e. the multivector fields) and the differential is the Schouten bracet with $\pi$. In the symplectic case, this boils down to the deRham complex via the musical isomorphim. -You can find a detailed discussion of this in the (by the way very: nice) book of Ana Cannas da Silva and Alan Weinstein on "Geometric Models for Noncommutative Algebras".<|endoftext|> -TITLE: Maximal euler characteristic of surfaces bounding two fixed curves -QUESTION [15 upvotes]: Let $\gamma_0$ and $\gamma_1$ be two simple closed curves in a closed surface $S$. - -What is the maximum Euler characteristic of a compact properly embedded surface $\Sigma \subset S\times [0,1]$ such that $\partial \Sigma = \gamma_0 \times \{0\} \cup \gamma_1 \times \{1\}$? - -Of course, in order for such a surface $\Sigma$ to exist, the two curves $\gamma_0$ and $\gamma_1$ must represent the same class in $H_1(S,\mathbb Z_2)$. Note that $\Sigma$ may be non-orientable. -If $2n$ is the minimum geometric intersection number of $\gamma_1$ and $\gamma_2$, it is easy to construct a $\Sigma$ with $\chi(\Sigma)\geqslant \chi(S) - n$. Is there a converse estimate of this kind? Do we have $\chi(\Sigma) \leqslant -n$ when $S$ is a torus? - -REPLY [7 votes]: Here is a graph $\mathcal{G}(S,x)$ associated to a surface and homology class (similar to the 1-skeleton of the curve complex): For a fixed class in $x\in H_1(S;\mathbb{Z}_2)$, consider isotopy classes of embedded multicurves representing the homology class $x$ (one may assume no components are parallel and there are no trivial components). Make this collection the vertices of the graph $\mathcal{G}(S,x)$. Connect two vertices $A, B$ to be adjacent in the graph $\mathcal{G}(S,x)$ if $A\cup B$ are disjointly embedded, and after removing all parallel curves of $A\cup B$, the remaining components bound a pair of pants or a twice-punctured projective plane (this second can only happen if $S$ is non-orientable). I'm not sure if such a complex has been defined before, but there is a somewhat analogous complex defined in the integral homology case by Bestvina, Bux, and Margalit. Also, this is related to a technique of Hatcher-Thurston to undertand surfaces in two-bridge knot complements. -I claim that the maximal Euler characteristic of a surface bounding $\gamma_0\times 0 \cup \gamma_1\times 1$ is the negative of the distance between $\gamma_0$ and $\gamma_1$ in $\mathcal{G}(S,x)$. Put the product metric on $S\times [0,1]$, and make $\Sigma$ into a minimal surface with respect to this metric (Theorem 6.12 of Hass-Scott). Then $S\times t, t\in [0,1]$ gives a foliation by totally geodesic surfaces, and by the maximum principle, they can be tangent to $\Sigma$ in only saddle tangencies (see an argument of Hass, we will assume things are perturbed to be generaic). Thus, for all but finitely many $t$, $S\times t$ meets $\Sigma$ in a finite collection of curves, giving rise to a vertex of $\mathcal{G}(S,x)$. As one passes through a tangency point $S\times t_0$ (assuming things are generic), the intersection changes by a saddle move, giving a surface in $S$ of Euler characteristic $-1$ bounding the curves before and after the tangency. There can never be a closed trivial curve occurring, because this would give rise to a center tangency. Thus, each saddle tangency gives an edge between the adjacent vertices of $\mathcal{G}(S,x)$, and therefore $\Sigma$ gives a path between $\gamma_0$ and $\gamma_1$ in $\mathcal{G}(S,x)$. Conversely, any such path gives rise to a surface. -Of course, there will be many geodesics connecting $\gamma_0$ and $\gamma_1$ in $\mathcal{G}(S,x)$, given by any Morse function on $(\Sigma, \gamma_0,\gamma_1)$ with only index 1 critical points, and I don't expect the distance function to be easy to compute (probably one should use normal surface theory to compute it).<|endoftext|> -TITLE: Proposition 3.93 of Harris-Morrison (rational classes on Deligne-Mumford moduli stack vs. rational classes on Deligne-Mumford moduli space) -QUESTION [6 upvotes]: Proposition 3.93 of Harris-Morrison's "Moduli of Curves" describes the relationship between rational divisor classes on the Deligne-Mumford moduli stack and rational divisor classes on the Deligne-Mumford moduli space. -Does this proposition generalize to Chow classes, K-theory classes, cohomology classes? -I assume the answer is "yes" -- is there a reference or an easy proof? - -REPLY [9 votes]: It does generalize to Chow classes and to (classical or étale) rational cohomology. For Chow classes this is part of the basics of intersection theory of stacks, for example, in my paper Intersection theory on algebraic stacks and on their moduli spaces. -For cohomology, I don't know a reference off the top of my head, but the argument is not difficult, here is a sketch. If $\mathcal X$ is a Deligne-Mumford stack (with finite inertia) and $\pi: \mathcal X \to \mathbf X$ is its moduli space, you need to show that $\mathrm R^i \pi_* \mathbb Q_{\mathcal X}$ (for the classical case) is zero fr $i > 0$, and concides with $\mathbb Q_{\mathbf X}$ for $i = 0$. This is an étale local problem on $\mathbb X$, so we may assume that $\mathcal X$ is of the form $[U/G]$, where $G$ is a finite group acting on a scheme $U$. Then it is a standard fact that the rational cohomology of $[U/G]$, which is the $G$-equivariant cohomology of $U$, coincides with the $G$-invariants in the classical cohomology of $U$, and this makes the result clear. The argument for étale cohomology is similar. -On the other hand, this is most definitely false for K-theory. For example, if $G$ is a finite group acting a point, the K-theory ring of the quotient stack $[\mathrm{pt}/G]$ is the representation ring of $G$, which is in general very far from being trivial. The connection between the K-theory of the stack and the K-theory of the space is complicated, and is given in its general form by Toën's Riemann-Roch. -[Edit] I guess I was really answering a different question, on whether there is an isomorphism between Chow groups, and so on, of the stack, and the corresponding groups of the moduli space. The Proposition in Harris-Mumford is actually more refined. The answer is still positive for Chow classes; for cohomology I don't know how to make sense out of it, because I don't know how to interpret the order of the stabilizer of the generic point.<|endoftext|> -TITLE: dependence of eigenvalues on parameters -QUESTION [7 upvotes]: Let $f$ be a positive real-analytic function on the closed unit disk. Consider the eigenvalue problem $\Delta \phi = \lambda f \phi$, -with $\phi = 0$ on the boundary. There exists a sequence of eigenvalues $\lambda_n$. Now suppose $f$ depends real-analytically on a parameter $t$ for $t$ in some interval containing $0$. Let $n$ be given and let $k$ be the dimension of the eigenspace of $\lambda_n$. Do there exist $k$ functions of $t$, defined on some interval about $t=0$, that are at least $C^1$ in $t$ and give $k$ eigenvalues of $t$, all of which equal $\lambda_n$ when $t=0$? -Remarks: (1) Courant-Hilbert, Methods of Mathematical Physics, vol. 1, page 419 proves continuity (i.e. $C^0$ dependence). You would think if $C^1$ dependence were known (in 1953) they would mention it. -(2) We do not require that the functions preserve the order of the eigenvalues. For example it might happen that $\lambda_2 < \lambda_3$ for $t < 0$ and $\lambda_2 > \lambda_3$ for $t > 0$, and maybe they cross transversally, though I do not know if this can really happen. -(3) Perhaps one can even get real-analytic dependence on $t$. -(4) There is a thick book by Kato on perturbation theory with a lot of theorems in it, one or more of which possibly contains the answer. - -REPLY [7 votes]: In your case I think you can apply Rellich's theorem, that is Theorem VII.3.9 in Kato's book (p.392 in my edition). The result states that, whenever you have a family of selfadjoint operators with compact resolvent, depending analytically on a real parameter on some open interval of the reals, with a common domain independent of the parameter (this is what Kato calls a family of type (A)), then you can enumerate both eigenvalues and eigenfunctions in such a way that they are analytic functions of the parameter in the same interval.<|endoftext|> -TITLE: Is there a Riemann-Roch for smooth projective curves over an arbitrary field? -QUESTION [8 upvotes]: Let $X$ be a smooth projective curve over a field $k$. We let $\omega$ be the canonical -line bundle of $X$ and we denote by $F$ the field of $k$-valued rational functions on $X$. -(1) When $k$ is algebraically closed then $\omega$ is a dualizing sheaf for $X$. From there it is -easy to prove Riemann-Roch for regular (holomorphic) line bundles $L$ over $X$: By this I mean -a precise formula which computes the Euler characteristic of $L$ in terms of the degree -of $L$ and the genus of $X$ (I think of both as being topological invariants). -(2) When $k$ is a finite field then one may consider the topological ring $\mathbf{A}_F$, -the ring of Adeles of $F$. Doing Fourrier analysis on this self-dual locally compact abelian group and doing a counting argument one may deduce Riemann-Roch. -Q1: Is it possible to generalize Riemann-Roch to other fields? What about real and $p$-adic numbers? -Q2: Is $\omega$ a dualizing sheaf when $k$ is finite? If not -(I guess that in general one has to replace the notion of dualizing sheaf by some kind -of complex in a derived category) -Q3: Is there a way to prove simultaneously $(1)$ and $(2)$? -Q4: Is there some notion that would encompass simultaneuously $\mathbf{A}_F$ and $\omega$? - -REPLY [25 votes]: Dear Hugo, the wonderful formalism of schemes allows us to have a Riemann-Roch theorem for a projective curve $X$ over an arbitrary field $k$, even without any assumption of smoothness. -It says, like in the good old times of Riemann surfaces, that for a Cartier divisor $D$ on $X$ we have - $$\chi (\mathcal O_X(D))= deg(D)+ \chi (\mathcal O_X)$$ -There is a dualizing sheaf $\omega$ and Serre duality yields the formula -$$ h^0(X,\mathcal O_X(D))-h^0(X,\omega \otimes\mathcal O_X(-D))=1-p_a(X)+deg D$$ -where $p_a(X)=1-\chi(\mathcal O_x)$ is the so called arithmetic genus of the curve. -Everything is in our friend Qing Liu's fantastic book Algebraic Geometry and Arithmetic Curves but I bet he's too modest to give you this obvious answer! -Edit The first displayed formula is actually valid in even greater generality: it holds for any projective curve $X$ (smooth or not) over an arbitrary artinian ring $k$. The proof is on page 164 of -Altman, A.; Kleiman, S., Introduction to Grothendieck duality theory. Lecture -Notes in Mathematics No. 146, Springer-Verlag, Berlin-New York, 1970 -Complement As an answer to Hugo's question in his comment below, let me add that indeed, in the case of a smooth projective curve over a field $k$, the dualizing sheaf $\omega$ is nothing else than the canonical sheaf . More generally for a smooth projective variety of dimension $r$ over $k$, the dualizing sheaf is just the canonical sheaf $\omega=\Omega^r_{X/k}$. This is (a special case of) Theorem I.4.6, page 14 in Altman-Kleiman's monograph.<|endoftext|> -TITLE: Writing "Semi-Formal" Proofs -QUESTION [23 upvotes]: I am very interested in proofs. I have taken an undergraduate course -called "Logic and Set Theory" which I found very interesting, but ultimately -unsatisfying. My biggest disappointment has to do with the language in which -proofs are expressed. It seems to me that we have all of the symbols necessary -to express a proof in "pure math". By which I mean, only using symbols and a -few specialized words (iff, let, ...). And yet most proofs that I have seen are -just walls of English text, interpolated with mathematical symbols. -When I read a complex proof, I find myself needing to transcribe it into pure -symbols before I have any chance at understanding it. I have talked to a -professor about this, and he informed me that my "pure math" proofs were -actually considered informal, and not proper proofs at all! He seemed skeptical -that anyone would actually prefer symbols to English. -I have searched Wikipedia and Google for more information, and I see that there -is something called a "Formal Proof" (although I have heard this term used in -other situations, and so I am not quite sure it means what I think it means) -which uses a computer to verify a proof written in a special programing -language. As fascinating as that is, it seems to be a step further than what I -am looking for. -Is there a well known method for writing and sharing proofs of mathematical -statements that uses only mathematical symbols and is not a full blown -programming language? And if not, why is this considered "taboo" or "informal"? -Thanks, ---jc -EDIT: -I guess this turned out to not be a real question? Strange, I checked, it definitely ends in a question mark. Thanks everyone for the help, advice, and links. I appreciate your input. - -REPLY [4 votes]: N. G. de Bruijn is known for, among many things, his work on the Automath project. Automath was a formal language for writing proofs that had a big influence on many of the languages used today for computer-aided mathematical proof (such as Coq and Mizar). However, de Bruijn also spent some time developing a system for "semi-formal" proof, which he called the "Mathematical Vernacular" (MV): - -N.G. de Bruijn, The Mathematical Vernacular, A Language for Mathematics with Typed Sets - -Here is some explanatory text from the introduction: - -1.2. The word "vernacular" means the native language of the people, in contrast to the official, or the literary language (in older days in contrast to the latin of the church). In combination with the word "mathematical", the vernacular is taken to mean the very precise mixture of words and formulas used by mathematicians in their better moments, whereas the "official" mathematical language is taken to be some formal system that uses formulas only. [...] -1.4. Many people like to think that what really matters in mathematics is a formal system (usually embodying predicate calculus and Zermelo-Fraenkel set theory), and that everything else is loose informal talk about that system. Yet the current formal systems do not adequately describe how people actually think, and, moreover, do not quite match the goals we have in mathematical education. Therefore it is attractive to try to put a substantial part of the mathematical vernacular into the formal system. One can even try to discard the formal system altogether, making the vernacular so precise that its linguistic rules are sufficiently sound as a basis for mathematics. [...] -1.6. The idea to develop MV arose from the wish to have an intermediate - stage between ordinary mathematical presentation on the one hand, and fully coded presentation in Automath-like systems on the other hand. One can think of the MV texts being written by a mathematician who fully understands the subject, and the translation into Automath by someone who just knows the languages that are involved. [...] -1.13. One might think of direct machine verification of books written in - MV, but this will be by no means so "trivial" as in Automath. Checking books in MV may require quite some amount of artificial intelligence. In the first place MV allows us to omit parts of proofs, at least as long as no definitions are suppressed. - -You can find an example of an "MV book" in section 18 (alas, it is a bit difficult to typeset here...).<|endoftext|> -TITLE: Immerse an affine schemes into $A^n_S$ -QUESTION [6 upvotes]: Suppose $f: X\rightarrow S$ is of finite type, S is Noetherian. Now X=Spec B is affine, but the morphism f is not an affine morphism. S is not affine (or really f does not factor through any affine subscheme on S), nor is S of finite type over $\mathbb Z$. Now my question is, can we have an immersion of $X\rightarrow \mathbb{A}^n_S$ as S schemes? -Remark: If S is affine this is trivial. Moreover when f is an affine morphism I think it can also be done. When S is of finite type over $\mathbb Z$ by transitivity of finite type we get a map $X\rightarrow \mathbb{A}^n_{\mathbb Z}$ therefore by universal property of fiber product we get a map $X\rightarrow \mathbb{A}^n_S$. - -REPLY [8 votes]: As noticed by Mattia, you have to suppose $f$ affine. Now under the further assumption that $S$ is separated and $X$ is affine, there exists a closed immersion as you want (Mattia's arguments then work): for any affine open subset $V$ of $S$, the canonical morphism -$$f^{-1}(V)=X\times_S V\to X\times_{\mathbb Z} V$$ -is a closed immersion because $S$ is separated. As $X$ is also affine, we get a surjective map $O(X)\otimes_{\mathbb Z} O(V)\to O(f^{-1}(V))$. Choose sufficiently many elements $a_1,\dots, a_n\in O(X)$ so they generate $O(f^{-1}(V_i))$ over $O(V_i)$ for a finite affine covering of $S$. Then the natural map $O_S[x_1,\dots, x_n]\to f_*O_X$ sending $x_i$ to $a_i$ is surjective and induces a closed immersion of $X$ in $\mathbb A^n_S$. -[EDIT] I just remembered that if $S$ is separated and $X$ is affine, then the above reasoning implies that any morphism $X\to S$ is affine. I should say that I learned this fact from V. Berkovich. -[EDIT2] The case when $S$ is not necessarily separated. - -Let $S$ be a quasi-compact scheme and let $f : X\to S$ be a morphism - of finite type from an affine scheme $X=\mathrm{Spec}(B)$ to $S$. Then there exists an - immersion of $S$-schemes $X\to \mathbb A^n_S$. - -Proof: First note that $f$ is separated because $X$ is separated, and $f$ is -quasi-compact by hypothesis. So $f_*O_X$ is quasi-coherent on $S$. -We are going to construct a morphism of quasi-coherent algebras -$O_S[x_1,\dots, x_n]\to f_*O_X$ which will induce an immersion -$X\to {\mathrm{Spec}}(f_*O_X)\to \mathbb A^n_S$. -Let $V$ be an affine open subset of $S$. Then $X_V$ can be covered -by a finite number of principal open subsets $D_X(b_1),\dots, D_X(b_m)$ -of $X$ and there exist $c_1,\dots, c_r\in B$ such that -$B_{b_i}=O(V)[c_1,\dots, c_r, b_i^{-1}]$ for all $i\le m$. -By taking a finite affine covering of $S$, we find $a_1,\dots, a_n\in B$ -such that for any open subset $V$ of this covering, -$X_V$ is covered by some $D_X(a_i)$'s and each of these $O(D_X(a_i))$ is -generated over $O(V)$ by the $a_j$'s and $a_i^{-1}$. -Use $a_1, \dots, a_n$ to construct a morphism $O_{S}[x_1,\dots, x_n]\to f_*O_X$ -sending $x_i$ to $a_i$. Let us show that the induced morphism -$\tau : X\to \mathbb A^n_S$ is an immersion. It is enough to check this -over an affine -open subset $V$ in the above covering of $S$. So consider -$\tau_V : X_V\to \mathbb A^n_V$. Suppose that $X_V$ is covered by -$a_1,\dots, a_m$. Then -$$\tau_V(X_V)\subseteq \cup_{1\le i \le m} D_{\mathbb A^n_V}(x_i).$$ -Moreover, $\tau_V^{-1}(D_{\mathbb A^n_V}(x_i))=X_V\cap D_X(a_i)=D_X(a_i)$ and -$O(D_{\mathbb A^n_V}(x_i))\to D_X(a_i)$ is surjective by -construction. So $\tau_V$ is a closed immersion into -$\cup_{1\le i \le m} D_{\mathbb A^n_V}(x_i)$ and $\tau$ is an immersion -(closed immersion into an open subscheme and also an open immersion into -a closed subscheme because $f$ is quasi-compact).<|endoftext|> -TITLE: Geodesics for a Homogeneous Space? -QUESTION [9 upvotes]: Is there a specific formula/method to find geodesics for a Homogeneous space? (excluding general methods applicable to arbitrary riemannian manifold) - -REPLY [10 votes]: One method is to use the generalization of the principle of conservation of momemntum and -angular momentum. If $\phi_t$ is any one parameter group of isometries of any Riemannian manifold, with time derivative expressed as a vector field $X$, then for any geodesic $g$, -the inner product -$\left \langle X, \dot g \right \rangle $ is constant along the geodesic, where $\dot g$ denotes the unit tangent vector to $g$ (assuming $g$ is parametrized by arclength). In Euclidean space, -applied to 1-parameter groups of translations, this gives the principle of conservation of -momemntum for a geodesic; applied to 1-parameter groups of rotations, it gives the principle -of conservation of angular momentum. -In a homogeneous space, there are at least enough such vector fields to form a basis at each point. In an $n$-dimensional homogeneous space, if there are more than $n$ linearly independent infinitesimal isometries, then the equations, for a typical initial vector, are consistent only in a proper subset, and on that subset, they typically define a vector field. -This may give you all the information you need, without even integrating the vector field. -But in any case, it cuts the differential equation down to many fewer degrees of freedom.<|endoftext|> -TITLE: Is there an explicit example of such a real number with the following property? -QUESTION [16 upvotes]: In Diophantine approximation, for a given positive real number $\alpha$ let $[a_0, a_1, \cdots]$ denote its continued fraction expansion and let $p_n/q_n = [a_1, \cdots, a_n]$. Then it is known that $q_n$ grows at least exponentially. In 1935, Paul Levy proved that in fact for almost all real $\alpha$, we have $\displaystyle \lim_{n \rightarrow \infty} q_n^{1/n} = \exp(\pi^2/12 \log 2)$. The result was proved using Ergodic theory. Now my question is, does there exist a single known example of a real number $\beta$ such that $\beta = [b_0, b_1, \cdots]$, $p_n/q_n = [b_0, \cdots, b_n]$, and $\displaystyle \lim_{n \rightarrow \infty} q_n^{1/n} = \exp(\pi^2/12 \log 2)$? -For example, one might expect $\beta = \exp(\pi^2/12 \log 2)$ to do the trick... - -REPLY [5 votes]: Here is a sort-of-but-not-really explicit answer, following up on Helge's idea. Set $h = e^{\pi^2/(12 \log 2)}$. Recursive define the $b_i$ and $q_i$ as follows: $q_{-2} =1$, $q_{-1}=0$ and $q_i = b_i q_{i-1} + q_{i-2}$. If $q_{i}^{1/i} > h$, then $b_{i+1} = 2$, otherwise, $b_{i+1}=4$. This is clearly a well defined recursion; you can decide whether or not you think the result of this process is explicit. -Lets first study, in general, the situation where every $b_i$ is either $2$ or $4$. Write $r_i = q_i/q_{i-1}$. Then $r_i = b_i + r_{i-1}^{-1}$ so $b_i < r_i < b_i + 1$. Note that $3 < h \approx 3.28 < 4$. -So, our algorithm has the property that, if $q_i^{1/i} < h$, then $r_{i+1}>h$ and vice versa. -Set $s_n = \log q_n = \sum_{i=0}^{n-1} \log r_i$. So, if $s_n < (\log h) n$, then $s_{n+1} = s_n + \log r_i \in [s_n+\log 4, s_n + \log 5]$ and we see that $s_{n+1} - (\log h) (n+1) \in $[(s_n - (\log h) n) + \log 4 - \log h, (s_n - (\log h) n) + \log 5 - \log h]$. Writing $t_n = s_n - (\log h) n$, we see that - -If $t_n <0$, then $t_{n+1} \in [t_n + (\log 4 - \log h), t_n + (\log 5 - \log h)]$. - -Similarly, - -If $t_n > 0$, then $t_{n+1} \in [t_n + (\log 2 - \log h), t_n + (\log 3 - \log h)]$ - -In particular, once $t_n$ gets into the interval $[\log 2 - \log h, \log 5 - \log h]$, it stays there. I leave it to you to check that it gets there. -So $t_n$ is bounded, $\log q_n = (\log h) n + O(1)$ and $\lim_{n \to \infty} q_n^{1/n} = h$.<|endoftext|> -TITLE: Example of a variety with $K_X$ $\mathbb Q$-Cartier but not Cartier -QUESTION [23 upvotes]: I know the definition of $K_X$ on a normal, singular variety, but I don't have a good set of examples in my mind. What's an example of a variety where $K_X$ is $\mathbb Q$-Cartier but not Cartier? Are there any conditions under which an adjunction formula lets me compute the canonical class of a singular divisor? - -REPLY [6 votes]: Here is a more algebraic perspective on your question. If $X=\text{Spec} (R)$ is affine and $R$ is a Cohen-Macaulay algebra over some field (the following is true in more general setting), then $K_X$ is Cartier is equivalent to $R$ is Gorenstein. On the other hand, $K_X$ is $\mathbb Q$-Cartier is the same as the class of $K_X$ is torsion in the divisor class group (assuming $R$ is normal). -So to find a class of examples, you need normal Cohen-Macaulay rings with torsion class group but not Gorenstein. If $R$ is a Veronese $S^{(d)}$ of $S=\mathbb C[x_1,\cdots, x_n]$ then $Cl(R)$ is always torsion, but $R$ is Gorenstein if and only if $d|n$ (Sandor's example is indeed the simplest one in this class, with $d=2, n=3$).<|endoftext|> -TITLE: Virasoro action on the elliptic cohomology -QUESTION [17 upvotes]: I'm trying to understand better the mathematical notion of elliptic cohomology. Note that I only know the physics definition of the elliptic genus given in Witten's paper. -Let $X$ be a Calabi-Yau manifold. (The elliptic genus can be defined for any $X$ with less properties, but the physics is nicest when $X$ is a CY. So let me assume that.) -In the paper quoted above, a sequence of sheaves $F_k$ on $X$ ($k=0,1,2,$...) were constructed (by taking suitable tensor powers of the tangent bundle), such that $\Phi(q)=\sum_k q^k\chi(F_k)$ gives the elliptic genus of $X$. -Now, the physics construction says that, before taking the Euler characteristic, there is an action of (N=2 superconformal) Virasoro algebra on $\oplus_{i,k} H^i(X,F_k)$; this is the basis of the modularity of the elliptic genus. I presume this action has already been geometrically constructed in the mathematical literature, given the fact that Witten's paper is from 1980s. -So, my question is, where can I find it? - -REPLY [2 votes]: I would like to add that (cohomology of) chiral de Rham complex has to be viewed as the large Kahler limit of halftwisted theory. Indeed, it does not use the Kahler data on the CY manifold and does not see the instanton count. So it is not the final answer in any sense, but is a very important step towards rigorous understanding of type II superstring theories.<|endoftext|> -TITLE: inferring the slope of a digitized line -QUESTION [8 upvotes]: Given real numbers $a$ and $b$, and an integer $n \geq 2$, let $f(n,a,b)$ be the minimum of $(nint(ja+b)-nint(ia+b)+1)/(j-i)$ (for $1 \leq i < j \leq n$) minus the maximum of $(nint(ja+b)-nint(ia+b)-1)/(j-i)$ (for $1 \leq i < j \leq n$), where $nint$ is the nearest integer function. What is known about the rate at which $f(n,a,b)$ goes to 0 as $n$ goes to infinity, for "generic" real numbers $a, b$? (That is, real numbers whose continued fractions have convergents that grow as dictated by Khinchin's law.) Experiments suggest that $f(n,a,b) = O(1/n^2)$. -This is related to the question of how accurately one can infer the slope of a line from a digitized version of the line; if there is any literature on this question, I would be very interested in pointers. -I am also interested in knowing how one can most efficiently compute the minimum of $(nint(ja+b)-nint(ia+b)+1)/(j-i)$ (for $1 \leq i < j \leq n$) and the maximum of $(nint(ja+b)-nint(ia+b)-1)/(j-i)$ (for $1 \leq i < j \leq n$). -For context, I'll mention that several recent postings of mine (especially sums of fractional parts of linear functions of n) approach this question from another angle. Specifically, my posting from last week is related to an estimator of the slope of the line that can be computed in linear (as opposed to quadratic) time but has typical error $O(1/n^{3/2})$. - -REPLY [2 votes]: Not an answer, more like an overgrown comment. The nearest integer to $x$ is the floor of $x+(1/2)$, and you can incorporate the 1/2 into $b$, so you're talking about things like $([ja+b]-[ia+b]+1)/(j-i)$. Now floor is identity minus fractional part, so we get to $(ja+b-\lbrace ja+b\rbrace-ia-b+\lbrace ia+b\rbrace+1)/(j-i)$, which simplifies to $a+(\lbrace ia+b\rbrace-\lbrace ja+b\rbrace+1)/(j-i)$. So you're looking at the minimum of $(\lbrace ia+b\rbrace-\lbrace ja+b\rbrace+1)/(j-i)$ minus the maximum of $(\lbrace ia+b\rbrace-\lbrace ja+b\rbrace-1)/(j-i)$, which may be a little more tractable than what you started out with. -EDIT: As to whether someone has already done this, the paper by Melter, Stojmenovic, and Zunic, A new characterization of digital lines by least square fits, Pattern Recognition Letters 14 (1993) 83-88, available at http://www.site.uottawa.ca/~ivan/F29-digital-line-lsf.PDF may be relevant or may point you to some useful work. -FURTHER EDIT: Another paper that looks like it might be useful is MR2323394 (2008e:68149) -Uscka-Wehlou, Hanna, -Digital lines with irrational slopes, Theoret. Comput. Sci. 377 (2007), no. 1-3, 157–169.<|endoftext|> -TITLE: Is there an alternative characterisation of vector bundles with vanishing characteristic classes? -QUESTION [16 upvotes]: This question came up yesterday during our index theory seminar. -Let $M$ be a 1-connected smooth manifold and let $E \to M$ be a finite-rank complex vector bundle over $M$. If all the Chern classes of $E$ vanish, what else can one say about $E$? In other words, is there an alternative characterisation of such $E$? -(For a similar question in the case of $M$ having nontrivial fundamental group, see this previous question.) - -REPLY [20 votes]: Since no one has mentioned it yet, let me point out one possibly interesting observation. If the base manifold $M$ is compact and has no torsion in its integral cohomology, then a vector bundle $E$ with vanishing Chern classes is stably trivial. This was pointed out to me by Robert Lipshitz. The reason is as follows: from looking at the Atiyah-Hirzebruch spectral sequence, one can see that there can't be any torsion in the complex K-theory of $M$. Looking at the Chern character, one concludes that $[E]$ must be trivial in $\widetilde{K}^0(M)$, i.e. $E$ is stably trivial.<|endoftext|> -TITLE: Which areas of arithmetic algebraic geometry can be learned as "black boxes" and are there any references where they are treated as such? -QUESTION [12 upvotes]: In Matthew Emerton's comment on Terry Tao's blog, he speaks about learning etale cohomology or the theory of Neron models as "black boxes". By this he means that you can learn what the theory is about and how to use it, without going into the detailed proofs of why they can be used. -Which theories (e.g. etale cohomology) can be learned as black boxes? -And where would one go (e.g. find lecture notes) to learn something like that? -Notes on something like this would ideally give you an idea of what is going on, give examples, and most importantly illustrate how they would be used to solve problems. I am mainly interested in arithmetic algebraic geometry and algebraic number theory, so I would especially like to know about "black boxes" in this direction, though "black boxes" in other areas might also be worth knowing about. - -REPLY [8 votes]: I find Hodge theory pretty scary stuff with its compact inclusions of Sobolev spaces, pseudodifferential operators and parametrixes for elliptic differential operators. However it is very easy to use the results of Hodge theory as emanating from a black box. I remember how exhilarated I was by the argument that a Hopf surface, homeomorphic to $S^1 \times S^3$, could not be Kähler, and much less projective, just because its first Betti number is $b_1=1$. Whereas by Hodge theory a compact Kähler manifold $X$ has betti numbers $b_q(X)$ which are even whenever $q$ is odd.<|endoftext|> -TITLE: Realizing braid group by homeomorphisms -QUESTION [18 upvotes]: Markovich and Saric proved the following remarkable theorem. Let $S$ be a compact surface of genus at least $2$ and let $MCG(S)=\pi_0(Homeo^{+}(S))$ be the mapping class group of $S$. There is then no right inverse to the natural projection $Homeo^{+}(S) \rightarrow MCG(S)$. This should be contrasted with the solution to the Nielsen realization problem by Kerckhoff, who proved that any finite subgroup of $MCG(S)$ can be realized by homeomorphisms of the surface. -My question is whether this is known for braid groups. Let me make this a little more precise. Let $X_n$ be a $2$-dimensional disc with $n$ punctures. By $Homeo^{+}(X_n)$, we mean homeomorphisms of $X_n$ that are the identity on the boundary and extend over the punctures (but are allowed to permute the punctures). The group $\pi_0(Homeo^{+}(X_n))$ is then the $n$-strand braid group $B_n$. My question is if it is known whether or not there is a right inverse to the natural projection $Homeo^{+}(X_n) \rightarrow B_n$. -Since $B_1=1$ and $B_2=\mathbb{Z}$, the first nontrivial case is $n=3$. -I would find it shocking if this is not known for $n=3$. The braid group then has two generators $a$ and $b$ and only one relation $aba=bab$. The problem is then asking whether or not we can find homeomorphisms $f,g \in Homeo^{+}(X_n)$ in the homotopy classes of $a$ and $b$ such that $fgf=gfg$. Surely this cannot be open! - -REPLY [8 votes]: Hot off the presses: a new preprint by Nick Salter and Bena Tshishiku proves that the braid groups cannot be realized by diffeomorphisms for $n \geq 5$.<|endoftext|> -TITLE: is $\nabla \cdot ( c^2 \nabla)$ a Laplace-Beltrami operator? -QUESTION [6 upvotes]: Someone mentioned, in passing, to me that $u \mapsto \nabla \cdot ( c^2 \nabla u)$ is a Laplace-Beltrami operator. Does anyone have some insight into this? From my understanding, the Laplace-operator generalizes the Laplacian to Riemannian manifolds, by taking the trace of the Hessian. I don't really see the connection to that and the above operator, unless c=1. -This operator comes from the wave equation, where $\partial^2_t u -\nabla \cdot ( c^2 \nabla u) = f$. There may or not be some smoothness conditions on $c$, and all these are functions on subsets of $\mathbb{R}^n$. -Thanks - -REPLY [10 votes]: As it was mentioned by @Denis Serre, for dimensions $\ne 2$, your operator is proportional to the Beltrami-Laplace operator. The proof is essentially in the comment of Denis Serre; one can also obtain it following suggestion of @Piero D'Ancona and writing this operator in local coordinates (instead of covariant derivative the usual derivatives appear by the price the determinant of the metric appears twice as factors). -For nontrivial $c$, the operator in NOT the Beltrami-Laplace of any metric. I will explain it in the case your metric and $c$ are generic and dimension is $>2$, but it seems that the proof works without this assumption. - -Proof of (2). Consider the symbol of your operator. It is a well-defined (2,0)-tensor and it is equal to $c^2 \cdot g^{ij}$. Would you operator be the Beltrami-Laplace of some $g'$, -its symbol will be covariantly constant, and in the case of generic $g$ this would imply that the metric $g'$ is a constant factor of $\frac{1}{c^2}g_{ij}$ (I assume $n>2$). Thus, we have only one choice for $g'$. Calculating the Beltrami-Laplace for $g'$, we see that it does not coincide with your operator (if $c\ne \mathrm{const}$).<|endoftext|> -TITLE: Computing H^2(X, T_X(-\log D)) -QUESTION [8 upvotes]: Let $X\subset \mathbb P^3$ be a smooth variety and $D \subset X$ be an effective, reduced, irreducible divisor. My question is the following. -If I know the defining equations of $X$ and $D$ then is there any software that can compute $H^2(X, T_X(-\log D))$? - -REPLY [3 votes]: Computing cohomology of the logarithmic tangent sheaf $T_X(-\log D)$ is a usually easier once you know the cohomology of the sheaves $T_X, \quad T_X(-D)$ and $N_D$, the normal bundle of $D$ in $X$. To get $H^i(X,T_X(-\log D))$ one can use the exact sequences -$$ -0 \to T_X(-D) \to T_X(-\log D) \to T_D \to 0 -$$or -$$ -0 \to T_X(-\log D) \to TX\to N_{D}\to 0. -$$These sequences make sense even when $D$ is singular (see Sernesi's book on deformation theory). These sheaves and their cohomology groups can be handeled in, say, Macaulay2, if you have the explicit equations. -EDIT: Let me try to explain the above in more detail: If one is in the lucky position that some of the cohomology groups of, say $N_D$, vanish (e.g., a line $L$ on a cubic surface has $h^0(L,N_L)=h^1(L,N_L)=0$, you can calculate the cohomology almost immediately. In the general case, you will have to write down the maps and calculate the cohomology using say, Cech cohomology. Macaulay2 have built-in routines for this, see for example Francesco Polizzi's answer to this question.<|endoftext|> -TITLE: Why is the dual of a torus the same as its fundamental group? -QUESTION [13 upvotes]: The set of continuous homomorphisms from a torus ${\mathbb T}^n = ({\mathbb R}/{\mathbb Z})^n \to {\mathbb R}/{\mathbb Z}$ can be identified with ${\mathbb Z}^n$ if we assign to each $k = (k_1, \ldots k_n) \in {\mathbb Z}^n$ the character $x \mapsto k \cdot x$. -The fundamental group of homotopy classes of loops ${\mathbb R}/{\mathbb Z} \to {\mathbb T}$ can also be identified with ${\mathbb Z}^n$ because each equivalence class (with base point at the origin) can be represented by $x \mapsto (k_1 x, k_2 x, \ldots, k_n x)$ for some $k \in {\mathbb Z}^n$. -My question is basically how much of a coincidence this isomorphism is. For one thing, it can't be too natural because the fundamental group pushes forward under a map, whereas the character group pulls back. So the natural question should be whether there is a natural relationship at the level of the first cohomology group instead of the fundamental group. -Of course, totally disconnected groups have interesting dual groups even though their cohomology is uninteresting as far as I know. And ${\mathbb R}^n$, being isomorphic to its dual but contractible, does not seem to exhibit a similar relationship. -But for a Lie group, say, I would like to know if there's a natural relationship between its representation theory (e.g. irreducible representations) on the one hand and its topology (e.g. cohomology) on the other hand. It might be no deeper than "well, the cohomology groups are representations". - -REPLY [5 votes]: This is just a minor elaboration on David Ben-Zvi's answer. You can see the duality between the fundamental group of $T$ and the character lattice by composing based loops $\mathbb{R}/\mathbb{Z} \to T$ with characters $T \to \mathbb{R}/\mathbb{Z}$. You end up with based loops in $\mathbb{R}/\mathbb{Z}$, whose homotopy classes are completely determined by an integer invariant, namely the degree. You can view the degree algebraically as the induced homomorphism on $\pi_1(\mathbb{R}/\mathbb{Z},0) \cong H_1(\mathbb{R}/\mathbb{Z}, \mathbb{Z})$, or geometrically as the winding number if you choose an isomorphism with an oriented circle in the plane. This yields a perfect pairing between $\pi_1(T,0)$ and $X^*(T)$. -In the nonabelian situation, one can see that the fundamental group of a connected topological group doesn't determine that much about the representation theory. There are plenty of simply connected groups with fairly complicated representation theory. However, the K-theory of the classifying space can say a lot about representations. -Regarding the title of the question, one should be careful about calling two groups "the same", even if they are isomorphic.<|endoftext|> -TITLE: Decomposability of positive maps -QUESTION [9 upvotes]: By results of Størmer and Woronowicz, every positive map $\Phi \colon \mathcal{M}_{d \times d} \rightarrow \mathcal{M}_{d' \times d'}$ for $dd' \leq 6$ can be decomposed as a convex combination -$$\Phi = p \phi + (1-p) ~ T \circ \psi$$ -where $\phi$, $\psi$ are completely positive maps and $T$ is the transposition map. -For higher dimensions, this is in general false. -Does there however (for fixed $d$, $d'$) exist a finite set of positive maps $(P_i)$ such that every general positive map $\Phi$ is a convex combination -$$\Phi = \sum p_i P_i \circ \phi_i$$ -where the $\phi_i$ are suitably chosen completely positive maps? - -REPLY [8 votes]: I'm quite late answering this question, but I figured it still deserves an answer. The answer to your question is "no": when $d,d^\prime \geq 3$ there does not exist such a finite (or even countable) set of positive maps. -In arXiv:1209.0437, we considered the following problem: given a set of positive maps $\mathcal{Q}$ from $\mathcal{M}_d$ to $\mathcal{M}_{d^\prime}$, what is the set -$$\mathcal{C}_\mathcal{Q} \stackrel{\text{def}}{=} \{\sum_i \psi_i \circ P_i \circ \phi_i : P_i \in \mathcal{Q}, \psi_i \text{ and } \phi_i \text{ are completely positive for all $i$}\}?$$ -As you noted, Størmer and Woronowicz showed that if $\mathcal{Q} = \{T\}$ (where $T$ is the transpose map) and $dd^\prime \leq 6$, then $\mathcal{C}_\mathcal{Q}$ is the set of all positive maps. -We didn't prove it in the paper, but as an offshoot Łukasz Skowronek proved in the $d,d^\prime\geq 3$ case that if $\mathcal{C}_\mathcal{Q}$ is the set of all positive maps then $\mathcal{Q}$ must be uncountable. I don't believe he ever did end up formally writing up the result, but he at least gave a talk about it (slides available here). -The slides go through the proof in pretty good detail: the rough idea is that there is a known uncountably infinite set of positive maps on $\mathcal{M}_3$ that are each exposed in the set of positive maps and are all indecomposable (i.e., can not be written in the form $\phi_1 + T \circ \phi_2$ for some completely positive $\phi_1,\phi_2$) due to Ha and Kye (see arXiv:1108.0130). Łukasz showed that there is no countable set $\mathcal{Q}$ such that $\mathcal{C}_\mathcal{Q}$ contains even just this particular (relatively small) set of positive maps. -Update [May 23, 2016]: It seems that Łukasz finally got around to formally writing up this result, and it is now proved explicitly in this new paper).<|endoftext|> -TITLE: Which cohomology theories are real- and complex-orientable? -QUESTION [16 upvotes]: A complex-oriented cohomology theory $E^*$ is a multiplicative cohomology theory with a choice of Thom class $x\in\tilde{E}^2(\mathbb{C}P^\infty)$ for the universal complex line bundle (which can be used to define generalised Chern classes for all complex vector bundles). -A real-oriented cohomology theory $F^*$ a multiplicative cohomology theory with a choice of Thom class $x\in \tilde{F}^1(\mathbb{R}P^\infty)$ for the universal real line bundle (which can be used to define generalised Stiefel-Whitney classes for all real vector bundles). -Question 0: Is this correct? -Question 1: Are there any examples of cohomology theories which are both real and complex orientable? -Question 2: (Assuming a yes to Question 1) Are there any results/papers where the interaction between a real and complex orientation is used in an essential way? (I'm thinking perhaps about non-immersion results for real projective spaces.) -Thanks. -Update: Neil's answer and Johannes' comments have answered my original questions: every real oriented cohomology theory is complex oriented, and in fact is a wedge of $H\mathbb{Z}/2$'s. Then let me ask a follow up question. Are there any complex-oriented theories which are not real-orientable but have $E^1(P^\infty)\neq 0$? - -REPLY [17 votes]: A real-orientable ring spectrum $F$ admits a ring map from $MO$, and there is a straightforward ring map $MU\to MO$, so $F$ is also complex orientable. Moreover, $MO$ is a wedge of $H/2$'s, so $MO\wedge F$ is also a wedge of $H/2$'s, and $F$ is a retract of $MO\wedge F$ (by using the ring structure etc) so it is again a wedge of $H/2$'s. Thus, you don't expect to learn anything about (non)immersion from $F$ that you could not already learn from $H/2$ (although some kinds of bookkeeping may be simplified). -For the follow-up question: -The mapping spectrum $E=F(S^1_+,MU)$ is complex-orientable but not real-orientable and has -$$ \tilde{E}^1(\mathbb{R}P^\infty) = - \tilde{MU}^1(\mathbb{R}P^\infty) \oplus \tilde{MU}^0(\mathbb{R}P^\infty) -$$ -Here $MU^{\ast}(\mathbb{R}P^\infty)=MU^*[[x]]/{[2]}(x)$ with $|x|=2$ and $MU^{-2}=MU_2\simeq\mathbb{Z}$ so the first summand is zero but the second is not.<|endoftext|> -TITLE: Non-isomorphic two-transitive permutation groups with isomorphic point stabilizers -QUESTION [12 upvotes]: The permutation groups $A = PSL(2,7)$ with its natural action on the projective line $\mathbb{P}^1(\mathbb{F}_7)$ and $B = A\Gamma L(1,8)$ with its natural action on the affine line $\mathbb{F}_8$ have the interesting property that $A$ and $B$ are non-isomorphic, but the point stabilizers $A_x$ and $B_y$ are isomorphic permutation groups (namely a Frobenius group of order 21 acting on 7 points). - -Are there other examples of non-isomorphic two-transitive permutation groups with isomorphic point stabilizers? - -Note that I really require the point stabilizers to be isomorphic as permutation groups, not just as abstract groups. -(My original motivation for this question is the fact that from every such pair of groups, one can construct an example of two non-isomorphic totally disconnected locally compact groups with isomorphic compact open subgroups. In the meantime, it is already known that there are many examples of this phenomenon, but at that time, this was unknown. I nevertheless believe that the question might still be interesting in its own right.) - -REPLY [10 votes]: There are lists of known finite 2-transitive groups, and for examples other than those of affine type (i.e. those with a regular normal elementary abelian subgroup), it should not be hard to show that there are no more examples. -But I would expect there to be examples of affine type, and by searching through the Magma database of primitive permutation groups I found an example of degree 343. These are -PrimitiveGroup(343,49) and PrimitiveGroup(343,50) in Magma. Irritatingly, the numbering is different in GAP, and (hoping I have got this right), they are numbers 67 and 68 in the GAP list. -The two groups in question have order $343 \times 342$ - so they are sharply 2-transitive. Their stabilizers are isomorphic nonabelian groups of order 342, with centres of order 6. -I would conjecture that there are infinitely many examples of a similar type. -Added: The group ${\rm A}\Gamma{\rm L}(1,7^3)$ has structure $7^3:(7^3-1):3$, and has four subgroups of index 3. The examples here are two of those subgroups. One of the other two is ${\rm AGL}(1,7^3)$, which is also sharply 2-transitive, but has cyclic point stabilizer. The fourth does not act 2-transitively. There are corresponding subgroups of ${\rm A}\Gamma{\rm L}(1,p^3)$ for all prime $p \equiv 1 \bmod 3$, so there are indeed infinitely many examples. More generally, for $p,q$ primes with $p \equiv 1 \bmod q$, there will be $q-1$ non-isomorphic sharply 2-transitive groups of degree $p^q$ with isomorphic stabilizers.<|endoftext|> -TITLE: Commuting matrices in GL(n,Z) -QUESTION [5 upvotes]: Suppose $M$ is a "hyperbolic" matrix in $GL(n,\mathbb Z)$, i.e., that its characteristic polynomial $p$ is irreducible over $\mathbb Z$ and has no roots of modulus 1. -Is there a closed description of the set of elements of $GL(n,\mathbb Z)$ which commute with $M$? -I have a vague recollection that it is somewhat similar to the Dirichlet theorem on the units of an algebraic field, but it is really vague so a reference would be appreciated. -The case I'm most interested in is when $p$ has only one root of modulus greater than 1. Can $M$ commute with another matrix $M'$ with the same property (and $M, M'$ not being powers of the same matrix in $GL(n,\mathbb Z)$)? - -REPLY [11 votes]: This has very little to do with being hyperbolic; the key point is that the characteristic polynomial is irreducible. It is convenient to "close in" on $\mathbb{Z}$ be thinking about easier rings. - -Let $M$ be a matrix in $GL(n, \mathbb{C})$ with distinct eigenvalues. Then I claim that the set of $n \times n$ matrices which commute with $M$ is $\mathrm{Span}_{\mathbb{C}} (\mathrm{Id}, M, M^2, \cdots, M^{n-1})$. -Proof: This statement is clearly invariant under change of basis, so we may assume that $M$ is diagonal, with diagonal entries $\lambda_i$. Then $A M = M A$ if and only if $\lambda_i A_{ij} = \lambda_j A_{ij}$. Since the $\lambda$'s are distinct, this forces $A$ to be diagonal. Since the Vandermonde matrix is invertible, the span of the first $n$ powers of $M$ is precisely the diagonal matrices. - -Let $M$ be a matrix in $GL(n, \mathbb{Q})$ whose characteristic polynomial has distinct roots. Then the set of $n \times n$ rational matrices which commute with $M$ is $\mathrm{Span}_{\mathbb{Q}} (\mathrm{Id}, M, M^2, \cdots, M^{n-1})$. -Proof: Let $A$ be "the set of matrices which commute with $M$" and let $B$ be "the linear span of the first $n$ powers of $M$". Both of these are rational subvector spaces of $\mathrm{Mat}_{n \times n}(\mathbb{Q})$. The previous section shows that $A \otimes \mathbb{C}$ and $B \otimes \mathbb{C}$ are the same subspace of $\mathrm{Mat}_{n \times n}(\mathbb{C})$. A standard lemma is that, if $U$ and $V$ are both $K$-subspaces of a $K$ vector space $W$, and $U \otimes L = V \otimes L$ as subspaces of $W \otimes L$, then $U=W$. -Now, let $p$ be the characteristic polynomial, and assume furthermore that $p$ is irreducible. The $\mathbb{Q}$-span of the powers of $M$ is isomrophic, as a ring, to $\mathbb{Q}[t]/p(t)$. Since $p$ is irreducible, this is some number field, call it $K$. So the set of $\mathbb{Q}$-matrices which commute with $M$ is isomorphic to a number field. Since every element in a field, other than 0, is invertible, we get that the matrices in $GL(n, \mathbb{Q})$ which commute with $M$ are isomorphic to $K^*$. - -Let $M$ be a matrix in $GL(n, \mathbb{Z})$ whose characteristic polynomial is irreducible. Let $K$ be the field of $\mathbb{Q}$-matrices which commute with $M$, as discussed above. The set of the matrices whose entries are in $\mathbb{Z}$ forms a lattice $\mathcal{O}$, of rank $n$, in $K$, which is also a subring. Such a subring of a number field is called an order; I don't think there is much to say about this order which is not true of general orders. -Finally, you want to understand those matrices of $\mathcal{O}$ which are in $GL(n, \mathbb{Z})$, meaning that their inverses are also in $\mathcal{O}$. This is the unit group of $\mathcal{O}$. And, indeed, Dirichlet's unit theorem applies to orders: if $K$ has $r$ real places and $s$ complex places, then $\mathcal{O}^*$ is a finite group times $\mathbb{Z}^{r+s-1}$. - -Finally, you want to know whether or not you can have $M$ hyperbolic, $N$ hyperbolic commuting with $M$, but $N$ not a power of $M$. The answer is YES. I'll first give a theoretical proof, and then sketch an actual computation. Let the eigenvalues of $M$ be $\lambda_1$, ..., $\lambda_n$, with $|\lambda_1|>1$. Let $N=f(\lambda_i)$, with $f$ a polynomial with rational coefficients. Note that the $\lambda$'s are the Galois orbit of $\lambda_1$. For any Galois automorphism $\sigma$, we have $f(\sigma(\lambda_1)) = \sigma(f(\lambda_1))$. So the eigenvalues of $N$ are the Gaois orbit of $f(\lambda_1)$. In short, your question is equivalent to the following: - -Let $\mathcal{O}$ be an order in a number field. Suppose that $\lambda$ and $\mu$ are units such that $|\lambda|$ and $|\mu|>1$, but their Galois conjugates are less than $1$. Can this hapen without $\mu$ not a power of $\lambda$? - -It certainly can. I'll give the conceptual proof, then sketch a computation. If you recall the standard proof of Diricihlet's unit theorem, it goes as follows: Map $\mathcal{O}^*$ to $\mathbb{R}^{r+s}$ by $u \mapsto (\log |u|, \log |\sigma_2(u)|, \cdots, \log |\sigma_{s+r}(u)| )$, where the inputs to the logs are the Galois orbit of $u$. Clearly, the image lands in the hyperplane where the coordinates sum to $0$. One proves that the image is a discrete lattice, of rank $r+s-1$, in this hyperplane. -In particular, we are interested in units where $\log |u|>0$ but where all the other coordinates are negative. This is the intersection of our discrete lattice with a full dimensional cone; once $r+s-1>1$, this will be larger than the one dimensional sublattice of the powers of any single unit. -If you want an explicit example, lets take $K=\mathbb{Q}[\cos (2 \pi/7)]$. The Galois action permutes $(\cos (2 \pi/7), \cos (4 \pi/7), \cos (6 \pi/7))$ cyclically. (Note that $\cos (4 \pi/7) = 2 \cos^2 (2 \pi/7) -1$, so it is in the same field, and similarly for $\cos (8 \pi /7 ) = \cos (6 \pi /7 )$. -Set $u=(1-\cos(2 pi/7))/(1-\cos (4 \pi/7))$, $v=(1-\cos(4 pi/7))/(1-\cos (6 \pi/7))$ and $w=(1-\cos(4 pi/7))/(1-\cos (8 \pi/7))$. So the Galois group permutes $(u,v,w)$ cyclically and $u*v*w=1$. You can check that $u$, $v$ and $w$ obey the equation -$$t^3 - 6 t^2 + 5t -1=0$$ -so $u$, $v$ and $w$ are units. I think they generate the unit group; in any case, the have finite index in it. Also note that $|u|$ and $|v| < 1$, while $|w|>1$. So any matrix with eigenvalues $u$, $v$ and $w$ is hyperbolic. -Numerical experimentation reveals that $|u w^4|>1$, while $|v u^4|$ and $|w v^4|<1$. -So, write down explicit matrices for the action of $w$ and $u w^4$ on the ring of integers of $K$. Then these will be two hyperbolic matrices which commute, but where neither is a power of the other.<|endoftext|> -TITLE: Spectral properties of finite metric sets -QUESTION [13 upvotes]: Given a finite metric set $S=\{P_1,\dots,P_n\}$, one gets a real symmetric matrix $M=M(S)$ -with rows and columns indexed by elements of $S$ by setting -$M_{i,j}=d(P_i,P_j)$. -It is easy to see that $M$ has at least one strictly positive and one strictly negative eigenvalue -if $S$ contains at least $2$ points. For metric sets of three points, the matrix $M$ has always -signature $(1,2)$ (one strictly positive and two strictly negative eigenvalues). -(This holds in fact for any symmetric $3\times 3$ matrix with zero diagonal -and strictly positive off-diagonal coefficients.) -In particular, we have always at least two strictly negative eigenvalues if $n\geq 3$. -It seems quite difficult to have more than one strictly positive eigenvalue -if $n$ is small (I have an example with $n=9$). -Given an integer $d\geq 2$, what is the smallest number $n=n(d)$ such that there exists a finite -metric space $S$ with $n$ elements giving rise to a matrix $M$ having $d$ non-negative -eigenvalues? -So far, all I know is $3< n(2)\leq 9$. -Update: $n(2)=4$, realized by the metric space with two pairs of points $A,B$ and $C,D$ at distance $2$, all other distances between distinct points are $1$. (The corresponding matrix $M$ has eigenvalues $-2,-2,0,4$). -I have an example with five points having two strictly positive eigenvalues. -Other bounds: $n(3)\leq 6$ (my example has however $0$ as an eigenvalue and only for $n=7$ do I have an example with three strictly positive eigenvalues), $n(4)\leq 9$ and $n(5)\leq 12$. - -REPLY [4 votes]: Here's an answer if you make a further assumption on your metric space: if $M$ is of strictly negative type, then it has $n-1$ negative eigenvalues, according to Lemma 3.6 of this paper. This condition means that -$$ -\sum_{i,j} M_{ij} x_i x_j < 0 -$$ -whenever $\sum_i x_i = 0$ and the $x_i$ are not all $0$, and it holds, for example, if your metric space is a subset of Euclidean space. -Added: it's maybe worth adding (although it doesn't directly bear on your question) that if $M$ is just of negative type (only $\le$ holds in the inequality above), then Lemma 3.6 also shows that $M$ has exactly one positive eigenvalue. -NB: There's a statement about 4-point spaces in the abstract of that paper which appears to contradict your 4-point example. The actual result (Proposition 6.1) includes an extra hypothesis that excludes that example. -Added, part 2: I realized I wrote the above in a way that emphasizes the wrong points. Lemma 3.6 in that paper is of course trivial. The real point is that the terminology strictly negative type is worth knowing, so that you can find out which metric spaces have that property (and subsets of Euclidean space are among such spaces).<|endoftext|> -TITLE: Coin flipping and a recurrence relation -QUESTION [11 upvotes]: How can one solve the following recurrence relation? -$f(n) = 1 + \frac{1}{2^n} \sum_{k = 0}^n {{n}\choose{k}} f(k)$ -$f(0) = 0$ -As it happens, I can show $f(n) = \Theta(\log n)$ through other means (see below). But I'd like to know how to solve the recurrence "directly". -The recurrence relation comes from the following coin flipping problem. There are $n$ independent, unbiased coins, and we toss all of then for a number of rounds. Let $T(n)$ be the first round when each coin has got head at least once (ie, $T(n) = \text{arg} \min_t \text{s.t.} H_t(i) \geq 1; \forall i \in [n]$, where $H_t(i)$ is the number of heads the $i^{th}$ coin has got in the first $t$ rounds). Then one can see that $E(T(n))$ fulfills the recurrence relation mentioned above. -To see that $E(T(n)) = O(\log n)$, note that we reduce the number of coins who haven't gotten head yet by a factor of 2, in expectation. The $\log n$ bound follows routinely from that. On the other direction (ie, $E(T(n)) = \Omega(\log n)$), let $S = \log n/20$. Then at time $S$ with high probability, a large number of coins will still be "headless", from which the lower bound follows. - -REPLY [6 votes]: Sums like -$f_n = \sum_{k \ge 0} (1 - (1-2^{-k})^n)$ -(and even much more terribly looking ones) have been considered at many places in the literature. There are methods for evaluating them asymptotically at arbitrary precision (i.e. up to any error of order $n^{-c}$). One such method is the Mellin Transform, see -http://algo.inria.fr/flajolet/Publications/mellin-harm.pdf -for an excellent survey by Flajolet et al. There, $f_n$ serves as one of several "running examples", and the authors show that -$f_n \sim \log_2 n + \gamma/\ln 2 + 1/2 + Q(\log_2 n) + O(n^{-1/2})$ -where $Q$ is an explicitly given Fourier series. Interestingly, this series remains bounded, but oscillates forever.<|endoftext|> -TITLE: An inequality for cosine of n -QUESTION [26 upvotes]: Can anyone provide a proof of the following inequality? -If $n$ is a positive integer, $n\geq2$, then $$\cos(n) \leq 1 - 2^{-n}.$$ -This is satisfied if $n$ is not within about $2^{-n/2}$ of a multiple of $2\pi$. -This inequality is sufficient for something else I am trying to prove but I and others have been unable to prove it. - -REPLY [32 votes]: It is well known that -$$|\tfrac{n}{k}-\pi|\ge \tfrac1{p(k)}$$ -where $p$ is a polynomial of some very finite degree. -From this your estimate follows for all sufficiently large $n$. -The polynomial $p$ can be written explicitly; all you need to find it in the literature and check first few values of $n$ by hands... -P.S. According to K.Mahler 1956, one can take $p(k)=k^{42}$; i.e.; it is sufficient to check all $n\le 1000$. -So, if you trust Martin Brandenburg, your inequality holds for all $n$. ;-)<|endoftext|> -TITLE: Why and how did preschemes become schemes? -QUESTION [24 upvotes]: Originally (e.g., in the first edition of EGA and in Mumford's Red Book), what are now called "schemes" were referred to as "preschemes." The word "scheme" was reserved for what are now called "separated schemes." Presumably (although admittedly I am guessing here), the original idea was that all schemes should be separated by analogy with manifolds, which are by definition Hausdorff; however, unlike in the case of manifolds, in algebraic geometry one cannot introduce a notion of "separated" until the category of (pre)schemes has already been defined, hence the need for a term for "not necessarily separated scheme." -I can think of at least two possible scenarios that might have contributed to the revision of the terminology: -1) Perhaps preschemes that are not separated, or at least not easy to show separated, came up sufficiently often that people decided they should be included in the "fundamental objects of study." (If so, I would be interested to know some examples of these non-separated schemes.) -2) Perhaps, when Grothendieck and Dieudonne carefully wrote EGA so as to assume the weakest reasonable hypotheses for every proposition, it was discovered that there were far more propositions about preschemes than about schemes, and people (most notably, Grothendieck himself) decided that preschemes, not separated schemes, were the more fundamental objects. -Unfortunately, both of these scenarios are largely speculative. What actually happened? In particular, what was the motivation for the shift in terminology? - -REPLY [12 votes]: Dear Charles, Dieudonné and Grothendieck themselves changed their terminology in the second edition of EGA I, published by Springer Verlag in 1971. At the end of their Avant-propos, on page 3, they write: -Signalons enfin, par rapport à la première édition, un changement important de terminologie: le mot schéma désigne maintenant ce qui était appelé "préschéma" dans la première édition, et les mots "schéma séparé" ce qui était appelé "schéma". -As to the suggestion "it was discovered that there were far more propositions about preschemes than about schemes, and people decided that this was ridiculous": considering the God-like status of Grothendieck and the awe he inspired, this sounds to me as plausible as courtesans telling Louis XIV "hey, this royalty business is pretty ridiculous. Why not name our country The Democratic Libertarian Republic of France?"<|endoftext|> -TITLE: Commutative diagrams for groups -QUESTION [5 upvotes]: You can present a group in a Cayley-like manner, replacing colors by explicit assignment of nodes to edges: while in a Cayley graph $x \circ y = z$ is presented like this: -     -you can also present it like this: -     -Now the group axioms can be stated like this: - -For each $x, y$ there is a unique $z$ with $x \circ y = z$, or: such that the following diagram holds ("commutes"): - -         - -There is an $e$ such that for all $x$ it holds that $x\circ e = e \circ x = x$, or: such that the following two diagrams commute: - -         -and for each $x$ there is a $x^{-1}$ such that $x \circ x^{-1} = x^{-1} \circ x = e$, or: such that the following diagram commutes: -         - -For each $x, y, z$ it holds that $x \circ (y \circ z) = (x \circ y) \circ z$, or: such that the following diagram commutes: - -         -The last diagram is somewhat ugly, even when drawn in this most balanced way (I didn't find a more appealing and symmetric one). -But an astonishing symmetry arises, when we consider Abelian groups. Commutativity is expressed by the diagram: -     -and associativity becomes: -     -In the presence of commutativity, associativity seems to be related to commutativity (some sort of "second level commutativity"). - -Can any use be made of this kind of diagrams, or is it - just vain baublery? - -REPLY [5 votes]: These pictures are a way of writing a group as an algebra over an operad. The little square is the 2-ary operation, and the arrows are an indicator that makes the inputs distinguishable. -John Baez has written about operads using pictures similar to yours. See for example TWF week 191. -The relation between associativity and commutativity is similar to a fact seen in some books on vertex algebras, where one starts with an axiom like $x(yz) = y(xz)$ and after some power series manipulations deduces $x(yz) = (xy)z$.<|endoftext|> -TITLE: Relationship between "different" quantum deformations -QUESTION [17 upvotes]: This is a generic question, a good answer to it may be a reference to a corresponding paper\textbook, but any useful comments would be okay too. -Let $\mathfrak{g}$ be a (simple) Lie algebra and $U_q(\mathfrak{g})$ be its q-deformation of its universal enveloping algebra. For example, for $\mathfrak{g} = \mathfrak{su}(2)$ with positive $\mathfrak{e}$, negative $\mathfrak{f}$ roots and Cartan element $\mathfrak{h}$ the commutation relation read -$[\mathfrak{e},\mathfrak{f}] =\mathfrak{h},\quad [\mathfrak{h},\mathfrak{e}] = 2\mathfrak{e},\quad [\mathfrak{h},\mathfrak{e}] = -2\mathfrak{f}.$ -After q-deformation the first relationship above turns into -$[\mathfrak{e},\mathfrak{f}] =[\mathfrak{h}]_q :=\frac{q^\mathfrak{h}-q^{-\mathfrak{h}}}{q-q^{-1}}$, -where $q \in\mathbb{C}^\ast$. -Let us now consider Lie group $G$ such that $\mathfrak{g}$ is its Lie algebra. Thinking in terms of differential geometry, Lie group can be seen as a smooth (real) manifold. Let some coordinate patch be parametrized by local coordinates $x_1,\dots x_n$. Here we can implement a noncommutative deformation of the coordinates. There are two examples I'm aware of: $xy-yx=\hbar$ and $xy = t yx$ for some nonzero $t$. -My goal is to understand relationship (if any) between the above mentioned deformations: q-deformation of the Lie algebra and noncommutative deformation of the coordinates on the corresponding Lie group. Returning back to $\mathfrak{su}(2)$, we have $SU(2)=S^3$ as its Lie group. The question is, does the noncommutative $S^3$ have anything to do with $U_q(\mathfrak{su}(2))$? - -REPLY [13 votes]: There is certainly a way to quantize the algebra of functions on a Lie group in a way that is compatible with the $q$-deformation of the universal enveloping algebra of its Lie algebra. The standard way to do it is this: let $G$ be a simple, connected, simply connected complex Lie group with Lie algebra $\mathfrak{g}$. Construct $U_q(\mathfrak{g})$ as you have done in your example, using the Cartan data of $\mathfrak{g}$. -Then it is a theorem that, up to some characters acting on the Cartan part of $U_q(\mathfrak{g})$ (i.e. on the $K_i$'s) the finite-dimensional representation theory of $U_q(\mathfrak{g})$ is the same as that of $\mathfrak{g}$. That is, the representations have the same dimensions, the same weight structures, the same decompositions of tensor products, etc. -For any finite-dimensional representation $V$ of $U_q(\mathfrak{g})$, one can define its matrix coefficients: for $v \in V$ and $f \in V^*$, the matrix coefficient is the linear functional -$$ c^V_{f,v} : U_q(\mathfrak{g}) \to \mathbb{C}$$ -given by -$$ c^V_{f,v}(a) = f (a\cdot v).$$ -Denote by $\mathcal{O}_q(G)$ the span of all of the $c^V_{f,v}$ inside the dual of $U_q(\mathfrak{g})$. With the algebraic operations defined by duality from those in $U_q(\mathfrak{g})$, this is a Hopf subalgebra of the finite dual $U_q(\mathfrak{g})^{\circ}$ which separates points of $U_q(\mathfrak{g})$. This is the analogue of the algebra of polynomial functions on $G$. For some values of $q$, you can introduce a $*$-structure and a norm and take a completion to get the analogue of the algebra of continuous functions. -Basically this is just mimicking the Peter-Weyl decomposition of the algebra of functions on a compact group. You can do lots of stuff with this, for instance look at quantum homogeneous spaces, building analogues of the de Rham complex, etc. -A good reference is Klimyk and Schmudgen, Quantum Groups and Their Representations.<|endoftext|> -TITLE: Exponential bounds for the number of lattice animals with a given boundary. -QUESTION [11 upvotes]: Hi all, -I am doing a work in collaboration with other mathematicians about phase transition in the Ising model and we need to know if exponential upper bounds exist for the number of lattice animals with boundary of size $n$. -To be precise, consider the square lattice $\mathbb{Z}^2$ as graph where the edges are pairs of points in the lattice having distance one from each other, where the distance is induced by the norm $\|(z_1,z_2)\|=|z_1|+|z_2|$. -We call a lattice animal the set of vertices of any connected subgraph of the square lattice. Given an animal $A$, we denote the boundary of $A$ by $\partial A$, that is, the set of vertices of distance one from $A$. -Fix a site $z\in \mathbb{Z}^2$ and let be -$$ -f(n)=\sharp \{A\ \text{is lattice animal}; A\ni z\ \text{and}\ |\partial A|=n\} -$$ -Is it known if $f(n)=O(e^{k n})$ ? -I learned from google that this problem is also known in the combinatorics community as enumeration of polyominoes with a given site-perimeter. -All the papers I found about the upper bounds at some point have to impose some geometric hypotheses on the polyominoes such as convexity, starcase shape or bargraph shape. -I don't know yet if those hypotheses are being used in order to get sharp upper bounds or if they are the only ones available. -If the question about exponential upper bound is not yet solved, is there a specialist in this area who could tell me what they think about the upper bound for this problem. - -REPLY [5 votes]: EDIT: I see that Steve has more or less the same construction above. I should have read his answer more carefully before I posted. -I don't believe it's true. Let's say you have a square polyomino with $n/2$ perimeter (so also $n/2$ site-perimeter) and you remove $n/2$ sites from its interior. Not all ways of doing this will give you a lattice animal with site-perimeter $n$, but if you just remove sites where both coordinates are even, you get a lattice animal (i.e., the polyomino will still be connected). The number of ways of doing this are roughly $cn^2 \choose n/2$, which grows as $e^{O(n \log n)}$. -ADDED MATERIAL: -There is an $e^{C n \log n}$ upper bound as well. Again, let's think about polyominos with site perimeter $n$. If we can specify the boundary of such a polyomino with $O(n \log n)$ bits, this gives a $e^{O(n \log n)}$ bound on how many of these there are. We will specify the boundary in two stages. First, let's look at the exterior edges (all the edges which can be connected to $\infty$ by a path of squares not in the polyomino). We can specify these exterior edges by a list of directions: e.g., EESENESSW$\ldots$, which is only $O(n)$ bits. -Now, let's look at the interior boundary edges. There are at most $cn$ of these for some constant $c$, and there are at most $n^2$ edges in the interior of the polynomial (the biggest it can be is an $n/4 \times n/4$ square), so we can specify these by creating some canonical list of the interior edges, and specifying which ones we have. This takes at most $\log_2 \sum_{k=0}^{cn} {n^2 \choose k} = O(n \log n)$ bits, and thus we get an $e^{C n \log n}$ upper bound on the number of these lattice animals. -This leaves the question open of what is $$C=\lim_{n\rightarrow \infty} \frac{\log f(n)}{n \log n}$$ (although it might even be difficult to prove rigorously that this limit exists).<|endoftext|> -TITLE: Concrete interpretations of higher (sheaf) cohomology groups -QUESTION [9 upvotes]: $H^1$ has an interpretation as torsors. But what about the higher $H^i$ (in the setting of algebraic geometry, and étale or flat cohomology)? For example $H^2(X, \mathbf{G}_m)$ is (often) isomorphic to the Brauer group, i.e. equivalence classes of Azumaya algebras. - -REPLY [3 votes]: I would say have a look at Duskin's paper '$K(\pi,n)$-torsors and the interpretation of "triple" cohomology' (pdf) and his student Glenn's paper 'Realization of cohomology classes in arbitrary exact categories' J. Pure Appl. Algebra, vol. 25, (1982) pp. 33-105. Also Duskin's 'Higher dimensional torsors and the cohomology of topoi : The abelian theory' Lecture Notes in Mathematics, Volume 753 (1979), pp255-279. -I'm afraid you have to move beyond just working with schemes to simplicial schemes and what-not.<|endoftext|> -TITLE: When is an extension of characters de Rham? -QUESTION [8 upvotes]: Let $G$ be the abolute Galois group of $\mathbb Q_p$, let $\delta_1, \delta_2: G\rightarrow L^{\times}$ be continuous characters, where $L$ is a finite extension of $\mathbb Q_p$. Assume that $\delta_1\delta_2^{-1}$ is neither trivial nor the cyclotomic character then -$Ext^1_{G}(\delta_2, \delta_1)$ is one dimensional. Hence there exists a unique non-split extension: -$0\rightarrow \delta_1\rightarrow V\rightarrow \delta_2\rightarrow 0$. -When is $V$ de Rham? -I believe that the answer is if and only if both $\delta_1$ and $\delta_2$ are de Rham and the Hodge-Tate weight of $\delta_1\delta_2^{-1}$ is $\ge 1$ (at least if the Hodge-Tate weights of $\delta_1$ and $\delta_2$ are distinct) and I guess I could x it out by using Bloch-Kato's paper in Grothendieck Festschrift, bu the answer must be well -known and maybe even written down somewhere. -Ideally, I would like to be able to quote a reference, where this has been worked out. - -REPLY [8 votes]: Dear Vytas, -lemma 6.5 of my 2002 inventiones paper says that if $V$ is any de Rham representation all of whose HT weights are at least 1, then any extension of $Q_p$ by $V$ is itself de Rham. This holds for reps of $G_K$ where the residue field $k$ of $K$ can be any perfect field (not merely finite). -If $k$ is finite, then this was well-known before and follows from the results of Bloch and Kato (which I think you should quote). See proposition 1.28 of Nekovar's "On $p$-adic height pairings" where this is stated explicitly and proved using BK's computations. -EDIT: see also the "Proposition" on page 196 of Perrin-Riou's "Représentations $p$-adiques ordinaires". It predates Nekovar's paper, and although the result is less strong, it's enough for what you need.<|endoftext|> -TITLE: Is an abelian variety with a Galois invariant, rank one submodule of its Tate module, CM? -QUESTION [12 upvotes]: Let $A$ be an absolutely simple abelian variety over a number field $K$. Assume that, for some prime $p$, the Tate module $T_p A$ has a submodule of rank one, invariant under the absolute Galois group of $K$. Does it follow that $A$ is has CM? -For elliptic curves, I guess this follows from Serre's open image theorem. That's all I know. I would be surprised if there was a counterexample as it would be a way of constructing abelian extensions of $K$ using non-CM abelian varieties, which would be surprising. - -REPLY [13 votes]: Yes. This follows from the main result of the following paper of Zarhin. - -MR0885780 (88h:14046) -Zarhin, Yu. G. -Endomorphisms and torsion of abelian varieties. -Duke Math. J. 54 (1987), no. 1, 131–145. - -His result, specialized to the $K$-simple case, is the following (fantastic) theorem. -Let -$A$ be a $K$-simple abelian variety defined over a number field $K$. The following are equivalent: -(i) $A(K^{\operatorname{ab}})[\operatorname{tors}]$ is infinite. -(ii) $A$ is of CM-type over $K$. -Your hypotheses imply that there is infinite torsion over the abelian extension cut out by -the action of Galois on the one-dimensional subspace (the Galois group is contained in $\mathbb{Z}_{p}^{\times}$), so Zarhin's theorem applies.<|endoftext|> -TITLE: Nerve: Groupoids-> Kan Complexes. Nerve: Bicategories w. adjoints -> ? -QUESTION [8 upvotes]: If you take the nerve of a groupoid, you get a Kan complex. -Question: -Take a bicategory that has adjoints for 1-morphisms, which is one notion of 'weak' groupoid (if all 2-morphisms are isomorphisms, then such a bicategory is a 2-groupoid), and take its nerve. -Is there a name for a bisimplicial set arising in this way? Does it have some nice properties? For example, is there a model structure on $\mathbf{ssSet}$ such that these are fibrant? - -REPLY [2 votes]: I would recommend checking through the various papers by Cegarra and Remedios (look on the archive) They have done a lot of work in this area, but I am not sure if they have an answer for your question. The Duskin nerve as suggested by David Roberts is related to the bisimplicial approach and the relation is explored in various other papers from Granada, e.g. one by Manolo Bullejos and coauthors.<|endoftext|> -TITLE: Finite-dimensionality for de Rham cohomology -QUESTION [11 upvotes]: I was browsing through the litterature, hoping to find sufficient and necessary conditions for a smooth manifold to have finite-dimensional de Rham cohomology, but I can't find any satisfactory answer. I wonder if anyone has ever encountered a paper, or a book, answering (possibly in part) the question. I am especially interested in real-coefficient cohomology, but I would appreciate answers related to cohomology with coefficients in any abelian group. -Obviously, I don't expect "compact manifold" as an answer; although this is a sufficient condition, it is far from answering the question. - -REPLY [17 votes]: We can restrict your problem to the case of open manifolds. It turns out that "finite dimensional" integral singular homology (i.e., finitely generated in each degree) is almost the same thing as the manifold being the interior of a compact manifold with boundary. For example, if $M$ is a 1-connected and open manifold of dimension $>5$, then the Browder-Levine-Livesay theorem says that $M$ is the interior of a compact manifold with boundary -(where the boundary is also 1-connected) iff the homology of $M$ is finitely generated and -$M$ is $1$-connected at infinity. -This result was later generalized in Siebenmann's thesis to the non-simply connected case. -Addendum. Here's a link to Siebenmann's thesis: -www.math.uchicago.edu/~shmuel/tom-readings/Siebenmann%20thesis.pdf<|endoftext|> -TITLE: Greatest common divisor of algebraic integers -QUESTION [16 upvotes]: I have several questions concerning some properties of algebraic numbers. The first concerns the folowing statement: -Given algebraic integers $\alpha$ and $\beta$ they have a unique greatest common divisor modulo asociates. ie there is an algebraic integer $\delta$ with $\delta \vert \alpha$ and $\delta \vert \beta$ and such that for any other integer $\gamma$ such that $\gamma \vert \alpha$ and $\gamma \vert \beta$ we also have $\gamma \vert \delta$; any other algebraic integer with the same properties is an associate of $\delta$. -This result has a simple proof using class field theory, ie if $H$ is the Hilbert class field of $\mathbb{Q}[\alpha,\beta]$, then by the principal ideal theorem, the ideal $(\alpha,\beta)$ becomes principal in $H$ say $\alpha {\mathcal O}_H + \beta {\mathcal O}_H = \delta {\mathcal O}_H$. -I have been looking for a simpler proof in several books in the subject, the nearest I found is theorem 98 in Hecke's Lectures ..., but I think it is not enough: it finds an integer $A$ such that the set of multiples of $A$ in $\mathbb Q[\alpha,\beta]$ coincides with the ideal $(\alpha,\beta)$ but it does not follow that the same is true in the bigger field $\mathbb Q[\alpha,\beta,A]$. So my question is: - -Is there a proof of the statement not using class field theory? - -My "ideal" proof will only use elementary properties of algebraic numbers so the statement could be proved just after the introduction of algebraic integers and units in a classical introduction to the subject, but I fear I'm asking too much. - -REPLY [6 votes]: The proofs of the statement more or less all go like this: let $A$ denote the ideal generated by $\alpha$ and $\beta$ in the ring of integers of a number field $K$, and set $A^h = (\mu)$, where the exponent $h$ of the ideal class of $A$ divides the class number of the field $K$. -Then $A$ becomes principal in $L = K(\gamma)$ with $\gamma^h = \mu$ since $A {\mathcal O}_L = (\gamma)$. -As for solving the Bezout equation $\alpha \rho + \beta \sigma = \gamma$, dividing through -by $\gamma$ shows that we have to solve $\alpha_1 \rho + \beta_1 \sigma = 1$ in ${\mathcal O}_L$. Over the rationals this is done by the Euclidean algorithm, which we do not have in general number fields. One option I can see is the following: factor $\beta_1$ into prime ideals and compute Euler's Phi function $m = \Phi(\beta_1)$; then $\rho = \alpha_1^{m-1}$ has the property that $\alpha_1\rho-1$ is divisibly by $\beta$. To keep the size of $\rho$ under control you can always reduce the coefficients of $\rho$ with respect to an integral basis by $N(\beta)$.<|endoftext|> -TITLE: faithful unipotent representations of (finite) $p$-groups -QUESTION [10 upvotes]: The title pretty much summarizes the question: does every $p$-group have a faithful unipotent representation (with coefficients in $\mathbb{F}_p$ or some finite extension thereof)? - -REPLY [2 votes]: Here is a slightly different approach. If $G$ is a $p$-group (infinite or finite) and $k$ is a field of characteristic $p$ then in the group algebra $k[G]$ we have $(x-1)^{p^r}=x^{p^r}-1$ for all $x \in k[G]$. (Actually, if elements $a$ and $b$ of any $k$-algebra do commute then $(a-b)^{p^r}=a^{p^r} - b^{p^r}$, thanks to divisibility properties of binomial coefficients.) Applying it to $x=g$ where $g$ is an element of $G$ of order $p^r$, we conclude that $(g-1)^{p^r}=g^{p^r}-1=1-1=0$ in $k[G]$. Since every representation space $V$ of $G$ over $k$ is a module over $k[G]$, we conclude that $g-1$ acts on $V$ as a nilpotent operator. -An example of a faithful representation of $G$ is provided by the regular representation where $V$ is the space of all $k$-valued functions on $G$. Another example is provided by its $G$-invariant space $V_0$ of all functions $f: G \to k$ with finite support (i.e., vanishing at all but finitely many points of $G$). Notice that for each $g \in G$ the space $V_0$ splits into a direct sum of $g$-invariant finite-dimensional subspaces (that correspond to finite left cosets of the cyclic group generated by $g$).<|endoftext|> -TITLE: For what reductive groups $G$ over $K$ are the inner forms classified by $H^1(K, G^{ad})$? -QUESTION [10 upvotes]: Suppose $G$ is a connected reductive algebraic group over an arbitrary field $K$; let $Z$ be the center of $G$. The inner automorphisms of $G$ are given by $\operatorname{Inn}(G) = G / Z = G^{\operatorname{ad}}$. Set $\operatorname{Out}(G)$ to be the quotient $\operatorname{Aut}(G) / G^{\operatorname{ad}}.$ The forms of $G$ are parameterized by $\operatorname{H}^1(K, \operatorname{Aut}(G))$, and the inner forms are those in the image of -$$\operatorname{H}^1(K,G^{\operatorname{ad}}) \rightarrow \operatorname{H}^1(K,\operatorname{Aut}(G))$$ -So we can recast the classification of inner forms of $G$ to: - -What conditions can we put on $G$ to guarantee that the map - $$\operatorname{H}^0(K,\operatorname{Aut}(G)) \rightarrow \operatorname{H}^0(K,\operatorname{Out}(G))$$ - is surjective? - -I'm primarily interested in the connected reductive case here, but I would be curious about the more general case as well. -On a related note, I've frequently seen the claim that for quasisimple $G$, the group $\operatorname{Out}(G)$ is given by the automorphism group of the Dynkin diagram of $G$. This holds for some reductive $G$ (such as $\operatorname{GL}_n$) and not others (most nontrivial tori will have a nontrivial outer automorphism group and a trivial dynkin diagram). - -How can we extend the description of $\operatorname{Out}(G)$ from the case of quasi-simple $G$ to connected reductive $G$? - -REPLY [2 votes]: I want to address your first question, regarding what conditions you can put on $G$ to guarantee that the map $Aut(G)(K) \rightarrow Out(G)(K)$ is surjective. I am only going to talk about the case where is $G$ semisimple. If $G$ is quasi-split, then the map has a section so it is surjective, as you can see from the references given by fherzig or Victor. -But if $G$ is not quasi-split, there can be obstructions coming from both the Tits index and Tits algebras. One example to consider is $G = Spin(6,2)$ (of type $D_4$) over the real numbers. In that case $Out(G)(K)$ is the symmetric group on 3 letters, but the image of your map has order 2, corresponding to the outer automorphism given by a hyperplane reflection. -If $G$ is absolutely simple, you can hope that the Tits algebras provide the only obstruction to the surjectivity of your map. But this seems to be open. For more details on these obstructions and some positive results (that in some cases the Tits algebras are the only obstruction), see section 2 of my recent paper `Outer automorphisms...'.<|endoftext|> -TITLE: Symplectic structures on $M \times S^{2n}$ -QUESTION [7 upvotes]: For $n > 1$, $2n$-dimensional sphere $S^{2n}$ does not admit symplectic structures. Then how about the product with a manifold? Are there any results about the symplectic structures on $M \times S^{2n}$? - -REPLY [8 votes]: If $M$ is symplectic, noncompact and connected, then there is a symplectic structure on $M \times S^{2n}$, for each $n$. The proof can be given with the h-principle for symplectic structures. It says the following: let $M$ be a connected noncompact (''open'' in the sequel) manifold, $a \in H^2 (M; \mathbb{R})$ and $J$ an almost complex structure on $M$. Then there exists a symplectic structure $\omega$ on $M$, such that the cohomology class of $\omega$ is $a$ and such that there is a compatible almost complex structure $I$ with $I$ homotopic to $J$. Therefore, any open almost complex manifold has a symplectic structure. -Now I claim: $M$ open and almost complex, then $M \times S^{2n}$ is almost complex (and of course open). -Step 1: $M$ has a vector field without zeroes. To see this, take a vector field $X$ with isolated zeroes and let $p$ be a zero. Choose an ''escape path'', i.e. an embedding $u:[0,\infty] \to M$ with $u$ proper and $u(1)=p$. Moreover, $u$ should avoid the other zeroes. Pick a tubular neighborhood of $u$. The result is that you extend $u$ to a proper embedding $U=D^{n-1} \times [0,\infty) \to M$. Let $\phi_t:U \to U$, $t \in [0,1]$ be an isotopy of embeddings $\phi_1=id$ and $\phi_0([0,\infty)) \subset [0,1/2]$. This istopy should be constant near $t=0,1$ and near $x=0$. Then define $\psi:U \to U$ by the formula $\psi(v,x):= (v, \phi_{|v|^2} (x))$. This is an embedding $U \to U$ whose image does not contain $0$. Extend to a self-diffeomorphism $\psi$ of $M$. Then $\psi^* X$ is a vector field without the zero $p$. -This decomposes the tangent bundle into $TM = \mathbb{R} \oplus V$. -EDIT: A vector field without zeroes can also be found using obstruction theory: if $M$ is open of dimension $m$, then $M$ is homotopy equivalent to an $m-1$-dimensional CW complex. -Step 2: Use the trivial factor to show that $T (M \times S^{2n}) \cong TM \times \mathbb{R}^{2n}$. Therefore, the tangent bundle of $M \times S^{2n}$ has a complex structure and $M \times S^{2n}$ is an almost complex manifold.<|endoftext|> -TITLE: Example of a projective module which is not a direct sum of f.g. submodules? -QUESTION [13 upvotes]: This semester I am teaching a graduate course in commutative algebra, and I have been taking the occasion to try to look at the proofs of some the results in my basic source material (Matsumura, Eisenbud, Bourbaki...) which I skipped as a little too complicated the first $N$ times around. -Recently I got the chance to read and understand I. Kaplansky's big theorem on projective modules, i.e., that a(n even infinitely generated) projective module over a local ring is free. En route to establishing this, he proves another result which is interesting but rather technical: -Theorem (Kaplansky, 1958): Every projective module is a direct sum of countably generated projective submodules. -For my take on this result, see $\S 3.10$ of these notes. In particular, it raises two natural questions: - -Question 1: Is there a ring $R$ and an $R$-module $M$ which is not a direct sum of countably generated submodules? -Question 2: Is there a ring $R$ and a projective $R$-module $P$ which is not a direct sum of finitely generated submodules? - -I was able to look up that the answer to Question 1 is "yes". In particular, I found work of L. Fuchs which says that for every infinite cardinal $\kappa$ there is an indecomposable (i.e., not expressible as a nontrivial direct sum) commutative group $G$ of cardinality $\kappa$. I would however be interested in hearing other examples or other takes on Question 1. -My real question is Question 2: presumably the answer is either yes or unknown, or people would mention the stronger result when Kaplansky's Theorem is discussed. A theorem of Bass that M. Reyes pointed out to me in his answer to another recent question of mine on modules is relevant in this regard: obviously an affirmative answer to Question 2 must involve an infinitely generated projective module, and if $R$ is Noetherian and connected then every infinitely generated projective module is free, hence a direct sum of singly generated submodules! - -REPLY [2 votes]: Let me give one more (elementary) example for Question 1. Let $k$ be an uncountable field (e.g. $\mathbb R$ or $\mathbb C$ will do) and $R = k[x]$ the ring of polynomials. Let $M = k(x)$, i.e. the field of fractions of $R$, which is naturally an $R$-module. -$M$ is indecomposable, since even more holds: Every two non-trivial submodules intersect non-trivially (every two non-zero elements have a common non-zero $R$-multiple). -Finally, why is $M$ not countably generated (in a slightly informal way): Each element of $M$ has only finitely many distinct irreducible polynomials in its denominator (in lowest terms), so among countably many elements, only countably many distinct irreducible polynomials appear in the denominators. Addition and multiplying by elements of $R$ cannot produce "new" irreducible factors of denominators. However, you have to generate (uncountably many) fractions of the form $1/(x-a)$ for each $a \in k$ and since the polynomials $x-a$ are clearly irreducible, we are done.<|endoftext|> -TITLE: Are there uncountably many essentially inequivalent versions of Mathematics? -QUESTION [7 upvotes]: Hi everyone, -Disclaimer 1: logic and set theory are a long way from my field, so apologies in advance if I demonstrate extreme ignorance or stupidity, and please correct me if (when?) I write stupid things. But hopefully my basic meaning should be fairly clear to everyone even if I get some details wrong. -Disclaimer 2: I admit this question might be slightly subjective. But I feel it's not too subjective, and is fairly natural and interesting to most mathematicians, out of mere curiosity. -Framework: Throughout, let's assume that standard ZF set theory is consistent, and take it as our basic mathematical foundation. (I don't necessarily think this is best, but I prefer to pin down the discussion). -We all know that Mathematics comes in several distinct flavours: e.g. you can believe or disbelieve the Continuum Hypothesis, and both points of view are (equally?) valid; they are really just matters of opinion. Thus there are at least 2 different versions. Of course we have infinitely many different versions: each number $m=1,2,3,\ldots$ gives a different flavour of Mathematics, given by the axiom $2^{\aleph_0} = \aleph_m$. -Subquestion Does the value of $m$ really matter very much? $2^{\aleph_0} = \aleph_1$ seems a particularly special case; but I find it hard to believe there'd be very much meaningful distinction (in terms of theorems anyone would want to consider) between the axioms $2^{\aleph_0} = \aleph_{103}$ or $2^{\aleph_0} = \aleph_{275}$, for example. -If desired, we could regard these different versions of Mathematics as essentially equivalent (in a rough sense): the axioms all look very similar, given by a single parametrisation. We could also throw in versions with $2^{\aleph_\alpha}$, etc. -Alternatively, we could remove these difficulties completely by not even considering cardinals beyond $\aleph_2$ or $\aleph_3$, say; (or any $\aleph_m$ with finite $m$). -It would be really amusing if we could do the following, for then we would have (at least) $2^{\aleph_0}$ different flavours of Mathematics! (Although I suppose there might be technical difficulties with nonconstructive infinite 0,1 strings...!) We'd have an explicit injective function $f$ from $[0,1]$ into the class of all possible versions of Mathematics! -Main question -Can we find (or prove the existence of) an infinite sequence of axioms $A_1, A_2, A_3, \ldots$, for which every sequence of true/false assignments is consistent? (e.g. the infinite string 1011001110... would mean that $A_j$ is true for $j=1,3,4,7,8,9,\ldots$ and false for $j=2,5,6,10,\ldots$; we want every string to be consistent). -If so, can it be done with $A_1, A_2, \ldots$ all being essentially different kinds of axioms? [maybe it's stupidly optimistic to hope for this]. Can it be done without ever considering $\aleph_k$ for $k>3$, say (or 4, or any fixed finite number)? -If not, what's a reasonable known lower bound $K$ on the number of $A_1, \ldots, A_K$ which are known to exist, so that we have at least $2^K$ essentially different versions of Mathematics? - -REPLY [17 votes]: Your question is essentially asking about the structure of the Lindenbaum–Tarski algebra of ZF. Jason gave a concrete example showing that one can embed the free Boolean algebra on countably many generators inside the Lindenbaum–Tarski algebra of ZF. In fact, it can be shown that the Lindenbaum–Tarski algebra of ZF is a countable atomless Boolean algebra. (There is nothing very special about ZF here, one only needs that the theory is consistent, recursively axiomatizable, and that it encodes a sufficient amount of arithmetic.) Since there is only one countable atomless Boolean algebra up to isomorphism, this completely determines the structure of the Lindenbaum–Tarski algebra of ZF.<|endoftext|> -TITLE: complement of a totally disconnected closed set in the plane -QUESTION [23 upvotes]: While preparing a course in complex analysis, I stumbled over a remark in Dudziak's book on removable sets, namely that any totally disconnected $K \subset\subset {\mathbb C}$ must have a connected complement; a remark, "that verifying the reader may find one of those exercises in 'mere' point-set topology that is a wee bit frustrating". Out of curiosity I spent an evening with this question. The assertion turned out to be a simple consequence of Theorem 14.2 ("If $x$ and $y$ are separated by the closed set $F$ in the open or closed plane they are separated by a component of $F$.") in an old book "Elements of the Topology of Plane Sets of Points" by M.H.A. Newman, Cambridge 1951. There, the proof is based on an lemma by Alexander (used in his proof of the Jordan-Brouwer separation theorem; Trans. AMS 23, 333-349, 1922), stating that, for disjoint closed sets $F_1$ and $F_2$ in the plane, two points which are connected in the complement of $F_1$ and in the complement of $F_2$ are connected in the complement of $F_1 \cup F_2$. This lemma fails for more general surfaces (Newman gives a counterexample for the torus) and is proved by homological methods. So, here are my questions: -(a) is there a more modern reference to these kind of results (Newman's book uses quite an ideosyncratic terminology and notation); -(b) is there a simpler proof not using Alexander's lemma (or similarily deep results); -(c) how about the connectedness of the complement of a totally disconnected closed set in surfaces (or topological spaces) more general than the plane? - -REPLY [10 votes]: I know this is an old question that already has an excellent answer (although it may apply only to compact sets). However, let me respond to the original question, concerning the following results: - -Lemma. If two points $\newcommand{\R}{\mathbb{R}}x,y\in\R^2$ are disconnected from each other by the closed set $F$, then they are disconnected by a connected component of $F$. - -It is clear (as noted) that the specific topology of the plane has to be used in some way, as e.g. this is no longer true on the torus. The key is indeed the following. - -Janiszewski's Theorem. If two compact sets $A,B\subset S^2$ do not separate $S^2$, and $A\cap B$ is connected, then $A\cup B$ also does not separate $S^2$. - -This result goes back to Janiszewski (1912), rather than the article of Alexander you mention. It is key to proving many "obvious" facts in plane topology (in particular it can be used to give a topological proof of the Jordan curve theorem). For example, a quick search found these notes giving a direct proof of Janiczewski's theorem online. I also seem to recall that there is a referene to a topological proof in Pommerenke's "Boundary behaviour of conformal maps", but I don't have the book to hand at the moment. -In any case, it should hopefully be clear that a direct argument can be furnished (e.g. by modifying a curve connecting two given points by replacing certain chords with those of simple closed curves sufficiently close around the two sets), but it is always going to be a bit fiddly. -Remark 1. The result in the form you quoted: - -Corollary. If two compact sets $A,B\subset S^2$ do not separate the points $x,y\in S^2$, and $A\cap B$ is connected, then $A\cup B$ also does not separate $x$ and $y$. - -follows easily from Janiczewski's theorem (filling in the components of $S^2\setminus A$ that do not contain $x,y$, and similarly for $B$). -Remark 2. A proof of the lemma above (which turns out to be very useful), in slightly different formulation, can be found e.g. in my paper Connected escaping sets of exponential maps, DOI:10.5186/aasfm.2011.3604 where it was given for completeness, although in that sketch the relevance of Janiczewski's theorem is brushed under the carpet in the last line.<|endoftext|> -TITLE: Categorification request -QUESTION [6 upvotes]: Possible Duplicate: -Can we categorify the equation (1 - t)(1 + t + t^2 + …) = 1? - -Can you give a categorification of the geometric series identity: -$$1/(1-x)=1+x+x^2+...$$ -Categorifications of partial sum identities -$$(1-x^{n+1})/(1-x)=1+x+x^2+...+x^n$$ -would also be nice. - -REPLY [15 votes]: I tried to discuss this geometric series example of categorification in one of my answers to another MO question by Jan Weidner, here. I can't tell whether this reply was considered unsatisfactory, but what one considers satisfactory would have to depend on what one is looking for (especially as "categorification" is a vague term -- intentionally so). -Qiaochu has already given one interpretation, rewriting the linear fractional transformation $L = \frac{1}{1-x}$ in the form $L = 1 + xL$ and categorifying that. There are general ways of "categorifying" fixed points of functions, replacing endofunctions by endofunctors and equations by isomorphisms, but one is generally interested in a canonical solution. To illustrate this in the present case, one may categorify the endofunction $f: s \mapsto 1 + xs$ (on $\mathbb{R}$, say) to an endofunctor $F: S \mapsto 1 + X \times S$ on the category of sets. Now, there will generally be many "fixpoint solutions" of endofunctors (meaning a set $L$ together with an isomorphism $F(L) \cong L$), but many people (for example, those who like to talk about datatypes from a categorical perspective) tend to favor a canonical fixpoint solution that arises by applying the following result of Joachim Lambek. - -If $F: C \to C$ is an endofunctor, define an $F$-algebra to be an object $c$ of $C$ together with a morphism $F(c) \to c$. Morphisms are defined in the obvious way (involving a commutative square). Theorem (Lambek): if $(c, \alpha: F(c) \to c)$ is initial in the category of $F$-algebras, then $\alpha$ is an isomorphism. - -For $F(S) = 1 + X \times S$ on $Set$, the initial $F$-algebra turns out to be the free monoid on $X$ as already indicated by Qiaochu. Another canonical fixpoint is obtained by dualizing Lambek's theorem, referring instead to terminal coalgebras of endofunctors. The first type of solution is typically recursive and algebraic; the second solution co-recursive and coalgebraic. -But perhaps this interpretation is not considered fully satisfactory if one is after a direct categorification of division or reciprocation which does not fall back on rewriting an equation multiplicatively. For example, when a topologist writes, in categorification mode as it were, -$$"BG = 1//G"$$ -for the classifying space (take '1' here to be $EG$ which is homotopy equivalent to a point, and divide out by the action of $G$ on $EG$), he clearly doesn't mean $G \times BG \cong EG$. People like Baez and Dolan have thought about what it means to categorify reciprocals; in the decategorification direction, they define the cardinality of a groupoid $G$, when $G$ is equivalent to a disjoint sum of finite groups $G_x$, where $x$ ranges over the set of connected components, to be -$$card(G) = \sum_{x \in \pi_0(G)} 1/|G_x|$$ -so that for example, the cardinality of the groupoid of finite permutations is e. In particular, the cardinality of a finite group is the reciprocal of its order. -In general, as far as I understand things, categorified division doesn't involve dividing by a set, but by a suitable (usually free) group action. Hints of this can be seen in my first answer to the other categorification request linked to above, where the categorified term $X^n/\mathbf{n!}$ means dividing by the usual action of the symmetric group $\mathbf{n!}$ on a tensor power $X^n$. A thorough discussion of this point would lead to considerations in $(\infty, 1)$-category theory, but to give a taste, one may think of a "space" $BG = 1/G$ (taking $G$ for now to be discrete) as given by the topos -$$1/G = Set^G$$ -where the '1' here is the one-point "space" given by the topos $Set$; here one can take advantage of an equivalence -$$EG = Set^G/G \simeq Set$$ -where the middle term is a slice topos (note: the notation for a slice should not be interpreted as division!), and define a "bundle projection" between toposes: -$$Set^G/G \to Set^G$$ -which is the geometric morphism right adjoint to pulling back along $G \to 1$ in $Set^G$; concretely, it takes a morphism $p: X \to G$ in $Set^G/G$ to the internal object of sections. -To return now to the question, where one is attempting to categorify $\frac1{1-x}$, one needs somehow to construe $1-x$ as a group $G$ with a suitable action on a contractible object playing the role of 1. The question is: what is $x$ here (what does it categorify to)? My best attempt at an answer (which I tried to give, maybe not very successfully, in one of my answers to the other MO question) is to write $x = 1 - G = -(G - 1)$, interpreting here $G$ actually as a group object $\mathbb{Z}G$ (more precisely, a cocommutative Hopf algebra, which is a group object in the cartesian category of cocommutative coalgebras), then interpreting $G - 1$ as the so-called augmentation ideal $IG$ fitting in the exact sequence -$$0 \to IG \to \mathbb{Z}G \stackrel{\pi}{\to} \mathbb{Z} \to 0$$ -where $\pi$ is an augmentation map which sits at the right end of a bar construction for $EG$ (in an abelian sense, meaning a free resolution for computing cohomology of $G$), and finally interpreting the additive inverse $-(G-1)$, or rather additive inversion generally, as the odd degree shift functor on the category of $\mathbb{Z}_2$-graded abelian groups (I'll call it $\Sigma$, for suspension; it takes a a graded object $(V_0, V_1)$ to $(V_1, V_0)$; I tried to explain why this is sensible here). The point I had alluding to is that there are various models for the contractible simplicial object $EG$, but the relevant one for purposes of this categorification problem seems to be where -$$EG_n = \mathbb{Z}G \otimes (\Sigma IG)^{\otimes n}$$ -(more precisely, the free resolution $EG$ is taken to be a normalized bar resolution; see the reference to Hilton-Stammbach in my earlier answer); here $EG_n$ lives in degree $n \pmod 2$. Dividing the total graded space $EG$ by the action of $\mathbb{Z}G$, one is left with the model -$$BG = \sum_{n \geq 0} (\Sigma IG)^{\otimes n}$$ -and this ultimately is how I am interpreting the categorification of the equation -$$\frac1{1-x} = \sum_{n \geq 0} x^n$$ -This was quite a long reply! I'm having trouble previewing; let's see how this looks...<|endoftext|> -TITLE: Properly Discontinuous Action -QUESTION [63 upvotes]: When looking definition, and theorems related to Properly discontinuous action of a group $G$ on a topological space $X$, it is different in different books (Topology and Geometry-Bredon, Complex Functions-Jones, Three Dimensional Geometry and Topology- Thurston). Therefore, it will be clarified, if we write these definitions separately, and see which is stronger or which are equivalent? I will name them as "Type A", "Type B"..) -Let $X$ be a topological space and $G$ be a group acting on $X$. -Definition 1: The action is of "Type A" if the map $G \times X \rightarrow X \times X$, given by $(g,x)\mapsto (x,g.x)$ is proper, i.e. inverse image of any compact set under this map is compact. -Definition 2: The action is of "Type B" if for any compact set $K\subseteq X$, $K\cap g.K=\phi$ for all but finitely many $g\in G$. -Definition 3: The action is of "Type C" if for each $x\in X$ has an open neighbourhood $U$ such that $g.U\cap U=\phi$ for all but finitely many $g\in G$. -Definition 4: The action is of "Type D" if for each $x\in X$, there exist an open neighbourhood $U$ of $x$, such that $g.U\cap U\neq \phi$ for $g\in G$ implies $g.x=x$. -Definition 5: The action is of "Type E" if each $x\in X$ has a neighbourhood $U$ such that the set {$ g\in G \colon g.x\in U $} is finite. -Q.1 Which type of actions imply which other type of action? -Q.2 If $X$ is Hausdorff, then under which type action, the quotient $X/G$ is Hausdorff? -(These are required, when studying action of a group on a compact Riemann surface, its quotient, whether quotient map is branched or unbranched, etc,) -(This question may be not applicable to post for MO; but when reading a paper related to enumeration of equivalent coverings of a space, with given (finite) transformation group, I came across this notion, and when looked into details, the different definitions puzzled.) - -REPLY [10 votes]: Theo Buehler did a great job relating Types A, B, C, and E. Going off of Stefan Witzel's new answer, I'd like to point out that in Munkres's Topology in section 81 (page 505 of that link), he defines an action to be properly discontinuous if for all $x\in X$ there is a nbhd $U$ s.t. $g(U)\cap U = \emptyset$ unless $g=1$. So if we tweak Type D from the original question to conclude $g=1$ rather than just that $g$ fixes $x$, then we recover Munkres' definition and it's no longer as trivial as Theo's answer showed the old Type D definition was. I think Munkres' definition is the one which I think point-set topologists would use. It's nice because you don't need to assume a topology on $G$, but of course you could just put the discrete topology on it. Perhaps the other definitions are more popular in the literature of Riemann surfaces, and the difference may be because of standing hypotheses in that field, since they often care most about the case of Fuchsian groups. Certainly Munkres' definition implies Type E and Type C -Munkres also points out that the quotient map $\pi: X\rightarrow X/G$ is a covering map iff the action of $G$ is properly discontinuous. -An exercise in section 81 gives: Let $X$ be locally compact Hausdorff and let $G$ act freely (i.e. fixed-point-free). Suppose that for each compact $C \subset X$ there are only finitely many $g\in G$ s.t. $C\cap g(C) \neq \emptyset$. Then the action of $G$ is properly discontinuous and $X/G$ is locally compact Hausdorff. So this tells you when Type B implies Munkres' definition. -Now let's relate Munkres' definition to Type A and Theo's answer. Using Theo's various propositions and corollaries it's not hard to see that if $X$ is locally compact Hausdorff space and $G$ is any group (which we'll equip with the discrete topology) then Munkres' definition implies Type A. Conversely, if $X$ is locally compact then a proper action of a discrete group must be of Type B (by Theo's comment) and this implies Munkres' definition because local compactness lets us get from $g(K)\cap K = \emptyset$ to $g(U)\cap U = \emptyset$.<|endoftext|> -TITLE: Description of quasi-coherent modules on a product -QUESTION [8 upvotes]: Let $R$ be a ring and $X,Y$ two $R$-schemes, which you may assume to be noetherian or anything reasonable you like. Is it possible to "construct" $\text{Qcoh}(X \times_R Y)$ out of $\text{Qcoh}(X)$ and $\text{Qcoh}(Y)$ in the $2$-category of all cocomplete $R$-linear tensor categories? -Perhaps it is the $2$-coproduct? So the question is if for every cocomplete $R$-linear tensor category $C$ the canonical functor -$\text{Hom}(\text{Qcoh}(X \times_R Y),C) \to \text{Hom}(\text{Qcoh}(X),C) \times \text{Hom}(\text{Qcoh}(Y),C)$ -$F \mapsto (F \circ (p_X)^*, F \circ (p_Y)^*)$ -is an equivalence of categories. This is satisfied if $X,Y$ are affine, but I think also when $X,Y$ are projective over $R$ (EDIT: Yes, now I've proved this in detail, should I write it up?). Actually for my purposes it would be enough to prove that the functor is conservative, i.e. reflects isomorphisms. -Here was a similar question on MO, but it adresses (as with the answer by David Ben-Zvi) only the derived setting, but I want to work with the usual category of quasi-coherent modules. - -The question is answered affirmatively here: https://arxiv.org/abs/2002.00383 - -REPLY [2 votes]: More generally, I have proven that for quasi-compact and quasi-separated schemes $\mathrm{Qcoh}(X \times_S Y)$ is the bicategorical pushout of $\mathrm{Qcoh}(X)$ and $\mathrm{Qcoh}(Y)$ over $\mathrm{Qcoh}(S)$ in the bicategory of cocomplete linear tensor categories. The technique of the proof has many other applications as well. -Localizations of tensor categories and fiber products of schemes (arXiv:2002.00383)<|endoftext|> -TITLE: Terminology: "cocompact" -QUESTION [6 upvotes]: Let $M$ be a Riemannian manifold such that its isometry group $G=\textrm{Iso}(M)$ is a Lie group, and let $\Gamma$ be a subgroup of $G$. - -1) What does the phrase "$\Gamma$ is a cocompact group of isometries of $M$" mean? Does it mean that the quotient space $M/\Gamma$ is compact, or does it mean that the coset space $G/\Gamma$ is compact ? - -And - -2) Is there an example as above in which $M/\Gamma$ is compact and $G/\Gamma$ is not, or viceversa ? - -Also, - -3) Is there any difference between "$\Gamma$ is a (discrete) cocompact group of isometries of $M$" and "The (discrete) group $\Gamma$ acts on $M$ cocompactly by isometries" and "The action $\Gamma\times M \rightarrow M$ is cocompact, where $\Gamma$ is a (discrete) group of isometries" ? - -REPLY [6 votes]: 1) from wikipedia "In mathematics, an action of a group G on a topological space X is cocompact if the quotient space X/G is a compact space." -2) Let $\bar{M}$ be a noncompact manifold, $\Gamma$ its fundamental group, and $M$ its universal cover. Given a riemannian metric on $\bar{M}$, we can lift the metric to $M$ and the group $\Gamma$ acts as a group of isometry on $M$. For a generic metric on $\bar{M}$, there is no other isometries on $\bar{M}$ than the one provided by $\Gamma$. In which case $G/\Gamma$ is trivial (hence compact) whereas $M/\Gamma$ is isometric to $\bar{M}$, hence noncompact. As an example, take an euclidean cylinder, give a good (asymetrical) kick in it, unroll it and you are done. -Commenting on Scatt Carnahan answer, the example of the hyperbolic upper half plane is misleading, because its isometry group is transitive. This is exceptional in the world of riemannian geometry. Just take some non compact quotient of the upper half plane, say $H/\Gamma(2)$, make a small bump on the quotient, lift the resulting metric, et voila you have killed -most isometries on $H$ with respect to this new metric.<|endoftext|> -TITLE: Colimits in a bigger universe -QUESTION [5 upvotes]: Fix a universe $\mathcal{U}$. Call a category $\mathcal{U}$-complete if every diagram indexed by a $\mathcal{U}$-small category has a limit, and a functor $\mathcal{U}$-continuous if it preserves $\mathcal{U}$-small limits. Usually, when one fixes a universe, one calls this simply complete and continuous. -Now assume we are given $\mathcal{U}$-complete categories $C,D$ and a $\mathcal{U}$-continuous functor $F : C \to D$. Does then $F$ also preserve limits (which exist in $C$) which are not necessarily $\mathcal{U}$-small? - -REPLY [4 votes]: The answer is yes if you put in some additional smallness/accessibility/presentability assumptions. For example, if $C$ is the $U$-cocompletion of a $U$-small subcategory, each of whose objects are ($U$-small)-presentable, then any $U$-cocontinuous functor is a left adjoint, and hence absolutely cocontinuous.<|endoftext|> -TITLE: Is there a bound on arithmetic genus of a variety in projective n-space in terms of dimension and degree? -QUESTION [10 upvotes]: Suppose you are given a closed subvariety $V$ of projective space $\mathbb{P}^n_k$. Let's say we fix the degree and the dimension of $V$. Can we then bound the arithmetic genus of $V$, or does no such bound exist? - -REPLY [3 votes]: I believe it was first proved by Kleiman (you only need to assume fixed dimension and degree, no need to assume a subvariety of some fixed $\mathbb P^n$), see Corollary 6.11 -S. Kleiman, Exp XIII in A. Grothendieck et al., Theorie des Intersections et Theoreme de -Riemann-Roch (SGA 6), Lecture Notes in Math No. 225, Springer-Verlag, Heidelberg (1971).<|endoftext|> -TITLE: Counting card distributions when cards are duplicated -QUESTION [6 upvotes]: If we have a deck of 48 different cards and 4 players each get 12 cards, it is well known how to calculate the number of possible distributions: take fac(48) and divide 4 times by fac(12). -In a german card came (Doppelkopf) there are 24 different card types, but 2 copies of each type, that is 48 cards at all. -How many distributions when there are 4 players? My first approach was to assume there are 48 different cards, and then to divide by 2 and by 2 (... 24 times). But this underestimates the real number of distributions. Because: Say player 1 has both spade kings. This has been ruled out already by dividing by fak(12). But if player 1 and player 2 both have a spade king, then we must divide by 2. -How to calculate the number of different distibutions? -Is there a closed formula as in the single card deck case? - -REPLY [6 votes]: Here is an elementary approach which works. It is based on Tony Huynh's method of dealing out the hands one at a time, but using transition matrices/dynamic programming instead of sums. -Let the state of the deck be $(p,s)$ where $p$ is the number of pairs it contains and $s$ is the number of singletons. Dealing out a hand changes the state of the deck, e.g., from the starting deck $(24,0)$ we can go to $7$ possible states: $(18,0),$ $(17,2)$, ..., $(12,12)$. -The number of ways to go from $(p_1,s_1)$ to $(p_2,s_2)$ by dealing one hand only depends on the labels, and is a single sum over the number of pairs $r$ in the new hand: $\sum_r {p_1 \choose r,p_2,p_1-r-p_2}{s_1 \choose 12-r-p_1+p_2}.$ -This gives us transition matrices which are $1\times 7$, $7\times 13$, $13 \times 7$, and $7 \times 1$, whose product in the proper order, as a scalar, is the number of distinct deals, $2248575441654260591964$. -Here is some Mathematica code which implements this: -transitions[p1_, s1_, p2_, s2_] := - Sum[Multinomial[r, p2, p1-p2-r] Binomial[s1, 12+p2-r-p1], {r,0, p1-p2}] -M1 = Table[transitions[24-i, 2i, 18-j, 2j], {i, 0, 0}, {j, 0, 6}] -M2 = Table[transitions[18-i, 2i, 12-j, 2j], {i, 0, 6}, {j, 0, 12}] -M3 = Table[transitions[12-i, 2i, 6-j, 2j], {i, 0, 12}, {j, 0, 6}] -M4 = Table[transitions[6-i, 2i, 0-j, 2j], {i, 0, 6}, {j, 0, 0}] -(M1.M2.M3.M4)[[1, 1]] - -Final output: -2248575441654260591964<|endoftext|> -TITLE: Are infinite dimensional constructions needed to prove finite dimensional results? -QUESTION [27 upvotes]: Infinite dimensional constructions, such as spaces of diffeomorphisms, spectra, spaces of paths, and spaces of connections, appear all over topology. I rather like them, because they sometimes help me to develop a good mental picture of what is going on. Emmanuel Farjoun, in the first lecture of an Algebraic Topology course about a decade ago, described "becoming comfortable with the idea of infinite dimensional manifolds" as being "one of the main conceptual advances in topology in the latter half of the 20th century". But, as I realized in a discussion yesterday, I don't understand whether infinite dimensional spaces are needed, or whether they are merely an intuitive crutch. - -Are there situations in which a significant finite dimensional result strictly requires an infinite dimensional construction in order to prove it or in order to properly understand it? -If the answer is no, then at least are there important finite-dimensional theorems for which infinite-dimensional proofs are "clearly" the easiest and the most natural? (for some conceptual reason which you can explain; not just "a finite-dimensional proof isn't known yet"). - -A closely related question is this, although its focus is somewhat different, and none of the answers there apply here; however Andrew Stacey's answer which argues that infinite-dimensional constructions are usually not strictly necessary, is relevant. -Edit: Your favourite finite-dimensional result proved by infinite-dimensional means answers this question only if you can explain to me why I should not expect a finite-dimensional proof to exist or to be anywhere near as "good". - -REPLY [11 votes]: Douady's prooof that there exists a space parametrizing all compact analytic subspaces -of a given analytic space uses infinite dimensional analytic spaces. -(This space is the analogue for analytic spaces of Grothendieck's Hilbert scheme, -and is now referred to as the Douady space. ) -So Douady first proves that the Douady space exists as a Banach-manifold, and then shows -that this space is in fact finitely dimensional. -See: Le problème des modules pour les sous-espaces analytiques compacts d'un espace analytique donné. Annales de l'institut Fourier, 16 no. 1 (1966), p. 1-95 -By the way, this paper is very famous for its first lines, whose English translation could read as: - -Let $X$ be a complex analytic space. -The goal of this work is to endow his author with the title of Doctor in Mathematical sciences and the set $H(X)$ of all compact analytic subspaces of $X$ with the structure of an analytic space. -To formulate in a more precise way the second problem,...<|endoftext|> -TITLE: what can be said about the choice of a prior in Bayesian statistics? -QUESTION [11 upvotes]: When reading about the Bayesian approach to statistics, priors are an important component of the whole methodology. -Yet, it seems like priors are chosen without any specific theoretical motivation. There is the theory of conjugate priors, which is motivated mostly computationally, I believe, but still, I haven't seen a result in the line of "the choice of a certain prior will lead to faster convergence rate" or something similar to that. -Is there a good reference that analyzes the choice of a prior somehow, instead of always assuming that it is given, and assuming that it is completely the modeler's choice? - -REPLY [3 votes]: The key point is to show that your analysis does not depend much on the prior in the first place. -You could try - -a conjugate prior (if one exists) -a (possibly improper) uniform prior and -a Jeffreys prior (justified by its invariance under re-parameterization) - -and if you get answers that are reasonably close then that should provide some support that your analysis does not depend on arbitrary choice of prior. -BUGS and similar systems give quite a bit of flexibility regarding the prior but otherwise, from a practical viewpoint, easy computability and the availability of software may restrict your choice of prior. -See: - -http://en.wikipedia.org/wiki/Conjugate_prior -http://en.wikipedia.org/wiki/Uniform_prior -http://en.wikipedia.org/wiki/Jeffreys_prior<|endoftext|> -TITLE: Bad behaviour of perverse sheaves over 'general' bases? -QUESTION [10 upvotes]: Could one define $\mathbb{Q}_l$-perverse etale sheaves over more or less general (excellent, separated) base scheme by combining the results of Gabber and Ekedahl? Would their functoriality properties (i.e. left and right $t$-exactness of various image functors, mostly those coming from affine morphisms) be similar to those described in BBD? Are any of these properties difficult to prove/ wrong in general? What could cause more difficulties: a 'large' base scheme, or a mixed characteristic one? -Upd. Any evidence or references concerning this question would be very interesting. In particular, did anybody study perverse sheaves (or something similar) over $\mathbb{Z}_p$ (cf. the discussion in comments)? - -REPLY [6 votes]: The answer is very likely "yes", but you will need to put together some technical articles (and unpublished results) that may not have yet been put together. Here are the key ingredients, as I see it : -(1) General definition of perverse t-structure, not using stratifications (using stratifications is probably a bad idea in general), not using finiteness conditions, no need for a base field either : see Gabber's article "Note on some t-structures". Section 8 is of particular interest, since this is where he defines perverse t-structures on étale n-torsion sheaves on a noetherian scheme X, provided that X admits a dualizing complex (a priori in the sense of SGA 5 I). He also gives a condition on the perversity that guarantees that the perverse truncation functors will respect constructibility. -(2) So you want dualizing complexes. Deligne has proved that they exist if X is of finite type over a regular basis S of dimension $\leq 1$ (SGA 4 1/2, [Th finitude], by the way the dualizing complex is what you might expect i.e. the exceptional inverse image of the constant sheaf on S). More generally, Gabber's recent (mostly unpublished) results include the existence of a dualizing complex over any noetherian excellent scheme that admits a dimension function (where "dualizing complex" has a slightly weaker definition that in SGA 5 I, but I don't think it matters for the application to (1)). A remark : for the first case (X of finite type of S regular of dimension $\leq 1$, the dimension function on X would simply be Artin's rectified dimension, if I remember well. In general, I think that the existence of a dimension function on X is equivalent to the fact that X is universally catenary (which is part of the definition of "excellent", so I am confused). -(3) In any case, suppose you're okay with (1) and (2), so you have your dualizing complex, your perverse t-structure, you chose the perversity function so the truncation functors are compatible with constructibility, and let's say you even chose the self-dual perversity, that is, the perversity is related to the dimension function in the usual way. Then your category of constructible perverse sheaves should be stable by duality, and artinian (use theorem 8.3 of the article of Gabber I cited in (1)). -(4) Now you want to study the exactness properties of the 4 operations, so probably you'll be happy to use Gabber's finiteness theorems so constructible things stay constructible, so you'll restrict to morphisms of finite type (and the schemes already have to be noetherian excellent). The exactness properties of direct images by affine maps also follow from results of Gabber (he generalized SGA 4 XIV; if you're of finite type over a trait, the result was proved by Gabber too, but much earlier). I think these are the central results, the t-exactness of shifted $f^*$ for $f$ smooth should just come from duality. -(5) And finally, you want to go $\ell$-adic. Then I don't know anything better than Ekedahl's results (maybe they could be generalized to other schemes using Gabber's finiteness theorem ? I haven't tried so I can't be sure). So you'll finally need to restrict to the case where X is of finite type over S regular of dimension $\leq 1$. Note that in this case the existence of the dualizing complex and the finiteness results are already in SGA 4 1/2, the construction of the t-structures is in a published paper of Gabber, so the only possible problem is affine Lefschetz (the exactness properties of affine direct images), which as far as I understand has been known for more than 15 years in this case, but I'm unable to cite a reference right now. -(6) Well, I guess you'd probably be interested in the nearby and vanishing cycles functors next. Over a dvr or a general base ? ;) Good luck anyway, I don't think I know any reference (published or not).<|endoftext|> -TITLE: vector to diagonal matrix -QUESTION [19 upvotes]: For any column vector we can easily create a corresponding diagonal matrix, whose elements along the diagonal are the elements of the column vector. -Is there a simple way to write this transformation using standard linear algebra operations (such as matrix multiplication, etc.), instead of explicitly writing it as $diag(\mathbf{x})$? -For example $M \mathbf{x}$ cannot work for any matrix M, since the result will be a vector, not a diagonal matrix. But maybe there is some more elaborate expression that yields the diagonal matrix. - -REPLY [14 votes]: I'm not sure whether it answers your question, but here is a "matrix procedure" to transform the column vector $v$ into a diagonal matrix $D$: -Let $E_i$ be the $n \times n$ matrix with a $1$ on position $(i,i)$ and zeros everywhere else; similarly, let $e_i$ be the $1 \times n$ row matrix with a $1$ on position $(1,i)$ and zeros everywhere else. Then -$$D = \sum_{i=1}^n E_i v e_i .$$<|endoftext|> -TITLE: What is random real forcing? -QUESTION [9 upvotes]: Hi all, -Can someone please explain the idea and the main steps in a random real forcing? -- what makes it (the new real) different from adding a Cohen real? -- is there a good reference for it? -Thanks. - -REPLY [14 votes]: This is relatively easy to answer. One of main differences between Cohen forcing and random real forcing is that random real forcing does not add an unbounded real. -That is, every function $f:\omega\to\omega$ in the forcing extension is bounded by -function $g:\omega\to\omega$ in the ground model, in the sense that for all $n\in\omega$, -$f(n)\leq g(n)$. -Let me sketch you a proof of this fact. Let $p$ be a set of positive measure $\varepsilon$ and assume that $p$ forces $\dot f$ to be a name for a function from $\omega$ to $\omega$. -Find a maximal antichain $p_n$, $n\in\omega$, below $p$ such that -each $p_n$ decides $\dot f(0)$, i.e., for some $m\in\omega$ forces $\dot f(0)$ to be $m$. -Now finitely many of the $p_n$ will union up to a subset of $p$ that has a measure close to the measure of $p$. Call this union $q_0$. Now $q_0$ does not decide $\dot f(0)$, but it gives you an upper bound, the maximum of the values forced on $\dot f(0)$ by the finitely many $p_n$ that union up to $q_0$. -Iterating this procedure we get a decreasing sequence $q_i$ of conditions such that -$q_i$ forces upper bounds on all $\dot f(k)$, $k\leq i$. -If each $q_{i+1}$ is a sufficiently large subset of $q_i$ in terms of measure, -then the intersection $q$ of the $q_i$ will be of positive measure and gives us a ground model function that is forced by $q$ to be an upper bound for $\dot f$. -Combining this with the usual density argument tells you that every new function is bounded by a ground model function. -This property is known as $\omega^\omega$-boundingness and can be formulated in terms of -a distributivity property of the corresponding Boolean algebra (the completion of the forcing notion, in this case the measure algebra on the reals). Jech elaborates on that -(i.e., proves the equivalence of these two properties) in his book. -Since Cohen forcing adds an unbounded real, the $\omega^\omega$-boundingness of random real forcing shows that it does not add Cohen reals, or algebraically speaking, -that the measure algebra has no countable atomless regular subalgebra. -Also, it can be shown the Cohen forcing does not add a random real. -A good source for these questions is Bartoszynski and Judah: Set Theory of the Real Line.<|endoftext|> -TITLE: Singular complex = cohomology ring + Steenrod operations? -QUESTION [10 upvotes]: Fix a prime $p$ and consider everything mod $p$. Steenrod operations arise somehow from the loss of information passing from the singular complex of a space to its cohomology ring. Are they exactly this gap, i.e. can I get the singular complex back from the cohomology ring of a space and its structure as a module over the Steenrod algebra? - -REPLY [4 votes]: It depends of what structure you want to consider on the complex of singular cochains. If you want to look at it just as a complex, then the cohomology groups are enough. If you want it as a differential graded algebra, then you would need the cohomology groups with an A-infinity algebra structure, etc. Steenrod operations are something which come from the E-infinity structure on cochains, but they are weaker.<|endoftext|> -TITLE: Differential of a nilpotent or semisimple element -QUESTION [5 upvotes]: Let $G$ be an algebraic connected subgroup of $GL_n(\mathbb{C})$ and let $\chi : G \to \mathbb{C}^*$ a character. Consider $d_e\chi : \mathfrak{g} \to \mathbb{C}$ the differential of $\chi$ at the identity of $G$. Is it true that $d_e\chi(\nu)=0$ if $\nu$ is a nilpotent element?If not, is it true under some assumptions? Can we say something if $\nu$ is semisimple? - -REPLY [6 votes]: The question should be formulated more precisely. The assumption seems to be that $G$ is an algebraic group and that $\chi$ is a morphism of algebraic groups (?) If so, assuming as we may that $\chi$ is nontrivial, $\chi$ induces an isomorphism of a 1-dimensional quotient of $G$ (a torus) onto the image . The corresponding 1-dimensional quotient of the Lie algebra is then isomorphic to the Lie algebra of the 1-torus and thus consists of semisimple elements. In particular, all nilpotent elements of $\mathfrak{g}$ must lie in the kernel of -$d_e\chi$. (This uses the algebraic theory, with Chevalley's version of Jordan decomposition in both the group and its Lie algebra along with good behavior of quotients. There are some similar ideas in prime characteristic, but with added complications.) -P.S. If the setting is supposed to be algebraic, the tag algebraic-groups is appropriate here. -SOURCES: In characteristic 0, Chevalley first investigated how the traditional Lie group correspondence between the groups and their Lie algebras would work for algebraic groups. This got combined with Jordan decomposition ideas by Borel (partly in collaboration with Springer), working over more general fields. In the expanded second edition of his older lecture notes, published as Springer GTM 126, see Section 7 for a discussion of characteristic 0, and combine this with Remark 4.9 in the section on Jordan decomposition. The remark points forward to two general results 11.8 and 14.26, which characterize the semisimple and nilpotent elements in the Lie algebra of a connected algebraic group.<|endoftext|> -TITLE: Universal cocompletion without leaving our universe -QUESTION [6 upvotes]: Let $\mathcal{U}$ be a universe and $\mathcal{U}^+$ a universe with $\mathcal{U} \in \mathcal{U}^+$. Denote by $\text{Cat}(\mathcal{U})$ the $\mathcal{U}^+$-category of all $\mathcal{U}$-categories, and by $\text{Cat}_c(\mathcal{U})$ the full subcategory consisting of $\mathcal{U}$-cocomplete categories, i.e. in which $\mathcal{U}$-small colimits exist. Consider $\mathcal{U}$ as the $\mathcal{U}$-category of all sets in $\mathcal{U}$. -Question. Does the inclusion $\text{Cat}_c(\mathcal{U}) \to \text{Cat}(\mathcal{U})$ have a left adjoint? -Remark that if $C \in \text{Cat}(\mathcal{U})$, then $\widehat{C} = \text{Hom}(C^{\text{op}},\mathcal{U})$, what is usually called the universal cocompletion of $C$, is a $\mathcal{U}^+$-category (the universum jumps!) which is $\mathcal{U}$-cocomplete, which satisfies the following universal property: $\mathcal{U}$-Functors $C \to D$, where $D$ is a $\mathcal{U}$-cocomplete $\mathcal{U}^+$-category, correspond to $\mathcal{U}$-cocontinuous $\mathcal{U}^+$-functors $\widehat{C} \to D$. Thus we have no adjointness property in the usual sense, right? -Actually I doubt that there is a left adjoint in the usual sense. But often universal cocompletions are quite useful and seem to be used without any discussion of the set-theoretic problems above (for example one restricts to small categories). Is that because it's not bad that we leave the universe, at least in some contexts? -Or is it possible to repair this? Is there a universe $\mathcal{U}$ which is big enough so that the above jumps "stabilize" below $\mathcal{U}$? Perhaps we can use Mahlo cardinals? Or can we repair this by restricting to finite colimits? Isn't this more reasonable since the definition of a universe is finitary? How does the left adjoint to $\text{Cat}_{fc}(\mathcal{U}) \to \text{Cat}(\mathcal{U})$ look like explictly, if it exists? -So my question is basically about the set theoretic subtletlies behind the universal cocompletion of a category. Feel free to write anything you know about them ... - -REPLY [3 votes]: you can generalizing the $Ind(\mathcal{C})$ construction (see SGA.4-I) for general small diagrams (no neccessarly directed), using the general criterion that characterizes the purpose of a diagram $ D: I \to \mathcal{C} $ in terms of connecting category paragraph -$ X \downarrow D, \ X \in \mathcal{C} $). In this way you get a cocompletion of a (big) category $\mathcal{C}$ relatively to the small colimits (without escape from base universe). -Is this generalization $Ind(\mathcal{C})$ is equivalent to the category $\mathcal{C}^>$ of presheaves on $\mathcal{C}$ formed by the elements $P\in \mathcal{C}^>$ such that the comma category (said also the Grothendieck costruction) $\mathcal{C}\downarrow P$ ha a small final subcategory. -You can find this construction in: -Shape Theory: Categorical Methods of Approximation di J. M. Cordier, T. Porter. -Or if you can read Italian language, I send you a my script about Category theory (involving aspect of completions of categories).\ -This a Sketch: -Then give a morphism $\underline{f}: P \to Q$ we can represent it chosing for any $i\in I$ a lifting $h_{X_i} \to h_{Y_j(i)}$ i.e. $X_i \to Y_j$ of the morphism $h_{X_i} \to P \to Q$ then we have a map $\phi : |I|_0 \to |J|_0$ and a family of morphism $f_i: X_i \to Y_{\phi (i)}$ such that for a arrow $i \to i'$ in $I$ the morphisms $f_i: X_i \to Y_{\phi(i)}$ and $X_i \to X_{i'} \to Y_{\phi(i') }$ are connected as objects of $X_i \downarrow \underline{Y}$, and a such data make a unique (but Iso) morphism $P \to Q$. -We call $(\phi , (f_i)_i) $ a representation of $\underline{f}$, and two representation $(\phi , (f_i)_i), \ (\phi' , (f'_i)_i) $ of $\underline{f}$ correspond to the some morphism $P \to Q $ IFF for every $i\in I$ the morphisms $ f_i: X_i \to Y_{\phi(i) },\ f'_i: X_i \to Y_{\phi'(i) } $ are connected in - $X_i \downarrow \underline{Y}$. -To a composition $\underline{g}\circ \underline{f}: P \to Q \to R$ we have in terms of representation: -$(\psi, (g_j))\circ(\phi,(f_i)) = (\psi\circ \phi,(g_{\eta(i)}\circ f_i)_i)$ -Then we have the category of representations data as above, and call it $Ind_0(\mathcal{C})$ the (generalized Ind-category os $\mathcal{C}$, and a equivalence $Ind_0(\mathcal{C}) \sim P_0(\mathcal{C})$. -Now is $\mathcal{A}$ has small colimits any funtors $F: \mathcal{C} \to \mathcal{A}$ has a unique (but ISo) extention to a colimits preserving funtors: $P_0(F): \mathcal{C} \to \mathcal{A}$ where $P_0(-)$ is the (puntual) left-Kan-extention , this is "computable" in terms of Ind-representation: let give $\underline{f}: P \to Q$ in $P_0(\mathcal{C})$ as above, then we can define $P_0(P)= {\underrightarrow{lim}}_{i\in I} F(X_i) $, $P_0(Q)= {\underrightarrow{lim}}_{j\in J} F(Y_j) $, and $P_0(\underline{f})$ inducted by the family $F(f_i): F(X_i)\to F(Y_{\phi (i)}$ (follow a well defined morphism $P_0(\underline{f})$ for the connection relation described above). -Excuse my poor English..<|endoftext|> -TITLE: Game involving 'asking questions about a real' -QUESTION [26 upvotes]: Consider the following game, played by two players, -called Q and A, in a time frame t = 1, 2, .... -At every time point i, Q mentions some $Q_i \subset \mathbb{R}$, -after which A mentions $A_i$ such that either $A_i = Q_i$ or -$A_i = Q_i^c$. -Define $C(i) = \bigcap_{k \lt i} A_i$ and -$C(\infty) = \bigcap_{k \in \mathbb{N}} A_i$. -Player Q is declared winner if either: -1) $C(i)$ has only one element for some $i < \infty$ -or -2) $C(\infty)$ is empty. -(Player A wins in all other cases.) -Question: -Which player (if any) has a winning strategy? -Edit: -Some additional explanatory remarks: -This question occurred to me while thinking about Brouwer style choice sequences in connection with Borel sets. -Before posting, I was not 100% sure yet, but pretty confident that player A has a winning strategy when Q is restricted to giving Borel sets. This has been confirmed, in the meantime, in one of the answers below. (More generally A has a winning strategy when Q is restricted to sets having the Baire property.) -The general case is still open, though. -Some easy observations that might help to clarify the puzzle: -a) Player Q obviously has a strategy to make sure that $C(\infty)$ has at most one element. But that's not enough to win the game. -b) Whenever $C(i)$ is countable for some (finite) $i$, player Q can win the game, by (from time $i$ onwards) eliminating the elements of that countable set, one by one. -c) Player A has an easy strategy to make sure that $C(i)$ is uncountable for all (finite) $i$. But that's not enough to describe a winning strategy for player A. For example, when $Q_i = A_i = (0, 1/i)$ for every $i$, then $C(i)$ is uncountable for every (finite) $i$, but $C(\infty)$ is empty, and therefore A loses the game. - -REPLY [2 votes]: I don't know enough logic to know if the reals possess a countably complete nonprincipal ultrafilter, but A has a winning strategy if the game is played on a set which has one. Choosing such an ultrafilter that contains the co-finite filter, A can always make their choice such that $\bigcap_{k\leq i} A_k$ is in the ultrafilter, since either $Q_i$ or $Q_i^c$ is in the ultrafilter and the filter is closed under finite intersections. Since it is countably complete, it is closed under countable intersections and so $\bigcap_{k \in \mathbb{N}} A_i$ lies in the ultrafilter as well. The intersection has infinitely many elements since the ultrafilter contains the co-finite filter, thus Q never does not win when A uses the strategy "pick whichever set is in the ultrafilter". -Edit: Thanks to those below who pointed out that $\mathbb{R}$ does not, in fact have a countably complete nonprincipal ultrafilter. This gives a cute proof of another partial result: Call a winning strategy for A strong if the choice of $A_i$ depends only on $Q_i$ and not on the previous state of the game. Then A has no strong winning strategy, for if A did, then the sets A would choose form a countably complete nonprincipal ultrafilter (the axioms are fairly easy to prove from the definition of the game). -If one could prove that a winning strategy for A implied the existence of a strong winning strategy for A, this would then show A has no winning strategy, but this seems beyond me.<|endoftext|> -TITLE: The canonical divisor of the Hilbert scheme $Hilb^n P^2$? -QUESTION [5 upvotes]: Hey everyone, -I was wondering if anyone knows what the canonical divisor of the Hilbert scheme $Hilb^n P^2$ is --$Hilb^n P^2$ is the Hilbert scheme of degree-n zero dimensional subschemes of the projective plane $P^2$. Any references? -Many thanks in advance. - -REPLY [3 votes]: $n=1$ already tells you that the anticanonical divisor is going to be nicer than the canonical, in that it's effective. There, the divisor given by the three coordinate lines is anticanonical. -Next step, look at the Chow variety of $n$ points in $P^2$, with an anticanonical given by "some point is on some coordinate line". -Then use the fact that the morphism from Hilb to Chow is crepant, to say that we can pull the anticanonical back. So: the divisor given by "some point is on some coordinate line" is again anticanonical up on the Hilbert scheme.<|endoftext|> -TITLE: A comparison question for non-positively curved disks -QUESTION [13 upvotes]: Let $A$ and $B$ be two closed, 2-dimensional, non-positively-curved Riemannian disks (not necessarily with convex boundary). Suppose that their boundaries $\partial A$ and $\partial B$ have the same length, and identify $\partial A$ with $\partial B$ with a length-preserving diffeomorphism of circles. Suppose also that for every pair of points $p,q \in \partial A = \partial B$, there is a distance inequality $d_A(p,q) \le d_B(p,q)$. Does it follow that the area of $A$ is less than or equal to the area of $B$, with equality only when $A$ and $B$ are isometric? -This is a variation of a question that arose in a joint paper that I am trying to finish. Another version which is more relevant to the paper is about $\mathrm{CAT}(0)$ disks which are tiled by Euclidean unit equilateral triangles. - -REPLY [13 votes]: The answer is yes. Moreover you don't need to assume that $B$ is nonpostively curved. (And, if you are not interested in the equality case or can afford a convex boundary, the nonpositive curvature of $A$ can be replaced by a weaker assumption that the geodesics in $A$ have no conjugate points). -This follows from a filling inequality that I have proved some time ago (and quite proud of it) and boundary rigidity results of Croke, Otal, or Pestov-Uhlmann. -Inequality. -There are two papers covering it: -1) S.Ivanov, On two-dimensional minimal fillings. Algebra i Analiz 13 (2001), no. 1, 26-38 (Russian); English translation in St. Petersburg Math. J., 13 (2002), no.1, 17-25. A preprint is here. In this paper the inequality $area(A)\le area(B)$ is proved under the assumptions that $A$ is free of conjugate points and has convex boundary (sorry;)) and for arbitrary $B$ such that $d_A(p,q)\le d_B(p,q)$ for all $p,q$ on the boundary. -2) S.Ivanov, Filling minimality of Finslerian 2-discs. arXiv:0910.2257, to appear in Trudy Mat. Inst. Steklov (= Proc. Steklov Inst. Math). In this paper the same thing is proved for Finslerian metrics $A$ and $B$ and without convex boundary assumptins. This is the maximum generality I can imagine. But the paper is harder to digest if you don't like Finsler metrics. -The equality of areas implies that the boundary distances of $A$ and $B$ coincide, and then you can apply boundary rigidity results mentioned below. -Rigidity. -The case you are asking for is covered by C.Croke, Rigidity for surfaces of nonpositive curvature, Comment. Math. Helv. 65 (1990), no. 1, 150–169. He proves that if $A$ is a nonpositively curved and $B$ is arbitrary Riemannian such that $d_A(p,q)=d_B(p,q)$ for all $p,q$ on the boundary, then $A$ and $B$ are isometric. He carefully works through the details of non-convex boundaries. -I have not read J.-P. Otal's paper mentioned by Igor Rivin, but from mathscinet review it seems that he assumes that both $A$ and $B$ are strictly negatively curved. -For the same result without curvature assumptions but with convex boundary, see L.Pestov and G.Uhlmann, Two dimensional compact simple Riemannian manifolds are boundary distance rigid. Ann. of Math. (2) 161 (2005), no. 2, 1093–1110.<|endoftext|> -TITLE: Why semigroups could be important? -QUESTION [17 upvotes]: There is known a lot about the use of groups -- they just really appear a lot, and appear naturally. Is there any known nice use of semigroups in Maths to sort of prove they are indeed important in Mathematics? I understand that it is a research question, but may be somebody can hint me the direction to look on so that I would see sensibility of semigroups, if you see what I mean (so some replies like look for wikipedia are not working as they are anti-answers). - -REPLY [4 votes]: Toric varieties in Algebraic Geometry!! Indeed, the category of normal toric varieties is equivalent with the dual of the category of finitely generated, integral semigroups.<|endoftext|> -TITLE: Card game / options pricing / Brownian bridge question -QUESTION [14 upvotes]: We play a game. I shuffle a deck of cards and start dealing them face up. After any card you can say "stop", at which point I pay you 1 dollar for every red card dealt and you pay me 1 for every black card dealt. What is your optimal strategy, and how much would you play to pay this game? - -Clearly the game is worth at least 0, since you can follow the strategy 'wait until all the cards are dealt' and get paid 0. -We can conclude that at any stage of the game, we will ask for another card if the expected value of continuing to play is greater than the amount of money already won; otherwise we will say stop. Mathematically, if $r$ is the number of red cards remaining and $b$ is the number of black cards remaining, and $f_{r,b}$ is the expected value of the game at this point, then the expected value of taking another card is -$e_{r,b} = \frac{r}{r+b} f_{r-1,b} + \frac{b}{r+b} f_{r,b-1}$ -and hence the value of the game at this point, assuming we play the optimal strategy, is -$f_{r,b} = \max ( \frac{r}{r+b} f_{r-1,b} + \frac{b}{r+b} f_{r,b-1}, b-r )$ -with $f_{r,0} = 0$ (if there are only red cards remaining, we have a guaranteed payout of 0) and $f_{0,b} = b$ (if there are only black cards remaining we won't take any more cards). - -Solving numerically gives an expected value for this game of 2.624 with 52 cards. I'm interested in what the value is for a deck of $2n$ cards, and if the optimal strategy can be expressed analytically as a function of $r$ and $b$ (ie keep taking cards until you have made $g(r,b)$ dollars, then stop). -I've tried changing coordinates to $u=b-r$, $v=b+r$, and I've tried approximating the difference equation as a PDE, but no luck with an analytical solution yet - or even any decent approximations. - -REPLY [2 votes]: This specific game seems very familiar, and I'm sure I have seen it discussed before along with a discussion of the optimal strategy, although I can't remember where. It might have been the Project Euler forums, perhaps in one of the forums restricted to those who have solved a particular problem. Using Google to search for the exact answer for $52$ cards might find it if it is in an open forum. -I believe the simulations said you should stop when there are about $k$ cards left in the deck and you are ahead by $c\sqrt k$, where $c$ did not seem to depend on $n$. -In an interesting variant, you can choose the fraction of your bankroll that you wager on whether each card is red or black. You do not need to bet only on red. It turns out that as long as you bet on the right side, and you bet your whole bankroll when you have a lock, your expected bankroll is multiplied by $\prod_{k=1}^n \frac {2k}{2k-1}$. Betting according to the Kelly criterion gives you that result with $0$ variance.<|endoftext|> -TITLE: module of differentials of formal power series ring and of its field of quotiens -QUESTION [5 upvotes]: For any $A$-algebra $B$ ( commutative ring with 1 ), we have the existence of $\Omega_{B/A}$, the module of relative differentials of $B$ over $A$, which can be defined by an universal property. In the case $A = k$ being a field and $B = k[[X]]$ being the formal power series ring over $k$, $\Omega_{B/A}$ is not always a finite $B$-module. ( And I don't know if it is a free $B$-module. ) For example, if char$(k) = 0$, since $ k[[X]] $ has a infinite subset whose elements are algebraic independent over $k$, one can show that $\Omega_{B/A}$ is not a finite $B$-module. I have seen another notion in a book as following: the "universal finite differential module" $\Omega^f_{B/A}$ is a $B$-module with an $A$-derivation $d : B \rightarrow \Omega^f_{B/A}$ such that for any $A$-derivation $d^{'} : B \rightarrow M$ with $M$ being a finite generated $B$-module, there exists a $B$-module homomorphism $\phi : \Omega^f_{B/A} \rightarrow M$ such that $ d^{'} = \phi \circ d$. With this definition, one can show that, in the case of formal power series ring, $\Omega^f_{B/A}$ is a free $B$-module of rank $1$ with $dX$ as a basis, and for any $f \in k[[X]], df = f' dX$, here $f'$ is defined by the natural way, which is a result I can't deduce for $\Omega_{B/A}$. One need to use Krull Intersection Theorem or the structure theorem for finitely generated modules over a $PID$. -Now I try to compute $\Omega^f_{B/A}$ for $B = k ((X))$, the field of quotients of $k[[X]]$, but I can't get the result. The problem is that the $A$-derivation $d'$ is not necessary continuous with respect to the $(X)$-adic topology. -My question is that: do we have $\Omega^f_{B/A}$, for $B = k((X))$, is a free $k((X))$-module of rank $1$ with $dX$ as a basis? - -REPLY [4 votes]: This is not a direct answer to your question, but a few comments about the topic. -First, in general it is not true that $\Omega^1_{B/A}$ is a free module. In fact, assuming $A \to B$ is flat and finitely generated, $\Omega^1_{B/A}$ is locally free if and only if $A \to B$ is a smooth ring map. -If in your situation (or any other adic formally smooth situation) you take the completion of $\Omega^1_{B/A}$ then you are guaranteed to get a projective module of the correct rank (and in your particular case you would indeed get the free module of rank 1 you expected). This is proved in EGA 0.IV, theorem 20.4.9. I think that the module of finite derivations should coincide with this completion. I recommend you check the book "Kahler Differentials" by Ernst Kunz which discusses the module of universal finite derivations. -Edit: remark 1.8 in http://arxiv.org/PS_cache/alg-geom/pdf/9510/9510007v4.pdf claims that indeed this completion coincides with the module of universal finite derivations, thus providing a positive answer to your question.<|endoftext|> -TITLE: Topos theory reference suitable for undergraduates -QUESTION [19 upvotes]: I am a third year undergraduate who has just learnt the rudimentals of category theory. -My specialization is computer science, not mathematics. As part of my course work I want to write an essay on Topos theory. My professor says that it is possible to do so with my level (very little) of mathematical maturity, but I am not able to find any sources that treat this theory at anywhere near my level. Any suggestions? - -REPLY [2 votes]: The internal language of the effective topos can be understood with requiring barely any technology and is lots of fun! For instance, Andrej Bauer observed that if you construct the effective topos using infinite-time Turing machines instead of ordinary Turing machines, then internal to the topos there is an injection $\mathbb{N}^\mathbb{N} \to \mathbb{N}$. See these slides for undergraduates.<|endoftext|> -TITLE: The effective topos - by Hyland -QUESTION [6 upvotes]: I have one other reference request (cf. my previous question: Topos theory reference suitable for undergraduates) -Are there any references on effective topoi that are better than Hyland's original paper: -The effective topos J.M.E. Hyland, in: Troelstra and Van Dalen (eds), The L.E.J.Brouwer Centenary Symposium, North-Holland 1982, 216. - -REPLY [4 votes]: Andrew Pitts’ note “Tripos Theory in Retrospect” sheds some useful light on $\mathcal{Eff}$, from a slightly different angle than most other books do. It’s available at his publications page, and also at doi:10.1017/S096012950200364X (paywalled but potentially more durable). -For my part, even as quite a toposophile, $\mathcal{Eff}$ (and realizability toposes in general) took me a while to get comfortable with — a lot longer than any of the other genres, sheaves, syntactic ones, etc. In the end it must have taken about four or five attempts to get to grips with them, over several years — spending a little time getting a little way on each attempt, understanding one step in the construction (e.g.: the tripos-to-topos step in general), then waiting a few months while that sank in, before coming back for another crack at the next step. This certainly isn’t everyone’s experience, of course, but I’ve talked to at least a couple of other people who had a similar time.<|endoftext|> -TITLE: History of connections -QUESTION [17 upvotes]: Are there any good detailed historical sources about development of connections on vector/principal bundles over the last 100 years? -The best source I am aware of is Michael Spivak's 5 volume opus, but this is not detailed enough for the project I have in mind (I am intending to set this as a topics essay for my first year graduate differential geometry class, and I want to make sure I know enough myself first!). - -REPLY [2 votes]: Try searching "A Historical Overview of Connections in Geometry". This is a wonderfully accurate account of the history of connections.<|endoftext|> -TITLE: Does Zariski's Main Theorem come with a canonical factorization? -QUESTION [22 upvotes]: Zariski's Main Theorem (EGA IV, Thm 8.12.6): Suppose $Y$ is a quasi-compact and quasi-separated scheme, and $f:X\to Y$ is quasi-finite, separated, and finitely presented. Then $f$ factors as $X\xrightarrow{g} Z\xrightarrow{h} Y$, where $g$ is an open immersion and $h$ is finite. - -Is there a canonical choice for the factorization $f=h\circ g$, at least under some circumstances? - -For example, suppose $f$ factors as $X\to U\to Y$, where $X\to U$ is finite étale and $U\to Y$ is a Stein open immersion (i.e. the pushforward of $\mathcal O_U$ is $\mathcal O_Y$). Then I'm pretty sure the Stein factorization $X\to \mathit{Spec}_Y(f_*\mathcal O_X)\to Y$ witnesses Zariksi's Main Theorem (i.e. is an open immersion followed by a finite map). - -In general, when does the Stein factorization witness ZMT? In the cases where it fails to witness ZMT (e.g. $X$ finite over an affine open in $Y$), is there some other canonical witness? - -REPLY [5 votes]: I realized that I completely missed the second part of the question (the example). Note that ZMT implies that $f$ is a quasi-affine morphism. Then $X\to \mathit{Spec}(f_*\mathcal O_X)$ is always an open immersion (see stack project, chapter 21, Lemma 12.3). So the Stein factorization witness ZMT if and only if $f_*\mathcal O_X$ is finite over $\mathcal O_Y$. -Some comments: one should note that in general, the quasi-coherent algebra $f_*\mathcal O_X$ is not finite over $\mathcal O_Y$ and even worse, the morphism $\mathit{Spec}(f_*\mathcal O_X)\to Y$ may not be of finite type (take $Y$ an algebraic variety and $f$ an open immersion. Then $\mathcal O(X)$ is or not finitely generated is related to Hilbert's 14th problem). Now consider a ZMT factorisation $X\to Z\to Y$. If the complementary of $X$ in $Z$ only consists in points of depth at least 2 (see discussions here), then $f_*\mathcal O_X=h_*\mathcal O_Z$ is finite and we are happy. This happens when $X$ is normal (or with non-normal locus finite over $Y$) and surjective to $Y$ with complementary in $Z$ of codimension at least 2. But I don't have a general criterion.<|endoftext|> -TITLE: On-the-nose commutative cup product $\Longrightarrow$ characteristic $0$? -QUESTION [11 upvotes]: I was discussing with a student today the nature of the non-commutativity of cup-product at the level of cochains. In trying to explain what happens, I came up with the following statement. - -Theorem. - Fix a commutative ring $R$. - Suppose $F:\mathrm{Top}^{\mathrm{op}}\to \mathrm{cDGA}_R$ is a contravariant functor - from spaces to commutative DGAs over $R$, such that $X\mapsto H^*(F(X))$ is ordinary - cohomology with coefficents in $R$. Then $R$ contains $\mathbb{Q}$ as a subring. - -I'm pretty sure this "Theorem" is true. But I don't have a proof at hand. -The question is: does anyone have a proof, or know of one in the literature? I'm particularly interested in seeing a proof which is relatively "elementary", in the sense of not requiring much more heavy machinery than is needed in order to make the statement. -Added. Tyler points out in his answer that this can't be true as stated. We should add a hypothesis, such as: F takes homotopy pushouts of spaces to homotopy pullbacks of chain complexes. - -REPLY [6 votes]: Your theorem needs at least one further assumption. Otherwise, we can let F be the functor on spaces sending $X$ to the commutative DGA $H^*(X;R)$, with zero differential. -I'm not sure what statement to add. The thing I initially wanted to write is that your functor should take homotopy pushouts to homotopy pullbacks. However, commutative DGAs aren't closed under homotopy pullback. -EDIT: Let's suppose you make this assumption. Take the diagram $* \leftarrow \mathbb{CP}^\infty \rightarrow *$, form the homotopy pushout $\Sigma \mathbb{CP}^\infty$, and apply your functor $F$ to the associated square pushout diagram. You get a commutative square of commutative differential graded algebras, which is in particular a commutative square of $E_\infty$-algebras. The category of $E_\infty$-algebras has homotopy pullbacks, and so you can construct by this method a weak equivalence $F(\Sigma \mathbb{CP}^\infty) \rightarrow P$ of $E_\infty$-algebras, where $P$ is the homotopy pullback. -However, in any characteristic the Steenrod operations are stable operations and an invariant of weak equivalence of $E_\infty$-algebras. If $pR = 0$, the element $x \in H^2(F(\mathbb{CP}^\infty))$ supports nonzero power operations (namely, powers) at any prime, and so the associated element -$$\sigma x \in H^3(P) \cong H^3(F(\Sigma \mathbb{CP}^\infty))$$ -would also support nonzero power operations. But for commutative DGAs those operations are automatically zero. -Ideally one could make this for more general $R$ where $p$ is not invertible using a secondary operation and $B\mathbb{Z}/p$ instead, but the kids were up early and I'm too tired to figure out how right now.<|endoftext|> -TITLE: t-structures on the derived category of finitely generated abelian groups -QUESTION [7 upvotes]: is it possible to explicitly parametrise all the t-structures -on the derived category of finitely generated abelian groups? - -REPLY [2 votes]: The question and Sasha's answer can be generalized to a Noetherian ring $R$. The parametrization is in terms of functions from the integers to a specialization closed subset of $Spec R$. I do not have access to precise references at the moment, but see the paper -'Invariants of t-structures and classification of nullity classes' by Don Stanley and further references therein.<|endoftext|> -TITLE: Invariance of $\mathbb{Z}[x]$ under a self-equivalence of the category of commutative rings with 1 -QUESTION [26 upvotes]: Let $\mbox{Rings}$ be the category of commutative rings with $1$. -Is there an equivalence of categories $F: \mbox{Rings} \to \mbox{Rings}$ such that -$$F(\mathbb{Z}[x])\not\cong \mathbb{Z}[x]?$$ - -REPLY [12 votes]: Here is a new, short proof, that the category $\text{CRing}$ of commutative rings is rigid: -Lemma: Let $R,S$ be commutative rings, such that $\text{CAlg}(R), \text{CAlg}(S)$ are equivalent as categories. Then $R,S$ are isomorphic. -Proof: $R$ is the initial object of $\text{CAlg}(R)$ and the comma category $\text{CAlg}(R) / R$ is (via unitalization) equivalent the category $\text{CAlg}'(R)$ of commutative $R$-algebras which are not assumed to be unital. Now $0$ is a zero object of $\text{CAlg}'(R)$, thus null morphisms are defined. A morphism $\alpha : A \times A \to A$ in $\text{CAlg}'(R)$ with $\alpha (\text{id}_A,0) = \text{id}_A, \alpha (0,\text{id}_A) = \text{id}_A$ exists if and only if $(x,y) \mapsto x+y$ is multiplicative, i.e. the multiplication on $A$ is trivial, that is $A$ is just a $R$-module. Thus we have reconstructed $\text{Mod}(R)$ categorically from $\text{CAlg}(R)$. Finally, we can reconstruct $R$ as the center of $\text{Mod}(R)$. $\checkmark$ -Theorem: Every self-equivalence of $\text{CRing}$ is isomorphic to the identity. -Proof: Let $F : \text{CRing} \to \text{CRing}$ be an equivalence. Now $\text{CAlg}(R)$ is just the comma category $R \backslash \text{CRing}$, so we get an equivalence $\text{CAlg}(R) \cong \text{CAlg}(F(R))$. The lemma implies $F(R) \cong R$, in particular, $F(\mathbb{Z}[x]) \cong \mathbb{Z}[x]$ as commutative rings. But then we get also an isomorphism of corings and then $F \cong \text{id}$, as explained in my previous proof. $\checkmark$ -Remark, however, that this proof is not applicable for automorphisms of $\text{CAlg}(R)$.<|endoftext|> -TITLE: What proofs cannot be relativized -QUESTION [18 upvotes]: I am afraid this post may show my naivety. At a recent conference, someone told me that there are some arguments in computability theory that don't relativize. Unfortunately, this person (who I think may be an MO regular) couldn't give me any examples off-hand. -My naive understanding is that one can take any proof, fix an oracle and replace "Turing machine" with "Turing machine using that oracle". (Similarly, replace "computable" with "computable w.r.t. this oracle" and so on.) But I am probably missing something. -So is it possible to relativize any proof? Is it not possible? Or is it technically possible to relativize any proof, but the relativization isn't what you may expect? -If it is not always possible, can someone provide me a good example or reference? Also, any discussion on what makes a proof relativizable or not would be of help. -Actually, while writing this, I may have come up with an example. It's known that $P=NP$ and $P \neq NP$ for different oracles. But this is a relativization of a theorem (conjecture), not it's proof. So this brings up related questions. Is relativizing the statement of a theorem different from relativizing the proof? What if the theorem/proof has nothing to do with the actual model of computation (e.g. the proof doesn't refer to ideas such as time-complexity, only computability)? -Thanks! - -REPLY [23 votes]: Early in the history of recursion theory, the realization that all known proofs in the subject could be relativized in the manner you indicate led Hartley Rogers to make what is called the homogeneity conjecture. -Let $\mathcal{D}$ be the structure of the Turing degrees with the partial order of Turing reducibility $\leq_T$. Let $\mathcal{D}(\geq \mathbf{x})$ be the structure of the Turing degrees that are $\geq_T \mathbf{x}$ also with the partial order $\leq_T$. The homogeneity conjecture says that for any Turing degree $\mathbf{x}$, $\mathcal{D}$ is isomorphic to $\mathcal{D}(\geq \mathbf{x})$. -Richard Shore refuted the homogeneity conjecture in an elegant 1979 paper -- it's only one page long (though it relies on earlier work coding models of arithmetic in $\mathcal{D}$). A couple years later, Harrington and Shore showed that you can do even better. Not only are there Turing degrees $\mathbf{x}$ for which $\mathcal{D}$ and $\mathcal{D}(\geq \mathbf{x})$ aren't isomorphic as structures, there are Turing degrees $\mathbf{x}$ so that $\mathcal{D}$ and $\mathcal{D}(\geq \mathbf{x})$ aren't elementarily equivalent. -So this means that there is a first order sentence $\varphi$ in the language only containing $\leq_T$ which is true about the Turing degrees, but which is false if you relativize the sentence to the Turing degrees $\geq_T \mathbf{x}$ for some $\mathbf{x}$ (and of course, working in the Turing degrees $\geq_T \mathbf{x}$ is equivalent to giving all Turing machines access to $\mathbf{x}$ as an oracle). Hence, the proof that $\varphi$ is true can't be relativized to $\mathbf{x}$. -A somewhat misleading and oversimplified explanation of why this occurs is that you can code models of arithmetic into $\mathcal{D}$ (or $\mathcal{D}(\geq \mathbf{x})$) which can then interpret unrelativized concepts like being arithmetically definable or hyperarithmetically definable. -A particularly spectacular form of this phenomenon would occur if Slaman and Woodin's biinterpertability conjecture is true. The conjecture says that the following relation (on $\overrightarrow{\mathbf{p}}$ and $\mathbf{d}$) is definable in $\mathcal{D}$: "$\overrightarrow{\mathbf{p}}$ codes a standard model of first order arithmetic and a real $X$ such that $X$ is of degree $\mathbf{d}$". -All that being said, just about every proof in recursion theory that I know of relativizes. I'd be very interested in a proof which doesn't relativize and doesn't factor through the coding machinery I've mentioned above. -By the way (and somewhat ironically), the Baker-Gill-Solovay theorem that you mentioned does itself relativize. The relativized version says that for any oracle $X$, there are oracles $A$ and $B$ so that $X$ is poly-time reducible to both $A$ and $B$, and $P^A = NP^A$ and $P^B \neq NP^B$. (We've relativized to $X$ here. The unrelativized result simply says that there are oracles $A$ and $B$ so that $P^A = NP^A$ and $P^B \neq NP^B$). Of course, the real point is that a proof that $P \neq NP$ can't use a technique that relativizes.<|endoftext|> -TITLE: Least sum squares given constraints on subcomponents -QUESTION [5 upvotes]: Hi all, -I recently encounter a difficult problem. -I wish to minimize in $ \mathbf{x} $ the sum $\min \sum_{i=1..n} (\mathbf{x}^T \mathbf{A}_i \mathbf{x})^2$ given the constraints on the norms of all $\mathbf{x}$'s subcomponents (let's say three 3-by-1 vectors) $|\mathbf{x}_1| = 1, |\mathbf{x}_2| = 1, |\mathbf{x}_3| = 1$. $\mathbf{A}_i$ may not be positive-definite. -Yes, it's quartic expression that we want to minimize. I'm not sure if any one has worked on this or similar problem in the math community. I search the literature for sometimes but no use. My question may be similar but actually much more difficult than this Least square given constraint on subcomponents -The 4th-order and constraints on all subcomponents makes it really hard for me to handle. -Any idea to a numerical/analytical solution, is greatly appreciated. Thanks for reading. -p/s: $\mathbf{x} = [\mathbf{x}_1^T , \mathbf{x}_2^T, \mathbf{x}_3^T]^T$. By "subcomponents", I mean the subvectors, as shown in the equation. - -REPLY [3 votes]: As a practical matter, a lot depends on $n$ and the dimension of $x$. If the problem is small enough then you might not be in deep trouble. If the problem is large (e.g. $x$ might have thousands of components), then it could be very hard. -If the problem is quite small, then you might consider using an approach that exploits the polynomial structure of your optimization problem. There are convex relaxations of such polynomial optimization problems that provide very tight lower bounds and software can often use these lower bounds to find a globally optimal solution. See for example the Gloptipoly2 software: -http://homepages.laas.fr/henrion/software/gloptipoly2/<|endoftext|> -TITLE: Algorithm for detecting ribbon or slice links? -QUESTION [11 upvotes]: A link in $S^3$ is said to be slice if it bounds a collection of flat disks into the $4$-ball. Here "flat," means that there is a (locally) trivial normal bundle. This condition can be strengthened to "smoothly slice," meaning that the link bounds a smooth collection of disks. Even more restrictive is the condition of "ribbon," which means that the link bounds a collection of disks with only local maxima, and no local minima, as one moves into the $4$-ball. Thus we have an increasingly subtle series of questions. Is the link topologically slice? Is it smoothly slice? Is it ribbon? As far as I know, there are no algorithms for all three of these questions. My question is whether this is really true. For example, is there an algorithm to detect whether a link is ribbon? One interesting and venerable open problem in the field is whether all slice links are ribbon. This paper gives a sequence of examples of links which are smoothly slice but not obviously ribbon. If there were an algorithm for detecting "ribbonness," then one could apply it to these examples, and possibly disprove the slice=ribbon conjecture. It seems to me that detecting ribbon disks should not be that hard, perhaps using ideas akin to normal surface theory. You would just be looking for disks in the link complement with ribbon singularities. Has anyone thought about this? -Edit: As Ryan Budney points out, if a slice disk exists, it can be found algorithmically by iteratively subdividing a triangulation, but in the absence of upper bounds on how many times you need to do this, the algorithm can't return a negative answer that a link is not ribbon or slice. -Also, I meant to mention in my original post that the question of whether the Whitehead double of the Borromean rings is topologically slice is related to whether surgery works in 4 dimensions. Given how hard that question is, it would seem highly unlikely that there is an algorithm for the topological case! - -REPLY [6 votes]: In the special case that your knot in $S^3$ is a fibered knot, there is an algorithm to determine if it is ribbon. It comes from a sequence of papers, due to Darren Long and Andrew Casson. The last paper in the sequence is: - -Algorithmic compression of surface automorphisms. Invent. Math 81 295--303 (1985). - -As far as I know we're still some ways from implementing this algorithm but it would be interesting to see someone try.<|endoftext|> -TITLE: Torsion points in Abelian varieties over number fields -QUESTION [14 upvotes]: Hello, -Suppose $A$ is an Abelian variety of dimension $g$ over a number field $k$. Then using height functions one can show that there are non-torsion points in $A(\bar k)$. This looks like an overkill. Is there an easy, elementary way to see this? -Thanks! -Ramin - -REPLY [5 votes]: Jan Denef once pointed out to me that this is a simple consequence of the Manin-Mumford Conjecture, i.e., Raynaud's Theorem that if $A$ is an Abelian variety defined over a number field and $C$ is a curve on $A$ that is not a coset of an abelian subvariety then $C$ contains only finitely many torsion points. -In this problem we may assume that $A$ has no proper abelian subvarieties. If $A$ has dimension -at least 2, take $C$ any curve on $A$ defined over $\bar k$, then $C(\bar k)$ is infinite, but -contains only finitely many torsion points. If $A$ is an elliptic curve, let $C$ be any curve -of genus at least 2 on $A\times A$. Again, $C(\bar k)$ contains only finitely many torsion points.<|endoftext|> -TITLE: Importance of large gaps between zeros of zeta function? -QUESTION [10 upvotes]: I have noticed that there are quite a few publications, many of them recent, on trying to determine the supremum of the gaps (normalized) between zeros of $\zeta \left(\frac{1}{2} + i t \right)$. Several make use of Wirtinger's inequality. I've been studying some of these papers in an attempt to broaden my knowledge of the zeta function. -My question is this - why are people interested in determining information about the gaps? I assume that there must be some information encoded in the gap sizes about the distribution of prime numbers. Is there a publication or book in which I could read about the reasons for studying the gaps? -Thanks, -Tom - -REPLY [8 votes]: Here are two papers that show a connection between the spacing of the zeros of the zeta function and the class number problem for imaginary quadratic fields: -Conrey, J. B., and H. Iwaniec, “Spacing of zeros of Hecke L-functions and the -class number problem.” -Montgomery, H. L., and P. J. Weinberger, “Notes on small class numbers.” -And of course the study of the pair correlation of the ordinates of the zeros of the zeta-function inspired the connection between the zeta function and random matrix theory. I believe some results about the gaps are ways to check that the zeros are spaced according to predictions that come from random matrix theory.<|endoftext|> -TITLE: Good books in Modular Representation Theory -QUESTION [11 upvotes]: Hi every one! -I am reading some paper and it uses Modular Representation Theory. I even dont really know about Representation Theory and I am looking for a good book for beginner. Could you please give me some suggestion for beginning Modular Representation Theory? -Thanks in advance. - -REPLY [3 votes]: David Benson: Modular Representation Theory. New Trends and Methods. Lecture Notes in Mathematics 1081, 1984 (2nd print 2006) -is also a good book on the topic.<|endoftext|> -TITLE: Quantum mechanics formalism and C*-algebras -QUESTION [43 upvotes]: Many authors (e.g Landsman, Gleason) have stated that in quantum mechanics, the observables of a system can be taken to be the self-adjoint elements of an appropriate C*-algebra. However, many observables in quantum mechanics - such as position, momentum, energy - are in general unbounded operators. Is there any way to reconcile these two apparently contradictory statements? -I have looked at the notion of affiliation for a C*-algebra in the sense of Woronowicz. However, I can't see how you would extend states to elements affiliated with a C*-algebra and so it doesn't seem to solve the problem. - -REPLY [3 votes]: I wrote a paper about this at an REU I attended two years ago. At the time, I had the goal of making the transition from classical mechanics to quantum mechanics as natural as possible, motivating the axioms of quantum mechanics from those of classical mechanics, which are a lot more intuitive. Unfortunately, given the time constraints of the REU (just a couple of months) and being relatively new to the subject, I found myself mostly following other sources (in particular, Strocchi), which made use of the $C^*$-algebraic formalism. I have to admit, at the time, I did feel a bit uncomfortable throwing away all the unbounded operators, because, having studied "physicist's" quantum mechanics before, these are some things I naturally wanted to include. -Come two years later at the same REU, I tried to tackle the same problem, but this time I wanted to develop the axioms of quantum mechanics so to as allow unbounded operators in the theory. This led me to develop the notion of what I call an $F^*$-algebra ($F$ for Fréchet). I am actually still working on the paper at the moment, but, after reading your post, I decided to upload a preliminary copy to my academia.edu account. You should read it, check it out, and see what you think. I'd be very happy to hear any comments you have. -Be warned though, I am still in the process of editing and revising it, so dare I say, there may be some errors. Read it with a skeptical eye and let me know if you catch anything. -Cheers! -Jonny Gleason -Here's the link: https://drive.google.com/open?id=0B6xfgYpCM4U3UXRNVkhPTGc5RDA -EDIT: I actually did find a couple of errors in the paper at some point, but I had since lost the .tex file and never got around to just rewriting the entire thing with corrections in place. I don't remember exactly where they are at this point (it's been a couple of years), but if I recall they're not subtle, so if you just make sure to check the proofs you should be okay.<|endoftext|> -TITLE: Ultraproducts and the empty set -QUESTION [11 upvotes]: If $I$ is a set, $U$ a nonprincipal ultrafilter on $I$ and $E=(E_i)_{i\in I}$ a family of sets indexed by $I$, then the ultraproduct $E^*$ of $E$ is generally defined as the quotient of $\prod_{i\in I}E_i$ by the equivalence relation "equality on a subset of $I$ which belongs to $U$". -However, this definition is "wrong": this $E^*$ is nonempty if and only if all the $E_i$'s are nonempty, while the expected condition is "if and only if $\{i\in I\vert E_i\neq\emptyset\}\in U$". In fact, Łoś' theorem is false with this definition since "nonempty" can be defined by the formula $(\exists x)(x=x)$. -So I guess the right definition is $E^*=\varinjlim_{J\in U}\prod_{i\in J}E_i$ where $U$ is ordered by reverse inclusion and the transition maps are the projections. If each $E_i$ is nonempty this is equivalent to the standard form, which explains why the latter is used since in most applications (at least in algebra) the $E_i$'s carry some algebraic structure which excludes emptiness. -Of course there is a sheaf-theoretic version of this: let $I^\vee$ be the Stone-Čech compactification of the discrete space $I$, $j:I\to I^\vee$ the canonical inclusion. Then $(E_i)$ defines a sheaf of sets $\mathcal{E}$ on $I$. Put $\mathcal{E}^\vee:=j_*\mathcal{E}$. Then $E^*$ is the stalk of $\mathcal{E}^\vee$ at hte point corresponding to $U$. (In the case of ultraproducts of rings, this is explicitly stated in Schoutens' LNM 1999 book: see 2.6.2 there). -My question: are there accessible references where this issue is correctly addressed? - -REPLY [4 votes]: If you allow, as one way of correctly addressing the issue, simply assuming that the factors of the ultraproduct are nonempty, then the issue is correctly addressed in, for example, Chang and Keisler's "Model Theory", Bell and Slomson's "Models and Ultraproducts", and Comfort and Negrepontis's "Theory of Ultrafilters". The same can undoubtedly be said for almost all other treatments of model theory; the three I listed were just the first three I happened to pull off my bookshelf. -If, on the other hand, you require that empty factors be permitted, then you may need to go to the category-theoretic literature, where empty structures are not so cavalierly excluded from consideration.<|endoftext|> -TITLE: Uncertainty principle for non-commutative groups -QUESTION [7 upvotes]: Is it true that for every group $G$ and $f\in \mathbb C[G]$ it holds that $$\dim(\mathbb C[G]*f)\mathop{supp}(f)\geq |G| ?$$ -Here, $\mathbb C[G]$ is the group algebra, and by $\mathbb C[G]*f$ I mean left ideal of the group algebra $\mathbb C[G]$ generated by $f$. -Essentially this is uncertainty principle for non-commutative groups. -Since $supp \hat {f} = dim C[G]*f$ in case $G$ is abelian. - -REPLY [9 votes]: The answer is yes, this always holds. -Note that -$$\dim(im(f)) \cdot \|f\|^2 \cdot | {\rm supp}(f)| \geq \tau(f^*f) \cdot |{\rm supp}(f)| \geq |G| \cdot \|f\|^2_1.$$ -Here, $\tau \colon \mathbb C[G] \to \mathbb C$ is the non-normalized trace on $\mathbb C[G]$, coming from the inclusion $\mathbb C[G] \subset M_{|G|} \mathbb C$. It is best decribed by $$\tau(\sum_{g \in G} a_g g) = |G| \cdot a_e.$$ Also, $f \mapsto f^*$ denotes the usual involution, i.e. $$(\sum_{g \in G}a_g g)^{*} = \sum_{g \in G} \bar a_g g^{-1}.$$ The first inequality, follows since $\tau(f^*f)$ is just the sum of the eigenvalues of $f^*f$, which is obviously bounded by the number of non-zero eigenvalues (which equals the dimension of image of $f$) times the size of the largest eigenvalue (which equals $\|f\|^2$). Note that by direct computation -$$\tau\left((\sum_{g \in G} a_g g)^* (\sum_{g \in G} a_g g)\right)=|G| \cdot \sum_{g \in G} |a_g|^2.$$ The second inequality follows from this observation and the Cauchy-Schwarz inequality applied to $f\cdot \chi_{{\rm supp} f}$, where the product is here the pointwise product of coefficients and $\|f\|_1$ denotes the usual 1-norm on $\mathbb C[G]$. -Now, since each group element acts as a unitary (and hence with operator norm $1$) on $M_{|G|} \mathbb C$, we get $\|f\|_1 \geq \|f\|$ and hence -$$\dim(im(f)) \cdot | {\rm supp}(f)| \geq |G|.$$ -This even has an extension to all (possibly infinite) groups with essentially the same proof. The appropriate statement is then that for the normalized Murray-von Neumann dimension (with respect to the group von Neumann algebra $LG$) of the closure of the image of $\lambda(f)$ acting on $\ell^2 G$ via the left-regular representation $\lambda$, we have -$$\dim_G \left (\overline{im(\lambda(f))} \right) \cdot |{\rm supp}(f)| \geq 1.$$<|endoftext|> -TITLE: Random walk on a simple finite network -QUESTION [9 upvotes]: Consider a graph $\Delta_N = \lgroup (x,y)\in\mathbb{Z}^2| x+y\leq N-1, x\geq 0,\ y\geq 0 \rgroup$ (set of edges is defined in a natural way): see here ). -Take a random walker that wonders around this network (transition probabilities are given as an inverse of the degree of a given node) . I am interested in the probability $\mathbb{P}(q)$ that a walker starting at point $p\in\Delta_N$ would reach point $\mathcal{O}=(0,0)$ before reaching the "bottom" of the network : $D=\lgroup (x,y)\in\Delta_N| x+y= N-1 \rgroup$. I introduce new pair of coordinates (X,Y) on $\Delta_N$ - see here. I want to find an easy proof of the following fact: - -Let $p,q\in(\Delta_N\setminus D)$ be such that $\rho(p,\mathcal{O})=\rho(q,\mathcal{O})$, >$\rho(p,q)=2$ and $|X|(p)>|X|(q)$. Under those conditions $\mathbb{P}(p)<\mathbb{P}(q)$. - -(In the above $\rho$ is a standard "Manhattan" metric on $\Delta_N$) -I managed to prove this property. Yet, proof is very long, difficult and "ugly". I want to use above result in a physics article so I want it to be as simple and concise as possible. -A friend of mine suggested the following argument that is much simpler then that of mine (unfortunately it is not complete) : -Consider sites in the interior of $\Delta_{N}$ lying on the bisection $\mathcal{B}$ of a line segment connecting $p$ and $q$ (see here). We label these points as $b_{1},b_{2},\ldots,b_{k}$. - By definition of $p$ , all trajectories leading from $p$ to $\mathcal{O}$ , without touching $D$ , must touch one of b 's at one point. This is clearly not the case for trajectories that start from $q$ . Let $\mathbb{P}^{(FP)}(p\rightarrow b_{i})$ ($\mathbb{P}^{(FP)}(q\rightarrow b_{i})$ ) denotes the probability that a random walker that was initially in $p$ (respectively in $q$ ) will reach $b_{i}$ before reaching any other b 's or points laying on $D\cup \mathcal{O}$. Therefore one can write: -$\mathbb{P}(p)=\sum_{i=1}^{i=k}\mathbb{P}^{(FP)}(p\rightarrow b_{i})\mathbb{P}(b_{i})$ -$\mathbb{P}(q)>\sum_{i=1}^{i=k}\mathbb{P}^{(FP)}(q\rightarrow b_{i})\mathbb{P}(b_{i})$ -Each $\mathbb{P}^{(FP)}(p\rightarrow b_{i})$ ($ \mathbb{P}^{(FP)}(q\rightarrow b_{i}) $) is a sum of probabilities corresponding to different trajectories $\gamma(p\rightarrow b_{i})$ ($\gamma(q\rightarrow b_{i}) $) that connect $p$ ($q$ ) with $b_{i}$ without touching other b 's and and $D$. Probability of a given $\gamma(p\rightarrow b_{i})$ is a product of probabilities that correspond to choices that a random walker makes on its trajectory. For every $\gamma(p\rightarrow b_{i})$ of this type we can find $\tilde{\gamma}(q\rightarrow b_{i})$ - trajectory connecting $q$ with $b_{i}$ being a mirror reflection of $\gamma(p\rightarrow b_{i})$ with respect to the bisection $\mathcal{B}$ (see here). Yet, converse is not true - there are trajectories connecting $p$ and $b_{i}$ (without touching $D$, $\mathcal{O}$ or otther b's) that cannot be obtained in this way. As long as $\gamma(p\rightarrow b_{i})$ does not touch the "edge" of $\Delta_N$ (i.e. as long as all nodes on the path have degree 4) we have equality of probabilities that correspond to $\gamma(p\rightarrow b_{i})$ and $\gamma(q\rightarrow b_{i})$. Yet, this is not the case when $\gamma(p\rightarrow b_{i})$ touches the edge and for such points one encounters bigger transition probabilities (they are equal $\frac{1}{3}$) then for points laying on the mirror reflection of this trajectory (they are equal $\frac{1}{4}$). -Without this problem one clearly have the inequality desired by me. Unfortunately, so far, I was unable to handle this problem properly.. - -REPLY [12 votes]: Here's an argument based on coupling. -First, note that $\mathbb{P}$ does not change if we consider instead the random walk that is lazy along the edges of $\Delta$, moving in each direction with probability $1/4$, and staying in place with probability $1/4$. -Couple the random walks from $p$ and $q$ so that (initially) they move in the same direction at every step. Eventually one of the following happens: - -They reach a position where $X(p)=1$ and $X(q)=-1$. In this case obviously they have the same probability of reaching $0$ before $D$. -They reach $D$ (together). -There is a time at which $p$ is on the boundary and $q$ moves towards the boundary. In this step $p$ is lazy, so after the step $p$ and $q$ are two adjacent points along the boundary with $q$ nearer to $0$ then $p$. Thus it suffices to show that $\mathbb{P}$ is decreasing along the boundary when moving away from $0$. This is done by continuing the coupling in exactly the same way, and now $p$ can only reach $0$ after $q$. - -In short, the coupling is that $p,q$ move in the same direction until either - one reaches $0$ or $D$ or until -they become symmetric, in which case they preserve the symmetry henceforth, or one of them reaches $0$ or $D$. With this coupling, $q$ reaches $0$ no later than $p$, and $p$ reaches $D$ no later than $q$ does.<|endoftext|> -TITLE: Why should I believe the Mordell Conjecture? -QUESTION [84 upvotes]: It was Faltings who first proved in 1983 the Mordell conjecture, that a curve of genus 2 or more over a number field has only finitely many rational points. -I am interested to know why Mordell and others believed this statement in the first place. What intuition is there that the statement must hold? Without reference to any proof, why should the conjecture 'morally' be true? Supposing one had to give a colloquium (to a general mathematically literate audience) on this, how could one convince them without going into details of heights or étale cohomology? -Answers I'm not looking for will be of the form "You fool, because it's been proved already", or even "Read Faltings' proof". - -REPLY [29 votes]: Since summer is here, I will bore people a bit with my own ahistorical view -of Mordell. The main point is that Serge Lang's strategy, mentioned by several people, is -essentially correct. Nevertheless, we might add that -you should believe the Mordell conjecture because most local -principal bundles should not extend to global principal bundles. -I've spoken about this in one form or another many times, but -maybe this particular sentence hasn't received enough emphasis. So -I will explain it. -Jean-Benoit Bost has pointed out that this story is not entirely devoid of historical context, since Andre Weil's famous 1938 paper on vector bundles can be construed as giving motivation roughly of this nature. - -Recall that Lang's idea is to consider the diagram -$$ -\begin{array}{ccc} -X(F)&\longrightarrow & X(\mathbb{C}) \end{array}$$ -$$\begin{array}{lcr} -\downarrow \ \ \ \ & & \ \ \ \ \downarrow\end{array}$$ -$$\begin{array}{ccc} -J(F) &\longrightarrow & J(\mathbb{C})\\ -\end{array}$$ -for a smooth projective curve $X$ of genus $\geq 2$ with Jacobian -$J$ and an embedding $F\hookrightarrow \mathbb{C}$ of the number -field $F$ into the complex numbers. Lang suggested that -$$X(\mathbb{C})\cap J(F)\subset J(\mathbb{C})$$ -should be finite. -This is indeed plausible and turns out to be true after Faltings -proof. Of course we are considering its status as motivation -rather than corollary, following the original question posted. -In fact, the plausibility is strengthened when we replace the diagram above by a refinement -$$\begin{array}{ccc} -X(F)\ \ \ \ \ \ \ \ \ \ &\longrightarrow &\ \ \ \ \ \ \ X(F_v) \end{array}$$ -$$\begin{array}{ccc} -\downarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ & &\ \ \ \ \ \ \ \ \ \ \ \downarrow j \end{array}$$ -$$\begin{array}{ccc} -H^1_f(G_S, \pi_1(\bar{X}, b)) &\stackrel{loc}{\longrightarrow} & -H^1_f(G_v, \pi_1(\bar{X}, b)) -\end{array}$$ and try to prove that -$$Im(j)\cap Im(loc) \subset H^1_f(G_v, \pi_1(\bar{X}, b))$$ -is finite. Here, $$\pi_1(\bar{X}, b)$$ is some -$\mathbb{Q}_p$-algebraic fundamental group of $\bar{X}$ with -base-point $b\in X(F)$. The completion $F_v$ should be taken to -have degree one over $\mathbb{Q}_p$. The $H^1$'s are moduli spaces -of (locally constant) principal bundles for $\pi_1(\bar{X}, b)$, a -global one$$H^1_f(G_S, \pi_1(\bar{X}, b))$$ consisting of -principal bundles over some $Spec(O_F[1/S])$ - and a local one -$$ H^1_f(G_v, \pi_1(\bar{X}, b)) \simeq \mathbb{A}^N_{\mathbb{Q}_p},$$ -(almost naturally) isomorphic to affine space, consisting of -principal bundles on $Spec(F_v)$ (satisfying some technical -condition). -Thus, we are replacing $\mathbb{C}$ by a non-Archimedean -completion and the Jacobian (a moduli space of line bundles) by a -moduli space of principal bundles. The vertical maps assign to a -point $x$ the principal bundle of paths $$\pi_1(\bar{X};b,x).$$ -This framework turns out to refine considerably the intuition that -$J(F)$ and $X(\mathbb{C})$ have very different natures inside -$J(\mathbb{C})$. - -Now, why should the Mordell conjecture be true? -There are two steps. -A. The easy one: The map $j$, being a non-Archimedean period map, -is highly transcendental, and maps $X(F_v)$ to a Zariski-dense -compact analytic curve in $H^1_f(G_v, \pi_1(\bar{X}, b))$, which -therefore meets any proper subvariety in finitely many points. One -proves this by showing that certain transcendental functions on -$X(F_v)$ (the coordinates of the map) are algebraically -independent. Meanwhile, the localization map -$$ loc: H^1_f(G_S, \pi_1(\bar{X}, -b))\longrightarrow H^1_f(G_v, \pi_1(\bar{X}, b)),$$ is algebraic, -and hence, has constructible image in the Zariski topology. These -are the different natures alluded to in the previous paragraph: -$$\begin{array}{ccccc} -H^1_f(G_S, \pi_1(\bar{X},b)) &\stackrel{\scriptstyle -\mbox{algebraic}}{\longrightarrow} & H^1_f(G_v, \pi_1(\bar{X},b)) - & \stackrel{\scriptstyle -\mbox{dense analytic}}{\longleftarrow}& X(F_v)\end{array}$$ -Therefore, it suffices to show: -B. The hard step: $Im(loc)$ is not Zariski dense, that is, the -localization map - is not dominant. -Now, why should this be true? Well, the moduli space -$$ H^1_f(G_v, \pi_1(\bar{X}, b))$$ -consists of local principal bundles, while -$$ H^1_f(G_S, \pi_1(\bar{X}, b))$$ -is a moduli space of global principal bundles. So it makes sense -that most local bundles should not extend to global ones when the -group $\pi_1(\bar{X}, b)$ is sufficiently large and non-abelian. -(Perhaps you will disagree...) -These thoughts were actually inspired by Yang-Mills theory: We -have something like local solutions to the Yang-Mills equation, -that is, on a small annulus (or a handle-body) embedded in a -Riemann surface (or a three-manifold). It seems they should not -all extend to global solutions. Natural enough, but quite hard to -prove in general. Galois cohomology, on the whole, appears to be -harder than Yang-Mills theory. One motivation for posting this -answer is the hope that some bright young person will have an idea. - -Added: As mentioned by Matthew Emerton, the strategy outlined above is an extension of Chabuaty's method, which quite likely inspired Lang's conjecture as well (according to one reading of the notes to Fundamentals of Diophantine Geometry). There, the analogue of -$H^1_f(G_v, \pi_1(\bar{X},b))$ is -$$T_eJ(F_v),$$ -the Lie algebra of the Jacobian, while the role of the global moduli space is played by -$$J(F)\otimes_{\mathbb{Z}}F_v.$$ -My own feeling is that the 'local vs. global perspective' that emerges out of the principal bundle interpretation is somehow critical to understanding the Mordell conjecture, and constitutes a natural generalization not just of Chabauty's method, but of the arithmetic theory of curves of genus zero and one. In this sense, the essential motivation for Mordell's conjecture should not just be probabilistic, but something rather precise coming out of class field theory. It's fairly clear that this couldn't have been Mordell's reason for believing in it, but it is plausible, as mentioned above, that it was Weil's reason, in spite of his eventual non-committal assessment.<|endoftext|> -TITLE: Degenerations of smooth projective varieties -QUESTION [9 upvotes]: Vague question. Is there anything special about degenerations of smooth projective varieties (separating them from arbitrary projective schemes)? -Precise setup. Let $f:X\to Y$ be a projective flat morphism of algebraic schemes (say, over $\mathbb{C}$), -where $Y$ is an irreducible variety. Suppose that the generic fiber of $f$ is smooth and connected. Consider special fibers $X_y=f^{-1}(y)$, $y\in Y$. Of course, $X_y$ may be singular, reducible, and/or non-reduced. Zariski's main theorem implies $X_y$ is connected. -Can we say anything else? -Specific Questions. Can $X_y$ have embedded components? Can $X_y$ have non-constant regular functions? (Since $X_y$ is connected, such a regular function would -have to be nilpotent.) -Reformulation. Consider a Hilbert scheme of closed subschemes in ${\mathbb P}^N$ -(with fixed Hilbert polynomial). There is an open subset in it corresponding to smooth -connected projective varieties. The question concerns the closure of this set. -Remark. I ran across this in a concrete situation, where the goal is to understand $Rf_*O_X$. But it seems that the question is quite natural, so perhaps the corresponding statements or counterexamples are well known... Any comments would be helpful! - -REPLY [4 votes]: A partial answer: if $X$ is normal and $Y$ is smooth of dimension $1$, then $X_y$ only have constant regular functions ($X\to Y$ is cohomologically flat in relative dimension $0$). This is proved in Raynaud: Spécialisation du foncteur de Picard, Prop. 6.4.2 (use the characteristic 0 hypothesis here, otherwise it is false even when $X$ is also smooth.) It is also proved in the begining of ''Surfaces fibrées en courbes de genre deux'', Lecture Notes in Math. 1137 (1985) by Gang Xiao. -Add: and of course in this situation $X_y$ has no embedded point as $X$ is (S$_2$).<|endoftext|> -TITLE: Doing explicit computations with coordinate rings -QUESTION [8 upvotes]: Suppose that we are given an integral $k$-algebra $A$ of finite type explicitly, by which I mean that we are given the generators of the defining ideal $J$ where $A = k[x_1,...,x_n]/J$. What kinds of tools are out there to compute the integral closure of $A$? I would like the answer as explicitly as possible i.e. generators of the defining ideal. -While suggestions of computer programs are welcome, I want to be able to do these on my own, so I am looking for results which let me prove the answer. I'm asking from the perspective of someone who knows very little computational commutative algebra. -As a related question, if I have two such rings $A$, $B$ given explicitly as above, together with an explicit homomorphism between them, how can I go about determining the kernel and cokernel explicitly? Also, how about if we localize everything at a maximal ideal? -Broadly speaking, I would like to know about what kinds of computational methods are available for rings which arise from studying complex algebraic varieties. Have people out there settled these kinds of computations completely, or is this a hard question in general? - -REPLY [3 votes]: Have you looked at the Swanson-Huneke book on integral closure? Especially chapter 15 which is a discussion of various methods of computing integral closure (including Stolzenberg's method mentioned in Steven Landsberg's answer above). -It is available online HERE -Another recent algorithm is due to A. Singh and I. Swanson, Click HERE. This also has some history discussion, and it is apparently implemented as well. -Since you did ask for computer implementations, please see HERE, an algorithm implemented in Macaulay2 apparently based off T. de Jong's algorithm (mentioned in the sources above). -Finally, I should note that sometimes blowing-up the conductor helps, see section 7 of Greco-Traverso, ``On seminormal schemes''. -EDIT: With regards to your other questions finding kernels and cokernels of maps of rings, you should see a book on computational commutative algebra. For example, you could try the section of Eisenbud's book "Commutative Algebra with a view towards algebraic geometry'' on Groebner bases, another common source is ``Ideals, Varieties and Algorithms" by Cox, Litlle and O'Shea. Again, these things are also implemented in Macaulay2 amoung other places.<|endoftext|> -TITLE: Compactness of the Unit Ball in a Superreflexive Space -QUESTION [5 upvotes]: The unit ball is compact in the weak topology iff the space is reflexive. Is there an analogous topology under which the unit ball is compact iff the space is super-reflexive? -(I know a space is super-reflexive iff the unit ball is super weakly compact, but I'm not aware of a topology which makes super weak compactness equivalent to compactness in that topology.) - -REPLY [3 votes]: I am guessing the answer is No. Because if you take, for instance, the product of a family of $\ell_p$ spaces for $p>1$ and $p$'s tends to 1, then it won't be superreflexive anymore but the unit ball of the product under the sought topology would be compact. -Edit: This was a quick late night answer and it doesn't work. By product I had $\ell_2$ direct sum in mind but either way it doesn't solve the problem (see the comments below).<|endoftext|> -TITLE: L1 distance from a trigonometric susbspace -QUESTION [5 upvotes]: How to check, whether the $L^{1}$ distance between a finite exponential sum $S_{F}(x)=\sum\limits_{n\in F} \exp(inx)$ and the $L^{1}$-closure of subspace $\mathrm{span}\left(\exp(inx): n\in \mathbb{Z}\setminus F \right)$ is less or equal than $1$? -This problem I have found oryginally stated as follows: -For which subset $F\subset\mathbb{Z}$ there exists a function $f\in L^{1}$ such that $\widehat{f}(k)=1$ for $k\in F$ and $\| f \|_1=1$. -I have known some estimates of $L^{1} $ norm of $S_F$, however, such estimates gives an error which leads to many cases left. - -REPLY [2 votes]: I'm not sure about your version of the problem, but the following result of Helson from 1955 partially answers the original problem: -Claim: Let $\mu$ be a measure on $\mathbb{T}$ such that $\hat{\mu}(n)=0$ or $1$ for $n=0,\pm1,\pm2,\ldots.$ Then $\{n\in\mathbb{N}\colon\hat{\mu}(n)=1\}=(F_{1}\cup V)\setminus F_{2}$ with $F_{1},F_{2}$ finite sets and $V$ a periodic set, i.e. a finite sum of arithmetic progressions. -Proof: We argue by contradiction. Suppose the conclusion is false. Then without loss of generality, we can find two increasing sequences $n(t),m(t)$ such that $n(t),m(t)\to\infty$ as $t\to\infty$, $\hat{\mu}(n(t)-j)=\hat{\mu}(m(t)-j)$ for $j=1,\ldots,t-1$, but $\hat{\mu}(n(t))\neq\hat{\mu}(m(t))$. Write $\mu=fdm+\mu_{s}$ with $dm$ Lebesgue measure and $\mu_{s}$ singular. By our assumptions (ignoring trivial cases, e.g. when $\mu$ is a finite trigonometric polynomial) $\mu_{s}\neq0$. (If $\mu_{s}=0$, the Fourier coefficients would vanish at infinity by Riemann-Lebesgue.) Now, consider the measure -$$\nu_{t}=(e^{-im(t)}-e^{-in(t)})\mu_{s}$$ -As $t\to\infty$, we get a weak$^{*}$ limit $\nu\in L_{1}(|\mu_{s}|)$ by Banach-Alaoglu. Note that $\nu$ is a singular measure, since it is the weak$^{*}$ limit of singular measures. However, $\hat{\nu}(n)=0$ for $n<0$, so $\nu$ is absolutely continuous, by the F.-M. Riesz Theorem. This contradiction gives our desired result.<|endoftext|> -TITLE: Teichmuller volume of moduli space -QUESTION [8 upvotes]: Someone asked me this question, and I was embarrassed to not know the answer: is the volume of Moduli space with respect to the Teichmuller metric finite? The answer is "yes" when we replace Teichmuller metric with Weil-Petersson metric, but the geometry of the two spaces is quite different. - -REPLY [9 votes]: The answer is YES, the volume of the moduli space is finite with respect to the Teichmuller metric. -The reason is the theorem of Royden, that the Kobayashi metric on Teich(S) coincides with the Teichmuller metric, and the fact that the moduli space $M(S)$ associated to S has a nice compactification $\overline{M(S)}$, the Deligne-Mumford compactification. -The argument goes as follows: -For a stable curve Z in $\overline{M(S)}$ with k nodes, you can find a neighborhood U of Z such that U is locally $\Delta^n /G$, where $\Delta$ is the unit disc in $\mathbb{C}$ and $G$ is a finite group. Then $U\cap M(S)$ is locally isomorphic to $((\Delta^{*})^{k} \times \Delta^{n-k} )/ G$. -The volume of $(\Delta^{*})^{k} \times \Delta^{n-k} $ near the origin is finite in the Kobayashi metric. Since inclusions contract the Kobayashi metric it follows that there is a small neighborhood V of $Z \in \overline{M(S)}$ such that volume of $V\cap M(S)$ is finite. The result now follows by compactness of $\overline{M(S)}$. -You can look at Curt McMullen's paper : -http://www.math.harvard.edu/~ctm/papers/home/text/papers/kahler/kahler.pdf -for more details and references. (Proof of Theorem 8.1) - -REPLY [5 votes]: The answer is again yes. See the proof of theorem 8.1 in Curtis McMullen's paper "The moduli space of Riemann surfaces is Kahler hyperbolic".<|endoftext|> -TITLE: Understanding the countable ordinals up to $\epsilon_{0}$ -QUESTION [37 upvotes]: in a recent MO question, link, discussing the current foundations of mathematics, the author linked a video lecture by Prof. Voevodsky, which argues against the principle of $\epsilon_{0}$-induction used in Gentzen's proof of the consistency of PA. -In discussions arising from the question, some people commented that imagining an infinite descending chain in $\epsilon_{0}$ is "crazy". -I would like to understand better this ordinal, since I actually don't know exactly how to depict it in my mind. -I have clear in my mind the order associated with the finite ordinals. I use in my mind a notation of the following kind: -$1 = I$ -$2= II$ -$3= III$ -$4= IIII$ -$\omega = (III\dots)$ -$\omega+1= (III\dots)I$ -$\omega +2 = (III\dots)II$ -$\omega + \omega= \omega \cdot 2= (III\dots)(III\dots)$ -In general I understand $\alpha + \beta$ as the juxtaposition of the two representations. -$\omega\cdot 3 = (III\dots)(III\dots)(III\dots)$ -$\omega\cdot \omega = \omega^{2} = \big( (III\dots)(III\dots)(III\dots)\dots\big)$ -In general I understand $\alpha \cdot \beta$, by replacing each $I$ symbol in $\beta$ with the representation of $\alpha$. So -$\omega^{3}=\omega^{2}\cdot \omega = \big( \omega^{2} \omega^{2} \omega^{2} \dots \big)$ -This allows me to visualize every ordinal of the form $\omega^{n}\cdot m + k$, with $n,m,k$ naturals (i.e finite ordinals). So far I have absolutely no doubt that there are no infinite descending chain in ordinals of the form $\omega^{n}\cdot m + k$. -However I start having problem with the ordinal $\omega^{\omega}= \bigsqcup_{n<\omega}\omega^{n}$. Do you have any idea on how to visualize $\omega^{\omega}$ is a way consistent with the representation used above (which i actually found here) ? -Anyway, looking at wikipedia, I still manage to visualize $\omega^{\omega}$ as the set of infinite strings of natural number, having only finitely many digits different from $0$. -Still I have no doubt that there are no infinite descending chain in $\omega^{\omega}$. -Perhaps i might be able to understand $\omega^{\omega^{\omega}}$, namely the set of infinite strings labeled with elements of $\omega^{\omega}$, having only finitely many elements different from $0$. -Or (i guess) equivalently a $\omega\times\omega$ square labeled with naturals, where only finitely many columns are different from $0^{\omega}$, and all of these non constant-$0$ columns, contains only finitely many digits different from $0$. -However I do not know how to visualize $\epsilon_{0}$. I mean I know that the elements of $\epsilon_{0}$ can be represented by finite-branching finite trees labeled with natural numbers, but that doesn't give me a strong intuition about the fact that no infinite chain exists, so I guess its not a great picture (or at least I do not understand it properly, yet). -Questions -A) Could you suggest a way to visualize $\omega^{\omega^{\omega}}$? It should be in such a way to convince me about the fact that there are no infinite down-chain. -B) Could you suggest a way to visualize $\epsilon_{0}$, again arguing that it should be very clear that there are no infinite down-chain. -C) Could you please state your opinion about Prof. Voevodsky, which argues against the principle of $\epsilon_{0}$-induction used in Gentzen's? -This shouldn't be a duplicate of the previous, wider thread link, I'm only interested in this little bit of Voevodsky's talk. -Thank you in advance, -bye -matteo - -REPLY [5 votes]: In the course of writing an expository article on the consistency of arithmetic, I was led to try to explain $\epsilon_0$ in as "finitary a manner" as possible. Here is what I came up with, inspired greatly by the account in Torkel Franzen's book Inexhaustibility: A Non-Exhaustive Account, as well as the LISP programming language. -Define a -list -to be either an empty sequence—denoted -by ( ) and referred to as the empty list—or, recursively, a -finite nonempty sequence of lists. If we write a list by separating the -constituent elements by commas and enclosing the entire sequence with parentheses, -then for example, -(( ), ( ), ( )) and ((( ), ( )), ((( ), (( ), ( ))), ( ))) are lists. -The number of constituent lists is called its length -(it is -zero for the empty list). If -$a$ -is a nonempty list, then we -write $a[i]$ for the -$i$th constituent list of $a$, - where $i$ ranges -from 1 to the length of $a$. -Next, recursively define a total ordering $\le$ -on lists as -follows (it is essentially a lexicographic ordering). Let -$a$ and $b$ -be lists, with lengths -$m$ and $n$ -respectively. If $m\le n$ and $a[i] = b[i]$ for all $1\le i \le m$ -(this condition is vacuously -satisfied if - $m=0$), then $a\le b$. Otherwise, there exists some $i$ -such that -$a[i] \ne b[i]$; let $i_0$ -be the least such number, and -declare $a< b$ if $a[i_0] < b[i_0]$. -Finally, recursively define a list $a$ -to be an -ordinal (or more precisely, an ordinal below $\epsilon_0$, but I will just say ordinal for short) -if all -its constituent lists are ordinals and $a[i] \ge a[j]$ whenever $i -TITLE: Geometric interpretation of $BN$-pairs -QUESTION [5 upvotes]: My question is relative to a geometric interpretation of the $BN$-pairs that arise in Tits' theory of buildings. Here is a definition that comes from an article by G. Stroth (Nonspherical spheres). -$[\ldots]$ -Let $\mathcal{P} = \{P_1, \ldots, P_n\}$ be a minimal parabolic system for a group $G$, $B=P_1 \cap \ldots \cap P_n$ the Borel subgroup. A subgroup $N$ of $G$ is called a Weyl group for $\mathcal{P}$ iff -1) $N= \langle x_1, \ldots, x_n \rangle, x_i \in P_i-B, x_i^2 \in B$. -2) $B \cap N$ is a normal subgroup of $N$. -3) $N \cap P_i = (B \cap N) \langle x_i \rangle , i = 1, \ldots, n$. -$[\ldots]$ -If additionally we have -(A) $G = BNB$ and -(B) $BgBhB \subset (BgB) \cup (BghB)$ for all $g,h \in N$ -then we have a $BN$-pair. -Geometrically, the Weyl group $N$ is the stabilizer of an apartment $\Delta$ of the geometry $\Gamma$ defined from $\mathcal{P}$, and $B$ is the stabilizer of a chamber of $\Delta$. I am trying to get a clear geometric view of these objects. -Here is my question. What could be a geometric interpretation of condition (B)? - -REPLY [9 votes]: As a further commentary beyond what Ben has said, I'd emphasize that Stroth (along with Ronan and other finite group theorists) has modified some of the BN-pair and building formalism introduced by Tits. My understanding is that this is done partly in an attempt to unify the study of sporadic simple groups and simple groups of Lie type within the kind of "geometric" setting formulated first for the latter groups. -In any case, the broader notion of "parabolic system" in a group as used here is motivated by the earlier Lie structure but requires some experimentation with additional axioms beyond what Tits did. As Ben points out, the condition (B) you quote from that 1990 Durham conference article by Stroth goes beyond the conventional BN-pair axiom. In that conventional setting, which is close to the geometry of buildings and apartments, the Weyl group and its length function play a vital role in talking about distances in the geometry, etc. This is partly encoded in the usual version of condition (B). -Whether or not the structures studied by Stroth really add "geometric" flavor to the finite groups of interest is more than I can judge, but this does get outside the conventional framework of buildings with finite Weyl groups. By the way, it can do no harm to digest some of the original Tits thinking about the subject formulated as a detailed series of exercises for Section 2 of Chapter IV in the 1968 Bourbaki Chapters IV-VI of Groupes et algebres de Lie (later published in English translation by Springer). Naturally much of this shows up in his own Springer lecture notes on the "spherical" case as well as in later books on buildings.<|endoftext|> -TITLE: Vanishing cycles in a nutshell? -QUESTION [29 upvotes]: To quote one source among many, "the general reference for vanishing cycles is [SGA 7] XIII and XV". Is there a more direct way to learn the main principles of this theory (i.e. without the language of derived categories), in particular as it applies to the study of certain integral models of curves? - -REPLY [16 votes]: Another good summary of vanishing cycles and Lefschetz pencils is found in Sections 4 and 5 of P. Deligne's "La conjecture de Weil, I" (available online). He discusses briefly both the theory over the complex numbers and the version in étale cohomology; and then immediately proceeds to putting these to pretty good use.<|endoftext|> -TITLE: Amenable exponential growth -QUESTION [7 upvotes]: Dear forum members, -Does anyone have a clear example of an amenable group with exponential growth? -Is real that if G is virtually amenable (has an amenable subgroup of finite index) then it is amenable? -I am sort of novice in advanced group theory. All your comments are more than welcome. -Many thanks - -REPLY [3 votes]: The lampligher group is a nice example, see e.g. the answer to this question by Jeremy Voltz, in which he wrote: - -The lamplighter group, defined as the wreath product $\mathbb Z/2\mathbb Z\wr \mathbb Z$, is amenable yet has exponential growth. It can be thought of as a bi-infinite sequence of street lamps, each of which can be turned on and off, and a lamplighter who moves along the sequence. The three generators of the group are to move the lamplighter right or left, and to switch the state of the lamp he is positioned in front of. With this picture in mind, it is easy to show the group has exponential growth.<|endoftext|> -TITLE: Lower bound of the number of relatively primes(each-other) in an interval -QUESTION [7 upvotes]: I am trying to find lower and upper bounds for the number of integers that are coprime in pairs in an interval of length n. - -What are the best bounds that we have? -Is that true that in any interval of length $n$ there is a set with at least $π(n)$ integers that are relatively prime to each other? Here $π(n)$ is the number of primes less or equal to $n$. - -REPLY [4 votes]: Let's turn the question around and let $f(n)$ be the number of consecutive integers required to guarantee that $n$ of them are pairwise relatively prime. E.g., $f(4)=6$ because you can find a set of 5 consecutive integers no 4 of which are pairwise relatively prime ($\lbrace2,3,4,5,6\rbrace$ will do) but given any set of 6 consecutive integers there must be 4 that are relatively prime (the three odd ones are pairwise relatively prime, and there will be an even that's not a multiple of 3 or 5, and it will be relatively prime to each of the odds). -I think that $f(n)=h(n-1)$, where $h(n)$ is (based on) the Jacobsthal function: $h(n)$ is the number of consecutive integers required to guarantee that one of them will not be a multiple of any of the first $n$ primes. E.g., $h(3)=6$ because you can find a set of 5 consecutive integers each of which is divisible by (at least) one of 2, 3, or 5 ($\lbrace2,3,4,5,6\rbrace$ will do) but given 6 consecutive integers only 3 can be even, and of the three odds, only one can be a multiple of 3, and only one can be a multiple of 5. -Now, why should $f(n)=h(n-1)$? Well, if you have $n$ pairwise coprime integers, it must be the case that (at least) one is not divisible by any of the first $n-1$ primes, for if each of your $n$ integers is divisible by one (or more) of the first $n-1$ primes, then two of them must be divisible by the same prime, hence, not relatively prime. Thus, $f(n)\ge h(n-1)$. I can't quite see my way through a proof that $h(n)\ge f(n+1)$, so there's still some work to be done here. -Anyway, the point is, lots of work has been done on the Jacobsthal function, including estimates. A good reference is Thomas R Hagedorn, Computation of Jacobsthal's function $h(n)$ for $n\lt50$, Math Comp 78 (2009) 1073-1087. As the title indicates, the paper is mostly concerned with computing Jacobsthal's function, but the author does summarize the history and gives many references.<|endoftext|> -TITLE: First Chern class of a flat line bundle -QUESTION [20 upvotes]: A referee asked me to include a reference or proof for the following classical fact. It's not hard to prove, but I'd prefer to just give a reference -- does anyone know one? -Let $X$ be a nice space (eg a smooth manifold, or more generally a CW complex). The topological Picard group $Pic(X)$ is the set of isomorphism classes of $1$-dimensional complex vector bundles on $X$. The set $Pic(X)$ is an abelian group with group operation the fiberwise tensor product, and the first Chern class map -$$c_1 : Pic(X) \longrightarrow H^2(X;\mathbb{Z})$$ -is an isomorphism of abelian groups. -Now make the assumption that $H_1(X;\mathbb{Z})$ is a finite abelian group. One nice construction of elements of $Pic(X)$ is as follows. Consider $\phi \in Hom(H_1(X;\mathbb{Z}),\mathbb{Q}/\mathbb{Z})$. Let $\tilde{X}$ be the universal cover, so $\pi_1(X)$ acts on $\tilde{X}$ and $X = \tilde{X} / \pi_1(X)$. Let $\psi : \pi_1(X) \rightarrow \mathbb{Q}/\mathbb{Z}$ be the composition of $\phi$ with the natural map $\pi_1(X) \rightarrow H_1(X;\mathbb{Z})$. Define an action of $\pi_1(X)$ on $\tilde{X} \times \mathbb{C}$ by the formula -$$g(p,z) = (g(p),e^{2 \pi i \psi(g)}z) \quad \quad \text{for $g \in \pi_1(X)$ and $(p,z) \in \tilde{X} \times \mathbb{C}$}.$$ -Observe that this makes sense since $\psi(g) \in \mathbb{Q} /\mathbb{Z}$. Define $E_\phi = (\tilde{X} \times \mathbb{C}) / \pi_1(X)$. The projection onto the first factor induces a map $E_{\phi} \rightarrow X$ which is easily seen to be a complex line bundle. The line bundle $E_{\phi}$ is known as the flat line bundle on $X$ with monodromy $\phi$. -Now, the universal coefficient theorem says that we have a short exact sequence -$$0 \longrightarrow Ext(H_1(X;\mathbb{Z}),\mathbb{Z}) \longrightarrow H^2(X;\mathbb{Z}) \longrightarrow Hom(H_2(X;\mathbb{Z}),\mathbb{Z}) \longrightarrow 0.$$ -Since $H_1(X;\mathbb{Z})$ is a finite abelian group, there is a natural isomorphism $\rho : Hom(H_1(X;\mathbb{Z}),\mathbb{Q}/\mathbb{Z}) \rightarrow Ext(H_1(X;\mathbb{Z}),\mathbb{Z}) $. We can finally state the fact for which I am looking for a reference : -$$c_1(E_{\phi}) = \rho(\phi).$$ - -REPLY [10 votes]: I noticed that someone voted this up today. Since this might indicate that someone else is interested in the answer, I thought I'd remark that Oscar Randal-Williams and I worked out a proof of this when I visited him earlier this year. A version of this proof can be found in Section 2.2 of my paper -The Picard group of the moduli space of curves with level structures, -to appear in Duke Math. J. -which is available on my webpage here. -(marked community wiki since it feels weird to get reputation for answering my own question)<|endoftext|> -TITLE: Fermat's Last Theorem in the cyclotomic integers. -QUESTION [26 upvotes]: Kummer proved that there are no non-trivial solutions to the Fermat equation FLT(n): $x^n + y^n = z^n$ with $n > 2$ natural and $x,y,z$ elements of a regular cyclotomic ring of integers $K$. -I am looking for non-trivial solutions to the Fermat equation FLT(p) in the cyclotomic integer ring $\mathbb{Z}[\zeta_{p}]$ for irregular primes p or any information about how the solutions must be (as a step toward constructing them). -George Lowther pointed out in an earlier discussion that by Kolyvagin's criterion any solution in $\mathbb{Z}[\zeta_{37}]$ must be in the second case. - -REPLY [2 votes]: Very late response but since it is still unresolved, I will answer your question. By Tauno's answer, any solution must belong to the second case, which in $\mathbb{Z}[\zeta]$ looks like $x^p+y^p=z^p$ with $1-\zeta \mid z$ and $x,y$ both coprime with $1-\zeta$. The second case also has no solutions in this ring by another criteria of Kolyvagin (from a different paper: " Fermat Equations over Cyclotomic Fields"). - The general criteria is a bit involved to write up here but the prime $p=37$ satisfies a simpler criteria (which applies to both the first and second case): 1) If the index of irregularity $=1$ with $p \mid B_i$ and 2) there is a prime $l \equiv 1 (\text{mod} \ p)$ for which $x^p+y^p=z^p$ has only trivial solutions modulo $l$ and $(1-\zeta(l))^{(l-3)}$, $(1-\zeta(l))^{(i)}$ are not $p$-th powers modulo $l$, then the Fermat equation has no solutions in $\mathbb{Q}(\zeta_p)$. Here $\zeta(l)$ is a primitive $p$-th root of unity modulo $l$ (which exists since $p \mid (l-1)$) and $a^{(j)}$ is the $j$-th component of $a \in \mathbb{F}_l^{\times}/(\mathbb{F}_{l}^{\times})^p$ relative to the Eigenspace decomposition of this group for the operators $\varepsilon_j = -\sum_{a=1}^{p-1}a^j\sigma_{a}^{-1}$. -Since the index of irregularity for $37$ is $1$ with $37 \mid B_{32}$ and $x^{37}+y^{37}=z^{37}$ has only trivial solutions modulo $l=149=1+4\times 37$, then by taking $\zeta(149)=16$, we compute that $(1-16)^{(34)}$ and $(1-16)^{(32)}$ are not $37$-th powers modulo $149$. Therefore $p=37$ satisfies the criteria.<|endoftext|> -TITLE: Are representations of a linearly reductive group discretely parameterized? -QUESTION [12 upvotes]: Suppose $G$ is a linearly reductive group over a field (say $\mathbb C$). Does somebody know of a proof that any flat family of finite-dimensional representations of $G$ must be locally constant? -In other words, I want to prove that if $A$ is a commutative $\mathbb C$-algebra (without idempotents) and $\rho:G\to GL_n(A)$ is an algebraic group homomorphism (roughly, a family of representations parameterized by $Spec(A)$), then after conjugating by some element of $GL_n(A)$, the image of $\rho$ is actually contained in $GL_n(\mathbb C)\subset GL_n(A)$. - -Remark 1: A finite-dimensional representation of $G$ is completely determined by the dimensions of its highest weight spaces. For a long time I thought this "discrete" parameterization somehow proved the result, but it doesn't. For example, nilpotent matrices are "discretely" parameterized by Jordan type, but it's obviously possible to have a flat family of nilpotent matrices in which Jordan type jumps, like $\begin{pmatrix}0&t\\ 0& 0\end{pmatrix}$ over the affine line with coordinate $t$. -Remark 2: Another reason I thought this was clear is that deformations of a representation $V$ are controlled by cohomology groups $H^{>0}(G,V\otimes V^*)$, which all vanish when $G$ is linearly reductive. This implies that any formal family—any flat family over an Artin ring—of representations has to be constant. However, this isn't enough to say that any family has to be locally constant. For example, deformations of $G$, as a group, are controlled by $H^{>0}(G,Ad)$, which vanish when $G$ is linearly reductive. So any formal family of linearly reductive groups has to be constant, but it is possible to have a flat family of linearly reductive groups which is not locally constant. Specifically, the affine line with a doubled origin is a flat group over the affine line. The fiber over the origin is $\mathbb Z/2$, but all the other fibers are trivial. - -REPLY [11 votes]: I don't think this is literally true. For example, suppose that $G$ is a finite cyclic group of order 2 generated by $s$, suppose that $L$ is an non-trivial 2-torsion invertible $A$-module over a Dedekind ring $A$. Then $L^{\oplus 2}$ is isomorphic to $A^{\oplus 2}$; you can let $G$ act on $L^{\oplus 2}$ with the rule $s(a,b) = (a, -b)$. This gives a family of representations that does not come from $\mathbb C$. You can give a similar example in which $G$ is a torus. -On the other hand, what you want is true Zariski-locally; would that be enough for your needs? -[Edit] Let me add a proof that the statement is Zariski-locally true. The key point is the following: given a locally free sheaf on a scheme $X$ with an action of $G$, the subsheaf of invariants is also locally free. This is easy: if $X = \mathop{\rm Spec} A$ and $E$ corresponds to a projectve $A$-module $M$, then $M^G$ is a direct summand of $M = M^G \oplus M_G$, because $G$ is linearly reductive (and $M$ is a locally finite representation of $G$, this is standard); and $M_G$ is also an $A$-submodule, so $M^G$ is a projective $A$-module. Now, let $E$ be a locally free sheaf with an action of $G$. Let $p$ be a rational point of $X$, and $V$ be the representation of $G$ appearing as the fiber $E(p)$ of $E$ at $p$. Consider the locally free sheaf $Hom_{\mathcal O_X}(V \otimes \mathcal O_X, E)$; the sheaf of invariants, that is, the sheaf of $G$-equivariant homomorphisms $V \otimes \mathcal O_X \to E$, is locally free. At the point $p$ we have a section of its fiber, given by the tautological isomorphism $V \simeq E(p)$; since the sheaf is locally free, this extends to a section in a neighborhood. By further restricting the neighborhood, this sectioin will be an isomorphism, and this concludes the proof.<|endoftext|> -TITLE: Status of Ihara's lemma for Shimura curves over totally real fields? -QUESTION [7 upvotes]: What is the status of Ihara's lemma for Shimura curves over totally real fields? -In particular, why is it not implicit in the exact sequence of Rajaei, "On the levels of mod $l$ Hilbert modular forms" (Crelle 2001), Theorem 3 (3.18)? -To be a little bit more precise (or at least to give the main idea), let $F$ be a totally real field of degree $d$. Let $M(\mathfrak{M}^+, \mathfrak{M}^{-})$ be the Shimura curve of level $\mathfrak{M}^+ \subset \mathcal{O}_F$ associated to the indefinite quaternion algebra of discriminant $\mathfrak{M}^{-} \subset \mathcal{O}_F$, and $M(\mathfrak{m};\mathfrak{M}^+, \mathfrak{M}^{-})$ the Shimura curve with maximal level at primes dividing $\mathfrak{m} \subset \mathcal{O}_F$. Fix two coprime ideals $\mathfrak{N}^+, \mathfrak{N}^{-} \subset \mathcal{O}_F$ such that $\mathfrak{N}^{-}$ is the squarefree product of a number of primes ideals congruent to $d \mod 2$. Suppose that -$\mathfrak{N}^{-}$ has at least one prime divisor $\mathfrak{q}$ say. Fix a prime $v \subset \mathcal{O}_F$ that does not divide $\mathfrak{N}^+\mathfrak{N}^{-}$. Let ${\bf{M}}(\mathfrak{q}; \mathfrak{N}^+, \mathfrak{N}/\mathfrak{q})$ and ${\bf{M}}(v\mathfrak{q}; \mathfrak{N}^+, \mathfrak{N}^{-}/\mathfrak{q})$ be the good reduction integral models over $\mathcal{O}_{(\mathfrak{q})}$ -of $\bf{M}(\mathfrak{q}; \mathfrak{N}^+, \mathfrak{N}/\mathfrak{q})$ and $M(v\mathfrak{q}; \mathfrak{N}^+, \mathfrak{N}^{-}/\mathfrak{q})$ respectively. -Ihara's lemma for Shimura curves over totally real fields, at least as I understand it, is the assertion that for any non-Eisenstein maximal ideal -$\mathfrak{m}$ in the algebra $\mathbb{T}(v\mathfrak{q}; \mathfrak{N}^+, \mathfrak{N}^{-})$ generated by Hecke operators acting on -$M(v\mathfrak{q}; \mathfrak{N}^+, \mathfrak{N}^{-}/\mathfrak{q})$, the map $$\begin{align*} H^1({\bf{M}}(\mathfrak{q}; \mathfrak{N}^+, -\mathfrak{N}/\mathfrak{q}), \mathcal{F})_{\mathfrak{m}}^{2} & -\longrightarrow H^1({\bf{M}}(v\mathfrak{q}; \mathfrak{N}^+, \mathfrak{N}^{-}/\mathfrak{q}), \mathcal{F})_{\mathfrak{m}} \\ -(f_1, f_2) & -\longmapsto 1_* f_1 +\eta_{v, *} f_2 \end{align*}$$ -is injective, where $\eta_v = \left( \begin{array}{cccc} 1 & 0 \\ 0 & \pi_v \end{array}\right)$. Here, $\mathcal{F}$ is the usual sheaf defined by Carayol and Jarvis, and $\pi_v$ is a fixed uniformizer at $v$. Rajaei shows that there is an injection of the associated dual character groups, -$$\begin{align*} \widehat{\mathcal{X}}_{\mathfrak{q}}(\mathfrak{q}; \mathfrak{N}^+, -\mathfrak{N}/\mathfrak{q}))_{\mathfrak{m}}^{2} &\longrightarrow \widehat{\mathcal{X}}_{\mathfrak{q}}(v\mathfrak{q}; \mathfrak{N}^+, \mathfrak{N}^{-}/\mathfrak{q})_{\mathfrak{m}} \\(f_1, f_2) &\longmapsto 1_* f_1 +\eta_{v, *} f_2 .\end{align*}$$ Sorry if the question is naive! It is related to a previous question I asked here (Soft proof of multiplicity one for character groups of Shimura curves?), to the effect of whether or not proofs of these types of results can be streamlined for the case of parallel weight $2$. In particular, consider the Ribet/Rajaei exact sequence \begin{align*} \mathcal{X}_{v_2}(\mathfrak{N}^+, -\mathfrak{p}v_1 v_2 \mathfrak{N}^{-}) \longrightarrow \mathcal{X}_{\mathfrak{p}}(\mathfrak{p}v_2; \mathfrak{N}^+; v_1 \mathfrak{N}^{-}) -\longrightarrow \mathcal{X}_{\mathfrak{p}}(\mathfrak{p}; \mathfrak{N}^+, v_1 \mathfrak{N}^{-})^2.\end{align*} Here, in the notations of the previous -question, $v_1, v_2, \mathfrak{p} \subset \mathcal{O}_F$ are distinct primes that do not divide the product $\mathfrak{N}^+ \mathfrak{N}^{-}$. Rajaei -shows that the localization of the natural map $$\begin{align*} -\widehat{\mathcal{X}}_{\mathfrak{p}}(v_2; \mathfrak{N}^+, v_1 \mathfrak{N}^{-})^2 &\longrightarrow -\widehat{\mathcal{X}}_{\mathfrak{p}}(\mathfrak{p}v_2; \mathfrak{N}^+, v_1 \mathfrak{N}^{-}) -\end{align*}$$ given by $1_* \oplus {\eta_{\mathfrak{p}}}_*$ at any non-Eisenstein maximal ideal of the Hecke algebra $\mathbb{T}(\mathfrak{p}v_2; \mathfrak{N}^+, v_1\mathfrak{N}^{-})$ is injective. On the other hand, given the diagram in the previous question (so beautifully compiled by Dror Speiser), we have identifications -$$\begin{align*} -\mathcal{X}_{\mathfrak{p}}(\mathfrak{p}v_2; \mathfrak{N}^+, v_1 \mathfrak{N}^{-}) &\cong -\operatorname{Div}^0 \left({\bf{M}}(\mathfrak{p}v_2; \mathfrak{N}^+, v_1 \mathfrak{N}^{-})^{ss} \otimes \kappa_{\mathfrak{p}} \right) \\ -\mathcal{X}_{\mathfrak{p}}(\mathfrak{p}; \mathfrak{N}^+, v_1 \mathfrak{N}^{-}) &\cong -\operatorname{Div}^0 \left({\bf{M}}(\mathfrak{p}; \mathfrak{N}^+, v_1 \mathfrak{N}^{-})^{ss} \otimes \kappa_{\mathfrak{p}} \right) -\end{align*}$$ Can we not then just take ${\bf{Z}}$-duals to deduce the result? This is the naive question. - -REPLY [3 votes]: Okay, so the answer to the naive question is most likely no for several reasons. For instance, the identifications explained in the second paragraph show that the injection in Rajaei is really an analogue of Ihara's lemma for totally definite quaternion algebras (via Carayol's description of supersingular points). As well, the cocharacter groups only give information about the p-new automorphic forms. In general, it seems very unlikely that Ihara's lemma for Shimura curves over totally real fields can ever be reduced to the totally definite case, at least via naive methods. As for the status of Ihara's lemma itself, there does appear to be at least one work towards a generalization to totally real fields of the method of Diamond and Taylor, for instance in the preprint of Chuangxun Cheng (Ph.D. student of Emerton) ...<|endoftext|> -TITLE: Spaces with a quasi triangle inequality -QUESTION [8 upvotes]: How do you call a space with a function which is symmetric, non negative, positive definite and which satisfies a quasi-triangle inequality: -$d(x,z) \leq C( d(x,y)+d(y,z) )$ -for all $x,y,z$ and some $C > 1$? -That is, it satisfies all the axioms of a metric space except for the triangle inequality, which is replaced by the one above. -Can anyone provide any reference on these spaces? -Thanks. - -REPLY [5 votes]: Here is a negative answer for your additional remark concerning Banach's fixed point theorem: Consider $d(x,y)=(\int_0^1|x-y|^p)^{1/p},$ $0 -TITLE: Question about the dimension of a Contact (Symplectic) manifold -QUESTION [5 upvotes]: I am reading about contact geometry and I have a question: Why do we only consider contact structure of an odd-dimension manifold? and the same question for definition of symplectic geometry? -I think for the contact geometry case, a reason is that we want $\alpha \wedge (d\alpha)^n$ to be a volume form. Am I right? I am not sure about that. -Thank you so much for your help. -P.S there is no tag for Contact-geometry. - -REPLY [6 votes]: And by popular request, here's my comment as an answer :) -Your guess is correct. Contact structures are structures associated to a one-form $\alpha$ with maximal rank. There are two cases: - -for odd rank, you want $\alpha∧(d\alpha)^k$ to be nowhere vanishing for the largest possible $k$ allowed by dimension, and -for even rank, you want the same condition on $(d\alpha)^k$. - -In the former case you have a contact structure and in the latter an exact symplectic structure. -More generally, symplectic forms are nondegenerate by definition. You can understand nondegeneracy of a 2-form $\omega$ pointwise, where it turns into the statement that an antisymmetric matrix has nonzero determinant. This can only happen if the dimension is even. -I'm assuming finite-dimensionality throughout. There is a reasonably well-developed theory of infinite-dimensional symplectic manifolds and presumably also of contact manifolds.<|endoftext|> -TITLE: Extension of the formality theorem? -QUESTION [6 upvotes]: The following question came up in a discussion the other day and I have been wondering whether something is known about it. Everything below takes place over $\mathbb{C}$. I don't have the expertise to know if this is trivial or of interest. Suppose a commutative dga has a free-commutative model $(\wedge V , d)$ where V is a finite dimensional vector space. -Recall that $T^{poly}$ is the Lie-algebra of polyvector fields on $\wedge V$ (yes, everything is superized as V will be in general graded) with Schouten bracket. Part of Kontsevich's formality theorem says that the HKR map $ T^{poly} \to HC^*$(Hochschild cochains) is the first Taylor coefficient in an $L_\infty$ quasi-isomorphism between the two. -We can think of the derivation $d$ as corresponding to a vector-field $v$. It follows from a spectral sequence argument that the HKR map gives a quasi-isomorphism: -$$ (T^{poly},[v,-]) \to HC( \wedge V,d)$$ -Question: Can this map be upgraded to a map of $L_\infty$ algebras? -Certainly, the Taylor coefficients in the usual formality map must be doctored. -A related statement that does seem to be true and standard is that there is an $L_\infty$ quasi-isomorphism $(T^{poly}[[t]],[tv,-]) \to HC^*(\wedge V[[t]],td )$ Thus, the question is in some reasonable sense about convergence of this isomorphism. Maybe one can prove the claim by a close inspection of Kontsevich's integral formulas. Based upon these facts, however, it seems plausible to me that that the statement is in general false, but I was unable to come up with a counterexamples or an a priori reason (I didn't try too hard however). Is it true for some more restrictive group of commutative dg algebras, for example pure Sullivan algebras? -Update: Having finally looked at the Kontsevich formulas, I'm beginning to think there are some simple counting reasons that make the above formula converge, but am not sure that $f_1$ stays the same (though I believe it remains a quasi-iso). Any confirmation or help would be great. Otherwise, I'll keep thinking and update again. - -REPLY [2 votes]: I think this does only work for so-called homological vector fields, i.e. vector fields of degree 1 which self-commute. Then you have $[v,v]=0$, which is the Maurer-Cartan equation in $T_{poly}$, and which insures that the corresponding derivative $d$ squares to zero. -Dealing with a Maurer-Cartan element you can simply use it to twist Kontsevich's formality $L_\infty$-quasi-isomorphism. When the Maurer-Cartan element is a vector field we know the explicit form of the first Taylor component of the twisted $L_\infty$-morphism: it is not HKR, but involve Bernoulli numbers (see e.g. https://www.math.ethz.ch/u/calaqued/research/LecturesDufloETH.pdf). -Anyway, the $k$-th taylor component of the $v$-twisted $L_\infty$-morphism will be given by the series -$$ -\phi_v^{(k)}(u_1,\dots,u_k):=\sum_{l\geq0}\frac{t^{l+1}}{(l+1)!}\phi^{(k+l)}(u_1,\dots,u_k,\underbrace{v,\dots,v}_{l~times}) -$$ -To conclude one just have to observe that $\phi^{(k)}$ preserves the grading given by the arity minus $2$ (arity means the number of arguments for poly-vectors and poly-differential operators). Therefore each time $v$ appears in the formula it decreases this degree by $1$. -EDIT: I include a comment into the main answer about the shape of $\phi^{(1)}_v$ in specific cases. Let me denote coordinates by $u^k$, and write $\partial_k=\frac{\partial}{\partial u^k}$ and $v=\sum_iv^i\partial_i$. We consider a matrix-valued one-form $\Xi$ given by -$$ -\Xi_i^j=\sum_k\partial_k\partial_iv^jdu^k, -$$ -and define -$$ -\Theta=\sum_{n>0}c_n\iota_{tr(\Xi^n)}. -$$ -Here $c_n$ are rational coefficients that do not matter. -Then $\phi_v^{(1)}$ is given by the precompsition of HKR with $e^{\Theta}$. -Now if I split the coordinates into even $x^i$ and odd $e^i$, and if I assume that -$v=\sum_iv^i(x^1,\dots,x^m)\frac{\partial}{\partial e^i}$, then the matrix $\Xi$ has non-zero entries only in the right-up block. In particular it is upper triangular, the trace of any power of it is therefore zero, and thus $\phi^{(1)}_v=HKR$.<|endoftext|> -TITLE: Witten's QFT and Jones Poly paper -QUESTION [10 upvotes]: Data: $M$ is an oriented 3-dim manifold, $E$ is a $G$-bundle over $M$, with $G$ compact -simple Lie group. -Question: How does $\pi_3(G)\cong \mathbb{Z}$ imply that there exists non-trivial -gauge transformations (i.e., continuous maps $M\rightarrow G$ which are not homotopic to -the trivial map)? -If anyone would like to read it from the source, check the paragraph leading up to -equation 1.4. - -REPLY [4 votes]: @kwl1026. Gauge transformations are sections of the Ad bundle $P\times_{Ad} g$ where $P\to M$ is the principal $G$ bundle; $g$ the lie algebra. When $G$ is abelian the adjoint action is trivial so, e.g. the $U(1)$ gauge group is always $Map(M,U(1))$ whether or not $P$ is trivial. Its homotopy classes are then $[M, U(1)]= H^1(M;Z)$, which is zero (for $M$ a closed 3-manifold) if and only if $M$ is a rational homology sphere. -An elementary answer to your original question for $SU(2)=S^3$ is that obstruction theory shows that the primary obstruction gives an isomorphism $[M,S^3]\to H^3(M;Z)$. An induction using the fibration $SU(n)\to SU(n+1)\to S^{2n+1}$ and cellular approximation shows that $[M,SU(n)]=[M,SU(2)]$. Other tricks can get you there for other $G$. It is true that the differnence in Chern-Simons invariants (suitably normalized) coincides with this isomorphism (composed with $H^3(M;Z)\to Z$), as indcated by Konrad. For $SU(2)$ it also agrees with the degree, as mentioned by Peter. -If $P$ is non-trivial you have to work a little harder, since you are asking what is the set of homotopy classes of sections of the fiber bundle $P\times_{Ad} g$. A useful reference is Donaldson's book on Floer homology.<|endoftext|> -TITLE: Symmetric matrices as a module over the skewsymmetric ones -QUESTION [6 upvotes]: I'm trying to understand the Cartan decomposition of a semisimple Lie algebra, $\mathfrak g=\mathfrak k \oplus \mathfrak p$, where $[\mathfrak k,\mathfrak p] \subseteq \mathfrak p$, cf. the wikipedia article on Cartan decomposition. -I posted the following question on math.stackexchange.com, where Darij suggested to repost the question here as an answer is not completely obvious, I suppose. -Let $\mathfrak {so}_{n}$ denote the skew-symmetric complex $n \times n$-matrices and let $M$ denote the symmetric $n \times n$-matrices of trace 0. -Then $M$ is a module over the Lie algebra $\mathfrak {so}_n$ (this comes from the Cartan decomposition of $\mathfrak {sl}_n$). - -What is the decomposition of $M$ into irreducible $\mathfrak {so}_n$-modules? - -The standard representation of $\mathfrak {so}_n$ has dimension $n$, the adjoint representation has dimension $\frac 1 2 n \cdot (n-1)$ and there are two spin representations of small dimension. But I don't see a way how these, together with trivial representations, should add up to the dimension of $M$, which is $\frac 1 2 n \cdot (n+1)-1$. - -REPLY [2 votes]: I'm still not totally sure that this question is appropriate to the site, but since it seems to have generated some activity, perhaps I should expand on my cryptic comment concerning the irreducibility in terms of tensors. -The Lie group $SO(n)$ is the subgroup of the general linear group $GL(n)$ of $\mathbb{R}^n$ which preserves the "dot" product and the associated volume form. -Suppose now that $V$ is an irreducible representation of $GL(n)$. It will decompose into irreducible representations of $SO(n)$ as $V = V_1 \oplus \cdots \oplus V_n$, where $n\geq 1$. The projector $V \to V_i$ is $SO(n)$-invariant (as an element in $\mathrm{Hom}(V,V_i)$) and hence it has to be built of the "dot" product and the volume form, these being the elementary $SO(n)$-invariant tensors. -Let $V$ denote the standard $n$-dimensional representation of $SO(n)$. It is also irreducible under $GL(n)$. The tensor square of $V$ decomposes into two $GL(n)$ irreducible representations: -$$ - V \otimes V = S^2V \oplus \Lambda^2 V -$$ -Under $SO(n)$, $\Lambda^2 V$ is the adjoint representation, which is irreducible if $\mathrm{dim} V \neq 4$. For $\mathrm{dim} V = 4$, the volume element defines an $SO(n)$-equivariant endomorphism of $\Lambda^2$ which decomposes it into two three-dimensional irreducible representations. This is nothing but the fact that $SO(4)$ is not simple. -How about $S^2V$? Using the dot product we can take the "trace" of a symmetric matrix. This gives an $SO(n)$-equivariant map $S^2 V \to \mathbb{R}$, whose kernel is the representation in the OP's question. It is irreducible because using the available invariant tensors one cannot extract any further components. -Now the above, as written, is not a proof. To show that the representation is irreducible, it is probably best to resort to roots and weights. However for small representations, the sort of argument above gives a good feel for their (ir)reducibility.<|endoftext|> -TITLE: What is the group of additive operations on topological K-theory? -QUESTION [5 upvotes]: Let us view topological K-theory as a functor $K$ from the cateory of compact pairs (that is, a compact Hausdorff spaces with a distinguished closed subset) to the category of $\mathbb Z/2$-graded Abelian groups. We could also restrict to second countable spaces and thus countable groups. -An additive operation on topological K-theory is just a natural transformation from $K$ to $K$. These natural transformations form an Abelian group under addition. -Question: What is the isomorphism class of this group? -Examples of operations are discussed in Efton Park's book and in Max Karoubi's book, but I cannot find discussion of the collection of all (additive) operations. -Edit: After looking at Karoubi's book again, I have to state that it actually contains a very satisfactory treatment of operations in K-theory. - -REPLY [5 votes]: The most explicit answers are in work of Sarah Whitehouse and her collaborators. You could start with this paper and its references: -http://www.ams.org/journals/proc/2010-138-06/S0002-9939-10-10237-8/home.html<|endoftext|> -TITLE: Do homotopy groups "always" commute with filtered colimits? -QUESTION [14 upvotes]: It is well-known that homotopy groups, of, say, simplicial sets, commute with filtered colimits. -However, I could not find a reference for an analogous result for homotopy groups of spectra, or, under which hypothesis the same result would hold for an "arbitrary" simplicial model category. -More precisely, let $\cal{M}$ be a simplicial model category. For a fibrant object $X \in \cal{M}$ its homotopy groups with coefficients in a cofibrant object $W\in \cal{M}$ may be defined as -$$ -\pi_n (X; W) = [\Sigma^nW, X] = \pi_n \mathrm{map}(W,X) \ , -$$ -where $\mathrm{map} $ denotes the simplicial mapping space from $W$ to $X$. - -So my first question is: which hypothesis do I have to assume for $W$ to obtain an isomorphism -$$ ->\mathrm{colim}_i \pi_n (X_i;W) = \pi_n (\mathrm{colim}_i X_i;W) \ ? ->$$ -And the second one: in which kind of model category such an isomorphism holds for every cofibrant object $W$ -or, at least, for "sufficiently" many cofibrant objects $W$? - -The reason behind my question is the following (and explains the meaning of that "sufficiently"): I have a filtered category $I$, functors $X_\bullet, Y_\bullet : I \longrightarrow {\cal M}\_f$ and a natural transformation $f_\bullet : X_\bullet \longrightarrow Y_\bullet$, such that, for every cofibrant object $W$, $f_\bullet$ induces isomorphisms -$$ -\mathrm{colim}_i \pi_n (X_i ; W) = \mathrm{colim}_i \pi_n (Y_i ; W) \ , -$$ -for every $n$. And I want to conclude that the induced map between the colimits -$$ -\mathrm{colim}_i X_i \longrightarrow \mathrm{colim}_i Y_i -$$ -is a weak equivalence. Which would be true if - -I could commute colimits and homotopy groups, at least for -"enough" cofibrant objects $W$ -in case of simplicial sets, one point $W = *$ is enough. - -I suspect the answer involves words like "smallness / compactness" and "cellular model category". For instance an answer like: "You can do that in no matter what simplicial cellular model category" -in which every cofibrant object is compact- would be fine. Nevertheless, as long as I can understand, commutations like -$$ -\mathrm{colim}_i {\cal M} (W, X_i ) = {\cal M} (W, \mathrm{colim}_i X_i) -$$ -hold for $W$ small and $\lambda$-sequences; that is, when the domain of the functor $X : \lambda \longrightarrow \cal{M}$ is an ordinal; in particular, a totally ordered set, which my filtered $I$ needs not to be. -So any references of a result along these lines, even just for spectra, are welcome. - -REPLY [12 votes]: The condition you're looking for is called combinatoriality (and local presentability). A model category is combinatorial provided it satisfies some complicated conditions involving accessibility, but at least when the underlying category is a presheaf topos and the cofibrations are exactly the monomorphisms, we can actually systematically construct these model structures using the framework of Denis-Charles Cisinski in his book Les Préfaisceaux comme modèles des types d'homotopie (Astérisque 308). This, in particular gives us the combinatoriality of the model structure on simplicial sets essentially for free. -For a general reference on combinatoriality and Jeff Smith's theorem, I suggest you take a look at the papers Sheafifiable Homotopy Model Categories by Tibor Beke and On Left and Right Model Categories and Left and -Right Bousfield Localizations by Clark Barwick. (The original theorem is due to Jeff Smith, but he has neglected to publish it (it was announced at a conference and the main ideas were given, I believe), although he has assured us (not me specifically!) many times that he is in the process of publishing a book).<|endoftext|> -TITLE: Consistent hierarchy of axiomatic systems -QUESTION [9 upvotes]: First of all, I am not an expert in model theory. I just want to get my personal view on the foundations of mathematics straight. -I just learned in Sergey Melikhov's answer to another question something about the Axiom of Determinacy (AD). This Axiom is equivalent to the statement: -$\forall G \subseteq Seq(S):$ -$$\forall a \in S :\exists a' \in S :\forall b \in S :\exists b' \in S :\forall c \in S :\exists c' \in S ... : (a,a',b,b',c,c'...) \in G$$ -or -$$\exists a \in S :\forall a' \in S :\exists b \in S :\forall b' \in S :\exists c \in S :\forall c' \in S ... :(a,a',b,b',c,c'...) \notin G$$ -where $Seq(S)$ is the set of all $\omega$-sequences of some countable set $S$. It is thus some infinitary generalization of de Morgan's rule and seems rather natural. $ZF+AD$ seems to be rather realistic (in my opinion) version of set theory, in the sense that it avoids many unphysical paradoxes (such as non-measurable sets, paradoxical decompositions, non-continuous linear functions on Banach spaces etc.) Still, it seems to be strong enough to reproduce enough infinitary mathematics, so that a development a lot of mathematics and of theoretical physics etc. is possible. -It is a fact that $ZF+AD$ can prove consistency of $ZF$. Basically, the question is whether one can continue this process. The question is more precisely: - -Question: Is there a hierarchy, which consists of an axiom $AD_{\alpha}$ for each ordinal some ordinals $\alpha$ (inspired maybe by some version of de Morgan's laws for ordinal sequences of quantifiers applied to sets of sequences in sets of some cardinality), such that $AD=AD_{1}$ and $ZF + \cup_{\beta \leq \alpha} ZF_{\beta}$ can prove consistency of $ZF + \cup_{\beta < \alpha} ZF_{\beta}$. - -If that is the case, why not taking $ZF + \cup_{\alpha} AD_{\alpha}$ as the axiomatic foundation of mathematics. The bad thing about this would be that the axioms do not form a set, the payoff would be that it proves the consistency of itself and is philosophically sound. -EDIT: From Emil Jeřábek's comment I understand that the question does not make sense as stated since there are only countably many formulas and hence there cannot be uncountably many axioms. So the right (and more modest) question to ask would maybe be the following: - -Question: Does there exist an equally natural axiom $AD'$ (based again on some infinitary version of a well-known principle like de Morgan's laws) which proves (together with $ZF+AD$) consistency $ZF + AD$? - -Maybe there is also some infinitary version of set theory which allows for sentences of arbitrary length like the one which was used above to describe the meaning of $AD$. This could be the place, where the first question could have an answer. - -REPLY [3 votes]: There is a generalization of the compactness theorem to infinitary logics that sounds somewhat close to what you want. Specifically, the compactness theorem tells us that every finitely satisfiable theory of the usual formulas from a first-order language $L$ is satisfiable. For any uncountable cardinal $\kappa$, we can extend $L = L_{\omega, \omega}$ to the infinitary language $L_{\kappa, \kappa}$ by closing the usual formulas under $\bigwedge_{\xi < \lambda}\varphi_{\xi}$, $\bigvee_{\xi < \lambda}\varphi_{\xi}$, $\exists \langle x_{\xi}| \xi < \lambda\rangle$, and $\forall \langle x_{\xi}| \xi < \lambda\rangle$ for all $\lambda < \kappa$. We then define $\kappa$ to be strongly compact if an analogous theorem holds for arbitrary $L_{\kappa, \kappa}$, mainly that if every collection of fewer than $\kappa$ many statements from a theory in $L_{\kappa, \kappa}$ is satisfiable, then the theory is satisfiable. -Now let me emphasize that while ZF + AD may sound natural, it does much much more than prove the consistency of ZF. In fact, ZF + AD proves the consistency of ZFC + "There exists a Woodin cardinal", and Woodin cardinals are quite high up in the large cardinal hierarchy (i.e., a sufficiently stronger theory than ZF that is more likely to be inconsistent than ZF alone). Also, since set theorists tend to like having choice, the usual assumption is not that AD holds in our universe. Instead, we assume that it holds in the minimal transitive ZF model containing all of the ordinals and all of the reals, i.e., $L(\mathbb{R})$. -Now from ZFC + "There exists a strongly compact cardinal", we can prove that ZF + AD holds in $L(\mathbb{R})$. Moreover, we can prove that for every $\alpha$, there exists $\beta > \alpha$ such that AD holds in the set $L_{\beta}(\mathbb{R})$. In particular, ZFC + "There exists a strongly compact cardinal" proves CON(ZF + AD), CON(ZF + AD + CON(ZF + AD)), CON(ZF + AD + CON(ZF + AD + CON(ZF + AD))), etc. -The above result illustrates how assuming a strongly compact cardinal is also a very strong large cardinal hypothesis, but it is only one (admittedly significant) jump higher in our large cardinal hierarchy.<|endoftext|> -TITLE: Fundamental group of a semiabelian variety -QUESTION [6 upvotes]: Let $K$ be an algebraically closed field. Assume for simplicity that $K$ has characteristic zero. Let $A/K$ be a semiabelian variety. Let $n$ be an integer coprime to $char(K)$. Denote by $\pi_1(A)$ the -etale fundamental group of $A$. Is it true that -$Hom(\pi_1(A), {\mathbb{Z}}/n)$ is isomorphic to $A[n]^\vee$ (edit)? What happens, if I replace $A$ by an arbitrary connected algebraic group over $K$? -Remark: The answer is ``yes'' in the case where $A/K$ is proper. (The key ingredient to the proof of this fact is the theorem of Serre-Lang, that every finite etale cover $X/A$ carries the structure of an abelian variety, provided $A/K$ itself is an abelian variety.) - -REPLY [4 votes]: I can give a partial answer, or at least a strategy towards my question, myself. I hope that this will also put my question into the right context, and increase its chances of being answered completely. (I admit that it was not so well-formulated at the beginning.) -Let $K$ be algebraically closed of characteristic zero (for simplicity). Let $X/K$ be a separated connected algebraic $K$-scheme. Then we have $H^1(X, \mathbb{Z}/n)=Hom(\pi_1(X), \mathbb{Z}/n)$. On the other hand we have an exact sequence -$$1\to \Gamma(X, {\cal{O}}_X)^\times/n\to H^1(X, \mathbb{Z}/n)\to H^1(X, {\mathbb{G}}_m)[n]\to 1\ (*),$$ -associated to the short exact sequence of etale sheaves -$$1\to \mu_n\to \mathbb{G}_m\to -\mathbb{G}_m\to 1$$ -(and noting that $\mu_n=\mathbb{Z}/n$ because $K$ is algebraically closed). -Furthermore $H^1(X, \mathbb{G}_m)=Pic(X)$, the group of invertible sheaves on $X$. If $X$ is proper in addition, then $\Gamma(X, {\cal{O}}_X)^\times=K^\times$. Hence there is an isomorphism -$$Hom(\pi_1(X), \mathbb{Z}/n)=H^1(X, \mathbb{Z}/n)\cong Pic(X)[n]$$ -for every connected proper $K$-scheme. -Now consider the special case where $A/K$ is an abelian variety. Then -$$Pic(X)[n]=A^\vee[n]=A[n]^\vee=Hom(A[n], \mathbb{Z}/n);$$ -hence we obtain in fact a canonical isomorphism -$$Hom(\pi_1(A), \mathbb{Z}/n)=Hom(A[n], \mathbb{Z}/n).$$ -This is where the "well-known" isomorphism $\pi_1(A)\cong \prod_\ell T_\ell(A)$ comes from. -(The fact that $\pi_1(A)$ is abelian has to be shown in addition.) -One also sees that $H^1(A, Z_{\ell})$ is dual to the Tate module $T_\ell(A)$ ($\ell$ a prime number). -Now let $B$ be a semiabelian variety. -I asked the above question, because I wanted to know, whether the situation is similar in the case of a semiabelian variety. For example, I wanted to know: -i) Is $Hom(\pi_1(B), \mathbb{Z}/n)$ (canonically) isomorphic to $Hom(B[n], \mathbb{Z}/n)$? -ii) Is there a useful relation between $\pi_1(B)$ and $\prod_\ell T_\ell(B)$, where $T_\ell(B)=lim_i B[\ell^i]$ is the Tate module (defined in a naive way analoguos to the proper case). Are these groups canonically isomorphic? -iii) Is there a useful relation between $H^1(B, Z_\ell)$ and $T_\ell(B)$? Is the first $\mathbb{Z}_\ell$-module canonically isomorphic to the dual of the second? -I think, this case of semiabelian varieties is somewhat different, because -$\Gamma(B, {\cal{O}}_B)^\times/n$ does not vanish any more, unless $B$ is proper. -But nevertheless these questions still make sense to me. Further comments / answers are appreciated very much.<|endoftext|> -TITLE: Is there a 'nice' interpretation of virtual representations? -QUESTION [27 upvotes]: Let $G$ be a compact group and let $R(G)$ be the representation ring of $G$. Additively, $R(G)$ is generated by the irreducible representations of $G$. Usually one only deals with those representations which are nonnegative integer combinations of the irreducible representations. However, often one has formulas which apply to an arbitrary element of $R(G)$, and so includes virtual representations which (for this question at least) are the elements of $R(G)$ whose decomposition into irreducible components includes negative coefficients. - -Question: Is there any 'natural' interpretation of virtual representations? In particular, aside from the obvious interpretation as the elements of the formal completion of the semiring of ordinary representations to a ring, is there a natural way to view these objects? Especially helpful would be pictures/ideas others use to assign meaning to virtual representations (if any). - -Motivation: Often in decomposing various formulas involving characters, virtual representations arise in one way or another. For example, in Lie groups, the notion of a highest weight representation $\rho_\omega$ can be extended to arbitrary weights $t$ of $G$ via: -$\rho_{w(t)} = (-1)^w\rho_t$ -In particular, if $t$ is not a dominant weight of $G$ then there is a unique $w$ in the Weyl Group of $G$ such that $w(t)$ is a dominant weight so that $\rho_t$ is defined for arbitrary weights $t$; note that the Weyl Dimension Formula agrees with this extension. If the length of $w$ is odd, then $\rho_t$ will have negative dimension from the dimension formula, hence is virtual. -Another simple example is that when considering the action of the Adams operation $\psi^k$ on $R(G)$, one has that $\psi^k(\rho)$ is in general a virtual representation. -It happens that from time to time I come across other instances of virtual representations appearing in equations I am considering, and I always work with them ignoring whether they have a physical interpretation or not, but at the same time it would be more satisfying if I could understand the equations as manifestations of some deeper structure. -Edit: Per Qiaochu's comment, yes, virtual representations can be fit into the framework of super-representations. If it is the case that virtual representations are often viewed as super-representations, then perhaps someone could elaborate on why super-representations are so natural and how one works around the dimension mismatches between virtual representations and super-representations, i.e. $dim(\rho_1\ominus\rho_2) = dim(\rho_1)-dim(\rho_2)$ but the dimension of the corresponding super-representation is $dim(\rho_1)+dim(\rho_2)$. - -REPLY [5 votes]: Andre's answer above does a perfect job of explaining some of the abstract theory of "virtual" representations (of an arbitrary group). I'd like to give one answer to the special case you bring up, when $G$ is a compact group. Or, even better, I'll work with the case when $G = \mathfrak g$ is a (finite-dimensional) semisimple Lie algebra over $\mathbb C$. -Then one place that the signs you talk about come up is in the Weyl character formula. Pick a triangular decomposition $\mathfrak g = \mathfrak n^- \oplus \mathfrak h \oplus \mathfrak n^+$. For any weight $\lambda$, let $V(\lambda)$ denote the Verma module with heighest weight $\lambda$ (recall that $V(\lambda) = {\rm U}\mathfrak g \hspace{1ex}\otimes_{\rm U\mathfrak b}\hspace{1ex} \mathbb C_\lambda$, where $\mathfrak b = \mathfrak h \oplus \mathfrak n^+$ is the standard Borel, and $\mathbb C_\lambda$ is the one-dimensional $\mathfrak b$-module on which $\mathfrak h$ acts by $\lambda$ and $\mathfrak n^+$ acts by $0$). Let $L(\lambda)$ denote the quotient of $V(\lambda)$ by its maximal proper ideal; it is the unique irreducible $\mathfrak g$-module with heighest weight $\lambda$. Recall that $\mathfrak h$ acts semisimply on any sufficiently nice $\mathfrak g$-module (including $M(\lambda)$ and its quotients and tensorands and summands; "sufficiently nice" is codified, for the present purposes, by "in the category $\mathcal O$"), and that the K-group of the semisimple $\mathfrak h$-modules is precisely the group ring of $\mathfrak h^*$ (the space of weights). The character $\operatorname{ch}(M)$ of a $\mathfrak g$-module $M$ is its image after first forgetting from the $\mathfrak g$-action to just a $\mathfrak h$-action, and then looking at the element it represents in $\mathbb Z[e^{\mathfrak h^*}]$ (or rather in some completion; for category $\mathcal O$, I want to take the adic completion for the ideal generated by $e^{-\lambda}$ for simple roots $\lambda$). -The Weyl Character Formula asserts that, for $\lambda$ a dominant integral weight: -$$ \operatorname{ch}(L(\lambda)) = \sum_{w\in \mathfrak W} \operatorname{sign}(w) \; \operatorname{ch}(V(w(\lambda+\rho)-\rho)), $$ -where $\rho$ is half the sum of the positive roots, $\mathfrak W$ is the Weyl group, and $\operatorname{sign}: \mathfrak W \to \{\pm 1\}$ is $w \mapsto \det_{\mathfrak h}w = (-1)^{\operatorname{length}(w)} $. -If there were not the signs there, you might hope that this formula came from some statement in the representation theory of $\mathfrak g$. Recall that taking K-groups of a category turns exact sequences into addition equations. Then a proof of the Weyl Character Formula can follow this outline: - -Write down a matrix $b_{\lambda\mu}$ for the coefficient of $\operatorname{ch}(V(\mu))$ in $\operatorname{ch}(L(\lambda))$. -Then the inverse matrix $(b^{-1})_{\mu\lambda}$ expresses the coefficient of $\operatorname{ch}(L(\lambda))$ in an expansion of $\operatorname{ch}(V(\mu))$. -Recognize that each $V(\mu)$ is an extension of some $L(\lambda)$s — this is actually an equation in the representation theory of $\mathfrak g$ — and by studying how $V(\mu)$ is built out of $L(\lambda)$s, understand enough of the structure of $(b^{-1})_{\mu\lambda}$ to conclude the theorem. - -The central point is that the inverse matrix $b^{-1}$ has only nonnegative integer entries, and so has a chance of descending from some extension problem in the representation theory. -So, what about the Weyl character formula itself? Since it has negative coefficients, it cannot express that $L(\lambda)$ is some extension of Verma modules. Instead, it descends from the fact that $L(\lambda)$ has a resolution in Verma modules, the so-called BGG resolution: -$$ 0 \to M(w_0(\lambda+\rho)-\rho) \to \cdots \to \bigoplus_{w\in W \text{ s.t. }\operatorname{length}(w) = k} M(w(\lambda + \rho)-\rho) \to $$ -$$ \cdots \to \bigoplus_{s \text{ a simple reflection}} M(s(\lambda + \rho) - \rho) \to M(\lambda) \to 0 $$ -has homology only in the last spot, where the homolgy is $L(\lambda)$. (I think the differentials are just the obvious inclusions of Verma modules. My reference for all of this is my collection of lecture notes from UC Berkeley's Lie Theory course (chapter 6.1), but it doesn't go into detail.) -So this might be the "nice interpretation" you're looking for. In Andre's language, the point is to realize virtual modules as objects in the derived category of modules; i.e. chain complexes. Then the chain complex above is isomorphic (in the derived category) to its homology $L(\lambda)$, by the projection $M(\lambda) \to L(\lambda)$, and the Weyl Character Formula is the decategorification of this isomorphism. In general, the philosophy is that any naturally-occurring alternating sum of dimensions ought to come from a chain complex. Conversely, any sum of dimensions out to come from an extension.<|endoftext|> -TITLE: Approximating derivatives between gridpoints -QUESTION [6 upvotes]: Suppose we have a grid (possibly irregular) of $N$ function/value pairs, $(x_i, f_i)$, $i=1...N$. The function is differentiable everywhere at least twice (perhaps more). -What would be a good way to find, for example, a polynomial (or other) approximation for the derivative of the function in an interval $(x_j, x_{j+1})$, or perhaps $(x_{j-1}, x_{j+1})$, for a given $j$? -Say, cubic splines would provide the derivative, but it seems too wasteful to compute a spline approximation over all $N$ points, when all is needed is a neighborhood of just one point $j$. -On the other hand, piecewise linear approximation only needs two points, but is rather imprecise, since we know that there are no jumps in the true derivative. What would be good alternatives that would need only a small number of points around $j$? - -REPLY [7 votes]: In general you could try out spectral methods. Spectral methods also do a good job of capturing derivatives. If you want a polynomial approximation, Chebyshev polynomials do a good job and are probably the best when it comes to polynomial interpolations. (No Runge phenomenon as in the case of Lagrange interpolation)<|endoftext|> -TITLE: Density question in algebraic group -QUESTION [12 upvotes]: Suppose G is an algebraic group defined over F, the algebraic closure of F is K. Consider the Zariski topology on G(K), is G(F) Zariski dense in G(K)? - -REPLY [5 votes]: A (smooth, connected) unipotent group $U$ is said to be $k$-split if there is a filtration by $k$-subgroups for which the successive quotients are isomorphic to $\mathbf{G}_{a/k}$. -The examples mentioned in comments (e.g. the subgroup of $\mathbf{G}_a^2$ defined by $y^p - y = tx^p$) are non-split unipotent groups. Any $k$-split unipotent -group $U$ is even a rational variety (in fact, $k$-isomorphic as a variety to $\mathbf{A}^n$) so it is clear that $U(k)$ is Zariski dense in $U(k_{alg})$ when $k$ is infinite. -More generally, let $G$ be a (smooth) linear algebraic group over $k$ and assume that the unipotent radical of $G$ is defined and split over $k$ (both of these conditions can fail). Then as a $k$-variety, $G$ is just the product of its reductive quotient $G_{red}$ and its unipotent radical (result of Rosenlicht). In particular, $G$ is unirational and if $k$ is infinite, $G(k)$ is dense in $G(k_{alg})$. -Of course, this observation isn't that interesting -- in some sense, it just "identifies" the problem.<|endoftext|> -TITLE: Proper way to deal with papers you've already refereed. -QUESTION [52 upvotes]: This question is anonymous for obvious reasons. -I referee what feels like a decent number of papers (though I don't know how many is normal!), and I try to take it seriously. Sometimes, based on something I explicitly said or something implicit in my review, the journal rejects the submitted paper. Usually this seems to be because the editor feels that the paper isn't up to the journal's standards. -In an event that has proven less rare than I would like, I am sometimes asked - by a different editor and journal entirely - to review the same paper again. My record is currently three such rotations. -Usually I try to beg off, because I feel like the author deserves to get a new referee who is less biased and might bring new insight to their review. What I've found in practice is that editors tend to persist, and solicit my opinion more informally. I understand this, because I know that referees are hard to find, and so I've usually done so. The editor usually will say that they will seek out an additional referee, though I rarely hear the final story. In some instances I have gotten the impression, for reasons that I don't want to get into, that an author is starting to feel persecuted due to repeated rejections by prestigious journals. -What is the best policy on requests to repeat a refereeing job? - -REPLY [10 votes]: Note throughout that there are two only-partly related issues: literal correctness, and "importance", and that the latter is tangled up with "status" and "relative prestige" of journals. And, don't forget, all journals get far more excellent papers than they can fit into their "pages", so they will reject many excellent ones, for essentially random reasons... because, looked at bluntly, the job of an editor is to reject papers (not to accept). -It is also the job of editors to maintain or enhance the reputation of their journal, while the real reason mathematicians "need" to publish is to maintain or enhance their reputation, and there's a delicate dance done to see who benefits and who "sacrifices", reputation-wise. In that context, I'd tend to bet that confessing prior rejection wouldn't help anything at all, since it resembles telling someone that you didn't really want to go with them to the prom, but you'd already asked other people and were refused. -In that context, I think one should decline to referee a paper a second time for a different journal, if only on the principle that it might be that one gives a negative opinion for (accidentally) subjective reasons, e.g., that the author's priorities are not what they ought to be. One may truly believe this, but declaring that the author should have written a different paper entirely is a hard objection to meet. If the objections are arguably "objective", that there're serious tangible errors, mistatements of fact, disregard of prior art, etc., that's of course a different matter, but I think these pseudo-objective concerns are not the usual watershed for publication-or-not. It's status/reputation.<|endoftext|> -TITLE: Grothendieck Riemann Roch involving Higher K ? -QUESTION [17 upvotes]: As we know, Grothendieck Riemann Roch only involves $K_{0}$. Is there any work generalizing this formula to (Quillen's Higher K)? If there is, what is the meaning for such kind of formula? -Thanks in advance - -REPLY [8 votes]: There is the recent paper Algebraic K-theory, $A^1$-homotopy and Riemann-Roch theorems by Riou, which gives a different proof of the results by Gillet referenced in the answer by profilesdroxford54. -Abstract: - -In this article, we show that the combination of the constructions done in SGA 6 and the $A^1$-homotopy theory naturally leads to results on higher algebraic K-theory. This applies to the operations on algebraic K-theory, Chern characters and Riemann-Roch theorems. - -To give you a rough idea, algebraic K-theory is representable in the stable homotopy category arising from $A^1$-homotopy theory of schemes over a regular scheme. As far as I understand it, Grothendieck-Riemann-Roch is just a shadow of the maps from the spectrum representing algebraic K-theory to the Eilenberg-MacLane spectrum. But the paper above considers more than just stable homotopy.<|endoftext|> -TITLE: Reference for an edge-matching problem -QUESTION [6 upvotes]: Perhaps not surprisingly, a variation of a recreational math puzzle (a so-called edge-matching puzzle or scramble square) is equivalent to a combinatorics question of interest (in this case, about "Quantum knots and mosaics", S.J. Lomonaco, L.H. Kauffman, question #9). -In a traditional edge-matching puzzle, you are given $n^2$ tiles, each tile square in shape and bearing a design, with the goal of arranging the tiles in an $n\times n$ grid so that the designs on the side of adjacent tiles "match". For instance, here's a puzzle with 24 possible tiles (4 sides, 3 colors, 2 halves) of which at most 9 actually appear: - (source) -For what it's worth, solving a general edge-matching puzzle is NP-complete (see article of Demaine). The combinatorics problem is phrased in a slightly different fashion. You begin with a finite collection of designs (such as quadruples of colored halves of butterflies) and for each design an ample supply of square tiles bearing that design. The problem is to calculate the number of arrangements of these tiles in an $n\times n$ grid so that, as in the game described above, the designs on sides of adjacent tiles match. The number of arrangements should be in terms of the size of the grid and the collection of designs on the tiles. - -Is anyone aware of results along these lines or, even better, able to provide a quick calculation of the number of arrangements of tiles into an edge-matched grid? - -My suspicion is that the number of arrangements goes like $\lambda^{n^2}$ where $\lambda$ is determined from the collection of designs on the tiles. - -REPLY [4 votes]: In many systems, there is a constant entropy per area, and the constant is not easy to compute. See, for example, the hard square constant. -If you specify boundary conditions, you can express some interesting combinatorial problems this way. For example, alternating sign matrices correspond to square ice configurations with specified boundary configurations. I don't recall the asymptotics of the number of alternating sign matrices, $\prod \frac{(3k-2)!}{(n-1+k)!}$, but I think there is a positive entropy per area and it shouldn't be too hard to calculate it from the exact formula. It should be lower than the entropy in the square ice model with no boundary conditions, computed by Lieb, since there is an Arctic region. -It is possible to have a number of tilings in a $d$-dimensional cube which grows like $\lambda ^{n^{d-1}}$, and not just by forcing the tiling to be essentially constant in one direction. For example, one collection of tiles means every tiling corresponds to an order ideal: a partition for $d=2$, a plane partition for $d=3$, etc.<|endoftext|> -TITLE: Self-taught undergrad math: ordering of topics? -QUESTION [8 upvotes]: After some initial research on math topics, it seems there are about 4 main streams as follows: -1) calculus -> analysis -> complex variables -2) linear algebra -> abstract algebra -> topology -3) discrete mathematics -> number theory -4) statistics -By "->", I mean "seems to be a good foundation for". -So is studying the above 4 "streams" in parallel a good way to self-school in undergrad math? - -REPLY [7 votes]: If you intend to study on your own the best approach is to follow a structured sequence just like in an ordinary Math degree.... but nevertheless not forgetting that everything is interconnected and prerequisites and applications are highly nonlinear among different subjects (like remarked in some comments above). A more detailed list could be this one (each column to be learned simultaneously within the rest of topics): -(Analysis Undergrad.) -Calculus (one variable) -> Vector Calculus -> Functions of One Complex variable -> Measure Theory ---------------------------------> Ordinary Diff. Eq. -> Partial Diff. Eq. -> Variational Calculus -> Integral Eq. -(Algebra & Discrete Undergrad.) -Linear & Multilinear Algebra -> Group Theory -> Rings & Modules -> Intro to Representation Theory -Combinatorics & Graph Theory -Elementary Number Theory -(Geometry & Topology Undergrad.) -Affine & Euclidean Geometry -> Projective Geometry -> Differential Curves & Surfaces -----------------------------------------> Point Set Topology -> Introduction to Elementary Algebraic Topology -(Probability & Statistics Undergrad.) -Elementary Statistics --> Elementary Probability -> Advanced Statistical Methods -(Analysis Grad.) -Real Analysis -> Functional Analysis -> Complex Analysis (several variables) -Dynamical Systems (and Chaos) -Partial Differential Equations (general theory) -(Algebra Grad.) -Commutative Algebra -> Homological Algebra -> Category Theory -Lie Algebras -> Representation Theory -(Geometry & Topology Grad.) -Smooth Manifolds -> Algebraic Topology ---------------------------> Differential Topology ---------------------------> Algebraic Geometry ---------------------------> Riemannian Geometry -> Complex Geometry -> Symplectic Geometry -I do not know about advanced statistics and probability, and graduate number theory should should deal with analytic number theory and algebraic number theory with class field theory up to diophantine and arithmetic geometry. -May be you could make your own list according to your tastes looking up some course sequences and syllabus offered by good universities.<|endoftext|> -TITLE: Generic points and local entropies -QUESTION [18 upvotes]: Let $X=\{1,\dots,p\}^\mathbb{N}$ be the space of sequences on a finite alphabet with a metric inducing the product topology, and let $\sigma\colon X\to X$ be the shift map. Let $\mu$ be a $\sigma$-invariant Borel probability measure on $X$. -A point $x\in X$ is generic for $\mu$ if $\frac 1n S_n\phi(x) \to \int \phi\\,d\mu$ for every $\phi\in C(X)$, where $S_n \phi(x) = \phi(x) + \phi(\sigma x) + \cdots + \phi(\sigma^{n-1}x)$. Denote by $G_\mu$ the set of $\mu$-generic points. -Fact #1. If $\mu$ is ergodic, then Birkhoff's ergodic theorem implies that $\mu(G_\mu)=1$. - -The local entropy of a point $x\in X$ is $h_\mu(x) = \lim_{n\to\infty} -\frac 1n \log \mu([x_1\dots x_n])$, where $[x_1 \dots x_n] = \{ y\in X \mid y_i = x_i \\,\forall 1\leq i\leq n\}$, provided the limit exists. Denote by $Z_\mu$ the set of points $x$ for which $h_\mu(x)$ exists and is equal to the measure-theoretic entropy $h_\mu(\sigma)$. -Fact #2. If $\mu$ is ergodic, then Shannon-McMillan-Breiman implies that $\mu(Z_\mu)=1$. - -The measure $\mu$ is a Gibbs measure if there exists a function $\phi\in C(X)$ and constants $K,P>0$ such that -$$ -K^{-1} \leq \frac{\mu([x_1\dots x_n])}{e^{-nP + S_n \phi(x)}} \leq K -$$ -for every $x\in X$ and $n\in \mathbb{N}$. -Fact #3. If $\mu$ is a Gibbs measure, then $G_\mu \subset Z_\mu$. That is, the local entropy of a point $x$ with respect to $\mu$ is "what it should be" provided the Birkhoff averages of continuous functions along the orbit of $x$ are "what they should be". -(Actually, even more is true: for a Gibbs measure the local entropy $h_\mu(x)$ of any point $x$ is completely determined by the Birkhoff averages $\frac 1n S_n \phi(x)$ of a single function.) - -Question. What is the broadest class of measures for which the inclusion $G_\mu \subset Z_\mu$ holds -- that is, for which genericity for Birkhoff averages of continuous functions implies genericity for local entropies? Does this hold for all ergodic measures? If it does not, is there a natural class of measures beyond the Gibbs measures (and various notions of weak Gibbs measures) for which it does hold? -Related question. Gibbs measures (and weak Gibbs measures) have the property that there exists a function $\phi\in C(X)$ such that $h_\mu(x)$ is completely determined by $\frac 1n S_n \phi(x)$ for every $x\in X$. (Not just for a full measure set of $x$ -- this is true for all ergodic measures.) Is there an example of a measure $\mu$ such that there is no single function $\phi\in C(X)$ whose Birkhoff averages determine $h_\mu(x)$, but there exist two function $\phi_1, \phi_2\in C(X)$ such that $h_\mu(x)$ is completely determined by $\frac 1n S_n \phi_i(x)$ for $i=1,2$? - -REPLY [3 votes]: My feeling is that there exists an ergodic measure $\mu$ for which $G_\mu \setminus Z_\mu$ is nonempty. It is sufficient to find a uniquely ergodic subsystem which admits exceptional points for the Shannon-McMillan-Breiman theorem. I think that one can be constructed symbolically without too much difficulty by the following method. -Pick a real number $h$ lying strictly between 0 and $\log 2$, and consider a sequence $x$ in the 2-shift with the following properties: -1) For every $n \geq 1$, the sequence contains precisely $e^{nh + o(n)}$ distinct words of length $n$. (For reasons of subadditivity the $o(n)$ term is necessarily positive). -2) Every word which occurs in $x$ occurs with a well-defined frequency which is not equal to 0 or 1. -The orbit closure $X$ of such a sequence is then a uniquely ergodic subsystem of the shift with topological entropy equal to $h$. An explicit procedure for constructing such a sequence was given by Grillenberger in the 1970s (in my opinion it's not particularly hard). In particular, $X$ supports a unique invariant measure $\mu$ and $G_\mu$ includes the whole of $X$. Now, suppose that the word $x$ also satisfies the property: -3) There exists a nested sequence of subwords of $x$ such that the frequency of each of these words is less than $e^{-n(h+\varepsilon)}$ for some $\varepsilon>0$. -This implies that there is a nested sequence of cylinder sets in $X$, containing some point, such that the measures of these cylinder sets decrease at a rate faster than the "standard" local entropy $h$, and hence the point in the intersection of the cylinders belongs to $G_\mu$ but not to $Z_\mu$. -I think that there shouldn't be any problem in reconciling all three of these criteria with one another, but I will admit that I haven't attempted to write a proof of that. I think it sounds reasonable that for a larger class of measures than Gibbs measures we should have $G_\mu \subseteq Z_\mu$, but I don't have much to contribute to that end of the question...<|endoftext|> -TITLE: The volume of the “unit ball” in $\mathbb{R}^{m\times n}$ with respect to the cut norm -QUESTION [9 upvotes]: This question is inspired by the question “ε-nets with respect to the cut norm” by the user Aaron, which had been reposted to cstheory.stackexchange.com. -The cut norm ||A||C of a matrix A=(aij)∈ℝm×n is defined as the maximum of |∑i∈I,j∈Jaij| over the subsets I⊆{1,…,m} and J⊆{1,…,n}. The “unit ball” in ℝm×n with respect to the cut norm is the convex polytope P(m, n) = {A∈ℝm×n: ||A||C≤1}. Let V(m, n) be the volume of this polytope P(m, n). -Since P(m, n) contains [0, 1/mn]m×n, we have that V(m, n) ≥ 1/(mn)mn. In other words, (V(m, n))1/mn ≥ 1/mn. -Question. Is this lower bound on (V(m, n))1/mn tight up to a constant factor? In other words, does there exist a constant c>0 such that for every m,n≥1, it holds that (V(m, n))1/mn ≤ c/mn? -This lower bound is indeed tight up to a constant factor if one of m and n is bounded by a constant. This can be shown as follows. In an answer on cstheory.stackexchange.com, I gave a sketch of a proof that V(1, n) = (2n)!/(n!)3. Using this, we have that V(m, n) ≤ (V(1, n))m = ((2n)!)m/(n!)3m. By using Stirling’s formula, we obtain that there exists an absolute constant d>0 such that for every m and n, it holds that (V(m, n))1/mn ≤ d min{1/m, 1/n}. -A positive answer to this question improves the lower bound on Aaron’s question to match the upper bound up to a constant factor. - -REPLY [3 votes]: A standard argument (Lemma 3.1 here) shows that the cut-norm is 4-equivalent to the operator norm from $\ell_{\infty}^m$ to $\ell_1^n=(\ell_\infty^n)^*$. Therefore the volume you ask is almost the same (up to a factor $4^{mn}$) as the volume of the unit ball in the projective tensor product $\ell_{\infty}^m \otimes_\pi \ell_\infty^n$. This is exactly this question (at least for $m=n$), and I gave an answer saying basically that (from general theorems) your lower bound on $V(m,n)^{1/mn}$ is sharp up to a logarimthic factor. -I guess this logarithm is not necessary, but I don't know how to remove it.<|endoftext|> -TITLE: Notion of generalized function/distribution for functional derivatives? -QUESTION [8 upvotes]: Is there any work on defining something analogous to a generalized function for functions whose domain is a Hilbert or Banach space? Is there an extension of the notion of Frechet/Hadamard/Gateaux derivatives that somehow resembles the theory of distributions? -I am hoping to take something like a Taylor expansion of a non-differentiable function from a Hilbert space to the real line. - -REPLY [2 votes]: There is a substantial literature on spaces of distributions on infinite dimensional spaces, in particular function spaces, so much so that there is a subsection of the AMS classification scheme devoted to it (35R15). This might be relevant for your question. Related subjects can be found in 46G. This work is mainly motivated by questions in quantum theory, especially quantum field theory. A possible starting point would be the work of Y.M. Berezanskii, e.g., his monograph "Self-adjoint operators in spaces of functions in infinitely many dimensiona". - -REPLY [2 votes]: Yes there is a notion of generalised function in infinite dimensions, see http://www.encyclopediaofmath.org/index.php/White_noise_analysis and the references contained therein. These functions are called "Hida distributions".<|endoftext|> -TITLE: Do the results of (1/n)-surgery determine the link?... -QUESTION [11 upvotes]: Knowing the result of knot surgery is often not enough to determine the knot. Indeed, there are 3-manifolds admitting an infinite number of descriptions as surgery on a (1-component) knot in $S^3$. However, if I know many surgeries, perhaps I can recover the knot? Let me be specific: -Suppose I have a 2-component link $U_1 \cup U_2$ inside the 3-sphere $S^3$ which has linking number $0$ and such that each component $U_1$, $U_2$ is the unknot. -I'm interested in knowing how much surgery tells us in this situation. If I do $1/n$-surgery on $U_2$ I get back $S^3$, but now $U_1$ sits inside $S^3$ as a knot $K(n)$. -Does the sequence $K(1), K(2), K(3), \ldots$ determine the original link $U_1 \cup U_2$ ? Would it even be expected to?. - -REPLY [6 votes]: If the orginal link $U_1 \cup U_2$ was hyperbolic, the answer is yes. For large enough $n$, $S^3 \setminus K(n)$ will also be hyperbolic, and will approach $S^3 \setminus (U_1 \cup U_2)$ in the Gromov-Hausdorff topology; so the sequence $K(n)$ determines the complement of $U_1 \cup U_2$. The complement doesn't determine the link in general (unlike for knots), but we also have the marking of the component $U_1$ by its meridian, which I believe is enough. -The answer in general is also almost certainly yes, but I haven't thought through all the cases. Note that this operation has a simple geometric description: arrange $U_1 \cup U_2$ so that $U_2$ sits as a flat unknot in a plane. Then to get $K(n)$, remove $U_2$ and twist the bundle of strands that passed through $U_2$ by $n$ full twists. -(This is all much easier than Lackenby's result mentioned above.)<|endoftext|> -TITLE: Null-homotopy of diagonal map -QUESTION [6 upvotes]: For a sphere $S^n$, the diagonal map $\Delta:S^n\to S^n\wedge S^n$ sending $x\mapsto x\wedge x$ is null-homotopic. This is the homotopy group $\pi_n(S^n\wedge S^n)=\pi_n(S^{2n})$ is trivial. -I'm wondering if there are other examples of this phenomenon. That is, is there a non-contractible based space $X$ which is not a sphere, such that the diagonal map $\Delta:X\to X\wedge X$ is null-homotopic? - -REPLY [7 votes]: To add to the above: an $r$-connected CW complex $X$ of dimension $\le 3r$ has the homotopy type of a suspension $\Leftrightarrow$ its reduced diagonal map $X \to X \wedge X$ Is null-homotopic. This is the Berstein-Hilton-Ganea theorem.<|endoftext|> -TITLE: different Shimura data with common underlying group? -QUESTION [6 upvotes]: A pure Shimura datum is of the form $(G,X)$ with $G$ a connected reductive $\mathbb{Q}$-group, and $X$ a homogeneous space under $G(\mathbb{R})$, subject to Deligne's conditions in terms of Hodge types, Cartan involutions, etc. cf.Deligne, "Varietes de Shimura: interpretation modulaire, et techniques de constructions de modeles canoniques" -One may ask to a given connected reductive $\mathbb{Q}$-group, how many pure Shimura data could one obtain of the form $(G,X)$. As remarked by M.Borovoi, if $G$ is a compact $\mathbb{Q}$-torus, then any homomorphism $h:\mathbb{C}^\times/\mathbb{R}^\times\rightarrow G_{\mathbb{R}}$ defines a pure Shimura datum $(G,\{h\})$. Thus for a finiteness answer, one should restrict to the case where $G$ is semi-simple. -For simplicity one even assume that $G$ is of adjoint type. Then as is pointed out in Deligne's article, the existence of $X$ is characterized by the special nodes in the Dynkin diagram of $G_\mathbb{C}$ (plus certain condition so that the node gives rise to an $\mathbb{R}$-homomorphism $\mathbb{S}\rightarrow G_\mathbb{R}$. For the adjoint $G$ chosen, there are at most finitely many special nodes (possibly empty in certain cases). And thus the finiteness is clear. -My first question is: for a given adjoint $\mathbb{Q}$-group $G$ and some pure shimura datum $(G,X)$, how many Shimura data can one obtain to be of the form $(G,X')$ (with the same $G$)$ up to isomorphism? When is it unique? (added: M.Borovoi has answered this in detail, see below). -Secondly, what about Shimura subdatum in a fixed $(G,X)$ can one find to share the common underlying $\mathbb{Q}$-group? By a Shimura subdatum is meant a pair $(G_1,X_1)$ where $G_1$ is a $\mathbb{Q}$-subgroup, $X_1$ a $G_1(\mathbb{R})$-orbit in $X$ such that $(G_1,X_1)$ is a Shimura datum itself. it is also easy to check that to obtain such a aubdatum, it suffices to (1) find a point $x$ in $X$ such that the corresponding homomorphism -$h:\mathbb{S}\rightarrow G_\mathbb{R}$ has image in $G_{1, \mathbb{R}}$. -And my second question is: if a connected reductive $\mathbb{Q}$-subgroup $G'$ of $G$ is given, how many subdatum of $(G,X)$ can one find to be of the form $(G',X')$? If there are such subdata, are they unique up to isomorphism (or isomorphism induced by inner automorphism of $G$)? -thanks a lot! - -REPLY [7 votes]: The question is essentially about ${\mathbf{R}}$-groups, so we shall assume that $G$ is defined over $\mathbf{R}$. -It is not true that for a given connected reductive ${\mathbf{R}}$-group $G$, there are at most finitely many Shimura data -of the form $(G,X)$ up to isomorphism. -Indeed, assume that $G$ is a compact (i.e. anisotropic) ${\mathbf{R}}$-torus. -Then for any homomorphism $h\colon \mathbf{C}^* / \mathbf{R}^* \to G$, the pair $(G,h)$ is a Shimura datum. -Now assume that $G$ is adjoint and simple. -I assume that the question is about classification of Shimura data $(G,X)$ up to isomorphism which is the identity on $G$. -The possible Shimura data $(G,X)$ are described in Section 1.2 of the paper: -P. Deligne, -Variétés de Shimura: interprétation modulaire, et techniques de construction de modèles canoniques. -Automorphic forms, representations and $L$-functions, Part 2, pp. 247–289, -Proc. Sympos. Pure Math., XXXIII, Amer. Math. Soc., Providence, R.I., 1979. -To a $G({\mathbf{R}})$-conjugacy class $X$ of homomorphisms -$h \colon \mathbf{C}^* / \mathbf{R}^* \to G$ -satisfying conditions 1.2.1(i),(ii),(iii) of the paper, -Deligne associates a special vertex $s=s(X)$ -of the Dynkin diagram $D$ of $G_{\mathbf{C}}$. -In Proposition 1.2.7 he says that the set of all such homomorphisms has two connected components, interchanged by $h\mapsto h^{-1}$. -In Corollary 1.2.8 he says: -(i) If $s(X)$ is not fixed by the opposition involution, then $G({\mathbf{R}})$ and $X$ are connected (so I conclude that there exist two such conjugacy classes). -(ii) If $s(X)$ is fixed by the opposition involution, then $G({\mathbf{R}})$ and $X$ have two connected components (so I conclude that there is only one conjugacy class). -Table 1.3.9 shows that there is one conjugacy class for the following adjoint ${\mathbf{R}}$-groups of Hermitian type: -$PSU(p,p)$, $B_l$, $C_l$, $D_l^{\mathbf{R}}$, $D_{2l}^{\mathbf{H}}$, $E_7$. Note that for $D_{2l}^{\mathbf{H}}$ the opposition involution is trivial, -see e.g. A.L. Onishchik and E.B. Vinberg, Lie Groups and Algebraic Groups, Springer-Verlag 1990, Table 3 on page 298. -EDIT: I answer the second part of the question. -Let $G=PGL_{2,\mathbf{R}}$, and let $T\subset G$ be a compact maximal torus. -There exists exactly one conjugacy class $X$ of homomorphisms -satisfying conditions 1.2.1(i),(ii),(iii) of Deligne's paper. -Let $i \colon T \hookrightarrow G$ be the inclusion homomorphism. -There exists a homomorphism -$h \colon \mathbf{C}^* / \mathbf{R}^* \to T$ -such that $i \circ h \in X$. -Then $i \circ h^{-1}$ also satisfies conditions 1.2.1(i),(ii),(iii) of Deligne's paper, -hence $i \circ h^{-1}\in X$. -We see that $(T,h)$ and $(T,h^{-1})$ are different Shimura subdata of $(G,X)$ with the same subgroup $T$.<|endoftext|> -TITLE: Duals and Tensor products -QUESTION [28 upvotes]: Let $A$ be a commutative ring with a unit element. Let $M$ and $N$ be $A$-modules. Let $M^v$ and $N^v$ be the dual modules. In general, do we have $M^v \otimes N^v \cong (M\otimes N)^v$? It is definitely true when M and N are free. I believe (though haven't worked out the details) that it is true when M and N are projective. Both Atiyah & Macdonald and Lang don't say anything on the matter. -I came up with this question while studying $Pic(A)$: defined as the group of isomorphism classes of projective modules of rank 1. If I and J are projective $A$-modules of rank 1, then the fact that $Pic(A)$ is a group immediately implies $I^v \otimes J^v \cong (I\otimes J)^v$, as both are the inverse of $I\otimes J$. - -REPLY [4 votes]: I may as well add the following. -One can always consider the dual map: -$$(M^\vee\otimes N^\vee)^{\vee} \leftarrow ((M\otimes N)^\vee)^{\vee}.$$ -I think that under moderate hypotheses (maybe just requiring the modules to be finitely generated is ok, but maybe one needs more: it's certainly ok if additionally the ambient ring is normal) this is an isomorphism. -Here's a brief discussion as to why. Both sides of the equality are reflexive because they are both duals of something. In the normal ring case, the statement reduces to codimension 1 (because reflexive modules in a normal ring are determined in codimension 1, let me know if you need references). Then you are done because the ring becomes a DVR (and thus a PID) and Neil's original answer gives you an isomorphism even without the dual.<|endoftext|> -TITLE: Factorizing polynomials in $\mathbf{Z}[[x]]$ -QUESTION [12 upvotes]: Let $f(x)\in\mathbf{Z}[x]$ be a non-constant, irreducible polynomial, and let $\alpha \in\mathbf{C}$ be a root of $f(x)$. Denote by $\varphi_\alpha:\mathbf{Z}[x]\rightarrow\mathbf{C}$ the ring homomorphism sending $x$ to $\alpha$. Let now $\nu$ be a $p$-adic place of the number field $K=\mathbf{Q}(\alpha)$ such that $\alpha$ has strictly positive valuation with respect to $\nu$, i.e. such that $\alpha$ belongs to the maximal ideal of the integers of $K$ defined by $\nu$. -There are only a finite number of such places $\nu$, they are precisely those occurring with positive exponent in the prime factorization of the principal fractional ideal of $K$ generated by $\alpha$. -Then $\varphi_\alpha$ can be extended uniquely to a continuous map -$$\varphi_{\alpha,\nu}:\mathbf{Z}[[x]]\rightarrow K_\nu,$$ -where $K_\nu$ denotes the completion of $K$ at the place $\nu$ (the topology considered here on $\mathbf{Z}[[x]]$ is the $x$--adic one). Let -$f_\nu(x)\in\mathbf{Z}[[x]]$ be a power series generating the kernel of $\varphi_{\alpha,\nu}$ (such ideal should indeed be principal, right?). -Is it true that the $f_\nu(x)$ can be chosen so that -$$f(x)=\prod_{\nu}f_\nu(x),$$ -where $\nu$ ranges through the places of $K$ considered above? -EDIT: As suggested below, it would be more correct to ask that $f(x)$ be equal to the product $\prod_{\nu}f_\nu(x)$ only up to units, and for any choices of the $f_\nu(x)$. - -REPLY [4 votes]: I think the answer to the question is yes. Here is the idea of a proof. -Let us assume that $\mathcal{O}_K=\mathbf{Z}[\alpha]$. I hope this is not too restrictive for you (hopefully, someone can extend the argument). -Put $(\alpha) = \mathfrak{p}_1^{a_1} \cdots \mathfrak{p}_t^{a_t}$ where the $\mathfrak{p}_i$ are distinct maximal ideals of $\mathcal{O}_K$. Let $\nu_i$ be the place of $K$ associated to $\mathfrak{p}_i$. Let $K_i$ be the completion of $K$ with respect to $\mathfrak{p}_i$, and let $\mathcal{O}_i$ be the ring of integers of $K_i$. We have isomorphisms -\begin{equation*} -\mathbf{Z}[[X]]/(f) \cong \varprojlim_{n \geq 1} \mathbf{Z}[X]/(X^n,f) \cong \varprojlim_{n \geq 1} \mathcal{O}_K/(\alpha^n) -\end{equation*} -\begin{equation*} -\cong \varprojlim_{n \geq 1} \prod_{i=1}^t \mathcal{O}_K/\mathfrak{p}_i^{a_i n} \cong \prod_{i=1}^t \mathcal{O}_{i}. -\end{equation*} -Now, convince yourself that the composition is just $(\varphi_{\alpha,\nu_1},\ldots,\varphi_{\alpha,\nu_t})$, using your notations (this is because $X$ is sent to $\alpha$). -Recall that $\mathbf{Z}[[X]]$ is a UFD. We already know (see my comment to the question) that $\ker(\varphi_{\alpha,\nu_i}) = (f_i)$ with $f_i \in \mathbf{Z}[[X]]$ irreducible. The ideals $(f_i)$ are pairwise distinct (they are the preimages of distinct ideals of $\prod_{i=1}^t \mathcal{O}_{\nu_i}$). So we get $(f)=\cap_{i=1}^t (f_i)=(f_1 \cdots f_t)$.<|endoftext|> -TITLE: formal differences? -QUESTION [6 upvotes]: Hi, -given a local ring $A$ with maximal ideal $m$ which are differences between $Spec(\hat{A})$ ($\hat{A}$ completion of $A$ along $m$) and $Spf(A)$? - -REPLY [4 votes]: There is a nice introduction to formal schemes in Luc Illusie's section of the book ``Fundamental Algebraic Geometry (Grothendieck's FGA Explained)''. Recommend!<|endoftext|> -TITLE: NP not equal to SPACE(n) -QUESTION [10 upvotes]: Exercise 3.2 of Computational Complexity, a Modern Approach states: -Prove: NP != SPACE(n) [Hint: we don't know if either is a subset of the other.] -I don't know how to solve this problem. -It's in the diagonalization chapter. -I've looked around google a bit, but it basically ends up linking back to the Arora/Barak book. -Anyone know how to attack this? -Thanks! -More generally: to prove a language to be uncommputable, I can use diagonalization -- but to prove that two sets of languages (Space(N) and NP) are different, when it's not known that either is contained in the other -- what techniques are there for these proofs? -Thanks again! - -REPLY [12 votes]: I think that a common technique for proving such statements is for example the following type: -One class shares a closure property, while the other cannot because of a hierarchy theorem. -Thus they cannot be equal. -In this particular case a proof could proceed along these lines: Since NP is closed under polynomial time reductions, so would SPACE(n), if they were equal. Then deduce that polynomial time reductions would imply that SPACE(n^2) is contained in SPACE(n), which is impossible by the space hierarchy theorem. - -REPLY [8 votes]: Scott Aaronson has a blog post, Sidesplitting Proofs, which is a highly recommended read for a variety of reasons. The first proof in the list, which is said to be folklore, is that E, the class of problems solvable in $2^{O(n)}$ time, is not equal to PSPACE. The key is again padding: if the two are equal, then E=EXP, and we derive a contradiction. Like in your case, which is bigger (if one is even contained in the other) is not known.<|endoftext|> -TITLE: Is a connected separable locally euclidean Hausdorff topological space second countable? -QUESTION [6 upvotes]: This question arose from considering for a connected smooth Hausdorff manifold the (possible) equivalence of the following properties: -(1) paracompact, -(2) metrizable, -(3) second countable, -(4) countable at infinity, -(5) $\sigma$−compact, -(6) Lindelöf, -(7) separable. -I know proofs for the equivalence of the first six, and that they imply (7), but it is problematic, whether this implies the others. By countable at infinity I mean existence of a sequence of compact sets $K_i$ whose union covers the space and which satisfies $K_i\subseteq{\rm Int\ }K_{i+1}$ . Of course, locally euclidean means that each point has a neighbourhood homeomorphic to $\mathbb R^k$ with the standard topology and $k\in\mathbb N$ . - -REPLY [3 votes]: See my article on nonmetrizable manifolds in the Handbook of Set-Theoretic Topology. Examples 3.7 and 3.9 include separable, nonmetrizable manifolds; neither is due to me and the second is due to R.L. Moore in a 1942 paper. After reading the descriptions, you might try going back to Gauld's smooth example.<|endoftext|> -TITLE: Valuations and separable extensions -QUESTION [11 upvotes]: Let $R$ be a valuation ring containing a field $k$, with residue field $F$ and quotient field $K$. Assume $F/k$ is separable. Is $K/k$ separable? -I have convinced myself that (for a positive answer) it is enough to treat the case of a rank one valuation ("height one" in Bourbaki's terminology), with $F=k$ and $K/k$ finitely generated. (Instaed of the latter condition, we may assume $K$ complete). -The answer is yes for a discrete valuation, because then (if $F=k$) the completion of $R$ is isomorphic to $k[[t]]$. - -REPLY [5 votes]: What about this: we have to prove that $K$ and every purely inseparable extension $l/k$ are linearly disjoint over $k$. -Let $x_1,\ldots ,x_r\in l$ be $k$-linearly independent elements and assume $0=a_1x_1+\ldots +a_rx_r$ for some elements $a_i\in K$. We can divide by the coefficient $a_j$ with the least value and thus assume $a_1,\ldots ,a_r\in R$ with at least one coefficient being a unit of $R$. -There is a unique valuation ring $S$ of $K.l$ dominating $R$. Taking residues modulo the maximal ideal of $S$ yields a linear combination $0=\overline{a_1}x_1+\ldots +\overline{a_r}x_r$ in the field $F.l$. The separability of $F/k$ yields $\overline{a_i}=0$ for all $i$, a contradiction.<|endoftext|> -TITLE: What is the state of research on Horn Angles? -QUESTION [5 upvotes]: The ancient Greeks struggled with the concept of a horn angle, the "angle" formed by the intersection of two curves. The only information I find in Mathworld is that horn angles are examples of non-Archimedean geometry. Where can I find out more about them? What work has been done on them, and what is the current state of knowledge on them? -Since they are part of non-Archimedean geometry, can they be understood using the hyperreal numbers or some other non-standard model of the first-order theory of real closed fields? -Any help would be greatly appreciated. Thank You in Advance. - -REPLY [3 votes]: A quick perusal of Math Reviews leads one to suspect that not much research goes on. (A better perusal might show otherwise). For Euclid a "line" was a curve and an "angle" was a Horn Angle. But he actually only looked at angles formed by lines except for Book III proposition 16 where he essentially commented that the angle formed by a circle and a tangent line is positive but less than any linear angle. Maybe he did not do much more because Horn angles are pretty tough; Unlike linear angles, it is hard to know how to add them or even double one (in a geometrically meaningful way). I'm not even sure how to say when one contains another. In modern times Knasner studied them using somewhat advanced notions of differential geometry (the references above are good) with latest paper in 1951 (dealing with trisection of Horn Angles). After that there is a paper from 1968 and the mentioned one from 2008 (along with some historical papers). -I think that Horn Angles might inform non-standard models more than vice versa. One does not need non-standard models to talk about the order on rational functions (or more general classes of functions) based on their behavior as $x\to\infty$. It is essentially the same to look at behavior as $x\to 0^+$ and that is pretty much the order of Horn Angles.<|endoftext|> -TITLE: Orbits of SL_n acting on matrices of determinant p -QUESTION [9 upvotes]: Fix a positive integer $n$ and let $S$ be the set of $n$ by $n$ matrices -with entries in $\mathbf{Z}_p$ (the $p$-adic integers) whose determinant is $p$. -The group $G:=\mathrm{SL}_n(\mathbf{Z}_p)$ acts freely on $S$ via left multiplication. - - Is it possible to write down an explicit list of representatives for the orbits of - $G$ on $S$? If so, what is this list? If not, is there an algorithm for computing - a complete list of orbit representatives? - -For example, for $n=2$, I think that this is basically the "Hecke Operator at $p$" computation, and the $p+1$ orbits of $G$ on $S$ have representatives -$$ -\left(\begin{matrix} p & b \\\ 0 & 1 \end{matrix}\right),\ b=0,1,\ldots, p-1\ \text{and}\ -\left(\begin{matrix} 0 & -1 \\\ p & 0 \end{matrix}\right) -$$ -This is probably a question whose answer is well-known to many people, so I apologize -in advance for my ignorance! - -REPLY [7 votes]: To complement Qiaochu Yuan's answer, check out Lemma 9.3.2 in Goldfeld's book Automorphic Forms and L-functions for the group GL(n,R). On the two sides of (9.3.3) you can see the decompositions of the set of integral $n\times n$ matrices of determinant $N$ into double cosets and right cosets of $\mathrm{SL}_n(\mathbb{Z})$. The right hand side corresponds to your question, and yes, it gives rise to the $N$-th Hecke operator, see (9.3.5). Of course this is over $\mathbb{Z}$, but localizing at $p$ gives what you need: in that case $N$ is a $p$-power without loss of generality, since other prime powers are $p$-adic units.<|endoftext|> -TITLE: K.Uhlenbeck's preprint "A priori estimates for Yang-Mills fields" -QUESTION [7 upvotes]: Does anyone have a copy of the unpublished preprint of Karen Uhlenbeck A priori estimates for Yang-Mills fields from around 1986? -It appears to have circulated for some time, and it is quoted in several papers in the field (Uhlenbeck-Yau, Daskalopoulos-Wentworth, De Bartolomeis-Tian,...) sometimes with precise references to theorem numbers. However, I could not find it anywhere on the internet. -The results covered in this preprint are probably proved in other places by now, but I'd still be interested in reading it. - -REPLY [10 votes]: Sorry it took so long. The scanned version of Uhlenbeck's preprint is available here: -http://www.math.uwaterloo.ca/~karigiannis/uhlenbeck-preprint.pdf -I can leave it there for a few months, at least.<|endoftext|> -TITLE: Expectation of a random sum -QUESTION [30 upvotes]: Let $X_1, X_2, X_3,\dots$ be an i.i.d. sequence of random variables with finite mean. Write $S_n=X_1+X_2+\dots+X_n$. -Let $N$ be a non-negative integer-valued random variable with finite mean. $N$ may not be independent of the sequence $(X_i)$. -Is it necessarily the case that $S_N$ has finite mean? -Of course, it's true if $N$ is independent of the sequence $(X_i)$. Then $E(S_N)=E(N)E(X_1)$. It's still true if $N$ is a stopping time for the sequence $(X_i)$. -It's also true if the $X_i$ have finite variance. Then for any $c>E(X_i)$, -the quantity $R_c=\sup(S_n-cn)$ has finite mean, and $E(S_N)\leq cE(N)+E(R_c)$. - -REPLY [13 votes]: Edit: I made it a bit clearer and simpler. -No. You start with noticing that there is no "linear" estimate for the mean of $S_N$ in terms of the mean of the sample $X$ under the assumption that the mean of $N$ is small. To this end just take $X$ be $0$ with probability $1-p$ and $A$ with probability $p$ so that $Ap$ is small. Now, once we have a sequence $X_i$ of independent copies of $X$, define $N$ to be $m$ if at least one of $X_1,\dots,X_m$ is not $0$ and $0$ otherwise. Then $ES_N= Apm$ and $EN\le m^2p$ and $EX=Ap$. -Now choose $A_j,p_j,m_j$ so that $\sum A_jp_j<+\infty$, $\sum A_jp_jm_j=+\infty$, $\sum m_j^2p_j<+\infty$. For instance, take $m_j=2^j$, $p_j=2^{-3j}$, $A_j=2^{2j}$. -Define the random samples $X^{(j)}$ and random interval lengths $N^{(j)}$ as above using $A_j,p_j,m_j$ in place of $A,p,m$. Put $X=\sum_j X^{(j)}$, $N=\sum_j N^{(j)}$. Then $EX$ and $EN$ are finite. Let $X_i$ be independent copies of $X$. We have $E\sum_1^N X_i\ge E\sum_j \sum_1^{N^{(j)}}X_i^{(j)}=\sum_j A_jp_jm_j=+\infty$.<|endoftext|> -TITLE: Predicates of infinite arity -QUESTION [10 upvotes]: Infinitary logic considers languages being infinite by infinite conjunctions and disjunctions. -I wonder why it not considers languages being infinite by relations and functions of infinite arity. -Relations of finite arity $n$ over a base set $A$ can be seen as unary predicates of functions $f:[n] \rightarrow A$. Nothing prohibits us to consider more general functions $f:\mathbb{N} \rightarrow A$ or even $f:\mathbb{R}^+_0 \rightarrow A$. - -Is there a model theory assuming a language that allows for - relations and functions of infinite - and even uncountable arity? - -I asked this question at MSE but did get no feedback. - -REPLY [7 votes]: I have several thoughts about this question. -First, to my way of thinking, there is little difference between -an infinite-ary relation $R(a_0,a_1,\dots)$ on a set $X$ and a -unary relation on a suitable power of that set, such as $X^\omega$ -or $X^\alpha$. For example, an $\omega$-ary relation on $\{0,1\}$ -is essentially the same thing as a unary relation on Cantor space -$2^\omega$. In the one case, we have $R(a_0,a_1,\dots,a_n,\dots)$, -and in the second case we have $R(\langle -a_0,a_1,\dots,a_n,\dots\rangle)$. It is a mere stylistic -difference without substantive difference. An $\omega$-ary -relation on the natural numbers $\omega$ is essentially the same -as a unary relation on Baire space $\omega^\omega$. A binary -relation on Baire space is the same as an $(\omega+\omega)$-ary -relation on $\omega$. -For these reasons, it seems to me that mathematics is filled with -abundant examples of infinite-ary relations. The lexical $<$ -relation on the Cantor set is essentially the same as an -$(\omega+\omega)$-ary relation on the two-element set $\{0,1\}$. -Binary relations on Baire space $\omega^\omega$ are essentially -the same as $(\omega+\omega)$-relations on the natural numbers -$\omega$. -I believe that we prefer in these cases to think of the -infinite-ary relation as a unary or finite-ary relation on the -higher-order space of sequences, for several reasons. First, it is -easier to think of the relation as a unary relation in the higher -order space of sequences, simply because we don't mind so much -going to a higher-order and we are used to finite-arity relations. -Secondly, the move to the higher order space allows us to be more -precise about exactly which sequences are allowed to be -considered. If one has an infinite-arity relation, but doesn't -specify the extent of the second-order sequences that are to be -considered (from which model of set theory will they be drawn?), -then the ontological meaning of that relation is a little -ambiguous. But when we think of the relation as a finite-ary -relation on a certain space of sequences, specified by a set of -sequences, then the extensional nature is more clear. -To give an example, usually one views the axiom of determinacy as concerned -with games on $\omega$, so that the players construct a play of -the game $a_0,a_1,a_2,\dots$, and the winning condition of a game -is a unary condition on Baire space $R(\langle -a_0,a_1,\dots\rangle)$. But one could just as easily view the -winning condition as an $\omega$-ary relation on $\omega$, as -$R(a_0,a_1,\dots)$. And this wouldn't really make any difference; -it is an inessential stylistic syntactic difference. -But lastly, let me also point out that there is a literature on -infinite-ary functions, undertaken for example by Addison, as in -his theory of infinitary Boolean operations. I once had the -pleasure of taking a seminar on the topic that he offered in -Berkeley on the topic, and he considered many different Boolean -operations $f:\{0,1\}^\omega\to\{0,1\}$, and investigated their -nature.<|endoftext|> -TITLE: Elementary mathematical books -QUESTION [15 upvotes]: I understand that this is a bit of offtopic but mathoverflow is my last resort, as google did not help. -I am about to publish an English translation of my Russian book for high school students. The problem is that many of our references are in Russian and not translated. So, I would be very grateful if people can recommend: -1) An introduction to group theory suitable for high-school students. Ideally it would emphasize the idea of the symmetry, discuss permutations in details and maybe prove something like Cayley's theorem in the end. -2) A book on the Fundamental Theorem of arithmetics. -3) An elementary book on Galois theory -4) Something very elementary about topology, like Mobius bundles, classification of surfaces, knots, maybe, a little bit of general topology, like Cantor set. -5) A few books for younger students, that is, books on mathematical puzzles or simpler olympiad problems. - -REPLY [2 votes]: I read a book called abstract algebra by dummit and foote second edition. I read it when i was 16 and i found it excellent because it explained everything very well.<|endoftext|> -TITLE: Approximate analytic solution of Schroedinger equation with arbitrary power potential -QUESTION [5 upvotes]: I'm solving the following Schroedinger equation in the domain $r>0$ -$\psi''(r) + \left(E-\frac{a}{r^b}\right)\psi(r)=0 $, -where $0 < b < 2$ and $a, E$ are positive constants. Primarily I'm interested in the asymptotical power behavior of the solution as $r\to 0$. To be complete in the description of the problem, I fix my boundary conditions at $r\to +\infty$ as a plane wave ansatz. -I did a lot of DSolve with Mathematica and found out that $\psi(r)\to const$ as $r\to 0$. It gave me a hint for the power series solution, that one of the terms in the expansion -$\psi(r) = \sum\limits_i a_i r^{\alpha_i}$ -for some (noninteger) $\alpha_i>0$ should cancel with $a/r^b$. However, even with this assumption there are a bunch of terms which do not cancel. Power expansion does not seem to work in this case as well as the WKB approximation (double checked numerically). What are the other known methods to find an approximate asymptotic behavior in this case? A good reference would be great too! - -REPLY [4 votes]: As $r$ approaches zero, the coefficient of $\psi$ becomes dominated by the contribution of $-a/r^b$. This means that in the vicinity of zero your solution is dominated by solution of the following equation -$$ -\psi''-\frac{a}{r^b}\psi=0. -$$ -This can be demonstrated more rigorously (and, also, refined to the higher accuracy) by scaling into the vicinity of $r=0$. This equation possesses an explicit solution in terms of the modified Bessel functions: -$$ -\psi=C_1\sqrt{r}K_{\frac{1}{2-b}}\left(\frac{2\sqrt{a}}{b-2}r^{1-b/2}\right)+C_2\sqrt{r}I_{\frac{1}{2-b}}\left(\frac{2\sqrt{a}}{b-2}r^{1-b/2}\right). -$$ -The latest edition of NIST Handbook tells us that for $z\to 0$ -$$ -I_{\nu}(z)\sim (\frac{1}{2}z)^{\nu}/\Gamma(\nu+1) -\quad\mbox{and}\quad -K_{\nu}(z)\sim \frac{1}{2}\Gamma(\nu)(\frac{1}{2}z)^{-\nu} -$$ -so the actual constant at $r\to0$ depends only on $C_1$, unless $C_1$ vanishes. It seems pretty clear that there are no singularities here; typically you should expect that -$$ -\psi\sim \frac{C_1}{\Gamma(\frac{3-b}{2-b})}\left(\frac{\sqrt{a}}{b-2}\right)^{\frac{1}{2-b}} -\quad\mbox{for}\quad r\to 0. -$$ -Specific values of $C_1$ and $C_2$ are less trivial to obtain, but they can be found by matching this limiting solution to another solution that is valid further away from $r=0$. -The method of matched asymptotic expansions is often used to solve this type of problems asymptotically, see e.g. Kevorkian & Cole (1996) "Multiple Scale and Singular Perturbation Methods", Springer. In this method you would need to construct two solutions: "inner" solution, valid as $r\to 0$, (it is the solution derived above) and "outer" solution, valid as $r\to\infty$ (which you took to be a plane wave). Some ingenuity may be needed to ensure that the ranges of validity of these two solutions overlap, so that they can actually be matched.<|endoftext|> -TITLE: Finite nonabelian groups with few positive real character values? -QUESTION [11 upvotes]: This question was just raised by a colleague (who shall for the moment remain anonymous). It may or may not have a reasonable answer. - -For which finite nonabelian groups $G$ do all irreducible complex characters of degree $>1$ have at most one strictly positive real value (namely the degree)? - -One small example is the alternating group $A_4$. This could be viewed as a sort of degenerate Chevalley group, whose Steinberg character of degree $3$ has other values -$-1, 0, 0$. But for most finite groups of Lie type over a finite field of order $q$, the Steinberg character of degree a power of $q$ will take on more than one positive integral value. While an enumeration of finite groups having the stated property may well be out of reach, it would be of interest to know: - -Are there infinitely many groups $G$ with the special property stated above? - -REPLY [3 votes]: Here is an argument that resticts the structure of such groups of even order, which expands on ARupinksi's comments above (inspiration came from a nice argument in this paper). -Namely, suppose that $1\neq t$ is a real element in $G$, where $G$ is as in the question. Form the following normal subgroups: -$$ K = \langle t \rangle^G = \bigcap_{\chi(1)=\chi(t)} \ker \chi - \quad \text{and} \quad - L = \bigcap_{\chi(t) \neq 0} \ker \chi < K , $$ -where the intersections run over irreducible characters $\chi$ of $G$. Then we have -$$ t^G = K \setminus L,$$ -that is, the elements in $K\setminus L$ are all conjugate in $G$ to $t$. In particular, it turns out that $t$ is rational in the sense that it is conjugate to every generator of $\langle t \rangle$, which is equivalent to every character having rational value at $t$. -Why is this true? Well, column orthogonality yields $\DeclareMathOperator{\Irr}{Irr}$ -$$ 0 = \sum_{\chi \in \Irr G} \chi(t)\chi(1) = |G:K| + \sum_{\chi(t)<\chi(1)}\chi(t)\chi(1). $$ -Now let $y\in K$ and suppose that $t$ and $y$ are not conjugate. Then plugging in $y$ instead of $1$ in the last formula, we see that -$$ \sum_{\chi(t)<\chi(1)} (-\chi(t))\chi(y) = |G:K| - =\sum_{\chi(t)<\chi(1)} (-\chi(t))\chi(1) .$$ -Since $-\chi(t)>0$ and $|\chi(y)|\leq\chi(1)$, it follows that $y$ is in the kernel of every irr char $\chi$ with $\chi(t)< 0$, that is, $y\in L$. Thus $K\setminus L = t^G$. -(Added later:) If $t$ is an involution, then it follows that $x^t=x^{-1}$ for all $x\in L$, in particular $L$ must be abelian, and elements of $L$ are real. -The "dual" argument (exchanging the roles of characters and conjugacy classes) shows the following: Suppose $1\neq \chi$ is real valued, and let $V= \operatorname{\mathbf{V}}(\chi)$ be the vanishing-off group of $\chi$, generated by all group elements on which $\chi$ is non-zero. Then -$$ \Irr( G/\ker \chi ) = \{ \chi \} \cup \Irr(G/V). $$ -From this it follows easily that $V/\ker\chi$ is a conjugacy class of $G/\ker\chi$ and that the only value of $\chi$ besides $0$ and $\chi(1)$ is $-\chi(1)/(|V/\ker\chi|-1)$. (In particular, any real character is rational.) Groups with such an character have been studied by Zhmud, where more information can be found. -I suppose there is also literature on groups having normal subgroups $L\subset K$ such that $K\setminus L$ is a conjugacy class of $G$. The notion of a Camina pair/group seems to be related (see this paper and papers that refer to it).<|endoftext|> -TITLE: Baire Category Theorem Application -QUESTION [6 upvotes]: In Antoine Henrot Michel Pierre - -Variation et optimisation de formes, Une analyse geometrique, a book I'm studying I found an interesting problem. The problem is listed below. The first 3 points of the problem are pretty easy, and I solved them. The 4-th seems a little harder. The only indication I get is to use point 3) and the Baire theorem for $(\Sigma,\delta)$. - -Denote by $\Sigma$ the quotient space - of the family of Lebesgue measurable - sets of $\Bbb{R}^N$ by the equivalence - relation $E_1 \sim E_2 \Leftrightarrow\chi_{E_1}=\chi_{E_2} a.e.$. Denote by - $|X|$ the Lebesgue measure of the - measurable set $X$. -1) Prove that - $\delta(E_1,E_2)=\arctan( |E_1 \Delta E_2|)$ is a distance on $\Sigma$. -2) Prove that given $(E_n)_{n \geq 1}, E$ measurable sets in $\Bbb{R}^N$ the - following three properties are - equivalent. - -$\delta(E_n,E) \to 0$; -$\chi_{E_n}-\chi_E \xrightarrow{\sigma(L^1,L^\infty)} 0$; -$\chi_{E_n}-\chi_E \xrightarrow{L^1} 0$. - -3) Prove that $(\Sigma,\delta)$ is a - complete metric space. -4) Given the sequence $ (f_n)$ of - integrable real valued functions on - $\Bbb{R}^N$, such that for any - measurable set $E$ of $\Bbb{R}^N$ - there exists $\displaystyle \lim_{n\to \infty}\int_E f_n$, prove that if - $|E| \to 0$ then $\displaystyle\sup_n\int_E |f_n| \to 0$. - (Hint: Use the Baire category theorem for $(\Sigma,\delta)$) - -The question is: How can I apply Baire theorem to solve the 4-th point in the problem? - -REPLY [2 votes]: By the way, in a posted solution we used only following properties of functionals $f(E)=\int\limits_{E} f$: - -Continuous respect to $|E|$ (Lebesgue measure) -Additive (on disjoint sets) -Finite - -Therefore, our problem can be reformulated in a following way: -Suppose, that $\mu_n$ is a sequence of finite measures, absolutely continuous w.r.t. the Lebesgue measure. Then $\sup\limits_{n} \ |\mu_n|(E) \to 0$ if $|E|\to 0$. -So we have proved Vitali-Hahn-Saks theorem :-)<|endoftext|> -TITLE: Is (n,m)=(18,7) the only positive solution to n^2 + n + 1 = m^3 ? -QUESTION [14 upvotes]: It's hard to do a Google search on this problem. -If I was using Maple correctly, there are no other positive solutions with n at most 10000. -I know some of these Diophantine questions succumb to known methods, and others are extremely difficult to answer. - -REPLY [3 votes]: I realise this is a massive revival/refresh of this question, but I just found a completely elementary solution of this problem, outlined in this MSE question and my own answer. -Does anyone know if there are other completely elementary solutions?<|endoftext|> -TITLE: Is it possible to do intersection theory in the derived category of a scheme? -QUESTION [8 upvotes]: Let's say we are given a smooth scheme $X$ over $\mathbb{C}$, is it possible to do intersection theory in the (bounded) derived category of coherent sheaves? -I want to know if there is a way to extend the naive thought that for two subschemes $V$ and $Z$ of $X$, $\mathcal{O}_V \otimes^L \mathcal{O}_Z$ is the "intersection" of $V$ and $Z$. Reference guides are also welcome. - -REPLY [2 votes]: This is certainly possible--in fact, it's essentially Grothendieck's original approach to defining the intersection product on the (rational) Chow groups of regular schemes. Grothendieck shows (see these nice notes of Gillet) that there is a graded isomorphism $$\bigoplus_k CH^k(X)_\mathbb{Q}\to Gr^*_{\gamma} K_0(X)$$ where $\gamma$ is a certain filtration on $K_0(X)$; this map is obtained via a sort of Chern character. (As Leo Alonso notes, this may be found in SGA 6.) -How does the derived category come in? Well first, one may recover $K_0(X)$ from the derived category of (bounded) complexes of coherent sheaves on $X$, by taking the free abelian group on objects modulo the relation that $[X]+[Z]=[Y]$ if $$X\to Y\to Z\to \Sigma X$$ is a distinguished triangle. It's not too hard to see that this is the usual $K_0(X)$ (see e.g. 3.1.4 in these notes of Schlichting). The product in $K_0(X)$ is precisely given by $[F]\cdot [G]=[F\otimes^L G]$. -Combining these two results, we exactly realize your dream of doing intersection theory (in the usual sense) via the derived category. (Of course for actual applications, one should check that this product on $CH^k(X)_\mathbb{Q}$ agrees with a more geometric definition, e.g. via deformation to the normal cone as in Fulton.)<|endoftext|> -TITLE: List of Classifying Spaces and Covers -QUESTION [58 upvotes]: I am looking for a list of classifying spaces $BG$ of groups $G$ (discrete and/or topological) along with associated covers $EG$; there does not seem to be such cataloging on the web. Or if not a list, just some further fundamental examples. For instance, here are the ones I have off the top of my head: -$B\mathbb{Z}_n=L_n^\infty$ with cover $S^\infty$ ($B\mathbb{Z}_2=\mathbb{R}P^\infty$) -$B\mathbb{Z}=S^1$ with cover $\mathbb{R}$ -$BS^1=\mathbb{C}P^\infty$ with cover $S^\infty$ -$B(F_2)=S^1\vee S^1$ with cover $\mathcal{T}$ (infinite fractal tree) -$BO(n)=BGL_n(\mathbb{R})=G_n(\mathbb{R}^\infty)$ with cover $V_n(\mathbb{R}^\infty)$ -$B\mathbb{R}=\lbrace pt.\rbrace$ with cover $\mathbb{R}$ -$B\langle a_1,b_1,\ldots,a_g,b_g\;|\;\prod_{i=1}^g[a_i,b_i]\rangle=M_g$ with cover $\mathcal{H}$ (hyperbolic plane tiled by $4g$-sided polygon) -And of course, $B(G_1\times G_2)=BG_1\times BG_2$, so I do not care that much about ''decomposable'' groups. -**The "associated cover" is the [weakly] contractible total space. -[Edit] I should make the comment that $BG$ will be different from $BG_\delta$, where $G_\delta$ denotes the topological group with discrete topology. For instance, the homology of $B\mathbb{R}_\delta$ has uncountable rank in all degrees (learned from a comment of Thurston). - -REPLY [8 votes]: If $G$ is a topological group, let $\Omega(G)$ be the group of loops in $G$ based at $e$, and $PG$ the space of paths in $G$ starting from $e$. Then $\Omega(G)$ acts on $PG$ freely and $PG$ is contractible, so $G$ is a classifying space for $\Omega(G)$.<|endoftext|> -TITLE: Semisimplicity of étale cohomology representations -QUESTION [9 upvotes]: Let $K$ be a number field and $G=Gal(\overline{K}/K)$ the absolute Galois group of $K$. Let $\ell$ be a prime number. -Let $A/K$ be an abelian variety. Then the representation of $G$ on $V_\ell(A)$ is semisimple. This is the famous theorem of Faltings (Invent. Math. 73). -Now let $X/K$ be a smooth projective variety and $0\le q\le 2\dim(X)$, and define $\overline{X}=X_{\overline{K}}$. -Question. Is it known that the representation of $G$ on $H^q(\overline{X}, \mathbb{Q}_\ell)$is semisimple? -Remark. The answer is yes for $q=1$, because $H^1(\overline{X}, Q_\ell)$ is dual to -$V_\ell(A)$ where $A$ is the Albanese variety of $X$. -I would also be interested in the case where the number field $K$ is replaced by a global function field (say), and $\ell$ is assumed to be coprime to the characteristic. - -REPLY [4 votes]: In https://arxiv.org/pdf/1709.04489.pdf, Moonen proves that for finitely generated fields of characteristic $0$, the Tate conjecture (surjectivity of the cycle class map $\mathrm{CH}^r(X) \otimes \mathbf{Q}_\ell \to \mathrm{H}^{2r}(\bar{X},\mathbf{Q}_\ell)^{G_K}$) implies the semisimplicity conjecture. -(For the Tate conjecture, see e.g. http://www.math.columbia.edu/~chaoli/docs/TateConjecture.html or Tate's article in the Motives volume.)<|endoftext|> -TITLE: characters on a finite group with `extremal' behaviour -QUESTION [8 upvotes]: The following question is a bit technical, and I haven't got to grips with it enough to be able to present it as a well-focused question. However, my hope is that the collective group-theoretic expertise on MO may spot something in there that follows from standard results, or which is likely to be intractable by way of connection to open problems. -Let $G$ be a finite group, equipped with normalized counting measure. If $\chi$ is an irreducible character on $G$ then standard theory/calculations show that it has $\ell^2$-norm equal to $1$; it follows easily from this that the $\ell^1$-norm of $\chi$ is $\geq d^{-1}$ where $d$ is the degree of $\chi$. If equality actually holds (which, as I'll discuss below, seems to be very restrictive) then let's say that $\chi$ is $1$-minimal (non-standard terminology). -$1$-minimal characters of degree $>1$ give rise to some interesting ideals in the centre of (algebras built from) $\ell^1(G)$, which I and some colleagues have been looking at. We have run into the following question which seems to lie out of our main expertise. -Question 1. for which groups $G$ are all the irreducible characters $1$-minimal? -Call such a group $1$-minimal, for want of a better term at present. -Every abelian group is $1$-minimal, as is the dihedral group of order 8 (to my initial surprise). It is not hard to show that products of $1$-minimal groups also have the property. -Here is the small amount I can show so far. A little calculation shows that if $\pi$ is an irrep of $G$, then its trace $\chi$ is a $1$-minimal character if and only if $\pi(x)$ is either traceless or a scalar multiple of the identity. Writing $d$ for the degree of $\pi$, it follows that the support of $\chi$ is a normal subgroup $K$ of $G$, containing $Z(G)$, with the index of $K$ in $G$ being $d^2$. This already seems to preclude many groups from being $1$-minimal. -The above shows that $d\chi$ is induced from some 1-dimensional representation of $K$, but I haven't succeeded in using this to get much leverage. Other partial results: if $\pi$, $\chi$, $d$, $K$ are as above, and every non-linear character of $G$ has degree $\geq d$, then one can play around with the conjugation representation associated to $\pi$ to show that $G/K$ is abelian. Unfortunately, restricting attention to cases where $d$ is small makes it harder to get control on $K$... -So far, the only examples of $1$-minimal groups that I have found are products of abelian groups with copies of $D_4$. All these examples are solvable and have non-trivial centre, which leads me to some more focused questions. -Question 2. Can we characterize those solvable, finite groups which are $1$-minimal? -Question 3. Do there exist $1$-minimal groups with trivial centre? -A final remark/piece of context: some foraging via Google and MathSciNet suggests that this problem might be connected with the old problem of characterizing the finite groups of "central type", namely those groups $G$ for which there is an irreducible character of degree $d$ where $d^2= | G: Z(G) |$ ( we always have $d^2\leq |G:Z(G)|$ for any group). However, it is not clear to me that the present question is either contained in, or contains, this older one. - -Update 5th Jan. 2012: Thanks to the arguments given below by F. Ladisch and M. Isaacs, we have the following: every $1$-minimal group is nilpotent (so that Question 3 has a negative answer if we exclude the trivial group) and every nilpotent group of class $2$ is $1$-minimal. -For an explanation of why I and my colleagues were interested in the condition of $1$-minimality, see Theorem 4.2 of our preprint 1110.6683. Note for the record that parts of the paper form part of the PhD thesis of the first author, which is at time of writing still in progress. - -REPLY [7 votes]: All finite groups that are nilpotent with nilpotency class at most 2 are 1-minimal in the sense of this question. Let $\chi \in {\rm Irr}(G)$ and write $Z = {\bf Z}(\chi)$. By Theorem 2.31 of my character theory book, $|G:Z| = \chi(1)^2$ if $G/Z$ is abelian. This condition always holds if $G$ has nilpotence class $2$, because in that case, $G/{\bf Z}(G)$ is abelian, and we certainly have ${\bf Z}(G) \subseteq Z$. By Corollary 2.30, it follows that $\chi$ vanishes on $G - Z$, and we know that this condition is equivalent to saying that $\chi$ is 1-minimal.<|endoftext|> -TITLE: delooping under Dold-Kan and simplicial delooping -QUESTION [8 upvotes]: What maps of simplicial sets exist between - -the image under the Dold-Kan correspondence of a chain complex shifted up in degree -and the image under the right adjoint to simplicial looping of the DK-image of the unshifted complex - -? -Here is the same question in detail: -Write -$$ - (G \dashv \bar W) : sGrp \stackrel{\leftarrow}{\underset{\bar W}{\to}} sSet_0 \hookrightarrow sSet -$$ -for the adjunction between simplicial groups and reduced simplicial sets whose left adjoint is the simplicial loop group functor (as for instance in Goerss-Jardine, chapter V); -and write -$$ - Ch_\bullet^+ \overset{\Xi}{\to} sAbGrp \hookrightarrow sGrp \overset{U}{\to} sSet -$$ -for the Dold-Kan correspondence, where in both cases I care about the images as simplicial sets. -Then for $V \in Ch_\bullet^+$ a chain complex and $V[1]$ (or $V[-1]$ if you prefer) its shift up in degree (its delooping as a chain complex) the two simplicial sets -$$ - U \Xi (V[1]) -$$ -and -$$ - \bar W (\Xi V) -$$ -should have the same homotopy type. What nice natural maps of simplicial sets do we have between them? - -REPLY [3 votes]: There's an explicit natural isomorphism between the two functors. -Rick Jardine says as much, but for the image of the functors in the category of chain complexes (i.e. after applying the normalization). You can find this in Goerss, Jardine Remark III.5.6, or in greater depth in section 4.6 of Jardine's book on Generalized Etale Cohomology. -The combinatorics for the isomorphism in simplicial abelian groups means that the isomorphism takes a little longer to state, but I could send you a pdf with everything written out if this would be useful.<|endoftext|> -TITLE: Non degenerate representations for C*-algebras -QUESTION [7 upvotes]: Hi! -While studying C*-algebras I found 2 different definitions for non degenerate representations (-homomorphisms $\pi:\mathcal{A} \rightarrow B(\mathcal{h})$ where $\mathcal{A}$ is a C-algebra and $B(\mathcal{h})$ is the space of bounded linear operators on some Hilbert space $\mathcal{h}$): -1) For every non-zero $\xi \in \mathcal{h}$ there exists $a \in \mathcal{A}$ such that $\pi(a)\xi \neq 0$; -2) The set $\{\pi(a)\xi \quad a \in \mathcal{A}, \xi \in \mathcal{h}\}$ is dense in $\mathcal{h}$. -Are they equivalent? -Thanks, -Alessandro - -REPLY [6 votes]: In fact, for unital $C^*$-algebras non-degeneracy just means $\pi(1) = 1$. In the non-unital case there is even a sharper statement than your item (2): One can find for every $\phi$ and every $\epsilon > 0$ another vector $\psi$ and a positive algebra element $a \in \mathcal{A}^+$ with -\begin{equation} - \phi = \pi(a)\psi - \quad - \textrm{and} - \quad - \|\phi - \psi\| < \epsilon. -\end{equation} -This is nice as it shows that we do not just get a dense subspace and we get in some sense as close as possible to $\pi(1) = 1$. I found this in Blackadars encyclopedia book in Theorem II.5.3.7 and in II.6.1.5. Might be worth a look :)<|endoftext|> -TITLE: Atiyah class for non-locally free sheaf -QUESTION [6 upvotes]: Let $E$ be a holomorphic vector bundle over a compact complex manifold (or projective algebraic variety) $X$. -The Atiyah class of $E$, $a(E)\in Ext^1(T_X,End(E))$, is defined to be the class of the extension -$$ -0 \rightarrow End(E) \rightarrow \mathcal{D}(E) \rightarrow T_X \rightarrow 0 -$$ -where $\mathcal{D}(E)$ is the bundle of differential operators from $E$ to $E$ of order $1$ and scalar symbol, the map to the tangent being the symbol map. -It is a theorem of Atiyah that $a(E)$ generates the characteristic ring of $E$. -My question is: what can be said if $E$ is not a vector bundle, but just a coherent torsion free $\mathcal{O}_X$-module? Could a similar statement be true? -One has anyway the characteristic ring of $E$. -To me looks like (although I may be wrong) that one can construct $\mathcal{D}(E)$ that fits the same exact sequence. -The problem is that in Atiyah's theory is essential that $E$ is locally free, since he proves the result through the curvature of connections on $E$, and these does not exist if $E$ is not locally free. -Is there any technique (from K-theory?) that would help? Or my problem is senseless? - -REPLY [14 votes]: It is better to define the Atiyah class as an element of $Ext^1(E,E\otimes\Omega^1)$. Then it is defined for all coherent sheaves, and even for all objects of the derived category. The most convenient definition is the following. Look at $X\times X$, let $\Delta:X \to X\times X$ be the diagonal, and $I$ --- the ideal sheaf of the diagonal. Then we have an exact sequence -$$ -0 \to I/I^2 \to O/I^2 \to O/I \to 0 -$$ -on $X\times X$. Since $I/I^2 \cong \Delta_*\Omega^1_X$, it gives a morphism $\Delta_*O_X \to \Delta_*\Omega^1_X[1]$ in the derived category $D(X\times X)$. Now denote $p,q:X\times X \to X$ the projections, take any $E \in D(X)$, tensor this morphism by $p^*E$ and apply $q_*$. We will get a morphism -$$ -q_*(p^*E \otimes \Delta_*O_X) \to q_*(p^*E \otimes \Delta_*\Omega_X^1)[1]. -$$ -The projection formula shows that the first term is $E$, and the second is $E\otimes\Omega^1_X[1]$. So, we constructed an element in -$$ -Hom(E,E\otimes\Omega^1_X[1]) = Ext^1(E,E\otimes\Omega^1_X). -$$ -This Atiyah class has all the nice properties of the classical one. For example, one can express the coefficients of the Chern character as traces of its powers.<|endoftext|> -TITLE: Number field analogue of the Goldbach Conjecture -QUESTION [8 upvotes]: Is there a generalization of Goldbachs conjecture for prime ideals in number fields? - -REPLY [16 votes]: Googling the obvious, «Goldbach number fields», comes up with a bunch of relevant results, In particular, a link to [Takayoshi MITSUI, -On the Goldbach problem in an algebraic number field I. J. Math. Soc. Japan, Vol 12, No. 3, 1960] here<|endoftext|> -TITLE: What is the intuition behind the Freudenthal suspension theorem? -QUESTION [27 upvotes]: The Freudenthal suspension theorem states in particular that the map -$$ -\pi_{n+k}(S^n)\to\pi_{n+k+1}(S^{n+1}) -$$ -is an isomorphism for $n\geq k+2$. -My question is: What is the intuition behind the proof of the Freudenthal suspension theorem? - -REPLY [3 votes]: Edit October 2013: I have given more information on the Blakers-Massey triad theorem on the ncatlab here, referring specifically to the algebraic determination of the critical, i.e. first non vanishing, group. This is described as a tensor product of relative homotopy groups, and of course a tensor product is trivial if one of the factors is trivial. So we get an intuitive view of the connectivity result and so for the Freudenthal Suspension Theorem. Indeed, they are both seen as a consequence of a higher order homotopy Seifert-van Kampen Theorem. -Original answer: -John Klein's answer refers to the Blakers-Massey triad theorem; this was generalised to an $n$-ad connectivity theorem by Barratt and Whitehead, and of which Goodwillie has given a geometric proof, using general position arguments. -In these results there is interest in describing the critical group, an $n$-ad homotopy group, which may be nonabelian in general. Some abelian cases have been described by Barratt and Whitehead, following Blakers-Massey, with simple connectivity assumptions. -This problem was solved in Theorem 3.7 of -Ellis, G.J. and Steiner, R. "Higher-dimensional crossed modules and the homotopy groups of $(n+1)$-ads." J. Pure Appl. Algebra 46 (2-3) (1987) 117--136. -which uses crucially the van Kampen Theorem for $n$-cubes of spaces in -Brown, R. and Loday, J.-L. "Van Kampen theorems for diagrams of spaces", Topology 26~(3) (1987) 311--335. With an appendix by M. Zisman. -Of course connectivity results follow from the algebraic results, but the proof in this last paper is by induction with connectivity in one dimension implying surjectivity in the next. -One intuitive point is that the description of the critical group of an $n$-ad will involve generalised Whitehead products coming from sub $r$- and $s$-ads of the given $n$-ad, and an elaborate algebra (crossed $n$-cubes of groups, as defined by Ellis-Steiner) is needed to handle all this. The proofs do not involve general position at all. -For a fairly recent application of these ideas, see -Ellis, G.~J. and Mikhailov, R. "A colimit of classifying spaces". Advances in Math. (2010) arXiv: [math.GR] 0804.3581v1.<|endoftext|> -TITLE: What does a homogeneous space of a linear algebraic group know about the group? -QUESTION [7 upvotes]: Let $X=G/H$, where $G$ is a connected linear algebraic group over the field $\mathbf{C}$ of complex numbers -and $H\subset G$ is an algebraic subgroup. -In general, we can write the algebraic variety $X$ as $X=G'/H'$ with $G'$ non-isomorphic to $G$. -What can be said about $G'$ (and $H'$)? -Of course, we can always take any surjective homomorphism $G'\to G$, then $X$ is naturally a homogeneous space of $G'$. -Conversely, if the action of $G$ on $X$ is not effective, we can take a normal subgroup $N$ of $G$, contained in $H$ -(say, the kernel of the action) and set $G'=G/N,\ H'=H/N$, then $X=G'/H'$. I am interested in less trivial examples. -I would be happy to get a general answer, but will be also grateful for special cases, examples, comments, etc. -I am obliged to Günter Harder for this nice title of my question. - -REPLY [4 votes]: To add perspective to Dave's answer, it seems that homogeneous spaces realized as quotients of a connected reductive group $G$ by parabolic subgroups are often natural and rich in structure besides having close connections with $G$. But I'm less confident about other types of homogeneous spaces. There are lots of arbitrary constructions, e.g., start with a product of connected 1-dimensional groups and factor out the product of one or more of them. At that extreme the homogeneous space has little connection with the given group. -Even though factoring out a parabolic subgroup (especially a maximal one) tends to produce a relatively "small" variety, its automorphism group is often close to $G$ (as seen in Demazure's paper). The only other type of maximal closed connected proper subgroup in $G$ is itself reductive, as follows from Borel-Tits theory in arbitrary characteristic: see for instance Section 30.4 of my Springer text GTM 21. But then you get an affine homogeneous space. In fact, R.W. Richardson's classic theorem in Bull. London Math. Soc. 9 (1977) shows that $G/H$ is affine iff the closed subgroup $H$ of the reductive group $G$ is itself reductive. -All of which leads me to wonder how realistic it is in these affine cases to expect the homogeneous variety to know much about the given group? In other words, I'm uncertain about how broad the statement of the original question should be.<|endoftext|> -TITLE: Mathematics TV clips -QUESTION [23 upvotes]: I have just come out of a talk on presenting science to the public via mainstream media. I became jealous of what other sciences, such as astrophysics, have achieved in these media. -This is tremendously borderline for MO, so feel free to close. But I'd like to solicit ideas for 2-3 minute TV clips about mathematics which could successfully compete with "private lives of celebrities" for the public's attention. Maybe TV producers could then glance through someday, and turn one or more of our fantasy TV clips into real TV clips. -Criteria are: - I'd like us to focus on 2-3 minute pieces. - No "talking heads". People will change channel rather than watch two people talking. Everything has to be visual. - The mathematical idea has to play lead role in some sense ("oh the thinks you can think!"), and has to be communicated to some degree. It shouldn't be precise, but it shouldn't be wrong either. - The scene has to be set, and the viewer's interest must be grabbed- not just some sitting at a desk. - No "green peas"- nobody "should know". It should be popularization as opposed to education. - The target audience has to be the general public, as opposed to people predisposed to watching Discover Channel and to reading scientific magazines. - -Here's my concept as I was listening to the talk, to start things off: -[A 2-minute presentation on basic knot theory. No words needed- it can (and maybe should) be a music video.] -A poor boy in India sees a street conjurer making various knots by a wrist movement on a rope with a weight at the end [there's an old Indian gentleman in Oakland who can do this]. It fills him with wonder, and he resolves to upgrade the trick by knotting a knot no knotter has knotted- a bodacious knot trick to shock the world. -Sinking into his imagination [effect], you see him throw the rope, which wraps itself in an amazing Bollywood way into something which looks like really complicated (from this MO question). He shows it to the conjurer, who untangles it in a second without touching the ends [maybe cut to a computer animation of this, which would be cleaner]. He tries again and fails again, maybe a few times. Eventually he brings one of Haken's gordian unknots, which is also untangled. -He returns dejected, and then suddenly "inspiration strikes". He throws a superhuman Bollywood knot confidently in front of the conjurer, who, try as he may, cannot untie it (without touching the ends). Overlaying animation, a tricolouring flashes over the knot like a dragon, which moves and changes with the knot as the conjurer struggles with it, never changing. Maybe "Can't get rid of the colours!" flashes on-screen, to complement the visual cue of what is going on. The boy wins. -However, the conjurer unknots it to a trefoil, which is an insufficiently bodacious knot, and the boy tries a final time and produces a better knot, which isn't a trefoil either because it's 5-colorable. Again, this is shown with visual effects, blending animation with live action. With more time one might try more colourings. -The boy laughs, and the conjurer stares at him in wonderment- the finest knotter who ever has knotted! Maybe then it returns to the real world, and the boy walks away. - -REPLY [6 votes]: I would be interested to make a short video to show how we can use geometry to explain some things related to an optical illusion. The "spinning dancer" is a rotating silhouette, which appear at the second sight that you can't tell in which direction rotates. At first sight it seems that everybody choses a particular direction. -The script is something like this - -introduce the illusion -let the viewer decide in which direction she thinks the ballerina rotates -explain why two people out of three see the dancer spinning clockwise -explain why the correct answer is that the dancer spins counter-clockwise - -(I explained the last two points here, using simple geometry in space and some elementary notions on perspective) - -REPLY [2 votes]: I want for years to do a short video based on an idea I had in 2004: -Polyhedral groups -which goes like this: - -introducing the Platonic solids -showing by animations the corresponding polyhedral groups -showing by animations what dual polyhedra are, and why their groups are the same. - -So far, this was known stuff. Here is the new part: - -showing that we can label the vertices and edges with permutations, so that we can use two identical polyhedra to compose permutations. I describe this idea here.<|endoftext|> -TITLE: Epimorphisms have dense range in TopHausGrp? -QUESTION [10 upvotes]: Consider the category of Topological Groups with continuous homomorphisms. Then a continuous homomorphism $f:G\rightarrow H$ with dense range is an epimorphism. Is the converse true? If not, what about for locally compact groups? -Even for groups, without topology, this is not trivial-- Wikipedia points me to a simple proof given by Linderholm, "A Group Epimorphism is Surjective", The American Mathematical Monthly Vol. 77, No. 2 (Feb., 1970), pp. 176-177 see http://www.jstor.org/pss/2317336 It is far from obvious to me that this argument extends to the topological case (but perhaps it does). -Edit: As suggested in the comments, I really was to ask about Hausdorff topologies. - -REPLY [13 votes]: Google, MathSciNet and some ferreting lead me to - -MR1235755 (94m:22003) - Uspenskiĭ, Vladimir(D-MNCH) - The solution of the epimorphism problem for Hausdorff topological groups. - Sem. Sophus Lie 3 (1993), no. 1, 69–70. - -where the review indicates that the answer is negative in general, but positive for locally compact groups; this latter case was apparently treated in - -MR0492044 (58 #11204) - Nummela, Eric C. - On epimorphisms of topological groups. - Gen. Topology Appl. 9 (1978), no. 2, 155–167. - -The case of compact groups had been done earlier by Poguntke: - -MR0263978 (41 #8577) - Poguntke, Detlev - Epimorphisms of compact groups are onto. - Proc. Amer. Math. Soc. 26 1970 503–504. - -and this apparently inspired the authors of the following paper - -MR1338245 (96c:46054) - Hofmann, K. H.(D-DARM); Neeb, K.-H.(D-ERL-MI) - Epimorphisms of $C^∗$-algebras are surjective. - Arch. Math. (Basel) 65 (1995), no. 2, 134–137.<|endoftext|> -TITLE: How to/Can you get a PhD position when you need more experience first? -QUESTION [11 upvotes]: My friend loves mathematics and wants to continue research as a -mathematics PhD student, but this doesn't seem possible! -She did a prestigious but inadequate bachelors program (almost no math) and -because of this she is having a very hard time getting good grades for her master courses. -Thus she feels she has inadequate knowledge for a PhD position; which is probably right. -Her thesis adviser told her that she cannot continue a career in -research mathematics: no university would consider her given the situation now nor if she took an extra year or two to study more. - -Question: Is this true? Does it really stop here? -I'm interested in similar experiences, opinions and suggestions (from -experienced professionals). -I'm particularly not looking for a rosy picture! - -In my opinion she is very talented, motivated, original and quick! She -would flourish given more time to take specific foundational bachelor -and deepening masters courses. Until now she performed extremely well -(near perfect score) and this is confirmed by for example two scholarships she -won (BA and MA). -PS. I tried to keep it terse but please ask for further details if needed. -Thanks a lot! - -Update: Thank you all very much for the responses and useful comments. -It turns out that there is much more "staff rotation" in applied PhD programs and - it seems plausible that she can find a nice position there; from which to continue - her studies. -Thanks again. - -REPLY [2 votes]: Does she really need to become PhD Student in Mathematics? Why doesn't she try with something different? -I mean, a mathematician is, in my experience, not just a person who studies abstract mathematics. There are plenty of other occupations that involve the same skills and duties as pure mathematics. There is a lot of mathematics needed into other similar subjects, like computer science. Moreover, the interdisciplinarity of computer science should help her in getting such a position. Finally, original thinking and ability to formalize and being a hardworker are skills really needed and rewarded in this field! -To say my experience, I graduated in Mathematics (in the field of arithmetic geometry) but now I'm a PhD Student in computer science... and despite my lack of knowledge in specific topics, like programming, I feel I'm really helpful here! -Moreover, in some areas of Mathematics the research borders are quite far, so starting "late" could imply facing with difficulties in the future... or at least, I feel that this is what the most mathematicians think about this, and probably is the reason why she's having hard times in finding a PhD position in Maths...<|endoftext|> -TITLE: Fixed points of Group Endomorphisms -QUESTION [5 upvotes]: Suppose $G$ is a finitely presented group with generators $a_1, \ldots, a_n$. Suppose $f \colon G \to G$ is a group endomorphism specified by defining $f(a_1), \ldots, f(a_n)$. As expected, we define a fixed point of $f$ to be any element $g \in G$ such that $f(g) = g$ and, as $f(\mathop{id}) = \mathop{id}$, we say that $\mathop{id}$ is the trivial fixed point. -For example, let $G = \langle a | \rangle$ and $f$ and $g$ be defined by $f(a) = \mathop{id}$ and $g(a) = a^2$. Note in both cases $f$ and $g$ have no non-trivial fixed points and for this particular group we can determine that an endomorphism $f$ has a non-trivial fixed point if and only if $f(a) = a$. - -For what groups is it possible to determine whether or not any given endomorphism has a non-trivial fixed point? - -I am particularly interested in the question of: - -Is $\langle a, b, c | \rangle$ such a group? - -REPLY [5 votes]: For the free group an algorithm is here: Sykiotis, Mihalis -Fixed points of symmetric endomorphisms of groups. -Internat. J. Algebra Comput. 12 (2002), no. 5, 737–745.<|endoftext|> -TITLE: Well-foundedness and elementary embeddings -QUESTION [5 upvotes]: I have been reading about elementary embeddings in set theory and there is a question that has been nagging me: -Typically, one looks at elementary maps $j:V\to M$ with $M$ well-founded. Without any assumptions on the large cardinal strength of $V$, we cannot give examples of such maps. Also, typically, $j$ comes from an ultrapower construction. -If no non-principal ultrafilter is $\sigma$-complete, no ultrapower embedding $j:V\to M$ can even have $\omega^M$ standard (unless the ultrafilter is principal, of course). -Now, the question(s): - -Is there any strength in the assumption that there is a (non-trivial) $j:V\to M$ elementary with $\omega^M$ standard? - -It could be that this already ensures measurable cardinals, or it could be that the (consistency) strength increases with the well-foundedness of $M$ (I mean, with the standard part of $ORD^M$). -Note I am not assuming that $j$ comes from an ultrapower. If it matters, say the discussion takes places in NBG (or whatever is appropriate) so we can argue freely about classes. - -REPLY [6 votes]: One natural interpretation of the question does give a measurable cardinal. Namely, suppose that $j:V\to M$ is an elementary embedding for which $M$ is an $\omega$-model. More precisely, we have a membership relation $E$ on $M$ and $j:\langle V,\in\rangle\to\langle M,E\rangle$ is an elementary embedding of these structures. Suppose that $j$ is nontrivial in the sense that it is not an isomorphism of every set with the $E$ predecessors of its image. In this case, I claim that there is a measurable cardinal. Fix any set $A$ for which $j$ is not an isomorphism of the elements of $A$ with the $E$-predecessors of $j(A)$. In this case, there must be an element $a\mathrel{E} j(A)$ which is not in the range of $j$. We may define a measure $\mu$ on $A$ by $X\in\mu\iff a\mathrel{E} j(X)$ for $X\subset A$. This is easily seen to be a nonprincipal ultrafilter on $A$. Furthermore, the fact that $M$ is an $\omega$-model will give us countable completeness of $\mu$, for if $X_n\in\mu$ for each $n$, the fact that $\omega^M$ is actually order type $\omega$ implies that every $E$ element of $\omega^M$ is $j(n)$ for some $n$, and from this it follows that the $E$ members of $j(\cap_n X_n)$ are precisely those in every $j(X_n)$. Thus, $a\in j(\cap_n X_n)$ and so $\cap_n X_n\in \mu$. So we have constructed a countably complete nonprincipal ultrafilter on a set $A$, and this implies the existence of a measurable cardinal. (The degree of completeness of $\mu$ is a measurable cardinal.) -Update. But there is another interpretation for which the property is weaker than measurability. Namely, it might happen that $j$ is an isomorphism of every set with the $E$-predecessors of its image, but $j$ is not an isomorphism of $\langle V,\in\rangle$ with $\langle M,E\rangle$. Thus, $j$ maps $V$ isomorphically to an initial segment of $M$, but $M$ has new objects at ranks above the ordinals of $V$. So essentially, $V$ is an elementary initial segment of $M$, with $j$ being essentially the inclusion map. This kind of situation can occur in general, since it is possible that $V_\delta\prec V_\kappa$, for example, when $\kappa$ is inaccessible, there are many such $\delta$. But for your question, we would have the situation where $M$ and $E$ are classes in $V$ that $V$ sees to be ill-founded. This is a more subtle situation, but one can build such an example starting only with a weakly compact cardinal, which is strictly weaker than measurability in consistency strength. Suppose $\kappa$ is weakly compact. Let $M_0$ be a transitive set of size $\kappa$, with $\kappa\in M$ and $V_\kappa\subset M$. Since $\kappa$ is weakly compact, there is an elementary embedding $j_0:M_0\to M_1$ to a transitive set $M_1$ of size $\kappa$ with critical point $\kappa$. Applying weak compactness again, we get a map $j_1:M_1\to M_2$, again with critical point $\kappa$. Iterating this, we build $j_n:M_n\to M_{n+1}$ with critical point $\kappa$ each time. Thus, we get an elementary map $j$ from $M_0$ to the direct limit model, which is a structure $\langle N,E\rangle$ that includes $V_\kappa$ in its well-founded part, but which is ill-founded above $\kappa$ and indeed, it has no $\kappa$-th element (since the critical point was $\kappa$ each time, new elements were inserted below any given element of the thread above $\kappa$). The ill-foundedness of the model exists below $j(\kappa)$. Thus, the structure $\langle V_\kappa,\in\rangle$ has an elementary extension to the $j(\kappa)$-rank initial segment of the direct limit. That is, we have $\langle V_\kappa,\in\rangle\prec\langle M,E\rangle$. The structure $M$ and relation $E$ on $M$ have size $\kappa$ and can therefore be coded using subsets of $\kappa$. But now the key point is that since $\kappa$ is inaccessible, we may equip $\langle V_\kappa,\in\rangle$ with all its subsets and still have a GBC model, indeed, a model of Kelly-Morse set theory. This model can see the elementary extension of $V_\kappa$ to the ill-founded model $M$, which is well-founded up to the height of $V_\kappa$. So this is an example of an elementary embedding of the $V$ of $V_\kappa$ into a class model that is an $\omega$-model, but still ill-founded, and is nontrivial in the sense that it is not an isomorphism, but one cannot extract any measurable cardinal, because it was built merely from a weakly compact cardinal. -(See the edit history for my earlier answer, which included some related information, which seems less relevant to me now.)<|endoftext|> -TITLE: Non finitely generated graded ring of a divisor in dimension >2 -QUESTION [8 upvotes]: I spent some time looking for an example of non finitely generate graded ring -$R=\oplus H^0(X,mD)$ where $D$ is a divisor on a variety $X$ of dimensison $>2$. -I know there are several such examples (e.g Zariski's), but they are all on surfaces. -I believe there must exist many examples. Do you know any? - -REPLY [9 votes]: I give Siu's analytical method to test that the ring $$R(X,L)=\bigoplus_{m=1}^\infty H^0(X,mL)$$ is not finitely generated -Let $X$ be a compact complex manifold and the ring $R(X,L)=\bigoplus_{m=1}^\infty H^0(X,mL)$ is finitely generated and let $s^{(m)}_1, … , s^{(m)}_{q_m}\in H^0 (X,mL) $ -be a basis over $\mathbb C$. Let -$$ \Phi =\sum_{m=1}^\infty \epsilon_m( ∑_{j=1}^{q_m} |s^{(m)}_j|^2)^{1/m} $$ -where $\epsilon_m$ is some sequence of positive numbers decreasing fast enough to guarantee convergence of the series. Then all the Lelong numbers of the closed positive $(1,1)$-current $$T=\frac{\frak{\sqrt{-1}}}{2\pi}\partial\bar\partial \log Φ$$ are rational numbers. -So one of ways to find some examples is to show that the Lelong number of $T$ is not rational at some point, hence the ring $R(X,L)=\bigoplus_{m=1}^\infty H^0(X,mL)$ is not finitely generated -Definition: Let -$W\subset \mathbb C^n$ -be a domain, and $\Theta$ a positive current of degree $(q,q)$ on -$W$. For a point $p\in W$ -one defines -$$\mathfrak v(\Theta,p,r)=\frac{1}{r^{2(n-q)}}\int_{|z-p| -TITLE: Morphisms of Banach spaces -QUESTION [8 upvotes]: What is the standard name in English for bounded linear maps $f:E\to F$ between Banach spaces such that the kernel $\ker(f)$ has a complement, and $\text{im}(f)$ is closed, and has a complement? -Apparently, when $f$ is injective or surjective, it is sometimes referred to as an admissible monomorphism or epimorphism, respectively. Is this terminology standard? - -REPLY [10 votes]: I don't know if there's a universally accepted name for this notion. Of course, if the kernel and the complement of the image are finite-dimensional, such a map is a Fredholm operator. I would suggest to call them pseudo-invertible (in reference to the Moore-Penrose pseudo-inverse), because $f:E \to F$ has complemented kernel and image if and only if there is a $g:F \to E$ such that $fgf = f$ (and $gfg = g$). -Since the question is tagged homological algebra, let me point out that these morphisms are precisely the morphisms factoring as $E \twoheadrightarrow I \rightarrowtail F$, where $\twoheadrightarrow$ stands for a split epimorphism and $\rightarrowtail$ for a split monomorphism. These are the admissible monomorphisms and admissible epimorphisms for the split exact structure (in the sense of Quillen) on the additive category of Banach spaces and bounded linear maps, see here for more on this. However, there are several exact structures on the category of Banach spaces (at least three of interest), so admissible monomorphism/epimorphism alone is not good enough from this point of view. -I've seen several names for the maps you're asking about in the literature: - -In Borel-Wallach, Continuous cohomology, discrete subgroups and representations of reductive groups, Annals of Math. Studies 94, Princeton University Press (1980), IX 1.5, they are called $s$-morphisms (actually for Hausdorff locally convex spaces). -Guichardet, Cohomologie des groupes topologiques et des algèbres de Lie, Textes Mathématiques 2, Fernand Nathan, Paris (1980), Appendice D, Def. D.1 calls them strong (again in the setting of Hausdorff locally convex spaces). -Monod, Continuous Bounded Cohomology of Locally Compact Groups, Springer Lecture Notes in Mathematics, 1758 (2001), Definition 4.2.2, calls them weakly admissible (the weakly refers to the fact that his notion of admissibility requires that if $f$ is of norm $1$, there is a pseudo-inverse $g$ of norm $1$, at least for monomorphisms).<|endoftext|> -TITLE: Exotic principal ideal domains -QUESTION [23 upvotes]: Recently I realized that the only PIDs I know how to write down that aren't fields are $\mathbb{Z}, F[x]$ for $F$ a field, integral closures of these in finite extensions of their fraction fields that happen to have trivial class group, localizations of these, and completions of localizations of these at a prime. Are there more exotic examples? Is there anything like a classification? - -REPLY [4 votes]: A commutative algebra is a PID if and only if it is a UFD and all nonzero prime ideals are maximal. This leads to an interesting method to construct PID's: Let $R$ be a UFD and let $S \subset R$ be a multiplicative set such that, for any prime $\mathfrak{p} \subset R$ of height $\geq 2$, there is some $f \in S$ with $f \in P$. Then $S^{-1} R$ will be a PID, because localizations of UFD's are UFD's and the poset of prime ideals in $S^{-1} R$ is obtained from the poset of prime ideals in $R$ by deleting those ideals containing an element of $S$. -This can be useful for building counterexamples, because $S^{-1} R$ is the forward limit of $f^{-1} R$ over all $f \in S$, and each of the $f^{-1} R$ will be a UFD but not a PID, so one can take counterexamples in UFD's and make them into PID counterexamples by this trick. Speaking vaguely, although $S^{-1} R$ has Krull dimension $1$, it often acts more like a ring of dimension equal to the Krull dimension of $R$. -I learned about this construction from Grayson's paper "$SK_1$ of an interesting principal ideal domain". The PID in question is to take $R = \mathbb{Z}[T]$ and $S = \{ T \} \cup \{ T^n-1 : n > 0 \}$, and the interesting property is that $SL_n(S^{-1} R)$ is not generated by elementary matrices. -I can't resist showing off: After I read Grayson's paper, I come up with the following simpler example. Let $R = \mathbb{R}[x,y]$ and let $S$ be the set of nonzero polynomials in $\mathbb{R}[x^2+y^2]$. Then $S^{-1} R$ is a PID by the above argument. I claim that $M= \left[ \begin{smallmatrix} x/(x^2+y^2) &y/(x^2+y^2) \\ -y&x \end{smallmatrix} \right]$ is not a product of elementary matrices. Suppose that $M=E_1 E_2 \cdots E_n$. Then the denominators of the $E_j$ only contain finitely many elements of $S$, so all the $E_j$ lie in $f(x^2+y^2)^{-1} R$ for some nonzero polynomial $f$. Choose some real number $r$ so that $f(r^2) \neq 0$, then each of the $E_j$ is a well defined continuous function on the circle $x^2+y^2 = r^2$. So $M=E_1 E_2 \cdots E_n$ gives a map from this circle to $SL_2(\mathbb{R})$. Consider the class of this map in $H_1(SL_2(\mathbb{R})) \cong \mathbb{Z}$. Rescaling each off diagonal entry of the $E_j$ by a real number $t$ and sliding $t$ from $1$ to $0$ is a homotopy to the trivial map, so this class is $0$. On the other hand, $\left[ \begin{smallmatrix} x/(x^2+y^2) &y/(x^2+y^2) \\ -y&x \end{smallmatrix} \right]$ represents the generator of $H_1$, a contradiction. The same argument shows that the block matrix $\left[ \begin{smallmatrix} M & \\ & \mathrm{Id}_{n-2} \end{smallmatrix} \right]$ in $SL_n(S^{-1} R)$ is also not a product of elementary matrices (this time we have $H_1(SL_n(\mathbb{R}))\cong H_1(SO_n(\mathbb{R})) \cong \mathbb{Z}/2$, and we need spin groups to compute the class in $H_1$, but I think it still works.).<|endoftext|> -TITLE: Infinite Grassmannians and their coordinate rings -QUESTION [8 upvotes]: I'm currently thinking about some combinatorics associated to an infinite analogue of the coordinate rings of the Grassmannians $Gr(2,n)$. The combinatorics should be thought of as relating to Plucker coordinates $\Delta^{ij}$ but with $i < j$ arbitrary integers, rather than restricted to $\{1,\ldots ,n\}$. So I've been trying to find the right infinite Grassmannian to have this coordinate ring (or, if the regular functions are a bit more complicated in the infinite case, to at least have these Plucker coordinates in there). I've looked (briefly) into: -(a) Kac's construction of infinite Grassmannians ([Kac, Infinite dimensional Lie algebras, 3rd ed.], Exercise 14.32, p.339) -(b) taking the union of the finite Grassmannians to get a classifying space for $O(n)$ or $U(n)$ -(c) infinite Grassmannians coming from Hilbert spaces ([Pressley and Segal, Loop groups]) -but none of these seem to quite describe what I want. Some of these are working with $\mathbb{N}$-dimensional space rather than $\mathbb{Z}$-dimensional space (i.e. something more like $\mathbb{C}[t]$ than $\mathbb{C}[t,t^{-1}]$), usually from a colimit of the finite ones, and I can't see how to alter the definition and be sure of keeping the theorems. And with the others that do work with something like $\mathbb{C}[t,t^{-1}]$, I can't see a description that corresponds to planes in that space (and I definitely need just the planes). -I feel sure this is well-known so does anybody know a reference for both the construction I want and also enough information about its coordinate ring? -Edit: Having thought about this a little more, I want to formulate the question more specifically as: - -Let $V=\mathbb{C}[t,t^{-1}]$ and define $Gr(2,V)$ to be the set of 2-dimensional subspaces of $V$. Does the finite-dimensional machinery of the Plucker embedding work in this setting and give Plucker coordinates in $\mathbb{C}[Gr(2,V)]$ of the form $\Delta^{ij}$ for integers $i < j$? - -REPLY [3 votes]: I have put a brief note that might be helpful at -http://neil-strickland.staff.shef.ac.uk/research/fock.pdf<|endoftext|> -TITLE: Example of a compact homogeneous metric space which is not a manifold -QUESTION [17 upvotes]: A metric space $(X,d)$ is isometrically homogeneous if its isometry group acts transitively on points, i.e., for every $x,y \in X$ there is an isometry $\varphi:X\to X$ with $\varphi(x) = y$. I'd like to know an example of a compact isometrically homogeneous metric space which is not a manifold (a space with finitely many points counts as a 0-dimensional manifold). -Googling a bit I've discovered enough recent literature on this general subject to be sure there must be classical examples known to experts, but I haven't managed to find them written down. For example, Theorem 1.2 of this paper implies: - -A compact isometrically homogeneous metric space is a finite-dimensional manifold if and only if it is locally contractible. - -So equivalently, I'd like an example of a compact isometrically homogeneous metric space which is not locally contractible. -Added: Pete and Neil both gave very nice answers. I'm accepting Neil's since, as Pete points out, it essentially contains Pete's answer as a special case. - -REPLY [22 votes]: Take $X=\prod_{n=0}^\infty S^1$ with $d(x,y)=\sum_n|x_n-y_n|/2^n$. Then the metric topology is the same as the product topology, which is compact by Tychonov. There is an obvious group structure by pointwise multiplication, and multiplication by any fixed element is an isometry, so the space is isometrically homogeneous. It is path-connected but not locally contractible. -More generally, I guess you can take any sequence of compact isometrically homogeneous spaces $X_n$, rescale the metric so that $d(x,y)\leq 2^{-n}$ for all $x,y\in X_n$, and then take $X=\prod_nX_n$ with $d(x,y)=\sum_nd(x_n,y_n)$. - -REPLY [13 votes]: The ring of $p$-adic integers $\mathbb{Z}_p$ with its standard metric seems to be an example of what you want. -Of course this space is a "$p$-adic analytic manifold", so you may want to see another example. E.g., how about a space which satisfies all of your properties and is also connected?<|endoftext|> -TITLE: Surface automorphisms and conformal automorphisms -QUESTION [5 upvotes]: Given a closed orientable surface $S$ and a topological automorphism $\sigma$ of $S$, it is not in general possible to find a conformal structure $\Sigma$ on $S$ so that $\sigma$ is isotopic to a conformal automorphism of the Riemann surface $(S,\Sigma)$. For example by the theorem of Hurwitz that the conformal automorphism group is finite, while $\sigma$ on the other hand may be of infinite order in the mapping class group. But by a theorem of Colin Maclachlan in "Modulus space is simply connected", Proc. Amer. Math. Soc. 29 (1971), 85–86, every surface automorphism is isotopic to the composition of finitely many conformal automorphisms (for varying complex structures on $S$). For being isotopic to a conformal automorphism is equivalent to being isotopic to an topological automorphism of finite order (one direction by Hurwitz, the other by averaging a metric). Maclachlan proved that the mapping class group is generated by elements of finite order. -I am interested in the minimal number $m(\sigma)$ of conformal structures required, especially for the torus, where the mapping classes have a nice explicit description. Unfortunately when I tried to use this explicit description, it translated into some obscure number theory with a Diophantine flavor. I could not even show that for the torus in general arbitrarily many conformal structures would be needed, i.e. that $m(\sigma)$ is unbounded for $T^2$. This is my question. An upper bound for $m(\sigma)$ for $T^2$ in terms of the explicit description of $\sigma$ by a 2 by 2 integer matrix would also be interesting. Perhaps the higher genus case could be worth looking at after the torus case. I got stuck on $T^2$ and gave up quite a long time ago. But now that Math Overflow is here, I can ask this as a question. -CLARIFICATION: For the case of a topological torus I am concretely asking how many conformal automorphisms (relative to various complex structures on the topological torus) I need to compose together to have enough freedom to represent a topological automorphism up to isotopy. I tend to agree with Sam Nead that for every positive integer $K$ there will be some topological automorphism $\sigma$ that will require at least $K$ conformal automorphisms in order to be so represented. But I don't know how to prove this, though Sam Nead's comment on proving it seems reasonable. - -REPLY [5 votes]: Complete edit, after talking to a colleague - -Suppose that $G$ is the mapping class group of a surface $S$. Then you are asking: - -Is there a number $K$, depending only on $S$, with the following property? For every $\sigma \in G$ there are torsion elements $\tau_i \in G$ so that $\sigma = \Pi_{i = 1}^K \tau_i$. - -As Henry Wilton puts it - you are asking if the mapping class group is boundedly generated by torsion. The answer is "No". This follows from a paper of Bestvina and Fujiwara "Bounded cohomology of subgroups of mapping class groups". They show that the group $G$ admits unbounded quasi-homomorphisms. See the first five pages of their paper. -Edit - to give a few details. A $D$-quasi-homomorphism is a map $\phi$ from $G$ to the reals so that for all $g,f \in G$ we have $|\phi(gf) - \phi(g) - \phi(f)| < D$. It is an exercise to show that if $g$ is torsion then $|\phi(g)| < D$. Thus, if $G$ was boundedly generated, say with constant $K$, then we would have, for all $g \in G$, that $|\phi(g)| < 2KD$. This is a contradiction. - -REPLY [4 votes]: It was proven in -J. MacCarthy and A. Papadopoulos. Involutions in surface mapping class groups. Enseign. Math. 33 (1987), 275–290. -that the mapping class group is generated by the conjugates of a single involution (for $g$ large, I think at least $2$). Call this involution $\tau$ and let $S$ be the Riemann surface that $\tau$ acts on. Then the conjugate $\tau^f$ by $f$ in the mapping class group acts on the Riemann surface $f_{\ast} (S)$, which is isomorphic to $S$. In other words, up to isomorphism you only need one conformal structure.<|endoftext|> -TITLE: Nonvanishing of central L-values of quadratic twists? -QUESTION [14 upvotes]: Let $\pi$ be a cuspidal automorphic representation of GL(2) over a number field (if you want, assume it's $\mathbb Q$ and $\pi$ comes from a holomorphic modular form). -In the case $\pi$ has trivial central character, the epsilon factor determines the parity of order of vanishing of $L(1/2,\pi)$. If $\chi$ is a quadratic character, and $\pi$ in fact comes from an elliptic curve, then one expects $L(s,\pi \otimes \chi)$ to have rank 0 half the time and rank 1 half the time (Goldfeld's conjecture). - -(1) Is there a precise generalization of Goldfeld's conjecture to more general $\pi$ (assume what you need)? - -I know there are several nonvanishing results and bounds on proportions for rank 0 and rank 1 if $\pi$ has trivial central character. However if $\pi$ does not have trivial central character (and is not self-dual), then I know little more than that Friedberg-Hoffstein says $L(s,\pi \otimes \chi)$ has rank 0 infinitely often. - -(2) Is anything else known/expected when $\pi$ is not self-dual? - -I know nothing about Katz-Sarnak philosophies and Random Matrix Models, but do they apply for non-self-dual representations? - -REPLY [7 votes]: (1) In a "true" Katz-Sarnak context, i.e., over finite fields, non self-dual situations definitely make sense, exist, and are among those studied in their book (chapters 9 and following). In fact the basic question of proportion of vanishing v.s. non-vanishing is really an application of Deligne's Equidistribution Theorem. Basically, when the size of the field goes to infinity, if you have a known (compactifo-complexified) monodromy group, the proportion of vanishing at the central point will converge to the Haar probability -- in the monodromy group -- of matrices with $1$ as an eigenvalue. In most cases this is computable once the monodromy is known. However, this story has the usual limitation (field goes to infinity can not be replaced, for the moment, with size of matrices goes to infinity). -(2) For L-functions over $\mathbf{Q}$, the analytic methods which give a "decent" proportion of non-vanishing for quadratic twists (not positive proportion, but about $1/(\log D)^A$ for twists by quadratic characters of size up to $D$, for some fixed $A$) work whether the form is self-dual or not. But one definitely expects that twists of a fixed form which is not self-dual should have "density" zero of order of vanishing at least $1$ (this is as precise as Goldfeld's Conjecture). In many cases, it might well be that no twist vanishes at $1/2$, or at most finitely many.<|endoftext|> -TITLE: Frobenius splitting of affine flag varieties -QUESTION [7 upvotes]: NOTE: I am very much not an expert on affine Kac-Moody groups or ind-varieties, so the following may be vague. -The first question is: Has anyone developed a theory of Frobenius splitting for ind-varieties in positive characteristic? (This might be very easy for all I know). Following from that, my main question is: If there is a reasonable theory of Frobenius splitting of ind-varieties, has anyone thought about whether affine flag varieties or affine Grassmannians corresponding to semisimple algebraic groups in positive characteristic are Frobenius split? - -REPLY [8 votes]: If you have a strict ind-scheme (inductive limit of schemes with maps closed imersions), for example affine flag varieties as a limit of schubert cells, then you can ask that all these varieties are compatably Frobenius split, so the notion of a Frobenius splitting makes sense. -Then you can ask if the affine flag variety admits a Frobenius splitting. The asnwer is yes. I learnt this under mild conditions from a course of Xinwen Zhu, from which the original reference appears to be Faltings' paper "Algebraic loop groups and moduli spaces of bundles". Here the affine flag variety is defined in a particular way, and of course if you want to define it in a different way (eg using Kac-Moody) groups, then you must do some work to show that definitions agree. -This result is harder than in finite type since you don't have the smooth Schubert cell associated to the long word as a crutch to get you started.<|endoftext|> -TITLE: Why does homotopy behave well with respect to fibrations and homology with respect to cofibrations? -QUESTION [17 upvotes]: (I apologize that this is a vague question). -I seems to me somehow that homotopy groups behave well with respect to (Serre)-fibrations. For example you get a long exact sequence of homotopy groups from it. On the other hand cofibrations and homotopy groups seems to be no good friends at all (e.g. $S^1\to D^2\to S^2$). -But then again, the situation in homology seems to be the other way round. They behave well with respect to cofibrations (you get a long exact sequence) and fibrations are harder to investigate (Serre spectral sequence etc.). -My question is: What is the intuition behind this difference? (in particular with respect to the fact that homology is just homotopy of another space.) - -REPLY [29 votes]: I agree that the long exact sequence in homotopy groups of a fibration follows from the fact that fibrations are defined using a mapping property in which the fibration is the target. -One way to understand why homology behaves well with respect to cofibrations is to spell out your remark that -"homology is just homotopy of another space". This is true, but not obvious. There are a number of constructions of ordinary homology which take the following form. -One finds a functor $F$ from (pointed) spaces to (pointed) spaces which takes cofibrations to quasifibrations. (A quasifibration is something for which you have a long exact sequence of homotopy groups, for example a Serre fibration). And then $H_* (X) \cong \pi_* F(X).$ If $X\to Y \to Z$ is a cofibration (maybe a cofibration of CW complexes), then $\dots \to \pi_* F(X) \to \pi_* F(Y) \to \pi_* F(Z) \to \ldots$ is the long exact sequence in homology associated to the cofibration. -Here are several contexts in which one can describe such a functor $F$. -First, a formal approach. Let $\mathbf{S}$ denote the category of spectra: it is connected to the category $\mathbf{T}$ of spaces by adjoint functors $\Sigma^\infty: \mathbf{T} \to \mathbf{S}$ and $\Omega^\infty: \mathbf{S} \to \mathbf{T}.$ There a spectrum called the "Eilenberg-MacLane" spectrum, denoted $H\mathbb{Z}$: its job is to represent singular cohomology, and one can take $F(X) = \Omega^\infty ((\Sigma^\infty X) \wedge H\mathbb{Z})$. -Why does $F$ have the cofibration-to-quasifibration property? Well, the way that this is set up, $\Sigma^\infty$ preserves cofibrations of CW complexes, and $\Omega^\infty$ preserves fibrations of fibrant objects, and in the category of spectra every cofibration is equivalent to a fibration. -To be more explicit about $H\mathbb{Z}$, you can define $F(X) = \lim (\ldots \Omega^k(X \wedge K(\mathbb{Z},k)) \to \Omega^{k+1} (X\wedge K(\mathbb{Z},k+1))a \ldots),$ where the limit is a colimit and the maps defining the system arise from the maps $K(\mathbb{Z},k) \simeq \Omega K(\mathbb{Z},k+1)$. -Second, the Dold-Thom theorem says that one can take $F(X) = Sp^\infty (X).$ Here $Sp^n(X) = X^n/\Sigma_n$, and $Sp^\infty(X) = \lim \ldots Sp^n(X) \to Sp^{n+1}(X) \ldots$, again the limit is a colimit. -Third, if you're willing to allow $X$ to be a simplicial set, then one can take then one can take $F(X) = \mathbb{Z}X$, the simplicial set whose $n$ simplices are the free abelian group on the $n$-simplices of $X$. (This approach is due to Dan Kan; see the proceedings of the Hurewicz conference) -All of this is to focus attention on functors which take cofibrations to quasifibrations. In fact all $-1$-connected generalized homology theories (at least the ones associated to cohomology theories: are there homology theories which are not? I don't know) are of the form $E_* X = \pi_* G(X)$, where $G$ is a functor which takes cofibrations to quasifibrations. Indeed one takes $G(X) = \Omega^\infty (\Sigma^\infty X \wedge R)$, where $R$ is the spectrum representing the cohomology theory. This approach goes back to G. W. Whitehead. -One of the more compact discussions of such a functor, which I like, is in an article by G. Segal in Springer LNM 575; he gives a construction of connective real $K$-homology there. Really he's showing how to generalize the work of Dold and Thom: Segal's argument applies just as well to $Sp^\infty(X)$. -I apologize that throughout I have done a poor job of saying how to handle basepoints.<|endoftext|> -TITLE: What are the epimorphisms in the category of schemes? -QUESTION [24 upvotes]: Is there a known characterization of epimorphisms in the category of schemes? -It is easy to see that a morphism $f : X \to Y$ such that the underlying map $\lvert f\rvert$ is surjective and the homomorphism $f^\# : \mathcal{O}_Y \to f_* \mathcal{O}_X$ is injective, is an epimorphism. But there are other examples, too: if $Y$ is reduced and locally of finite type over a field, the obvious morphism from $X=\bigsqcup_{y\in Y_\text{cl}} \operatorname{Spec}(k(y))$ to $Y$ is an epimorphism (see #8(b) in Mark Haiman's Homework Set 9 for Math 256AB). -If this is is not possible, what about regular, extremal, or effective epimorphisms? Here, again, I know only some examples. -My background is that I want to know if there is a categorical characterization of the spectra of fields in the category of schemes. In the full subcategory of affine schemes, they are characterized by the property: $X$ is non-initial and every morphism from a non-initial object to $X$ is an epimorphism. But I doubt that this characterization takes over to the category of schemes. EDIT: Kevin Ventullo has shown below that the characterization takes over. Thus my original question has been answered (and I wonder if it's appropriate to accept it as an answer). But of course every other hint about the characterization of epimorphisms of schemes is appreciated. - -REPLY [9 votes]: $\DeclareMathOperator\Spec{Spec}$Actually, your suggested categorical characterization of spectra of fields does work. -Edit: (I had written something incorrect here) -By Martin's comment below, we just have to show that maps from affines into $\Spec(k)$ are epis in the full category. But if we had two maps $f,g: \Spec(k) \rightarrow Y$ which agreed on some affine mapping into $\Spec(k)$, then first of all $f$ and $g$ would have to be the same topological map. Then both would land inside some affine $\Spec(R)\subset Y$, and now we're -reduced to the affine situation where we know it holds. -Conversely, suppose $X$ is not the spectrum of a field. If every point is dense, $X$ is affine -and we are done by what we know about the affine subcategory. Otherwise, we can find an open subscheme $U\subsetneq X$. Then the inclusion of $U$ into $X$ is not an epi as is witnessed by the two inclusions -$$X \rightrightarrows X\sqcup_U X,$$ -where the last object is $X$ glued to itself along $U$.<|endoftext|> -TITLE: Has the following kind of (minimum degree $d$) random graph been studied? -QUESTION [6 upvotes]: The following random construction is simple enough that I am guessing it must have been studied. Fix $d \ge 3$, and let $n > d$. For each of the $n$ vertices, pick exactly $d$ other vertices to connect it to, uniformly over all ${n-1 \choose d}$ possible choices, and making this choice independently over all $n$ vertices. Some edges may get put in twice, and that's fine, but we will consider the final graph to not have multiple edges. - -Has this model been studied, and if so does it have a name? - -Motivation: Erdos-Renyi random graphs $G(n,p)$ must have average vertex degree greater than, before they become connected and have nice expansion properties, etc. But this is because they still have isolated vertices. Here we don't have the problem of isolated vertices. (I think I can show that these graphs are a.a.s. connected.) I am aware of the large literature on random $d$-regular graphs, but I thought this might be an alternative model if one does not care about regularity. -I am especially wondering about spectral properties of these graphs. - -REPLY [5 votes]: Yes, this model has been studied. For some early results, see -Fenner, T. I.; Frieze, A. M. On the connectivity of random $m$-orientable graphs and digraphs. Combinatorica 2 (1982), no. 4, 347–359. -So there it is called "random $m$-orientable graph" - -REPLY [5 votes]: The main aspect of the underlying graph of a random $m$-out graph (each vertex is given a random set of $m$ out-edges) that has been studied is the fact that they are a.s. Hamiltonian. -Denoting this graph by $D_m(n)$, it is known that $D_2$ almost surely contains a vertex adjacent to three vertices of degree two, and in the paper "On the existence of Hamiltonian cycles in a class of random graphs" Fenner and Frieze proved that $D_{23}$ is a.s. Hamiltonian. Finally Bohman and Frieze recently proved that $D_3$ is almost surely Hamiltonian in "Hamiltonian cycles in $3$-out".<|endoftext|> -TITLE: A precise statement of the categorical version of geometric Langlands conjecture -QUESTION [28 upvotes]: The statement of the ordinary non-categorical version of geometric Langlands conjecture, which was proven for GL(n) in around 2002 by Frenkel, Gaitsgory and Vilonen, is quite well-known and is easy to find in the literature. -Recently, by talking to some students of Dennis Gaitsgory and postdocs working in this area, I understand that a stronger categorical version of the geometric Langlands conjectures has been in circulation for at least the past few years, and with recent advances Dennis is perhaps even close to proving it (at least in type A). My question is: what is the precise statement of the categorical version of geometric Langlands? -I understand on the left you have something related to the category of $D$-modules on $Bun_G(X)$. After you take the moduli stack of $G^{\vee}$-local systems, on the right you have some category in between the categories of coherent sheaves on this stack, and the category of quasi-coherent sheaves on this stack. I also hear that the categorical version uses ideas from work of Lurie ( $(\infty,1)$-categories, $DG$-categories, etc). -Am I correct that on the left we have the full category of D-modules on $Bun_G(X)$? (Where-as in the simpler non-categorical version we simply request a Hecke eigensheaf for each irreducible local system). - -REPLY [30 votes]: For context for Tom's answer, -let me state the naive version of the conjecture, which has been around since around 1997 I think (due to Beilinson-Drinfeld). It calls for an equivalence of (dg) categories -$$D(Bun_G(X))\simeq QC(Loc_{G^\vee}(X))$$ -between (quasi)coherent $D$-modules on the stack of $G$-bundles on a curve $X$, and (quasi)coherent sheaves on the (derived) stack of flat $G^\vee$ connections on the curve. -Moreover (and this is where most of the content lies) this equivalence should be an equivalence as module categories for the spherical Hecke category $Rep(G^\vee)$ acting on both sides for every choice of point $x\in X$. The action on the right is given by simple multiplication operators (tensor product with a tautological vector bundle on $Loc_{G^\vee}(X)$ attached to a given representation of $G^\vee$ and point $x$. On the right hand side the action is by convolution (Hecke) functors, associated to modifications of $G$-bundles at $x$ (relative positions at $x$ of $G$-bundles are labeled by the affine Grassmannian, and in order to formulate this statement we use the geometric Satake theorem of Lusztig, Drinfeld, Ginzburg and Mirkovic-Vilonen). -The equivalence can be further fixed uniquely by using "Whittaker normalization", -a geometric analog of the identification of L-functions in the classical Langlands story. -In this form the conjecture is a theorem for $GL_1$, using the extension of the Fourier-Mukai transform for D-modules. There are no other groups for which it's known (yet - though perhaps Dennis already has $SL_2$), and few curves (one can do $P^1$ for example). Also (as Scott points out) this is only the unramified case, and there are natural conjectures to make with at least tame ramification (a "parabolic" version of the above). -There's also a close variant of this conjecture which comes naturally from S-duality for N=4 super-Yang-Mills, thanks to Kapustin-Witten. [Edit: in fact the physics suggests a far more refined version of the whole geometric Langlands program.] -It is a theorem of Beilinson-Drinfeld if we restrict on the left hand side to D-modules generated by D itself, and on the right to coherent sheaves living on opers. -It is also well known that this conjecture as stated is too naive, due to bad "functional analysis" of the categories involved (precisely analogous to the analytic issues appearing eg in the Arthur-Selberg trace formula). On the D-module side, the stack $Bun_G$ is not of finite type, and one might want to make more precise "growth conditions" on D-modules along the Harder-Narasimhan strata. On the coherent side, $Loc$ is a derived stack and singular, and one ought to modify the sheaves allowed at the singular points -- in particular at the most singular point, the trivial local system. [Edit:One approach to correcting this involves the "Arthur $SL_2$" -- roughly speaking looking at local systems with an additional flat $SL_2$-action that controls the reducibility..this comes up really beautifully from the physics in work of Gaiotto-Witten, one of the first points where the physics is clearly "smarter" than the math.] These two issues are very neatly paired by the duality. If one restricts to irreducible local systems -and "cuspidal" D-modules, these issues are completely avoided (though the conjecture even on this locus remains open except for $GL_n$, where I believe it can be deduced from work of Gaitsgory following his joint work with Frenkel-Vilonen, see his ICM). -In any case, Dennis has now given a precise formulation of a conjecture, which was at some point at least on his website and follows the outline Tom explained. It is very clear that homotopical algebra a la Lurie is crucial to any attempt to prove this, and Dennis and Jacob have made (AFAIK) great progress on this. The basic idea of the approach is the same as that carried out by Beilinson-Drinfeld and suggested by (old) conformal field theory (in particular Feigin-Frenkel) and (new) topological field theory (Kapustin-Witten and Lurie) -- i.e. a local-to-global argument, deducing the result from a local equivalence which comes from representation theory of loop algebras. The necessary machinery for "categorical harmonic analysis" is now available, and I'm looking forward to hearing a solution before very long.. -Edit: In response to Kevin's comment I wanted to make some very informal remarks about ramification. -First of all one needs to keep in mind the distinction between reciprocity (which is what the above conjecture captures) and functoriality. In the geometric setting the former is strictly stronger than the latter, while in the arithmetic setting most of the emphasis is on the latter. It is in fact quite easy to formulate a functoriality conjecture in the geometric setting with arbitrary ramification. Namely, fix some ramification and look at the category of D-modules on the stack of G-bundles with corresponding level structure. Then this is a module category over coherent sheaves on the stack of $G^\vee$ connection -with poles prescribed by the ramification (eg we can take full level structure and allow arbitrary poles). Then one can conjecture that given a map of L-groups $G^\vee\to H^\vee$ the corresponding automorphic categories are given simply by tensoring the module categories from $QC(Loc_{G^\vee}(X))$ to $QC(Loc_{H^\vee}(X))$ (everything here must be taken on the derived level to make sense). It is not hard to see this follows from any form of reciprocity you can formulate. And there is also a geometric version of the Arthur-Selberg trace formula in the ramified setting (under development). -Second, one can make a reciprocity conjecture with full ramification, though you have todecide to what extent you believe it. In the "completed"/"analytic" form of geometric Langlands that comes out of physics such a conjecture in fact appears in a paper of Witten on Wild Ramification. Roughly speaking in the above reciprocity rather than just looking -at the module category structure for D-modules over QC of local systems, you can ask for them to be equivalent... not stated anywhere since maybe I'm too naive and this is known to be too far from the truth, but I think more likely people haven't thought about it very much. -Third, the local story: of course in the p-adic case Kevin discusses one restricts to $GL_n$. There are two things to point out: first of all in the geometric setting one is interested in all groups, and $GL_n$ is not much simpler as far as our understanding of the local story goes. Second, while everything is much harder and deeper in the p-adic setting than the geometric setting, it's worth pointing out that a formulation of a local geometric Langlands conjecture is a much more subtle proposition, involving the representation theory of loop groups on derived categories which is only beginning to be within the reach of modern technology (even at the level of formulation of the objects!) -That being said, there are rough forms of local geometric Langlands conjectures developed -by Frenkel, Gaitsgory and Lurie. It is best understood in the so-called "quantum geometric Langlands program", a deformation of the above picture involving the representation theory of quantum groups, where Gaitsgory-Lurie give a precise general local conjecture and make progress on its resolution. The usual case above is a bit degenerate and one needs to be more careful. In any case the rough form of the local conjecture is an equivalence of 2-categories (again everything has to be taken in the appropriate derived sense) between "smooth" LG-actions on categories and quasicoherent sheaves of categories over the stack of connections on the punctured disc.. ---THAT being said we don't have a proof of this for $GL_n$ so again the number theorists win! just wanted to give some sense that there is a reasonable understanding of full ramification.<|endoftext|> -TITLE: Historical Question about Simplicial Sets -QUESTION [7 upvotes]: I have a pretty easy historical question about simplicial sets. Unless I am mistaken, simplicial sets first came out of topology, explicitly from combinatorial topology and the study of simplicial complexes. However, simplicial sets, as we all know, are intimately linked with the study of categories. On one hand, a la Joyal we know that they provide a model for quasicategories, or in Lurie's terminology, $\infty$-categories. And on the other hand, some machinery of Mark Weber takes the free category monad and spits out the category $\Delta$ on which simplicial sets are based, showing that they belong just as much to category theory as they do to topology. My question is, when was the connection of simplicial sets to the study of categories first noticed, and by whom? - -REPLY [7 votes]: I once heard a talk by Ezra Getzler in which he attributed to Grothendieck (in the 1950s) the observation that a small category is the same thing as a simplicial set with unique fillings for inner horns. I believe I've seen a citation since then, but I can't locate one now: google searches involving the words "Grothendieck" and "Category" aren't particularly effective. - -REPLY [4 votes]: According to Mac Lane (see p19 of this paper) they were introduced by Eilenberg-Zilber in 1950 under the name complete semisimplicial complexes.<|endoftext|> -TITLE: About irreducible trinomials -QUESTION [13 upvotes]: This question was inspired by this one. For every $n>m>0$ consider the polynomial $p_{m,n}=x^n-x^m-1$. -For which $m,n$ is $p_{m,n}$ irreducible over $\mathbb Q$? -In particular, if $m$ is odd, is it always irreducible? - -REPLY [12 votes]: Ljunggren's result on $\pm 1$ trinomials $X^n \pm X^m \pm 1$ amounts to saying that every such trinomial has at most one non-cyclotomic irreducible factor. Since $\{\zeta,\mu\} = \{e^{2\pi i/3},e^{4\pi i/3}\}$ are evidently the only roots of unity solutions to $1 + \zeta + \mu = 0$, the precise factorization is easily derived from this. -Since Ljunggren's proof follows a case by case analysis that may not be very illuminating, I thought I would add an answer outlining a more conceptual proof due to Schinzel. I was reminded of it on rereading Smyth's survey on the Mahler measure, which recounts Schinzel's proof in section 14.1. -Since the logarithmic Mahler measure $m(P) = \int_{S^1} \log{|P(z)|} \, d\theta$ is manifestly additive ($m(PQ) = m(P) + m(Q)$), the proof that the trinomial has not more than a single non-cyclotomic factor is an almost immediate consequence of two general, if not easy, facts about polynomials: - -Arithmetic component: Smyth's theorem that a non-reciprocal integer polynomial $P \in \mathbb{Z}[X]$ with $P(1) \neq 0$ has $m(P) \geq \log{\rho} = 0.281\ldots$, where $\rho^3 = \rho+1$ (the Plastic number). -Analytic component: Goncalves's inequality. With $z_1,\ldots,z_d$ any ordering of the complex roots of a monic polynomial $P(X) = X^d + a_{1}X^{d-1} + \cdots + a_d \in \mathbb{C}[X]$, this states that $|z_1\cdots z_t|^2 + |z_{t+1} \cdots z_d|^2 \leq \|P\|_2^2 = 1 + \sum_1^d |a_i|^2$ for all $t = 1,\ldots,d$. - -In terms of Mahler measure, Goncalves's inequality is expressed as $e^{2m(P)} + e^{-2m(P)} \leq \|P\|_2^2$. Applied to our trinomial $P(X) = X^n \pm X^m \pm 1$, which has $\|P\|_2^2 = 3$, its conclusion is $m(P) \leq \log{\varphi} = 0.4812\ldots$, where $\varphi^2 = \varphi+1$ (the Golden ratio). We had the trivial bound of $\log{2}$, which is not quite enough to conclude with Smyth's theorem. But since $\log{\varphi} < 2\log{\rho}$, while it is readily seen that the only reciprocal factors are cyclotomic (If $\alpha$ and $1/\alpha$ both satisfy the equation then $\ldots$), the desired conclusion follows.<|endoftext|> -TITLE: What are the monomorphisms in the category of schemes? -QUESTION [21 upvotes]: Someone recently asked what the epimorphisms in the category of schemes are; the other day I had been wondering about the similar question: what are the monomorphisms in the category of schemes? I am often frustrated working with schemes because, unlike a lot of other categories, it is not immediate that you have left cancellation of morphisms when you know the underlying map on sets is injective--and I think it must not be true in general, though I don't have an example in mind. Are there nice situations or additional conditions that guarantee that one may safely cancel morphisms of schemes on the left? - -REPLY [49 votes]: In EGA IV, 17.2.6 the following characterization of monomorphisms is given: - -Let $f : X \to Y$ be a morphism locally of finite type. Then the following conditions are - equivalent: -a) $f$ is a monomorphism. -b) $f$ is radicial and formally unramified. -c) For every $y \in Y$, the fiber $f^{-1}(y)$ is either empty or isomorphic to $\text{Spec}(k(y))$. - -Also note that (due to the adjunction) a morphism between affine schemes is a monomorphism (in the category of schemes) if and only if the associated homomorphism of rings is an epimorphism (in the category of rings) and the latter ones can be characterized in many ways. See, for example, this Samuel seminar and this MO discussion. Monomorphisms of noetherian schemes are treated in detail in Exposé 7 by Daniel Ferrand. -Further examples: -1) Immersions are monomorphisms; this follows from the universal property of a closed resp. open immersion. -2) A morphism $X \to Y$ is a monomorphism if and only if the diagonal $X \to X \times_Y X$ is an isomorphism. In particular, every monomorphism is separated. -3) In EGA IV, 18.12.6 it is shown that proper monomorphisms are exactly the closed immersions.<|endoftext|> -TITLE: Singularity structure of integrals of rational functions -QUESTION [8 upvotes]: Suppose I have a convergent integral of the form $\int_0^1dx_1\dots\int_0^1 dx_n \frac{P(x_i)}{Q(x_i)}$, where $P$ and $Q$ are polynomial functions of $n$ nonnegative real variables $x_i$. Let the coefficients of the various monomials in $Q$ be $a_k$. Depending on the structure of $Q$, the integral could develop singularities in the limit that some of the coefficients $a_k$ vanish. For example, the integral $\int_0^1 dx\frac 1 {a+x}$ goes like $-\log a$ as $a\to 0$. Is there a systematic way to extract the structure of these singularities for general multidimensional integrals? -The particular example I'm interested in is the following: I'd like to determine the singularities of -$\begin{eqnarray} -\int_0^1 d\lambda_1\int_0^1 d\lambda_2 \int_0^1d\alpha_1 \int_0^1d\alpha_2 \frac{\alpha_1\alpha_2}{(\lambda_2\alpha_1\alpha_2+\lambda_1\alpha_1\alpha_2+a\alpha_2\lambda_1^2+b\alpha_1\lambda_2^2+c\alpha_1\alpha_2)^2}, -\end{eqnarray}$ -as $a,b,$ and $c$ tend to zero. If necessary, it's alright to assume they all tend to zero at the same rate. I naively expect a leading singularity that looks like $\log a \log b \log c$, followed by sub-leading singularities that look like $\log a \log b$ times some convergent integral, $\log b \log c$ times some other convergent integral, etc.. For my purposes, these $\log$-squared terms suffice, but I'd be interested to know how to systematically go further. -Edit: The above integral is an example of a particular Feynman integral, written in terms of Feynman parameters. - -REPLY [4 votes]: OK. It is going to be long and extremely boring, but, as promised, here goes. We'll get the precision $O(mL^2)$ where $m=\max(a,b,c), L=\max(\log\frac 1a,\log\frac 1b,\log\frac 1c)$, which should be enough for any reasonable interpretation of "$a,b,c$ going to $0$ at about the same rate". -We start with the approximation of an auxiliary integral: -$$ -I(a,b)=\iint_{[a,+\infty)\times[b,+\infty)}\frac{1}{(1+x+y)^2}\frac{dx}{x}\frac{dy}{y} -$$ -as $a,b\to 0+$. -First, we have -$$ -\frac{1}{(1+x+y)^2}-\frac 1{(1+x)^2}\frac 1{(1+y)^2}=\frac{xy(2+2x+2y+xy)}{(1+x+y)^2(1+x)^2(1+y)^2} -\\ -\le -\frac{2xy}{(1+x)^{2}(1+y)^{2}}\,, -$$ -which means that $I(a,b)=J(a)J(b)+D_1+O(a+b)$ where -$$ -J(a)=\int_{a}^\infty\frac{1}{(1+x)^2}\frac{dx}x\,, -$$ -and -$$ -D_1=\iint_{(0,\infty)^2}\frac{(2+2x+2y+xy)dx\,dy}{(1+x+y)^2(1+x)^2(1+y)^2} -$$ -(below we will always denote by $D_k$ various numerical constants whose values are expressed by some convergent integrals. Some of them are easy to evaluate and some aren't, but that distinction won't bother us here). -Now -$$ -J(a)=\int_a^\infty\left[\frac 1{(1+x)^2}-\chi_{[0,1]}(x)\right]\frac{dx}{x}+\log\frac 1a\,. -$$ -Since for $x<1$, we have $\frac 1{1+x}-1=-\frac x{1+x}$, we conclude that -$$ -J(a)=\log\frac 1a+D_2+O(a) -$$ -with -$$ -D_2=\int_0^\infty \left[\frac 1{(1+x)^2}-\chi_{[0,1]}(x)\right]\frac{dx}{x} -$$ -Multiplying out, we get -$$ -I(a,b)=\log\frac 1a\log\frac 1b+D_2(\log\frac 1a+\log\frac 1b)+D_3 -\\ -+O\left((a+b)(\log\frac 1b+\log\frac 1a)\right)\,. -$$ -Now we can return to our quadruple integral. Let us rewrite the integrand in the form -$$ -\frac{1}{(\lambda_1+\lambda_2+c)^2}\frac{1}{\left(1+\frac{a\lambda_1^2}{\lambda_1+\lambda_2+c}\frac 1{\alpha_1}+ -\frac{b\lambda_2^2}{\lambda_1+\lambda_2+c}\frac 1{\alpha_2}\right)^2} -\,d\lambda_1\,d\lambda_2\,\frac{d\alpha_1}{\alpha_1}\,\frac{d\alpha_2}{\alpha_2} -$$ -For fixed $\lambda_1,\lambda_2$, we can now happily integrate $\alpha_1,\alpha_2$ out using the computation above to reduce our integral to -$$ -\iint_{[0,1]^2} -\frac 1{(\lambda_1+\lambda_2+c)^2}\left[\log\frac{\lambda_1+\lambda_2+c}{a\lambda_1^2}\log\frac{\lambda_1+\lambda_2+c}{b\lambda_2^2} -\\+ -D_2\left(\log\frac{\lambda_1+\lambda_2+c}{a\lambda_1^2}+\log\frac{\lambda_1+\lambda_2+c}{b\lambda_2^2}\right)+D_3\right]\,d\lambda_1\,d\lambda_2+E_1 -$$ -where the error term $E_1$ is controlled by -$$ -\iint_{[0,1]^2}\frac{1}{(\lambda_1+\lambda_2+c)^2}\left[ -\left(\frac{a\lambda_1^2}{\lambda_1+\lambda_2+c}+ -\frac{b\lambda_2^2}{\lambda_1+\lambda_2+c}\right) -\\ -\left(\log\frac{\lambda_1+\lambda_2+c}{a\lambda_1^2}+\log\frac{\lambda_1+\lambda_2+c}{b\lambda_2^2}\right) -\right]\,d\lambda_1\,d\lambda_2 -$$ -Now we do not need a special investigation of $E_1$ because the first factor in the brackets is bounded by $a+b$ and if we look at the rest, it is just one of the terms in the main asymptotic integral. Next, once we open the parentheses and expand logarithms of products, we see that we just need to approximate three double integrals: -$$ -I_2=\iint_{[0,1]^2} -\frac {d\lambda_1\,d\lambda_2}{(\lambda_1+\lambda_2+c)^2}\log\frac{\lambda_1+\lambda_2+c}{\lambda_1^2}\log\frac{\lambda_1+\lambda_2+c}{\lambda_2^2}\,, -$$ -$$ -I_1=\iint_{[0,1]^2} -\frac {d\lambda_1\,d\lambda_2}{(\lambda_1+\lambda_2+c)^2}\log\frac{\lambda_1+\lambda_2+c}{\lambda_1^2}\,, -$$ -and -$$ -I_0=\iint_{[0,1]^2} -\frac {d\lambda_1\,d\lambda_2}{(\lambda_1+\lambda_2+c)^2} -$$ -IT is not hard to convince yourself that in all $3$ cases the integral over the domain $\lambda_1+\lambda_2>1$ tends to its value at $c=0$ and the deviation from that value is $O(c)$, so we can replace those parts by appropriate constants. In the other part it is convenient to make the change of variable $\lambda_1=v\lambda,\lambda_2=(1-v)\lambda$ ($0 -TITLE: Is there a discrete Cerf theory? -QUESTION [20 upvotes]: Towards the end of the 1990's, Robin Forman developed a discrete version of Morse theory, which concerns maps from a simplicial complex to $\mathbb{R}$ satisfying a combinatorial analogue to the condition that critical points be non-degenerate. The theory has proven quite useful in that it has allowed Morse theoretic arguments to be used in discrete settings, and it forms a part of computational topology. -My attention was recently drawn to the technique as a result of reading a preprint of Conant, Schneiderman, and Teichner. Therefore, this question might well be hopelessly naïve (it's also possible that it's open). - -Is there a discrete version of Cerf Theory? Are there at least partial results in this direction? Conversely, is it known that no such theory can exist? - -My motivation is that I would imagine that a discrete proof of Kirby's Theorem, among other results which use Cerf Theory, might prove quite valuable in quantum topology (I'd love such a result at my fingertips!). I know that this doesn't (yet?) exist, or I would surely have heard about it. Additionally, a topological "machine" can only be used by a computer if it requires only finite information. - -REPLY [8 votes]: I realize that I am several months late to the Cerf theory party, but this paper of Chari and Joswig might be of interest to the original poster and certainly deserves a mention in the context of this question. At the very least, their construction provides various interesting topological avenues of investigating the relationship between two discrete Morse functions on the same complex. -Start with a simplicial complex $\Delta$ and let $M(\Delta)$ be the set of all possible discrete Morse functions $\mu:\Delta \to \mathbb{R}$. Note that two such functions $\mu$ and $\mu'$ are considered equivalent if they induce the same discrete vector field. We can assume without loss of generality that $M(\Delta)$ has been quotiented by this obvious equivalence relation: $\mu \sim \mu'$ if and only if $\mu(\sigma) > \mu(\tau) \leftrightarrow \mu'(\sigma) > \mu'(\tau)$ for each facet relation $\sigma \prec \tau$ in $\Delta$. Thus, we may as well assume that $M(\Delta)$ is simply the collection of acyclic partial matchings on $\Delta$. -It turns out that $M(\Delta)$ itself can be canonically endowed with the structure of a simplicial complex. Each facet relation $\sigma \prec \tau$ of $\Delta$ is a vertex, and a $D$-dimensional simplex spans $D+1$ facet relations $\sigma_d \prec \tau_d$ if and only if pairing $\sigma_d$ to $\tau_d$ for each $1 \leq d \leq D+1$ creates an acyclic matching on $\Delta$. -The reason that $M(\Delta)$ is interesting with regards to your question, is that each simplex corresponds to a unique acyclic matching on $\Delta$. So, among other things you can try, just impose a Morse function $f:M(\Delta) \to \mathbb{R}$ on $M(\Delta)$ itself so that the two acyclic matchings you wish to compare are critical cells. Now each gradient path between these critical cells corresponds to a "deformation" from one to the other in the set of acyclic matchings on $\Delta$.<|endoftext|> -TITLE: Quasi-isometries vs Cayley Graphs -QUESTION [8 upvotes]: The following questions might be trivial, however, I couldn't solve them: -Let $G$ be generated by a finite symmetric set $S$. Suppose that $\Gamma(G,S)$ is the corresponding right Cayley graph of $G$. Let $X$ be a metric space (or, maybe, a topological space with some nice structure). -(1) Is there a way to check the following property: $X$ is not quasi-isometric to a space $Z$ which is quasi-isometric to a (hence, every) Carley graph $\Gamma(G,S)$ of some f.g. group $G$. -I.e., if we partition the space of spaces up to quasi-isometric equivalence, then does every equivalence class contain a space which is quasi-isometric to a Cayley graph of some f.g. group $G$? -(2) By Stalling's theorem, the number of ends is a geometric property of the group. Does this mean that the number of ends is a quasi-isometric invariant of the spaces which are quasi-isometric to Cayley graphs? -If the answer of question (2) is affirmative and if the question (1) about equivalence classes has a negative answer, i.e., there is an equivalence class whose elements are not quasi-isomorphic to any Cayley graph, then what is an example of spaces $W_1, W_2$ which are not quasi-isometric to any Cayley graph, but $W_1$ is quasi-isometric to $W_2$ ,however, the number of ends of $W_1$ is different from the number of ends of $W_2$? - -REPLY [7 votes]: There was a conjecture by Woess that every infinite vertex-transitive graph is quasi-isometric to a Cayley graph. A slightly more sophisticated counter-example for your first question is the counter-example to this conjecture that was proposed by R.Diestel and I. Leader in "A conjecture concerning a limit of non-Cayley graphs". It was later proved by A. Eskin, D. Fisher, and K. Whyte in "Quasi-isometries and rigidity of solvable groups".<|endoftext|> -TITLE: Representations of central extensions -QUESTION [10 upvotes]: Let $G$ be central extension of an abelian group $A$ by some group $H$. -Is it possible to characterize all irreducible representions of $G$ -in terms of irreducible representations of $A$ and $H$? - -REPLY [6 votes]: The question is somewhat unprecise. I assume you mean irreps over $\mathbb{C}$ and finite groups. Then the answer is "no" as long as you aren't more specific about the central extension in question. For example, four of the five nonisomorphic groups of order $p^3$ can be obtained as central extensions of $C_p$ by $C_p\times C_p$, and the irreps of the abelian and the nonabelian ones are quite different. (By the way, it seems more usual to call $G$ a central extension of $H$ by $A$, assuming you want to have $A\subseteq \operatorname{\textbf{Z}}(G)$.) -However, the projective representations of $H$ tell you something about representations of central extensions, and here Schur's theory of the Schur multiplier and the "Darstellungsgruppe" of $H$ is helpful. You should look for these keywords in books on representation theory (e. g., the appropriate sections in Huppert, Endliche Gruppen I, Kapitel V, contain much information).<|endoftext|> -TITLE: A puzzle about finding three points $(x,y)$, $(x,z)$ and $(y,z)$ in a subset of a square. -QUESTION [24 upvotes]: I was asked (by myself) to give a proof of the following seemingly simple geometric statement, but after thinking a little I now suspect it could be less elementary than I thought (or am I being silly?). Does anybody know it, and can give an answer or a reference to it? Of course, I'm quite sure it should fit within a larger theory in combinatorics or in probability, but an elementary answer would be appreciated. - -Let $S$ be a (say open) subset of a square - $[0,1]^2$ with Lebesgue measure - $|S|>1/2$. Then, there exists a rectangle with a vertex on the diagonal, and the other three vertices in $S$ (in other words, there are three points of $S$ of the form - $(x,y)$, $(x,z)$ and $(y,z)\\ $). - -The constant $1/2$ cannot be lowered, as the example of the subset $S^* :=(0,1/2)\times(1/2,1)\cup(1/2,1)\times(0,1/2)$ shows (for any three points $x,y,z$ in $[0,1]$, at least 2 of them are both either smaller or larger than $1/2$, so the corresponding pair is not in $S^*$. - -REPLY [30 votes]: Let $S(x)=\{y\mid(x,y)\in S\}$ and $S^{-1}(y)=\{x\mid(x,y)\in S\}$, and let $\lambda_n$ denote the Lebesgue measure on $[0,1]^n$. We have -$$\begin{align*} -\int_S(\lambda_1S(x)+\lambda_1S^{-1}(y))\,dx\,dy -&=\int_S\lambda_1S(x)\,dx\,dy+\int_S\lambda_1S^{-1}(y)\,dx\,dy\\ -&=\int(\lambda_1S(x))^2\,dx+\int(\lambda_1S^{-1}(y))^2\,dy\\ -&\ge\left(\int\lambda_1S(x)\,dx\right)^2+\left(\int\lambda_1S^{-1}(y)\,dy\right)^2\\ -&=2(\lambda_2S)^2>\lambda_2S=\int_S1\,dx\,dy, -\end{align*}$$ -hence there exists $(x,z)\in S$ such that $\lambda_1S(x)+\lambda_1S^{-1}(z)>1$. This implies $S(x)\cap S^{-1}(z)\ne\varnothing$, i.e., there exists $y$ such that $(x,y),(y,z)\in S$.<|endoftext|> -TITLE: Symmetric polynomial from graphs -QUESTION [20 upvotes]: Let $g$ be a directed, connected multigraph, on $n$ vertices, without loops. -Define -$$P_g(x_1,\dots,x_n) := Sym\left[ \prod_{(i,j) \in g} (x_i-x_j) \right]$$ -where $(i,j)$ is the directed edge from $i$ to $j,$ and $Sym$ denotes the symmetrization, -that is, sum of all permutations of variables in the argument. -Now, for some multigraphs $g,$ we have that $P_g$ is identically zero. -A sufficient condition is that if we can change the direction of an odd number of edges in $g$ -and obtain a graph isomorphic to $g,$ then $P_g$ is identically 0. -This is however not necessary, as the graph with edges (1,2),(2,3),(3,4),(2,4),(2,4) -will give a polynomial that is identically 0. -The number of connected multigraphs with n edges that yields a zero polynomial -are, for n=1,..,6, equal to 1,0,3,2,19,20, and this sequence gives no hits in Sloane. -What I am asking for is a necessary and sufficient condition of a graph that gives the zero polynomial, as defined by the procedure above. -EDIT: Note that if $g_1$ and $g_2$ are isomorphic as undirected graphs, -then $P_{g_1} = \pm P_{g_2}.$ Changing the direction of one single edge changes the sign of the associated polynomial. - -REPLY [20 votes]: I am afraid this innocuous-looking question is in fact extremely hard, and I would be surprised if one could find a necessary and sufficient criterion which is more useful than the definition itself. Here is why: -The question pertains to the classical invariant theory of binary forms. -Firstly, suppose your graph is $v$-regular, i.e., all vertices have the same valence $v$. -Then if the $x_1,\ldots,x_n$ are interpreted as the roots of a polynomial of degree $n$, -or after homogenization as a homogeneous polynomial of degree $n$ in two variables, -i.e., a binary form, your sum defines an $SL_2$ invariant for such a binary form. -The nonregular case likewise corresponds to what 19th century mathematicians called -covariants. -Here is a fact. In the regular case, the product $nv$ has to be even since this is twice the -number of edges. Take $n=5$ and $v$ even but not divisible by 4 such that $v<18$. Then for every graph satisfying -this condition the polynomial is identically zero. -Likewise you can take $v=5$ and impose $n$ even but not divisible by 4 and $n<18$ and the result also is that -all graphs of this kind give zero. This is a nontrivial fact which has to do with -the invariants of the binary quintic: there is no skew invariant before degree 18 which is -the degree of Hermite's invariant. -Another example of your question is the following. Take $n=m^2$, and arrange the vertices -into an $m\times m$ square array. Take for the graph $g$ the following: -put an edge between two vertices if they are in the same row or in the same column. -It is trivial to see that the polynomial will vanish if $m$ is odd. -Now if you can prove that the graph polynomial is nonzero (for any even $m$) -then you would have proved the Alon-Tarsi conjecture, which implies the even case -of the Rota basis conjecture. -As far as I know these are widely open problems. -A reference on the general graph polynomial vanishing question is: -G. Sabidussi, Binary invariants and orientations of graphs, Disc. Math 101 (1992), pp. 251-277. -This very question was historically important for the development of graph theory since it motivated Petersen's work.<|endoftext|> -TITLE: Matrices: characterizing pairs $(AB, BA)$ -QUESTION [27 upvotes]: Let $A$ be an $m\times n$-matrix and $B$ an $n \times m$-matrix over the same field. Consider the matrices $C=AB$ and $D=BA$. It is probably well known (and not difficult to show) that the only difference between the canonic rational forms of $C$ and $D$ are nilpotent blocks (blocks with minimal polynomial $x^k$). (In particular, these compensate the different dimensions of $C$ and $D$.) -I'm interested in the converse question: - -Given an $m\times m$-matrix $C$ and an $n\times n$-matrix $D$, what are necessary and sufficient conditions that there exist matrices $A$ and $B$ such that $C=AB$ and $D=BA$? - -One may assume without loss of generality that $C$ and $D$ are both nilpotent. I'm thinking of a characterization in terms of the Jordan normal forms of $C$ and $D$. These in turn are characterized by their block sizes. In fact, an equivalent version can be stated as a question on partitions: - -Suppose $\lambda=(\lambda_1 \geq\lambda_2\geq \dotsc )$ and $\mu= (\mu_1 \geq \mu_2 \geq \dotsc)$ are partitions of the integers $m$ and $n$. When are there matrices $A$ and $B$ such that the blocks in the jordan normal forms of $AB$ and $BA$ belonging to the eigenvalue $0$ have sizes $\lambda_1, \dotsc$ and $\mu_1, \dotsc$, respectively? - -These are not arbitrary, for example, the quotient of the minimal polynomials must be in $\{1, x^{\pm 1}\}$, meaning that $|\lambda_1-\mu_1| \leq 1$. -This problem seems so natural that I think it has been addressed somewhere (not in the linear algebra books I looked into, however), so in particular I would appreciate a reference. -EDIT: I have now seen that $|\lambda_i -\mu_i|\leq 1$ for all $i$ is sufficient (and this is easy, since we may assume $C$ and $D$ in Jordan form, and then reduce to the case of one Jordan block). I guess it's necessary, too. Does anyone know a reference for this? And are there any nontrivial mathematical applications of this situation? - -REPLY [12 votes]: By accident, I came across a note in the American Mathematical Monthly (C. R. Johnson and E. A. Schreiner, The Relationship between AB and BA, vol. 103 (MR)), where this is treated. It seems that this was first treated in the following article: - -Harley Flanders, Elementary divisors of $AB$ and $BA$, Proc. Amer. Math. Soc. 2 (1951), p.871--874 (MR). - -In fact, $\lvert\lambda_i-\mu_i\rvert\leq 1$ is the exact condition; this is sometimes called "Flanders's theorem". This gives a keyword to search for on Math Reviews, etc., but probably this has been rediscovered more than once. (Note that the titles above don't contain very specific keywords.)<|endoftext|> -TITLE: On a decomposition of L^1(G) -QUESTION [7 upvotes]: [EDITED by Y. Choi - I have attempted to paraphrase the original question into something a bit terser and more precise; if this is not what the original poster intended, they should make corrections themselves.] -Let $G$ be a locally compact abelian (LCA) group and let $f\in L^1(G)$. Can we always find $g\in L^2(G)$ such that $h=f-g$ lies in $L^1(G)\cap B(G)$, where $B(G)$ is the Fourier-Stieltjes algebra of $G$? -($B(G)$ consists of all Fourier transforms of complex-valued regular Borel measures on $\Gamma$, the dual group of $G$.) -If there are counterexamples, are there counterexamples with $G={\mathbb R}^n$? -In the case $G={\mathbb R}^n$, as we know, the Calderon-Zygmund decomposition theorem asserts that every $f\in L^1({\mathbb R}^n)$ is the sum of its good part $g$ and bad part $b$. -Since $g$ is bounded and belongs to $L^1({\mathbb R}^n)$, it is not hard to verify that $g$ belongs to $L^p({\mathbb R}^n)$ for every $p\ge 1$. But it is easy to see that there exists an $f$ such that the inversion formula of Fourier transform fails for $b$. That is to say, the Calderon-Zygmund decomposition is not the decomposition of $L^1({\mathbb R}^n)$ that I want. - -REPLY [5 votes]: I'm answering Yemon's version of the question. -The answer is trivially yes for discrete $G$ since $\ell^1(G) \subset \ell^2(G)$, so let me focus on the non-discrete case. -The first observation to make is that $B(G)$ is contained in the bounded (and uniformly continuous) functions of $G$. So the question asks in particular if every integrable function on $G$ is the sum of a bounded function and a square-integrable function. -This is clearly false for compact infinite $G$: For such $G$ we have the strict inclusions $L^\infty \subsetneqq L^2 \subsetneqq L^1$ so $L^\infty + L^2 \subset L^2$, and hence every function in $L^1 \smallsetminus L^2$ provides a counterexample to the question. -Since the question asks for a counterexample in $\mathbb{R}^{n}$, I'll give one for $\mathbb{R}$ which is easily adapted to the higher-dimensional case and with a little care should also gives a counterexample for any non-compact and non-discrete locally compact abelian group. -Take $f = \sum_{n=1}^{\infty} n \cdot [n,n+\frac{1}{n^{3}}]$. This is a function in $L^1 \smallsetminus L^2$. For a bounded function $h$ we have for all $n \geq \Vert h \Vert_{\infty}$ and all $x \in [n,n+\frac{1}{n^3}]$ that $|f(x) - h(x)| \geq n- \Vert h \Vert_{\infty}$, which implies that $g = f - h \notin L^2(\mathbb{R})$ by a straightforward estimate.<|endoftext|> -TITLE: polyhedra with equilateral pentagons faces -QUESTION [9 upvotes]: In page http://loki3.com/poly/isohedra.html around six polyhedra with equilateral pentagons as faces are shown: a pyritohedron, icositetrahedrons... Is there a complete list of this kind of polyhedra? How to compute the angles of those pentagons? - -REPLY [8 votes]: The pentagonal isohedra with sides of equal length listed on http://loki3.com/poly/isohedra.html are the regular dodecahedron, non-convex equilateral pyritohedron, equilateral pentagonal icositetrahedron, non-convex equilateral pentagonal icositetrahedron, non-convex equilateral pentagonal hexecontahedron and non-convex equilateral pentagonal hexecontahedron. - -(source)   - -(source)   - -(source)   - -(source)   - -(source)   - -(source)   -(I should note that the page jolumij references, is a page I built to summarize my findings. I had tried to learn about the isohedra from pages such as http://mathworld.wolfram.com/Isohedron.html, but all sources I could find were very incomplete when it came to describing the pentagonal isohedra. They offer names such as "octahedral pentagonal dodecahedron" without describing how to construct them or mentioning they may represent an infinite family of shapes.) -I assume by "this kind of polyhedra," you're referring to isohedra with pentagonal faces. Mathworld offers this definition of an isohedron: -An isohedron is a convex polyhedron with symmetries acting transitively on its faces with respect to the center of gravity. -For my list of isohedra, I relax the definition to include non-convex polyhedra. The isohedral transforms can also be used to create polyhedra with intersecting faces "with symmetries acting transitively on [their] faces with respect to the center of gravity." -As to whether those 6 are the only isohedra with equilateral pentagonal faces, I believe the list is complete, but I haven't rigorously proven it. What I have done is start from the tetrahedral, octahedral and icosahedral symmetry groups and applied the isohedral pentagonal transform to them. This transform has two degrees of freedom. Then I explored the space for equilateral pentagons as well as other interesting symmetries or patterns. I haven't seen references to many of the shapes I found (including 5 of the 6 shapes listed here), so I'd be interested if other people know of any other references. -As to how to compute the angles of those pentagons, http://loki3.com/poly/transforms.html#penta gives a description of the transform and what my notation means. You can use the parameters and transform to derive the angles. For example, the non-convex equilateral pyritohedron is 4p(0.09549150, 0.6605596), which means you apply the isohedral pentagonal transform (p) to a tetrahedron (4) using the parameters 0.09549150 and 0.6605596. In this case, you get two 36 degree angles and three 108 degree angles.<|endoftext|> -TITLE: Is every Kahler form on a Stein manifold exact outside a compact set? -QUESTION [5 upvotes]: Is it true that on a smooth Stein manifold (or smooth affine variety or smooth complete intersection in $\mathbb{C}^{n}$), every Kahler form is exact outside a compact set? - -REPLY [9 votes]: This seems extremely false to me: I can't think of any example of a positive dimensional Stein manifold $W$ and a compact subset $K$ such that $H^2(W) \to H^2(W \setminus K)$ has nontrivial kernel. That means there should be an abundance of counter-examples: Just take any Kahler form representing a nontrivial class in $H^2(W)$. -In any case, here is a specific counter-example. Take $W = (\mathbb{C}^*)^2$ with coordinates $z_1$ and $z_2$. Take a $(1,1)$ form of the form: -$$\omega := a \frac{dz_1 \wedge d\overline{z_1}}{z_1 \overline{z_1}} + b \frac{dz_1 \wedge d\overline{z_2}}{z_1 \overline{z_2}} + c \frac{dz_2 \wedge d\overline{z_1}}{z_2 \overline{z_1}} + d \frac{dz_2 \wedge d\overline{z_2}}{z_2 \overline{z_2}}$$ -This is obviously closed. If I haven't dropped any signs, it is Kahler if and only if $\left( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right)$ is positive definite Hermitian. (It is easiest to do this computation after setting $z_i = e^{w_i}$, so $d z_i/z_i = d w_i$.) -Now, let $T$ be the torus $|z_i|=r_i$. Irrespective of what $r_1$ and $r_2$ are, $\int_T \omega = (b-c) (4 \pi^2)$. (Again, if I haven't dropped any signs.) So, if we take $\left( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right)$ a positive definite Hermitian matrix with $b-c \neq 0$, then we will have a Kahler form $\omega$ for which $\int_T \omega \neq 0$. For example, $\left( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right) = \left( \begin{smallmatrix} 2 & i \\ -i & 2 \end{smallmatrix} \right)$ would do. -Now, let $K$ be any compact set. For $(r_1, r_2)$ large enough, the torus $T$ is disjoint from $K$. So we have exhibited a Kahler form on $W$ such that, for any $K$, there is a $2$-cycle $T$ in $W \setminus K$ with $\int_{T} \omega \neq 0$. So $\omega$ is not exact on any $W \setminus K$.<|endoftext|> -TITLE: What notions are used but not clearly defined in modern mathematics? -QUESTION [103 upvotes]: "Everyone knows what a curve is, until he has studied enough mathematics to become confused through the countless number of possible exceptions." -Felix Klein - -What notions are used but not clearly defined in modern mathematics? - -To clarify further what is the purpose of the question following is another quote by M. Emerton: - -"It is worth drawing out the idea that even in contemporary mathematics there are notions which (so far) escape rigorous definition, but which nevertheless have substantial mathematical content, and allow people to make computations and draw conclusions that are otherwise out of reach." - -The question is about examples for such notions. -The question was asked by Kakaz - -REPLY [9 votes]: The concept of turbulence is still vaguely or ill defined such as applied to too many phenomena. -Examples from Is there a mathematically precise definition of turbulence for solutions of Navier-Stokes? and elsewhere - -In the Ptolemaic Landau–Hopf theory turbulence is understood as a cascade of bifurcations from unstable equilibriums via periodic solutions ([the Hopf bifurcation][2]) to quasiperiodic solutions with arbitrarily large frequency basis. -According to [Arnold and Khesin][3], in the 1960's most specialists in PDEs regarded the lack of global existence and uniqueness theorems for solutions of the 3D Navier–Stokes equation as the explanation of turbulence. -Kolmogorov suggested to study minimal attractors of the Navier-Stokes equations and formulated several conjectures as plausible explanations of turbulence. The weakest one says that the maximum of the dimensions of minimal attractors of the Navier–Stokes equations grows along with the Reynolds number Re. -In 1970 Ruelle and Takens formulated the conjecture that turbulence is the appearance -of global attractors with sensitive dependence of motion on the initial conditions in the phase space of the Navier–Stokes equations ([link][4]). In spite of the vast popularity of their paper, even the existence of such attractors is still unknown. -Existence of energy cascades (eg. Big vortices feeding on smaller vertices). This reflects the physical notion that mechanical energy injected into a fluid is generally on fairly large length and time scales, but this energy undergoes a “cascade” whereby it is transferred to successively smaller scales until it is finally dissipated (converted to thermal energy) on molecular scales -Von Karman: “Turbulence is an irregular motion which in general makes its appearance in fluids, gaseous or liquid, when they flow past solid surfaces or even when neighboring streams of the same fluid flow past or over one another.” -Hinze: “Turbulent fluid motion is an irregular condition of the flow in which the various quantities show a random variation with time and space coordinates, so that statistically distinct average values can be discerned.” -Chapman: “Turbulence is any chaotic solution to the 3-D Navier–Stokes equations that is sensitive to initial data and which occurs as a result of successive instabilities of laminar flows as a bifurcation parameter is increased through a succession of values.” -Criteria listed in McDonough's notes: - -nonrepeatability (i.e., sensitivity to initial conditions); -disorganized, chaotic, seemingly random behavior -extremely large range of length and time scales -enhanced diffusion (mixing) and dissipation (both of which are mediated by viscosity at molecular scales) -three dimensionality, time dependence and rotationality (hence, potential flow cannot be turbulent because it is by definition irrotational); -intermittency in both space and time.<|endoftext|> -TITLE: Subfields of a function field -QUESTION [14 upvotes]: Is there an algorithm for generating (some or all) subfields of a certain genus of a given function field (even a random one,I mean for example generating a random elliptic subfield of a certain given function field). I did a quick search and it seems to me that the problem is heavily treated in the case of cyclic and Hermitian function fields, but I was wondering what do we know in general case. Is there something that I can do in Magma? -On the other hand, do we have an algorithm to check if $F$ is a subfield of $E$, When $F, E$ are function fields (of one variable)? Florian Hess told me that somebody developed such an algorithm using his automorphism algorithm but I don't have much luck finding it. -In order to stick to the tradition, I give a motivation also: Subfields of function fields with a rich automorphism group are subject to cover attack in cryptography when they are not one of those few which are fixed by an automorphism of the cover. -Thank you very much indeed! - -REPLY [5 votes]: The algorithm to embed function fields, ie. to test if a function field E can be embedded into a function field F has been developed (and implemented in Magma) by a student of Florian Hess: Gerriet Möhlmann as -part of his Diploma work. His thesis (in German) can be found at -http://www.math.tu-berlin.de/~kant/publications/diplom/moehlmann.pdf. The method is an extension of Florian's automorphism algorithm, in Magma, it is available through the Inclusions command. - -To generate function fields of a given genus there are a few possibilities, none of them worked out completely. If the field can be obtained as an Abelian extension (eg. (hyper)elliptic curves have a degree 2 model) then class field theory can be used to generate all such fields. Similar, soluble exxtensions can -be constructed this way. For general extensions, one could use Hunter's theorem to get bounds on the valuations of a primitive element and then enumerate all polynomials that might have such roots. Both methods have in common that they produce too many field extensions that correspond to isomorphic curves. The class field theoretic approach has the advantage of being available through Magma.... (I can provide details if anyone is interested)<|endoftext|> -TITLE: Recent progress on Bochner-Riesz conjecture -QUESTION [8 upvotes]: Consider the family of operators $T_\delta$, $\delta \geq 0$, defined on $\mathbb{R}^n$ by -$ -\widehat{T_\delta f}(\xi) = (1-|\xi|^2)_+^\delta \widehat{f}(\xi). -$ -($(1-|\xi|^2)_+^\delta$ are known as Bochner-Riesz multipliers.) We are interested in the $L^p$ boundedness of $T_\delta$. The case $\delta = 0$ has been solved since 1971, once Fefferman provided a proof that $L^p$ boundedness fails in dimension $n \geq 2$ when $p \neq 2$. For general $\delta > 0$, a Theorem due to Herz shows that a necessary condition for boundedness is that -$ -|\frac{1}{p}-\frac{1}{2}| < \frac{2\delta + 1}{2n}. -$ -It's thus natural for one to conjecture that this is also a sufficient condition. -Here's what I think I know about current progress, from jotted down notes: - -Holds for $n \leq 2$ -Holds for $p = 2$ -$T_\delta$ is bounded when $\delta > \frac{n-1}{2}$ (by Young's inequality) -$T_\delta$ is bounded when $|\frac{1}{p}-\frac{1}{2}| < \frac{\delta}{n-1}$ -Holds if $\delta > \frac{n-1}{2(n+1)}$ (Can't remember the reference) -Holds for $n \geq 3$ when $p \geq \frac{2(n+2)}{n}$ or $p \leq \frac{2(n+2)}{n+4}$ (Found in Tao's Recent progress on the Restriction conjecture) - - -Question: What is the most recent - progress on this conjecture? I'm - curious about the general case and - also specific values of $n$, such as - $n = 3$. - -REPLY [5 votes]: The results stated in your post are improved in the recent work of Bourgain and Guth, in dimensions 5 and higher. The numerology is the same as for the restriction problem for the sphere (see the statement of theorem 1 in that paper). In the case of the restriction problem for the sphere Bourgain and Guth improved the 3 dimensional result to $p > 3 + 3/10=3.3$, which is slightly better than the $p>10/3=3.333...$ from Tao's bilinear estimates (which, combined with Lee's work, gives the corresponding result for Bochner-Riesz.) This is the "same" $p>10/3$ you mention in your post. It seems likely that Bourgain and Guth's argument will give a similar improvement for the 3 dimensional Bochner-Riesz problem, but they do not work this out in the paper. They do write (page 5) "Thus in principle, one could expect the proof of Theorem 2 to carry over and lead to the validity of the Bochner-Riesz conjecture for $max(p, p′) ≥ 3 +3/10$ , if n = 3. We do not pursue the details of this matter here."<|endoftext|> -TITLE: What is a Specht module? -QUESTION [8 upvotes]: I'm studying the structure of the Specht module for $S_n$ and I would like to know if there is some generalizations of this structure for Weyls groups or Coxeter groups. -Also, I'm interest to know about category-theoretics way of study this module. I just know one article about this subject: -http://www.math.uni-bonn.de/people/stroppel/Specht.pdf - -REPLY [7 votes]: The short answer to your question about Specht modules for other types than the symmetric group is "yes". The long answer is that you have to dig into the extensive literature built up around cyclotomic Hecke algebras and the like. One place to look is the arXiv, where the papers of Andrew Mathas (Sydney) and his collaborators from recent years can be found. A recent example is here, but there are many related papers by other people including Meinolf Geck and his collaborators. -Though I haven't followed all of these developments closely, my perception is that the emphasis tends to shift from the finite Coxeter groups such as $S_n$ to their Iwahori-Hecke algebras and related module categories. Combinatorics is a major subtheme throughout, as well as categorification.<|endoftext|> -TITLE: When do isometric actions exist? -QUESTION [10 upvotes]: Let $X$ be a metrizable topological space and $G$ be a locally compact group. Given a continuous (left) action of $G$ on $X$, is there a metric on $X$, compatible with the topology, for which the action of $G$ becomes an isometric action? Conversely, given a metric on $X$, is there a nontrivial action of $G$ on $X$ that preserves the metric? -I am looking for the most general necessary and sufficient conditions and any possible obstructions. For the first question, the answer is obviously positive when $G$ is compact: one chooses a metric $d$ on $X$ and simply does an "averaging process along the orbits" by defining -$$ \rho(x,y) = \int_{G} d(g^{-1}\cdot x,g^{-1}\cdot y) dg . $$ -I suspect that a similar idea would work more generally using a "cut-off" function on $X$ when the action of $G$ on $X$ is proper. Any connections to amenability (of the group or the group action) would also be interesting. - -REPLY [9 votes]: I am sorry That I am getting involved so late. I was away at a meeting for two days. I have an old (1961) paper in the ANNALS OF MATH called "On the Existence of Slices for Actions of Non-Compact Lie Groups" which is quite relevant. In particular, on page 318 you can find the following concerning proper actions of an arbitrary Lie group $G$. -Theorem 4.3.4. Every seperable, metrizeable, proper $G$-space $X$ admits an invariant metric. ... -There are a large number of other theorems there that show that the theory of proper G-spaces for G an arbitrary Lie group is similar to the theory of G-spaces for a compact Lie group. You can find the paper here: http://vmm.math.uci.edu/ExistenceOfSlices.pdf<|endoftext|> -TITLE: When is a holomorphic submersion with isomorphic fibers locally trivial? -QUESTION [12 upvotes]: A justly celebrated theorem by Ehresmann states that a proper smooth submersion $\pi: X\to S$ between smooth manifolds is locally trivial in the sense that every point $s\in S$ downstairs has a neighbourhood $ U$ such that $\pi ^{-1} (U) $ is $S$-diffeomorphic (=fiber preserving diffeomorphism) to $U\times F$ for some manifold $F$, called the typical fiber. Of course the holomorphic analogon is completely false: deformation theory might be described as the study of this failure! -To give a concrete example, consider the family $X \subset S\times \mathbb P^2 (\mathbb C) $ of elliptic curves $y^2z=x(x-z)(x-\lambda z)$ with $\lambda \in S=\mathbb C \setminus \{0,1\} $ and the corresponding proper holomorphic submersion $\pi: X \to S: (\lambda , [x:y:z]) \mapsto \lambda $. This $\pi$ is certainly not locally trivial downstairs because its fibers are not mutually isomorphic. -However, in the proper case, non-isomorphy of fibers is the only obstruction to being locally trivial. Indeed, Fischer and Grauert proved that a proper holomorphic submersion having all its fibers isomorphic is locally trivial downstairs. I wonder what can be salvaged of their theorem in the non-proper case: -Question: Is there a class $\mathcal C$ of non-compact complex manifolds such that the following holds. If a holomorphic submersion $\pi: X\to S$ between complex manifolds has all its fibers $\pi^{-1}(s)\;(s\in S)$ isomorphic to the same $F \in \mathcal C$, then $\pi$ is locally trivial downstairs with typical fiber $F$. -Bibliography Fischer and Grauert unfortunately published their result in a rather confidential journal: W. Fischer, H. Grauert, Lokal-triviale Familien kompakter komplexer Mannigfaltigkeiten, -Nachr. Akad. Wiss. G¨ottingen Math.-Phys. Kl. II (1965), 89–94. -(Please note that the Fischer here is Grauert's doctoral student Wolfgang Fischer, not the complex geometer Gerd Fischer. ) - -REPLY [5 votes]: Here is a variant of Jason's example with a proof that it is not even topologically locally trivial. Let $T$ be a (complex) manifold that admits a morphism $\phi$ onto $\mathbb P^1=\mathbb P^1_{\mathbb C}$ (or $S^2$ if you prefer) and there exists a point $a\in T$ with $b=\phi(a)\in \mathbb P^1$ such that $\{a\}=\phi^{-1}(b)$. This is satisfied for example if $\phi={\rm id}_{\mathbb P^1}$. Let $\Gamma\subset T\times \mathbb P^1$ be the graph of $\phi$ and let and $c\in \mathbb P^1, c\neq b$. -Now let $X=T\times \mathbb P^1\setminus \bigg( \Gamma\cup \big(T\times \{b\}\big)\cup \{(a,c)\}\bigg)$ with the natural projection $\pi:X\to T$. Then every fiber of $\pi$ is isomorphic to $\mathbb P^1\setminus \{0,\infty\}\simeq \mathbb C^*\sim S^2\setminus \{P,Q\}$ (for two points $P,Q\in S^2$). -Claim $\pi$ is not topologically locally trivial near $a\in T$. -Proof Suppose $a$ has a neighbourhood $U\subseteq T$ such that $Y:=\pi^{-1}U\simeq U\times S^2\setminus \{P,Q\}$. Then there exists a projection $p:Y\to S^2\setminus \{P,Q\}$. Consider a circle in $S^2\setminus \{P,Q\}$ that's non-trivial in $H_1(S^2\setminus \{P,Q\}, \mathbb Z)$. Since $p$ is an isomorphism between $\pi^{-1}(a)$ and $S^2\setminus \{P,Q\}$, the same circle lives in $\pi^{-1}(a)$ as well. Then the homology class of the circle can be represented by a "small" circle around the point $(a,c)$ (this point is not in $X$!). Next take a "small" ball inside $T\times \mathbb P^1$ with center at $(a,c)$ that contains the previous "small" circle. By the construction of $X$, the intersection of this ball with $X$ is the entire ball except its center $(a,c)$. Therefore the homology class of that "small" circle in $X$ is trivial. However, this is a contradiction, because it was chosen in a way that its image via $p_*$ would be a nontrivial homology class. $\qquad {\rm Q.E.D.}$ -Remark I suppose a similar proof works to show that Jason's example is also not locally trivial.<|endoftext|> -TITLE: p-adic representations of a quaternion algebra over a local field -QUESTION [10 upvotes]: How to determine a complete set of isomorphism class representatives of the irreducible algebraic representations of $D^{\times}/F$ (where $D$ is a quaternion algebra over a local field $F/\mathbb{Q} _p$) on $E$ (also a finite extension of $\mathbb{Q} _p$)? -Answer for other (algebraic) groups would also be welcome as well as any references to the literature. - -REPLY [3 votes]: If $E$ is an algebraic closure of $F$, then $D\otimes_F E\simeq M_2(E)$. (In fact this is also true if $E$ is taken to be, say, the unramified quadratic extension field of $F$.) We get an algebraic representation -$$\phi\colon D^\times\hookrightarrow (D\otimes E)^\times=\text{GL}_2(E).$$ -And then for each $a\geq 0$ and $b\in \mathbf{Z}$ we get the representation $\text{Sym}^{a}\phi\otimes (\det\phi)^b$. My feeling is that these exhaust the irreducible algebraic representations of $D^\times$, but I'm afraid I don't have a proof at the ready. -As the other answerers show, the question of classifying the admissible representations of $D^\times$ (with complex coefficients) is a far more subtle issue!<|endoftext|> -TITLE: learning crystalline cohomology -QUESTION [30 upvotes]: From which sources would you learn about crystalline cohomology and the de-Rham-Witt complex? - -REPLY [19 votes]: With enough enthusiasm, I would try to learn about crystalline cohomology and the de-Rham-Witt complex from the homonymous article by Illusie: -Illusie, Luc. Complexe de deRham-Witt et cohomologie cristalline. (French) Ann. Sci. École Norm. Sup. (4) 12 (1979), no. 4, 501--661. MR0565469 (82d:14013) -Fortunately, it is publicly available at: -http://archive.numdam.org/ARCHIVE/ASENS/ASENS_1979_4_12_4/ASENS_1979_4_12_4_501_0/ASENS_1979_4_12_4_501_0.pdf -But this is most usefully read as needed after one is acquainted with the following also relevant references (perhaps in this order): -S. Bloch, Algebraic K-theory and crystalline cohomology, Publ. Math. Inst. Hautes Etudes Sci. 47 (1977), 187–268. -O. Hyodo and K. Kato, Semi-stable reduction and crystalline cohomology with logarithmic poles, in Periodes p-adiques (Bures-sur-Yvette, 1988), Asterisque 223 (1994), 221–268. -O. Hyodo, On the de Rham–Witt complex attached to a semi-stable family, Compositio Math. 78 -(1991), 241–260. -O. Hyodo, A cohomological construction of Swan representations over the Witt ring. I, Proc. Japan Acad. Ser. A Math. Sci. 64 (1988), 300–303.<|endoftext|> -TITLE: homotopy pushout of spaces homotopic to finite CW complexes -QUESTION [7 upvotes]: Does anyone know a reference for the fact that a homotopy pushout (double mapping cylinder) of spaces which are homotopy equivalent to finite CW complexes is also homotopy equivalent to a finite CW complex? - -REPLY [4 votes]: I don't have a reference, but here is an easier argument, based, like John's, on the homotopy -invariance of the homotopy pushout. -The invariance implies that you can replace the maps $i: A\to B$ and $j: A\to C$ with homotopic maps and get the same homotopy pushout, up to homotopy type. So assume they are cellular. It is easy to give the inclusions $A\hookrightarrow M_i$ and $A\hookrightarrow M_j$ finite CW structures so that $A$ is a subcomplex of each, and then the union $M_i \cup_A M_j$ inherits a finite CW structure.<|endoftext|> -TITLE: How equivalent are the theories of reduced and groupal $\infty$-groupoids? -QUESTION [7 upvotes]: I hope that my question is sufficiently trivial that someone will be able to give me a pedantic answer, and not so trivial that no one takes the time to give an answer. My motivation for asking this question is that my category number has been hovering somewhere around $2$, and I'd like to increase it, but $\infty$ is often easier than $3$. -Suppose that I have some familiarity with the following words (meaning, feel free to "remind" me what the correct definitions are): - -Some version (Stasheff associahedra?) of $A_\infty$ monoids. -Kan simplicial sets as $\infty$-groupoids. - -One can then define the following: A groupal $\infty$-groupoid is an $A_\infty$ monoid $G$ in $\infty$-groupoids, such that the map $(g,h) \mapsto (g,gh)$, $G \times G \to G\times G$ is an equivalence of $\infty$-groupoids. (If this isn't quite the right definition, please let me know.) -One could instead talk about $\infty$-groupoids for which the set of $0$-morphisms is a point. I think this is what are called reduced. -I'm under the impression that these should be "the same". If I were working not with $\infty$-groupoids but rather at a low categorical level, I would understand how they are the same: a groupoid with one object is "the same" as a group or groupal set (an associative monoid such that the map $(g,h) \mapsto (g,gh)$ is an isomorphism). -More precisely, there should be functors $\Omega$ and $\rm B$ between the $(\infty,1)$-categories of reduced $\infty$-groupoids and groupal $\infty$-groupoids, and I would assume that these are an equivalence, in the appropriate sense. -I almost understand these functors: - -Given a reduced $\infty$-groupoid $G$, I would try to define $\Omega G = \hom(S,G)$, where $S$ is some $\infty$-groupoid version of the circle, say a particular $S = \mathrm B\mathbb Z$ that I might construct by hand. The "$\hom$" is just the hom of $\infty$-groupoids (reduced $\infty$-groupoids are full in $\infty$-groupoids), and in particular it takes values in $\infty$-groupoids; on the other hand, letting $\vee$ denote the coproduct in reduced groupoids, there is a distinguished map $S \to S\vee S$ which winds around the outside of the figure-eight, and pulling back along this map gives the groupal structure on $\Omega G$. Left to check is that this really is a groupal structure, but that should be easy. -Given a groupal $\infty$-groupoid $G$, I should try to define $\mathrm B G$ in the same way that I would if I were just starting with a group. But a priori I only see how to define $\mathrm B G$ as a simplicial object in $\infty$-groupoids. So my biggest difficulty here is that don't know how to collapse what I'm modeling as a "double simplicial set" into a "single simplicial set". Writing $\Delta$ for the category whose objects are finite totally-ordered sets and whose morphisms are non-decreasing maps (so that a simplicial set is a functor $\Delta^{\mathrm{op}} \to \mathrm{Set}$), maybe there is a nice map $\Delta \to \Delta^{\times 2}$ along which I can pull back? If so, then there only remains to check the Kan condition. -Oh, and I'd need to check that $\Omega,\mathrm B$ are inverse (up to ...) to each other. - -After a rambly introduction, my questions are: - -Is this all correct? What is the $\mathrm B$ construction? What's the precise statement of the equivalence between groupal and reduced $\infty$-groupoids? - -I assume that this type of thing is carefully spelled out somewhere in the literature. So maybe my real question is: - -What is a good reference that will take my hand and walk me through this part of category theory? - -REPLY [8 votes]: To answer your first set of questions in order (I'm going to use the word "space" for "$\infty$-groupoid"): -Yes, this is all correct. -You seem to be familiar with how to construct $BG$ as a simplicial space via a bar construction. To turn this into an actual space, you just have to form the geometric realization of this simplicial space. This is just like the geometric realization of a simplicial set: if $X_*$ is a simplicial space, its realization is the quotient of $\bigsqcup X_n \times \Delta^n$ given by gluing together simplices according to the face (and degeneracy) maps. More succinctly, it is the coend of the functor $X:\Delta^{op}\to Spaces$ along the functor $\Delta \to Spaces$ sending $[n]$ to the standard $n$-simplex. More abstractly, it is the (homotopy) colimit of the diagram $X:\Delta^{op}\to Spaces$ in the $(\infty,1)$-category of spaces. If you model spaces as simplicial sets, it can also be described as the pullback of your bisimplicial set along the diagonal functor $\Delta\to\Delta^2$ (to show this, you need to only verify it for single "bisimplices" (ie, representable functors on $\Delta^2$), which amounts to the fact that the product of two representable functors on a category $C$ is the pullback of the associated representable functor on $C^2$ along the diagonal $C\to C^2$). -I don't know the best way of precisely formulating and proving this, but one way is to construct an explicit Quillen equivalence between the categories of simplicial groups and the category of reduced simplicial sets (simplicial groups can be used instead of $A_\infty$-groups because $A_\infty$-spaces can be rigidified). For this, the delooping functor is exactly just taking the geometric realization of the bar construction. The looping functor is subtler--one has to be careful to get a functor which actually lands in simplicial groups. Even if you were happy to land in $A_\infty$-groups, just taking the simplicial mapping space $Maps(S^1,X)$ would not work because for non-fibrant objects $X$ this does not have a multiplication. The classical solution to this is called the "Kan loop group", the exact details of which I don't remember but are described on nlab in the generalized setting of a not-necessarily-reduced simplicial set (in which case you get a simplicial groupoid, rather than a simplicial group). In any case, this is just a specific simplicial model for the loopspace of a reduced simplicial set which happens to actually be a group. -As for a reference, I learned the Kan loop group and the Quillen equivalence it gives from Goerss-Jardine's book Simplicial Homotopy Theory, but they say essentially nothing about thinking about this from an $\infty$-categorical perspective.<|endoftext|> -TITLE: Amenable groups not containing free semigroups -QUESTION [10 upvotes]: It is known that all amenable groups do not contain free subgroups (of rank>1). But there are amenable groups containing free semigroups. Which amenable groups cannot contain free semigroups? - -REPLY [18 votes]: This is the answer to the question asked by Henry. The wreath product $\mathbb Z_2 {\rm wr} G$, where $G$ is the Grigorchuk (torsion) group of subexponential growth, obviously has exponential growth and is amenable and torsion. In particular, it has no free subsemigroups. -For elementary amenable (in particular, solvable) groups, existence of non-cyclic free subsemigroups is equivalent to exponential growth [C. Chou, Elementary amenable groups, Illinois J. Math. 24 (1980), 3, 396-407].<|endoftext|> -TITLE: Equivalence of Branched Coverings -QUESTION [7 upvotes]: For equivalence of unbranched coverings of topological spaces, there is a criteria: -Two coverings (unbranched) $p_1\colon Y_1\rightarrow X$ and $p_2\colon Y_2\rightarrow X$ are equivalent iff for some $q\in X$ and $\bar{q_1}\in p_1^{-1}(q)$ and $\bar{q_2}\in p_2^{-1}(q)$, the induced subgroups $p_*\pi_1(Y_1,\bar{q_1})$ and $p_*\pi_1(Y_2,\bar{q_2})$ are conjugate in $\pi_1(X,q)$. -Is there any criteria (particularly group theoretic) for equivalence of branched coverings of Riemann surfaces? - -REPLY [8 votes]: The answer is yes: the equivalence class of the covering is detected by the monodromy representation of the fundamental group of the base minus the branch locus, up to conjugacy. -More precisely, let $f \colon X \to Y$ be a (possibly branched) covering of degree $d$ of Riemann surfaces. Choosing a point $y_0 \in Y$ not lying in the branch locus $B$, there is a monodromy representation $$\rho \colon \pi_1(Y-B, y_0) \to S_d \quad (*)$$ -whose image is transitive (since $X$ is assumed to be connected). Moreover, if we choose a different base point it is easy to check that the map $\rho$ varies only up to conjugacy in $S_d$. -Then there is the following well-known - -Theorem. - There exists a one-to-one correspondence between branched coverings $f \colon X \to Y$ of degree $d$ whose branch points lie in $B$ and group homomorphisms of type $(*)$ with transitive image (up to conjugacy in $S_d$). - -For further details, see Miranda's book Algebraic curves and Riemann surfaces, especially Chapter 4.<|endoftext|> -TITLE: Existence of Non-measurable sets -QUESTION [5 upvotes]: Existence of non-measurable sets requires the use of Axiom of choice. And one can construct models eg. Solovay model without using the Axiom of choice where there are no non-measurable sets. - So, Can one start with the the hypothesis that non-measurable sets exist and then 'prove' the Axiom of choice? - -REPLY [8 votes]: The answer to your question is "no". -Using the technique of forcing one can easily violate the axiom of choice in a variety of ways (essentially, by focusing on "large enough sets") without affecting the fact that there are non-measurable sets of reals. This includes leaving a bit of choice to make sense of the construction of Lebesgue measure and development of its basic properties (the principle of "dependent choice for relations on reals" seems more than enough for this). -As far as I know, there are no known "choice or choice-like principles" even at the level of the reals known to be equivalent or to reasonably follow from the assumption that there are non-measurable sets. -(By the way, this is form 93 in "Consequences of the axiom of choice" by Howard and Rubin. The companion website does not reveal any known implications.)<|endoftext|> -TITLE: sum of the character of the symmetric group -QUESTION [5 upvotes]: Suppose $\mu$ is a fixed partition of $n$ of length $l(\mu)$, and I was encountered with the following sum, namely -$\sum_{\nu} \chi_{\nu}(\mu)$. -I did some calculation using the character table that I can find (mainly Fulton & Harris's book, they have the character table up to $S_5$), and found that the sum does not vanish only if $\mu$ has an even number of even parts(someone call such $\mu$ an orthogonal partition). -This is actually very simple to prove, only use the fact that -$\chi_{\nu^t}(\mu)=(-1)^{n-l(\mu)} \chi_{\nu}(\mu)$, if $\mu$ has an odd number of even parts, then $n-l(\mu)$ is odd. -But my calculation indicates more: the sum is nonzero only if every even part of $\mu$ occurs even times.(someone also call such partition an orthogonal partition, and I donot know which is the correct definition...can anyone help?) -I checked this for $n \leq 6$ and also for $n=11$ (I found the charater table of $S_{11}$ in some paper...) -I donot know whether this is just an coincidence or this is always true. -Since my knowledge of symmetric group is very limited, I donot hesitate to ask for help on MO. Hopefully, someone will give me an answer. Thank you all! -p.s. (1) My second question, which is quite related to the above one. We know that -$\sum_{\nu} s_{\nu}(x) s_{\nu}(y)=\prod_{i,j} \frac{1}{1-x_i y_j}$, where $s_{\nu}$ is the Schur function. -Is there a similar expression for $\sum_{\nu} s_{\nu}$? -(2) My third question: Is there a similar expression for $\sum_{\nu} (\frac{|\nu|!}{dim R_{\nu}})^k s_{\nu}$? Here $R_{\nu}$ is the irreducible representation indexed by $\nu$, and $k$ is a positive integer. - -REPLY [10 votes]: Your conjecture is correct; as a matter of fact, it is possible to compute these sums $C(\mu):=\sum_\nu\chi_\nu(\mu)$ exactly for any $\mu$: -Suppose $\mu$ has $m_i$ parts of length $i$, for $i=1,2,\ldots$. Then $C(\mu)=\prod_{i>0} c_{i,m_i}$, where $c_{i,m_i}$ is the coefficient of $t^{m_i}/({m_i}!)$ in $\exp(t+\frac{1}{2}it^2)$ if $i$ is odd, or in $\exp(\frac{1}{2}it^2)$ if $i$ is even. -A first proof can be found in Macdonald's Symmetric functions and Hall polynomials, ex.11 p.122; it relies on symmetric function techniques. A second one is given in Stanley's Enumerative Combinatorics Vol. 2, ex. 7.69, and is based on the fact that, from a general character theory result, $C(\mu)$ is the number of square roots of a given permutation of cycle-type $\mu$.<|endoftext|> -TITLE: When does a polynomial have all pure imaginary roots? -QUESTION [6 upvotes]: Let $P(x)=x^{n}+a_{1}x^{n-1}+\cdots+a_{n-1}x+a_{n}$, where $a_1, a_2, \dots, a_n$ are intergers. -Question 1. When does the polynomial $P(x)$ have its zeros all being pure imaginary or zero(here 0 is a root of the given polynomial)? -Question 2. Does there exist a characterization that $P(x)$ have its zeros all being pure imaginary or zero? -Please point me to some references if this has already been studied. Thanks for your time! - -REPLY [3 votes]: There is a method due to Sturm which allows the determination of the number of roots of a polynomial in any real interval, in particular you may apply this to find the number of negative roots. The following Wikipedia article may serve as an initial reference: -http://en.wikipedia.org/wiki/Sturm%27s_theorem<|endoftext|> -TITLE: How to be updated with current advances in mathematics -QUESTION [17 upvotes]: Hello, -I wanted to know that how researchers in mathematics keep updates in their field of interest. -(original question by Rahul Gupta) - -REPLY [5 votes]: The same as everyone else: arxiv, conferences, in my case "Number theory web", talking to friends and colleagues, etc. But I've come to realize that all of this keeps me updated about a very small portion of mathematics. To remedy this, once every couple of months I walk down to the math periodicals section of our library and spend an hour or so browsing the most recent issues of good journals: read abstracts, scan keywords, etc. I somehow find this very helpful; granted, by the time papers appear in print, they are for the most part outdated, but this still gives me a sense, though with some delay, as to what's going on in the mathematical world. In the same vein, I find the Notices of the American Mathematical Society very helpful.<|endoftext|> -TITLE: What is the best way to study Rational Homotopy Theory -QUESTION [13 upvotes]: I studied basic algebraic topology elements: -fundamental group, higher homotopy groups, fibre bundles, homology groups, cohomology groups, obstruction theory, etc. -I want to study Rational Homotopy Theory. -Specifically, I want to study Sullivan's model. -What is the short way and what is the complete way to study Sullivan's model? - -REPLY [4 votes]: Here are some video lectures that John Morgan gave at Stony Brook -http://www.math.sunysb.edu/Videos/dfest/ -This also has many other nice mathematical videos.<|endoftext|> -TITLE: Applications of the notion of of Gromov-Hausdorff distance -QUESTION [12 upvotes]: I am looking for applications of the notion of Gromov-Hausdorff convergence to prove theorems that a priori have nothing to do with it. Examples that I am aware of (thanks to wikipedia and google): - -Gromov's theorem -The wikipedia page links to a paper that uses GH convergence to prove a stability result in cosmology. - -What are more examples? Ideally they would be along the lines of Gromov's theorem, or proofs of geometric facts, but I'm interested to hear about anything. -As a subquestion, are there interesting applications of Gromov's compactness theorem to prove results about manifolds with bounded Ricci which have nothing to do with GH convergence? - -REPLY [5 votes]: Another graph-theoretic application. -Given an undirected finite graph $G$, the Colin de Verdière graph invariant $\mu(G)$ is defined as the maximum multiplicity of the second smallest eigenvalue of a matrix $M\in O_G$, provided that this eigenvalue is structurally stable. Here, $O_G$ is an interesting (for $G$) subset of the symmetric matrices called the Schrödinger-like operators on $G$. -It turns out that this invariant is very nice in the following sense: if $H$ is a minor of $G$, then the invariant verifies $\mu(H) \leq\mu(G)$. A trick used to prove this result is to put weights on the edges of $G$ to turn it into a metric space (with the obvious distance of smallest sum of weight in a path). Then, any minor $H$ can be obtained as a Gromov-Hausdorff limit of $G$ with suitable weights (contracted edges see their weight go to zero and deleted edges have their weight go to infinity), and conversely, any sequence of weights on the edges of $G$ that converges gives a minor. -Now that I wrote it down, it does not seem like such a powerful remark, but, as I recall, this shift in point of view is (at the very least) very convenient to prove the result.<|endoftext|> -TITLE: Verlagerung made "explicit" -QUESTION [5 upvotes]: Suppose given a finite extension $L/K$ of number fields. I would like to develop a better intuition for the Verlagerung giving an embedding $Ver : Gal(K^{ab}/K) \to Gal(L^{ab}/L)$. -For example, let $L'/L$ be a finite abelian (Galois) extension. Define $\phi:Gal(K^{ab}/K)\to Gal(L'/L)$ to be the composition of the Verlagerung $Gal(K^{ab}/K) \to Gal(L^{ab}/L)$ followed by the projection $Gal(L^{ab}/L) \to Gal(L'/L)$. Then there is a finite, abelian (Galois) extension $K' / K$ such that $Gal(K'/K) \cong Gal(K^{ab}/K)/Ker\phi$. -Is it true that $K' \subset L'$ or even $K' = K^{ab}\cap L'$? (we think of every field being embedded into $L^{ab}$) -Further I would be happy about literature dealing with the Verlagerung in this context. -Thank you very much in advance! -Summary: -As shown below, in the answer of GH plus comments, it is not alway true that $K' \subset L'$. But, if we restrict to strict ray class fields $L^{\mathfrak m}$ of $L$, for $\mathfrak m$ an ideal of $L$, one can show that for every $\mathfrak m$ there is an ideal $\widetilde {\mathfrak m}$ of $L$ with $\mathfrak m | \widetilde {\mathfrak m}$ such that $ K^{\widetilde{\mathfrak m}}$ $\subset L^{\widetilde{\mathfrak m}}$. -Moreover, it seems that it is generally not true (except for trivial cases) that $K^{\widetilde{\mathfrak m}} = L^{\widetilde{\mathfrak m}} \cap K^{ab}$. -But, nevertheless, the $K^{\mathfrak m}$ exhaust the maximal abelian extension $K^{ab}$ of $K$, i.e. $$\bigcup_{\mathfrak m}K^{\mathfrak m} = \ \ \bigcup_{\mathfrak m}K^{\widetilde{\mathfrak m}} \ = \ K^{ab}$$ - -REPLY [5 votes]: I am no expert, but let me share an idea. For a number field $M$ let us denote by $C_M$ the idele class group of $M$. By class field theory, $K' \subset L'$ is the same as $N_{L'/K}(C_{L'})\subset N_{K'/K}(C_{K'})$, where $N$ stands for the norm map. Let us denote by $U$ the open subgroup $N_{L'/L}(C_{L'})$ of $C_L$, and by $j$ the natural injection of $C_K$ into $C_L$. Then, by the transfer theorem, the previous relation can be rewritten as $N_{L/K}(U)\subset j^{-1}(U)$, that is, $j(N_{L/K}(U))\subset U$. For certain open subgroups $U$ of $C_L$ this relation follows from ramification theory, and further local analysis might extend it to more general open subgroups. -EDIT: I think now that the conclusion $K' \subset L'$ fails when $L$ has two places $w$ and $w'$ above the same place $v$ of $K$ such that $L'/L$ is much more ramified at $w$ than at $w'$. Indeed, viewing $U$ as a subgroup of the ideles $J_L$, this assumption implies that $U_w:=U\cap L_w^\times$ is much smaller than $U_{w'}:=U\cap L_{w'}^\times$. However, $j(N_{L/K}(U))\subset U$ would imply that $N_{L_{w'}/K_v}(U_{w'})\subset U_w$ which is false when $U_w$ is sufficiently small in terms of $U_{w'}$. Perhaps this argument can be rewritten in terms of the transfer map and ramification theory only, i.e. avoiding class field theory as a whole.<|endoftext|> -TITLE: Probability of return vs. probability of return in minimal number of steps -QUESTION [9 upvotes]: Consider a random walk on a connected graph $G=(V,E)$. That is, associate to each neighbouring nodes $a,b\in V\ $ transition probabilities $\mathbb{P}(a\rightarrow b), \mathbb{P}(b\rightarrow a) $ that define a Marcov process with the state space $V$. Fix one point $\mathcal{O}\in V$ and a set $D\subset V$. Define $\mathbb{P}(x)$ as the probability that a random walker that was initially in $x\in V$ will reach $\mathcal{O}$ before reaching any points from $D$ (for simplicity assume that it is possible to reach $\\mathcal{O}$ from any point on a graph without passing trough $D$) . Let $d(.,.)$ denotes the "minimal path" metric on $V\setminus D$. Let $\mathbb{P}_{d}(x)$ denotes the probability that a walker starting from $x$ will reach $\mathcal{O}$ precisely in $d(x,\mathcal{O})$ steps. My question is the following: - -Take two points $x,y\in V\setminus B$ such that $d(x,\mathcal{O})=d(y,\mathcal{O})$ and $\mathbb{P}_d(x) > \mathbb{P}_d(y)$. Under what conditions we can say that $\mathbb{P}(x)>\mathbb{P}(y)$? - -Remark 1: It is easy to construct a counterexample for above statement when one considers a random walk on part of $\mathbb{Z}$ (in this situation $\mathcal{O}$ splits $V$ into two disjoint sets). Therefore one should certainly exclude the possibility that $\mathcal{O}$ splits $V$ into two independent parts. -Remark 2: This question is somewhat related to the one I previously asked. I give there a non trivial (I think) example in which above property holds. -Remark 3: Note that if there are some non trivial assumptions under which above assertion holds, this would give quite a lot information about $\mathbb{P}$ for relatively low cost. In particular, as there is a correspondence between "Kirchoff laws" and Marcov processes on graphs, it would give some information about discrete harmonic functions on network of interest. -Remark 4: Obviously, instead of the one "source point" $\mathcal{O}$, one can take some more general subset of $V$. - -REPLY [3 votes]: Your question is slightly ambiguous. -I think that there is almost no relation between $P(x), P(y)$ and $P_d(x),P_d(y)$. -Here is the reason: - You can take any graph $G(V,E)$ and add two edges to it edge from $x$ to $O$ and edge from $y$ to $O$ such that $Pr(x\mapsto O)=\varepsilon_1$ and $Pr(y\mapsto O)=\varepsilon_2$. -If $\varepsilon_i$ are small enough then it will not affect $P(x), P(y)$, but in this case $P_d(x)=\varepsilon_1$ and $P_d(y)=\varepsilon_2$. Therefore $P(x), P(y)$ are almost unrelated to $P_d(x),P_d(y)$. -Hope that this is helps.<|endoftext|> -TITLE: Proof of L^p Elliptic Regularity -QUESTION [8 upvotes]: Let $L = \sum_{i,j=1}^n -\frac{\partial}{\partial x^i} (a^{ij}(x)\frac{\partial}{\partial x^j}) + \sum_{i=1}^n b^i(x) \frac{\partial}{\partial x^i} + c(x)$ be a second order elliptic operator with smooth coefficients, $\Omega$ a bounded open domain with smooth boundary in $\mathbb{R}^n$, and $f$ be a function in $L^p(\Omega)$. We say that $u \in W_0^{1,p}(\Omega)$ (one weak derivative in $L_p$ and vanishing boundary values) is a weak solution of $Lu = f$ if for all $g \in W_0^{1,q}(\Omega)$ $(q = p^*)$ we have -$\int_{\Omega} \sum_{i,j=1}^n a^{ij}(x)\frac{\partial u}{\partial x^i}\frac{\partial g}{\partial x^j} + \sum_{i=1}^n b^i(x)\frac{\partial u}{\partial x^i}g(x) + c(x)u(x)g(x) = f(x)$. -The standard result is of course that all such weak solutions $u$ actually belong to $W_0^{2,p}(\Omega)$. -I am trying to complete the following proof of this statement: -(1) First we establish an a priori estimate for strong solutions $v \in W_0^{2,p}(\Omega)$ of $Lv = f$: -$\vert\vert v\vert \vert_{W_0^{2,p}(\Omega)} \leq C(\vert\vert f\vert\vert_{L^p(\Omega)} + \vert\vert v\vert\vert_{L^p(\Omega)})$ -This is non-trivial but can be established by proving the relevant estimate for the Laplacian with a Newton Potential argument and then using the freezing coefficients technique. -(2) Next we observe that if $L$ is injective on $W_0^{1,p}$, then we are done. This is because $L$ injective implies -$\vert\vert v\vert\vert_{L^p(\Omega)} \leq C\vert\vert Lv\vert\vert_{L^p(\Omega)}$ -One proves this by assuming it was false and then using Rellich compactness to produce a non-zero solution to $Lv = 0$. -Having established this estimate, we consider the smooth mollifications $f_{\epsilon}$ of $f$. By $L^2$ theory we can find smooth $v_{\epsilon}$ strong solutions of $Lv_{\epsilon} = f_{\epsilon}$. Since we have -$\vert\vert v_{\epsilon} - v_{\epsilon'}\vert\vert_{W_0^{2,p}} \leq C\vert\vert f_{\epsilon}-f_{\epsilon'}\vert\vert_{L^p(\Omega)}$ -The $v_{\epsilon}$ converge to some $v \in W_0^{2,p}(\Omega)$ which will solve $Lv = f$ strongly. Since strong solutions are clearly weak solutions, by the injectivity of $L$ on $W_0^{1,p}(\Omega)$ we conclude that $u = v \in W_0^{2,p}(\Omega)$ and we are done. -(3) We have no way to guarantee that $L$ is injective, for example $0$ might be an $L^2$ eigenvalue of $L$. However, if $p=2$ then we could guarantee that $L_{\lambda} = L + \lambda I$ is injective for some large $\lambda$. If we could establish this fact in the general case we would be done since $L_{\lambda}u = f + \lambda u \in L^p$ and $L_{\lambda}$ injective imply that (2) is applicable. -Question: What is the simplest way to prove that $L_{\lambda} = L + \lambda I$ is injective on $W_0^{1,p}(\Omega)$ for large $\lambda$? Do there exist weak $L^P$ maximum principles? -Of course, I would prefer that the proof not use $L^p$ regularity. - -REPLY [5 votes]: I stumbled across the book Second Oder Elliptic Equations and Elliptic Systems by Yah-Ze Chen which appears to contain an answer to my question. It is available on google books here -http://books.google.com/books?id=eQcbiPQPweQC&pg=PA49&dq=strong+solution+dirichlet+problem+Lp&hl=en&ei=byKiTeXXO6GG0QGG7dGgBQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCcQ6AEwADgK#v=onepage&q=strong%20solution%20dirichlet%20problem%20Lp&f=false -To save time for those who are interested, here is the relevant argument: -For large $\lambda > 0$ we want to show that $L_{\lambda} = L - \lambda I$ is injective on $W_0^{1,p}$. -Claim: Let $L^T_{\lambda}$ be the transpose of $L^{\lambda}$ with respect to the paring that defines weak solutions. Then we claim that $L^T_{\lambda}$ inective on $W_0^{2,p}$ implies that $L_{\lambda}$ is injective on $W_0^{1,p}$ -Proof: Suppose that $L^T_{\lambda}$ is injective on $W_0^{2,p}$. Then, by an argument contained in the original post above, for every $f \in L^p(\Omega)$ we can find $u \in W_0^{2,p}(\Omega)$ such that $L^T_{\lambda}u = f$. Now, suppose that $L_{\lambda}v = 0$ for some $v \in W_0^{1,p}$. After an integration by parts and the definition of weak solution, we see that $\varphi \in W_0^{2,q}$ implies that -$\int_{\Omega}uL^T_{\lambda}\varphi = 0$. -Now choose $\Omega'' \subset\subset \Omega' \subset\subset \Omega$ and a bump function $\rho$ identically one in $\Omega''$ with support in $\Omega'$. $\rho\text{sgn}(u)$ is in $L^q$, and we can find $g \in W_0^{2,q} $ such that $L^T_{\lambda}g = \rho\text{sgn}(u)$. Plugging this $g$ into the above equality gives -$\int_{\Omega''}|u| = -\int_{\Omega\setminus\Omega''}\rho |u|$ -Due to the arbitrariness of $\Omega''$, this implies that $\int_{\Omega} |u| = 0$ and hence $u$ is $0$ a.e. -Claim: For $\lambda$ large enough, $L_{\lambda}$ is injective on $W_0^{2,p}$. -Proof: Suppose $L_{\lambda}u = 0$ for $u \in W_0^{2,p}$. Let $\tilde{\Omega} = \Omega \times (-1,1)$, and $\tilde{\Omega'} = \Omega \times (-1/2,1/2)$. Let $(x,t)$ be the coordinates on $\Omega \times (-1,1)$. Then define $v(x,t) = \cos(\sqrt{\lambda}t)u(x)$. Let $\hat{L_{\lambda}} = L_{\lambda} + \partial_t^2$. We have $\hat{L_{\lambda}}v = 0$. The strong solution estimates give -$\vert\vert v\vert\vert_{W^{2,p}(\tilde{\Omega'})} \leq C\vert\vert v\vert\vert_{L^p(\tilde{\Omega})} \leq C\vert\vert u\vert\vert_{L^p(\Omega)} \Rightarrow $ -$\vert\vert \partial_t^2v\vert\vert_{L^p(\tilde{\Omega'})} \leq C\vert\vert u\vert\vert_{L^p(\Omega)} \Rightarrow $ -$\lambda\vert\vert u\vert\vert_{L^p(\Omega)}(\int_{-1/2}^{1/2}|\cos(\sqrt{\lambda}t|^p)^{1/p} \leq C\vert\vert u \vert\vert_{L^p(\Omega)} \Rightarrow$ -$\lambda^{1 - \frac{1}{2p}}\vert\vert u\vert\vert_{L^p(\Omega)}(\int_{-1/2}^{1/2}|\cos t|^p)^{1/p} \leq C\vert\vert u\vert\vert_{L^p(\Omega)}$ -Now taking $\lambda$ large enough implies that $u = 0$ almost everywhere.<|endoftext|> -TITLE: Hurwitz spaces and moduli spaces of marked elliptic curves -QUESTION [5 upvotes]: I would like to know if the moduli space $\mathcal M_{1,n}$ of genus $1$ curves with $n$ marked points can be realized as a Hurwitz space ? - -REPLY [4 votes]: This might depend on what you count as a "Hurwitz space." To me, any cover of M_{g,n} parametrizing covers branched at the marked points is a Hurwitz space; so I would say, in a tautological tone of voice, that M_{1,n} is a Hurwitz space parametrizing degree-1 covers of elliptic curves branched at the n marked points! -But maybe you really want M_{1,n} to be a moduli space of branched covers of P^1? This seems plausible. It might be a pain to do in practice. I suppose I would try to set it up as a moduli space of covers Y -> P^1 which factor as Y -> E -> P^1, and where Y -> E is branched at the n marked points. But then you'll have to worry about collisions between marked points and Weierstrass points... it sounds like a pain.<|endoftext|> -TITLE: Proof of a fact about traces -QUESTION [8 upvotes]: I'm following the open courseware content on Machine Learning from Stanford University. In the lecture notes, it is given that -$$\Delta_A \ tr(ABA^TC) = CAB + C^TAB^T$$ -which I tried but couldn't prove easily. It is not required to follow the course content but I just wondered and wanted to learn its proof. Any suggestions? -Update: $A$, $B$, and $C$ are matrices and $\Delta_A$ is the gradient operation on matrix $A$. - -REPLY [4 votes]: I'm going to decompose Fabian's answer into something a little more newb friendly. An important concept is the partial derivative of a matrix with respect to one of its elements. An example of this is -$$ -\partial_{\mathbf{A}_{31}} \mathbf{A} -= \partial_{\mathbf{A}_{31}} - \begin{bmatrix} - \mathbf{A}_{11} & \mathbf{A}_{12} \\ - \mathbf{A}_{21} & \mathbf{A}_{22} \\ - \mathbf{A}_{31} & \mathbf{A}_{32} - \end{bmatrix} -= \begin{bmatrix} - 0 & 0 \\ - 0 & 0 \\ - 1 & 0 - \end{bmatrix}. -$$ -In general, if we take the derivative with respect to the $(i,j)$ entry, then the $(m,n)$ entry of the resulting matrix is -$$ -\partial_{A_{ij}} A_{mn} = \delta_{im} \delta_{jn} -$$ -where $\delta$ is the Kronecker delta. -This is simply the basic statement of multivariate calculus: -namely $\partial_x x = 1$ and $\partial_x y = 0$. -In particular, -\begin{align*} -\sum_{mn} \partial_{A_{ij}} A_{mn} = \partial_{A_{ij}} A_{ij} = 1. -\qquad \text{(1)} -\end{align*} -To begin the proof, we consider matrix equations, which are -\begin{align*} -\partial_{\mathbf{A}} \text{tr}(\mathbf{A} \mathbf{B} \mathbf{A}^\text{T} \mathbf{C}) -&= \partial_{\mathbf{A}} \sum_{m} \left(\mathbf{A} \mathbf{B} \mathbf{A}^\text{T} \mathbf{C}\right)_{mm} -\qquad \text{(Trace[2])} \\ -&= \partial_{\mathbf{A}} \sum_{m} \Bigg(\sum_{n k \ell} A_{mn} B_{nk} \left(A^T\right)_{k \ell} C_{\ell m}\Bigg)_{mm} -\qquad \text{ (Matrix multiplication [3])} \\ -&= \partial_{\mathbf{A}} \sum_{m n k \ell} A_{mn} B_{nk} A_{ \ell k} C_{\ell m} -\qquad \text{ (Transpose [4])} \\ -&= \mathbf{C}^T \mathbf{A B}^T + \mathbf{C A B}, -\end{align*} -where we justify the last step component-wise. -For all elements $(i,j)$, it follows that -\begin{align*} -&\Bigg( \partial_{\mathbf{A}} \sum_{m n k \ell} A_{mn} B_{nk} A_{ \ell k} C_{\ell m} \Bigg)_{ij} \\ -&= \partial_{\mathbf{A}_{ij}} \sum_{m n k \ell} A_{mn} B_{nk} A_{ \ell k} C_{\ell m} -\qquad \text{(Scalar-by-matrix derivative[5])} \\ -&= \sum_{mnk \ell} \partial_{A_{ij}} A_{mn} B_{nk} A_{ \ell k} C_{\ell m} -\qquad \text{ (Linearity of differentiation [6])} \\ -&= \sum_{mnk \ell} (\partial_{A_{ij}} A_{mn}) (B_{nk} A_{\ell k} C_{\ell m}) + (A_{mn}) (\partial_{A_{ij}} B_{nk} A_{\ell k} C_{\ell m}) -\qquad \text{(Product Rule[7])} \\ -&= \sum_{mnk \ell} (\partial_{A_{ij}} A_{mn}) B_{nk} A_{\ell k} C_{\ell m} - + \sum_{mnk \ell} A_{mn} (\partial_{A_{ij}} B_{nk} A_{\ell k} C_{\ell m}) \\ -&= \sum_{k \ell} B_{jk} A_{\ell k} C_{\ell i} - + \sum_{mnk \ell} A_{mn} (\partial_{A_{ij}} B_{nk} A_{\ell k} C_{\ell m}) -\qquad \text{ (Eqn. 1)} \\ -&= \sum_{k \ell} B_{jk} A_{\ell k} C_{\ell i} - + \sum_{mnk \ell} A_{mn} \Big[(\partial_{A_{ij}} A_{\ell k}) (B_{nk} C_{\ell m}) + (A_{\ell k}) (\partial_{A_{ij}} B_{nk} C_{\ell m})\Big] -\text{ (Product R.)} \\ -&= \sum_{k \ell} B_{jk} A_{\ell k} C_{\ell i} - + \sum_{mnk \ell} A_{mn} (\partial_{A_{ij}} A_{\ell k}) B_{nk} C_{\ell m} \\ -&= \sum_{k \ell} B_{jk} A_{\ell k} C_{\ell i} - + \sum_{mn} A_{mn} B_{nj} C_{i m} -\qquad \text{ (Eqn. 1)} \\ -&= \sum_{k \ell} C_{\ell i} A_{\ell k} B_{jk} - + \sum_{mn} C_{i m} A_{mn} B_{nj} \\ -&= \sum_{k \ell} \left(C^T\right)_{i \ell} A_{\ell k} \left(B^T\right)_{kj} - + \sum_{mn} C_{im} A_{mn} B_{nj} -\qquad \text{(Transpose)} \\ -&= \left( \mathbf{C}^T \mathbf{A} \mathbf{B}^T + \mathbf{C} \mathbf{A} \mathbf{B} \right)_{ij}, -\qquad \text{(Matrix multiplication)} -\end{align*} -which completes the proof. - -[2]: https://en.wikipedia.org/wiki/Trace_(linear_algebra) -[3]: https://en.wikipedia.org/wiki/Matrix_multiplication#General_definition_of_the_matrix_product -[4]: https://en.wikipedia.org/wiki/Transpose -[5]: https://en.wikipedia.org/wiki/Matrix_calculus#Scalar-by-matrix -[6]: https://en.wikipedia.org/wiki/Linearity_of_differentiation -[7]: https://en.wikipedia.org/wiki/Product_rule<|endoftext|> -TITLE: What are some geometric reasons why a Dirac operator would have a gap in its spectrum? -QUESTION [23 upvotes]: My question is motivated by the following well-known computation. Let $M$ be an even dimensional Riemannian spin manifold and let $D$ be the spinor Dirac operator on $M$. Lichnerowicz showed that $D^2 = \nabla^* \nabla + \kappa/4$ where $\nabla$ is the spin connection on the spinor bundle and $\kappa$ is the scalar curvature of $M$. It is not hard to show that $\nabla^* \nabla$ is a positive operator, and thus if $\kappa > 0$ there is an interval containing $0$ in the real line which avoids the spectrum of $D^2$ (and therefore $D$). A corollary is the well-known fact that the Fredholm index of $D$ vanishes if $M$ has positive scalar curvature. -This example is a starting point for the entire theory of positive scalar curvature obstructions. The machinery involved gets more sophisticated, but in the end all one really needs about positive scalar curvature metrics on spin manifolds is the fact that they create a gap around 0 in the spectrum of the spinor Dirac operator. -So I am wondering if there are other geometric causes for gaps in the spectrum of a Dirac operator. Note that I do not want to restrict my attention to the spinor Dirac operator; I think the question is particularly interesting for the signature operator or the Dolbeault operator, for example. I am aware that elliptic operator theory of this sort helps produce a proof of the Weyl character formula, so conceivably the sort of answer that I'm looking for could come from representation theory. -Any ideas? - -REPLY [4 votes]: The area in dimension $2$ for the sphere (with abitrary metric): In C. Bär: Lower eigenvalue estimates for Dirac operators (Math. Ann. 293) it was shown that -$$\lambda^2 area\geq 4\pi.$$ -This can be generalized to higher genus surfaces if one restricts to eigenvalues with higher multiplicity.<|endoftext|> -TITLE: Conformal blocks vector bundles on $\overline{M}_{g}$ in terms of generalized theta functions? -QUESTION [27 upvotes]: Conformal field theory uses representation theory to produce various vector bundles on the Deligne-Mumford compactified moduli spaces $\overline{M}_{g}$ and $\overline{M}_{g,n}$, known as bundles of conformal blocks or covacua. Several authors (Beauville, Laszlo, Sorger, Pauly, Faltings, Kumar, Narasimhan, Ramanathan) proved in the late 80s and early 90s that the fiber over a point in the interior (i.e., a smooth curve) is naturally identified with a certain space of generalized theta functions. At the time this was proven not much had been said about moduli of vector bundles over singular curves, so I wonder if any of the more recent treatments of moduli of G-bundles allow for an extension of this theorem to the Deligne-Mumford boundary. -For instance, the conformal blocks bundle on $\overline{M}_g$ corresponding to $\mathfrak{sl}_r$, level $l$, has fiber over $C\in M_g$ equal to $H^0(SU_C(r),\mathcal{L}^l)$, where $SU_C(r)$ is the moduli space of semistable rank $r$ vector bundles with trivial determinant on $C$ and $\mathcal{L}$ is the determinant line bundle. Is there a similar interpretation for fibers over $C\in\overline{M}_g\setminus M_g$? - -REPLY [3 votes]: Under a certain natural compatibility assumption, there are counterexamples due to Belkale-Gibney-Kazanova. -The assumption essentially says that at a point $[\mathrm{C}]$ of the moduli space $\overline{\mathscr{M}}_{g,n}$, an algebra obtained from the fiber at $[\mathrm{C}]$ of the conformal blocks bundle is isomorphic to the algebra of sections of a line bundle $\mathscr{L}$ over some variety. (This holds whenever $[\mathrm{C}]$ is in $\mathscr{M}_{g,n}$ if one takes $\mathscr{L}$ to be the theta line bundle over some moduli space of sheaves on $\mathrm{C}$, all determined by the input data (level, type of Lie algebra, etc.) defining the conformal blocks.) -Consider the classical case $\overline{\mathscr{M}}_2$, $\mathfrak{sl}_2$, $\ell=1$. In case $\mathrm{C}$ is smooth, the moduli space of sheaves on $\mathrm{C}$ with trivial determinant is nothing but $\mathbf{P}^3$, and the theta line bundle is $\mathscr{O}(1)$ (theorem of Narasimhan and Ramanan). Belkale-Gibney-Kazanova show (example 4.2 in their paper) that if one takes $\mathrm{C}$ to be a curve with a separating node, one arrives at a counterexample by considering the first Chern class of the bundle of conformal blocks. -A formula for the Chern character of the bundle of conformal blocks can be found in a paper of Marian-Oprea-Pandharipande-Pixton-Zvonkine.<|endoftext|> -TITLE: paper rejected because not so general -QUESTION [7 upvotes]: Dear All, -thanks in advance to anyone that could give some suggestion. -Here's the situation: starting from a type of random graph I extend -the construction introducing a new class of random graphs. I studied -some properties and I write a paper that I submitted to the arxiv.org -successfully after the endorsment of one of the author cited in the -references. I have a math degree but I don't have any formal -affiliation to my university, so I submitted the paper as a private -without affiliation. -After some minor review I consider submitting it to a math journal. I -submitted it and after one month my paper was refused because of the -following reviewer motivation: -"The paper considers random graphs where roughly speaking one starts -with a precise graph and adds new vertices into random cycles. -Clearly, the graphs that are obtained have a very special form. The -presented results are neither interesting nor significant enough for -publication in our journal." -So it seems to me that even if the paper was correct it's not as -general results as required. -So here are my questions: -Considering that the paper seems ok, I mean no first read bad mistakes, would I submit it to another journal and see if is accepted? -In case I would try to generalize the process and would I have to -submit again to the same journal? -If I don't have any affiliation this could be somehow bad at the eyes -of a reviewer? -Thanks to everyone that could help. -Paolo - -REPLY [3 votes]: I think that you need a second (and perhaps third) opinion from a professional. If possible, write some individual emails requesting people to give a quick impression as to the publication-worthiness of your result. Since the result is on ArXiv, no question of precedence should arise. You can ask the author you asked before for names of other people to ask. It is important that you emphasize that you don't need them to spend time going over the paper (if all you need is a quick impression; for a more thorough review, you will want a different strategy of approach). -Please note: MathOverflow is a place for specific questions. If you have trouble with a particular proof, you can ask about that detail. Zev Chonoles is right in commenting that MathOverflow is not a place to solicit reviewers for your work. Pablo Shmerkin is (somewhat) right in the idea that it is smart to include a link to your ArXiv submission, in case someone is interested and volunteers to review it. (In short: Asking for review on MathOverflow, bad; making it possible using less than 100 characters without asking, not so bad, and possibly priceless.) -Gerhard "Will Rewrite Commercials For Barter" Paseman, 2011.02.27<|endoftext|> -TITLE: Explicit ring of differential operators for polynomial algebras over the integers? -QUESTION [14 upvotes]: Does anyone know of a reference or have any idea for an explicit description of the ring of differential operators for polynomial algebras over the integers? I'm hoping there is something analogous to the case of a polynomial ring over a field $K$, where if the field has characteristic $0$ an explicit description is given by the Weyl algebra. In particular, for $K[x_1,\ldots,x_n]$ this is the noncommutative $K$-algebra spanned by the symbols $x_1,\ldots,x_n$ and $d_1,\ldots,d_n$ where the $x_i$'s commute, the $d_i$'s commute and one has the Leibniz rule: $d_i x_j - x_j d_i = \delta_{i,j}$ where $\delta_{i,j}$ is the Kronecker delta. -If the field has characteristic $p > 0$, then one also has divided powers $\frac{1}{p!}(\frac{d^p}{dx_i^p})$, (Hasse derivatives or hyperdifferential operators) which are not in the Weyl algebra but can be constructed using Frobenius. Of course, for $\mathbf{Z}[x_1,\ldots,x_n]$ the divided powers $\frac{1}{t!}(\frac{d^t}{dx_i^t})$ for integers $t > 1$ are not in the Weyl algebra. Is the full ring of differential operators spanned by the Weyl algebra and these divided powers? Is there a finite generating set? - -REPLY [16 votes]: The answer to your first question is "yes". You can find a calculation of the full ring of differential operators on a suitably nice scheme here : Theoreme 16.11.2 on page 54 of EGA 4 IV, PIHES 32 (1967). Generally speaking, EGA 4 IV.16 is the original and definitive reference for differential operators on arbitrary schemes. -It follows from this result that the answer to your second question is also "yes". If $A$ is the polynomial algebra $K[x_1,\ldots, x_n]$ for any commutative ring $K$ and $m \geq 0$ is an integer, then the $A$-module $\mathcal{Diff}^m_{A/K}$ of all $K$-linear differential operators of order at most $m$ is free with basis precisely the divided powers $\frac{1}{\alpha !} \frac{\partial^\alpha}{\partial x^\alpha}$ as $\alpha$ varies over all multi-indices in $\mathbb{N}^n$ with $|\alpha| = \alpha_1 + \cdots + \alpha_n \leq m$. -The answer to your third question is "no". If there was a finite generating set for the algebra of full differential operators on $\mathbb{Z}[x_1,\ldots, x_n]$, then by passing to a field $k$ of positive characteristic $p$, we would obtain a finite generating set for $\mathcal{Diff}_{k[x_1,\ldots,x_n]/k}$. But it is known that any differential operator on $A = k[x_1,\ldots, x_n]$ is $A_r = k[x_1^{p^r}, \ldots, x_n^{p^r}]$-linear for some $r$ --- see the reference below. So if there was a finite generating set, then the whole differential operator algebra would be contained in the matrix algebra $End_{A_r}(A)$ for some fixed $r$; this can be seen not to be the case by considering a divided power of order $\geq p^r$. -A good reference for the computation of differential operators on $k[x_1,\ldots, x_n]$ is the following paper by S.P. Smith: "The global homological dimension of the ring of differential operators on a nonsingular variety over a field of positive characteristic", J. Algebra 107(1) 1987, 98---105. It should be possible to access it through www.sciencedirect.com . -I should perhaps mention at this point that Berthelot has developed a theory of so-called "arithmetic differential operators", which can be viewed as a refinement of Grothendieck's classical theory. You can find an introduction here.<|endoftext|> -TITLE: Is there a non-quotient stack with affine stabilizers whose good moduli space is a geometric point? -QUESTION [12 upvotes]: Definitions: One says that a map $\pi\colon\mathcal X\to X$ from an algebraic stack to an algebraic space is a good moduli space if $\pi$ is cohomologically affine and universal for maps to schemes. (A good (but not equivalent) way to think about a stack whose good moduli space is a single point is that for any two points, the closure of their orbits intersect.) -One says that a stack $\mathcal X$ is a quotient stack if it is the quotient of an algebraic space by a subgroup of GL_n (see for instance EHKV, which also gives a criterion for a stack to be a quotient stack in terms of vector bundles). -One says that a stack has the resolution property if every coherent sheaf is a quotient of some vector bundle; -Totaro's paper The resolution property for schemes and stacks relates the property of being a quotient stack to the resolution property. - -Question: Let $\mathcal X$ be a stack with a good moduli space $\mathcal X \to X$ such that X is a geometric point (i.e., X = Spec k, where k is a separably closed field). Suppose further that the stabilizers are affine linearly reductive groups. Is $\mathcal X$ a quotient stack? - -(See this answer for the definition of stabilizer of a point of a stack that isn't a quotient stack.) -Remarks: - -The condition on stabilizers excludes things like BE with E an elliptic curve. -The condition that k is separably closed excludes non-trivial gerbes. -I'd be just as happy with an answer to "Does the resolution property hold for $\mathcal X$ ?". - -REPLY [3 votes]: This is answered in general by Theorem 13.1 in our paper The étale local structure of algebraic stacks (arXiv:1912.06162). Let $\mathcal{X}$ be a stack with a good moduli space $X$ such that - -$\mathcal{X}$ has affine stabilizers, -$\mathcal{X}$ has separated diagonal, and -$\mathcal{X}$ is of finite presentation over an algebraic space. - -Then $\mathcal{X}$ has the resolution property étale-locally (and even Nisnevich-locally) on $X$. In particular, if $X$ is the spectrum of a field or a henselian local ring, then $\mathcal{X}$ has the resolution property. -Conditions 1 and 2 are necessary. Note that having the resolution property locally on $X$ implies that $\mathcal{X}$ has affine diagonal. -In general $\mathcal{X}$ does not have the resolution property Zariski-locally on $X$. There is an example in SGA 3, Exp X, §1.6 (cf. Remark 2.5 in our paper), of a 2-dimensional torus $G$ over the nodal cubic curve $C$ such that $G$ is not locally isotrivial. This means that $G$ cannot be embedded in $\mathrm{GL}_N$ for any $N$, not even Zariski-locally on $C$. It follows that $\mathcal{X}=BG$ does not have the resolution property Zariski-locally on its good moduli space $X=C$.<|endoftext|> -TITLE: The etale site of a closed subscheme and its etale Grothendieck subtopology -QUESTION [10 upvotes]: There is a very basic theorem for the Zariski topology. -Let X = Spec(R) and Y=Spec(R/I) for I some reduced ideal. Y obtains a topology two ways, one is the subspace topology as a subset of X and another as the spectrum of a ring. These topologies are the same by the correspondence between ideals of R containing I and ideals in R/I. -Is there a close statement to this in the etale toplogy? There are two natural ways to understand open sets on Y, those which come from etale neighborhoods of X base changed to Y and those which are etale neighborhoods of Y. -I did a computation today in a very special case and it seems that both of these topologies seem to be 'the same'. -Does anyone know if this statement is true in a general context and where I might locate this resource? -Thanks. - -REPLY [13 votes]: Here is one way of addressing your question. Let $X$ be a scheme, $Y$ a closed subscheme of -$X$, and $U$ its open complement. Let $i: Y \rightarrow X$ and $j: U \rightarrow X$ denote the inclusion maps. Then the pushforward functor $i_{\ast}$ determines a fully faithful embedding from the category of etale sheaves on $Y$ to the category of etale sheaves on $X$. The essential image of this embedding is the full subcategory spanned by those etale sheaves $\mathcal{F}$ on $X$ such that -$j^{\ast} \mathcal{F}$ is final (in other words, such that $\mathcal{F}(V)$ has a single element for every etale map $V \rightarrow X$ which factors through $U$). -In topos-theoretic language, this says that $i$ induces a closed immersion of etale topoi, which is complementary to the open immersion of etale topoi determined by $j$. (The above discussion makes sense in an arbitrary topos, and for a topos of sheaves on a topological space it recovers the usual notion of closed embedding). -I don't know a reference for the above statement offhand, but my guess would be that you can find a discussion in SGA somewhere. - -REPLY [8 votes]: The statement is at least true Zariski locally. That is, given an étale map $V\to Y$, there exists a Zariski open cover $X=\bigcup X_i$ so that the pullback of $V$ to $Y\cap X_i$ is the restriction of an étale neighborhood of $X_i$. -To see this, use the structure theorem for étale morphisms: Theorem 34.11.3 in the chapter on étale morphisms in the Stacks Project. - -It says that any étale morphism to $Y=Spec(R/I)$ is Zariski locally† an open subscheme $V$ of $Spec((R/I)[t]_{\bar f'}/(\bar f))$, where $\bar f\in (R/I)[t]$ is monic. Let $f\in R[t]$ be an arbitrary monic lift of $\bar f$. Then since the Zariski topoplogy of $Spec((R/I)[t]_{\bar f'}/(\bar f))$ is the restriction of the Zariski topology of $Spec(R[t]_{f'}/(f))$, there is some open subscheme $U$ of $Spec(R[t]_{f'}/(f))$ so that $V = U\cap Spec((R/I)[t]_{\bar f'}/(\bar f))$. This $U$ is étale over $X$ and pulls back to $V$. - -† I'm implicitly replacing $X=Spec(R)$ and $Y=Spec(R/I)$ by localizations $Spec(R_g)$ and $Spec(R_g/I\cdot R_g)$ in the rest of the argument. The open cover $X=\bigcup X_i$ consists of these $Spec(R_g)$'s and the complement of $Y$.<|endoftext|> -TITLE: Does there exist a family of curves (or abelian varieties) on the punctured line with specified monodromy on H^1? -QUESTION [8 upvotes]: Suppose I have a finite set of points in $\mathbb{P}^1$ (over the complex numbers), and suppose that at each point, I am given a [Edit: quasi-unipotent] conjugacy class in $Sp(2g,\mathbb{Z})$ for $g$ a fixed positive integer. Then near each point, I have an analytic neighborhood where I can construct a family of complex tori whose $H^1$ varies according to the monodromy. Furthermore, if there are representatives of the conjugacy classes whose product is $1 \in Sp(2g, \mathbb{Z})$, then I can at least glue these local families into a $C^\infty$ family of $2g$-dimensional tori over the punctured $\mathbb{P}^1$. -First question: If we have representatives whose product is identity, does there exist a family of abelian varieties over the punctured line whose $H^1$ has the specified monodromy at the points? I think we can do this by choosing a variation of Hodge structure on the corresponding local system of rank $2g$ groups, and taking a quotient, but it's all a bit cloudy to me. -Second question: Given a particular puncture and its assigned conjugacy class in $Sp(2g,\mathbb{Z})$, does there exist a family of genus $g$ curves over a small neighborhood of the puncture whose $H^1$ has monodromy in the specified conjugacy class? (I suppose I should just ask this about the punctured affine line.) In the $C^\infty$ world, this can be done with a mapping torus, but I don't know how it works holomorphically. -Third question: If the answers to the first two questions are "yes", can we make our family of abelian varieties out of Jacobians of genus $g$ curves? Up to some finite cover problem, I think this is basically asking if there exists a family of genus $g$ curves whose $H^1$ has the specified monodromy. -From David Brown's answer here, it seems too much to ask for an explicit minimal model curve over the punctures when $g > 2$, but I'm just hoping for existence away from those points. - -REPLY [4 votes]: I agree with Donu. Indeed, I think even the much weaker question of whether a mod-p representation of the fundamental group of the base on Sp(2g,Z/pZ) occurs as a monodromy representation might typically have a negative answer. Given such a representation rho, you get a fibration X_rho -> P^1, whose fibers are isomorphic to the moduli space of abelian g-folds with full p-level structure; this will be general type for p large. Any abelian g-fold A/C(t) with monodromy rho corresponds to a section from P^1 back to X_rho, and I don't see why there would be such a section in general. -Oh yeah, and; the answer to your second question is yes, I think. When the monodromy is of the form -I M -0 I -with M of full rank; you can construct an abelian variety over C((t)) with totally multiplicative reduction which has any desired monodromy, as in Mumford's paper "Degenerating abelian varieties...." -If M has smaller rank maybe you can just use a product of a constant a.v. with a totally multiplicative one of dimension rank(M)?<|endoftext|> -TITLE: Rigidity of the category of schemes -QUESTION [72 upvotes]: Call a category $C$ rigid if every equivalence $C \to C$ is isomorphic to the identity. I don't know if this is standard terminology. Many of the usual algebraic categories are rigid, for example sets, commutative monoids, groups, abelian groups, commutative rings, but also the category of topological spaces. The category of monoids (or rings) is not rigid because $M \mapsto M^{\mathrm{op}}$ is an equivalence which is not isomorphic to the identity. See [M] for a survey and the general strategy for proving rigidity. The case of commutative rings was discussed recently on MO here. The philosophy is that a category is rigid if every object can be defined in a categorical way, which is a quite interesting property. -Question: Is the category of schemes rigid? -Here is what I've done so far: The initial scheme is $\emptyset$ and the terminal scheme is $\text{Spec}(\mathbb{Z})$. Spectra of fields are characterized by the property that they are non-initial and and every morphism from a non-initial object to them is an epimorphism, see Kevin's answer here. The underlying set $|X|$ of a scheme is the set of equivalence classes of morphisms $Y \to X$, where $Y$ is the spectrum of a field. So this recovers $|X|$ from $X$ in a categorical manner. If $x \in |X|$, then $\text{Spec}(\kappa(x))$ is the terminal spectrum of a field which maps to $X$ and has (set) image $x$. -However, I'm not able to recover the topology from $X$. I don't know how to characterize open or closed immersions. They are exactly the étale resp. proper monomorphisms, see this MO question, but it seems to be hard to characterize étale and proper categorically. After all, if are able to characterize affine schemes, then we will be done, since the category of affine schemes is rigid and every scheme is the canonical colimit of the affine schemes mapping into it. -In order to characterize affine schemes, it is enough to characterize the ring object $\mathbb{A}^1_\mathbb{Z}$ in the category of schemes, since we can then define the ring of global sections of a scheme categorically and then say that affine schemes $Y$ are characterized by the property that for all schemes $X$ the map $Hom(X,Y) \to Hom(\mathcal{O}(Y),\mathcal{O}(X))$ is bijective. -Other approaches: 1. First show that the category of fields is rigid. I've already shown that the notions of prime field, $\mathbb{F}_p$, $\mathbb{Q}$, finite, characteristic, normal, separable, algebraic, galois, transcendent, transcendence degree are categorical, but this is not enough to distinguish, for example, $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$. If $F$ is a self-equivalence of the category of fields, then $F$ maps $K(X)$ to $F(K)(X)$, so taking automorphisms there is a natural isomorphism $\text{PGL}(2,K) \cong \text{PGL}(2,F(K))$, but I wonder if this already implies that $K \cong F(K)$ naturally. 2. Characterize local schemes as a special full reflective subcategory containing the spectra of fields. 3. Try to categorify cohomology theory and use Serre's criterion for affineness. -EDIT (May '11): I've restarted this project in the last days. If $k$ is a field with only trivial endomorphisms, then I can show that every self-equivalence of $\text{Sch}/k$ preserves $\text{Spec}(k[\epsilon]/\epsilon^2)$, but also $\text{Spec}(k[[t]])$. But I still have no idea how to approach $\text{Spec}(k[t])$ categorically. Even basic notions such as "closed point" or "quasicompact" remain unclear. -EDIT (Feb '12): Let's work with $\mathrm{Sch}/k$ for some algebraically closed field $k$. Then $F$ maps $\mathbb{A}^1_k$ to a ring object in $\mathrm{Sch}/k$. If we already knew that it is of finite type over $k$ and irreducible, then a Theorem by Greenberg (Cor. 4.4 in Algebraic Rings, Trans. AMS, Vol. 111, No. 3, pp. 472 - 481) will imply that the underlying scheme is just $\mathbb{A}^n_k$ for some $n$. Now using my question about factorization we should be able to conclude $n=1$. Of course, many details are missing here; for example it is not clear at all why $F$ should preserve schemes of finite type. -Any ideas concerning the categorical characterization of other properties / objects are appreciated. Feel free to add every piece as a single answer even if it does not answer the whole question. -[M] E. Makai jun, Automorphisms and Full Embeddings of Categories in Algebra and Topology, online - -REPLY [10 votes]: As requested by the OP in the comments of the (correct and complete) accepted answer of user131755: it's possible to say more. - -Theorem [Mochizuki 2004, vDdB 2019]. Let $S$ and $S'$ be schemes. Then the natural functor -$$\operatorname{Isom}(S,S') \to \mathbf{Isom}(\mathbf{Sch}_{S'},\mathbf{Sch}_S)$$ -is an equivalence of categories, where $\operatorname{Isom}(S,S')$ is a discrete category and $\mathbf{Isom}$ denotes the category whose objects are equivalences and whose morphisms are natural isomorphisms. - -The version where $\mathbf{Sch}$ denotes the category of locally Noetherian schemes with finite type morphisms is due to Mochizuki [Mochizuki 2004], and the general statement appears in a preprint of myself [vDdB 2019]. -In particular, taking $S = S' = \operatorname{Spec} \mathbf Z$ answers the question, since $\operatorname{Aut}(\operatorname{Spec} \mathbf Z) = 1$. - -Some ideas of the proof. -Here is a broad overview of the proof; more details can be found in [vDdB 2019]. As will become clear, most of the ideas were already present in some form, but there were some key tricks missing. -1. Underlying set. -The underlying set of $X \in \mathbf{Sch}_S$ is reconstructed as the set of isomorphism classes of simple subobjects. -2. Topology. -Although we don't know if regular monomorphisms in $\mathbf{Sch}$ are the same as (locally closed) immersions (see also this question), we do know: - -Every open immersion is a regular monomorphism; -Every closed immersion is a regular monomorphism; -Every regular monomorphism is an immersion. - -Thus, a morphism $f \colon X \to Y$ is an immersion if and only if it can be written as a composition of two regular monomorphisms. -Next, one shows: - -Proposition. Let $(X,x)$ be a pointed scheme. Then $(X,x) \cong (\operatorname{Spec} R, \mathfrak m)$ for a valuation ring $R$ with maximal ideal $\mathfrak m$ if and only if all of the following hold: - -$X$ is reduced and connected; -the category of immersions $Z \hookrightarrow X$ containing $x$ is a linear order; -there exists a subset $V \subseteq |X|$ that is the support of infinitely many pairwise non-isomorphic immersions $Z \hookrightarrow X$ containing $x$. - - -Together with the characterisation of immersions, this leads to categorical criteria for closed immersions and open immersions in $\mathbf{Sch}_S$. -3. Quasi-coherent sheaves. -A variant of the Beck cogroup argument (see also user131755's post) realises nilpotent thickenings $\mathbf{Spec}_X(\mathcal O_X \oplus \mathscr F) \to X$ as cogroups in $X/\mathbf{Sch}_X$. This gives (loosely speaking) a pseudofunctor -\begin{align*} -\mathbf{Sch}_S &\to \mathbf{Cat}^{\operatorname{op}}\\ -X &\mapsto \mathbf{Qcoh}(\mathcal O_X), -\end{align*} -reconstructed from $\mathbf{Sch}_S$ using only categorical data. -4. The structure sheaf. -Now we run an enhanced version of this argument of the OP (that took place in the ring setting). We would like to say that the '(pre)sheaf End' $\mathscr End(\mathbf{Qcoh}(\mathcal O_{-}))$ on $\mathbf{Sch}_S$ is isomorphic to the structure (pre)sheaf $\mathcal O$ on the big Zariski site $\mathbf{Sch}_S$. -This is possible, but the difficulty is to say what exactly this presheaf End (or really prestack End) should be (also since it all takes place on the big Zariski site $\mathbf{Sch}_S$, not just the small Zariski site $S$). -5. Proof of main theorem. -By 2 and 4 above, we have reconstructed from $\mathbf{Sch}_S$ the topology on $|S|$ together with its structure sheaf $\mathcal O_S$. This gives (roughly speaking) some sort of lax functor of $2$-categories -\begin{align*} -\{\text{categories equivalent to } \mathbf{Sch}_S \text{ for some } S\} &\to \mathbf{Sch}\\ -\mathbf{Sch}_S &\mapsto S. -\end{align*} -But in fact the reconstruction of the scheme $X \in \mathbf{Sch}_S$ (with its structure morphism $X \to S$) from categorical data in $\mathbf{Sch}_S$ is functorial in $X$. With some work, this shows that this lax functor is a lax inverse of $S \mapsto \mathbf{Sch}_S$. $\square$ -(Because I don't really speak $n$-category, I phrase the last part a little differently in my paper.) - -References. -[vDdB 2019] Remy van Dobben de Bruyn, Automorphisms of categories of schemes, 2019. Submitted. arXiv:1906.00921. -[Mochizuki 2004] Shinichi Mochizuki, Categorical representation of locally Noetherian log schemes. Adv. Math. 188.1, p. 222-246 (2004). ZBL1073.14002.<|endoftext|> -TITLE: Homotopy dimension of a mapping -QUESTION [6 upvotes]: The homotopy dimension $h \dim X$ of a space $X$ is the minimal dimension of a CW-complex $Z$ homotopy equivalent to $X$. -I am interested in the generalisation to maps $f\colon X\to Y$. Here's what I think it should be: - -The homotopy dimension of $f\colon X\to Y$ is the smallest $k$ such that $f$ factorises through a $k$-dimensional CW-complex up to homotopy (meaning there is a $Z$ of dimension $k$ and maps $g\colon X\to Z$ and $h\colon Z\to Y$ with $f\simeq h\circ g$). - -Question 1: Is this the "correct" generalisation? My hesitance stems from the fact that, with this definition, $h\dim 1_X$ (the homotopy dimension of the identity map $1_X\colon X\to X$) does not necessarily equal $h\dim X$. Indeed, the former is the smallest dimension of a CW-complex which dominates $X$, and (I believe) Wall has shown that there are spaces for which $h\dim 1_X$ $<$$\infty$ while $h\dim X=\infty$. -Question 2: I am sure that this is a well-known and well-studied notion, and that I am merely using the wrong search terms. Where should I look in the literature to learn more about this concept? - -REPLY [8 votes]: Regarding Question 1: No, I do not think that's correct. In my opinion, the -definition should be one of the following: -The relative homotopy dimension of $f: X \to Y$ is $\le k$ if and only if there is a factorization -of $f$ as -$$ -X \overset{f'}\to Y' \overset{g} \to Y -$$ -in which $f'$ is an inclusion, $Y'$ is obtained from $X$ by iterated cell attachments of dimension $\le k$, and $g$ is a weak homotopy equivalence. -The notion of dimension I am describing is internal to the category of spaces under $X$, i.e, -$X\backslash\text{Top}$, where $f$ is to be regarded as an object of that category. In this scheme, the homotopy dimension of the identity map is $\le -1$. -There is another variant of though: let define us say that the fiberwise homotopy dimension of -$f: X\to Y$ is $\le k$ iff if there is a weak homotopy equivalence $X' \to X$ such that -$X'$ is a cell complex of dimension $\le k$. -In particular $Y$ is homotopy equivalent to a cell complex of dimension $\le k$ if and only if the identity map of $Y$ has fiberwise dimension $\le k$. -Regarding Question 2: To a certain extent, I have written about both of these notions in the paper: Poincaré duality embeddings and fiberwise homotopy theory, Topology 38, 597$-$620 (1999), but this is by no means my concept, nor is my treatment to be regarded as definitive. -Added: The above notions generalize to a single notion as follows: let $f: A \to Y$ be any map of spaces and define $\text{Top}_f$ to be the category of factorizations of $f$ -the objects of this category are factorizations $A \to X \to Y$ and morphisms are maps -$X \to X'$ commuting with the given structure maps. -Then we can define dimension in this setting as follows: let's say that an object $X \in \text{Top}_f$ has dimension $\le k$ iff it is built up from the initial object (represented by $A$) by attaching cells over $Y$ of dimension at most $k$. -It's easy to see that the case $f:\emptyset \to Y$ gives the notion of fiberwise dimension, whereas the case when $f: A \to \text{pt}$ gives the notion of relative dimension.<|endoftext|> -TITLE: Entropy of nested compact invariant sets -QUESTION [6 upvotes]: Let $f$ be a homeomorphism on a compact metric space $X$. $K_1\supset K_2\supset\cdots \supset K$ are compact subsets of $X$ such that $f(K_n)=K_n$ and -$K=\bigcap K_n$. If $h(f, K_1)<\infty$, do we always have $h(f,K)=\lim h(f,K_n)$? -I can show that this is true for $C^\infty$ diffeomorphisms where the entropy map -of invariant measures is upper semi-continuous. -OK. I see the point. This is definitely false for the most general case. For example, take the union of countable hyperbolic toral automorphisms of shrinking size and add one point with Identity map on it. -However, what if $f$ is a diffeomorphism on a compact manifold? I still expect negative answer. - -REPLY [2 votes]: The strategy of Benoît Kloeckner fails for differentiable maps. Indeed if $K$ is a single point and $K_n$ invariant balls around $K$ it implies that $K$ is an attracting fixed point. Therefore the log of the differential of the diffeo $f$ should be close to zero near $K$ and so does the entropy. -However conter-examples in any finite smoothness ($C^r$ maps with $1\leq r<+\infty$) -were given by Misisurewicz in the early seventies : -Diffeomorphism without any measure with maximal entropy, Bull. Acad. Pol. Sci., Ser. sci. math., astr. et phys. 21 (1973), 903--910<|endoftext|> -TITLE: Deformation to the normal cone -QUESTION [16 upvotes]: Deformation to the normal cone appears in several places including Intersection theory and Verdier specialisation of construtible sheaves or D-modules. I'd like to understand what happens when we iterate the process: deform to the cone of $A_1 \subset X$ and then to the cone of $C_{A_1\cap A_2} A_2 \subset C_{A_1} X$. In particular is it commutative (can we exchange $A_1$ and $A_2$)? -Deformation to the normal cone: Let $A$ be a closed subscheme of $X$ with ideal $I \subset O_X$. One has the space of deformation to the normal cone $D_A X = Spec_X( R_I O_X)$ where $(R_I O_X)\subset O_X[t^{\pm 1}]$ is the Rees algebra of $I$ defined by -$$ - R_I O_X = \bigoplus_{k\in \mathbb{Z}} I^{-k} t^k -$$ -with $I^{k} = O_X$ if $k \leq 0$. The projection $t: D_A X \to \mathbb{A}^1$ is flat. For $t \neq 0$ the fiber is $X$ while for $t =0$ the fiber is the normal cone -$$ - C_A X = Spec \left( \bigoplus I^{k}/I^{k+1} \right) -$$ -hence the name. In a more geometric fashion, $D_A X$ is the complementary of $\mathbb{P}(C_AX) \subset \mathbb{P}(C_{A}X \oplus 1)$ inside the blow-up $B_{A\times 0} X\times \mathbb{A}^1$. The projection onto $X\times \mathbb{A}^1$ being given by the inclusion $O_X[t] \to R_I O_X$. -The construction is fonctorial. For $f:X\to Y$, $B\subset Y$ and $A = X\times_Y B$, we have -$$ - f^* \left( \bigoplus I_B^{-k} t^k \right) \to \bigoplus I_A^{-k} t^k -$$ -inducing a morphism $D(f):D_A X \to X\times_Y D_B Y$ compatible with the projections onto $\mathbb{A}^1$. If $f$ is a closed embedding then so is $D(f)$. -Multiple deformations: Now consider $A_1,A_2 \subset X$ with ideals $I_1,I_2$. -We can form the multi-Rees algebra -$$ - R_{(I_1,I_2)} O_X := \bigoplus_{k_1,k_2\in \mathbb{Z}} (I_1^{-k_1} \cap I_2^{-k_2}) t_1^{k_1} t_2^{k_2} -\subset O_X[t_1^{\pm 1},t_2^{\pm 1}] -$$ -Its Spec aver $X$ is a space $D_{(A_1,A_2)} X$ together with a morphism $D_{(A_1,A_2)} X \to X\times \mathbb{A}^2$ given by the coordinates $(t_1,t_2)$. -We also have a closed immersion $D_{A_1 \cap A_2} A_2 \subset D_{A_1} X$ with ideal -$$ - J_2 = \bigoplus_{k\in \mathbb{Z}} (I_2 \cap I_1^{-k}) t^{k} -$$ -So we can form the Rees algebra $R_{J_2} R_{I_1} O_X$ and its Spec over $D_{A_1} X$, we simply denote by $D_{A_2} D_{A_1} X$. It also has a canonical morphism -$$ - D_{A_2} D_{A_1} X \to D_{A_1}X \times \mathbb{A}^1 \to X\times \mathbb{A}^1 \times \mathbb{A}^1. -$$ -Question 1: Is it the same thing to - -Deform simultaneously using $D_{(A_1,A_2)} X$ -Deform to the normal cone of $A_1$ in $X$ and then to the normal cone of $C_{A_1\cap A_2} A_2$ in $C_{A_1} X$ - -Question 2: Do we have a canonical isomorphism -$$ - R_{(I_1,I_2)} O_X = R_{J_2} R_{I_1} O_X? -$$ -Does is it induce an isomorphism of $(X\times \mathbb{A}^2)$-schemes compatible with the $\mathbb{G}_m^2$-actions coming from the gradings? -Question 3: Do we have a canonical isomorphim -$$ - D_{(A_1,A_2)} X|_{t_1 = 0} = D_{C_{A_1\cap A_2} A_2} C_{A_1} X -$$ -i.e. -$$ - R_{(I_1,I_2)} O_X / (t_1) = R_{Gr_{I_1} I_2} Gr_{I_1} O_X -$$ -Question 4 What can we say if $A_1$ and $A_2$ are transverse subvarieties? What changes for the deformation spaces? -Note: I know that in this case the canonical morphism $C(i_2) : D_{A_1\cap A_2} A_2 \to A_2 \times_{X} D_{A_1} X$ is an isomorphism. -Question 5 Does any one know any good reference where basic functorial properties of Rees algebras are detailed? - -REPLY [3 votes]: Ok so the solution is to consider the product of ideals instead of their intersection. -Just like we have -$$ - B_{\widetilde{I}_1} B_{I_2} X = B_{I_1I_2} X = B_{\widetilde{I}_2} B_{I_1} X -$$ -with $\widetilde{I}_j$ the total transform of the ideal $I_j \subset O_X$, we have -$$ - D_{\widetilde{D}_1} D_{A_2} X = D_{(A_1,A_2)} X = D_{\widetilde{D}_2} D_{A_1} X -$$ -with $\widetilde{D}_j = A_i \times_X D_{A_j} X$ the total pullback of the normal cone deformation space to $A_i$. -With this definition, we have nice functoriality properties w/r to maps and direct products.<|endoftext|> -TITLE: A characteristic-free proof that the action of a connected algebraic group $G$ on the fundamental group of a $G$-variety is trivial -QUESTION [6 upvotes]: Let $G$ be an algebraic group (not necessarily linear) defined over an -algebraically closed field $k$, -acting on a smooth integral $k$-variety $X$. -Let $x_0\in X(k)$ and let $\pi_1(X,x_0)$ denote the étale -(Grothendieck's) fundamental group of $X$. -Assume that either $G$ fixes $x_0$ or the group $\pi_1(X,x_0)$ is abelian. -In both cases $G(k)$ acts on $\pi_1(X,x_0)$. -I need a proof that if $G$ is connected, then this action is trivial. -I know a proof in characteristic 0. -In this case by the Lefschetz principle we may assume that -$k=\mathbf{C}$, and we can consider -the action of $G(\mathbf{C})$ on the topological fundamental group -$\pi_1^{\mathrm{top}}(X(\mathbf{C}),x_0)$. -Let $g\in G(\mathbf{C})$. -Since $G$ is connected, we can connect $g$ with the unit element $e\in -G(\mathbf{C})$ by a continuous path. -We see that the automorphism $g_*\colon X(\mathbf{C})\to -X(\mathbf{C})$ is homotopic to the identity automorphism. -It follows that the induced automorphism of -$\pi_1^{\mathrm{top}}(X(\mathbf{C}),x_0)$ is the identity. -I would like to see a proof that the action is trivial in arbitrary -characteristic. - -REPLY [11 votes]: This is false as stated in positive characteristic. For example, suppose that the characteristic of $k$ is $p$; take $X = \mathbb A^1_k = \mathop{\rm Spec} k[t]$ and $G = \mathbb G_{\rm m}$. If $a \in k^*$ and $E$ is the standard étale cover $E = \mathop{\rm Spec} k[x,t]/(x^p - x - t) \to \mathop{\rm Spec} k[t]$ of $X$, its pullback through multiplication by $a$ is $ \mathop{\rm Spec} k[x,t]/(x^p - x - at) \to \mathop{\rm Spec} k[t]$, which will not be isomorphic to $E$ in general. -However, it is true when $X$ is proper over $ \mathop{\rm Spec} k$. The point is that in this case we have $\pi_1(X \times G, (x_0, 1)) = \pi_1(X, x_0) \times \pi_1(G, 1)$ (see, for example, Corollary 5.6.6 in Szamuely's wonderful book on the fundamental group); this uses the hypothesis that $G$ is connected. The embedding of $\pi_1(G, 1)$ corresponding in the decomposition above is induces by the embedding $G \subseteq X \times G$ defined by $g \mapsto (x_0, g)$. If $\alpha \colon X \times G \to X$ is the action, let us show that $\alpha_*\colon \pi_1(X \times G, (x_0, 1)) \to \pi_1(X, x_0)$ coincides with the homomorphism $\pi_1(X \times G, (x_0, 1)) \to \pi_1(X, x_0)$ induced by the first projection. From the formula above we see that it is sufficient to show that the composite $G \to X \times G \to X$, where the first map is $g \mapsto (x_0, g)$ and the second is the action, induces a trivial map on fundamental groups. But the map is in fact constant, because $G$ fixes $x_0$, so this is clear. -Now fix $g \in G(k)$. The composite $X \to X \times G \to X$, where the first map is $x \mapsto (x,g)$, and the second is the action, induces the action of $g$ on $X$. On the other hand, the composite $X \to X \times G \to X$, where the first map is $x \mapsto (x,g)$, and the second is the projection, induces the identity. From the previous fact, we have that these two maps induce the same homorphism on fundamental groups, and we are done.<|endoftext|> -TITLE: Can you tell whether a space is Banach from the unit ball? -QUESTION [21 upvotes]: Let $V$ be a real vector space. It is well known that a subset $B\subset V$ is the unit ball for some norm on $V$ if and only if $B$ satisfies the following conditions: - -$B$ is convex, i.e. if $v,w\in B$ and $\lambda\in[0,1]$ then $\lambda v+(1-\lambda)w \in B$. -$B$ is balanced, i.e. $\lambda B \subset B$ for all $\lambda \in [-1,1]$. -$\displaystyle\bigcup_{\lambda > 0} \lambda B = V$ and $\displaystyle\bigcap_{\lambda>0} \lambda B = \{0\}$. - -My question is: is there some simple way to determine from $B$ whether the resulting norm on $V$ will be complete? Keep in mind that $V$ does not yet have a topology. -Edit: I guess the word "simple" is a bit misleading. What I'm looking for is some geometric insight into how the shape of $B$ affects whether the result is a Banach space. When $V$ is finite dimensional, all sets $B$ satisfying conditions (1) - (3) give equivalent norms, so all $B$'s are somehow roughly the same shape. In what way do the shapes vary when $V$ is infinite-dimensional, and how does this affect the completeness of the resulting norm? - -REPLY [4 votes]: Unit balls with precisely the property that you are looking for have been studied under the rather awkward name of completant (presumably directly from the French) in the book on applications of bornologies to functional analysis by Hogbe-Nlend. I think that the only result of any substance that you will find is a variant of Grothendieck's completeness theorem which can be found there. One assumes that the ball is a closed bounded set in an ambient topological vector space which is complete. This, amongst others, provides what is probably the simplest and most transparent proof of the completeness of the $ \ell^p$ and $L^p $-spaces. -By the way the class of spaces of Bill Johnson's answer has also been investigated. They were introduced by Waelbroeck and called Waelbroeck spaces by Buchwalter. They form a concrete representation of the category opposite to that of Banach spaces---see Cigler-Losert-Michor on functors on categories of Banach spaces (available onine). A good example of their use is in the characterisation of von Neumann algebas as $ C^*$ algebras which, as Banach spaces, are Waelbroeck. This givea a useful pointer on how to form limits in the category of von Neumann algebras.<|endoftext|> -TITLE: Bounds for generalized harmonic numbers -QUESTION [6 upvotes]: Guo and Qi recently discovered sharp bounds for the harmonic numbers (qq.v. doi:10.1016/j.amc.2011.01.089). For example, they show that $$H_n < \ln(n) + \frac{1}{2n} + \gamma - \frac{1}{12n^2+\frac{6}{5}},$$ where $H_n$ is the $n$th harmonic number and $\gamma$ is the Euler-Mascheroni constant. -My question: are there any similar bounds on the generalized harmonic numbers? By "generalized", I mean $H_{n,r}$ where $$H_{n,r} = \sum_{k=1}^n \frac{1}{k^r}.$$ -I am specifically interested in the case when $r = \frac{1}{2}$. - -REPLY [10 votes]: Let $f(x)=1/\sqrt{x}$. Then by Euler-Maclaurin summation, -$$ \sum_{2\leq k\leq n} \frac{1}{\sqrt{k}} = \int_1^n \frac{dx}{\sqrt{x}}+\sum_{r=0}^m\frac{(-1)^{r+1}B_{r+1}}{(r+1)!} \big(f^{(r)}(n)-f^{(r)}(1)\big)+ R$$ -where $R$ is a remainder term of the form -$$R=\frac{(-1)^m}{(m+1)!}\int_1^n B_{m+1}(x)f^{(m+1)}(x)dx.$$ -Here $B_m$ is the $m$th Bernoulli number and $B_m(x)$ is a periodic function of period 1 that coincides with the $m$th Bernoulli polynomial on [0,1). -Taking $m=0$, for instance, since $B_1=-1/2$ and $B_1(x)=${x} - $\frac{1}{2}$ we get -$$ \sum_{2\leq k\leq n} \frac{1}{\sqrt{k}} \leq 2\sqrt{n}-2 + \frac{1}{2}\Big(\frac{1}{\sqrt{n}}-1\Big)+\frac{1}{3}\int_1^n x^{-3/2} dx.$$ -Now compute the integral and add back in the term $k=1$ to get an upper bound. -Sharper bounds are possible choosing $m=1,2,3,\ldots$.<|endoftext|> -TITLE: Proof of the "Neo-classical Inequality", a fractional extension of the binomial theorem -QUESTION [11 upvotes]: I came across the following inequality, dubbed the "Neoclassical Inequality" which holds uniformly in $p\geq 1$ and $n\in\mathbb N$: -$$\frac{1}{p^2}\sum_{j=0}^n\frac{a^{\frac{j}p}b^{\frac{n-j}p}}{\frac{j}p!\frac{n-j}p!}\leq \frac{(a+b)^{\frac{n}p}}{\frac{n}p!}$$ -Notice that if $p=1$, we get the binomial theorem. As well, the inequality is homogenous in $a$ and $b$ so wlog replace $a$ with $x$ and $b$ with $(1-x)$. -My question concerns the proof of this inequality, found in the paper 'Differential equations driven by rough signals', Rev. Mat. Iberoamericana 14 (1998) 215-310, on page 254, theorem 2.2.3. Also see Keisuke Hara and Masanori Hino, Fractional order Taylor’s series and the neo-classical inequality. -Without going into too much detail (see the first paper above for the gory details), the idea of the proof is to define functions $F_{\theta_j}(x,v)$ which more-or-less correspond to individual terms of the sum and reduce the proof of the inequality to showing that -$$\sum_{j=0}^nF_{\theta_j}(x,v)\leq 1 \qquad \qquad \ \ (\star)$$ -holds for all $v>1/n$ and all $x$. Now comes the key idea: we can prove $(\star)$ by showing that -$$\left(\frac{\partial}{\partial x}(x(1-x))\frac{\partial}{\partial x}-\frac{\partial}{\partial v}\right)F_\theta\geq 0,$$ -so that by the maximum principle for sub-parabolic functions, we can conclude that any positive linear combination of the $F_\theta$ attains its maximum over the region $v>1/n,x\in(0,1)$ on its parabolic boundary. In particular, we get $(\star)$. -This is a technique that's new to me. I was wondering where else one can use such maximum principles to prove inequalities. It looks like the general idea is to find an appropriate parabolic differential operator for the inequality which makes the terms in the inequality a subsolution. More importantly, I was wondering when one should consider the inspiration to apply such a technique to proving an inequality. - -REPLY [2 votes]: As far as I know this inequality was first proved by Terry Lyons in 90-s by standard method of Lagrange multipliers.<|endoftext|> -TITLE: D-modules on rigid analytic spaces -QUESTION [16 upvotes]: Is there a good notion of holonomic $D$-modules on rigid analytic spaces? - -REPLY [13 votes]: Yes. Although it is only beginning to be developed. -You probably want to start with Berthelot: D-modules arithmétiques I : Opérateurs différentiels de niveau fini and Introduction à la théorie arithmétique des D-modules, and other papers that can be found at http://perso.univ-rennes1.fr/pierre.berthelot/ Section 5 of the second paper I mentioned is perhaps most relevant. -There is also a recent paper of Caro which I cannot find online called 'Holonomie sans structure de Frobenius et criteres d'Holonomie' which removes the necessity of the Frobenius action from Berhelot's work. I suppose he would send you a copy of upon request. -Finally, in a piece of shameless self-advertising, Konstantin Ardakov and I recently put a preprint on the arXiv http://arxiv.org/abs/1102.2606 part of which seeks to find a framework to further develop the theory. -Update: Caro's paper mentioned above now seems to be available here: http://aif.cedram.org/item?id=AIF_2011__61_4_1437_0 although you need a subscription to access it. -Further update (10th Feb 2015): Apologies for the further self-advertising but Konstantin Ardakov and I now have two further preprints on the topic of D-modules on rigid analytic spaces http://arxiv.org/abs/1501.02215 and http://arxiv.org/abs/1502.01273. There is no mention of holonomicity in either of these but there seems to be a natural definition of the notion in the framework outlined in these. Whether this definition behaves as one might hope is likely to be discussed in future work. -Further update (1st May 2019): Once again apologies for self-advertising but Konstantin Ardakov, Andreas Bode and I now have preprint https://arxiv.org/abs/1904.13280 with a tentative definition of holonomicity for a D-module on a rigid analytic space (D-module in the sense of my previous two papers with Ardakov mentioned in the last update). -For completeness I should also point to the paper La théorie du polynôme de Bernstein–Sato pour les algèbres de Tate et de Dwork–Monsky–Washnitzer by Mebkhout and Narváez-Macarro http://www.numdam.org/item/ASENS_1991_4_24_2_227_0/ that we cite in our latest preprint but I hadn't been previously aware of. This includes a more classical theory of D-modules on rigid spaces than ours --- the main distinction between the two theories is that our sheaves of rings are completions of theirs so that infinite order differential operators appear in our setting. Which theory might be preferred will depend on what problem is being considered.<|endoftext|> -TITLE: About the proof of the proposition "there exists irrational numbers a, b such that a^b is rational" -QUESTION [6 upvotes]: What does the classical proof of the proposition "there exists irrational numbers a, b such that $a^b$ is rational" want to reveal? I know it has something to do with the difference between classical and constructive mathematics but am not quite clear about it. Materials I found online does not give quite clear explanations either. Could someone give a better explanation? - -REPLY [12 votes]: The proposition can also be proved without explicitly knowing any particular irrational number: -Fix a positive rational number $c$. Then for any positive irrational real number number $a$, -the equation $a^b = c$ has a solution, and the solutions for different $a$ are different. -Now assume that there would be no irrational numbers $a$ and $b$ such that $a^b$ is rational. -Then all these solutions would be rational, thus we would have an injective mapping from the -positive irrational real numbers to the rationals; however we know that the rationals are -countable and the positive irrational reals are not, so there is no such injection, and we -have a contradiction.<|endoftext|> -TITLE: What characteristic class information comes from the 2-torsion of $H^*(BSO(n);Z)$? -QUESTION [13 upvotes]: This is just a general curiosity question: -In the standard textbook treatments of characteristic classes, and in particular the treatment of universal Pontrjagin classes, it's standard to consider $H^\ast(BSO,Z[1/2])$ (or $H^\ast(BSO(n),Z[1/2])$) in order to kill the 2-torsion. But I'm curious about that 2-torsion, since it should still give us some extra characteristic class-type information about real oriented bundles. If nothing else, it would give a class that's characteristic in the sense that it behaves the right way with respect to pullbacks (though I imagine it might be too much for these to be stable or have any kind of product formulas). So I suppose my questions are: - -What's known about such classes? -Are they useful for anything? -Do these interact in any interesting way with the Stiefel-Whitney classes when everything is reduced mod 2? -Why are they usually ignored (or obliterated by coefficient change)? - -REPLY [32 votes]: The basic fact is that the 2-torsion all has order exactly 2, so it injects into the mod 2 cohomology, forming a subalgebra of the polynomial algebra on the Stiefel-Whitney classes. This subalgebra can also be described as the image of the mod 2 Bockstein homomorphism, so it is computable although the answer is not easy to state. The conclusion of all this is that there is no new information in the torsion in the integral cohomology, beyond what one gets from Stiefel-Whitney classes. -These facts are stated as an exercise in Chapter 15 of Milnor and Stasheff's book. A proof can also be found in my notes on Vector Bundles and K-Theory, Theorem 3.16, available on my webpage.<|endoftext|> -TITLE: What does the adjective "natural" actually mean? -QUESTION [35 upvotes]: Terms like "in the natural way" or "the natural X" are used frequently in mathematical writing. While it is certainly clear most of the time what is meant, on occasion, I have been confounded. The word "natural" seems to be one of the most ambiguous terms used in formal mathematics. I have never seen anyone actually define it. People just use it and expect others to understand it. -What exactly is meant by "natural" in mathematical writing? - -REPLY [5 votes]: One issue I have with all of the answers is that mathematical writing is writing by humans. Much of the language in a mathematical paper written by a human is plain language, intended to impart understanding from one human to another human, rather than formal language intended to codify a proof. In proving many existence theorems, one has to make a rather wild guess at the definition of the object whose existence is asserted, long before one is able to carry out the actual proof that the object satisfies all of the required properties of the theorem. Such wild guesses have many different motivations in humans. I often use the word "natural" to let my reader know what motivated me to guess at the right object. Sometimes the motivation was background knowledge or experience, sometimes it was just blind intuition which I cannot account for, and sometimes it is motivated by what felt "natural". I would not assert that there is any formal mathematical definition for this use of "natural". I suppose we'll have to wait to see what happens with this term once mathematical papers are written by computers to impart understanding from one computer to another.<|endoftext|> -TITLE: Must a linearly ordered, separable space be metrizable? -QUESTION [7 upvotes]: Let $X$ be a linearly ordered topological space with a countable dense subset. Does it necessarily follow that $X$ is metrizable? -EDIT: Apollo's comment int he answers implies the answer is negative. Let $X$ be the open unit interval $(0,1)$ and adjoin to every real number $x$ a "ghost number" $x'$ such that $x'$ is the immediate successor of $x$. The "real rationals" are dense in this space. Simply note that sets of the form $(y, x]$ with $x$ and $y$ real and $[x',y)$ with $x'$ ghost and $y$ real form a basis, and these sets all contain a real rational. This space cannot be metrizable, because the subspace topology on the set of all ghost reals is exactly that of the Sorgenfrey line. - -REPLY [8 votes]: You already found a (classical) counterexample: the double arrow ($[0,1] \times \{0,1\}$, ordered lexicographically), which is even compact and separable. There is however a nice metrization theorem for linearly ordered spaces (due to Lutzer): a linearly ordered space $X$ is metrizable (in the order topology) iff the diagonal $D = \{(x,x) : x \in X\}$ is a countable intersection of open subsets of $X \times X$ (a $G_\delta$). This condition is also necessary and sufficient for countably compact regular spaces as well, not just the ordered ones.<|endoftext|> -TITLE: Scheme-theoretic account of why every variety embeds in a complete variety -QUESTION [7 upvotes]: The standard reference for the statement that "any abstract variety is an open subscheme of a complete variety" is Nagata's 1962 paper Imbedding of an abstract variety in a complete variety. Unfortunately, this paper was apparently written before the language of schemes became standard, and uses Nagata's own language for "algebraic geometry over a Dedekind domain." Does anyone know of a translation of this proof (or another of the same statement) into scheme-theoretic language (or other language more comprehensible to the contemporary reader)? - -REPLY [10 votes]: Apart from Brian's, published as: -Deligne's notes on Nagata compactifications. J. Ramanujan Math. Soc. 22 (2007), no. 3, 205–257. -there are: -Lütkebohmert, On compactification of schemes. Manuscripta Math. 80 (1993), no. 1, 95–111. -and -Vojta: Nagata's embedding theorem, arXiv:0706.1907 -and, finally -Deligne: Le théorème de plongement de Nagata, Kyoto J. Math. 50, Number 4 (2010), 661-670. -All of them are worth reading. The issue is certainly subtle and important, at least for cohomological constructions.<|endoftext|> -TITLE: The natural inclusion of an infinite abelian group $G$ into $\widehat{\widehat{G}}$ -QUESTION [7 upvotes]: I was recently trying to think of a simple example that demonstrates that the natural inclusion of an abelian group $G$ into -$$\widehat{\widehat{G}}=\text{Hom}\_{\mathsf{Ab}}(\text{Hom}_{\mathsf{Ab}}(G,\mathbb{C}^\times),\mathbb{C}^\times)$$ -is not necessarily an isomorphism. Note that I'm looking at all homomorphisms; no topology on the group is involved (or, if you prefer, they are all discrete). -Obviously, $G$ has to be infinite. However, the only example I could think of (much less actually prove worked) was $G=\mathbb{Z}$, in which case -$\widehat{G}=\text{Hom}\_{\mathsf{Ab}}(\mathbb{Z},\mathbb{C}^\times)\cong\mathbb{C}^\times$, so that -$$\widehat{\widehat{G}}\cong\text{Hom}_{\mathsf{Ab}}(\mathbb{C}^\times,\mathbb{C}^\times)$$ -which is uncountable due to the existence of uncountably many automorphisms of the field $\mathbb{C}$, and therefore not isomorphic to $\mathbb{Z}$. However, (my understanding is that) the existence of the anything more than the two obvious automorphisms of $\mathbb{C}$ requires the axiom of choice. So my questions are, -Does the claim that $\mathbb{Z}\not\cong\text{Hom}_{\mathsf{Ab}}(\mathbb{C}^\times,\mathbb{C}^\times)$ require AC? -and, if the answer to the above is yes, -Does the claim that there exists some abelian group $G$ such that $G\not\cong\widehat{\widehat{G}}$ require AC? Is it equivalent? -I initially asked this on math.SE but I suspect it's harder than I initially thought. - -REPLY [8 votes]: Let me give a context where I think a modified version of your question will sound more natural. -For an abelian Hausdorff topological group $G$, let's write $\widehat{G}$ for the continuous homomorphisms from $G$ to $S^1$ and $X(G)$ for the continuous homomorphisms from $G$ to ${\mathbf C}^\times$. Both $\widehat{G}$ and $X(G)$ are groups. There is a natural containment of sets $\widehat{G} \subset X(G) \subset C(G,{\mathbf C})$, where the last space is all continuous functions from $G$ to the complex numbers. When $G$ is also locally compact and we give $C(G,{\mathbf C})$ the compact-open topology then it is Hausdorff and $\widehat{G}$ and $X(G)$ are closed subsets. The topology induced on $\widehat{G}$ and $X(G)$ makes them both Hausdorff topological groups. -Fourier analysis on locally compact Hausdorff abelian groups $G$ is concerned with the interplay between $G$ and $\widehat{G}$, not $G$ and $X(G)$. It is not the case that $X(G)$ for locally compact Hausdorff abelian $G$ is uninteresting, e.g., in Tate's thesis $X(G)$ plays a prominent role for various concrete choices of $G$. But one of the primary features of the construction of $\widehat{G}$, namely that for locally compact Hausdorff abelian $G$ the topological groups $G$ and $\widehat{\widehat{G}}$ are naturally isomorphic, is false for the construction of $X(G)$. That, I think, is the interesting question: what is an example of a locally compact Hausdorff abelian group $G$ such that $X(G)$ is locally compact (it's automatically Hausdorff and abelian) and the topological groups $G$ and $X(X(G))$ are not isomorphic by the natural map $G \rightarrow X(X(G))$? Three counterexamples arise quickly if we compute several examples of $X(G)$. -Example 1: $X({\mathbf Z}) = {\mathbf C}^\times$. -Example 2: $X({\mathbf R}) = {\mathbf C}$ by analysis: the continuous homs ${\mathbf R} \rightarrow {\mathbf C}^\times$ are parametrized by complex numbers via the formula $\chi_s(x) = e^{xs}$ for $s \in {\mathbf C}$. -Example 3: $X(G) = \widehat{G}$ for compact $G$, and in particular $X(G) \cong G$ for finite $G$ and $X(S^1) = \widehat{S^1}$, which is isomorphic to ${\mathbf Z}$ by the parametrization $\chi_n(z) = z^n$ for $n \in {\mathbf Z}$. -Example 4: Although ${\mathbf Q}_p$ is not compact, $X({\mathbf Q}_p) = \widehat{{\mathbf Q}_p}$ because of the $p$-adic topology: if $\chi \colon {\mathbf Q}_p \rightarrow {\mathbf C}^\times$ is a continuous homomorphism then $\chi({\mathbf Z}_p) \subset S^1$ by compactness of ${\mathbf Z}_p$ and for any $x \in {\mathbf Q}_p$ we have $\chi(x)^{p^n} = \chi(p^nx)$, which is in $S^1$ for $n$ large since $p^nx$ is in ${\mathbf Z}_p$ for $n$ large. Therefore $\chi(x) \in S^1$ for any $x$ in ${\mathbf Q}_p$. It is part of Fourier analysis that $\widehat{{\mathbf Q}_p} \cong {\mathbf Q}_p$, so we get $X({\mathbf Q}_p) \cong {\mathbf Q}_p$, which is a contrast to the case of ${\mathbf R}$ in Example 2: -$X({\mathbf R}) \not\cong {\mathbf R}$ as topological groups since ${\mathbf R}$ and ${\mathbf C}$ are not homeomorphic. -To make more computations, we record a result about products. For two locally compact Hausdorff abelian groups $G_1$ and $G_2$, the natural map $X(G_1 \times G_2) \rightarrow X(G_1) \times X(G_2)$ is a topological group isomorphism. -Example 5: $X({\mathbf R}^n) \cong {\mathbf C}^n$. -Example 6: $X({\mathbf R}^\times) \cong X(\{\pm 1\} \times {\mathbf R}) \cong X(\{\pm 1\}) \times X({\mathbf R}) \cong \{\pm 1\} \times {\mathbf R} \cong {\mathbf R}^\times$. -Example 7: $X({\mathbf C}^\times) \cong X(S^1 \times {\mathbf R}) \cong X(S^1) \times X({\mathbf R}) \cong {\mathbf Z} \times {\mathbf C}$. -Thus we find $X(X({\mathbf Z})) \cong X({\mathbf C}^\times) \cong {\mathbf Z} \times {\mathbf C}$, so ${\mathbf Z}$ and $X(X({\mathbf Z}))$ are not isomorphic as topological groups (or even as groups: one is countable and the other is not). Another counterexample is ${\mathbf R}$: $X(X({\mathbf R})) \cong X({\mathbf C}) \cong {\mathbf C}^2 \cong {\mathbf R}^4$. A third counterexample is $S^1$: $X(X(S^1)) \cong X({\mathbf Z}) \cong {\mathbf C}^\times$, so $S^1$ and $X(X(S^1))$ are not isomorphic as topological groups since one is compact and the other is not. -The natural isomorphism of $G$ with $\widehat{\widehat{G}}$ for all locally compact Hausdorff abelian $G$ depends crucially on the use of $S^1$ as the receiver of character values. By this I don't just mean that if we use ${\mathbf C}^\times$ as the receiver of character values then such a duality isomorphism breaks down in specific examples. There is a theorem somewhere in Pontryagin's Topological Groups (maybe it is also in Weil's book on integration on topological groups) which says something like the following: if a locally compact Hausdorff abelian group $A$ could be used as the receiver of character values for a duality isomorphism on all locally compact Hausdorff abelian groups then $A$ must be isomorphic to $S^1$. Perhaps someone else can track down the reference.<|endoftext|> -TITLE: What about stacks of categories in algebraic geometry? -QUESTION [35 upvotes]: Stacks qua moduli spaces were introduced to keep track of nontrivial automorphisms of the objects they parameterize. In essence they are groupoids of objects with some form geometric cohesion. The classic example is principal bundles/torsors, the whole category of which is actually a groupoid. But what about objects one might want to parameterise which have non-invertible maps between them, such as vector bundles, or coherent sheaves? One could imagine a stack of such objects, because they 'glue' as principal bundles do, but if one keeps track of all maps this thing should be a category, not a groupoid. It is certainly deserving of being promoted to something geometric, and so one could present it by a category in algebraic spaces or schemes, much as the stacks we are more familiar with are presented by algebraic groupoids. -Some might argue that we have classifying topoi or similar for these situations, and this is well and good, but what about some geometry on these topoi? I know of two different takes on classifying topoi of a small (internal) category, one approach involving flat functors and the other torsors for the groupoid of all invertible arrows of the category in question. Between these two competing definitions, there are arguments (at least in my own head, too fuzzy to unveil here) both ways, and concrete examples of where one uses stacks of categories in a geometric context would certainly push the balance in one direction or another. This isn't the only reason I would like to know an answer to this question, but it has some bearing - and I've rabbited on long enough. - -Question: Do stacks $\operatorname{Sch}^{op} \to \operatorname{Cat}$ of categories come up anywhere in algebraic geometry, such that one considers presentations by internal categories in $\operatorname{Sch}$? If yes, how is the presentation specified? If no, why not? And is there anything stopping us from doing so? (Technical reasons, terminological, or we have other techniques that are better and so on) - -REPLY [19 votes]: The category theoretical definition of stacks (as given for instance in Giraud: Cohomologie non-abélienne) allow for arbitrary categories as targets (the stack condition only involves isomorphisms however). A natural example is the category of quasi-coherent sheaves (which has the category of vector bundles as a subcategory). However, when you are talking about algebraic stacks (which are category theoretic stacks fulfilling extra conditions) they only involve isomorphisms. Note that given any stack restricting to isomorphisms gives a stack in groupoids. This is what one does when one considers the algebraic stack of vector bundles: Start with the stack of vector bundles with arbitrary morphisms. This is not an algebraic stack but restricting to isomorphisms gives one. -General stacks (with non-isomorphisms) are used extensively as they encode the idea of descent. This is somewhat orthogonal to algebraic stacks which try to encode the idea of a moduli problem. -Addendum: All morphisms in a descent datum are isomorphisms (this actually -follows and does not have to be assumed). However full descent means -that you can descend objects (a descent datum of objects comes from an object downstairs) but also arbitrary morphisms (a descent datum of morphisms comes from a morphisms downstairs). These two properties together can be formulated as an equivalence of categories between the category of obejcts downstairs and the category of descent data. -Addendum 1: Charles poses an interesting question. One answer can be based on the fact there seems to also be a philosophical difference between general stacks and algebraic stacks. General stacks are based on the idea that we have some objects and relations between them, the morphisms, that can be glued together over some kind of covering. Hence, usually the objects themselves are the things of main interest and the gluing condition is just an extra (though very important) condition on such objects. -Algebraic stacks on the other hand are things that themselves are glued. The relevant idea is that groupoids are a natural generalisation of equivalence relations. (One can more or less arrive at the idea of a groupoid by thinking of bereasoned equivalence relations, elements do not just happen to be equivalent but there are specific, in general several, reasons for them to be equivalent.) -Having said that, one could start with the fact that an algebraic is the stack associated to a smooth algebraic groupoid (i.e., source and target maps are smooth). This gives a candidate generalisation by just looking at smooth algebraic categories instead. However, no natural examples that are not groupoids comes (at least) to my mind. I think the reason might be the above philosophical distinction.<|endoftext|> -TITLE: Thin subbases for the primes? -QUESTION [5 upvotes]: Hi all, -My question concerns a general problem concern the Erdos-Turan conjecture on additive bases; that of finding thin subbases in a given basis. For a given $A \subset \mathbb{N}$, define $r_{A,h}(n)$ to be the number of ways to write $n$ as $h$ (not necessarily distinct) elements of $A$. We say $A$ is an additive basis if $r_{A,h}(n) > 0$ for all $n \in \mathbb{N}$. From some trivial counting arguments, one can easily deduce that -$$\displaystyle N \leq \sum_{n \leq N} r_{A,h}(n) \leq | A \cap [1,N]|^h \leq \sum_{n \leq hN} r_{A,h}(n).$$ -We wish to investigate "thin" bases; namely those such that the lower bound is essentially achieved... that is $|A \cap [1,N]| = O(N^{1/h + o(1)})$. Erdos showed in 1956 that such bases exist when $h=2$, and Erdos-Tetalli showed that such bases exist for all $h$ in 1990. Their argument uses the probabilistic method, and the result is essentially achieved by proving that additive bases $B$ exist with $r_{B,h}(n) = \Theta(\log n)$, which easily implies that $B$ is thin by the above inequality. -A question, then, raised by Erdos, Nathanson, and several others, is whether or not a given basis contains a thin subbasis. That is, if $A$ is an additive basis, does it necessarily contain a subset $B$ such that $B$ is an additive basis, and $r_{B,h}(n) = O(\log n)$. A partial answer was initially given by Choi, Erdos, and Nathanson who showed that the squares $\mathbb{N}^2$ contains a subbasis $B$ such that $r_{B,4}(n) = O(n^{1/3} + \epsilon)$. Relatively recently, Van Vu proved in 2000 that in fact $\mathbb{N}^k$ contains a thin subbasis for all $k \geq 2$ such that $r_{B,h}(n) = O(\log n)$. Trevor Wooley in 2003 gave that in this case, $h = O(k \log k)$. -A very natural question to ask which I have not seen any results or works in progress on, is to do this for the primes. The result seems to be highly plausible given the much stronger conjecture which asserts thin subbases exist for all additive bases, but also the fact that the primes are 'thicker' than $\mathbb{N}^k$ for any $k \geq 2$. However, the primes suffer the disadvantage of being much less structured than $\mathbb{N}^k$ and hence it is not so clear how to adopt the probabilistic method, still the only way to generate thin bases, would apply. -So the main question is this: -If $\mathcal{P}$ is the set of primes, then it is known (Goldbach-Shirnel'man) that $\mathcal{P}$ is an additive basis of order at most 6 (and hence all higher orders). Can one show, for some $h > 1$, that $\mathcal{P}$ contains a subset $B$ such that $B$ is an additive basis of order $h$ (that is, $r_{B,h}(n) > 0$ for all $n$ sufficiently large) and $|B \cap [1,N]| = O(N^{1/h + o(1)})$? -So I would greatly appreciate if anyone would shed more insight on this problem, the best would be a paper where someone has done relevant work on this topic. -Thanks in advance! - -REPLY [2 votes]: Two papers on this topic are: Granville's Refinements of Goldbach’s conjecture, and the generalized Riemann hypothesis and Wirsing's Thin subbases. -In Granville's paper it is shown that if a quantitative form of Goldbach's conjecture is true, then there exists a set of primes, B, such that $|B \cap [N]| \ll (N \ln(N))^{1/2}$ and every even integer is the sum of two elements of B. In Wirsing's paper it is shown that For any $k \geq 3$ there is a set $B_{k}$ of primes, such that $|B_{k} \cap [N]| \ll (N \ln(N))^{1/k}$, that is a basis of order k for large $n \equiv k mod 2$. All of these constructions are probabilistic. -If you only care that the sumset is almost all of the even/odd integers, one can remove the $\ln^{1/k}(N)$ term in the above results.<|endoftext|> -TITLE: Product of coordinates of a random point from Hamming sphere -QUESTION [5 upvotes]: Let us consider a boolean hypercube $C = \{-1, 1\}^n$. Let $S = \{x \in C \mid |\{i \mid x_i = -1\}| = \varepsilon n\}$ be a Hamming sphere in $C$ (here $\varepsilon$ stands for the fixed parameter from $(0, 1/2)$). Let us sample $X \in S$ uniformly at random. And we are interested in estimating $\mathrm{E}[X_1 X_2 \ldots X_{\alpha n}]$, where $\alpha$ is a sufficiently small constant. -We would like to argue that this expectation is somewhat close to $(1 - 2\varepsilon)^{\alpha n}$ (that could have been the case if $X_i$'s were independent). -Is there any way to avoid tedious computations here? It is tempting to say that $X_i$'s are "almost independent" and use some "limit theorem". -Or is it possible to replace $X_i$'s by properly correlated gaussians and argue that expectation doesn't change much? -After all, I feel that this question must be considered somewhere, so it would be nice to find an appropriate reference. - -REPLY [5 votes]: If I am not mistaken, this expectation equals the coefficient of $x^{\varepsilon n}$ in $(1-x)^{\alpha n}(1+x)^{(1-\alpha) n}$ divided by the corresponding coefficient in $(1+x)^n$ (which equals, of course, $\binom{n}{\varepsilon n}$). Such coefficient may be represented as integrals over unit circle (and so over $[0,2\pi]$) and their asymptotics may be calculated by some standard machinery. I do not write down more, because I do not understand, what exactly do you want (say, which small constant is more, $\alpha$ or $\varepsilon$? Is $n$ chosen large after these constants are already fixed, or $n$ tends to infinity simultaneously with constants tending to zero?). -Update. -Such statement seems to be true. Call two values log-equivalent, if their logarithms are equivalemt. -We choose $x_0:=\varepsilon/(1-\varepsilon)$, then $\binom{n}{\varepsilon n}$ is log-equivalent to $x_0^{-\varepsilon n} (1+x_0)^n$. Then interpret our coefficient as $(2\pi i)^{-1}\int f(z)dz/z$ for $f(z)=z^{-\varepsilon n} (1-z)^{\alpha n}(1+z)^{(1-\alpha)n}$, and integral is taken over the circuit around 0. Choose the circuit $|z|=x_0$ and note that for small $\alpha$ we will have $|f(z)|\leq f(x_0)$ (this is less or more clear: $|1+z|\leq 1+x_0$ and for $\alpha$ close to 0 this is most important, we have to be careful with neighborhood of $x_0$, but it is ok too). Then just apply this estimate for estimating integral.<|endoftext|> -TITLE: Probabilities independent of ZFC? -QUESTION [8 upvotes]: Hi guys, -is it possible to change the probability of an event via forcing? More precisely, is there an innocent looking question on the probability of "something" whose answer is independent of ZFC? -All the best, -Sebastian - -REPLY [19 votes]: There are several issues. -On the one hand, any set can be made countable by forcing, and this process will certainly affect the measure of the set, if it did not have measure zero in the ground model. -But in the context of the Lebesgue measure on the reals, say, it is natural to consider not the set itself, but the Borel description of the set, interpreted first in the ground model and then reinterpreted in the forcing extension. (For exampe, the "unit interval" of $V$ is not necessarily the same as the unit interval of a forcing extension $V[G]$, but we have a borel code that correctly picks out the unit interval when interpreted in any model of ZFC.) In this case, one gets a positive solution for preservation of measure. The reason is that the assertion that the measure of the set with Borel code $b$ is $x$ has complexity at most $\Sigma^1_2(b,x)$ and hence is absolute to all forcing extensions by the Shoenfield absoluteness theorem. In this sense, the measure of a measurable set cannot be affected by forcing. -Meanwhile, the use of other non-absolute descriptions can lead again to a negative answer, where the measure can be affected by forcing. For example, consider the set $X$ of all binary sequences $x$ whose sequence of digits is realized somewhere in the GCH pattern of cardinals, in the sense that there is an ordinal $\beta$ such that $x(n)=1$ iff $2^{\aleph_{\beta+n}}=\aleph_{\beta+n+1}$. If the Generalized Continuum Hypothesis holds, then $X$ has measure zero, since only one pattern is realized. But one can force the GCH pattern to realize all patterns, and so there are forcing extensions in which $X$ has full measure. -Here is another comparatively concrete example. Consider the set of reals that are constructible, in the sense of Gödel's constructible universe. This set has complexity $\Sigma^1_2$ in the descriptive set-theoretic hierarchy, which is just a step up from Borel. The set has full measure in the constructible universe, of course, but it is easily made to have measure zero in a forcing extension. Thus, the probability that a randomly chosen real number is constructible has an answer that is independent of ZFC, because in some models of set theory this probability is 1 and in others it is 0.<|endoftext|> -TITLE: Automorphisms of Riemann Surfaces -QUESTION [15 upvotes]: Izumi Kuribayashi and Akikazu Kuribayashi have classified all groups of automorphisms of compact Riemann surfaces of genus 3,4,5. (J. Pure Applied Alg.65(3)-Sept.1990, and J. Alg.134(1) Oct.1990) -What are the other values of $g\geq 6$, for which a complete classification of automorphisms of compact Riemann surfaces of genus $g$ has been obtained? Can one suggest references for it, please? - -REPLY [7 votes]: While this question was posed several years ago, I wanted to add one reference which was not mentioned above. -Marston Conder has computed lists of all “large” automorphism groups acting on curves up to genus 101, where by “large” we mean that if $G$ acts on a curve $X$ of genus $g$, then $|G|>4(g-1).$ (This means the genus of the quotient $X/G$ must be $0$ and there are at most $4$ branch points in the mapping $X \to X/G$.) His data is here: -https://www.math.auckland.ac.nz/~conder/BigSurfaceActions-Genus2to101-ByGenus.txt -He presents his groups as a list of generators so that the group is the quotient of the free group on $3$ or $4$ variables (depending on the number of branch points) by those generators. -Additionally, Thomas Breuer's data can be a bit hard to find. Since I regularly need the data for my own research, I put his full list of groups and signatures up to genus 48 here: -http://www.math.grinnell.edu/~paulhusj/Monodromy/groupsignaturedata -Each line of that file reads as follows: -[*genus, order of group, signature, group identification number *] - -where the group identification number is a tuple $(a,b)$ where the group $G$ is SmallGroup(a,b) in the database of small groups in GAP or Magma.<|endoftext|> -TITLE: Local structure of Deligne-Mumford stacks -QUESTION [10 upvotes]: Let $\mathcal{X}$ be a separated Deligne-Mumford stack over an algebraically closed field $k$ and let $X$ be the corresponding coarse moduli space, which we assume to exist. There is a map $p:\mathcal{X}\to X$ of stacks. Is it true that every point of $X$ has an etale neighborhood $U\to X$ such that its pullback under $p$ is the map $[V/G]\to V/G=U$ where $V$ is an affine variety over $k$ on which a finite group $G$ acts. While reading some papers I've got the impression that the authors are implicitly using this statement or a similar one, but I wasn't able to locate a precise statement or reference in the literature. So I would be grateful if someone points me to one. -In the example I'm interested in $\mathcal{X}$ is in fact a quotient stack, but I do not want to assume that $char(k)=0$ or that the orders of the stabilizers are coprime with $char(k)$ (unless this follows from the previous conditions?). - -REPLY [7 votes]: This is Lemma 2.2.3 of the paper -Abramovich-Vistoli: Compactifying the space of stable maps; -see also section 5.4 of the "Guide to the stacks literature" by Jarod Alper.<|endoftext|> -TITLE: Which principlal bundles are locally trivial? -QUESTION [22 upvotes]: If $H$ is a closed subgroup of a topological group $G$, then the orbit map $G\to G/H$ is a principal bundle, yet somewhat surprisingly, it need not be locally trivial. -In the wikipedia article on fiber bundles it is claimed that if $H$ is a Lie group, then $G\to G/H$ is locally trivial. Is the claim true, and if so, what is the reference? - Remarks: - -That $G\to G/H$ is a principal bundle is explained e.g. in Husemoller's "Fiber bundles", example 2.4 in the 3rd edition. In the same section one can also find a definition of a principal bundle (which does not require local triviality). -A simple example when $G\to G/H$ is not locally trivial can be found in -the paper -of Karube [On the local cross-sections in locally compact groups, -J. Math. Soc. Japan 10 1958 343–347]. In the example $G$ is the product of infinitly many circles, and $H$ is the product of their order $2$ subgroups; there can be no -cross-section because $G$ is locally-connected and $H$ is not, so $G$ is not even locally homeomorphic to $H\times G/H$. -In the same paper Karube proves that $G\to G/H$ is locally trivial -in a number of cases, including when $G$ is locally compact, and $H$ is a Lie group. - -UPDATE: If $H$ is a Lie group, Palais's paper mentioned in his answer actually -characterises the principal $H$-bundles that are locally trivial; -details are below. -For a topological group $H$ acting freely and by homeomorphisms on a space $X$, we -let $X^\ast$ be the subsets of $X\times X$ consisting of pairs -$(x,hx)$ where $x\in X$ and $h\in H$. -Since $H$ acts freely, there is a map -$t: X^\ast\to H$ given by $t(x,hx)=h$. -Theorem 4.1 of Palais's paper says that if the space $X$ is -completely regular, and -if $H$ is a Lie group, then the free $H$-space $X$ is locally trivial if -and only if the map $t$ is continuous. -Note that in the terminology of Husemoller's "Fiber bundles" book -continuity of $t$ is assumed in the definition of a $H$-principal bundle, -thus Husemoller's $H$-principal bundles are all locally trivial -(provided $H$ is a Lie group and $X$ is completely regular). -If $X$ is a topological group and $H$ is a subgroup, then -continuity of $t$ follows from continuity of multiplication and inverse in $X$. -It is fun to see why Palais's result doesn't show that -the $\mathbb Z$-action on $S^1$ by irrational rotation is a principal bundle: -here $X=S^1$, and $H$ is the subgroup $\{e^{in}: n\in \mathbb Z\}$ with the subspace topology. The map $t$ is continuous, but $H$ is not a Lie group. - -REPLY [25 votes]: ...if $H$ is a Lie group, then $G \to G/H$ is locally trivial. Is the claim true, and if so, what is the reference? - -Yes, it is true. See the Corollary in section 4.1 of: "On the Existence of Slices for Actions of Non-compact Lie Groups", which you can download here: http://vmm.math.uci.edu/ExistenceOfSlices.pdf -This is a paper originally published in the March 1961 Annals of Math. -The Corollary says that "If X is a topological group and G is a closed Lie subgroup of X then the fibering of X by left G cosets is locally trivial."<|endoftext|> -TITLE: Generalized Euclidean TSP -QUESTION [11 upvotes]: Suppose I have n sets $X_1,\dots,X_n$ consisting of $k$ points each, where all $nk$ points are i.i.d. uniform random samples in the unit square $[0,1]\times[0,1]$. Consider the shortest path that goes through at least one element of each set $X_i$. Is the asymptotic behavior of this, a la the Beardwood-Halton-Hammersley (BHH) theorem, well-known? (By "asymptotic behavior", I mean, assume that $k$ is fixed and look at what happens as $n$ becomes large) - -REPLY [10 votes]: You should be able to get $O(\sqrt{n/k})$ by choosing a smaller square of area $1/k$, which will contain one point from most of the point sets, and use the BHH theorem to find a TSP tour of this. Now, you have to show that adding the points from the point sets you left out doesn't increase the length of the tour much. It certainly won't increase it by more than a $\sqrt{\log n}$ factor, because a square $\log n$ times larger in area will with high probability contain a point from all the point sets, and choosing one point randomly from each of the point sets in that square gives points uniform in this square. I'd guess the tour length will still be $O(\sqrt{n/k})$, but you may have to do some work to prove this. On the other hand you certainly can't do better than this, because you need $n$ points in your tour, and the typical distance from one point to its nearest neighbor is $O(1/\sqrt{nk})$. -Here's a strategy for a proof of $O(\sqrt{n/k})$. First take a square of area $\alpha /k$, with $\alpha$ chosen so that this contains $9/10\, n $ of the classes of points. The BHH theorem shows you can find a TSP tour of this square with length $O(\sqrt{n/k})$. Now, we need to show adding the rest of the points doesn't increase the tour length very much. -We'll find subsquares $S_1$, $S_2$, $S_3$, $\ldots$, $S_{\log n}$ where the $i$'th square is big enough that the expected number of classes of points it contains is $(1-1/10^{i})n$. The area of the $i$'th square will be $\alpha i /k$ for some constant $\alpha$. To get a TSP of the $9/10^in$ new classes of points in the $i$'th square takes length $O\left(\sqrt{i 9/10^i n/k}\right)$. The series sums to $O(\sqrt{n/k})$.<|endoftext|> -TITLE: How many LLL reduced bases are there? -QUESTION [11 upvotes]: For a given $n$-dimensional lattice embedded inside $\mathbb R^n$ along with a given inner product, how many distinct LLL-reduced bases are there? -In this question, a lattice is the set of all $\mathbb Z$-linear combinations of a set of $n$ vectors from $\mathbb R^n$ that are linearly independent over $\mathbb R$. - -REPLY [10 votes]: Here is the answer. -Claim: Any given n-dimensional lattice has at most $2^{0(n^3)}$ LLL reduced bases. -Notice: the bound is a function of the dimension $n$ only, and does not depend on the determinant of the lattice. Here is a simple proof: -Proof: Fix a lattice $L(B)$ and let $\lambda$ be the minimum distance of the lattice. The first vector of an LLL reduced basis $B=[\vec b_1,\ldots,\vec b_n]$ has length at most $2^{O(n)}\lambda$. Since spheres of radius $\lambda/2$ centered around lattice points are disjoit, by a simple volume argument, the number of lattice points in a sphere of radius $r=2^{O(n)}\lambda$ is at most $(1+2r/\lambda)^n = 2^{O(n^2)}$. For any such first vector $\vec b_1$, let $\pi_1$ be the projection orthogonal to $\vec b_1$. By definition of LLL reduced basis, $\pi_1(B)$ is also LLL reduced. Using LLL size reduction conditions, each projected LLL reduced basis $\pi_1(B)$ thas a unique lift such that all its Gram-Schmidt coefficients are in the range $[-1/2,1/2)$. So, we can proceed by induction, and see that there are at most $2^{O((n-1)^2)}$ possible choices for $\vec b_2$, and so on. -Overall, the number LLL reduced bases for $L$ is at most $\prod_{k=1}^n 2^{O(k^2)} = 2^{O(n^3)}$. This concludes the proof of the upper bound. [Q.E.D.] - -Of course, the number of LLL bases for a given lattice can be much smaller, e.g., you can easily build lattices whose LLL basis is unique up to the sign of the basis vectors. (E.g., take an orthogonal lattice with longer and longer basis vectors.) So, the number of LLL reduced basis can be as low as $2^n$. Every lattice has at least these many LLL reduced basis because you can set the signs of the basis vectors arbitrarily. -However, my guess is that the upper bound is asymptotically optimal in the worst case, i.e., there are lattices with $2^{\Omega(n^3)}$ LLL reduced bases. It should be possible to construct such lattices starting from examples lattices that achieve LLL worst case approximation factor $2^{O(n)}$ on the length of the shortest vector, but I didn't check the details.<|endoftext|> -TITLE: Down-To-Earth Uses of de Rham Cohomology to Convince a Wide Audience of its Usefulness -QUESTION [46 upvotes]: I'm soon giving an introductory talk on de Rham cohomology to a wide postgraduate audience. I'm hoping to get to arrive at the idea of de Rham cohomology for a smooth manifold, building up from vector fields and one-forms on Euclidean space. However, once I've got there I'm not too sure how to convince everyone that it was worth the journey. What down-to-earth uses could one cite to prove the worth of the construct? - -REPLY [4 votes]: Here is my attempt to explain DeRham cohomology and Hodge theory to a group of first year grad students that have no knowledge of manifolds.<|endoftext|> -TITLE: Simple modules for $U_q(\mathfrak{sl}_n)$ at roots of unity -QUESTION [12 upvotes]: Can anyone point me to a classification/construction of the irreducibles for $U_q(\mathfrak{sl}_n)$, or the associated small quantum groups, when the parameter $q$ is a root of unity and $n>2$? Neither Jantzen or Lusztig's quantum group books seem to help. -Edit: perhaps the best way to clarify what I mean when I refer to the `small quantum group' is to give the definition: take the $\mathbb{C}$-algebra (other fields will do) generated by (for $n=3$) $E_1, E_2, F_1, F_2, K_1, K_2$ subject to the following relations. -$$ E_1^2 E_2 - [2] E_1E_2E_1 + E_2 E_1^2 =0 $$ -$$ E_2^2 E_1 - [2] E_2E_1E_2 + E_1 E_2^2 =0 $$ -and the same relations on the $F$s, where $[2]$ is the quantum integer $q+q^{-1}$ -$$ [E_i, F_j] = \delta_{ij} (K_i-K_i^{-1})/(q-q^{-1}) $$ -$$ K_i E_j K_i^{-1} = q^{a_{ij}} E_j $$ -$$ K_i F_j K_i^{-1} = q^{-a_{ij}} F_j $$ -where $[a_{ij}]$ is the Cartan matrix for $\mathfrak{sl}_3$. -$$E_i^N = F_i^N = 0, K_i^N=1$$ -and also define $E_{1+2} = qE_2E_1 - E_1E_2$ and $F_{1+2}$ similarly and impose $E_{1+2}^N=F_{1+2}=0$. I'm convinced this last is necessary for the algebra to be finite-dimensional, though I have seen papers omitting it --- the small quantum group above should be a finite dimensional Hopf algebra with dimension $N^8$, with a PBW basis described by Lusztig. $q$ is a primitive $N$th root of unity in the field used. - -REPLY [5 votes]: I have tracked down some results on explicit classifications of simple modules for $u_q(\mathfrak{sl}_3)$. The general picture is that the simple modules are bigraded by the root lattice and look like towers of concentric hexagons. -For the benefit of anyone else interested, there is a long series of papers by Dobrev: -Multiplet classification of highest weight modules over quantum universal enveloping algebras: the Uq(SL(3,C)) example in Groups St Andrews 1989 vol 1 LMS LNM #159 -Representations of Quantum Groups, Symmetries in Science V (Lochau 1990), 93–135, Plenum Press, NY, 1991. -A chapter from Lecture Notes in Physics, 1990, Volume 370, here -Dobrev-Truini Irregular Uq(sl(3)) representations at roots of unity via Gel’fand–(Weyl)–Zetlin basis -Dobrev-Truini Polynomial realization of the Uq(sl(3)) Gel’fand–(Weyl)–Zetlin basis -MR1182163, MR1191199, -...and many others. -also there is a paper by -Abdesselam, Arnaudon, Chakrabarti -and a discussion of dimensions by Pereira here -Some of the relevant material is hard to find and/or requires paying large sums of money to publishing corporations.<|endoftext|> -TITLE: Biography of Felix Hausdorff -QUESTION [14 upvotes]: Felix Hausdorff was of course a great mathematician, who had major effects on several branches of mathematics. However he also wrote literature and philosophy and was affiliated with important German musicians. When Nazism came to power, Hausdorff failed to escape in time, lost his job, and finally committed suicide in order to avoid being sent to concentration camp. -The above summary of Hausdorff's life comes from reading his wikipedia page and other similar documents. I would like to learn more about him, but for some reason I don't seem to be able to find a book-length biography. (Can there really not be one?) -Question: What is a good source for a detailed biography of Felix Hausdorff? - -REPLY [6 votes]: The book Paradoxes of Measures and Dimensions Originating in Felix Hausdorff's Ideas - By Janusz Czyż -contains a 40 pages biographical sketch. It is available on Google books, most of it can be read -free, including the biographical chapter.<|endoftext|> -TITLE: Are there any rational solutions to this equation? -QUESTION [15 upvotes]: I am not sure if this is an appropriate question, but I was asked this by a colleague today and do not know how to answer it. -1) Are there any rational solutions to the following equation: -$$x^3-8x^2+5x+1 = -7y^2(x-1)x$$ -2) Is it possible that this is an elliptic curve in disguise? I have noticed that after projectivizing, there are two points at infinity. Perhaps this is okay under some change of variables? (I plead ignorance on this.) - -REPLY [4 votes]: I think Dror's nice argument can be made more transparent as follows. Suppose that $y^2= -x(x+7)(x^3+56x^2+245x-343)$ with $x,y$ in $\bf Q$ and $x$ not $0$ or $-7$. One sees easily that at a prime $p$ other than $7$ neither $x$ nor $x+7$ can have odd ord, either negative or positive. It follows -that $x^3+56x^2+245x-343$ is either a square or $7$(square) in $\bf Q$. -As Dror says, the first case can't occur. Suppose we're in the second. Write: -$x=7a$, $x^3+56x^2+245x-343=343b^2$, so that $x(x+7)=343c^2$ for some $c$ in $\bf Q$. Then: -$a$ is not $0$ or $-1$, $a(a+1)=7c^2$, and $b^2=a^3+8a^2+5a-1$. The second equation shows that $a$ is in the $7$-adic integers and is $0$ or $-1\pmod7$. Furthermore $a$ is a square or $7$(square) in the -$7$-adic integers. So $a$ can only be $0 \pmod 7$. But then $b^2 = -1 \pmod 7$, which can't happen. -EDIT: This is quite wrong. But I think Dror made the same errors; see the comment I -attached to his answer. -FURTHER EDIT: I'll write out a less computer-dependent version of Dror's excellent answer. -The curve Y^2=X^3-8(X^2)+5X+1 has good reduction at each prime other than 2 or 7, and -cuspidal reduction at 7. So its conductor is 49(power of 2), and the same is true of the twists: -A: 7(Y^2)=X^3-8(X^2)+5X+1 -B:-7(Y^2)=X^3-8(X^2)+5X+1 -(1)... The transformation X-->X+3 takes A to 7(Y^2)=X^3+X^2-16X-29 which is isomorphic to -Y^2=X^3+7(X^2)-794X-9947. X-->X-2 takes this to Y^2=X^3+X^2-800X-8359. So A has the -minimal Weierstrass model (0,1,0,-800,-8359), and B has the minimal model (0,-1,0,-800,8359). Looking up Cremona's tables for curves of conductor 49(a power of 2) we find that -A is isomorphic to 392b, that B is isomorphic to 784d, and neither has any finite rational -points. -(2a)... Consider the curve 7(Y^2)=X(X^3-8(X^2)+5X+1). The transformation X-->1/X,Y-->Y/X^2 - takes this to 7(Y^2)=X^3+5(X^2)-8X+1. X-->X+1 takes this in turn to 7(Y^2)=X^3+8(X^2) - +5X-1. Then X-->-X gives B. So the only finite rational point on our curve is (0,0). -(2b)... Next consider -7(Y^2)=(X-1)(X^3-8(X^2)+5X+1). X-->X+1 takes this to -7(Y^2)= X(X^3-5(X^2)-8X-1), and X-->-X takes this in turn to -7(Y^2)=X^4+5(X^3)-8(X^2)+X. Applying X-->1/X, Y-->Y/X^2 we get B again. So (1,0) is the only finite rational point on our curve. -THE PROOF: The conclusions of 1,2a and 2b give Speiser's result: there are no rational -points (x,y) on Y^2=-7(X)(X-1)(X^3-8(X^2)+5X+1) other than (0,0) and (1,0). For let (x,y) be -such a point. Then at any odd prime other than 7 the ord of x cannot be an odd number, either positive or negative, and the same holds for the ord of x-1. So each of x, x-1 and x^3-8(x^2)+5x+1 has one of the forms: square, -(square), 7(square), or -7(square). The conclusion of 1 above tells us that x^3-8(x^2)+5x+1 can only be a square or -(a square). -Suppose that x^3-8(x^2)+5x+1 is a square. Then x(x-1) is -7(square). So x is integral at -p=7 and congruent to 0 or 1 mod 7 in the 7-adics. If x were congruent to 1, x^3-8(x^2)+5x+1 -would not be a square. So x is congruent to 0, and x-1, being congruent to-1, can only be --(square). So x=7(square), and the same holds for x(x^3-8(x^2)+5x+1) contradicting the -conclusion of 2a. -Suppose finally that x^3-8(x^2)+5x+1 is -(square). Then x(x-1)=7(square). So x is integral -at p=7 and congruent to 0 or 1 mod 7 in the 7-adics. If x were congruent to 0, x^3-8(x^2)+5x+1 would not be the negative of a square. So x is congruent to 1, and can only be a square. Then -x-1 is 7(square), and (x-1)(x^3-8(x^2)+5x+1) is -7(square). This contradicts the conclusion of 2b.<|endoftext|> -TITLE: k-form: sum of wedge products of 1-forms? -QUESTION [12 upvotes]: Let M be a smooth manifold. Can every k-form $\omega$ on M be written as a sum of k-forms, that are wedge products of 1-forms, i.e. $\omega = \sum_{i=0}^n \alpha_1^{(i)} \wedge \ldots \wedge \alpha_k^{(i)} $, where $\alpha_l^{(i)} \in \Omega^1(M) $ ? If M is compact, one can cover M be finitely many charts and use a partition of unity to see that this holds...but how about the general case? - -REPLY [5 votes]: Johannes' answer can be upgraded to the following statement: -Let $M$ be a second countable and Hausdorff manifold (who cares about others?) and $\pi_i\colon E_i \longrightarrow M$ vector bundles for $i = 1, \ldots, N$. Then the canonical map -\begin{equation} - \Gamma^\infty(E_1) \otimes_{C^\infty(M)} \cdots \otimes_{C^\infty(M)} - \Gamma^\infty(E_N) - \longrightarrow - \Gamma^\infty(E_1 \otimes \cdots \otimes E_N) -\end{equation} -is an isomorphism of $C^\infty(M)$-modules. The same holds if you do some symmetrization/antisymmetrization in tensor powers of a single vector bundle in addition. -My favorite argument uses the smooth version of Serre-Swan's theorem: the sections $\Gamma^\infty(E)$ are a finitely generated and projective module over $C^\infty(M)$ and any finitely generated projective module is (up to iso) of that form. This can e.g. be proved along the same lines as Johannes' argument by embedding the tangent bundle of $E$ into some big $\mathbb{R}^n$ and observe that $E$ is naturally identifiable with the vertical subbundle of $TE$ viewed as bundle over $M$. -Then the above statement is shown by noting that the map is clearly injective (and well-defined). The surjectivity is the harder part. But the tensor product on the left hand side is again finitely generated and projective, so it has to be the space of smooth sections of some vector bundle. Then the injectivity gives that this vector bundle can be included as a subbundle of $E_1 \otimes \cdots \otimes E_N$. Counting fibre dimensions shows that they coincide... -The main point is that Serre Swan also holds in the non-compact case. An alternative proof of this can be found e.g. in Well's book in complex differential geometry.<|endoftext|> -TITLE: What about stacks of categories in algebraic geometry? II -QUESTION [7 upvotes]: I've made this a new question, rather than expanding the first one. -Torsten gives a good answer, and it partially illustrates in practice the 'second approach' I outlined in my other question. (You don't need to know the fiddly details of my other question, but you are welcome to look at it). -I'm well aware of the machinery of stacks of categories, a la Giraud, but what I want to know is: why can't these stacks be given some geometry? And not just 'because we don't'. Here's a sketch of how it might work: -Consider the 2-category of stacks of categories over some site $S$, considered as functors $X \to S$. Call a stack representable if it is representable in the usual way (equivalent to $S/a$ for some $a\in S$). Call a map of stacks $p:X\to Y$ lax-representable if for every representable stack $a$ and map $f:a \to Y$ the comma object $(f/p)$ is representable (comma objects coincide the the usual 2-pullback one uses with stacks in groupoids when all the categories involved are groupoids). -A comma object fits into a 2-commuting square, but the 2-arrow filling the square isn't necessarily invertible. We then say as per usual that a lax-representable map has property P if all the projections $(f/p) \to a$ have property P for representable $a$. Then one says a stack in categories $X$ is geometric if given a notion of cover (like smooth maps of schemes) if it admits a lax-representable cover $j:u\to X$ by a representable stack $u$ (and the diagonal $X\to X\times X$ should be lax-representable as well). Then the comma object $(j/j)$ together with $u$ should form an internal category $(j/j) \rightrightarrows u$ in $S$. -This is all well and good, but what stops this happening in practice? Perhaps lax-representable maps just do not exist? Or not enough of them? One potential barrier, in the algebraic case is that the collection of not-necessarily-invertible maps between two objects might be too big to the algebraic. Consider the automorphisms of your favourite algebro-geometric object, such as a vector bundle, or algebraic curve. Such things can again be formed into geometric objects in nice cases. But the categories of such things may not be cartesian closed. I haven't thought too much about this, but it seems to be a possible barrier. -To be concrete, my question is this: - -What goes wrong/can go wrong with the above recipe? If you like, consider a concrete example, like the stack of quasicoherent sheaves, or the stack of vector bundles (with all morphisms of vector bundles). - - -One more thing... to make this theory analogous to that of ordinary geometric stacks (in groupoids), one thing that is lacking is the notion of stackification. The stackification $st(G)$ of a algebraic groupoid $G$ is essentially the category of torsors for that groupoid. Then $G_0$ (the objects of $G$) comes with a (fairly) canonical map $G_0 \to st(G)$ which is the presentation described above. -The problem is how one wants to stackify. What is a torsor for an algebraic category $C$? If one is thinking in terms of descent, it might be a torsor for the maximal algebraic groupoid contained in $C$. The hint I am getting is that this might be the right choice, but it really depends on coming up with working examples, as in my question. - -REPLY [7 votes]: I think the right thing to do with category-valued stacks is to keep the notion of "representability" the same (that is, use (pseudo) 2-pullbacks rather than comma objects), but to replace representability of the diagonal by representability of $X^2 \to X\times X$, where $X^2$ is the power (cotensor) of X by the free-living arrow (which is equivalent to X in case X is groupoid-valued). This does imply that $(j/j)$ is representable whenever the domain of $j$ is so, since $(j/j)$ is the pullback of $X^2 \to X\times X$ along $j\times j$. -One reason I think this is the right thing is that in the well-developed theory of "indexed categories" over a topos S regarded as "large categories relative to S as a universe of sets", representability of $X^2 \to X\times X$ is equivalent to the standard notion of "local smallness". (More generally, various kinds of "comprehensibility" for indexed categories can be rephrased as the representability of certain functors in the above sense.) Therefore, the resulting notion of "geometricity" would coincide with "essential smallness", as one would expect.<|endoftext|> -TITLE: derived critical locus -QUESTION [7 upvotes]: I am looking for discussion in the literature that properly formalizes the heuristic idea that a BV-BRST complex is a model for the "derived critical locus of a function on an $\infty$-Lie algebroid". -The kind of statement that I am after would be in the following style: -Pass to the ambient $\infty$-topos of $\infty$-sheaves on the $\infty$-site of formal duals to commutative cochain dg-algebras in non-positive degree over a field of characteristic 0 (for some topology, which I think won't matter much for the following question): the context of dg-geometry. There is then a derived functor $dgAlg^{op} \to Sh_\infty(dgAlg_-^{op})$ that interprets unbounded dg-algebras as objects in this $\infty$-topos, and this I shall make use of in the following. -In there we should have a canonical morphism -$$ - \theta : \mathbb{A}^1 \to \mathbb{L}\Omega^1_K(-) -$$ -from the line to the $\infty$-sheaf of cotangent complexes, that sends over $A \in dgAlg_-$ an element $a \in Q A \simeq \mathbb{A}^1(A)$ to $d a$, for $Q A$ a cofibrant replacement. -Now consider an $\infty$-Lie algebroid, for instance as a simple standard example the homotopy quotient of a Lie algebra action on an ordinary affine, for which sugestive notation would be $X//\mathfrak{g}$. The dg-algebra corresponding to this dually is the corresponding Chevalley-Eilenberg algebra / BRST complex (in non-negative degree). Then a morphism -$$ - S : X//\mathfrak{g} \to \mathbb{A}^1 -$$ -is a $\mathfrak{g}$-invariant "action functional". The composite -$$ -d S : X//\mathfrak{g} \stackrel{S}{\to} \mathbb{A}^1 \stackrel{\theta}{\to} - \mathbb{L} \Omega^1_K(-) -$$ -would be its differential. The derived critical locus of $S$ ought to be the homotopy fiber $hofib (d S)$ (over the global point given by the 0-forms). -Is the BV-BRST complex in $dgAlg$ of the data $(X, \mathfrak{g}, S)$ a model for $hofib (d S)$ ? -Or do you know writeups of details about statements of a similar flavor? - -REPLY [2 votes]: You might look at Costello-Gwilliam book (especially 'Derived Euler-Lagrange equations' and 'Derived critical locus' - in the Appendix) -http://math.northwestern.edu/~costello/factorization_public.html (Wayback Machine)<|endoftext|> -TITLE: On the cohomology of a finite covering map -QUESTION [5 upvotes]: So let $X$ be a "nice" topological space and assume that $G$ is a finite group which acts freely on $X$. -Q: Is there a simple relationship between the cohomology groups -$H^i(G,\mathbf{Z}), H^i(X,\mathbf{Z})$ and $H^i(X/G,\mathbf{Z})$? Does the Leray -spectral sequence simplifies in this special case? - -REPLY [8 votes]: There is a precise relation at the level of complexes: $C^\ast(X,\mathbb Z)$ is a $G$-complex and as such it is perfect (that is quasi-isomorphic to a finite complex consisting of projective modules) and furthermore $C^\ast(X/G,\mathbb Z)$ is quasi-isomorphic to the derived functor value $R\Gamma(G,C^\ast(X,\mathbb Z)$. The latter is mostly used through its consequent spectral sequence $H^\ast(G,H^\ast(X,\mathbb Z))\implies H^\ast(X/G,\mathbb Z)$ but (as with all spectral sequences) it contains some ambiguity which (somehow) has to be resolved. -This is mainly going from knowledge of the cohomology of $X$ to that of $X/G$. Going the other direction is more difficult as the $G$-cohomology kills a lot of information (if one works with rational coefficients instead, it just picks out the trivial representations). However, the fact that $C^\ast(X,\mathbb Z)$ is perfect helps out even though it can still be difficult to say something. -As an example of the relevance of perfectness consider the case when $G$ is cyclic (of order $n$, say) acting by fixed point free orientation preserving maps on the $k$-sphere. Then the cohomology of $X$ is the trivial representation in degrees $0$ and $k$. Such a complex is classified (this is essentially the Yoneda Ext-description) by an element $\alpha$ in $H^{k+1}(G,\mathbb Z)$. For $C^\ast(X,\mathbb Z)$ to be perfect we must have that $\alpha$ must have order exactly $n$. This excludes $k$ even as the order then is always $1$ and for $k$ odd $\alpha$ must be a generator of $H^{k+1}(G,\mathbb Z)=\mathbb Z/n$. It is now easy to compute $R\Gamma(G,C^\ast(X,\mathbb Z)$ (and its additive structure is actually independent of $\alpha$) but we have also obtained a (necessarily) non-trivial invariant of the action. When $k=3$ this is a well-known invariant of lens spaces.<|endoftext|> -TITLE: A remark of Connes -QUESTION [33 upvotes]: In an interview (at http://www.alainconnes.org/docs/Inteng.pdf) Connes remarks that - -I had been working on non-standard analysis, but after a while I had found a catch in the theory.... The point is that as soon as you have a non-standard number, you get a non-measurable set. And in Choquet’s circle, having well studied the Polish school, we knew that every set you can name is measurable; so it seemed utterly doomed to failure to try to use non-standard analysis to do physics. - -What does he mean; what is he referring to? - -REPLY [5 votes]: A recent article by Leichtnam and myself (arxiv) in the American Mathematical Monthly contains a "theorem" to the effect that, in the presence of a construction of the hyperreals, the following is true: as soon as you have a Connes infinitesimal, you get a non-measurable set.<|endoftext|> -TITLE: FLM-like construction of VOA for other simple groups -QUESTION [5 upvotes]: Frenkel, Lepowsky, Meurman constructed the vertex operator algebra (VOA) $V^\natural$ as a chiral orbifold CFT whose target space is $\mathbb{R}^{24}/\Lambda/\mathbb{Z}_2$. (Here the last $\mathbb{Z}_2$ is a group of order two which acts by sending $\vec x\to -\vec x$, and not the ring of 2-adic integers. Sorry for using physics notation.) -This construction makes (almost) manifest that the said VOA has $2^{1+24}_{+}\cdot Co1$ as part of its symmetry group. In addition, there is an operation $\sigma$ which mixes the twisted sector and the untwisted sector. Then, adding $\sigma$ to $2^{1+24}_+\cdot Co1$ gives rise to the monster simple group $M$. Then $M$ has another involution $z$ (possibly conjugate to $\sigma$) - such that $2^{1+24}_+\cdot Co1$ is the centralizer of $z$, $C_G(z)$. -(Honestly I don't really understand the construction, but that's the story I've heard.) -Now, there are many other simple groups $G$ which has a similar structure, i.e. there is an involution $z\in G$ such that the centralizer $C_G(z)$ has the structure $2^{1+n}.H$. -Then my question is: is it always the case that there is a lattice $L$ of rank $n$ whose symmetry is $H$ (and $-1$) such that the VOA based on $\mathbb{R}^{n}/L/\mathbb{Z}_2$ has the symmetry $G$, given by adjoining an operation $\sigma$ mixing the twisted and the untwisted sector, to the part $2^{1+n}.H$ which exists almost by construction? -Update: -I realized now that Jeff Harvey already asked almost the same question a few months ago, see this MO question. I even made a comment in that thread; I completely forgot about that. You see, I'm only slowly digesting the interesting problem Jeff raised... -But of course the moderators can close my question as an exact duplicate. Sorry about that. - -REPLY [2 votes]: I have not studied FLM in complete detail, but don't think the involution $\sigma$ is sufficiently natural to make the generation of $G$ an automatic process. That is to say, even if it does always exist as an automorphism of the VOA, it may be difficult to find without explicit combinatorial information about the lattice. -Perhaps the best place to look for similar constructions is John Duncan's paper on the Conway group, where he gives an explicit construction of a $N=1$ vertex superalgebra with a sporadic automorphism group. -Slightly off-topic: There is an alternative construction of the monster VOA as a $\mathbb{Z}/3$ orbifold outlined in a 1994 conference proceedings paper by Dong and Mason (The construction of the moonshine module as a $\mathbb{Z}/p$-orbifold), and there is an analysis of the orbifold theory with no claims about the automorphism group given in a paper by Montague. The Dong-Mason paper promises the full details of proving that the automorphism group is in fact the monster simple group in a forthcoming paper that has yet to appear. I believe it also uses an involution but the calculations are apparently even more complicated than the order 2 case.<|endoftext|> -TITLE: Why do filtered colimits commute with finite limits? -QUESTION [56 upvotes]: It's not hard to show that this is true in the category Set, and proofs have been written down in many places. But all the ones I know are a bit fiddly. -Question 1: is there a soft proof of this fact? -For example, here's a soft proof of the fact that filtered colimits in Set commute with binary products. If $J$ is a filtered category, and $R,S:J\to$ Set are functors, then -$$colim_{j\in J} R(j)\times colim_{k\in J} S(k) \cong colim_{j\in J} colim_{k\in J} R(j)\times S(k)$$ -$$\cong colim_{(j,k)\in J\times J} R(j)\times S(k) \cong colim_{j\in J} R(j)\times S(j) $$ -where the first isomorphism uses the fact that Set is cartesian closed, so that the functors -$X\times-$ and $-\times X$ are cocontinuous; the second isomorphism is the "Fubini theorem"; and the third isomorphism follows from the fact that the diagonal functor $\Delta:J\to J\times J$ is final. -Is there some way to extend this to deal with equalizers and/or pullbacks? (The case of the -terminal object is easy.) -For the sort of person who'd rather just prove the fact directly (which after all is not that hard), it's worth pointing out that this proof works not just in Set but for any cartesian closed category with filtered colimits. It works without knowing how to construct colimits in Set. -So another way to ask my question might be -Question 2: what is a class of categories in which you can prove that filtered colimits commute with finite limits (without first proving that this is true in Set)? -So yes, I know that the commutativity holds in any locally finitely presentable category, but the only proofs of this I know depend on the fact that it is true in Set. - -REPLY [4 votes]: One possible answer to question 2 is "categories in which finite limits distribute over filtered colimits". A general notion of distributivity described on the nlab by Dmitri Pavlov points out that the comparison morphism for a diagram $D:I\times K \to C$ -$$f\colon {\rm colim}_K {\rm lim}_I D \to {\rm lim}_I {\rm colim}_K D$$ -whose invertibility we describe by "$I$-limits commute with $K$-colimits", factors as a composite -$${\rm colim}_K {\rm lim}_I D \xrightarrow{g} {\rm colim}_{K^I} {\rm lim}_I D' \xrightarrow{h} {\rm lim}_I {\rm colim}_K D.$$ -where $K^I$ is the functor category and $D'\colon I\times K^I\to C$ is obtained from $D$ by precomposition with the functor $(\pi_1,{\rm ev}) : I\times K^I \to I\times K$ that sends a pair $(i,s)$ to $(i,s(i))$. The map $g$ is induced by the diagonal functor $K\to K^I$, while $h$ is defined by universal properties. -Thus, if the diagonal $K\to K^I$ is final, it follows that $f$ is an isomorphism if and only if $g$ is. The latter property is described as "$I$-limits distribute over $K$-colimits", and if $I$ is discrete it's easily equivalent to the usual notion of products distributing over colimits. -However, if $I$ is not discrete, it's not clear to me how to reformulate this distributivity condition in a more explicit way so as to deduce it from something like local cartesian closure, or even to prove directly that it holds in $\rm Set$. But maybe someone else sees how.<|endoftext|> -TITLE: Subgroups of p-groups -QUESTION [6 upvotes]: If $G$ is a (non-abelian) $p$-group, $|G|=p^n$, $n>3$, then it is elementary that $G$ contains a (normal) abelian subgroup of order $p^2$. It is also true that $G$ necessarily contains a normal abelian subgroup of order $p^3$ (Group Theory - W. R. Scott). -1) What is the largest possible value of $m$ such that any non-abelian group of order $p^n$ contains a normal abelian subgroup of order $p^m$? -2) What is the largest possible value of $m$ such that any non-abelian group of order $p^n$ contains an abelian subgroup of order $p^m$? -[Please suggest references.] - -REPLY [2 votes]: A. Yu. Olshanskii proved in 1978 -that, for any $n$ and any prime $p$, -there exists a $p$-group of order at least -$ -p^{{1\over 8}(n^2+4n-8)} -$ -having no abelian subgroups of order large than $p^n$. -Together with Miller's estimate $p^{{1\over2}n(n+1)}$ (see @m_t's answer), this gives quadratic upper and lower bounds. - -Similar questions for commutative subalgebras in (associative and Lie) algebras was studied by M.V.Milentyeva. The estimates are also quadratic in this case.<|endoftext|> -TITLE: Riemannian manifold of bounded geometry has a normal bundle of bounded geometry -QUESTION [13 upvotes]: Hi, -I asked this question already on math.stackexchange but got no answer (link: https://math.stackexchange.com/questions/22155). -Our setting: An Euclidean vector bundle $(E, h, \nabla^E)$ over a Riemannian manifold (M,g) is said to have bounded geometry, if the norms of the curvature tensor $R^E$ and of all its covariant derivatives are bounded. The manifold itself is said to have bounded geometry, if the tangent bundle TM, equipped with the manifold metric and the Levi-Civita connection, has bounded geometry and additionally the metric is complete and the injectivity radius fulfills $\operatorname{inj rad}(x) > \epsilon > 0$ for all x. -The question: We have a Riemannian manifold (M,g) of bounded geometry and some isometric embedding $\iota\colon M \to R^N$. Now we can look at the normal bundle NM over M, equipped with the pull-back metric and pull-back connection. Has this bundle bounded geometry? My intuition says "yes". -I tried it with local computations using the corresponding projection matrices but got nowhere. -I use the fact that a manifold has bounded geometry, if and only if the Christoffel symbols of the Levi-Civita connection and all their derivatives are uniformly bounded functions when computed in Riemannian normal coordinates (where the radii of the coordinate balls are the same for all points p). An analogous statement holds for vector bundles of bounded geometry, where the frames we use for the computation of the Christoffel symbols are acquired by choosing a orthonormal basis for the bundle in the point p and then parallel translate it along the radial geodesics in a normal coordinate ball (also with fixed radius for every point). -So if $\partial_{x_i}$ are the normal coordinates and $\{n_i\}$ is the orthonormal frame for the normal bundle we have the following expression: $\Gamma_{ij}^{k, TM} = g^{kl}\langle \nabla_{\partial_{x_i}} \partial_{x_j}, \partial_{x_l}\rangle$ and analogously $\Gamma_{ij}^{k, NM} = h^{kl}\langle \nabla_{\partial_{x_i}} n_j, n_l\rangle$, where $g^{ij}$ is as usually the inverse matrix of the matrix of the metric g (computed w.r.t. the normal coordinates), $h_{ij}$ the matrix of the metric of the normal bundle and $\langle \cdot, \cdot \rangle$ is the Euclidean metric of $R^N$ (we pushed the coordinates $\partial_{x_i}$ and the frame $\{n_i\}$ forward via the embedding $M \to R^N$). Since the frame we use for the normal bundle is orthonormal, we have $h_{ij}=\delta_{ij}$ and so the formula for the Christoffel symbols of the normal bundle reduces to $\Gamma_{ij}^{k, NM} = \langle \nabla_{\partial_{x_i}} n_j, n_k\rangle$. -For the matrices of the projections $p^{TM}: TR^N \to TM$, resp. $p^{NM}$ we get the following expressions w.r.t. the standard coordinates $\{e_i\}$ of $R^N$: $(p^{TM})_{ij} = g^{kl}\langle e_j, \partial_{x_l}\rangle \langle e_i, \partial_{x_k} \rangle$ and analogously $(p^{NM})_{ij} = h^{kl}\langle e_j, n_l\rangle \langle e_i, n_k \rangle$. -Now I want to deduce that if the Christoffel symbols of TM and all their derivates are uniformly bounded (i.e. the manifold has bounded geometry), then the entries of the projection matrix $p^{TM}$ and all their derivatives are uniformly bounded (which automatically gives the uniform boundedness of the entries of $p^{NM}$ and their derivatives). And from here I want to deduce the uniform boundedness of the Christoffel symbols of NM and all their derivatives. But I do not see how using the equations I got so far. -May we get further equation / information which give the desired results? Or maybe there is some other way to answer the question posed in the third paragraph (not using ugly local computation)? I would be happy with any solution. -Thanks, -Alex - -REPLY [17 votes]: Interesting question. The answer is no: surfaces with bounded geometry can have normal -bundles with unbounded curvature. -To set the stage, it's worth first noting that you can have a surface with extreme geometry isometrically embedded in $\mathbb E^3$, where the normal bundle, being one-dimensional, has a trivial connection; or include this into $\mathbb E^4$ (Euclidean 4-space) where the -normal bundle is 2-dimensional, but the curvature is still 0. This at least illustrates that bounded geometry of the normal bundle and tangent bundle are decoupled. -I'll now describe an isometric embedding of the $\mathbb E^2$ - into $\mathbb E^6$ where the connection -on the normal bundle has unbounded curvature. The embedding have local 1-parameter groups of symmetry, which makes it easier to keep track of curvature without needing to write down equations. -Start by visualizing - a helical curve in $\mathbb E^3$. The tangent vector to a helix goes repeatedly around a circle in its spherical image. The connection on the tangent bundle is induced by this -Gauss map from the connection on $S^2$, so the parallel translation of the normal bundle rotates the plane by an angle equal to the area enclosed inside this circle once every coil of the helix. -Now consider a similar curve in $\mathbb E^5$, thought of as $\mathbb E \times \mathbb E^2 \times \mathbb E^2$. In the $\mathbb E$-direction, the curve makes uniform progress, while going around circles of possibly different radii at possibly different rates in the $\mathbb E^2$ directions. If the term weren't otherwise engaged, one could call this a double helix. -It is invariant by a 1-parameter group of isometries of $\mathbb E^5$ that translates in -the $\mathbb E$ direction while spinning the two perpendicular planes at their own rates. -The normal bundle splits into two $\mathbb E^2$ subbundles, its intersection with -the two 3-dimensional $\mathbb E \times \mathbb E^2$'s. The connection preserves this splitting, rotating the two $\mathbb E^2$'s indendently. -Now add an extra "parameter" dimension, making the ambient space $\mathbb E^6$. -Modify the curve in $\mathbb E^5$ by increasing the radius of one helix while decreasing the -radius of the other, balancing the changes so the curve remains invariant by the same -1-parameter group, and its arc length remains constant (as measured by the time parameter of the 1-parameter group). It's easy to see, since it's -locally isometric to a surface of revolution because of the symmetry, that the -resulting surface is isometric to $\mathbb E^2$. We can make the circle in on -$\mathbb E^2$ go all the way to 0. After making sure it has $C^\infty$ contact to -a straight line in this projection, we can then start making this projection helical -again, but with a -steeper and tighter helix, adjusting by letting the other helical projection shrink to a line. The local symmetry group in $\mathbb E^6$ has changed, but the induced symmetry on - the surface remains the same. -We can go back and forth, alternating between helical effects in the two factors, -inexorably tightening the screws without distoring the surface. -The curvature of the induced connection on the normal bundle becomes arbitrarily high, -as you can see by following the connection around a small rectangle, with two edges -in the "parameter" direction, one edge where say the first helix has become straight -and the fourth edge where the first helix is wound in very small tight coils.<|endoftext|> -TITLE: Compatibility of Bloch-Kato and Beilinson-Bloch -QUESTION [12 upvotes]: Suppose $V/K$ is a smooth projective variety. Let $\mathrm{Ch}^{j}(V)_{0}$ -be the group of codimension-$j$ homologically trivial $K$-rational cycles on $V$, modulo rational equivalence. A conjecture of Beilinson and Bloch predicts that the dimension of $\mathrm{Ch}^j(V)_{0} \otimes \mathbf{Q}$ as a $\mathbf{Q}$-vector space is given by the order of vanishing of the L-function $L(s,H^{2j-1}_{\mathrm{et}}(V \times_{K} \overline{K},\mathbf{Q}_{\ell}))$ at its central critical point. On the other hand, the Bloch-Kato conjecture predicts that this order of vanishing is equal to the dimension of the Bloch-Kato Selmer group $H^1_f(G_K,H^{2j-1}_{\mathrm{et}}(V\times_{K}\overline{K},\mathbf{Q}_{\ell})(j))$. So it seems reasonable to ask: is there a natural map -$\phi_j : \mathrm{Ch}^j(V)_{0} \to H^{1}_{f}(G_K,H^{2j-1}_{\mathrm{et}}(V\times_{K}\overline{K},\mathbf{Q}_{\ell})(j))$ ` -which is "close" to being an isomorphism, which explains the compatibility of these conjectures? - -REPLY [9 votes]: The Bloch-Kato conjecture is actually more precise than that. As mentioned by Hunter Brooks, there is indeed an $\ell$-adic Abel-Jacobi map -$$\phi : Ch^j(V)_0 \to H^1(G_K,H^{2j-1}_{\mathrm{et}}(V\times_{K}\overline{K},\mathbf{Q}_{\ell})(j))$$ -The map $\phi$ is also called the cycle class map and is defined for any field $K$, say of characteristic $0$. -Now if $K$ is a number field, the conjecture of Bloch-Kato predicts that -(1) $\phi$ takes values in $H^{1}_f(G_K,\cdot)$. -(2) $\phi \otimes \mathbf{Q}_\ell$ is an isomorphism. -In fact (1) is a purely local question : this is really a question about the Abel-Jacobi map associated to a variety over $\mathbf{Q}_p$ (where $p$ can be equal to $\ell$ or not). The conjecture (2) together with the Beilinson-Bloch conjecture implies the statement you mention about the order of vanishing. -You can find a good survey on the $\ell$-adic Abel-Jacobi map and more details about what is known here : -J. Nekovar, p-adic Abel-Jacobi maps and p-adic heights, In: The Arithmetic and Geometry of Algebraic Cycles (Banff, Canada, 1998), 367 - 379, CRM Proc. and Lect. Notes 24, Amer. Math. Soc., Providence, RI, 2000. -See also the "Conjecture $\mathrm{Mot}_\ell$" in the following article : -M. Flach, The Equivariant Tamagawa Number Conjecture : A Survey.<|endoftext|> -TITLE: Picard group of reducible varieties -QUESTION [11 upvotes]: What's the strategy for computing the Picard group of a variety with more than one irreducible components? -For instance, consider the simple case where $X$ has two components $C$ and $D$, meeting transversely at one point. Then it seems that $\text{Pic}(X)=\text{Pic}(C)\times\text{Pic}(D),$ but I don't know how to prove it. -Thanks. -Edit: I'd like to see a formal proof (or a reference), i.e. using cohomology etc., instead of using "gluing", especially when we are not gluing along open intersections. The map $C\coprod D\to X$ is not flat, so there's no fppf gluing either. - -REPLY [16 votes]: I will only consider the case of connected projective (this is not really necessary) curves $X, C, D$. over an algebraically closed field $k$. The canonical injection $O_X\to O_C\times O_D$ induces an exact sequence of sheaves on $X$ -$$ 1 \to O_X^* \to O_C^* \times O_D^* \to F \to 1 $$ -where $F$ is a skyscrapper sheaf supported at the intersection points of $C$ and $D$. Passing to cohomology, -we get -$$ 1 \to k^* \to k^* \times k^* \to F(X) \to \mathrm{Pic}(X)\to \mathrm{Pic}(C) \times \mathrm{Pic}(D)\to H^1(X, F)=0.$$ -When $C, D$ intersect transversally at a single point (ordinary double point), a local computation shows that $k^* \times k^*\to F(X)$ is surjective, and you get your isomorphism. -As Steven said, this really depends on how $C$ and $D$ intersect (in fact, when $C$ and $D$ are smooth, your isomorphism implies that $C$ and $D$ intersect transversally at a single point; non transversal intersection point can give additive subgroup in $\mathrm{Pic}(X)$ and more transversally intersection points give subtori in $\mathrm{Pic}(X)$). Also in higher dimension $H^1(X, F)$ may not vanish, and the above methode does not work. -The general picture for proper curves can be found in Bosch, Lütkebohmert and Raynaud, Néron Models, Chap. 9., §2. They make a ''dévissage'' of $\mathrm{Pic}^0(X)$. -[EDIT] Let me rewrite Sándor's nice interpretation in cohomological terms. It will make explicit the sheaf $F$ above and gives a better understanding of what is going on, and in any dimension. Denote by $E=C\cap D$ the closed subscheme defined by the ideal $J_C+J_D$. Then we have an exact sequence of sheave on $X$: -$$ 1 \to O_X^* \to O_C^* \times O_D^* \to O^*_{E} \to 1 $$ -in the middle, the map is $(a, b)\mapsto a|_E (b|_E)^{-1}$. The exactness is checked locally. Passing to cohomology, we get -$$ O(C)^\star\times O(D)^\star\to O(E)^* \to \mathrm{Pic}(X) \to \mathrm{Pic}(C)\times \mathrm{Pic}(D)\to \mathrm{Pic}(E)$$ -and the last map is $(L, H)\mapsto L_{|E}\otimes (H_{|E})^{-1}$, therefore the exact sequence becomes -$$ O(C)^\star\times O(D)^\star\to O(E)^\star \to \mathrm{Pic}(X) \to \mathrm{Pic}(C)\times_{\mathrm{Pic}(E)} \mathrm{Pic}(D)\to 1.$$ -So Sándor's map $\mathrm{Pic}(X) \to \mathrm{Pic}(C)\times_{\mathrm{Pic}(E)} \mathrm{Pic}(D)$ is always surjective. It is injective if and only if $O(C)^\star\times O(D)^\star\to O(E)^\star$ is surjective. This is not always the case (consider two irreducible curves meeting at more than one point or meeting at a single point but not transversally which implies that $E$ is a non-reduced point), but is true if for instance $O(E)=k$ (e.g. $E$ is geometrically connected, geometrically reduced and proper). -[EDIT 2] Of course, in all this answer, $X$ is supposed to be reduced. Otherwise $O_X\to O_C\times O_D$ (and that one with invertible functions) would not be necessarily injective. If $X$ is not reduced, there is a dévissage from $\mathrm{Pic}(X)$ to $\mathrm{Pic}(X_{\mathrm{red}})$ in Bosch & al, op. cit.<|endoftext|> -TITLE: Which finite groups have faithful complex irreducible representations? -QUESTION [24 upvotes]: Obvious necessary condition is that the center must be a cyclic group. Is it sufficient (doubt here)? If not, is there any nice characterization in terms of group structure, without appealing to representations? - -REPLY [11 votes]: I thought I'd add a specific example of a finite group with cyclic centre (trivial, in fact), yet no faithful irreducible complex representation (the example is from problem 2.19 of Isaacs' Character theory of finite groups, MR460423). -The group $(C_2)^4\rtimes C_3$, where $C_n$ denotes the cyclic group of order $n$ and $C_3=\langle \sigma\rangle$ acts on $(C_2)^4=\langle \tau_1,\tau_2,\tau_3,\tau_4\rangle$ via -$$\begin{align*} - \sigma\cdot\tau_1=\tau_2 \hspace{0.5in}&\sigma\cdot\tau_2=\tau_1\tau_2 - \newline \sigma\cdot\tau_3=\tau_4 \hspace{0.5in}&\sigma\cdot\tau_4=\tau_3\tau_4. -\end{align*}$$<|endoftext|> -TITLE: Are generalized cohomology theories a set of complete homotopy invariants for spaces ? -QUESTION [9 upvotes]: In the same vein of this MO question, one can ask: - -If two spaces $X$, $Y$ have isomorphic generalized cohomology rings $\mathrm{h}^{\bullet}(X)\cong \mathrm{h}^{\bullet}(Y)$ for every multiplicative generalized cohomology theory $\mathrm{h}^{\bullet}$, do they have to be homotopically equivalent? -What if we also take cohomology operations into account? - -(It's just a spontaneous question that popped in my mind, as a non expert in topology) - -REPLY [15 votes]: No, they don't have to be homotopically equivalent. In fact: - -There is a map of CW-complexes $X \to Y$ which is an isomorphism on (co)homology, full stop, for every generalized (co)homology theory $h$, multiplicative or not. -This is, in fact, equivalent to the map $X \to Y$ being an isomorphism on integral homology. -Systematic examples of such maps are given by plus constructions. -However, if the spaces involved are connected and simply connected, such a map must be a homotopy equivalence by the homology Whitehead theorem. - -This leads to a more subtle question, which is: - -Are there two simply-connected spaces whose cohomology rings are isomorphic for all generalized cohomology theories $h$, but which are not homotopy equivalent? - -Then the answer is yes. The spaces $S^3 \vee S^5$ and $S^1 \wedge {\mathbb{CP}}^2$ have the same cohomology rings for all generalized cohomology theories, but they're not homotopy equivalent. This is the suspension of Neil Strickland's example in the other question, which has the effect of killing all the multiplication in the cohomology ring. -EDIT: In the new version, where cohomology operations are allowed, the question becomes more difficult because you're moving in the right direction: attaching more data like cohomology operations (the next to allow would be secondary operations). I think this example covers that case: you have two spaces, $S^3 \vee S^3 \vee S^8$ and another one formed by some Whitehead-product cell attachment, whose cohomology rings both have trivial multiplication. The cohomology operations on the Whitehead-product space are all trivial because the 8-cell is attached to both copies of $S^3$, but not to either one individually: specifically, if you collapse down either copy of $S^3$ to a point you're left with something homotopy equivalent to $S^3 \vee S^8$. - -REPLY [11 votes]: No – Knot complements provide a counterexample. The cohomology is that of $S^1$ and hence, by the Atiyah-Hirzebruch spectral sequence, the same holds for any generalized cohomology theory. Also there cannot be any interesting products. On the other hand, knot complements determine knot types by the Gordon-Luecke theorem, and there are certainly knots different from the unknot.<|endoftext|> -TITLE: Is the deformation limit of Ricci-flat Kahler manifolds Kahler? -QUESTION [9 upvotes]: Let $X$ be a compact complex Kahler manifold with first real Chern class $c_1 = 0$. Consider a family $\pi : \mathcal X \to \Delta$ over the unit disc in $\mathbb C$, where the fibers $X_s$ are compact Kahler with $c_1(X_s) = 0$ for $s \not= 0$. Do we know that the central fiber $X_0$ is Kahler with $c_1 = 0$? -Some remarks: -a) The condition on the Chern class is topological, so the question is really if the central fiber is Kahler. -b) This is true for complex tori and K3 surfaces, though for K3 surfaces being Kahler is a consequence of the topological condition of having even first Betti number. -c) Kuranishi constructed an example of non-Kahler deformations of projective manifolds, but the manifolds in question were not Kahler-Einstein so that example does not apply here. -d) There exist non-Kahler compact complex manifolds with $c_1 = 0$, like the Iwasawa manifold. The Iwasawa manifold does not have the right first Betti number to provide a counterexample to the question. However, I've heard physicists have found many examples of non-Kahler manifolds of Calabi-Yau type, and maybe one of those does at least not have topological obstructions to being a counterexample? -[edit] Two more remarks: -e) For the special case of Calabi-Yau manifolds, Popovici gives that the central fiber is Moishezon. One could hope that a Moishezon manifold with the Hodge numbers of a Calabi-Yau manifold is Kahler, but this is false by an example of Oguiso. His example is however not homeomorphic to a projective manifold, leaving the question open. -f) One could try looking at Calabi-Yau threefolds, of which many examples are apparently known (I only know of complete intersections of appropriate degree in projective space). The case of a Calabi-Yau hypersurface in $\mathbb P^4$ is uninteresting, as they are rigid, so the central fiber is isomorphic to the general fiber. - -REPLY [7 votes]: Let $\pi:X\to Y$ be a flat projective family of n-dimensional varieties with central fibre $X_0$ be Calabi-Yau variety with canonical singularities then all general fibres $X_y$ are Calabi-Yau varieties with at worst canonical singularites. See Fibration when central fibre is a Calabi-Yau variety with canonical singularities -The inverse of this statement also holds true with some additional assumption. -Let $\pi:(X,D)\to Y$ be a flat projective family of n-dimensional varieties with $K_{X/Y}+D\cong \mathcal O_X(D)$ (i.e. fibres are log Calabi-Yau varieties) and $(X,(X_0,D_0))$ be divisorially log terminal then the log Calabi-Yau central fibre $(X_0,D_0)$ has at worst canonical singularities if and only if there exists an irreducible component $N$ of $X_0$ with a resolution of singularities $\tilde N\to N$ with $H^0(\tilde N,K_{\tilde N})\neq 0$. In fact, if $f:(\tilde X_0,\tilde D_0)\to (X_0,D_0)$ is a resolution of singularities, then we have -$$K_{\tilde X_0}+\tilde D_0\thicksim \sum_{E}a_EE$$ with $a_E\geq 0$, it follow that $K_{\tilde X_0}+\tilde D_0$ is effective. Now it is not very hard by the Idea of Wang to show that the converse is also holds true and we just need to know $K_0+D_0$ is irreducible and trivial. -By the results of Shigeharu Takayama, Wang and V.Tosatti if $0$ lies at finite Weil-Petersson distance then central fibre $X_0$ is Calabi-Yau variety with canonical singularities. -Note that for polarized Calabi-Yau degeneration $f:X\to \mathbb D$, the central fobre $X_0$ is Calabi-Yau variety with canonical singularities if and only if the Tian's Kahler potential of Weil-Petersson metric be upper bounded. -$$\omega_{WP}=-\sqrt[]{-1}\partial_y\bar\partial_y\log\int_{X_y}|\Omega_y|^2$$<|endoftext|> -TITLE: What are the algebras over $\Omega^k\Sigma^k$ ? -QUESTION [6 upvotes]: Let $Ho(Spc)$ be the homotopy category of spaces. There is an adjoint pair -$$ -\Sigma^k \colon Ho(Spc) \leftrightarrows Ho(Spc)\colon \Omega^k, -$$ -where $\Sigma^k$ is the $k$-th supension functor and $\Omega^k$ is the $k$-fold loop space functor ($k\ge 1)$. This adjunction has as associated monad the functor $\Omega^k\Sigma^k$. My question is what are the algebras over this monad (a precise reference to this fact would be welcome). -If we denote by $Ho^s$ the homotopy category of spectra, then there is another adjunction -$$ -\Sigma^{\infty} \colon Ho(Spc) \leftrightarrows Ho^s\colon \Omega^{\infty}, -$$ -where $\Sigma^{\infty}$ is the suspension spectrum functor. My question is again what are the algebras over the monad $\Omega^{\infty}\Sigma^{\infty}$. - -REPLY [15 votes]: I know interesting answers to two questions that are not the same as the one asked, but are related. - -Consider the monad $T=\Omega^\infty L_{K(n)}\Sigma^\infty$ on the homotopy category of spaces. It is straightforward to construct a functor $$\Omega^\infty:Ho(\{K(n)-\text{local spectra}\})\to \{T-\text{algebras}\}.$$ One can show using the Bousfield-Kuhn functor and related ideas that this is actually an equivalence. -Consider the monad $Q=\Omega^\infty\Sigma^\infty$ on the category of based spaces (not up to homotopy). If we use spectra in the sense of Lewis and May, there is an evident functor $\Omega^\infty:\{\text{spectra}\}\to\{Q-\text{algebras}\}$. This is actually full and faithful (even on spectra whose homotopy groups are concentrated in negative degrees), which means that the point-set level $Q$-action carries a lot more information than you might naively guess. The key point in the proof is that we can use a trick with the Hopf map and space-filling curves to express $S^2$ as the coequaliser of two based maps from $S^3$ to $S^3$. This gives a natural way to express $\Omega^2X$ as the equaliser of two maps from $\Omega^3X$ to $\Omega^3X$, which allows us to do a bunch of things by induction.<|endoftext|> -TITLE: Coproducts of schemes ("gluing construction") ? -QUESTION [5 upvotes]: In this MO question it was raised the topic of "gluing constructions" in the category of schemes. I understand the phrase "gluing two schemes along maps to them" as "there exists a coproduct of the two schemes (with respect to the two given morphisms) in the category of schemes". -Let's consider the affines first. If $R'$, $R''$ and $R$ are rings, and $\phi': R' \to R$ and $\phi'':R''\to R$ are homomorphisms, then one can define the ring -$A=R'\times_{\phi',R,\phi''} R'':=\;${$(a,b)\in R'\times R''$ | $\phi'(a)=\phi''(b)$}. -The first question is: - -Is the ring $A$ so constructed always the fibered product in the category of rings of $R'$ and $R''$ along the prescribed maps $\phi'$ and $\phi''$ ? (I guess this may be answered by abstract nonsense alone) - -In case the answer to the above question is "yes", then one automatically gets the existence of fibered co-products (i.e. verifying the dual universal property than fibered products) in the category of affine schemes. -So one may ask: - -Under which assumptions does it carry over to the non-affine case? - -REPLY [4 votes]: Gluing produces a coproduct only when you glue along the empty subscheme. In general, if we glue $X$ and $Y$ along maps $U \to X , U \to Y$ together, we want to get the pushout (also called amalgamated sum) of the diagramm $X \leftarrow U \to Y$. -The answer to the first question is, of course, yes. You can verify the universal property directly. More generally, every limit can be constructed via products and equalizers, which is a fact from basic category theory. In particular, a fiber product of the diagram $A \to C \leftarrow B$ is equal to the equalizer of $A \times B \to A \to C$ and $A \times B \to B \to C$. -Also Sándor Kovács has already pointed you to Karl Schwede's paper in which the second question is answered, let me just say that this article shows that pushouts along closed immersions exist and are given locally by the fiber product of rings. But in general, pushouts of schemes do not exist at all. See this question.<|endoftext|> -TITLE: Computational software in Algebraic Topology? -QUESTION [30 upvotes]: I was wondering if there is any good software out there that allows you to do specific computations in algebraic topology. For example: - -Create a simplicial complex/set and ask questions about its homology, cohomology; -Build manifolds using handle decompositions; -Calculate homotopy limits, colimits. - -Something quite flexible and robust in the vein of MAGMA -Thank you. - -REPLY [2 votes]: Although it might now be exactly what you are looking for (e.g. lack of homotopy-theoretic constructions), but there is a nice computational package called javaPlex that "implements persistent homology and related techniques from computational and applied topology, in a library designed for ease of use, ease of access from Matlab and java-based systems, and ease of extensions for further research projects and approaches." -JavaPlex allows straightforward construction of chain complexes and things like the Mayer-Vietoris sequences, as well as computational techniques for persistent homology. -Link: http://git.appliedtopology.org/javaplex/<|endoftext|> -TITLE: Categories of descent data -QUESTION [5 upvotes]: Let us work over the etale site $\mbox{Aff}/S$ (for the sake of definiteness) for some fixed base scheme $S$, where the covers are jointly surjective etale maps $\{ U_i \rightarrow U\}_{i\in I}$ (and $I$ is finite if you like). Let us also consider a prestack $F$ fibred in groupoids. Recall the definition of a category of descent data $F(\{ U_i \rightarrow U\}_{i \in I})$ associated to a cover $\{ U_i \rightarrow U\}_{i \in I}$. The objects of this category are collections of elements $\xi_i\in F(U_i)$ together with morphisms $\phi_{ij}$ between their appropriate pullbacks satisfying the cocycle condition. This definition is the one appearing on p. 15 of "Champs algebriques" by Laumon & Moret-Bailly (among other sources, e.g., Vistoli's notes in "FGA explained"). -However, one can also consider coverings $U^\prime \rightarrow U$ consisting of one element only. In $\mbox{Aff}/S$ starting with any covering one can obtain one of such form by taking $U^\prime := \bigsqcup U_i$. However, it is not clear to me how this passage to a cover with a single morphism interacts with the associated categories of descent data. I suspect, they should be equivalent (in "Champs algebriques" for instance, the authors switch to the latter when exhibiting the stackification of a prestack in (3.2)) but on the other hand I don't see how is one supposed to get a single $\xi \in F(U^\prime)$ starting off with the $\xi_i$ as above, let alone an equivalence of categories between $F(\{ U_i \rightarrow U\}_{i \in I})$ and $F(\{ U^\prime \rightarrow U\})$. Are these categories equivalent and what is a functor exhibiting this equivalence? And if not, why is one allowed to consider only coverings of the form $U^\prime \rightarrow U$ when constructing the stackification? - -REPLY [7 votes]: This mistake seems to be made all over the place; see also this question. As David has said, the context in which this is usually done is an extensive category with a topology which includes the extensive topology (whose covering families are those of the form $(U_i \to \coprod_i U_i)$). -However, even in this case, the categories of descent data for a covering family and for the associated single cover are only equivalent if you already know that your prestack is a stack for the extensive topology. -At the nLab page on superextensive sites there is a proof that if you have a presheaf that is a sheaf for the extensive topology, and you sheafify it with respect to the singleton covers, then you get a sheaf with respect to all the covering families. I expect that this generalizes to stacks as well. But if your presheaf is not a sheaf for the coproduct families, then sheafifying it for singleton covers is not sufficient. -I'm not necessarily saying that any particular reference says something wrong; I don't have "Champs algebriques" in front of me so I can't check whether they have prestacks that are extensive-stacks already or are otherwise avoiding the issue. But in general it is something you have to worry about.<|endoftext|> -TITLE: How to compute the cohomology of the general linear group with integral entries -QUESTION [14 upvotes]: Q: So how does one compute the cohomology groups $H^*(GL_n(\mathbf{Z}),\mathbf{Z})$? -First note that $H^*(GL_n(\mathbf{Z}),\mathbf{Z})$ is isomorphic to $H_B^*(Y/GL_n(\mathbf{Z}),\mathbf{Z})$ (Betti cohomology) where $Y$ is any contractible space on which $GL_n(\mathbf{Z})$ acts freely. Maybe one should first ask to compute the cohomology -with rational coefficients and then deal with the torsion separately. -Secondly, note that $GL_n(\mathbf{Z})$ acts on $\mathbf{R}^n-\{0\}$. Unfortunately it does not act discontinuously on $\mathbf{R}^n-\{0\}$ so its quotient by $GL_n(\mathbf{Z})$ will be quite messy. Nevertheless it might be possible to use some version of the Leray spectral sequence on -$$ -G\rightarrow E\rightarrow E/G -$$ -where $G=GL_n(\mathbf{Z})$, $E=\mathbf{R}^n-0$. -By the way, does $E/G$ have a geometrical description? - -REPLY [2 votes]: The quotient $E/G$ is non-Hausdorff, I'm not sure there will be a nice geometric description. -There's a standard way to get $Y$. The symmetric space for $GL(n,\mathbb{R})$ is the symmetric space $Q$ of positive definite symmetric matrices of determinant $>0$, isomorphic to $GL(n,\mathbb{R})/O(n,\mathbb{R})$. Then $GL(n, \mathbb{Z})$ acts discretely on this space, but torsion elements have fixed points. Also, the torsion elements of $GL(n,\mathbb{Z})$ map non-trivially to $GL(n,\mathbb{Z}/p)$ for some prime $p$. One may take a $K(GL(n,\mathbb{Z}/p),1)=X$, then $GL(n,\mathbb{Z}/p)$ and therefore $GL(n,\mathbb{Z})$ acts on the universal cover $\tilde{X}$. Now, take the diagonal action of $GL(n,\mathbb{Z})$ on $Q\times \tilde{X}$. This action is free and discrete. Of course, this assumes that you have a nice way to construct $X$, which must be infinite dimensional!<|endoftext|> -TITLE: Filtrations generated by cadlag martingales. -QUESTION [8 upvotes]: Let $(\Omega,P,\mathcal{F})$ be a probability space with filtration $\mathbb{F} = (\mathcal{F}_t), t \in [0,T]$, where $T$ can be finite or infinite. Let $M$ be a cadlag (local) martingale with respect to $\mathbb{F}$, and let $\mathbb{F}^M$ be the filtration generated by $M$ and then completed with respect to $P$. -Question: Is $\mathbb{F}^M$ a right-continuous filtration? -Some facts: - -If $X$ is a strong markov process, then the completion of $\mathbb{F}^X$ is right-continuous. This is in Karatzas and Shreve. -A sort of converse: If $M$ is a local martingale in a right continuous and complete filtration, it has a right continuous modification. - -One possible idea: A continuous local martingale can be expressed as a time-changed Brownian Motion, which is strong markov. - -REPLY [11 votes]: No, that is not true. Consider the following, defined on a filtered probability space $(\Omega,\mathcal{F},\{\mathcal{F}_t\}\_{t\in[0,T]},\mathbb{P})$. - -$W$ is a standard Brownian motion. -$U$ is an $\mathcal{F}_0$-measurable Bernoulli random variable independent of $W$, with $\mathbb{P}(U=0)=\mathbb{P}(U=1)=1/2$. - -Then, set $M_t=UW_t$. This is a continuous martingale. If $\mathcal{F}^M_t$ is its completed natural filtration then $U$ is $\mathcal{F}^M_t$-measurable for all $t > 0$. Then, $U$ is $\mathcal{F}^M_{0+}$-measurable but is not measurable with respect to $\mathcal{F}^M_0$ (which only contains sets with probability 0 and 1). So $\mathcal{F}^M_{0+}\not=\mathcal{F}^M_0$. -Also, this is essentially the same as the example I gave in a previous answer of a Markov process which is not strong Markov. - -As another example to show that there is not really any simple way you can modify the question to get an affirmative answer, consider the following; a Brownian motion $W$ and left-continuous, positive, and locally bounded adapted process $H$. Then, $M=H_0+\int H\\,dW$ is a local martingale. Also, $M$ has quadratic variation $[M]=\int H^2_t\\,dt$ which has left-derivative $H^2$ for all $t > 0$. So, $H_t$ is $\mathcal{F}^M_t$-measurable, as is $W_t=\int H^{-1}\\,dM$. In fact, $\mathbb{F}^M$ is the completed natural filtration generated by $W$ and $H$. If $H$ is taken to be independent of $W$, then $\mathbb{F}^M$ will only be right-continuous if $\mathbb{F}^H$ is, and it easy to pick left-continuous processes whose completed natural filtration fails to be right-continuous.<|endoftext|> -TITLE: Nilpotent matrices related to Lie algebras of special orthogonal groups in characteristic 0 -QUESTION [6 upvotes]: In terms of matrix theory, the question I'm led to is the following: Start with an $n$-dimensional vector space over an algebraically closed field of characteristic 0 such as $\mathbb{C}$, which has a non-degenerate symmetric bilinear form. Consider all $n \times n$ skew-adjoint matrices $A$ relative to this form. - -Given $n\geq 5$, what is the least power $k$ for which all nilpotent matrices of this type satisfy $A^k=0$? - -For any nilpotent $n \times n$ matrix $A$, it follows from the Cayley-Hamilton Theorem that $A^n=0$. But in the special case here it seems plausible to expect a slightly smaller minimum: namely, $n-1$ if $n$ is odd and $n-2$ if $n$ is even. I also wonder what is written down in the literature along this line. -There is of course a hidden agenda, relative to simple Lie algebras attached to special orthogonal groups over a field like $\mathbb{C}$. In the classification of simple types $A_\ell-D_\ell$ of rank $\ell$, the respective Coxeter numbers are $\ell+1, 2\ell, 2\ell, 2(\ell-1)$. (Types $B, C$ share the same Weyl group.) Types $A, C$ are realized naturally as $n\times n$ matrices with $n=h$, but the other Lie algebras of orthogonal type yield $n=2\ell+1$ and $n=2\ell$. So my question for the latter types means: does every nilpotent element $e$ of this natural matrix Lie algebra satisfy $e^h =0$ as in types $A,C$? -Behind this question is a related prime characteristic question for restricted Lie algebras, motivated in part by Kostant's classical 1959 paper in Amer. J. Math. (Corollary 5.4). In the general setting of simple Lie algebras he showed that regular (= principal) "nilpotent" elements $e$ are characterized by a condition on their adjoint operators: $(\mathrm{ad} \:e)^{2q}\neq 0$ where $q$ is the sum of coefficients of the highest root expressed relative to simple roots. Moreover, the next power annihilates all regular nilpotents. Earlier he showed that $q+1 = h$ is the Coxeter number of the Weyl group. (But there is a misprint in that corollary.) -ADDED: As Victor points out, except for a small decrease in type $D$ the four classical families of simple Lie algebras have index of nilpotence in their natural representations given by the Cayley-Hamilton approach in type $A$. My question arose from passing to characteristic $p>0$ via a Chevalley basis -over $\mathbb{Z}$, then extending scalars. For $p \geq h$, results from the mid-1980s on cohomology of restricted Lie algebras and support varieties of modules (Jantzen, Friedlander-Parshall, ... ) reveal that for the built-in $[p]$ operation on such a Lie algebra one has $\text{ad}\: e^{[p]} = 0 = (\text{ad} \:e)^p$ for all nilpotents $e$. But in natural matrix representations like those for types $A-D$, the $[p]$ operation is the usual matrix power. Here the slightly modified Cayley-Hamilton power needed for vanishing agrees. -At the extreme of $E_8$ there is more divergence: Here the "natural" smallest faithful representation is given by the adjoint module with $n=248$, whereas $h=30$. For $p = 31$ this power of each $\text{ad} \:e$ vanishes, contrasting with Kostant's characteristic 0 result which requires a power at least $59$ when $e$ is regular. - -REPLY [4 votes]: Let $\lambda$ be a partition of $n.$ Then there exists a skew-symmetric nilpotent matrix whose Jordan blocks sizes are $\lambda_i$ if and only if every even parts has even multiplicity. This follows easily from the representation theory of $\mathfrak{sl}_2$ and is duly recorded in standard sources, e.g. Collingwood and McGovern. It follows that the maximum "nilpotence index" of a skew-symmetric $n\times n$ matrix is $n$ for odd $n$ and $n-1$ for even $n.$<|endoftext|> -TITLE: Why isn't the orbifold cohomology of $pt/G$ equal to the cohomology of $BG$? -QUESTION [13 upvotes]: The classifying space of a group $G$ is given by taking a contractible space $E$ equipped with a free $G$-action, and looking at the quotient, which we dub $BG$. The homotopy type of this space (and thus its cohomology) depend only on $G$, and this gives us one definition of group cohomology. -Now, we can also look at the orbifold $[pt/G]$, and compute its orbifold cohomology (as its regular cohomology is rather uninteresting). For finite groups $G$, we have the isomorphism (see Adem-Leida-Ruan) -$$ -H^*_{orb}\big([pt/G],\mathbb{C}\big) \cong Z(\mathbb{C}G). -$$ -We can easily see that for finite groups the cohomology obtained is not the same as we would get from the usual group cohomology. -Why? I understand that the actual constructions are very different, but this seems very unsatisfying to me. Morally this space is a contractible one modulo a free group action, and it is a model of $BG$ if we look at it in the appropriate way, so why shouldn't their cohomologies be the same? - -REPLY [33 votes]: Orbifold cohomology is not a model for the cohomology of the orbifold (or more generally stack) itself, but for that of its inertia stack (a.k.a. derived loop space), which parametrizes points of the stack together with automorphisms. The inertia of $BG$ is $G/G$ (which is also the correct homotopy type for the free loops in the classifying space of $G$), hence you're seeing class functions (aka the center of the group algebra). This is very natural from the point of view of topological field theory -- we're trying to capture the space of states of a topological field theory (a sigma model) on the circle, which is a measure of the space of maps from the circle into some target. In the case of a stacky target, these maps also can wrap around automorphisms of points, resulting in the inertia stack. In physics these are called "twisted sectors". -The orbifold cohomology can also be described in terms of Hochschild homology -of sheaves on the space itself (and Hochschild homology is an algebraic model of taking loops in great generality).<|endoftext|> -TITLE: Is there a solution/approximation for the non-linear difference equation $c_n = c_{n-1}+c_{\lceil \alpha n \rceil}$, where $0 < \alpha < 1$? -QUESTION [6 upvotes]: Is there a solution/approximation for the non-linear difference equation $c_n = c_{n-1}+c_{\lceil \alpha n \rceil}$, where $0 < \alpha < 1$? - -REPLY [6 votes]: As Aaron Meyerowitz mentions when $\alpha=\frac{1}{b}$ the sequence is related to the number of partitions of $bn$ into powers of $b$. The asymptotic value of this sequence was determined by de Bruijn (On Mahler's partition problem). I believe his methods can be used to get asymptotic values for general $\alpha$, though I haven't checked this carefully. By looking at the generating function you can reduce it to studying the asymptotics of the coefficients of a power series $F(x)$ (which should be approximately the generating function of your sequence) which satisfies -$$F(x)=\frac{1-x^{\lceil 1/\alpha\rceil}}{(1-x)^2}F(x^{\lceil 1/\alpha\rceil})$$ -this is a Mahler type functional equation, and there are standard methods in the literature to obtain asymptotics.<|endoftext|> -TITLE: fundamental groups of curves -QUESTION [6 upvotes]: I saw the following statement made without proof in a paper of Bogomolov and Tschinkel: -If $X$ is an algebraic surface, and $C$ is an ample smooth curve in $X,$ then the fundamental group of $C$ surjects onto that of $X.$ -I was wondering if someone could supply a reference, and perhaps some idea of what the most general version of this statement was... - -REPLY [8 votes]: There is also an algebraic proof which works over fields of positive characteristic. -See, e.g., Lemma 5.1 of the paper Zariski's conjecture and related problems by Madhav Nori (Annales scientifiques de l'École Normale -Supérieure, Sér. 4, 16 no. 2 (1983), p. 305-344). -The proof of that Lemma is more general than what you need. It goes as follows. Let $X$ be a proper smooth algebraic surface and let $A\subset X$ be an ample curve. That $\pi_1(A)$ surjects onto $\pi_1(X)$ means the following: any finite étale cover $f:Y\to X$ which is split over $A$ is split. So if $Y$ is connected, $\deg(f)=1$. -Let us prove that if $A$ is ample; in fact, we only need $A$ big and nef. -If $f$ has a section on $A$, one can write $f^{-1}(A)=B+R$ where $B\to A$ -is an isomorphism, and $R$ is disjoint from $B$. The Hodge index theorem -says that the intersection form restricted to the space generated by $B$ and $R$ has -at most one +-sign. Since $(B+R)^2=(f^*A)^2=\deg(f) A^2>0$, it has exactly one +-sign, -and the determinant -$$\begin{vmatrix} (B+R)^2 & (B+R)\cdot B \cr  B\cdot (B+R) & B^2\end{vmatrix} -$$ -is nonpositive. By the projection formula, one has -$$(B+R)\cdot B=f^*A\cdot B=A\cdot f_*B=A^2.$$ -Since $B$ and $R$ are disjoint, we obtain $B^2=A^2$. Then, -$$ \deg(f)A^2 = (B+R)^2=B^2+R^2, $$ -so $$R^2=(\deg(f)-1) A^2$$. -The above determinant is equal to -$ (\deg(f)-1)A^2$. Since $A^2>0$, $\deg(f)\leq 1$. -This proof generalizes to the so-called Ramanujam lemma according to which an effective divisor on a surface which is big and nef is numerically connected (doesn't decompose as the sum of two nonzero effective divisors with 0-intersection), hence connected. In our case, the existence of the section implies that $f^*A$ is not numerically connected; it is however big and nef because this property is stable under finite pull-back. -See also the paper of J-B. Bost, Potential theory and Lefschetz theorems for arithmetic surfaces (Annales scientifiques de l'École Normale Supérieure, Sér. 4, 32 no. 2 (1999), p. 241-312) where this argument is explained for surfaces and adapted for arithmetic -surfaces.<|endoftext|> -TITLE: Topology of plane curve complements after blow-ups -QUESTION [9 upvotes]: Let $C$ be a (singular, reducible) curve in the complex projective plane $\mathbb{P}^2$. An old problem is to study the topology of the complement $\mathbb{P}^2-C$. A famous result in this direction was originally stated by Zariski, but not proven until 1980 by Fulton. It states that if $C$ has only nodes for singularities (that is, double points with distinct tangents), then the fundamental group $\pi_1(\mathbb{P}^2-C)$ is abelian. -My problem is as follows: again, let $C$ be a (singular, reducible) curve in $\mathbb{P}^2$. Now, blow up the plane at one of the singular points of $C$. Denote the resulting surface $S_1$, the exceptional divisor by $E_1$, and the proper transform of $C$ by $\hat{C}_1$. Then blow up $S_1$ at a singular point of $\hat{C}_1$, and repeat this procedure to get a surface $S=S_n$, exceptional divisors $E_1,\ldots,E_n\subset S$, and the proper transform $\hat{C}\subset S$ of the original curve. I am interested in the topology of the complement $X=S-(\hat{C} \; \cup E_{i_1} \cup \ldots \cup E_{i_k})$. In other words, $X$ is obtained from $S$ by deleting $\hat{C}$ as well as some (but not all) of the exceptional divisors obtained from blowing up. In particular, I would like to prove that the fundamental group $\pi_1(X)$ is abelian. -If the $C$ has only nodes for singularities, then $\pi_1(\mathbb{P}^2-C)$ is already abelian, and it is not hard to see that $\pi_1(X)$ will also be abelian. However, I am more interested in some cases where $C$ has slightly more complicated singularities. For example, if $C$ has a triple point with distinct tangents, $\pi_1(\mathbb{P}^2-C)$ is non-abelian; but I suspect that if I blow up the plane at the triple point, then $\pi_1(S-\hat{C})$ will be abelian. Does anyone know of any techniques or references for dealing with this situation? -(Note: It is certainly not the case that $\pi_1(X)$ will always be abelian. I am interested in techniques that will allow me to prove it is abelian for certain specific curves and configurations of exceptional divisors.) - -REPLY [5 votes]: The following Theorem by Nori (Proposition 3.27 of this paper) is in the spirit of what you are looking for. - -Theorem. Let $D$ and $E$ be curves on - a smooth projective surface $X$. Assume that - -the only singularities of $D$ are nodes; -$D$ and $E$ intersect transversally; -every irreducible component $C$ of $D$ satisfies $C^2 > 2 r(C)$ where - $r(C)$ is the number of singularities of $C$. - -Then the kernel of the natural - morphism $$ \pi_1(X-D-E) -> \longrightarrow \pi_1(X-E) $$ is - abelian. - -In particular if you take an irreducible plane curve $C$ having a triple point with distinct tangents as its only singularity then the fundamental group of the complement -of its strict transform is abelian (apply Theorem above to $D = \hat C$ and $ E = \emptyset$ ). As Dmitri pointed out, if you drop irreducibility this is no longer true. -You might want to take a look at this survey. There you will find Nori's Theorem -in Section 2.3. - -Further comments added later: - -If $C \subset \mathbb P^2$ is an irreducible curve with only one singularity having -smooth branches intersecting pairwise transversely then -the complement of $C$ in $\mathbb P^2$ and well as the complement of its strict transform in the blow-up of $\mathbb P^2$ are abelian ($D = \hat C$, $E =$ exceptional divisor in the first case; and $D = \hat C$, $E= \emptyset$ in the second). -If you have a reduced connected curve $C = E_1 + \ldots + E_k$ with rational irreducible components, the intersection matrix $(E_i\cdot E_j)$ is negative definite, and the dual graph is a tree then the fundamental group of the complement of a neighborhood of $C$ has been determined by Mumford in this paper.<|endoftext|> -TITLE: Character of the Basic Representation for Affine E_8 in Terms of Jacobi Theta Functions -QUESTION [7 upvotes]: When $\mathfrak g$ is a complex, simple, simply laced Lie algebra of rank r then the (specialized) character of the basic representation for the corresponding affine Lie algebra $\hat {\mathfrak g}$ is given by $\chi_{\hat {\mathfrak g}}(q)=\frac{\Theta _{\mathfrak g}(q)}{\eta (q)^r}$, where $\Theta _{\mathfrak g} (q)$ is the theta function for the root lattice of $\mathfrak g$. It is well known that $\Theta _{E_8} (q)$ is equal to the normalized Eisenstein series of weight 4, $E_4 (q)$. We also have another basic result that $E_4(q)=\frac{1}{2}(\theta_2 (0,q)^8 +\theta_3 (0,q)^8 + \theta _4 (0,q)^8)$, where $\theta_2(x,q)$, $\theta_3(x,q)$, $\theta_4(x,q)$ are the three classical even Jacobi theta functions. Being odd, the other Jacobi theta function $\theta_1(x,q)$ vanishes at $x=0$. My question is in regards to the unspecialized form of the character; that is, I'm wondering if evaluating the character on a non-zero element of the Cartan subalgebra produces something of the form $$\chi _{\hat {E_8}}(x_1,...,x_8,q)=\frac{\frac{1}{2}(\prod_{i=1}^8\theta_2 (x_i,q) + \prod_{i=1}^8\theta_3(x_i,q) + \prod_{i=1}^8\theta _4 (x_i,q)+\alpha \prod_{i=1}^8\theta_1 (x_i,q))}{\eta (q)^8}$$ -for some $\alpha$? Moreover, in the case $\alpha = 0$, does it make sense that Taylor expanding the numerator of this expression and writing it in terms of elementary symmetric polynomials in $x_1^2,...,x_8^2$ one should find that everything below degree 8 can be written as quasimodular forms times only powers of $p_1(x_1,...,x_8)=x_1^2+...+x_8^2$? This is what I got when I plugged it into Mathematica and in fact the coefficients are similar to the Eisenstein-Jacobi series $E_{4,1}(z,q)$. What would probably be very relevant is if there's anything regarding Jacobi forms in several variables that has been studied that is similar to the "development coefficients" which are well studied in Eichler and Zagier "The Theory of Jacobi Forms". -I've tried looking through a good bit of the literature regarding both of these matters, but couldn't find what I was looking for. If anyone could point me in the right direction, I would be greatly appreciative. - -REPLY [3 votes]: Your (related?) question brought me here. I'm not sure if I'm following you right, but I guess your first question asks: is there {$h_1,...,h_8$},a basis of the Cartan subalgebra, such that $${\rm tr}_{L(\Lambda_0)} q^{L(0)-1/3} x_1^{h_1} \cdots x_8^{h_8}$$ equals the right hand side for some $\alpha$? (Here $L(\Lambda_0)$ is your basic representation). I assume you are flexible on the choice of a basis of the Cartan, right? For instance, your formula obviously doesn't hold if you take (for some purposes natural) $h_i=\omega_i$ (fundamental co-weights). -Let me ignore the denominator which doesn't change. -Start from the usual Euclidean $\mathbb{Z}^8$-lattice with orthonormal basis $\epsilon_i$. Then, $D_8$ lattice, as you know, consists of all $v \in \mathbb{Z}^8$ such that the sum of coordinates of $v$ is even. To get the $E_8$-lattice you also include all $v \in (\mathbb{Z}+\frac{1}{2})^8$ such that (again) the sum of coordinates of $v$ is even. -Set -$$\theta_3(q^{1/2},x)=\sum_{n \in \mathbb{Z}} q^{n^2/2} x^{n}$$ -$$\theta_4(q^{1/2},x)=\sum_{n \in \mathbb{Z}}(-1)^n q^{n^2/2} x^n $$ -$$\theta_2(q^{1/2},x)= \sum_{n \in \mathbb{Z}} q^{(n+1/2)^2/2} x^{n+1/2}$$ -$$\theta_1(q^{1/2},x)=\sum_{n \in \mathbb{Z}} (-1)^n q^{(n+1/2)^2/2} x^{n+1/2} $$ -So you will be summing over the $D_8$-lattice and another "shifted" $D_8$. -For the $D_8$ part, you get -$$\prod_{i=1}^8 \theta_3(q^{1/2},x_i)+\prod_{i=1}^8 \theta_4(q^{1/2},x_i)$$ -multiplied by $\frac{1}{2}$.The shifted part contributes with -$$\prod_{i=1}^8 \theta_2(q^{1/2},x_i)+\prod_{i=1}^8 \theta_1(q^{1/2},x_i)$$ -multiplied by $\frac{1}{2}$. The resulting formula looks very similar to yours (with $h_i=\epsilon_i$). -This construction is essentially decomposition -$L(\Lambda_0)=L_{D_8}(\Lambda_0) \oplus L_{D_8}(\Lambda_8)$ into -irreducible (standard) $D^{(1)}_8$-modules (I hope $\Lambda_8$ is correct here). All this is very natural if you are familiar with the spinor construction of ${E_8}^{(1)}$ (see this monograph ). -Just in case you prefer $${\rm tr}_{L(\Lambda_0)} q^{L(0)-1/3} x_1^{\omega_1} \cdots x_8^{\omega_8},$$ simply express $\omega_i$ as a linear combination of $\epsilon_j$'s, and rearrange $x_j$'s accordingly. -I hope this helps. My apologies if I said something wrong. -n.b. $x_i$ should be $e^{2 \pi i x_i}$ to be consistent with " $\theta_1$ vanishes at $x_i=0$". -EDIT: (Too long for a comment.) Regarding your second question and the fact that up to degree $8$ in $x_i$'s there is only contribution from the power sum $p_2(x)=x_1^2+ \cdots + x_8^2$. -Actually something similar happens if you consider one-point functions on the torus so it might be related. For a holomorphic rational CFT $V$ (resp. VOA) of central charge $c$, meaning that there is a unique irreducible sector (resp. module), the vector space of one-point functions (essentially traces of degree zero operators coming from vertex operators) -$$\langle Z(v,\tau), v \in V \rangle$$ -is a fairly nice space of modular forms with a mild pole at infinity. Basically after you multiply with $\eta^{c}(\tau)$ you expect to get all holomorphic modular form. -For the ($c=24$) moonshine module this was shown by Dong and Mason here , but I think it must also hold for $c=8k$ (shown maybe in this more recent paper.) . -Now back to $V=L_{E_8}(\Lambda_0)$, which is holomorphic with $c=8$. If you accept what I said, each $\eta^8(\tau) Z(v,\tau)$ is holomorphic (possibly zero) for $\Gamma(1)$ of weight $deg(v)$. So if you pick $v$ of degree $2,4,6$ you will be generating one-dimensional spaces spanned by $\frac{E_6(\tau)}{\eta^8}$, $\frac{E_8(\tau)}{\eta^8}$ and $\frac{E_{10}(\tau)}{\eta^8}$. However in degree $8$ you have a contribution from the cusp form $\Delta$, so the dimension jumps by one. If I remember correctly, up to degree $8$ you can use conformal vector to get those, but higher you also have to construct -some primary vectors coming from spherical harmonics to get a nonzero $Z(v,\tau)$, and then you apply a result of Waldspurger . So I agree with you, the Weyl invariant of degree $8$ (for $E_8$) is most likely relevant in your case. -Dong and Mason (and their collaborators/students) wrote several papers on this subject, so it might be useful to read some of their work. For instance, this paper , might be also useful for your purposes.<|endoftext|> -TITLE: homology classes as immersed submanifolds -QUESTION [5 upvotes]: Possible Duplicate: -Cohomology and fundamental classes - -Hello, -I would like to know if all homological classes in a smooth manifold can be represented as immersed submanifolds, or examples where this is not true and possible obstructions. -Thanks!! - -REPLY [19 votes]: It might be better to split the question into 2 cases and 2 steps. -Step 1: Which homology classes in $X$ can be represented by continuous maps of closed smooth manifolds (ie, which classes are $f_*[M]$ where $f\colon M\to X$ is a map from a closed manifold with fundamental class $[M]\in H_*(M)$)? This is the so-called 'Steenrod representability problem'. -Case 1: In the unoriented case (if you are asking about homology with mod $2$ coefficients) the answer is all of them. See Thom's paper "Quelques propriétés globales des variétés différentiables", or the book "Differentiable periodic maps" by Conner and Floyd. -Case 2: In the oriented case, we are asking which integral homology classes are represented by maps from closed orientable manifolds. The answer is not all of them, but positive multiples of all of them. More precise statements can be found in the papers of Thom ("Sur un problème de Steenrod" and "Quelques propriétés...") and their reviews. -Step 2: Now you have to ask which unoriented and oriented bordism classes contain immersions. To the best of my knowledge, this part is not completely known. But here is a reference to start looking: -Li, Gui Song "On immersions in bordism classes", Math. Ann. 291 (1991), no. 2, 373–382. -Note that in the introduction he quotes a result of Szűcs to the effect that, for any given oriented bordism class of sufficiently high codimension in a manifold, some positive multiple of it contains an immersion. -Update: Sorry to revive this old post, but I just wanted to shamelessly plug an article which arose directly from this question. In http://arxiv.org/abs/1111.0249 (to appear in Bulletin of the LMS) András Szűcs and myself show that in any codimension $k\ge 2$ there exists a closed smooth manifold $N^n$ and a mod $2$ homology class of dimension $n-k$ which cannot be realized by an immersion of a closed manifold. The proof employs explicit obstructions to realizability, involving Bocksteins and Steenrod squares (see Theorem 1.2).<|endoftext|> -TITLE: Subgroups of groups of Square-free order -QUESTION [7 upvotes]: If $G$ is a group of square-free order with at-least three prime factors, $|G|=p_1p_2....p_r$, $(2< p_i < p_{i+1})$, does $G$ contain a cyclic subgroup of composite order? -(As groups of square-free order are solvable, $G$ will necessarily have a proper subgroup of composite order.) - -REPLY [2 votes]: Adding a bit to Tom's very complet answer you can in fact find a cyclic subgroup of oder $pq$ where $p,q$ are among the three bigest primes in the factorization of $|G|$. First notice that $G$ contains a subgroup $H$ of order $p_{r-2}p_{r-1}p_r$. This follows inductively from: -Fact: If $|G|=pm$ where $(p,m)=1$ and $p$ is the smallest prime dividing $|G|$ then $G$ contains a subgroup, in fact normal, of order $m$. -Now you've got a group $H$ of order $p_{r-2}p_{r-1}p_r$, so you use Tom's argument with $H$ and obtain that $m$ or $n$ is $pq$. I believe that for groups of order $pqr$ one could actually show by hand what you want without invoking the known presentation of Z-groups.<|endoftext|> -TITLE: Kunen's use of Countable Transitive Models -QUESTION [10 upvotes]: Hi, -I have a doubt concerning Kunen's exposition of forcing in his classical book (arguably $the$ book on forcing). When dealing with Countable Transitive Models to set up the forcing machinery, Kunen is always very careful with letting this C.T.M. to model $only$ finite fragments of ZFC. I recently read in one of the answers to this MO question that the point is that CON(ZFC) cannot prove the existence of countable transitive models of ZFC, and I don't understand why not... wouldn't this be just a matter of taking a set model for ZFC (by consistency of ZFC, which we are assuming), which without loss of generality can be countable (by L\"owenheim-Skolem) and then apply the Mostowski collapsing lemma to this in order to get a C.T.M. of the $full$ ZFC? -Also, a professor once told me that Kunen did things this way in order to avoid assuming CON(ZFC), but I didn't understand this explanation either (isn't it pointless avoiding to assume CON(ZFC)... if the negation holds, everything would be provable anyways!!!) -I'm pretty sure there's something about this issue that I'm not taking into account, I would like to know what that is... I'm kindly asking for your help with that! - -REPLY [6 votes]: If there is any ordinal $\alpha$ such that $L_\alpha$ satisfies ZFC, then consider the least one. This is some countable ordinal $\beta$. It is easy to show that $L_\beta \vDash$ "There is no transitive model of ZFC." However, by absoluteness, it will still think that ZFC is syntactically consistent, and therefore has a (non-well-founded) model. Actually this also shows that any transitive model of ZFC with rank higher than this $\beta$ thinks that there is a transitive model of ZFC, so the existence of transitive models fails only for the very shortest ones.<|endoftext|> -TITLE: Discrete Fourier Transform of the Möbius Function -QUESTION [24 upvotes]: Consider the Möbius function $\mu (m)$. (Thus $\mu(m)=0$ unless all prime factors of $m$ appear once and $\mu (m)=(-1)^r$ if $m$ has $r$ distinct prime factors.) Next consider for some natural number $n$ the discrete Fourier transform -$$\hat \mu (k)= \frac1n \sum _{r=0}^{n-1} \mu(r)e^{2 \pi i rk/n}.$$ -So $\sum _{k=0}^{n-1} |\hat{\mu} (k)|^2$ is roughly $6/\pi ^2$; and the Prime Number Theorem asserts that $\hat \mu(0)=o(1)$; -My questions are: -1) Is it true (or even obvious based on PNT?) that the individual coefficients tend to 0? -2) Is it known that $\sum_{k=0}^{s}|\hat \mu (k)|^2=o(1)$, where say $s=\log n^A$? or perhaps for some other specific $s$ Fourier coefficients? (Here $A$ is suppose to be a large real even as large as some large power of $\log \log n$ but it can also be a small real.) -3) Is it perhaps known that the sum of squares of the largest (in absolute value) $s$ Fourier coefficients is $o(1)$? When $s=$ some power of $\log n$ as above? -4) What is expected to be the correct value of $s$ so that the sum of squares of absolute value of the highest $s$ Fourier coefficients are already not $o(1)$? -UPDATE and AND A FOLLOW-UP QUESTION: -The answers below give a very good picture regarding what is known for the individual Fourier coefficients unconditionally and under GRH. I am still curious if there are some cases that a bound on the contribution of a bunch of Fourier coefficients can be given which is better than what you obtain from individual coefficients. So here is a question: -5) Let $n$ be a positive integer. Let $\cal P$ be its prime factors and let $Q(n,r)$ be all divisors of $n$ that involve at most $r$ primes in ${\cal P}$. -Consider $$\sum _{k \in Q(n,r)} |\hat{\mu} (k)|^2.$$ -We now from the result on individual coefficients that this sum is o(1) for every $n$ if $r$ is bounded. Is it possible to prove that the sum is o(1) even when $r$ is a specific very slow growing function of n, where the bound of individual coefficients is not good enough? (We can ask a similar question about improving over the individual coeffient bound for $P(n,r)$ the set of all divisors of $n$ which are the product of at most $r$ primes in ${\cal P}$.) This example is geared towards the specific extension for the Walsh-Fourier coefficients. But there it seems that the knowledge regarding individual coefficients uconditionally and under GRH is weaker. END of UPDATE - -The motivation came from a certain computational complexity extension of the prime number theorem which is discussed here and here. For the conjecture we will need Walsh expansions rather than Fourier expansion and we will need (3) for Walsh coefficients corresponding to sets of size $(\log n)^{(\log \log n)^t}$. - -Here is a follow-up MO question about Walsh-Fourier transform]3, which has direct applications to the computational complexity question. "Transforming" the answers given here from the Fourier transform to the Walsh-Fourier transform will already have interesting consequences! - -REPLY [10 votes]: The high level of abstraction that number theorists sustain is a continual source of amazement to me ... doesn't anyone want to see what $\hat\mu(k,n)$ concretely looks like? -So let's do it! Mainly for fun (and as a gesture of respect for Gil), here is a density plot of $n^{1/2} \hat\mu(k,n)$ for all values $(n,k) \in 1,1024$ (note that a normalizing factor $n^{1/2}$ is included; this scaling yields near-uniform luminance in the plot): -(canonical Moebius plot) http://faculty.washington.edu/sidles/Moebius/Moebius__1024_canonical.png -As usual, the argument (phase) of $\hat\mu(k,n)$ is encoded as hue, and the magnitude as saturation and value. For details, see the Mathematica code that produced the above plot (for many people, the most interesting aspect of the code will be the idiom for exporting Mathematica graphics to png files; certainly this is by far the longest part of the code, and the toughest to debug). -Such codes code encourages us to do numerical experiments ... -What happens if we randomize the sign of the Möbius function? -(canonical Moebius plot) http://faculty.washington.edu/sidles/Moebius/Moebius__1024_randomized.png -Hmmm ... when we randomize the sign of the Möbius function, $\hat\mu(k,n)$ doesn't look much different, does it? -On the other hand, we know (from other numerical experiments) that when we compute a pseudo-Riemann function $\zeta^\prime(s) = (\sum_{r=1}^\infty \mu(r)/r^s)^{-1}$ using a randomized-sign Möbius function, then the resulting $\zeta^\prime(s)$ has a distribution of zeros that is grossly different from the distribution of zeros in $\zeta(s)$ (to see this, just try it). -In summary, the special properties of the distribution of primes (relative to randomized distributions) that are so plainly evident when we "look" at the zeros of $\zeta(s)$, are not immediately evident when we "look" at the magnitude and phase of $\hat\mu(k,n)$. -As for what this observation means (if anything at all) ... well ... that is for the number theorists to comment upon, not me! :)<|endoftext|> -TITLE: Heegaard splitting, equivalent homeomorphisms, mapping class group of genus n-torus -QUESTION [5 upvotes]: Given a Heegaard splitting of genus $n$, and two distinct orientation preserving homeomorphisms, elements of the mapping class group of the genus $n$ torus, is there a method which shows whether or not these homeomorphisms, when used to identify the boundaries of the pair of handlebodies, will produce the same $3$-manifold? - -REPLY [5 votes]: The original question was "Is there an algorithm that, given two genus g Heegaard splittings, decides if the resulting two manifolds are homeomorphic?" The answer to this question is "yes" but doesn't have anything to do with Heegaard splittings. -In theory the solution to the Geometrization Conjecture gives an algorithm to decide if two manifolds are homeomorphic. -In practice one converts the splitting to a triangulations and feeds the result to SnapPy. There are several other programs for three-manifold recognition, such as snap, orb, Regina, and the 3-Manifold Recognizer. However all use SnapPea as the engine for dealing with hyperbolic manifolds. See http://www.math.uiuc.edu/~nmd/computop/ for links. The Recognizer, for Windows, can be found at http://www.matlas.math.csu.ru/.<|endoftext|> -TITLE: Automorphisms of $SL_n(\mathbb{Z})$ -QUESTION [15 upvotes]: What is the group of outer automorphisms of $SL_n(\mathbb{Z})$. I wanted to understand semidirect products of the form $SL_n(\mathbb{Z})\rtimes_\varphi \mathbb{Z}$ and its isomorphism type depends only on $[\varphi]\in Out(SL_n(\mathbb{Z})$. There is always the conjugate inverse, which is clearly not an inner automorphism, as it doesn't preserve the minimal polynomial of the matrix (at least for $n\ge 3$). Are there any other outer automorphisms ? - -REPLY [7 votes]: See L. K. Hua and I. Reiner, "Automorphisms of the unimodular group", Trans. Amer. Math. Soc. -71 (1951), 331-348.<|endoftext|> -TITLE: Random points in a rectangular grid defining a closed path -QUESTION [5 upvotes]: Suppose we have a $n\times m$ rectangular grid (namely: $nm$ points disposed as a matrix with $n$ rows and $m$ columns). -We randomly pick $h$ different points in the grid, where every point is equally likely. -If only horizontal or vertical movements between two points are allowed, what is the probability that the points define at least one closed path? -ps: we can suppose $m=n$ to simplify -For example, let $n=m=4$ and $h=6$. -1 denotes a selected point, 0 a non-selected one. -These $6$ points define a closed path: -1 0 0 1 -0 0 0 0 -1 1 0 0 -0 1 0 1 -as these $6$ do (the $4$ in the bottom-right corner): -1 0 0 1 -0 0 0 0 -0 1 0 1 -0 1 0 1 -while the following $6$ points do not: -1 0 0 0 -0 0 0 1 -1 1 0 0 -0 1 0 1 -Substantially, the $h$ points define a closed path if and only if there exist a subset of these $h$ points such that every point in the subset has one other point of the subset on the same row and one on the same column. -Thanks for your help. -ALTERNATIVE DEFINITION: -Someone cleverly suggested me that my problem is equivalent to another one on graph-theory, which is more likely to have been solved. -Suppose you have a complete bipartite graph $K_{n,m}$. -Randomly choosing $h$ edges, what is the probability that these $h$ edges generate at least one cycle? -Thanks very much for any help. -Also a reference to some interesting work in the literature would be very appreciated. - -REPLY [10 votes]: Having screwed up the answer by getting the wrong answer for a really simple calculation the first time, I'm now going to try to redeem myself. -First, to make things easier, let each point be present with probability $h/(mn)$, and let these random variables be independent. This won't change the probabilities much. -You can calculate explicitly the expected number of squares. The expected number of squares is ${m \choose 2}{n \choose 2}(h/mn)^4$, since there are ${m \choose 2}{n \choose 2}$ possible squares, and each point is present with probability $h/mn$. This is roughly $1/4 (h^2/mn)^2$ (for large $mn$, and fixed $h^2/mn$). -Now, let's calculate the expected number of six-cycles. You can assume a $2k$-cycle lies in $k$ different columns and $k$ different rows (otherwise there is a smaller cycle), so we have ${m \choose 3}{n \choose 3}\alpha_3 (h/(mn))^6$ ways of getting a six-cycle. Here, $\alpha_3$ is the ways of getting a six-cycle in a 3 by 3 square, and it is easy to check that there are six of them. -There are four ways that it can look like this: -1.1 -.11 -11. - -and two ways it can look like this: -11. -1.1 -.11 - -Now, if we can compute $\alpha_k$, where $\alpha_k$ is the number of ways we can get a minimal $2k$-cycle in a $k\times k$ grid, we can get an infinite series for the expected number of cycles. For $\alpha_k$, we need to find a cycle which covers every row twice and every column twice. Let $\pi_1$ be the order that we visit the rows, and $\pi_2$ the order we visit the columns. We can assume that $\pi_1$ starts with $1$, and we can traverse each cycle in $2$ directions. This means that there are $(k-1)!$ possible $\pi_1$, $k!$ possible $\pi_2$, and we have to divide by $2$ because otherwise we count each cycle twice. Thus, the expected number of cycles when $m,n\rightarrow \infty$ is -$$ \sum_{k=2}^\infty \frac{1}{2}(k-1)! k!{m \choose k}{n \choose k}\left(\frac{h}{mn}\right)^{2k},$$ -which simplifies in the limit to -$$ \sum_{k=2}^\infty \frac{1}{2k}\left(\frac{h^2}{mn}\right)^k,$$ -or -$$-1/2 \ln(1 - x) - x/2, \mathrm{\ \ where\ }x=h^2/(mn).$$ -If objects are distributed with a Poisson distribution with expectation $r$, then the probability that there is an object is $1-e^{-r}$. Thus, if we assume that the closed paths are distributed according to the Poisson distribution, the probability that a closed path exists is -$$1-(1-x)^{1/2}e^{x/2}.$$ -As Kevin Costello remarks in the comments, you may be able to use Janson's theorem (also covered in Alon and Spencer's book) to prove that these closed paths obey Poisson statistics. - -And below, another argument that the probability of a closed path rises to 1 when $h^2 =mn$, salvaged from my first attempt at solving the problem. -Let's start at a single point $P_0$, and look at the tree of paths that results. The tree of paths is obtained by looking for points in the same row as $P_0$, then points in the same column as these points, and then rows, etc. (Also do the same procedure, but starting with columns.) If any two points in this tree are equal (that is, there are two distinct paths leading from the starting point to the same point) then there is a closed path. So we want to know the probability that this tree of paths is fairly large. In particular, if the tree of paths has size $\sqrt{h}$, then by the birthday paradox there should be a reasonable chance that two of these points are equal. -But, assuming all the points are different, this tree of paths is a Galton-Watson branching process, and we can calculate the probability that it's large. (You can calculate lots of properties of Galton-Watson branching processes.) In particular, if the expected number of children is greater than 1, there is a finite probability that the tree is infinite. Using this is a rough criterion, you will start to get closed paths when $mn \approx h^2$. You should be able to a much better approximation by analyzing the associated Galton-Watson branching process more carefully. -The Galton-Watson tree has two branches. On the first branch, the number of children on odd levels are distributed approximately as a Poisson process with expectation $h/m$, and the number of children on even levels as a Poisson process with expectation $h/n$. On the second branch, distributions on the odd and even levels are reversed.<|endoftext|> -TITLE: Eigenvalues of submatrices -QUESTION [12 upvotes]: I am interested in results on the eigenvalues of submatrices. -Given a symmetric and positive-semidefinite matrix $M$, denote the submatrix obtained by deleting the ith column and jth row as $[M]_{ji}$. -How does the spectra $\lambda([M]_{ji})$ relate to the spectra $\lambda(M)$? -I know when looking at principal submatrices (ie, $i=j$), we get an interlacing property. However, I can't seem to find such results for other submatrices. - -REPLY [2 votes]: Maybe this helps: -http://www.math.wm.edu/~ckli/ima/note-2.pdf<|endoftext|> -TITLE: How to estimate a time distribution -QUESTION [7 upvotes]: This likely isn't a research-level question, but it is at least a question of interest to this researcher. I'm happy with an answer that sends me somewhere (preferably online) to read about a well-known (to somebody) solution, if there is such a thing. First the question, then the motivation. - -Let $X_1,X_2,\dots,X_N$ be independent $N(\mu,\sigma^2)$ random variables - ($\mu$ and $\sigma^2$ are unknown), and let $Y_1,Y_2,\dots,Y_N$ be defined by - $\{X_1,\dots,X_n\}=\{Y_1,\dots,Y_N\}$ and $Y_1 < Y_2 < \cdots < Y_N$. - For $k < N$, how can one estimate the parameters $\mu,\sigma^2$ from $Y_1,Y_2,\dots,Y_k$? - -The application that I need this for is a strategy to buy light bulbs. I just purchased a stunning chandelier with 27 tiny light bulbs. The bulbs themselves are strange, and I'll have to special-order them. After the first several burn out, I'd like to be able to estimate how many will burn out in, say, the next 6 months. Although the bulbs are independent, the lifetimes of the first 3 to burn out are not. -This problem must come up all the time: medical researchers don't know how long their patients will last, they only know how long some of them didn't. - -REPLY [3 votes]: As requested, I'm making my comment an answer. -While this does not answer the problem as you've formalized it, I would argue that the correct method to consider this problem is actually Weibull analysis--generally failure times are not distributed normally. Essentially, plot the failure times on Weibull paper; the slope of the line you get will give you the information you want. -As Kevin O'Bryant has pointed out, this site does Weibull computations. -For a beautiful theorem explaining the relevance of the Weibull distribution, see the Fisher-Tippett-Gnedenko theorem. The model to keep in mind is the following: objects generally have more than one possible point of failure, and failure occurs when the first point of failure breaks. So in an ideal case one wants to compute the minimum of a collection of IID's. The Fisher-Tippett-Gnedenko theorem gives that the extreme value distributions, of which the Weibull distribution is an instance, are universal for this situation, just as the central limit theorem gives that the normal distribution is universal for the mean of a family of IID's.<|endoftext|> -TITLE: Must a surjective isometry on a dual space have a pre-adjoint? -QUESTION [6 upvotes]: Background: Let $X$ be a Banach space. We know a linear map $h$ is a surjective isometry of $X$ if and only if its adjoint $h^*$ is a surjective isometry of $X^*$. -In general, a linear map $g:X^* \to X^*$ need not have a pre-adjoint. But what if $g$ is a surjective isometry? Must there exist $f:X \to X$ such that $g=f^*$? If we identify $X$ with its embedding in $X^{**}$, this is equivalent to $g^*(X) \subset X$. -Sorry if this question is trivial; it seems like this should be well-known, but I haven't been able to find a reference or an easy counter-example. - -REPLY [6 votes]: Let $X$ be the space of sequences, indexed by the nonzero integers, that tend to $0$ at $-\infty$ and to an arbitrary finite limit at $\infty$, with sup norm (a direct product of $c_0$ and $c$). Then $X^*$ can be identified with $\ell^1$ of $\mathbb{Z}$. If $f$ is in $\ell^1$, then the corresponding functional on $X$ sends $x$ to -$\sum_{n\neq0}f_nx_{n} + f_0\cdot\lim_{n\to\infty}x_n$. The map on $\ell^1\cong X^*$ defined by $(f_n)_{n\in\mathbb Z}\mapsto (f_{-n})_{n\in\mathbb{Z}}$ is a linear surjective isometry with no pre-adjoint. (If there were a preadjoint, it would have to send $(x_n)_{n\in \mathbb Z\setminus\{0\}}$ to $(x_{-n})_{n\in\mathbb Z\setminus\{0\}}$.)<|endoftext|> -TITLE: Euler characteristic, Gauss-Bonnet, and a product formula -QUESTION [10 upvotes]: I know very little about the Pfaffian or how it works, and I'm new at Riemannian geometry in general. But I was wondering if there is some way to make this "intuitive" argument for the fact that a fiber bundle satisfies $\chi(\text{bundle}) = \chi(\text{fiber})\chi(\text{base})$ rigorous. - -Say you've got a surjective smooth submersion $\widetilde{M} \rightarrow M$ that's also a (smooth?) fiber bundle. (Assume all the manifolds are compact... though I wish there were some weaker condition than that for the Gauss-Bonnet theorem to apply. It'd be nice if one could make sense of it whenever the manifold had finitely generated cohomology). -EDIT: See update below. Slap Riemannian metrics on $\widetilde{M}$ and $M$ so that the submersion is a Riemannian submersion. (I just feel like one would need this in step 4. The other thing that would make sense is if we could somehow put a Riemannian metric on $\widetilde{M}$ that would be the 'twisted product' of Riemannian metrics on $F$ and $M$- whatever that means). -By the Generalized Gauss-Bonnet Theorem, $\int_{\widetilde{M}}Pf(\Omega_{\widetilde{M}}) = (2\pi)^n \chi(\widetilde{M})$. -And here's the thing that seems the hardest to make rigorous (though who knows, I may have already been wrong at step 2): I want to say that $\int_{\widetilde{M}}Pf(\Omega_{\widetilde{M}}) = \int_M\int_F Pf(\Omega_F)Pf(\Omega_M) = (2\pi)^k(2\pi^{n-k})\chi(F)\chi(M)$. - -In order for this to even start making sense there would have to be some sense in which an analog of Fubini's theorem works for twisted products. I'd also need the Pfaffian to satisfy the identity that I want it to satisfy in this case (which I think follows from the fact that if we have a direct sum of matrices, $A \oplus B$, then $Pf(A \oplus B) = Pf(A)Pf(B)$.) So: is there any way to actually make this argument stand? (I think that it would work for a trivial fibration, but that's no fun.) -So far I only know how to prove this product formula using a spectral sequence, so it'd be cool if this were a legitimate way of visualizing it in this special case. -Updates: First of all, it turns out that there is a Fubini's theorem for local products (see Generlized Curvature by Jean-Marie Morvan). Instead of Step 2 I think what we want is to put a metric on $M$, and then put metrics $h_x$ on $F_x$ (the fiber over $x$) that somehow 'varies smoothly with $x$' (hopefully that's possible) and then define a metric on $\widetilde{M}$ by breaking every vector into its horizontal and vertical pieces and applying the metric on $M$ and $F$ separately. Pretty much by definition this should make $\Omega_{\widetilde{M}}$ be the matrix $\Omega_M \oplus \Omega_F$ (I think?) and then the result should follow. -So I still have details to work out- and I'll keep thinking about it, but in the meantime if any of you have seen this before, then a reference would be really helpful! I'm afraid I've done something silly, or I'm assuming something I can't... - -REPLY [3 votes]: The Gauss-Bonnet theorem has, in my opinion, two parts. The first part is how to associate to a smooth oriented vector bundle $V \to X$ on a manifold a differential form representing the Euler class. For that, one needs a metric on $V$ and a metric connection. Take the curvature und plug it into the Pfaffian - that gives a closed differential form, the Euler form, whose de Rham class only depends on $V$ and its orientation. If $V=TM$ and $M$ is Riemannian, you can pick the Levi-civita connection, but that is not really relevant here. -The second step is that $\int_M e (TM)= \chi(M)$, and this is algeraic topology, involving Poincare duality. -Let $f:E \to B$ be a smooth fiber bundle with $d$-dimensional compact fibres (=proper submersion) and assume that $B$ is closed and oriented. Let $T_v E := ker (df)$ be the vertical tangent bundle. There is a short exact sequence of vector bundles $0 \to T_v E \to TE \to f^{\ast} TB \to 0$ and a splitting is given e.g. by a Riemann metric on $E$. -The Euler class $e$ of vector bundles satisfies $e(V \oplus W) = e(V) e(W)$, which follows for examples from direct sum formula for the Pfaffians (in fact, the above identity holds even for the differential form representatives). There is an integration over the fibre -$$\int_{E/B} \omega,$$ -which produces a $k-d$-form on $B$ out of a $k$-form on $E$. For a construction, see Berline, Getzler, Vergne ''Heat kernels and Dirac operators'' (it is in the background chapter 1). There is a Fubini principle for integration over the fibres, which reads: -$$\int_E e(TE) = \int_E e(T_v E) f^{\ast} e (TB) = \int_B (\int_{E/B} e(T_v E)) e (TB).$$ -The form $\int_{E/B} e(T_v E)$ is a $0$-form, i.e. a function, and you can test its values by restriction to the fibres. The value is the Euler characteristic of the fibre, by Gauss-Bonnet. Thus -$$ \int_B (\int_{E/B} e(T_v E)) e (TB)= \chi(F) \int_B e(TB)= \chi(F) \chi(B).$$<|endoftext|> -TITLE: Can formally differentiating give a derivative of a discrete function? -QUESTION [11 upvotes]: When I teach calculus, I really try to stress the importance of knowing the domain of a function. -One example that I sometimes like to use to show students the importance of inspecting the domain is the following: $f(x) = \sqrt{\sin(x)-1}$. I ask the student to differentiate this function, and they happily apply their formal rules to obtain $f'(x) = \frac{\cos(x)}{2\sqrt{sin(x) - 1}}$. Then I ask them to graph the original function. Hey, it is just a discrete set of points! I think at this point they realize that the differentiation they did was totally meaningless, and that paying attention to the domain might be a good idea. -One defect with this example is that the derivative really does fail to exist everywhere - it has no domain. I would like an example of a rule for a function whose "largest domain" (the domain where the rule can be evaluated) is a discrete set of points, but where formally differentiating the rule yields a derivative which has a nonempty "largest domain". I have thought off and on about this for a while now, and was hoping the collective MO community might be able to come up with something. Or perhaps someone might be able to prove that the standard "formal rules of differentiation" (say from Stewart's calc book), and the standard list of "common functions" are incapable of producing such an example. I would be happy either way. - -REPLY [9 votes]: I will show that Gerry's answer is typical, in some sense. -If $g(x)$ is analytic in a neighborhood of a point, say $x=0$, then its Taylor expansion is of the form $g(z) = \sum_{n=0}^{\infty} c_n z^n$ where $c_n = g^{(n)}(0)/n!$. Thus if $g$ is real-valued for real $x$ then all the $c_n$ are real. Apply this to $g(x) = f'(x)$ with $f$ as in the question. Then by integrating term by term we have $f(x) = C + \sum_{n=0}^{\infty} c_n x^{n+1}/(n+1)$, where $C$ is constant. The only way $f$ is not real is if $C$ is a non-real complex number. This is actually the case in Gerry's example of $f(x) = \log(-x^2) = \log(|x|^2) + i \arg(-1)$ (however you want to define the argument of $-1$). The trick to generating further examples is then to find disguised ways to write such a function $f$.<|endoftext|> -TITLE: Exactly how is 'the diagonal is representable' used for algebraic stacks... -QUESTION [9 upvotes]: ...apart from stating properties of $(s,t):X_1 \to X_0\times X_0$ for the a presenting algebraic groupoid $X_1 \rightrightarrows X_0$? -Once we know that given a stack $\mathcal{X}$ we have a smooth representable $X_0 \to \mathcal{X}$ where $X_0$ is a scheme, then we can talk about the algebraic groupoid $X_1 :=X_0\times_\mathcal{X} X_0 \rightrightarrows X_0$, which has source and target smooth maps. We thus have the map $(s,t)$, and can talk about its properties, such as being separated or whatever. We can specify its properties (such as having 'property P') by demanding that the diagonal $\Delta:\mathcal{X} \to \mathcal{X} \times \mathcal{X}$ is representable and has 'property P'. But where else is representability and property $P$ of $\Delta$ used, in a way that couldn't be otherwise derived from property $P$ of $(s,t)$? 'Everybody knows' that algebraic stacks and algebraic groupoids form the objects of two equivalent bicategories (there is a 1996 article by Dorette Pronk that makes this precise, and recent work by myself - available on the arXiv if you care - expands hers to be applied in more general situations). Thus I wonder what properties of the diagonal $\Delta$ are used that couldn't be instead derived from $(s,t)$ of a presenting groupoid. (Edit: or can all (stable under pullback) properties of the diagonal be so described - and also used?) -Pointers (in comments) to any relevant questions where example situations are discussed in detail would be appreciated. -(NB This is a spin-off from comments at this question.) - -REPLY [6 votes]: If $X_1 \rightrightarrows X_0$ is a groupoid and $\mathcal{X}$ is the associated stack, consider the atlas $p:X_0 \to \mathcal{X}$. If you form the weak $2$-pullback of $p \times p:X_0 \times X_0 \to \mathcal{X} \times \mathcal{X}$ against the diagonal $\Delta:\mathcal{X} \to \mathcal{X} \times \mathcal{X}$ you get by first projection $\left(s,t\right):X_1 \to X_0 \times X_0$. But $p \times p$ is an atlas for $\mathcal{X} \times \mathcal{X}$ so it follows that $\Delta$ is representable, and it has "property $P$" if and only if $\Delta$ does.<|endoftext|> -TITLE: "Skew Cohomology" of a Space -QUESTION [9 upvotes]: Let $X$ be a space. The symmetric group $\Sigma_{n+1}$ acts on the function space -$$ -X^{\Delta^n} -$$ -of continuous maps from the standard $n$-simplex to $X$. The action is induced -by permuting the vertices. -Let $\Sigma_{n+1}$ act on $\Bbb Z$ by means of the sign representation. -Then the singular $n$-cochains -$$ -S^n(X) := \text{map}(X^{\Delta^n},\Bbb Z) -$$ -inherits a $\Sigma_{n+1}$-action given by conjugating functions. -Definition. -The skew $n$-cochains on $X$ is given by the invariants -$$ -S^n(X)^{\Sigma_{n+1}} -$$ -that is, by equivariant functions $X^{\Delta^n} \to \Bbb Z$. -Then an elementary calculation shows that the usual singular coboundary operator $\delta$ -maps $S^n(X)^{\Sigma_{n+1}}$ into $S^{n+1}(X)^{\Sigma_{n+2}}$. -So we get a cochain complex: -$$ -S^0(X) \overset \delta \to \cdots \overset \delta \to S^n(X)^{\Sigma_{n+1}} \overset \delta \to S^{n+1}(X)^{\Sigma_{n+2}} \overset \delta \to \cdots -$$ -Define the skew cohomology of $X$ to be the cohomology of this cochain complex. -Questions: What is it? What properties does it have? Has it ever before been studied? - -REPLY [5 votes]: I do not have the book at hand to check, but maybe you do... -Loday defines in his book on Cyclic homology what a crossed simplicial group is and for each such gadget a corresponding homology theory. There is, for example, the "cyclic" crossed simplicial group, and the corresponding homology theory is cyclic homology. Well, there is a crossed simplicial groups---call it $\Delta S$---built from symmetric groups, and it turns out that the corresponding homology theory coincides with Hochschild homology. I am pretty sure your complex is the analogue of Connes' $\lambda$-complex corresponding to the homology theory for $\Delta S$, which is quasi-isomorphic to the complex which defines the actual $\Delta S$-homology rationally and, therefore, should just give you simplicial homology over $\mathbb Q$. -Integrally, you should not take invariants in each degree, but consider a double complex, each of whose columns are the complex which computes $H^\bullet(S_{n+1},\mathord-)$, much as the double complex which computes cyclic homology has columns which compute $H^\bullet(C_{n+1},\mathord-)$. -(I would love to have an example where taking invariants integrally does not work, by the way!)<|endoftext|> -TITLE: Why use Teichmuller representatives? -QUESTION [5 upvotes]: In p-adic mathematics, what is the advantage of using Teichmuller representatives over using just the numbers 0,1,2,...,p-1 ? -In either case, the norm is the same. -In either case, all the points are in the center. -Is is fair to say: "Teichmuller representives, just like the digits 0,1,2,...,p-1 , are just names of elements that are different in name but in no other characteristic. Although p-adic arithmetic seems more complex using Teichmuller representatives, more advanced mathematics is easier and more robust." - -REPLY [13 votes]: The Teichmüller lift can be seen as a map of multiplicative monoids from $\mathbb{F}_p$ to $\mathbb{Z}_p$, that is the unique multiplicative section of the mod $p$ reduction map. In particular, the lift produces roots of unity from nonzero inputs. You cannot say that about the integers from $1$ to $p-1$ (which tend to have infinite order). -Unlike writing integers up to $p-1$, taking the Teichmüller lift is a method that readily extends to more general settings. As KConrad mentioned, there is a more natural way to view the Teichmüller lift, as the embedding into the zeroth spot in the ring of Witt vectors. This works not only for the field with $p$ elements, but for any characteristic $p$ ring. In particular, finite extensions of $\mathbb{F}_p$ have Teichmüller lifts to unramified extensions of $\mathbb{Z}_p$. Natural operations like the Frobenius lift and Verschiebung can be expressed concisely in the Witt vector setting. -I mostly disagree with the quotation you offer. Integer representatives are better-suited to $p$-adic addition than Teichmüller representatives, and this makes explicit computations with polynomials and power series more tractable. Multiplication is handled by both types of representatives with roughly equal ease. If you want to prove a theorem (involving algebraic structures in the $p$-adics), you are likely to be better off with Teichmüller representatives.<|endoftext|> -TITLE: Tensor product of rings of Witt vectors -QUESTION [9 upvotes]: Let $A$, $B$, and $C$ be commutative rings such that $A\otimes_C B$ makes sense. If $W_n(A\otimes_C B), W_n(A), W_n(C),$ and $W_n(B)$ are the length $n$ Witt vectors of the rings $A,B,C,$ and $A\otimes_C B$. Is it true that -$$ -W_n(A\otimes_C B)\cong W_n(A)\otimes_{W_n(C)}W_n(B)? -$$ -It seems as though this is a sensible property for Witt vectors to have. The case I am particularly interested in is the case when $C$ is a field of characteristic $p$ (not necessarily perfect) and $A$ and $B$ are $C$-algebras, but any suggestions for the general case would be helpful as well. - -REPLY [14 votes]: When $B$ is étale over $C$ and $A$ or $B$ is finite over $C$, -then the result is known by Theorem (2.4) in my paper -Descent for the $K$-theory of polynomial rings.<|endoftext|> -TITLE: Lovasz theta function integrality -QUESTION [8 upvotes]: Is anything known about Lovasz Theta Function taking integral value in non-perfect graphs? In particular, does integral value of Lovasz theta always coincide with the size of largest independent set? -For instance, graph below is non-perfect, and its Lovasz theta function gives the independence number. -     -(source) -Its true for a few other non-perfect graphs I tried, here's a Mathematica package I used to compute it - -REPLY [4 votes]: No. For some examples see When the Lovász theta-function saturates its upper bound<|endoftext|> -TITLE: When should a supervisor be a co-author? -QUESTION [157 upvotes]: What are people's views on this? To be specific: suppose a PhD student has produced a piece of original mathematical research. Suppose that student's supervisor suggested the problem, and gave a few helpful comments, but otherwise did not contribute to the work. Should that supervisor still be named as a co-author, or would an acknowledgment suffice? -I am interested in two aspects of this. Firstly the moral/etiquette aspect: do you consider it bad form for a student not to name their supervisor? Or does it depend on that supervisor's input? And secondly, the practical, career-advancing aspect: which is better, for a student to have a well-known name on his or her paper (and hence more chance of it being noticed/published), or to have a sole-authored piece of work under their belt to hopefully increase their chances of being offered a good post-doc position? -[To clarify: original question asked by MrB ] - -REPLY [2 votes]: I believe it to be in bad taste for advisors to co-author a paper with their current student, no matter what their relative contributions. Of course, I am in pure mathematics, and understand that they do things differently in other parts of mathematics, and in other disciplines. Otherwise, there is the possibility of an advisor exploiting their position. This strict policy also obviates any soul-searching about how much contribution is required to be a co-author. -Of course, it is expected that the student will acknowledge in the paper, the help (if any) received from their advisor.<|endoftext|> -TITLE: Examples of $G_\delta$ sets -QUESTION [14 upvotes]: Recall that a subset $A$ of a metric space $X$ is a $G_\delta$ subset if it can be written as a countable intersection of open sets. This notion is related to the Baire category theorem. Here are three examples of interesting sets that are -$G_\delta$ sets. - - The set of continuity points of a function $f:X\rightarrow R$. - - The set of positively recurrent points of a continuous transformation $T: X \rightarrow X$. Recall that a point is recurrent if there exists some sequence $n_i\rightarrow \infty$ such that $T^{n_i}x \rightarrow x$. - The set of transitive points of a continuous transformation $T: X \rightarrow X$. Recall that a point is transitive if its orbit $\{T^n(x)\}_{n\in Z}$ is dense in $X$. - -Question : -Can you provide more examples of interesting $G_\delta$ sets ? - -REPLY [2 votes]: Here are a few $G_\delta$-sets related to measure theory. -Let $S$ be a $\sigma$-compact metric space -and denote by ${\cal M}(S)$ the set of Borel positive measures on $S$ endowed -with the weak topology. The following sets are $G_\delta$ sets: - -the set of non-atomic measures in ${\cal M}(S)$, -the set of measures in ${\cal M}(S)$ -singular with respect to a given measure, -the set of measures in ${\cal M}(S)$ whose support -contains a given closed set. - -This leads to the following surprising theorem. -Let $U$ be an open subset of ${\bf R}^d$, -then the set of singular continuous measures is a dense $G_\delta$ subset -of the set of ${\cal M}(U)$. (Lenz, Stollmann 2004). -There are related results in the theory of self-adjoint operators on a separable Hilbert space endowed with the strong resolvent topology. These are $G_\delta$ subsets: - -operators with no eigenvalues in a given closed subset of ${\bf R}$, -operators with spectrum containing a given closed subset of ${\bf R}$, -operators such that the spectral measure has no absolutely continuous component on a given open subset of ${\bf R}$. - -This leads to the following result. -The set of self-adjoint operators with spectrum $[-1,1]$ -and with purely singular continuous spectrum is a dense $G_\delta$ subset of the set of self-adjoint operators of norm $\leq 1$ on a separable Hilbert space, endowed with the strong topology. -This is a consequence of the Wonderland theorem of B. Simon, 1993.<|endoftext|> -TITLE: Completely positive maps as "positive operators" -QUESTION [11 upvotes]: Let $A$ be a unital $C^{*}$-algebra and $\phi:A \rightarrow A$ be a completely positive map, i.e. $\phi^{(n)}:M_{n}(A) \rightarrow M_{n}(A)$ preserves positivity for any natural number $n$, where $\phi^{(n)}((A_{ij})_{ij})=(\phi(A_{ij}))_{ij}$. It is well-known that the norm $\|\phi\|=sup_{\|x\|\leq 1}\{\|\phi(x)\|\}=\|\phi(I)\|$. -Also recall that, given an element $x$ in a $C^{*}$-algebra $A$, the element $\|x\|-x$ is always a positive element. -It is often mighty tempting to want to treat a unital c.p. map like it is an element of a $C^{*}$-algebra and say that $\|\phi\|id-\phi$ is a positive map. I've heard that there are many circumstances where this is doable. - -Question What are some natural circumstances where $\|\phi\|id-\phi$ is a positive map, with $\phi$ completely positive as above? - -For example, if $A$ is a finite von Neumann algebra, then sometimes the map $\phi$ naturally gives rise to a bounded operator on $L^{2}(A)$. If this operator happens to be a positive operator then we get what we want. (I'd be interested in knowing natural circumstances where we know that precisely this happens.) - -REPLY [7 votes]: This answer deals with the case that $\phi$ is non-unital. In this case, $\phi$ -must be of the form $\phi(a)=ha$, where $h$ is a positive element in the center -of $A$. -Unfortunately, the solution I've got is somewhat long (hopefully correct): Let us assume that $\phi$ is contractive (otherwise one can rescale). Since $\phi(a)\leq a$ for any positive element $a$, $\phi$ preserves orthogonality of positive elements. There's a nice structural result for such maps proven in ``Completely positive maps of order zero", by Winter and zacharias. By Theorem 2.3 of that paper $\phi$ has the form $\phi(a)=h\pi(a)$, where $h=\phi(1)$, -$$\pi:A\to M(C^*(\phi(A)))$$ -is a unital homomorphism into the multiplier algebra of $C^*(\phi(A))$ and $h$ commutes with the image of $\pi$. -From $\phi(a)\leq a$ one gets $h\pi(a)\leq a$ for any positive $a$. -Suppose that $\pi(b)=0$, with $b\in A_+$ positive contraction. Then $h=h\pi(1-b)\leq 1-b$. -Similarly, $h\leq 1-b^{1/n}$ for all $n$. This implies that $hb=0$. So $h$ is orthogonal to $\ker \pi$. In particular, $\pi$ is injective on $\overline{hAh}$. -Let us show that $\pi$ is the identity on $\overline{hAh}$. Taking $n$-root in $h\pi(h^n)\leq h^n$ we get $h^{1/n}\pi(h)\leq h$. Since $h$ is a strictly positive element of $C^*(\phi(A))$ and $\pi(h)$ is a multiplier for that algebra, we have that $h^{1/n}\pi(h)\to \pi(h)$ in the strict topology. So, $\pi(h)\leq h$, and so $\pi$ maps $\overline{hAh}$ into itself. Passing to the limit in $h^{1/n}\pi(a)\leq a$, with $a\in \overline{hAh}$, one gets $\pi(a)\leq a$ for all such $a$. But if $\alpha$ and $\beta$ are homomorphisms such that $\alpha\leq \beta$ on all positive elements, then $\alpha+\tilde\alpha=\beta$, where $\tilde\alpha$ is a homomorphism with orthogonal range to $\alpha$. (There's probably a reference for this; I'll skip the argument to keep this answer short.) In our case, since $\pi$ is injective, we get $\pi(a)=a$ for $a\in \overline{hAh}$. In particular, $\pi(h)=h$. -We have that -$$ -\phi(a)=h\pi(a)=\pi(h^{1/2} a h^{1/2})=h^{1/2} a h^{1/2} -$$for all $a\in A$. Finally, let us show that $h$ is in the center of $A$. -We have $\pi(ah-ha)=\pi(a)\pi(h)-\pi(h)\pi(a)=0$. So -$ha-ah\in \ker \pi$. It was argued above that this implies -$h(ha-ah)=0$. So $h^2a=ah^2$. So $h$ is in the center.<|endoftext|> -TITLE: Are units of rings of functions on algebraic varieties finitely generated (mod. constants)? -QUESTION [16 upvotes]: Hello, -Consider the following question. Let $A$ be a finitely generated reduced algebra over an algebraically closed field $k$. Consider the group of units of $A$, modulo the group $k^*$. Is this group always finitely generated? -Admittedly, I did not think of that question seriously, but I will be glad to hear the answer. -Thank you, -Sasha - -REPLY [11 votes]: I translate into English Lemma 6.5 from Sansuc's paper Groupe de Brauer et arithmétique des groupes -algébriques linéaires sur un corps de nombres, J. reine angew. Math. -327 (1981), 12-80. -Let $X$ be an algebraic variety over an arbitrary field $k$, -i.e a geometrically integral algebraic $k$-scheme. -We denote $ U(X) := k[X]^* / k^* $, -the group of units of the ring of regular functions $k[X]^*$ -modulo nonzero constants. - -Rosenlicht's theorem: Let $X$ and $Y$ be two algebraic $k$-varieties -and $G$ be a connected, smooth, linear algebraic $k$-group, . - - -(i) $U(X)$ is a finitely generated free abelian group; - - -(ii) $U(X\times_k Y)=U(X)\oplus U(Y)$; - - -(iii) $U(G)=\hat{G}(k)$ (the group of $k$-characters of $G$). - -Reference: M. Rosenlicht, Toroidal algebraic groups, -Proc. AMS 12 (1961), 984–988, -scan here. -For other references see Sansuc's paper. -(i) answers the question. -Rosenlicht's proof of (i) is close to Mohan's answer.<|endoftext|> -TITLE: Analogue of simplicial sets -QUESTION [18 upvotes]: This question is prompted by this one (and some of the comments that it drew). -Simplicial complex is to ordered simplicial complex as $X$ is to simplicial set. The question is about $X$. -Let $\text{Unord}$ be the full subcategory of $\text{Set}$ whose objects are the sets -$[n]=\lbrace 0,\dots ,n\rbrace$ for $n\ge 0$. An object of $\text{Set}^{\text{Unord}^{\text{op}}}$ is basically a simplicial set $X_\bullet$ with a suitable $\Sigma_{n+1}$-action on $X_n$ for all $n$. The obvious functor $\Delta:\text{Unord} \to \text{Top}$ (which creates a simplex with given vertex set) determines a functor -$$ -\text{Set}^{\text{Unord}^{\text{op}}} \to \text{Top} -$$ -This has a left adjoint analogous to the realization of simplicial sets. -Is the counit of this adjunction a weak homotopy equivalence for all spaces? -If so, is this adjunction a Quillen equivalence for some model structure on $\text{Set}^{\text{Unord}^{\text{op}}}$? -If not, is there something else along these lines that works? -(This must be known. It seems like an obvious question, and from some comments at the other question I gather that at least on the homology side this is something people thought about a long time ago.) - -REPLY [13 votes]: I have never thought of a counter example, but I would not bet on a positive answer to the first question. However, the answer to the second question is yes: this version of the singular functor is a right Quillen equivalence for a suitable model category on $\mathrm{Set}^{\mathrm{Unord}^{op}}$. -As mentioned above by Mike Shulman, this homotopy theory of symmetric simplicial sets is studied in the paper -J. Rosický and W. Tholen, Left-determined model categories and universal homotopy theories, Trans. Amer. Math. Soc. 355 (2003), 3611-3623. -In their paper, they construct the model structure on $\mathrm{Set}^{\mathrm{Unord}^{op}}$, but this has to be completed by -J. Rosický and W. Tholen, Erratum to "Left-determined model categories and universal homotopy theories", Trans. Amer. Math. Soc. 360 (2008), 6179-6179. -in which the authors explain that, after all, the class of cofibrations they considered is not the class of all monomorphisms (this is why I have some doubts about the fact that the topological realisation would be well behaved if we do not consider a cofibrant resolution in the picture). -You may find another construction of this model category (with an explicit description of what the cofibrations are) as well as its precise link with Kan's subdivision functor in section 8.3 of -D.-C. Cisinski, Les préfaisceaux comme modèles des types d'homotopie, Astérisque 308 (2006), -as an example of a test category in the sense of Grothendieck; see also this -MO question.<|endoftext|> -TITLE: Character tables and simple groups. -QUESTION [8 upvotes]: Is it known if there are two finite non-isomorpic non-abelian simple groups with the same character table? Does this answer change if the subsidiary information (like the orders and sizes of the conjugacy classes) is included? (There are no examples of this in the "ATLAS") - -REPLY [14 votes]: There are no such two non-isomorphic groups. As is pointed out in Neil Strickland's answer, up to some exceptions, a finite simple group is determined by its order. For these exceptions it is known that the smallest character degree larger than 1 is different -(For the infinite series $B_n(q)=O_{2n+1}(q)$ vs $C_n(q)=PSp_{2n}(q)$ for $q$ odd this follows from the results in Landazuri and Seitz (J. Algebra 32 (1974), 418–443, MR0360852 (50 #13299)). -This means that a finite simple group is determined in the class of finite simple groups by its character degrees with multiplicities, and led Huppert (Illinois J. Math. 44, 4 (2000), 828-842, MR1804317 (2001k:20009)) to the conjecture: - -Conjecture: If two finite groups $G$ and $H$ have the same set of character degrees (without counting multiplicities) and $G$ is nonabelian simple, then $H\cong G\times A$ for some abelian $A$. - -This has been verified for some simple groups, but is still open to the best of my knowledge. As mentioned in comments, Tong-Viet (MR2905242) has shown that finite simple groups are determined by their character degrees with multiplicities among all finite groups. Needless to say that all this depends heavily on the classification.<|endoftext|> -TITLE: Are most primes in a prime arithmetic progression of length at least 3? -QUESTION [8 upvotes]: Following the following two previous questions on mathoverflow: -Are all primes in a PAP-3? -and -Covering the primes by 3-term APs ? -I have attempted to show that infinitely many primes are in an arithmetic progression of length 3 in the primes following Ben Green's comment that one can do this using the circle method; but I have not found any success. Can anyone suggest (with more detail perhaps) a way to show that infinitely many primes are in an arithmetic progression of length at least 3? -Edit: In view of the comments, I have rephrased the question: My intention was to ask what can one do (Ben Green suggested the circle method, but gave no details) to show that 'most' primes (that is, the exceptional set has upper density 0) are in an arithmetic progression of at least 3 in the primes. - -REPLY [25 votes]: It is a theorem of Ben Green that every subset of the primes of positive relative density contains a progression of length three. As an immediate consequence, the set of primes $A$ which are not the first term in a progression of primes of length three has density zero (otherwise $A$ would contain a length three progression, a contradiction). -Ben's proof is, strictly speaking, an application of the circle method, but is probably overkill for your problem (roughly speaking, Ben wants to find a length three progression with all three elements in an arbitrary dense subset $A$ of the primes; for your problem, one only needs to study the simpler problem where the smallest term of the progression needs to be in $A$ but the other two elements lie in the set $P$ of primes). So a simpler proof (still by the circle method) is likely to exist. (Note, by the way, that a later paper of Ben and myself gives a slightly simpler proof of Ben's theorem (and, of course, we also have a more complicated proof as well).) -Another approach is to modify the argument of Montgomery and Vaughan (also, ultimately, based on the circle method) that shows that the number of even numbers that are not the sum $p_1+p_2$ of two primes is very low in density (much lower than the density of the primes, in particular). (Actually, an older and somewhat simpler paper of Vaughan already suffices for this.) The same argument should also show that the odd integers $n$ that are not the first term $2p_1-p_2$ of an 3-term AP whose other two terms $p_1,p_2$ are primes larger than $n$, also has density much smaller than that of the primes.<|endoftext|> -TITLE: Existence of an isometric embedding into Euclidean space with bounded second fundamental form -QUESTION [10 upvotes]: suppose we are given a complete, non-compact Riemannian manifold $(M,g)$. Is it possible to embed (or just immerse) it isometrically into some $R^N$, such that the second fundamental form is bounded? Maybe under some additional assumptions on our manifold and/or metric on it? -This question is a follow-up question to this one: Riemannian manifold of bounded geometry has a normal bundle of bounded geometry. -Thanks, Alex -edit: With "second fundamental form" I mean the quadratic form on the tangent space defined via taking the covariant derivative in $R^N$ and then orthogonally project it onto the normal bundle. So it is defined not only for hypersurfaces. -Anton Petrunin claimed in his answer that bounded curvature of $M$ and bounded injectivity radius are sufficient for the existence of such an embedding. I this is true, I would be grateful for a reference. - -REPLY [10 votes]: The curvature tensor can be expressed in terms of second fundamental form. -Therefore bounded curvature is a necessary condition. -Yet injectivity radius has to be bounded below. -These two conditions are sufficient. -It seems that this could be proved along the same lines as the Nash embedding theorem. -P.S. In the formulation, you had to say what you mean by "second fundamental form". -Most people think it is only defined for hypersurfaces, but you mean a quadratic form on tangent space with values in the normal space.<|endoftext|> -TITLE: Median largest-prime-factor -QUESTION [16 upvotes]: Let $P(n)$ denote the largest prime factor of $n$. For any integer $x\ge2$, define the median -$$ -M(x) = \text{the median of the set }\{P(2), P(3), \dots, P(x) \}. -$$ -Classical results of Dickman and de Bruijn show that the median is roughly $x^{1/\sqrt{e}}$. More specifically, I think that the Dickman-de Bruijn rho-function approach can show the following: for any function $f(x)$ tending to infinity with $x$, the median $M(x)$ is between $x^{1/\sqrt{e}}/f(x)$ and $x^{1/\sqrt{e}}f(x)$ for all sufficiently large $x$. -But I got to thinking the other day: is there a way to determine how the median compares to $x^{1/\sqrt{e}}$ specifically? In other words, which one of the following is true? - - For all sufficiently large $x$, we have $M(x) \lt x^{1/\sqrt{e}}$. - Each inequality $M(x) \lt x^{1/\sqrt{e}}$ and $M(x) \ge x^{1/\sqrt{e}}$ holds for arbitrarily large $x$. - For all sufficiently large $x$, we have $M(x) \ge x^{1/\sqrt{e}}$. - -REPLY [9 votes]: I posted a paper to the arXiv which deals with this question along with some other things. -Using some results regarding either the mean of $\omega(n)$, or integers without large prime factors, we can prove that $$M(x)=e^{(\gamma-1)/\sqrt{e}}x^{1/\sqrt{e}}\left(1+O\left(\frac{1}{\log x}\right)\right),$$ where $\gamma$ is the Euler Mascheroni constant. -More specifically, if we let $\text{li}_f(x)=\int_2^x \frac{\{x/t\}}{\log t} dt$, then $$M(x)=x^{\frac{1}{\sqrt{e}}\exp\left(-\frac{\text{li}_f(x)}{x}\right)}\left(1+O\left(e^{-c\sqrt{\log x}}\right)\right).$$ (In case the TeX is not readable, the exponent of $x$ in the above equation is $\frac{1}{\sqrt{e}}\exp\left(-\frac{\text{li}_f(x)}{x}\right)$.) -This function $\text{li}_f(x)$ has the asymptotic expansion $$\text{li}_f(x)=c_0 \frac{x}{\log x}+c_1 \frac{x}{\log^2 x}+\cdots+c_{k-1} \frac{(k-1)!x}{\log^k x}+O\left(\frac{x}{\log^{k+1}(x)}\right),$$ -where $$c_k=1-\sum_{j=0}^k\frac{\gamma_k}{k!},$$ and $\gamma_k$ denotes the $k^{th}$ Stieltjes constant.<|endoftext|> -TITLE: Analogue to Serre spectral sequence for cofiber sequences and homotopy -QUESTION [10 upvotes]: (This is a follow-up question to this one). -As it is nicely outlined in an answer to this question, homotopy groups behave well with respect to (Serre)-fibrations and (co)homology groups behave well with respect to cofibrations. The Serre spectral sequence calculates what happens with (Serre)-fibrations and (co)homology. My question is: Is there an analogue to the Serre spectral sequence for cofibrations and homotopy groups? - -REPLY [9 votes]: There is, sort of, and the idea was developed in the paper "Induced fibrations and cofibrations" by Tudor Ganea in a 1967 paper appearing in the transactions. -The idea is roughly this: -given a cofibration -$$ -A \to X \to C , -$$ -let us set $X_1 = \text{hofiber}(X \to C)$, where the latter means the homotopy fiber. -Then there is a preferred lift $A \to X_1$. Define $C_1$ to be the mapping cone of this map, so we have a cofiber sequence $A \to X_1 \to C_1$. We now iterate the forgoing to get a sequence of spaces -$$ -\cdots \to X_n \to \cdots \to X_1 \to X_0 = X -$$ -together with compatible maps $A \to X_n \to C_n$. Under mild hypotheses, the map -$$ -A \to {\text{holim}}_n \phantom{,} X_n -$$ -is a weak homotopy equivalence (for example, it will be the case when $A$ and $C$ are $1$-connected). -The homotopy groups tower above gives rise to a spectral sequence where the $E_1$-term is -given by $\pi_p(X_q,A)$ (I am being sloppy about the indexing conventions here). In a certain stable range (roughly, $p < 2(qa + c)$ if I recall, where $A$ is $a$-connected and $C$ is $c$-connected), these latter groups are identified with -$$ -\pi_p(A^{[q]}\wedge \Omega C) -$$ -where $A^{[q]}$ is the $q$-fold smash product of $A$ with itself. -A problem, which I'm not sure how to solve, is how does one identify the $k$-invariants of the tower to obtain a description of the $E_2$-term in this stable range? -Added: On second reflection, the above is not really an analogue of the Serre spectral sequence. The Serre spectral sequence for $F \to E \to B$ is given by filtering $B$ as a CW -complex and pulling back the fibration along skeleta. Dually, the thing to do would be to approximate the cofibration $A\to X \to C$ by a Postnikov tower $A^i$, and the take cobase change of $A \to X$ along $A\to A^{(i)}$ to get a tower $X^{(i)}$ with compatible maps from $X$ into it. Under mild hypotheses, this ought to converge to $X$ when taking the homotopy inverse limit. Then the thing to do would be to consider the homotopy spectral sequence of the tower $X^{(i)}$. However, I'm skeptical that there's a description of the $E_2$-term in terms of $A$ and $C$.<|endoftext|> -TITLE: Rate of convergence of smooth mollifiers -QUESTION [7 upvotes]: How does one figure out/prove the rate of convergence (in some norm) of mollifiers given a function bounded in some other norm (say Sobolev space, Besov space)? Also, is there a dimensional analysis heuristic which will predict what the rate will be? -For example, it is true that -$$ ||u - u^{(\epsilon)}||_{L^3} \leq C \epsilon^\alpha ||u||_{B_3^{\alpha,\infty}} $$ -where the norm on the right hand side is a Besov space norm. (This fact is used in Constantin, E and Titi's paper on Onsager's Conjecture for Solutions to Euler's Equation). - -REPLY [8 votes]: You can do something with simple scaling as long as you work on the full space. Assume that $|\cdot|$ and $\|\cdot\|$ are two shift-invariant (semi)norms and for the scaling operator $T_a f(x)=f(x/a)$ you have $|T_af|=a^t|f|$, $\|T_a f\|=a^s\|f\|$. Then, if you do the $\delta$-mollifying on $f$, it is equivalent to the $a\delta$-mollifying on $T_a f$, which means that if any inequality of the form $|f-f_\delta|\le C\delta^r\|f\|$ exists at all, you must have $a^t=a^r a^s$, i.e., $r=t-s$. -Now, to check that the inequality is there, you just need to check that $\|f\|$ dominates the deviation of $f$ from some faithfully reproduced function in the norm $|\cdot|$. Note that we need the homogeneous spaces for such scaling tricks, so the Sobolev norm will really mean the $L^p$ norm of the highest derivative involved, not the full sum of $L^p$-norms of the previous derivatives. -Example: $C$ and $C^k$ (with uniform bounds on the entire line). If we have any chance to get anything at all, the speed is $\delta^k$ by scaling. We have this chance realized if and only if our mollifier (which I assume to be compactly supported for simplicity) has first $k-1$ moments correct (i.e., reproduces the polynomials of degree up to $k-1$ precisely).<|endoftext|> -TITLE: Are Kahler differentials the same on the affine closure on a quasi-affine scheme? -QUESTION [6 upvotes]: Let $X$ be a quasi-affine scheme; that is, the natural map -$$X\rightarrow \overline{X}:=Spec(\Gamma(X,\mathcal{O}_X))$$ - is an inclusion. Each scheme has a quasi-coherent sheaf of Kahler differentials $\Omega$, and the above open inclusion induces a $\Gamma(X,\mathcal{O}_X)$-module map of global Kahler differentials -$$ \Gamma(\Omega_{\overline{X}})\rightarrow \Gamma(\Omega_{X})$$ -Is this map always an isomorphism? -Edit: Changed $\mathcal{O}_X$ to $\Gamma(X,\mathcal{O}_X)$. - -REPLY [5 votes]: The assumption implies that the natural embedding induces an isomorphism $\Gamma(X,\mathscr O_X)\simeq \Gamma(\overline X,\mathscr O_{\overline X})$. Then this means that the complement of $X$ has at least codimension $2$. -In addition assume that $X$ is noetherian and $S_2$ (for instance normal). -In this case if $\Omega_{\overline X}$ is a reflexive sheaf, then the restriction -$\Gamma(\overline X,\Omega_{\overline X})\to \Gamma(X,\Omega_{\overline X})=\Gamma(X,\Omega_{X})$ is an isomorphism. -More generally, let $Z:=\overline X\setminus X$. If $\mathrm{depth}_Z\Omega_{\overline X}\geq 2$ then the restriction $\Gamma(\overline X,\Omega_{\overline X})\to \Gamma(X,\Omega_{X})$ is an isomorphism. -This certainly holds if $\Omega_{\overline X}$ is a reflexive sheaf, but obviously it could hold "by accident" even if one of the above conditions fail, so I am not claiming that these are necessary conditions, but at least they sure seem to provide a natural set of conditions under which the required map is an isomorphism. -Sketch that if $X$ is $S_2$, then a reflexive coherent sheaf $\mathscr F$ is also $S_2$: -First observe that by the argument in this answer to another MO question we may assume that $X$ is affine and it is enough to prove that $H^i_x(X,\mathscr F)=0$ for $i=0,1$ for all $x\in X$ and it also follows that $\mathrm{depth}_Z\mathscr F\geq 2$ even if $Z$ is not contained in an affine piece of $X$. To do that write $\mathscr F^\vee$ as the quotient of a (locally) free sheaf ($X$ is affine!). Then $\mathscr F$ is a submodule of the dual of this locally free sheaf, let's call it $\mathscr E$, and the quotient $\mathscr E/\mathscr F$ is torsion-free. Therefore none of them have torsion and so $H^0_x(X,\mathscr F)=0$ and $H^1_x(X,\mathscr F)$ embeds into $H^1_x(X,\mathscr E)$. But the latter is $0$ by the assumption that $X$ is $S_2$. -EDIT 1 removed intro paragraph about the starting assumption. -EDIT 2 added "more generally" paragraph. -EDIT 3 added noetherian assumption. this is probably not necessary but without this one should possibly be more careful about the other conditions. -EDIT 4 Added Sketch above.<|endoftext|> -TITLE: Are there any non-linear solutions of Cauchy's equation $f(x+y)=f(x)+f(y)$ without assuming the Axiom of Choice? -QUESTION [24 upvotes]: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be s.t. $f(x+y) = f(x) + f(y), \ \forall x, y$ -It is quite obvious that this implies $f(cx)=cx$ for all $c \in \mathbb{Z}$ and even further: $\forall c \in \mathbb{Q}$ -It is also known that using an infinite dimensional basis for $\mathbb{R}$ over $\mathbb{Q}$, it is possible using the Axiom of Choice to construct a function which is not linear. -My question is whether there exists a known solution not invoking the Axiom of Choice? -Edit: My questions are: - -Is it not possible to find an -explicit expression for a non-linear -solution? Why? - -Can existence of non-linear -solutions be proven from ZF alone? - - -Reference: Cauchy's functional equation - -REPLY [52 votes]: It is not possible to provide an explicit expression for a non-linear solution. The reason is that (it is a folklore result that) an additive $f:{\mathbb R}\to{\mathbb R}$ is linear iff it is measurable. -(This result can be found in a variety of places, it is a standard exercise in measure theory books. As of this writing, there is a short proof here (Internet Archive).) -The point is that it is consistent that every set of reals is measurable, and therefore every function $f:{\mathbb R}\to{\mathbb R}$ is measurable. -The first to realize that it is possible using choice to construct a non-linear additive function was Hamel in 1905 (Hamel, G., Eine Basis aller Zahlen und die unstetigen Lösungen der Funktionalgleichung: (f(x+y) =f(x) +f(y))., Math. Ann. 60, 459-462 (1905). ZBL36.0446.04.); indeed, a Hamel basis of ${\mathbb R}$ over ${\mathbb Q}$ allows us to provide examples. Note that with a Hamel basis it is straightforward to build Vitali's standard example of a non-measurable set. -Solovay (Solovay, R. M., A model of set-theory in which every set of reals is Lebesgue measurable, Ann. Math. (2) 92, 1-56 (1970). ZBL0207.00905.) proved that, relative to the consistency of an inaccessible cardinal, via forcing we can prove that it is consistent with ZF + DC that all sets of reals are Lebesgue measurable (DC is the axiom of dependent choice, a weak version of the axiom of choice useful to develop the basic theory of analysis). It follows that, in Solovay's model, all additive functions are linear and therefore no "explicit" example is possible. -The other standard approach to models where all sets of reals are measurable is via determinacy. However, this requires a stronger commitment in terms of consistency strength (using $\omega$ Woodin cardinals we can force an inner model of determinacy). The two approaches are closely related: Assuming enough large cardinals, the inner model $L({\mathbb R})$ consisting of all sets constructible from reals is a model of determinacy, and a Solovay model. Note that this is a theorem (from enough large cardinals) rather than a consistency result, i.e., $L({\mathbb R})$ is a model of these statements, without needing to pass to a forcing extension. -Solovay's result does require an inaccessible cardinal, this is a result of Shelah (Shelah, Saharon, Can you take Solovay’s inaccessible away?, Isr. J. Math. 48, 1-47 (1984). ZBL0596.03055.). This leaves the question of whether (the consistency of) ZF suffices to produce a model where all additive maps are linear. But this is now easily solved: In the same paper, Shelah proved that it is relatively consistent with ZF that all sets of reals have the property of Baire. The standard proofs that additivity and measurability give linearity work with "Baire measurability" instead of Lebesgue measurability. -Let me close, however, by pointing out that, in appropriate models of set theory, we can "explicitly" build additive, non-linear maps. More precisely, in nice inner models, we have explicitly definable well-orderings of the reals, from which such maps can be built. For example, in Gödel's inner model L of constructible sets, there is an easily definable Hamel basis (see this MO question), from which we can easily define such a function. "Easily" is formalized in terms of the projective hierarchy. The Hamel basis we obtain is $\Pi^1_1$ and the function is $\Sigma^1_2$. -(By the way, the statement "there is a discontinuous additive function" is form 366 in Howard-Rubin "Consequences of the axiom of choice". The highly recommended companion website does not reveal any additional choice-like implications regarding this form. In particular, it only mentions Solovay's model for its negation, but not Shelah's. Sadly, the last link does not work currently, but see this Wayback Machine backup.)<|endoftext|> -TITLE: Does the derived category remember the homological dimension? -QUESTION [10 upvotes]: Question: -Let $\mathcal{A}$ be an abelian category and $D^?(\mathcal{A})$ be its derived category, where ? could be empty, +, - or b (for boundedness). Is it possible to recover the homological dimension of $\mathcal{A}$ from the derived category? -Here I'm using the term homological dimension in the sense of Gelfand-Manin, i.e. if for all $X,Y\in\mathcal{A}$, $\text{Ext}\_{\mathcal{A}}^i(X,Y):=\text{Hom}_{D(\mathcal{A})}(X[0],Y[i])=0$, then the homological dimension is said to be less than $i$. The homological dimension is the maximal $n$ such that there exists a non-vanishing $\text{Ext}^n(X,Y)$. -Note that in the derived category one could have all kinds of non-vanishing $\text{Ext}^n(X,Y)$, as $X,Y $ can be complexes shifted arbitrarily. Is it still possible to recover this information via other method? - -REPLY [13 votes]: There are lots of examples to be found in the theory of tilted algebras. -A tilted algebra $B$ is an algebra of the form $\operatorname{End}_A(T)$ with $A$ an hereditary finite dimensional algebra (over an algebraically closed field, say) and $T$ a tilting $A$-module. Then $B$ and $A$ have equivalent derived categories, but $B$ is usually not hereditary (that is, its global dimension is usually bigger than $1$; one does have $\operatorname{gldim}B\leq2$, though, so it does not blow up too much). -A concrete example: let $A$ be the path algebra of the quiver $$\bullet \leftleftarrows \bullet \leftarrow \bullet$$ Number the vertices $1$, $2$ and $3$ from left to right, and let $T$ be the direct sum of the simple $S(1)$ and the indecomposable injective modules $I(1)$ and $I(3)$, which is a tilting module. Then $B=\operatorname{End}_A(T)$ is the algebra given by the same quiver, but bound by the relations given by the two paths of length two. In particular, $B$ is not hereditary. -You will find all this discussed at length in Assem, Simson and Slowroński's book Elements of the Representation Theory of Associative Algebras, Vol. 1. -If one considers more generally tilting complexes as opposed to just modules, then the difference between the global dimensions of the algebras involved can be made as large as you want.<|endoftext|> -TITLE: How hopficity is preserved under finite index? -QUESTION [5 upvotes]: Dear All! I guess this should be well-known, but I cannot find it in literature: -Is it true that if $H$ is a subgroup of finite index in $G$, and $H$ is hopfian, then $G$ is also hopfian? -In case when the groups are finitely generated, the answer is "yes" by R. Hirshon. -(And the other way around question has answer no, as was kindly answered by Kate Juschenko, even for finitely generated case -- which is quite bizarre!!!) - -REPLY [9 votes]: Baumslag-Solitar group BS(12,18) is Hopfian, but it contains normal subgroup of finite index which is not Hopfian. -it is proved in the following paper: -Baumslag; Solitar: Some two-generator one-relator non-Hopfian groups.<|endoftext|> -TITLE: Ends of topological spaces. Why independent of choice of ascending sequence of compact subsets? -QUESTION [5 upvotes]: Quoting from http://en.wikipedia.org/wiki/End_(topology): -"Let X be a topological space, and suppose that -K1 ⊂ K2 ⊂ K3 ⊂ · · · -is an ascending sequence of compact subsets of X whose interiors cover X. Then X has one end for every sequence -U1 ⊃ U2 ⊃ U3 ⊃ · · · -where each Un is a connected component of X \ Kn. The number of ends does not depend of the specific sequence {Ki} of compact sets; in fact, there is a natural bijection between the sets of ends associated with any two such sequences." -How does one prove that the number of ends does not depend on the specific sequence of {Ki} of compact sets? -An expository proof for a relatively neophyte math student (no category theory...) would be much appreciated. - -REPLY [14 votes]: Neil has already given adequate reply; this answer is partly for Simon, and partly for those who do like category theory, and realize that its purpose is to make life simpler, not more complicated! -First, IMHO that's not a very good definition in the wikipedia article. A better definition is given in Spivak's A Comprehensive Introduction to Differential Geometry, Volume I, page 30: an end of a non-compact topological space $X$ is a function $e$ which assigns to each compact subset $K \subset X$ a nonempty component $e(K)$ of the complement $X - K$, in such a way that $K \subset K'$ implies $e(K') \subset e(K)$. This way of putting it circumvents having to choose a covering by interiors of compact sets at the outset, and then requiring a lemma that shows independence of choice. -In categorical language, the set of ends of $X$ is the inverse limit of sets -$$\lim_{K \subset X} \pi_0(X - K)$$ -where $K$ ranges over compact subsets. -Anyway, in answer to the question, the point is that any sequence of compact subsets whose interiors cover $X$ is cofinal in the directed set of all compact subsets. (A partially ordered set is directed if it is nonempty and if any two elements have an upper bound. A subset is cofinal if any element in the partial order is bounded above by an element in the subset.) -The point then is that the limit over a directed set is isomorphic to the limit over a cofinal subset (with partial order inherited from the order of the directed set): in the present case, the sequence $K_j$ is cofinal, and the map given by restriction -$$\lim_{K} \pi_0(X - K) \to \lim_j \pi_0(X - K_j)$$ -is a bijection. The inverse function takes a sequence of components $C_j$, and assigns to it the function whose value at $K$ is the unique component of $X - K$ which contains $C_j$, where $K_j$ is any compact subset containing $K$. This doesn't depend on $j$, and it is routine to show this does give the inverse function, according to what Neil has already explained. -But it is really just a special case of a much more general argument about cofinal functors; see Categories for the Working Mathematician, page 217, for a rather more general statement.<|endoftext|> -TITLE: finite flat commutative group schemes arising from Abelian varieties -QUESTION [9 upvotes]: How are the finite flat group schemes $\mathcal{A}[\ell^n]$ arising from an Abelian scheme $\mathcal{A}/S$ singled out among other finite flat commutative group schemes of exponent $\ell^n$? - -REPLY [3 votes]: I'm reminded of two papers by Maja Volkov, an erstwhile student of Jean-Marc Fontaine. They are -MR2148801 (2006a:14027) -Volkov, Maja -A class of $p$-adic Galois representations arising from abelian varieties over $\Bbb Q_p$. -Math. Ann. 331 (2005), no. 4, 889–923. -MR1837096 (2002d:11067) -Volkov, Maja -Les représentations $l$-adiques associées aux courbes elliptiques sur ${\Bbb Q}_p$. -J. Reine Angew. Math. 535 (2001), 65–101.<|endoftext|> -TITLE: Can we unify addition and multiplication into one binary operation? To what extent can we find universal binary operations? -QUESTION [72 upvotes]: The question is the extent to which we can unify addition -and multiplication, realizing them as terms in a single -underlying binary operation. I have a number of questions. - -Is there a binary operation $n\star m$ on the integers -$\mathbb{Z}$ such that both addition $+$ and multiplication -$\cdot$ can be expressed as specific composition -expressions using only $\star$? A more relaxed version of -the question would allow constants into the expressions; -for example, perhaps we can arrange that $a+b=0\star(a\star -b)$ and $ab=1\star(a\star b)$. -One obvious idea is to try somehow to use a pairing -function, -so that $a\star b$ codes up both $a$ and $b$ into one -number, which then can appear as one argument to $\star$, -whose other argument will signal whether or not addition or -multiplication is desired. But there is the difficulty of -making these two tasks not interfere with one another. -Note that if we allow a trinary operation, then we can -easily do it simply by defining $\star(0,a,b)=a+b$ and -$\star(1,a,b)=ab$ and extending this arbitrarily. Can we -get rid of the need for parameters here? -Can we prove that there is no associative such binary -operation $\star$ on the integers, from which both $+$ and -$\cdot$ are expressible by terms? -Does every ring have such a binary operation $\star$ -from which both addition $+$ and multiplication $\cdot$ are -expressible as $\star$-terms? Does it matter if the ring is -finite or infinite? -Can every countable family $F$ of finitary operations on -an infinite set $Z$ be unified by a single binary operation -$\star$ on $Z$, so that every function in $F$ is the same -function as that induced by some $\star$-term? - -This question arose from what I found interesting about a -recent question on -math.SE, -asked by someone who was interested in the phenomenon that -logical and and or are expressible from nand and also -from nor. - -REPLY [4 votes]: Here is an easy answer to question 1. -Let $\langle x,y\rangle$ be a pairing function with $L(\langle x,y\rangle)=x$ and $R(\langle x,y\rangle)=y$. -Define: -$$\begin{align} -a \star 1 &= 2a\\ -a \star 3 &= L(a)\\ -a \star 5 &= R(a)\\ -a \star 2b &= \langle a+b,ab\rangle -\end{align}$$ -and other values of $a\star b$ equal to 1. -Then: -$$a+b = L(a\star2b) = (a\star(b\star1))\star3$$ -$$ab = R(a\star2b) = (a\star(b\star1))\star5$$<|endoftext|> -TITLE: How to distinguish division algebras from matrix algebras? -QUESTION [14 upvotes]: Suppose $D$ is an explicitly given rank 9 central simple algebra over ${\mathbb Q}$ (or a number field). For example it could be specified by two cubic subfields ${\mathbb Q}(a), {\mathbb Q}(b)\subset D$ and one relation $ba=\sum a_i b_j$. What is the best way to determine whether $D$ is a matrix algebra or a division algebra? -Any ideas or references would be highly appreciated! I simply tried to look for random elements whose minimal polynomial is reducible (sort of like Parker's meataxe for group representations), but this is not terribly efficient, and I don't know how to do this cleverly. Perhaps one could also take many cubic subfields and record which primes split in them - if the union of split primes over all subfields seems to cover all prime numbers, this suggests that $D$ is a matrix algebra; again, I don't know how to make this into an actual algorithm. - -REPLY [8 votes]: This may be repeating what others have said as it essentially follows the maximal order approach, but have you looked at Nebe, Gabriele; Steel, Allan, Recognition of division algebras, J. Algebra 322 (2009), no. 3, 903–909? -http://dx.doi.org/10.1016/j.jalgebra.2009.04.026 -Preprint version and magma code available here: -http://www.math.rwth-aachen.de/~nebe/pl.html -(I know this should really be a comment, but I'm afraid that I don't have enough reputation yet.)<|endoftext|> -TITLE: Percolation in Cayley graphs of semigroups. -QUESTION [5 upvotes]: Percolation in Cayley graphs of groups are studied by many researchers. There are also the concept Cayley graphs for semigroups. Are there any research about percolation in Cayley graphs for semigroups? - -REPLY [2 votes]: Percolation on Cayley graphs of groups is a particular case of percolation on transitive graphs. Percolation on transitive graphs satisfies many important properties which do not hold in general (see this and -this papers. That provides a motivation. The Cayley graphs of semigroups are typically not transitive. Nevertheless some Cayley graphs of semigroups have been considered. For example percolation on the NE quadrant of the square lattice (which is the Cayley graph of the free commutative semigroup) is considered here.<|endoftext|> -TITLE: How many different possible simply graphs are there with vertex set V of n elements -QUESTION [8 upvotes]: Is V is a set with n elements, how many different simple, undirected graphs are there with vertex set V? - -REPLY [8 votes]: See http://oeis.org/A000088. This is the sequence which gives the number of isomorphism classes of simple graphs on n vertices, also called the number of graphs on n unlabeled nodes. You will also find a lot of relevant references here.<|endoftext|> -TITLE: Looking up the Mordell-Weil rank and generator(s) of a Weierstrass Equation -QUESTION [5 upvotes]: Is there a web site where one can look up a Weierstrass equation, by discriminant say, or coefficients of some readily derivable "standard" form, and find the rank of its solutions over Q neatly listed, along with a set of generators and torsion group? -After several web searches, the nearest I've come is the Cremona Tables. But they list by conductor, whatever that is, and there seems no obvious way for an amateur such as myself to translate the data on that site into a form usable for the above mentioned purpose. -Failing that, I'd be content with an eay to use Mathematica or Sage package to achieve this, with idiot-proof instructions. -Right now, I'm especially interested in the equation $ x^2 + x (x + 1)^2 = y^2 $, which obviously has some rational solutions, and I'm pretty sure has positive rank over Q. -But the answer for that specific equation, although useful at present, obviously won't help with others that may interest me in future - "Teach a man to fish" and all that .. -P.S. If the Cremona Tables can easily be used to look up equations such as the example above, I'd very much appreciate a simple walkthrough, using the above as an example, and I think others would also find this useful. - -REPLY [8 votes]: This is a very old question, and although the old answers are okay, there's now a great source to get information about specific elliptic curves, with multiples ways to search. This includes Cremona's tables, vastly extended. It's the L-Series and Modular Forms Database (LMFDB).<|endoftext|> -TITLE: A geometric reference for (affine) Gorenstein varieties and singularities -QUESTION [15 upvotes]: I would like to ask for a reference to some text that explains in relatively down to earth (if possible geometric) terms (for dummies) what is a Gorenstein singularity and Gorenstein variety (for a person whose knowlage of commutative algebra is minor). What standard things one could try to do to check if a given scheme is Gorenstein? -If the field of definition is $\mathbb C$ - it is even better. I know one source - Eisenbud "commutative algebra" but find it a bit hard. - -REPLY [10 votes]: Sorry, tough luck, but most first (and second) algebraic geometry courses don't even touch Cohen-Macaulay rings, let alone Gorenstein. Look, for example, at Definition 4.2 here. So it is unlikely you can find such reference. -Here is an explanation why you need both Cohen-Macaulayness and the fact that the canonical divisor is Cartier, as mentioned in Karl's answer. The trouble is that there are UFDs (so all divisors are Cartier) which are not Cohen-Macaulay (for instance, the invariant ring of $\mathbb Z_4$ acting by cyclically permuting the variables on the polynomial ring in four variables over a field of char 2). Such examples are not very well-known, I remember Sándor Kovács pointed out in a recent comment that most people don't even realize that it could be an issue. -But without Cohen-Macaulayness, the canonical sheaf would not be truly dualizing (see the comments here). This is perhaps where the real power of the property lies. -Finally, I would like to recommend the survey on Gorenstein rings by Craig Huneke: Hyman Bass and Ubiquity: Gorenstein Rings. You can pick up a lot about them from there, including the very interesting history. To quote from the Introduction: - -As we shall see, they could perhaps more justifiably be called Bass rings, or Grothendieck rings, or Rosenlicht rings, or Serre rings. - -Enjoy!<|endoftext|> -TITLE: A finitely generated $\mathbb{Z}$-algebra that is a field has to be finite -QUESTION [19 upvotes]: I was trying to understand completely the post of Terrence Tao on Ax-Grothendieck theorem. This is very cute. Using finite fields you prove that every injective polynomial map $\mathbb C^n\to \mathbb C^n$ is bijective. It seems to me that the only point in the proof presented in the post that is not explained completely is the following lemma: - -Take any finitely generated ring over $\mathbb Z$ and quotient it by a maximal ideal. Then the quotient is a finite field. - -Is there some comprehensible reference for the proof of this lemma? -In slightly different wording, the question is the following: assuming Nullstellensatz, can one really give a complete proof of Ax-Grothendick theorem in two pages, so that it can be completely explained in one (2 hours) lecture of an undergraduate course on algebraic geometry? - -REPLY [5 votes]: Let me add a self-contained answer which is completely elementary and avoids both the Nullstellensatz and Noether's normalization. It is a polished version of this article (already linked in Daniel Litt's comments). It partially overlaps with the answers of Daniel Litt and Guillermo Mantilla. - -Lemma 1. Let $R$ be a UFD with infinitely many primes. Then an algebraic field extension $L$ of $F:=\operatorname{Frac}(R)$ cannot be finitely generated as an $R$-algebra. -Proof. Assume $L=R[y_1,\dots,y_n]$, with each $y_j$ being a root of a certain monic polynomial $p_j\in F[x]$. Taking the common denominator $d\in R$ of the coefficients of these polynomials $p_j$, we get that $y_j$ is integral over $R':=R[1/d]\subseteq F$. But then, given a prime $p\nmid d$ (here we use that $R$ has infinitely many primes), the same holds for $1/p\in L=R'[y_1,\dots,y_n]$. However, since $R'$ is still a UFD, this implies $1/p\in R'$ (as a UFD is integrally closed), contradiction. $\blacksquare$ -Lemma 2. Given a field extension $K\subseteq L$, if $L$ is finitely generated as a $K$-algebra then the extension is algebraic, and in particular finite (meaning $\operatorname{dim}_K L<\infty$). -Proof. Assume $L=K[z_1,\dots,z_m]$. Since $L=K(z_1)[z_2,\dots,z_m]$, by induction $L$ is algebraic over $K(z_1)$. If $z_1$ is transcendental over $K$, then $R:=K[z_1]\cong K[x]$ satisfies the hypotheses of Lemma 1, which contradicts that $L$ is finitely generated as an $R$-algebra. So $z_1$ is algebraic over $K$, hence also $L$. (Note that the base case $m=1$ is obvious, as $1/z_1\in K[z_1]$ implies that $z_1$ is the root of some polynomial over $K$.) $\blacksquare$ -Theorem. If $L$ is a field which is finitely generated, then $L$ is a finite field. -Proof. $L$ is (isomorphic to) the quotient of the ring $\mathbb{Z}[x_1,\dots,x_n]$ by a maximal ideal $M$. Observe that $M\cap\mathbb Z$ is a prime ideal of $\mathbb Z$. If $M\cap\mathbb Z=\{0\}$, then $\mathbb Z$ embeds into $L$; but this contradicts Lemma 1 (with $R:=\mathbb Z$)! Hence $\mathbb F_p$ embeds into $L$ for some prime number $p$ and Lemma 2 gives that $L$ is finite. $\blacksquare$<|endoftext|> -TITLE: Examples in mirror symmetry that can be understood. -QUESTION [54 upvotes]: It seems to me, that a typical science often has simple and important examples whose formulation can be understood (or at least there are some outcomes that can be understood). So if we consider mirror symmetry as science, what are some examples there, that can be understood? -I would like to explain a bit this question. If we consider the article -"Meet homological mirror symmetry" http://arxiv.org/abs/0801.2014 -it turns out, that in order to understand something we need to know huge amount of material, including $A_{\infty}$ algebras, Floer cohomology, ect. -Here, on the contrary, is an example, that "can be understood" (for my taste): According to Arnold, the first instance of symplectic geometry was "last geometric theorem of Poincare". -This is the following statement: -Let $F: C\to C$ be any area-preserving self map of a cylinder $A$ to itself, that rotates the boundaries of $A$ in opposite directions. Then the map has at least two fixed points. -(this was proven by Birkhoff http://en.wikipedia.org/wiki/George_David_Birkhoff) -So, I would like to ask if there are some phenomena related to mirror symmetry that can be formulated in simple words. -Added. I would like to thank everyone for the given answers! I decided to give a bit of bounty for this question, to encourage people share phenomena related to mirror symmetry that can be simply formulated (or at least look exciting). Since there are lot of people in this area I am sure there must be more examples. - -REPLY [3 votes]: My master's thesis, An Introduction to Homological Mirror Symmetry and the Case of Elliptic Curves, might provide a piece of what you're looking for. It mostly concerns itself with the symplectic side of HMS (cause I have only a very superficial knowledge of algebraic geometry), but it includes a good amount background and some history. I tried to make it a useful document for other beginners to read... though then I let it rot on JSTOR for two years before posting it on arXiv :) -Oh well, it's up there now. I hope it helps!<|endoftext|> -TITLE: "Surprising" categorical equivalences -QUESTION [8 upvotes]: This is inspired by this question about the equivalence between the category of finite sets and non-negative integers. Now this question was (rightly, I guess) closed, but the fact was surprising to the OPer. I didn't know about this, but it was easy to verify. I had a little bit of difficulty understanding the nuances between categorical equivalences and category isomorphism until I thought about the analogy with homotopic spaces and homeomorphic spaces. -I was wondering about other equivalences that might unexpected and/or not so straightforward to prove. I would appreciate answers to a general audience...things that a beginner or an expert might find of interest. - -REPLY [3 votes]: I was pretty shocked by this one: if $V$ is the additive group of a finite-dimensional vector space over a $p$-adic field, the category of smooth (sometimes called algebraic or discrete) complex representations of $V$ is equivalent to the category of sheaves of complex vector spaces on the Pontryagin dual of $V$. -More generally, this is true of any Hausdorff, locally compact, and totally disconnected abelian group which is a filtered union of its compact open subgroups.<|endoftext|> -TITLE: Walsh Fourier Transform of the Möbius function -QUESTION [52 upvotes]: This question is related to this previous question where I asked about ordinary Fourier coefficients. -##Special case: is Möbius nearly Orthogonal to Morse -! -Harold Calvin Marston Morse (24 March 1892 – 22 June 1977), -August Ferdinand Möbius (November 17, 1790 – September 26, 1868) -Consider the sequence of values of the Möbius functions on nonnegative integers. (Starting with 0 for 0.) -0, 1, −1, −1, 0, −1, 1, −1, 0, 0, 1, −1, 0, −1, 1, 1, 0, −1, 0, −1, 0, 1, 1, −1, 0, 0, ... -And the Morse sequence -1, -1, -1, 1, -1, 1, 1, -1, -1, 1, 1, -1, 1, -1, -1, 1, -1, 1, 1, -1, 1, -1, -1, 1, 1, -1, -1, 1 -Are these two sequences nearly orthogonal? -Remark: This case of the general problem follows from the solution of Mauduit and Rivat of a 1968 conjecture of Gelfond. They show that primes are equally likely to have odd or even digit sum in base 2. See Ben Green's remark below.) -The Problems -Start with the Möbius function $\mu (m)$. (Thus $\mu(m)=0$ unless all prime factors of $m$ appear once and $\mu (m)=(-1)^r$ if $m$ has $r$ distinct prime factors.) Now, for a n-digit positive number $m$ regard the Mobius function as a Boolean function $\mu(x_1,x_2,\dots,x_n)$ where $x_1,x_2,\dots,x_n$ are the binary digits of $m$. -For example $\mu (0,1,0,1)=\mu(2+8)=\mu(10)=1$. -We write $\Omega_n$ as the set of 0-1 vectors $x=(x_1,x_2,x_m)$ of length $n$. We also write $[n]=\{1,2,\dots,n\}$, and $N=2^n$. -Next consider for some natural number $n$ the Walsh-Fourier transform -$$\hat \mu (S)= \frac{1}{2^n} \sum _{x\in \Omega_n} \mu(x_1,x_2,\dots,x_n)(-1)^{\sum\{x_i:i\in S\}}.$$ -So $\sum_{S \subset [n]}|\hat \mu (S)|^2$ is roughly $6/\pi ^2$; and the Prime Number Theorem asserts that $\hat \mu(\emptyset)=o(1)$; In fact the known strong form of the Prime Number Theorem asserts that -$$|\hat \mu (\emptyset )| \lt n^{-A} =(\log N)^{-A},$$ -for every $A>0$. (Note that $|\hat \mu (\emptyset)=\sum_{k=0}^{N-1}\mu(k)$.) -My questions are: - -Is it true that the individual coefficients tend to 0? Is it known even that $|\hat \mu (S)| \le n^{-A}$ for every $A>0$. - -Solved in the positive by Jean Bourgain (April 12, 2011): Moebius-Walsh correlation bounds and an estimate of Mauduit and Rivat; (Dec, 2011) For even stronger results see Bourgain's paper -On the Fourier-Walsh Spectrum on the Moebius Function. - -Is it the case that - -$$(*) \sum \{ \hat \mu ^2(S)~:~|S|<(\log n)^A \} =o(1), $$ -for every $A>0$. -(This does not seem to follow from bounds we can expect unconditionally on individual coefficients.) -Solved in the positive by Ben Green (March 12, 2011): On (not) computing the Mobius function using bounded depth circuits. (See Green's answer below.) - -The Riemann Hypothesis asserts that $$|\hat \mu (\emptyset )| < N^{-1/2+\epsilon}.$$ - -Does it follows from the GRH that for some $c>0$, -$$| \hat \mu (S)| < N^{-c},$$ for every $S$? -An upper bound of $(\log N)^{-{\log \log N}^A}$ suffices to get the desired application. - -##The Motivation -The motivation for these questions from a certain computational complexity extension of the prime number theorem. It asserts that every function on the positive integers that can be represented by bounded depth Boolean circuit in terms of the binary expansion has diminishing correlation with the Mobius function. This conjecture that we can refer to as the $AC^0$- prime number conjecture is discussed here, on my blog and here, on Dick Lipton's blog. The conjecture follows from formula (***) by a result of Linial Mansour and Nisan on Walsh-Fourier coefficients of $AC^0$ functions. -Question 3 suggestes that perhaps we can deduce the $AC^0$ prime number conjecture from the GRH which would be of interest. Of course, it will be best to prove it unconditionally. (Ben Green proved it unconditionally). -For polynomial size formulas, namely for functions expressible by depth-2 polynomial size circuits we may need even less. A result of Mansour shows that the inequality $|\hat \mu (S)| \le n^{-(\log \log n)^A}$ for every $A>0$, would suffice! Moreover, a conjecture of Mansour (which also follows from a more general conjecture called the Influence/Entropy conjecture, see this blog post for a description of both conjectures) implies that it will be enough to prove that -$$|\hat \mu (S)| \le n^{-A}$$ -for every $A>0$, to deduce the PNT for formulas.) - -Some background -Let me mention that the question follows to a large extent a line of research associating $AC^0$ formulas with number theoretic questions. See the papers by Anna Bernasconi and Igor Shparlinski and the paper by Eric Allender Mike Saks Igor Shparlinski, and the paper COMPLEXITY OF SOME ARITHMETIC PROBLEMS FOR BINARY POLYNOMIALS by Eric Allender, Anna Bernasconi, Carsten Damm, Joachim von zur Gathen, Michael Saks, and Igor Shparlinski. -Related MO question: Odd-bit primes ratio - -REPLY [14 votes]: Rather than updating the question, let me devote a separate answer to discuss the emerging knowledge. (Please corrent me if I make any mistake.) I will update this answer when necessary. -First, the Prime Number Theorem in its stronger known form asserts that -$$|\sum_{k=1}^X \mu(k)| \le X/e^{\sqrt \log X}. $$ -And the RH asserts that $$|\sum_{k=1}^X \mu(k)| \le X^{1/2+\epsilon}.$$ -Let $X=2^n$, the Prime number theorem deals with the Walsh coefficient -$\hat \mu(\emptyset)$. -(Remark: I am still a little confused about the situation, since the upper bound for the ordinary discrete Fourier coefficients in this answer by Matt Young are not as strong as the statement for the 0th coefficient given by the PNT. This is now clarified by Ben's remark below.) -##The second question and the $AC^0$-prime number conjecture (resolved by Ben Green, March 12). -Ben wrote a paper showing that -$$\hat \mu (S) \le X/e^{\sqrt \log X}, $$ whenever $|S| \le n^{1/2-\epsilon}$ using the Herman Katai's method. -This is more than enough to imply a positive answer to question 2. -Ben's positive answer to question 2 implies the $AC^0$- Prime Number Conjecture (a.k.a Sarnak-Kalai conjecture)! In my opinion, this is a very nice result. -Ben's number theoretic argument is rather delicate from it the implication is rather direct. Hastad Switching lemma implies that the total influence (a.k.a. average sensitivity) of an $AC^0$ Boolean function is polynomial in (log n) and this implies that most of the Fourier-Walsh coefficients are below the polylog level which together with an affarmative answer to question 2 gives the $AC^0$ PNC. The connection of $AC^0$ circuits and Walsh expansion was first explored by Linial, Mansour and Nisan and their full result (which was later improved a little by Hastad) asserts that the Fourier-Walsh coefficients decay exponentially above their expected value. The exponential decay does not play a role here, but it will imply stronger orthogonality consequence with better upper bounds on the Walsh coefficients of the Mobius functions. -##The first question (Update April 12, resolved by Jean Bourgain) -The first question was if $\hat \mu(S)$ tends to zero uniformly with X and at what rate. -A special case of interest was the correlation between the Mobius function and the Morse function (which is $\hat \mu ([n])$. Ben Green noted that the method by Mauduit and Rivat gives directly that -$$\hat \mu ([n]) \le X^{-c}, $$ -for some positive constant $c$. -Also according to Ben the results and methods of Harman and Kátai will give that $\hat \mu (S)$ uniformly tends to zero whenever $|S| \le n^{1/2-\epsilon}$ (in fact they give a stronger result mentioned below). -According to Ben, the technique of Mauduit-Rivat are likely to work unless S∩[n/3] is very ``thin'', and with more effort combining Mauduit-Rivat and Herman-Katai. -Update (April 12): Jean Bourgain proved (private communication) that for every Walsh function $W_S$ we have -$\sum_{m=1}^{X}\mu(m) W_S(m) \le X \cdot e^{-(\log X)^{1/10}}.$ -In other words, $\hat \mu(S) \le e^{-(\log X)^{1/10}}.$ -Jean also showed that under GRH -$\sum_{m=1}^X\mu(m)W_S(m) \le X^{1-(c/(\log\log X)^2)}.$ -In other words, $\hat \mu(S) \le X^{-(c/\log \log X)^2}.$ -This result suffices to show under the GRH the "monotone prime number conjecture." -Update Sept 14: Bourgain's paper is now arxived. -##The relation with known CS literature? -In the question there are links to several papers which deals with related question of the inability of $AC^0$ functions to compute certain number theoretic questions. These papers rely heavily on Fourier expansion of $AC^0$ circuits, Linial-Mansour-Nisan, Hastad etc. -It seems that the paper by Anna Bernasconi and Igor Shparlinski (Wayback Machine) and some papers cited there are most relevant. It looks that there is a proof there that much weight of the Fourier coefficients of a function expressing square-freeness (which is close to Mobius but seems easier). -##Follow up questions - -Give an affarmative answer to question (1) - -Extend the PNT when you consider functions expressed ACC[p] circuits, namely by Boolean depth circuits with mod p gates. Note that question (1) is a very special case of ACC2, It would be nice to "merge" the Rasborov-Smolensky method to deal with ACC[p] functions with some ANT. Now that Ben settled the PNC for $AC^0$ functions this will be a natural next step. I will ellaborate on this question below. - -Give an affermative answer to question 3. It will imply that under GRH the AC^0 PNT extends "almost" all the way to log-depth. (Update: The new result of Bourgain comes very close to that.) - - -Showing that $\hat \mu (S) \le X^{-1/3}$ will imply "The prime number conjecture for monotone Boolean functions" namely that the Mobius function is asymptotically orthogonal to every function described by a monotone Boolean function of the digits. (No complexity assumptions.) (Update: The new result of Bourgain appears to implies this under GRH.) -(This probably implies statement like: if you consider a randon sequance of integers $0=X_1,...,X_n$ so that $X_{i+1}$ is obtained from $X_i$ by switching a digit from 0 to 1 then the Mobius function will change sign many times on the sequence.) - -It would be interesting to see if appropriate low level complexity classes (Also allowing random inputs to the circuits) account for other known results about "Mobius randomness". Interesting examples: Standard L functions, the Green-Tao bracketed polynomials; non deterministic sequences in the sense of Peter Sarnak, - -There is no special reason to state the AC^0 Prime number theorem just based on the binary digit expressions. Can it be extended to expansions w.r.t. other p's? - - -Low degree polynomials over Z/2Z -The "Walsh-Fourier" functions considered in the question are high degree monomials over the real but they can be considered as linear functions over Z/2Z. For that, replace the values {-1,+1} by {0,1} both in the domain and range of our Boolean functions? What about low degree polynomials instead of linear polynomials? -If we can extend the results to polynomials over Z/2Z of degree at most polylog (n) this will imply by a result of Razborov Mobius randomness for AC02 circuits. (This is interesting also under GRH). -Update (12 April) While this question is still open. Jean Bourgain was able to prove Mobius randomness for AC0(2) circuits of certain sublinear size. Jean also noted that to show that the Mobius function itself is non-approximable by a AC0[2) circuit (namely you cannot reach correlation say of 0.99) can easily be derived from Razborov Smolensky theorem, since an easy computation shows that $\mu(3x)^2$ has correlation >0.8 with the 0(mod 3) function. -Moreover (As explained by Avi Wigderson), if we can show that certain functions -have very low correlations with all low degree polynomials over Z/2Z this will already be ground breaking result in computational complexity. (Say, correlation which is smaller than 1/n.) This will be interesting even under GRH. -Let me just say what low degree polynomials are. You have a bunch of sets of variables; -all the sets are small (smaller than $log n^t$), and your function is the parity of the number of sets for which all variables has value '1'. (If the sets are singletones we are back to the Walsh functions.) -##More updates, more questions (December 2011) -Update: Jean Bourgain has now proved that every monotone Boolean function is asymptotically orthogonal to the Mobius function, unconditionally. -Questions: In addition to questions mentioned above it will be interesting - -To relate these results with other recent results on Mobius randomness. - -To see if the results about Mobius randomness translate to result about primes. Namely, are results of the form - - - -(*) A $\pm$1 function f is -asymptotically orthogonal to the -Mobius function - -can be "transformmed into a result of the form: - -(**) There are infinitely many promes primes -such that f(p)=1 - -Of course, we will probably need also to assume that the density of ${n:f(n)=1}$ is not too small. See also this MO question.<|endoftext|> -TITLE: Dominant eigenvector of a real symmetric tridiagonal matrix -QUESTION [6 upvotes]: What is the most efficient way to calculate the dominant eigenvector of a real symmetric tridiagonal matrix? What's the corresponding time complexity bound? -Could someone give me a reference for this special structure? Thank you in advance. - -REPLY [3 votes]: The following paper proposes an $O(nk)$ algorithm called $MR^3$ (Multiple Relatively Robust Representations) to compute $k$ orthogonal eigenvectors of an $n \times n$ symmetric tridiagonal matrix. An implementation of this method is also available in LAPACK. - -I. S. Dhillon and B. N. Parlett. Multiple representations to compute orthogonal eigenvectors of symmetric tridiagonal matrices. (LAA 387, 2004). - -(The paper also includes details of an algorithm called "GetVec" for computing an eigenvector corresponding to an isolated eigenvalue). -Additionally, searching for the MRRR algorithm will turn up other relevant results for you<|endoftext|> -TITLE: Is it possible to improve the Whitney embedding theorem? -QUESTION [37 upvotes]: Edited to fix the example, as per Zack's suggestion. -Edit 2: So it turns out that when I think 'manifold' I tend to assume the nicest possible object. As I believe is standard, I would like to assume that all manifolds are 2nd countable and Hausdorff. Furthermore, let's say that our manifolds are connected and closed. -The Whitney embedding theorem states that any smooth $n$-manifold may be smoothly embedded into $\mathbb{R}^{2n}$. If we consider embeddings into more general $k$-dimensional manifolds, is it possible find a '$n$-universal' manifold of dimension less than $2n$? -For example, a non-orientable 2-manifold cannot be embedded into $\mathbb{R}^3$, demonstrating the sharpness of the Whitney embedding theorem. -However, there are 3-manifolds into which we can embed any surface, such as $M = \mathbb{RP}^3 \sharp \mathbb{RP}^3$. Indeed, by the classification of surfaces we know that any surface may be decomposed as a connected sum of copies of $\mathbb{RP}^2$ and tori. In fact, by the monoid structure of closed surfaces under connected sums we may take this sum to have at most 2 copies of the projective plane. Now, embed 2 disjoint copies of the projective plane into $M$ and arbitrarily many copies of the torus. Taking the connected sum of these we see that any closed surface is embeddable into $M$. -Can we do something similar in higher dimensions? - -REPLY [10 votes]: If we restrict our interest to manifolds which are $k$-connected, then Wall proved that any $n$-manifold (closed) $M$ admits a locally flat PL embedding in $\Bbb R^{2n-k}$, thereby improving on Whitney by $k$ dimensions. If in addition we assume the metastable range condition $2k < n$, then we can even take the embedding to be smooth. The latter theorem was also known to Haefliger and Hirsch and is historically earlier. -One further thing worth mentioning: the Hirsch Conjecture says that a stably parallelizable $n$-manifold is supposed to embed in $\Bbb R^m$, where $m = \lceil (3/2)n\rceil $. The conjecture is still open. Partial results are known: for example it's true when the manifold is $[n/4]$-connected.<|endoftext|> -TITLE: Classifying space for S1-bundle? -QUESTION [5 upvotes]: What is the classifying space for $S^1$-bundle? Here, $S^1$-bundle means a fiber bundle which doesn't mean that it is principal $S^1$-bundle. -I know that for a space $F$, -$\lbrace$the set of fiber bundles over $M$ whose fiber is $F\rbrace$ = $[M,B\operatorname{Homeo}(F)]$. -Therefore, my question can be rephrased as what is the homotopy type of $B\operatorname{Homeo}(S^1)$? - -REPLY [8 votes]: As mentioned by Tom Goodwillie, $Homeo(S^1)$ is homotopy equivalent to $O(2)=\mathbb Z/2\ltimes S^1$. We have a (split) short exact sequence of groups -$$ -S^1 \to \mathbb Z/2 \ltimes S^1 \to \mathbb Z/2 -$$ -which, upon applying applying the functor $B$, produces a (split) fibration sequence of classifying spaces -$$ -BS^1 \to B(\mathbb Z/2 \ltimes S^1) \to B\mathbb Z/2. -$$ -The latter can be rewritten as -$$ -\mathbb CP^\infty \to BHomeo(S^1) \to \mathbb RP^\infty. -$$ -The space $BHomeo(S^1)$ can be characterized by the facts that its $\pi_1$ is $\mathbb Z/2$, its $\pi_2$ is $\mathbb Z$, it has a non-trivial action of $\pi_1$ on $\pi_2$, and it has a trivial $k$-invariant in $H^3(\mathbb Z/2, \{ \mathbb Z \} )\cong\mathbb Z/2$. The last statement comes from the fact that the above fibration sequence is split.<|endoftext|> -TITLE: Existence of a sink in directed graphs with a certain structure -QUESTION [9 upvotes]: I'm not a mathematician (I'm an economist) but I hope that this problem is sufficiently non-trivial that someone here will find it interesting. -Motivation: -I'm trying to model how workers decide what "skills" to acquire when (a) they have different innate abilities for different skills but (b) they face competitive pressure from others that also choose to acquire those skills. -Problem: -Suppose we have $N$ workers that can choose to belong in any of $M$ different groups. Multiple workers can belong to the same group; a worker can be in one and only one group at a time. They can jump from any group to any other group. -A worker $i$ in group $j$ gets value $v_{ij}f(n_m)$ where $n_m$ is the number of workers in that group, and $f'(n_m) < 0$ and $f(1) = 1$ and as $n_m$ approaches infinity, $f$ approaches 0. $v$ is uniformly distributed. Workers jump between groups to try to maximize the value they receive. -Graph theory formulation: -I'm interested in the movement of workers between groups. I've modeled it as a directed graph, where each node is one possible configuration of workers among groups. Two nodes are connected if one worker changing states can convert one node to the other; edges point towards the greatest utility gain for the "jumping" worker. -In simulations, I've found that the system always reaches an equilibrium where no worker wants to jump and I haven't been able to construct a counter-example. -Conjecture: -My conjecture is that this is a general property of graphs with this structure, i.e., for any directed graph with the $(m,n)$ structure described above, there exists at least one "sink" with no outgoing edges and that this sink is reachable from all other nodes. -Ignoring values, it is possible to draw graphs without "sinks" but it leads to contradictions when I try to assign actual values to the worker-group pairings. None of the approaches I've tried so far seem promising enough to mention here. - -REPLY [12 votes]: Your conjectures are proven in the paper "Congestion Games with Player-Specific Payoff Functions" by I. Milchtaich, published in Games and Economic Behavior in 1996. -Usually, the term "congestion game" means a game in which players choose nonempty subsets of the set of resources. Each resource is assigned yields a certain utility (the same utility) to all players choosing it, and this utility depends only on the number of players, not their identities. Each player's total utility is the sum over all resources he has chosen. The general result is that such games are potential games, so they admit pure Nash equilibria (the sinks you consider) and any best-reply improvement path will reach one. -This particular paper considers a related notion of congestion game where players each only select one resource (as in your case). Again, their utility is based on the number of players choosing that resource, but different players choosing the resource may get different utilities. Under a monotonicity assumption slightly more general than the one you have made, the author shows that a pure Nash equilibrium still exists, and while not all best-reply improvement paths reach such an equilibrium, there exists one which can be reached in this way from any starting position.<|endoftext|> -TITLE: On the fundamental group of a finite CW complex -QUESTION [8 upvotes]: So let $X$ be a finite CW complex which is connected. -Q1: Is $\pi_1(X)$ necssarily a finitely presented group? -If the answer is yes, then how does prove it. I've tried to prove it using -an induction argument but I'm stuck... So every time one glues a cell then one needs -to show that this only throws in finitely many new generators and finitely many new relations... If the argument is too involved then I'd like to have a reference. -For every finitely presented group $G=\langle g_1,\ldots g_n| R_1,\ldots R_m\rangle$ -one may construct a connected finite CW complex $X$ having only cells of dimensions $\leq 2$ such that $\pi_1(X)\simeq G$. You have one circle for every generator and one 2-cell for every relation. -Q2: So did topologists try to prove results going in the other direction namely, say that $X$ is a finite connected CW complex with $n_i$ cells of dimension $i$ for $i\leq k$. Suppose that we know nothing about the incidence relations of these cells (except that $X$ is connected) then what can we say about the fundamental group of $X$ (outside the fact that it is finitely presented)? -One might ask similar questions where one imposes some incidence relations on the various cells etc. - -REPLY [19 votes]: Let $X$ be a CW-complex, and write $X_k$ for the $k$-skeleton. The cellular approximation theorem says that any based map $S^1\to X$ is homotopic to a cellular map, and that any two cellular maps that are homotopic are homotopic via a cellular homotopy. This means that the map $\pi_1(X_1)\to\pi_1(X_2)$ is surjective, and the map $\pi_1(X_2)\to\pi_1(X)$ is an isomorphism. Thus, we need only deal with $\pi_1(X_2)$. -Next, $X_1$ is just a connected combinatorial graph (with loops and multiple edges allowed). We can choose a maximal tree $T$ in this graph, and let $A$ be the set of edges not in $T$. For each edge $a\in A$ we have a loop $a'$ given by moving from the basepoint to the start of $a$ within $T$, then traversing $a$, then returning to the basepoint through $T$. It is a standard fact that $\pi_1(X_1)$ is freely generated by $\{a':a\in A\}$. -Next, $X_2$ is obtained from $X_1$ by attaching some $2$-cells. Each attaching map is an unbased map $S^1\to X_1$. After choosing a path from the image of the basepoint to the basepoint in $X$ we get an element of $\pi_1(X_1)$, whose conjugacy class is independent of this choice. An argument with van Kampen's Theorem (most easily done one cell at a time) shows that $\pi_1(X_2)$ is obtained from $\pi_1(X_1)$ by killing off these elements. This gives a finite presentation of $\pi_1(X_2)$. - -REPLY [3 votes]: What Richard Kent said, although it probably deserves to be expanded on. -Suppose for simplicity that $X'$ is obtained from a CW complex $X$ by attaching an $n$-disk $D^n = \{x: |x| \leq 1\}$ along a map $S^{n-1} \to X$, where $n > 2$. To see that the inclusion $X \to X'$ induces an isomorphism on $\pi_1$, use the van Kampen theorem. Namely, cover $X'$ by two open sets where one, $U$ is the image of $\{x: |x| < 2/3\}$ and the other, $V$, is the image of $X \cup \{x: |x| > 1/3\}$, both images being taken along the map to the pushout, $X \cup D^n \to X'$. -Notice that $X$ is a deformation retract of $V$, so $\pi_1$ applied to the inclusion $X \to V$ is an isomorphism. Obviously $\pi_1(U)$ is trivial, and so is $\pi_1(U \cap V)$ since $U \cap V \cong S^{n-1} \times (1/3, 2/3)$ and $\pi_1(S^{n-1})$ is trivial. So when you compute $\pi_1(X')$ as the pushout of the diagram of fundamental groups -$$\pi_1(U) \stackrel{\pi_1(i)}{\leftarrow} \pi_1(U \cap V) \stackrel{\pi_1(j)}{\to} \pi_1(V),$$ -according to the conclusion of van Kampen's theorem, the result is isomorphic to $\pi_1(X)$.<|endoftext|> -TITLE: Universal order type -QUESTION [10 upvotes]: Every countable order type, such as the countable ordinals, $\mathbb Z$, etc. can be embedded in $\mathbb Q$, so it is universal for countable order types. Is there a universal space for all linear orders of cardinality continuum? Or more generally, a universal space for all linear orders of any given cardinality? - -REPLY [6 votes]: In Joel Hamkins' remark, "If CH fails, however, then we cannot necessarily get a saturated model of size $\aleph_1$," one can actually drop the "necessarily." The existence of a saturated linear order of size $\aleph_1$ is in fact equivalent to CH. If $(L, <)$ is a line, then one can show that $L$ is $\aleph_1$-saturated if and only if - -For any countable increasing sequence $a_0 < a_1 < \ldots$ there exists $a \in L$ such that $a > a_n$ for all $n$. Similarly, any countable decreasing sequence $b_0 > b_1 > \ldots$ is bounded below by some $b$. This says there are no countable cofinal sequences in $L$. -If $a_0 < a_1 < \ldots < b_1 < b_0$ are sequences in $L$, then there exist points $a, b \in L$ such that $a_i < a < b < b_j$ for all $i, j$. This says that our line has no countable gaps. -If $a_0 < a_1 < \ldots < x < \ldots < b_1 < b_0$, then there exist points $a, b \in L$ such that $a_i < a < x < b < b_j$ for all $i, j$. This says that no $x$ in $L$ can be approached by a countable sequence from either side. - -From this, one can prove that the order $(2^{\omega_1}, <_{lex})$ embeds into the Dedekind completion of any $\aleph_1$-saturated $L$. If such an $L$ has cardinality $\aleph_1$ (that is, if $L$ is saturated in the full sense), then we have that $(2^{\omega_1}, <_{lex})$ has a dense subset of size $\aleph_1$. But any dense subset of $(2^{\omega_1}, <_{lex})$ must have size at least $2^{<\omega_1}$. The equality $\aleph_1$ = $2^{<\omega_1}$ gives CH. -The saturated order $L$ plays much the same role in $(2^{\omega_1}, <_{lex})$ that $\mathbb{Q}$ plays in $\mathbb{R}$. The combinatorial characterization of saturation given above allows one to naturally generalize Cantor's proof that any countable dense linear order without endpoints is isomorphic to $\mathbb{Q}$. Indeed, if $(L_1, <)$ and $(L_2, <)$ are two orders both satisfying conditions 1–3, and $|L_1| = |L_2| = \aleph_1$, then a Cantor style back-and-forth argument shows that $L_1 \cong L_2$. One can also see why any such $L$ will be universal: the conditions 1–3 allow any countable partial embedding of an $\aleph_1$-sized linear order into $L$ to be extended without trouble to a full embedding. -The generalization of the above argument to higher cardinalities also goes through, showing that GCH not only implies but is equivalent to the statement "for every cardinal $\kappa$, there exists a saturated linear order of size $\kappa^+$."<|endoftext|> -TITLE: Unramified (finite) extensions of fields complete with respect to a discrete valuation -QUESTION [5 upvotes]: Hello, -I've been reading the excellent online book on Algebraic Number Theory by J.S.Milne. In the section described above there is a footnote maintaining that the separability of the residue field extension implies the separability of the original field extension in the unramified case (base field complete w.r.t. a discrete valuation). Unfortunately, I haven't been able to find a reference nor been able to supply a proof of this myself. Any help would be greatly appreciated ! Kind regards, Stephan. - -REPLY [3 votes]: The statement that every unramified extension is separable holds in fact for any Henselian base field $(K,v)$, and Nakayama's lemma is unneccesary for the proof. It is sufficient to prove the assertion for finite extensions. So let $L|K$ denote a finite unramified extension, and let $\lambda|\kappa$ denote the corresponding finite field extension. Since $[\lambda:\kappa] = [L:K]$ is finite and $\lambda|\kappa$ is separable, we can find some $\bar{\alpha} \in \lambda$ such that -$\lambda = \kappa(\bar{\alpha})$ holds. Let $\alpha \in \mathcal{O}_L$ denote -a preimage. Since $\kappa(\bar{\alpha})$ is the residue class field of $K(\alpha)$ and every subextension of an unramified extension is unramified, we obtain -$$ -[L:K] = [\lambda:\kappa] = [\kappa(\bar{\alpha}):\kappa] = [K(\alpha):K] -$$ -and $L = K(\alpha)$. Let $f \in \mathcal{O}_K[x]$ denote the minimum polynomial -of $\alpha$ over $K$. Then the image $\bar{f} \in \kappa[x]$ is the minimum polynomial -of $\bar\alpha$ over $\kappa$. If $f$ were inseparable, then $f' = 0$ and -$\bar{f}' = 0$. But the latter contradicts the separability of $\bar\alpha$ -over $\kappa$.<|endoftext|> -TITLE: Are measurable homomorphisms $ (\Bbb{C},+) \to (\Bbb{C},+) $ or $ (\Bbb{C},+) \to (\Bbb{C},*) $ continuous, and do they admit an explicit description? -QUESTION [15 upvotes]: I am interested in generalizations of the following fact (known as automatic continuity, as pointed out below). I am especially looking for references to papers dating back to 1920’s. I feel that question like these have been well-studied when people were still interested in set-theoretic aspects of analysis... - -(Cauchy) Any measurable automorphism $ (\Bbb{R},+) \to (\Bbb{R},+) $ is necessarily a linear function, and any measurable homomorphism $ (\Bbb{R},+) \to (\Bbb{R}^{\times},\times) $ is necessarily an exponential function $ x \mapsto e^{a x} $. - -Is something similar true for homomorphisms $ f: (\Bbb{C},+) \to (\Bbb{C},+) $ or $ f: (\Bbb{C},+) \to (\Bbb{C}^{\times},\times) $ (assuming that $ \ker(f) = \Bbb{Z} $ in the latter)? (Yes, see answers below.) -That is, I am interested in facts which follow the following rough pattern: -If a map is not set-theoretically wild, e.g. it is measurable or Borel, and satisfies some identities, then it is in fact continuous, and, furthermore, it can be given by an explicit formula. - -REPLY [5 votes]: I just thought that I should mention a result that considerably strengthens André Weil’s result mentioned by François in his answer above. -Adam Kleppner, in his 1989 paper Measurable Homomorphisms of Locally Compact Groups, proved that any Borel-measurable homomorphism between locally compact (Hausdorff) groups is continuous. It does not matter whether or not any one of the groups is Polish.<|endoftext|> -TITLE: Minimizing determinant(Ztranspose.A.Z) -QUESTION [6 upvotes]: Let $A$ be a fixed positive semi-definite symmetric $m\times m$ matrix, and let $p$ be a fixed positive integer. Let $Z$ vary over all $m\times p$ matrices with orthonormal columns, and denote the transpose of $Z$ by $Z^T$. How does one prove that the minimum of $\mbox{det}(Z^T.A.Z)$ is the product of the $p$ smallest eigenvalues of $A$? -I've seen this quoted in a few places, but the references cited have turned out to be either vacuous ("a little algebra shows") or encrusted with so much generality that I can't see through to the core of the proof. Does anyone have a good reference or a proof that is short enough to be given in this forum? Not surprisingly, I can prove it for $p=1$ and for $p=m$. -And is there an analogous result when "minimum" is replaced by "maximum"? -Does anyone know to whom these results are due? They are probably pretty ancient. - -REPLY [7 votes]: A stronger fact holds: If $\lambda_1\le\lambda_2\le\dots\le\lambda_m$ are the eigenvalues of $A$, and $\mu_1\le\mu_2\le\dots\le\mu_p$ are eigenvalues of $Z^TAZ$, then $\mu_k\ge\lambda_k$ for all $k=1,\dots,p$. Update: This fact is well-known as Cauchy interlacing theorem. -Proof. Let $Q$ be the quadratic form on $\mathbb R^m$ defined by $A$ (that is, $Q(x)=x^TAx$ for all $x\in\mathbb R^m$), $L:\mathbb R^p\to\mathbb R^m$ the (isometric) linear map defined by $Z$. Then $Z^TAZ$ is the matrix of the quadratic form $Q'$ on $\mathbb R^p$ given by $Q'(x)=Q(L(x))$. I suggest you think of $Q'$ as the restriction of $Q$ to the subspace $L(\mathbb R^p)$ of $\mathbb R^m$. -Suppose that $\mu_k<\lambda_k$ for some $k$. Let $V$ be the $k$-dimensional subspace of $\mathbb R^p$ spanned by the first $k$ eigenvectors of $Q'$. Then $Q'(x)\le \mu_k |x|^2$ for all $x\in V$. Let $W$ be the $(m-k+1)$-dimensional subspace of $\mathbb R^m$ spanned by the eigenvectors corresponding to $\lambda_k,\lambda_{k+1},\dots,\lambda_m$. Then $Q(x)\ge\lambda_k|x|^2$ for all $x\in W$. The subspaces $W$ and $L(V)$ have nonzero intersection since the sum of their dimensions is greater than $m$. Hence there exists a nonzro vector $x\in V$ such that $L(x)\in W$. For this vector, we have $Q'(x)\le \mu_k |x|^2 <\lambda_k|x|^2$ but $Q'(x)=Q(L(x))\ge \lambda_k|L(x)|^2=\lambda_k|x|^2$, a contradiction. -Q.E.D. -Similarly (inverting all inequalities in the argument) one sees that the $k$-th largest eigenvalue of $Q'$ is no greater than the $k$-th largest eigenvalue of $Q$, answering your second question.<|endoftext|> -TITLE: Order of $\zeta(1+it)$ -QUESTION [10 upvotes]: What is known about the order of $\zeta(1+it)$? -Iwaniec-Kowalski gives (pp. 226 citing a result of Vinogradov-Korobov) -$|\zeta(1+it)| \lesssim (\log t)^{2/3},$ -and oppositely Titchmarsh gives (pp. 188 citing Bohr-Landau) -$|\zeta(1+it)| \gtrsim \log \log t$ -for infinitely many values of $t$. -Is this the limit of our knowledge? Is it conditionally known (or even expected) unconditionally known that -$|\zeta(1+it)| = e^{o(\log \log t)}$? -[As David points out below, on RH the result in Titchmarsh's book is optimal.] - -REPLY [8 votes]: This a classical problem which has been worked on by Littlewood, Titchmarsh, Levinson, and more recently Granville & Soundararajan. -Littlewood showed that the Riemann Hypothesis (RH) implies -$$\limsup_{t\to \infty} \frac{|\zeta(1+it)|}{\log\log t} \leq 2e^\gamma$$ -where $\gamma$ is Euler's constant. -Titchmarsh and later Levinson showed (unconditionally) that -$$\limsup_{t\to \infty} \frac{|\zeta(1+it)|}{\log\log t} \geq e^\gamma.$$ -Hence, on RH, a factor of 2 remains unresolved. Littlewood conjectured that -$$\limsup_{t\to \infty} \frac{|\zeta(1+it)|}{\log\log t} = e^\gamma$$ -but states "there is perhaps no good reason for believing ... this hypothesis." -A similar situation (a factor of 2 remaining unresolved) exits for large values of the function -$$1/|\zeta(1+it)|$$ -when $|t|\geq 1$, say. -I got this information from the introduction to Granville and Soundararajan's paper: http://arxiv.org/abs/math/0501232. It can also be found in the notes of Chapter 13 in Montgomery & Vaughan's book "Multiplicative Number Theory." I believe it is also discussed in Heath-Brown's notes in the end of one of the chapters of Titchmarsh's book on the zeta-function.<|endoftext|> -TITLE: "Probabilistic ultrafilters?" -QUESTION [11 upvotes]: A naive question. -Let $S$ be a set and let $[0,1]^S$ the set of functions from $S$ to the closed interval $[0,1]$. -Suppose given some function $P \colon [0,1]^S \to [0,1]$ satisfying the following three conditions: - -If $f \geq g$ everywhere on $S$, then $P(f) \geq P(g)$; -$P(\min(f,g)) \geq \min(P(f),P(g))$; -$P(1-f) = 1 - P(f)$. - -This is supposed to model a situation each point in $S$ has a "degree of belief" in some proposition, which yields a function $f$ in $[0,1]^S$; then $P$ is a process which takes all these degrees of belief and aggregates them into a "consensus" degree of belief $P(f)$. -Of course, this is meant to mimic the definition of an ultrafilter, which I think is given by the above definition with [0,1] replaced by {0,1}. -Certainly you have "principal" $P$, which just evaluate $f$ at some point $s$ of $S$. I suppose you could get other $P$ by sending $f$ to its limit with respect to some non-principal ultrafilter. -Is that it? -Added: Actually, the second condition above is perhaps too strong. I don't see an option for "hide question until I've thought about a bit more about what the best version of the question is" so I will just append this remark. -Added: Thanks, guys, for all the great answers. I now think the formulation of (2) was misguided (at least if the definition is meant to model consensus about degrees of belief) and I don't know what the "right" formulation is. One might well, for instance, want P to behave well when f and g refer to independent propositions; that would ask that P(fg) = P(f)P(g), which in the case of {0,1}-valued functions again agrees with the ultrafilter definition. This rules out averages but leaves in evaluation at ultrafilters. - -REPLY [5 votes]: Assume that $P$ satisfies the three conditions in the original question plus Noah Stein's additional requirement (declared " very wise" by the OP) that $P(s\mapsto c)=c$ for all constants $c$. I claim that then $P$ is the operation of limit along a certain ultrafilter $U$ on $S$. -As a first step, consider any function $f$ on $S$ that takes only the values 0 and 1; so $f$ is the characteristic function of a subset $A$ of $S$, while $1-f$ is the characteristic function of $S-A$. Conditions 1 and 2 in the question imply that $P$ commutes with min, i.e., that equality holds in condition 2. In particular, since $\min\{f,1-f\}=0$, we have that one of $P(f)$ and $P(1-f)$ is 0, and then, by condition 3, the other one is 1. In particular, the characteristic function of any $A\subseteq S$ is sent by $P$ to 0 or to 1. Then the conditions in the question immediately imply that $U=\{A\subseteq S:P(A)=1\}$ is an ultrafilter on $S$. Clearly, on functions that take only the values 0 and 1, $P$ agrees with limit along $U$; it remains to prove the same for arbitrary functions $f:S\to[0,1]$. -Let $f$ be such a function, and let $l$ be its limit along $U$. Consider an arbitrary $cc\}$ is in $U$. Let $g$ be the characteristic function of $A$, and notice that $f\geq\min\{s\mapsto c, g\}$. Since $P$ sends $s\mapsto c$ to $c$ and sends $g$ to 1, it must send $f$ to a value $\geq c$. Similar reasoning (using condition 3 to "turn the picture upside down") shows that $P(f)\leq d$ for all $d>l$. Therefore $P(f)=l$, as required.<|endoftext|> -TITLE: Anti-concentration about the mean for sum of Bernoulli random variables -QUESTION [9 upvotes]: Let $w\ll n$ (say $w=n^{0.1}$) and $a_1,\ldots,a_w$ be positive real numbers such that $\sum_{i \in w} a_i=n$. Also, let $x_1,x_2,\ldots, x_w$ be i.i.d. $\pm 1$ random variables. What is the best $t$ such that one always has -$$ -\Pr_{x_1,\ldots, x_w} [|\sum_{i\le w} a_i x_i| \ge t] \ge 0.01 -$$ - -REPLY [2 votes]: Okay, I used Paley Zygmund inequality and unless I messed up the calculation, one gets that for $Z= (\sum_{i=1}^w a_i x_i)^2$, -$$ -\Pr [ Z \ge \Theta E[Z]] \geq \frac{(1-\Theta)^2}{3} -$$ -Now, $E[Z] = \sum_{i=1}^w a_i^2 \geq n^2/w$. This implies that -$$ -\Pr \left[ Z \ge \frac{ n^2}{2w}\right] \geq \frac {1}{12} -$$ -Thus, -$$ -\Pr \left[ \left|\sum_{i=1}^w a_ix_i \right| \ge \frac{ n}{\sqrt{2w}}\right] \geq \frac {1}{12} -$$ -Up to constant factors, this is the best that can be said as putting all the $a_i$'s = $n/w$ and applying Berry Esseen theorem shows.<|endoftext|> -TITLE: Kodaira dimension and fundamental groups -QUESTION [13 upvotes]: In Riemannian geometry the non-negativity of the Ricci curvature $R$ of a manifold $X$ has strong implications on the size of the fundamental group $\pi_1(X)$: - -If $R>0$, then $\pi_1(X)$ is finite. -If $R=0$, it is known that $\pi_1(X)$ is almost abelian, i.e., it contains an abelian subgroup of finite index. Also, $\pi_1(X)$ has polynomial growth. - -In the case $X$ is a smooth complex projective variety, the positivity of Ricci curvature is related to ampleness properties of of $-K_X$, so it would be interesting to see whether analogous results of the above hold in algebraic geometry, with Ricci curvature replaced by the Kodaira dimension $\kappa(X)=\sup_n\dim\phi_{nK}(X)$. So my question is: - -What implications do non-positive Kodaira - dimension have for the fundamental - group of $X$? In particular, does some - versions of the above results hold - with Ricci curvature replaced by - Kodaira dimension? - -For example, if $X$ is a smooth projective variety with $\kappa(X)=0$, is $\pi_1(X)$ almost abelian? -One could also ask for refined versions of the above statements. For example, when $X$ is Fano it is well-known that $\pi_1(X)=0$. Does the same condition hold for all $X$ with big $-K_X$? - -REPLY [6 votes]: I realize this is an old question, but I just noticed it. Testa is correct that without stronger hypotheses, the fundamental group can be arbitrary. However, there are very strong results of Chenyang Xu that prove (following a conjecture of Kollár) that the local étale fundamental group of every KLT singularity is finite, and hence the étale fundamental group of the smooth (orbifold / stacky) locus of every log Fano variety is also finite. This uses the fantastic boundedness theorem of Hacon-McKernan-Xu. So if there is a birational model of your projective manifold that is a log Fano variety, the étale fundamental group is finite. -Also, recently Fujino has proved that for a quasi-log canonical variety that has anti-ample dualizing sheaf (i.e., is Fano), the variety has trivial étale fundamental group. That is simultaneously stronger than Xu's theorem -- because the group is trivial, not just finite -- yet also weaker, since this is not (obviously) a birational invariant: there is no reason that an étale cover of a birational model extends to an étale cover of the singular Fano variety, although it obviously extends to an étale cover of the smooth locus (by purity, etc.).<|endoftext|> -TITLE: Degeneracies for semi-simplicial Kan complexes -QUESTION [16 upvotes]: By a semi-simplicial set I mean a simplicial set without degeneracies. In such a thing we can define horns as usual, and thereby "semi-simplicial Kan complexes" which have a filler for every horn. Unlike when degeneracies are present, we have to include 1-dimensional horns: having fillers for 1-dimensional horns means that every vertex is both the source of some 1-simplex and the target of some 1-simplex. -I have been told that it is possible to choose degeneracies for any semi-simplicial Kan complex to make it into an ordinary simplicial Kan complex. For instance, to obtain a degenerate 1-simplex on a vertex $x$, we first find (by filling a 1-dimensional 1-horn) a 1-simplex $f\colon x\to y$, then we fill a 2-dimensional 2-horn to get a 2-simplex $f g \sim f$, and we can choose $g\colon x\to x$ to be degenerate. But obviously there are many possible choices of such a $g$. -I have three questions: - -Where can I find this construction written down? -Is the choice of degeneracies unique in some "up to homotopy" sense? Ideally, there would be a space of choices which is always contractible. -Does a morphism of semi-simplicial Kan complexes necessarily preserve degeneracies in some "up to homotopy" sense? (A sufficiently positive answer to this would imply a corresponding answer to the previous question, by considering the identity morphism.) - -REPLY [4 votes]: Kan wrote a very short note while the paper by Rourke and Sanderson (mentioned in the other answers) was in the publishing process: -Daniel M. Kan, Is an ss complex a css complex? Advances in Mathematics Volume 4, Issue 2, April 1970, Pages 170–171. -(as explained in the comments "ss" refers to semi-simplicial set -- without degeneracies -- and "css" to what is known as simplicial set nowadays). -The proposition is: "An ss complex $X$ which satisfies the extension condition can be completed (although, in general, in many diferent ways)." and he states afterwards that "It is clear that any two such completions will have the same homotopy type." -The proof proceeds by an inductive construction of a simplicial set $WX$ such that there is an isomorphism $FWX \to X$ where $F$ is the forgetful functor from simplicial sets to semi-simplicial sets. As a crucial ingredient he uses the geometric realization from Rourke and Sanderson (he gives a reference to [1,1.3], which should probably be Proposition 2.1. on p.325 in the published version of Rourke and Sanderson's On $\Delta$-sets, I).<|endoftext|> -TITLE: Example of noncomplete quotient of complete lcs mod closed subspace -QUESTION [11 upvotes]: The following statement is well-known: for a Fréchet space $V$ and a closed subspace $W \subseteq V$ the quotient $V / W$ is again complete and hence a Fréchet space. For the particular case of a Banach space the same statement holds true. -Beyond the metrizable case this is no longer correct. So my first question is about a good counter-example, i.e. a complete locally convex space $V$ with a closed subspace $W$ such that $V / W$ is no longer complete. -My second question is whether counter-examples necessarily arise beyond the metrizable case, i.e. does every complete lcs have a closed subspace with a non-complete quotient? In other words, does the above quotient property characterize Fréchet spaces? -My third question is how the situation looks like for sequentially complete lcs with sequentially closed subspace. Are there any positive results/situations where the quotient is at least sequentially complete again? -Thanks a lot. - -REPLY [4 votes]: Yet another answer: EVERY incomplete Hausdorff (locally convex) topological vector space is a counterexample! It is due to the late Susanne Dierolf who proved (Manuscripta Math., 17(1):73–77, 1975) that every topological vector space is a quotient of a complete one.<|endoftext|> -TITLE: Standard model of particle physics for mathematicians -QUESTION [46 upvotes]: If a mathematician who doesn't know much about the physicist's jargon and conventions had the curiosity to learn how the so called Standard Model (of particle physics, including SUSY) works, where should (s)he have a look to? -References (if they exist!) written for a mathematical target (so can assume e.g. basic differential geometry, basic Lie group theory...) in a "mathematical style" with rigorous definitions, theorems and proofs would be appreciated. - -REPLY [5 votes]: This was just posted to the arXiv today: - -M. J. D. Hamilton. - "The Higgs boson for mathematicians. Lecture notes on gauge theory and - symmetry breaking." - arXiv:1512.02632 [math.DG]. - - -"These notes form part of a lecture course on gauge theory. The material covered is standard in the physics literature, but perhaps less well-known to mathematicians. The purpose of these notes is to make spontaneous symmetry breaking and the Higgs mechanism of mass generation for elementary particles more easily accessible to mathematicians interested in theoretical physics. We treat the general case with an arbitrary compact gauge group G and an arbitrary number of Higgs bosons and explain the situation in the classic case of the electroweak interaction where G=SU(2)xU(1). Prerequisites are only a basic knowledge of Lie groups and manifolds. No prior knowledge of gauge theory or bundle theory is assumed."<|endoftext|> -TITLE: $p$-adic integrals and Cauchy's theorem -QUESTION [32 upvotes]: A short version of my question is: Is there a $p$-adic theory of integration? -Now let me expand a little further. In introductory texts such as Koblitz' book $p$-adic numbers,.. a bunch of $p$-adic analysis is developed. However, since all applications are towards number theory, the exposition stops at some point. In particular, there is no theory of integration developed for $p$-adic numbers. -By this I do not mean putting a measure on $\mathbb{C}_p$ and integrating real or complex valued functions, but instead putting a "$p$-adic measure"(whatever this may be) on it and integrating $\mathbb{C}_p$-valued functions on it. -To rephrase my question: Is there are an integration theroy for $\mathbb{C}_p$-valued functions on $\mathbb{C}_p$. In particular I would like to know if an analogue of Cauchy's theorem holds. Where can I read more about such a theory? - -REPLY [3 votes]: In recent decade, several number theorists for instance, professor T. Kim, H. Srivastava, Serkan Araci, extended $q$-integral concept for $p$-adic numbers and found several combinatorial identities for Bernoulli, Euler and Genocchi numbers by using this new method. -I try to briefly explain this method here. -Assume that $p$ be a fixed odd prime number. By $\mathbb Z_p$ we denote the ring of $p$-adic rational integers, $\mathbb Q$ denotes the field of rational numbers, $\mathbb Q_p$ denotes the field of $p$-adic rational numbers, and $\mathbb C_p$ denotes the completion of algebraic closure of $\mathbb Q_p$. -The $p$-adic absolute value is defined by $|p|_p = 1/p$. We assume $|q − 1|_p < 1 $ as an indeterminate. Let $UD(\mathbb Z_p)$ be the space of uniformly differentiable functions on $\mathbb Z_p$. For a positive integer $d$ with $(d, p) = 1$, set -$$X = X_d = \varprojlim_n \mathbb Z/dp^n \mathbb Z,$$ -$$X^∗ = \bigcup_{0< a< dp}^{(a,p)=1} a + dp \mathbb Z_p$$ -and $a + dp^n \mathbb Z_p = \{x \in X \mid x \equiv a \mod {dp^n}\}$, where $a \in \mathbb Z$ satisfies the condition $0 \le a < dp^n$. -Firstly, for introducing fermionic $p$-adic $q$-integral, we need some basic information which we state here. A measure on $\mathbb Z_p$ with values in a $p$-adic Banach spaceB is a continuous linear map $$f \mapsto \int f(x)\mu = \int_{\mathbb Z_p} f(x)\mu(x)$$ -from $\mathcal C^0(\mathbb Z_p,\mathbb C_p)$, (continuous function on $\mathbb Z_p$) to $B$. We know that the set of locally constant functions from $\mathbb Z_p$ to $\mathbb Q_p$ is dense in $\mathcal C^0(\mathbb Z_p, \mathbb C_p)$ so explicitly, for all $f \in \mathcal C^0(\mathbb Z_p, \mathbb C_p)$, the locally constant functions -$$f_n =\sum_{i=0}^{p^n−1} f(i)1_{i+p^n \mathbb Z_p} \to f$$ in $\mathcal C^0$. -Now, set $\mu (i + p^n \mathbb Z_p) = \int_{\mathbb Z_p}1_{i+p^n\mathbb Z_p}\mu$, then $\int_{\mathbb Z_p}f\mu$ is given by the following Riemannian sum, -$$\int_{\mathbb Z_p}f\mu = \lim_{n \to \infty} \sum_{i=0}^{p^n−1}f(i)\mu(i + p^n \mathbb Z_p)$$. -T. Kim introduced $\mu$ as follows: $\mu_{−q}(a + p^n \mathbb Z_p) ={(-q)^a} / {[p^n]_{-q}}$ for $f \in UD(\mathbb Z_p)$, which is famous to the fermionic $p$-adic $q$-integral on $\mathbb Z_p$ and you can find the applications of this definition in several papers which about $q$-Bernoulli numbers and polynomials. See here.<|endoftext|> -TITLE: Which Kahler Manifolds are also Einstein Manifolds? -QUESTION [26 upvotes]: Is it known which Kahler manifolds are also Einstein manifolds? For example complex projective spaces are Einstein. Are the Grassmannians Einsein? Are all flag manifolds Einstein? - -REPLY [11 votes]: One of the most well known classes of Kähler-Einstein manifolds, i.e. complex manifolds -which carry a Kähler metric $g$ such that $Ric_{g}= \lambda \cdot g$ $c\in\mathbb{R}$, are the generalized flag manifolds -$$G^{\mathbb{C}}/P\cong G/K$$ - of a compact connected simple Lie group. -Here $P$ is a parabolic subgroup of the complexification $G^{\mathbb{C}}$ of $G$, and $K=p\cap G$ is the centralizer of a torus $S\subset G$, i.e. $K=C(S)$. If $S=T=$maximal torus, then we obtain a full flag manifold $G/T$. -Inside the class of generalized flag manifolds, we find a very important subclass of Kähler-Einstein manifolds, the (isotropy irreducible) Hermitian symmetric spaces $M=G/K$ of compact type -(i.e. compact symmetric spaces endowed with a Hermitian structure invariant under the symmetries. In particular, this Hermitian structure is Kähler). It is well known that such a space $M=G/K$ admits a unique (as isotropy irreducible) Kähler-Einstein metric. Let me mention two basic facts for isotropy irreducible Hermitian symmetric spaces $M=G/K$: -1) The isotropy subgroup $K$ has an 1-dimensional center. -2) They are the only generalized flag manifolds which are the same time symmetric spaces. -A (generalized) flag manifold is a homogeneous Kähler manifold (the Kähler structure corresponds to the Kirillov-Kostant-Souriau symplectic form, since any flag manifold can be viewed as an adjoint orbit of an element in the Lie algebra of $G$). In particular, flag manifolds exhaust all compact and simply-connected homogeneous Kähler manifolds $M=G/K$ corresponding to a compact, connected, simple Lie group $G$. Their classification is based on the painted Dynkin diagrams. -Any coset $M=G^{\mathbb{C}}/P=G/K$ $(K=C(S))$ admits a finite number of invariant complex structures. Moreover, for any such complex structure we can define (a unique) homogeneous Kähler--Einstein metric, which is given in terms of the so-called Koszul form $$2\delta_{\frak{m}}=\sum_{\alpha\in R^{+}\backslash R_{K}^{+}}\alpha.$$ Thus, a flag manifolds admits a finite number of Kähler-Einstein metrics. Note that if some of the invariant complex structures are equivalent, then, the Kähler-Einstein metrics corresponding to these complex structures would be isometric. -More information about the geometry of flag manifolds, painted Dynkin diagrams, invariant Kähler-Einstein metrics, etc, can be found in the following articles: -D. V. Alekseevsky: Flag manifolds, -in Sbornik Radova, 11th Jugoslav. Geom. Seminar. Beograd 6 (14) (1997) 3--35. -D. V. Alekseevsky and A. M. Perelomov: - Invariant Kähler-Einstein metrics on compact homogeneous spaces, -Funct. Anal. Appl. 20 (3) (1986) 171--182.<|endoftext|> -TITLE: Statements forced by one condition of a poset, but not the whole thing -QUESTION [8 upvotes]: In order to get the relative consistency of some statement, it suffices to find a notion of forcing, and a condition $p$ in that forcing, such that $p$ forces the desired statement. It seems to be the case, most often, that the interesting statements we try to force end up being forced by the whole poset. -A sufficient property for a poset to possess to make the above phenomenon occur (in the case where all parameters in the forced statement are canonical names for objects in the ground model) is almost homogeneity: For every $p, q \in P$ there is an automorphism $i$ of $P$ such that $i(p)$ and $q$ are compatible. -It makes sense that if you're building a poset to force something, the whole poset forces it (there's also the ad hoc reason that you could throw out the part that doesn't force it). However, it might happen that in trying to force a particular statement, you build a poset where some, but not all, of the conditions happens to force a different interesting statement. Also, I haven't given this much deep thought, but it seems natural for most posets to be almost homogeneous. -My questions: Are there any interesting, instances of independence results forced by part, but not all, of some poset? Are there examples of commonly encountered posets which aren't almost homogeneous. -(If there ends up being a big list of answers, I'll add the "big-list" tag and make it community wiki) - -REPLY [7 votes]: Amit: A very interesting example of non-homogeneous posets are lotteries. Joel introduced them and has done a significant amount of work on them. -Given a collection $\{{\mathbb P}_i\mid i\in I\}$ of posets, their lottery is $\{(p,i)\mid p\in {\mathbb P}_i\}$ ordered by $(p,i)\le(q,j)$ iff $i=j$ and $p \le_{{\mathbb P}_i} q$. -The generic "randomly" selects a ${\mathbb P}_i$ and adds a generic for it. By starting with wildly different posets, we may end up with radically different extensions, depending on the generic. -This is particularly useful in iterations, where along the way we need to ensure a variety of posets are selected (in order to obtain, say, certain forcing axioms). It is a nice alternative to using Laver functions. -The standard reference here is J. D. Hamkins, "The lottery preparation", Ann. Pure Appl. Logic 101 (2000), 103–146. For a (sophisticated) recent application, see for example N. Dobrinen - S. Friedman, "The consistency strength of the tree property at the double successor of a measurable cardinal", Fundamenta Mathematicae 208 (2010), 123–153. (In this paper, we start with a ground model where GCH holds and $\kappa$ is weakly compact hypermeasurable, and a poset is described that preserves the measurability of $\kappa$ while forcing the tree property at $\kappa^{++}$. The poset is an Easton support iteration of a lottery of iterated Sacks forcing posets at different cardinals.) -A good example of a poset that is as far from being homogeneous as you may want is the Vopenka algebra, see Theorem 15.46 of Jech's "Set Theory". Woodin realized that Vopenka algebras are particularly useful in the study of models of determinacy, and this has been a key insight. -In the original application, Vopenka showed that if $A$ a set of ordinals, then $L[A]$ is a generic extension of its HOD. Conditions are (ordinals coding) ordinal definable sets of subsets of $\kappa$, where $A\subseteq\kappa$. The usefulness of this algebra and its variants for determinacy is that it allows us to argue about arbitrary sets of reals as if they where Borel sets. For a concrete application, see for example my recent paper with Ketchersid, "A trichotomy theorem in natural models of AD${}^+$", in "Set Theory and Its Applications", Contemporary Mathematics, vol. 533, Amer. Math. Soc., Providence, RI, 2011, pp. 227-258. - -REPLY [6 votes]: There are several issues. -First, of course, if a statement $\varphi$ is forced by a condition $p\in\mathbb{P}$, then of course it is forced by every condition in the poset $\mathbb{P}|p$, which restricts $\mathbb{P}$ to conditions below $p$. So whenever a statement is forced by any condition, then there is a poset such that every condition forces the statement. -Second, there are many statements that can be forced, but not be almost homogeneous forcing. For example, one of the most important consequences of almost homogeneous forcing is that one can limit the HOD of the forcing extension. Namely, if $V\subset V[G]$ is an almost homogeneous forcing extension by $\mathbb{P}$, then the $HOD^{V[G]}\subset HOD(\mathbb{P})^V$. Thus, one can never force $V=HOD$ by nontrivial definable almost homogeneous forcing. The natural class forcing of $V=HOD$ is ordinal definable, but not almost homogeneous. One way to do it is to proceed in an Easton support class iteration, which at each stage $\gamma$, forces either the GCH to hold at $\gamma$ or to fail at $\gamma$, letting the generic filter decide which is to be done. A simple density argument (replacing the bookkeeping argument that would have been used in earlier years) ensures that every new set of ordinals is coded into the GCH pattern on a block of cardinals, thereby ensuring GCH. But this forcing is definitely not almost homogeneous, since conditions that opt to force an instance of GCH at a cardinal cannot be automorphed to those that force the opposite. -Third, when every statement is forced by every condition in a partial order $\mathbb{P}$, whether or not the partial order is homogeneous, then the theory (restricted to any level of complexity) of the forcing extension must be an element of the ground model, since we can recover it from the forcing relation. Thus, one can make an easy counterexample. Start in $L$, add a Cohen real $c$ and then code $c$ into the continuum pattern on the $\aleph_n$'s. You get a model where the GCH pattern on the $\aleph_n$'s is $L$-generic. But there can be no partial order in $L$ forcing that statement, if all statements are to be forced by every condition. -There are many other examples of non-almost- homogeneous forcing notions. For example, the generic Souslin trees added by forcing (and constructed under $\diamond$) are rigid and therefore not almost-homogeneous as forcing notions. For example, you can arrange to have a Souslin tree such that forcing with it adds exactly one branch through the tree. -Another example is the lottery preparation, which I introduced. This is like the Laver preparation, but works with a variety of different large cardinals. At stage $\gamma$, the forcing consists of the lottery sum of a family of allowed forcing notions; this is the partial order consisting of $\{1\}\cup\{(\mathbb{Q},q)\mid q\in\mathbb{Q}\in F\}$, where $F$ is the family, and the order has $1$ above everything and otherwise $(\mathbb{Q},q)\leq (\mathbb{Q}',q')\iff \mathbb{Q}=\mathbb{Q}'$ and $q\leq_{\mathbb{Q}} q'$. So the generic filter in effect chooses a winning poset from the family and forces with it. This is clearly not almost homogeneous when the family is rich. The lottery preparation is a long iteration of such lotteries, and achieves various indestructibility results.<|endoftext|> -TITLE: Is every finite group a proper quotient of a finite primitive group? -QUESTION [11 upvotes]: Let $G$ be a finite group. Is there necessarily a finite primitive permutation group $P$ and a normal subgroup $N>1$ of $P$ such that $P/N \cong G$? -If not, what restrictions are there on quotients of finite primitive permutation groups? - -REPLY [14 votes]: Yes. We can assume that $G$ is a transitive permutation group. Let $S$ be any primitive finite simple group, such as $A_5$ in its natural representation. Now let $P$ be the wreath product of $S$ with $G$ using the product action, which has degree $d(P) = d(S)^{d(G)}$. This gives a primitive group, and the quotient of $P$ with the base group $S^{d(G)}$ of the wreath product is isomorphic to $G$. -Note that the primitive wreath product action of $S \wr G$ can also be described as its action by multiplication on the cosets of its maximal subgroup $T \wr G$, where $T$ is a point stabilizer in $S$.<|endoftext|> -TITLE: Strassen Algorithm 7 multiplications -QUESTION [9 upvotes]: Strassen Algoritm is a well-known matrix multiplication divide and conquer algorithm. -The trick of the algorithm is reducing the number of multiplications to 7 instead of 8. I was wondering, can we reduce any further? Can we only do 6 multiplications? -Also, what happens if we divide the NxN arrays into 9 arrays each of (N/3)x(N/3) instead of 4 arrays of (N/2)x(N/2). Can we then do less multiplications? - -REPLY [2 votes]: Of possible relevance to your question (though it might go into more geometric considerations than you care to read): -Generalizations of Strassen's equations for secant varieties of Segre varieties (J. Landsberg and L. Manivel, Comm. Algebra 2008)<|endoftext|> -TITLE: Orbits in modular arithmetic -QUESTION [9 upvotes]: Let $p$ be an odd prime number and consider the set of $p-2$ integers that is $\mathbb{Z}_p$ minus 0 and 1. Next define two bijective functions on this set -\begin{align} -f(x) &= 1-x \mod p -\end{align} -and -\begin{align} -g(x) &= x^{-1} \mod p \qquad \text{(the multiplicative inverse of $x$).} -\end{align} -One can view these two functions as group actions (generated by $f$ and $g$) on this set of integers and study the orbits. For example, if $p=7$, there are two orbits namely {2,4,6} and {3,5}. -For the general prime $p>3$, I can prove there is always an orbit consisting of $\{ 2,p-1, 2^{-1} \}$. -Question : I'm interested to find the number of orbits as a function of $p$. - -REPLY [4 votes]: Your transformations act on the projective line $\mathbb Z_p \cup\{\infty\}$, -preserving the three points $0$, $1$, $\infty$. The group they generate -is isomorphic to the group $\mathfrak S_3$ of permutations on these three points. -Anyway, you can enumerate the orbit of a point $x\in\mathbb Z_p\setminus\{0,1\}$ -and obtain -[ { x, 1-x, \frac 1x, \frac1{1-x}, \frac x{x-1} , \frac{x-1}x}. ] -This orbit usually has cardinality 6, unless if the stabilizer of $x$ -is nontrivial. When this happens, you have $x=1-x$, or $x=1/x$, or $x=1/(1-x)$, -or $x=x/(x-1)$, or $x=(x-1)/x$. -Assume now that $p>3$. (The cases $p=2$ and $3$ are trivial.) -The preceding analysis shows that there is exaclty one orbit with cardinality $3$, -namely $\{-1,2,1/2\}$, -and one orbit of cardinality $2$, $\{\alpha,1-\alpha\}$, where $\alpha$ -is an element of $\mathbb Z_p$ satisfying $\alpha^2-\alpha+1=0$. -Such an $\alpha$ exists if and only if $-3$ is a square in $\mathbb Z_p$. -Let $k$ be the number of orbits with cardinality $6$. -The total number of orbits is $k+2$ if $-3$ is a square, and $k+1$ otherwise. -Counting the number of elements in $\mathbb Z_p\setminus\{0,1\}$ gives -$p-2=6k+3+2$ in the former case, and $p-2=6k+3$ in the latter, -that is: $p=6k+7$, resp. $p=6k+5$. -Finally, we obtain that for $p\equiv 1\pmod 6$, $-3$ is a square modulo $p$ -and there are $(p+5)/6$ orbits, while for $p\equiv -1\pmod 6$, -$-3$ is not a square and there are $ (p+1)/6$ orbits. -NB. In both cases, the number of orbits is equal to $\lfloor(p+5)/6\rfloor$.<|endoftext|> -TITLE: Algebraic P vs. NP -QUESTION [30 upvotes]: I recently attended a lecture where the speaker mentioned that what he was talking about was connected to the algebraic version of the $P$ vs. $NP$ problem. Could someone explain what that means in a simple way or point me to a source suitable for a non-expert mathematician? Thanks. - -REPLY [25 votes]: I suspect that the question under consideration is whether or not $VP=VNP$; this is the problem directly studied by geometric complexity theorists, as I understand their work. This project is described in some detail here (this paper is by Burgisser, Landsberg, Manivel, and Weyman, describing work of Mulmuley and related people)--it is aimed at algebraic geometers; so you will likely be comfortable with it. The description of the complexity problem under consideration is in section 9. -The algebraic $VP$ vs. $VNP$ conjeture is due to Valiant, in this paper and his paper "Reducibility by algebraic projections" which I can't find online at the moment, unfortunately; these are references [63] and [64] in the paper I link to above. Valiant is, as I recall, a very clear writer, so hopefully you will find these papers readable as well. -Essentially, Valiant argues that some algebraic properties of the permanent and related varieties should have complexity-theoretic implications; a reasonable heuristic for this might be the many combinatorial interpretations of the permanent. Unfortunately, as far as I know there are few implications between these algebraic versions of P vs. NP and the problem itself; there are some results assuming GRH. See e.g. this paper by Burgisser. -Hopefully, this is the algebraic analogue of P vs. NP you were looking for.<|endoftext|> -TITLE: Lefschetz Hyper-plane theorem for singular projective varieties? -QUESTION [10 upvotes]: Lefschetz hyper-plane theorem for smooth projective varieties, $X\subset \mathbb{P}^{n+1}$ says: -For smooth hyperplane section $Y= X\cap H$, the restriction map -$H^i(X) \rightarrow H^i(Y)$ is an isomorphism for $0\leq i \leq n$ and an injection for $i=n$. -Similarly we get an statement for homologies. -For a singular variety projective variety $X$, lets consider the singular homology (or singular cohomology) -Here is the question: Is there a Lefschetz hyperplane theorem in for projective varieties with possible canonical singularities? -What which I expect is some thing like this: -$X$ as before and assume $X_{sing}$ (singular locus of $X$) has codimension at least $k$ in $X$; then for generic hyperplane section $Y$ we have an isomorphism: -$H_i(X)\cong H_i(Y)$ for $i$ less then some function of $k$ and $n$ !!! - -REPLY [4 votes]: I think what you are looking for is the Lefschetz hyperplane theorem for singular varieties from Goresky and MacPherson's " Stratified Morse theory" (part II, section 1.2). The range in which there is an isomorphism depends on the number of equations needed to define $X$ locally. The theorem says (after some deciphering) that if this number is $\leq k$ for the points of $X$ outside the hyperplane, then the hyperplane section map is an isomorphism in degrees $< N-k-1$ where $N$ is the dimension of the ambient projective space. -Also, for the middle perversity intersection homology the Lefschetz theorem is stated almost exactly as for smooth varieties and ordinary homology: for a generic hyperplane the hyperplane section map in homology is an isomorphism in degrees $<\dim X-1$ and is surjective in degree $\dim X-1$.<|endoftext|> -TITLE: Laplace-deRham operator for 1-forms on the sphere -QUESTION [5 upvotes]: What do the eigenforms of the 1-form Laplace-de Rham operator look like on the 2-sphere, seen as vector fields via the inner product? -For the standard Laplace-de Rham operator on 0-forms (functions) the simple answer is the spherical harmonics. What about for the 1-form operator? - -REPLY [8 votes]: If $f:S^2\to R$ satisfies $\Delta f=\lambda f$, then -$$\Delta(df)=(dd^*+d^*d)(df)=\lambda df$$ -and similarly -$$ \Delta(\ast df)=(dd^*+d^*d)(*df)=\lambda \ast df $$ -Since $H^1(S^2)=0$, these are all eigenvectors on 1-forms. Here $*$ is the Hodge * operator and $d^*=-\ast d \ast$. -The vector field is the unique $X$ so that $df(v)=(X,v)$. -On 2-forms all eigenvectors are of the form $\ast f$.<|endoftext|> -TITLE: Why are there no wild arcs in the plane? -QUESTION [33 upvotes]: On math.stackexchange it was asked whether all arcs in the plane are ambient-isotopic. I suggested that one could prove this by appealing to the Schönflies theorem, which you can do as long as you can extend your Jordan arc to a Jordan curve. That is, as long as you can extend an embedding of an interval to an embedding of a circle. However, I have to admit I don't readily see how to do this, and there are examples which show this is a subtle question. E.g. you can take an arc which spirals infinitely into a point. So my question is how one can show an embedding of an arc into $\mathbb R^2$ extends to an embedding of a circle into $\mathbb R^2$, or, failing that, if someone knows some other proof that all planar Jordan arcs are standard. -Edit: I want to highlight Bill Thurston's elegant answer (in a comment) to the question of whether you can extend an arc to a circle, even though I accepted his other answer using Caratheodory's theorem. Namely, assume your arc runs from $0$ to $\infty$ in $\mathbb C\cup\{\infty\}$. Then take the pre-image under the double branched cover $z\mapsto z^2$. The original arc can be identified with one of its two preimages, while the other preimage fills it in to a circle. Then one can apply the Schönflies theorem. - -REPLY [2 votes]: The desired result appears in M.H.A. Newman's classic book, Elements of the Topology of Plane Sets of Points (2nd ed., 1951), as Theorem 14.5 in Chapter VI, on pg. 164: -Theorem 14.5: Every simple arc in $X^2$ is an arc of a simple closed curve in $X^2$. -$X^2$ is Newman's notation for a space that is either the "open" plane, $R^2$, or the "closed" plane, $R^2 \cup \{\infty\}$. -N.B. The proof of Theorem 14.5 does not involve Schoenflies' Theorem. Also, Newman's definition of an arc is such that it has ends.<|endoftext|> -TITLE: How to cite math journals? -QUESTION [12 upvotes]: This is kind of embarrassing, but I never figured out how to cite journal names in the bibliography, especially when to abbreviate and how. For example, do we write "Adv. in Math." or "Advances in Math."? Or is "Trans. of AMS" OK? So: - -Is there any formal guideline on how to cite and abbreviate journals names in a math paper? - -I googled and found one resources which says: follow Math Sci Net. But it is not very convenient, since I would have to go to Math Sci Net and search for such journal every time I need to cite it. -And while we are at it, what is the protocol on arxiv links? Do we keep them in the citation after the paper is published, since it is still the most accessible source? -Thank you! - -REPLY [2 votes]: I agree with the other answers suggesting to use the standard abbreviations. The question is of course, Which are the standard abbreviations? Just recently (while arguing with the Springer Correction Team about, well, abbreviations of journal titles) I learned that for example Springer uses the abbreviations as given by the ISSN database. To me this seems indeed somewhat more standard and moreover more universal than any other list. Unfortunaetly, this list seems not accessible to mere mortals. But said page can, with some care, at least be partially useful, as it provides this accessible list of standard abbreviations of words that may occur in titles of serials.<|endoftext|> -TITLE: Listing lattice points in a simplex -QUESTION [8 upvotes]: Let $n \in \mathbf{Z}_{\geq 1}$. Is there an algorithm which, given a simplex $\Delta \subset \mathbf{R}^n$ specified as the convex hull of $v_0,\dots,v_n \in \mathbf{Z}^n$, computes the set $\Delta \cap \mathbf{Z}^n$ in time $O(\delta^t\mathrm{vol}(\Delta))$ for some explicit $t \in \mathbf{R}$, where $\delta=\max_{i,j} \log|v_{ij}|$ is the input size and $\mathrm{vol}(\Delta)$ is the (normalized) volume of $\Delta$? -Note here that the dimension is fixed. Implicit in this question is that $|\Delta \cap \mathbf{Z}^n|=O(\mathrm{vol}(\Delta))$, which I haven't been able to extract yet from the literature. -This question arises for me after trying to ask the same thing for a polytope $P \subset \mathbf{Z}^n$; the natural way to do this seems to be by computing a triangulation (or do people say tetrahedralization?), such as the Delaunay triangulation, which reduces the problem from $P$ to a bunch of $\Delta$s. -(And in fact, really what I need is to compute $d\Delta \cap \mathbf{Z}^n$ for $d=O(n)$, but $n$ is constant for now so this doesn't seem to help.) -People often just want to count the set $\Delta \cap \mathbf{Z}^n$ (use short rational generating functions) or to know that it is nonempty (reduces to an integer program), but I need to actually list every lattice point. -Translating by $v_0$ we may assume $v_0=0$ is a vertex, and then we want to compute the set of points $x=a_1v_1+\dots+a_n v_n$ such that $a_i \geq 0$ for all $i$ and $a_1+\dots+a_n \leq 1$. This should be doable by considering representatives for $\mathbf{Z}^n/\sum \mathbf{Z} v_i$ via a Hermite normal form, but this gives generators for the whole parallelopiped (wasteful, but perhaps not noticeably so in fixed dimension)? If this makes sense, is there a clean algorithm with a rigorous analysis of the running time (counting the time for integer arithmetic)? -Thanks very much for your help! - -REPLY [5 votes]: Hermite normal form (HNF) should work much as you suggest. -Translate $v_0$ to 0. Let $L = \bigoplus_{i=1}^n {\bf Z} v_i$. HNF gives an explicit decomposition of $G := {\bf Z}^n / L$ as a direct sum of cyclic groups. Both the computation and the resulting generators take time polynomial in the input size. This lets you run through representatives of the $|G|$ cosets of $L$ in ${\bf Z}^n$. Write each one uniquely as $\sum_{i=1}^n a_i v_i$ for some rational $a_i \in {\bf Q}$. Replacing each $a_i$ by its fractional part $\{ a_i \} = a_i - \lfloor a_i \rfloor$ gives the only candidate for the intersection of the coset with $\Delta$, which works iff $\sum_i \{ a_i \} \leq 1$. (Exception: the trivial coset has $n+1$ representatives, but you already know them: $0$ and the $v_i$.) The processing time per coset is again polynomial in the input size, and the number of cosets is $|G| = n! {\rm Vol}(\Delta)$; since for you $n$ is fixed, this answers your question. -Along the way we've obtained $|\Delta \cap {\bf Z}^n| \leq n! {\rm Vol}(\Delta) + n$, giving (again for fixed $n$) the result you "haven't been able to extract yet from the literature", with explicit constants. Equality holds at least when $n! |\Delta| = 1$. -I haven't yet given or cited a proof that HNF can in fact be computed in time polynomial in input size. This seems clear, though the first thing one might try may produce an algorithm that's too slow in practice. A brief Google search indicates that (not surprisingly) there has been considerable work on this question; one article that turns up often is "Asymptotically Fast Computation of the Hermite Normal Form of an Integer Matrix" by -Arne Storjohann and George Labahn (1996), which gives bounds whose dependence on $n$ is polynomial as well (in particular the exponent does not depend on $n$). There's also an HNF implementation in gp with several variations, see mathnf — that's MATrix HNF, not MATH something :-) — and probably other packages do this too. You might also try Smith normal form (matsnf), which seems to do much the same thing for this purpose.<|endoftext|> -TITLE: Is there the Whitehead theorem for cohomology theory? -QUESTION [12 upvotes]: We know that there are Whitehead theorem for homotopy and homology theory. -Is there the Whitehead theorem for cohomology theory for 1-connected CW complexes? - -REPLY [18 votes]: Conceptually, the following two theorems (both due to Whitehead) are Eckmann-Hilton duals. -Theorem. A weak homotopy equivalence between CW complexes is a homotopy equivalence. -Theorem. A homology isomorphism between simple spaces is a weak homotopy equivalence. -They don't look dual, but they are. See -J.P. May. -The dual Whitehead Theorems. -London Math. Soc. Lecture Note Series -Vol. 86(1983), 46--54. -The point is that the second statement is really about cohomology, and the standard -cellular proof of the first statement dualizes word-for-word to a ``cocellular'' proof -of the second. Cocellular constructions are what appear in Postnikov towers, and they -can be used more systematically than can be found in the literature. Yet another plug: -they are central in the upcoming book "More Concise Algebraic Topology'' by Kate Ponto -and myself.<|endoftext|> -TITLE: If a polynomial f is irreducible then (f) is radical, without unique factorization? -QUESTION [9 upvotes]: Is there a short way to prove that for each irreducible polynomial $f$ in $k[x_1,...,x_n]$ the principal ideal $(f)$ is radical without using unique factorization of polynomials? A short proof of this statement (contained, for example, in the "Primer on CA of Milne") uses the fact that polynomial ring is an UFD, but is it possible to give a reasonable proof without using this fact? - -REPLY [7 votes]: Here are some thoughts on why such a proof might be hard to find (and interesting). They are not totally rigorous, but I think they give a different perspective, so might be worth something. -I believe the existence of an irreducible element whose ideal is not radical might be related to non-trivial torsions in the class group (I will assume the ring is normal, one can avoid it by using Chow group of codimension one instead). Indeed, just from definition, you need such an element $x$, and some elements $y,z$ such that -$$xy=z^n$$ -but $z\notin (x)$. Now, if $P=(x,z)$ happens to be prime and $y \notin P$, then by computing the Weil divisor corresponding to the Cartier divisor $(x)$, one gets $n[P]=0$ in the class group. But $[P]$ is not principal: if it is, it would have to be generated by $x$ because $x$ is irreducible, but $z\notin (x)$ by assumption. -This is precisely what happened in the examples by Qiaochu ($\mathbb Z[-\sqrt{5}]$) and Gerry ($k[x,y,z]/(xy-z^2)$) (both have class group $\mathbb Z/(2)$). -So it seems to me the proof you want would rule out certain torsions in the class group but without showing that the group is trivial (which means our ring is a UFD). Unfortunately, understanding torsions in $\text{Cl}(R)$, especially over arbitrary fields, is a harder problem (think about elliptic curves!) -I hope this makes some sense.<|endoftext|> -TITLE: What spaces can be obtained from $\mathbb{R}^{n}$ by taking quotient spaces and subspaces? -QUESTION [13 upvotes]: Is there a good characterization of the smallest collection of topological spaces which contains $\mathbb{R}^{n}$ for each $n$, and is closed under taking subspaces and quotient spaces? -A bit of motivation: A friend of mine asked me to give an argument why the definition of a topological space is "right" or "natural", considered perhaps as a generalization of manifolds or cell complexes. While trying to answer him, I briefly wondered whether the collection of topological spaces is the closure of $\{ \mathbb{R}^{n} \}_{n \geq 0}$ under certain operations, say taking subspaces and quotient spaces. I quickly realized that this is false in general, though (there are counterexamples which have very large cardinality or don't satisfy first or second countability). - -REPLY [2 votes]: Let us recall that a family $\mathcal N$ of subsets of a topological space $X$ is called a $cs$-network if for any sequence $\{x_n\}_{n\in\omega}\subset X$ that converges to a point $x\in X$ and any neighborhood $U\subset X$ of $x$ there exists a set $N\in\mathcal N$ such that $x\in N\subset U$ and $x_n\in N$ for all but finitely many numbers $n\in\omega$. -The following theorem answers the problem posed by Sam Nolen. I thank Martin Sleziak for his comment containing the reference to the paper of Franklin and Rajagopalan. -Theorem. The smallest class of topological spaces that contains $\mathbb R$ and is closed under taking subspaces and quotient spaces coincides with the class of subspaces of sequential spaces that have cardinality $\le \mathfrak c$ and possess a countable $cs$-network. -Proof. We shall prove that $\mathcal R=S\mathcal K=SQSQS\mathbb R$, where -$\bullet$ $\mathcal R$ is the smallest class of topological spaces, which contains the real line and is closed under taking subspaces and quotient spaces; -$\bullet$ $\mathcal K$ is the class of sequential spaces of cardinality $\le\mathfrak c$ that possess a countable $cs$-network; -$\bullet$ $S\mathcal K$ is the class of subspaces of spaces in the class $\mathcal K$. -$\bullet$ $S\mathbb R$ is the family of subspaces of the real line; -$\bullet$ $QS\mathbb R$ is the class of quotient spaces of spaces in the class $S\mathbb R$; -$\bullet$ $SQS\mathbb R$ is the class of subspaces of spaces in the class $QS\mathbb R$; -$\bullet$ $QSQS\mathbb R$ is the class of quotient spaces of spaces in the class $SQS\mathbb R$; -$\bullet$ $SQSQS\mathbb R$ is the class of subspaces of spaces in the class $QSQS\mathbb R$; -The proof of the equality $\mathcal R=S\mathcal K=SQSQS\mathbb R$ is divided into six lemmas. -The first lemma was proved in the answer of Will Brian. -Lemma 1. The class $QS\mathbb R\subset\mathcal R$ contains all separable metrizable spaces. -Lemma 2. For any zero-dimensional Polish space $Z$ and any anti-discrete space $A$ of cardinality continuum the product $Z\times A$ belongs to the class $QS\mathbb R\subset\mathcal R$. -Proof. The anti-discrete space $A$ can be identified with the topological group $\mathbb R/\mathbb Q$ endowed with the quotient topology (which is anti-discrete). Then $Z\times A$ belongs to $QS\mathbb R$, being a quotient space of the separable metrizable space $Z\times \mathbb R$, which belongs to $QS\mathbb R$ by Lemma 1. -Lemma 3. $\mathcal K\subset QSQS\mathbb R\subset \mathcal R$. -Proof. Let $X$ be a sequential space of cardinality $\le\mathfrak c$ that has a countable $cs$-network $\mathcal N$. -Without loss of generality, we can assume that $\mathcal N$ is closed under finite unions and intersections. Endow the set $\mathcal N$ with the discrete topology and consider the zero-dimensional Polish space $\mathcal N^\omega$. -Let $X_a$ be the set $X$ endowed with the anti-discrete topology $\{\emptyset, X\}$. -In the space $\mathcal N^\omega\times\ X_a$ consider the subspace -$Z$ consisting of pairs $((N_k)_{k\in\omega},x)\in\mathcal N^\omega\times X_a$ such that for any open neighborhood $U\subset X$ of $x$ there exists $n\in\omega$ such that $x\in N_{k+1}\subset N_k\subset U$ for all $k\ge n$. -By Lemma 2, the space $Z$ belongs to the class $SQS\mathbb R$. -Let $q:Z\to X$, $q:((N_k)_{k\in\omega},x)\mapsto x$, be the projection on the second factor. It is easy to see that the map $q$ is surjective and continuous. To see that $q$ is quotient, we need to check that a set $F\subset X$ is closed if its preimage $q^{-1}(F)$ is closed in $Z$. Assuming that $F$ is not closed in the sequential space $X$, we can find a sequence $\{x_n\}_{n\in\omega}\subset F$ that converges to some point $x\in X\setminus F$. Let $\mathcal N'$ be the family of all sets $N\in\mathcal N$ such that $x\in N$ and $x_n\in X$ for all but finitely many numbers $n$. Let $(N_i')_{i\in\omega}$ be an enumeration of the countable set $\mathcal N'$. For every $k\in\omega$ let $N_k=\bigcap_{i\le k}N_i'$. -Taking into account that the $cs$-network $\mathcal N$ is closed under finite unions, we can show that $z:=((N_k)_{k\in\omega},x)\in Z$. It is clear that $z\notin q^{-1}(F)$. It can be shown that the pair $z$ belongs to the closure of $q^{-1}(F)$, which is not possible as $q^{-1}(F)$ is closed in $Z$. -This contradiction shows that $X$ is a quotient space of $Z$ and hence $X\in QSQS\mathbb R$. -Lemma 3 implies -Lemma 4. $S\mathcal K\subset SQSQS\mathbb R\subset \mathcal R$. -Lemma 5. The class $S\mathcal K$ is closed under quotient spaces. -Proof. Let $S$ be a subspace of a space $K\in\mathcal K$ and $q:S\to X$ be a quotient map of $S$ onto some topological space $X$. The map $q$ determines the equivalence relation $$E=\{(x,y)\in K\times K:x=y\}\cup\{(x,y)\in X\times X:q(x)=q(y)\}.$$ In Proposition 3.2 of this paper, Franklin and Rajagopalan prove that $X$ can be identified with a subspace of the quotient space $K/E$. -The following lemma ensures that the quotient space $K/E$ belongs to the class $\mathcal K$. -Lemma 6. Let $f:X\to Y$ be a quotient map. If $X$ is a sequential space with countable $cs$-network, then so is the space $Y$. -Proof. The sequentiality of the quotient space $Y$ is a well-known fact (that can be found in Engelking, I hope). So, it remains to prove that the space $Y$ has a countable $cs$-network. -Let $\mathcal N$ be a countable $cs$-network for the space $X$. Without loss of generality, we can assume that the family $\mathcal N$ is closed under finite unions. -Consider the family $\mathcal M:=\{q(N):N\in\mathcal N\}$ and for every $M\in\mathcal M$ let $\ddot M$ be the intersection of all open sets in $Y$ than contain $M$. We claim that the countable family $\ddot{\mathcal M}=\{\ddot M:M\in\mathcal M\}$ is a $cs$-network for the space $Y$. Since the family $\mathcal N$ is closed under finite unions, so are the families $\mathcal M$ and $\ddot{\mathcal M}$. -Fix a sequence $\{y_n\}_{n\in\omega}\subset Y$, convergent to a point $y\in Y$ and let $U\subset Y$ be a neighborhood of $y$ in $Y$. Let $\mathcal M'=\{M\in\ddot{\mathcal M}: M\subset U\}$. We claim that the family $\mathcal M'$ contains a set $M$ with $y\in M$. To find such set $M$, take any point $x\in q^{-1}(y)$ and find a set $N\in\mathcal N$ such that $x\in N\subset q^{-1}(U)$. Then $M=q(N)$ contains $y$ and the set $\ddot M$ contains $y$ and belongs to the family $\mathcal M'$. -So, we can choose an enumeration $\{\ddot M_k\}_{k\in\omega}$ of the countable family $\mathcal M'$ such that $y\in \ddot M_0$. We claim that for some $k\in\omega$ the set $\bigcup_{i\le k}\ddot M_i$ contains all but finitely many points $y_n$. Assuming that this is not true, we can construct an increasing number sequence $(n_k)_{k\in\omega}$ such that $y_{n_k}\notin \bigcup_{i\le k}\ddot M_i$. Taking into account that the set $\ddot M_0$ contains $y$, but $\ddot M_0$ does not contain the points $y_{n_k}$, we conclude that $y$ does not belong to the closure $\overline{\{y_{n_k}\}}$ of the singleton $\{y_{n_k}\}$ for all $k\in\omega$. Consequently, the set $B=(X\setminus U)\cup\bigcup_{k\in\omega}\overline{\{y_{n_k}\}}$ does not contain its accumulation point $y$ and hence is not closed in $Y$. -Since the map $q$ is quotient, the preimage $q^{-1}(B)$ is not closed in $X$. By the sequentiality of $X$, there exists a sequence $(x_m)_{m\in\omega}\in q_E^{-1}(B)$ that converges to some point $x\notin q^{-1}(B)$. It follows from $X\setminus q^{-1}(U)=q^{-1}(Y\setminus U)\subset q^{-1}(B)$ that $x\in q^{-1}(U)$. -By the definition of a $cs$-network, there exists a set $N\in\mathcal N$ such that $x\in N\subset q^{-1}(U)$ and $N$ contains all but finitely many points $x_m$. Find $k\in\omega$ such that $q(N)=M_k$. Since the closed set $F_k:=\bigcup_{i\le k}q^{-1}(\overline{\{y_{n_i}\}})$ does not contain $x$, it does not contain the points $x_m$ for all sufficiently large numbers $m$. So, we can find $m\in\omega$ so large that $x_m\notin F_k$. Then $x_m\in q^{-1}(\overline{\{y_{n_i}\}})$ for some $i>k$ and hence $q(x_m)\in \overline{\{y_{n_i}\}}$. Now observe that $q(x_m)\in q(N)=M_k$ and hence $y_{n_i}\in\ddot M_k\subset \bigcup_{j\le i}\ddot M_j$, which contradicts the choice of $y_{n_i}$.<|endoftext|> -TITLE: Correlated Brownian motion and Poisson process -QUESTION [17 upvotes]: Is there an (easy) way to construct, on the same filtered probability space,a Brownian motion $W$ and a Poisson process $N$, such that $W$ and $N$ are not independent ? -I first asked this question at math.stackexchange.com and was suggested to post it here : https://math.stackexchange.com/questions/25519/correlated-brownian-motion-and-poisson-process. -Note that I want $W$ and $N$ to be BM and Poisson for the same filtration $({\cal F}_t)$ (see Shai Covo's answer on math.stackexchange for a construction without this requirement). - -REPLY [11 votes]: To further elaborate on my comment, it is a theorem that if $X^1,X^2,\ldots,X^n$ are Lévy processes with respect to a common filtration, all starting from zero, then they are independent if and only if their quadratic covariations $[X^i,X^j]$ are all (almost surely) zero. This is stated as Theorem 11.43 of He, Wang & Yan, Semimartingale Theory and Stochastic Calculus. It's not difficult to prove with a bit of stochastic calculus, and I'll give a proof for two Lévy processes below. -In the situation described in the question, there are two Lévy processes $W$ and $N$, where $N$ is a pure jump process. So, the quadratic covariation is simply a sum over the jumps of the processes. -$$ -[W,N]\_t=\sum_{s\le t}\Delta W_s\Delta N_s. -$$ -But as Brownian motion is continuous, $\Delta W$ is zero, so the covariation $[W,N]$ is zero. Therefore, they are independent. -Now, let's show that if $X$, $Y$ are Lévy processes w.r.t. the filtration $\{\mathcal{F}\_t\}\_{t\in\mathbb{R}^+}$ with $X_0=Y_0=0$ and $[X,Y]=0$ then they are independent. The characteristic functions of $X$ and $Y$ can be written as -$$ -\begin{align} -&\mathbb{E}\left[e^{iaX_t}\right]=\exp(t\psi_X(a)),\\\\ -&\mathbb{E}\left[e^{iaY_t}\right]=\exp(t\psi_Y(a)). -\end{align} -$$ -Independence of the increments w.r.t. $\mathcal{F}\_{\cdot}$ implies that $M_t\equiv\exp(iaX_t-t\psi_X(a))$ and $N_t\equiv\exp(ibY_t-t\psi_Y(b))$ are martingales. As the jumps of the quadratic covariation equals the product of the jumps of the processes, $\Delta [X,Y]=\Delta X\Delta Y$, it follows that $X$ and $Y$ cannot jump simultaneously. So, $\Delta [M,N]=\Delta M\Delta N=0$. Also, the continuous part of the quadratic covariation $[M,N]^{c}$ is just an integral with respect to $[X,Y]^{c}$ (which follows from Ito's formula for non-continuous semimartingales). So, the covariation $[M,N]$ is zero. Using integration by parts, -$$ -d(M_tN_t)=M_{t-}dN_t + N_{t-}dM_t+d[M,N]\_t=M_{t-}dN_t + N_{t-}dM_t. -$$ -As a sum of integrals with respect to martingales, $MN$ is a local martingale. As it is also bounded at any time, this is a proper martingale. So, $\mathbb{E}[M_tN_t\mid\mathcal{F}\_s]=M_sN_s$ for $s < t$. Plugging in the definitions of $M$ and $N$, -$$ -\mathbb{E}\left[e^{iaX_t+bY_t}\;\Big\vert\;\mathcal{F}\_s\right]=\exp(iaX_s+(t-s)\psi_X(a))\exp(ibY_s+(t-s)\psi_Y(b)). -$$ -This determines the joint characteristic function of $(X_t,Y_t)$ conditional on $\mathcal{F}\_s$, showing that they are independent. As the distributions of $(X_t,Y_t)$ conditional on $\mathcal{F}\_s$ ($s < t$) determine all finite distributions, $X$ and $Y$ are independent. I'll leave the converse ($X,Y$ independent implies $[X,Y]=0$) as an exercise. It's not needed for the question anyway. -You can also compare this with the argument given by kakuritsu. It is essentially the same thing. Rather than working under the generality of Lévy processes, he (or she?) works directly with the Brownian motion and Poisson process, for which $\psi_W(a)=-\frac12a^2$ and $\psi_N(a)=\lambda(e^{ia}-1)$, and uses the moment generating rather than characteristic function (effectively, $a$ and $b$ above are imaginary).<|endoftext|> -TITLE: Isotropic subspaces in cohomology -QUESTION [5 upvotes]: Hello, -Here is a problem I encountered in the study of Kähler manifolds but there is a natural generalisation of this for topological spaces. -If $X$ is a topological space, denote by $g_\mathbb{R}$ the real genus of $X$, that is the maximal dimension of an isotropic subspace in $H^1(X,\mathbb{R})$ (isotropic means that the cup-product restricted to this space is $0$). We can in the same way define $g_\mathbb{C}$ (here we take the complex dimension). -Now the question is : $g_\mathbb{C} = g_\mathbb{R}$ ? -This seems totally obvious : if one has a real isotropic space, then its complexification is a complex isotropic subspace but conversely I don't see how to construct a real isotropic subspace from a complex one. -This is true if $H^2$ has dimension $0$ or $1$ : $0$ is clear and for $1$ one can see the cup-product as a standard symplectic form (maybe degenerate), but in general ? -Thank you for your answers and sorry if I just missed something obvious. - -REPLY [3 votes]: In the full linear algebra generality, the answer is no. -Take four generic $2$-planes, $V_1$, $V_2$, $V_3$ and $V_4$ in $\mathbb{R}^4$. Over $\mathbb{C}$, there are always two $2$-planes $W$ such that $W \cap V_i \neq (0)$ for $1 \leq i \leq 4$. (This is the first nontrivial Schubert calculus example.) Choose the $V_i$ such that the $W$ are NOT defined over $\mathbb{R}$. If you want a concrete example, take -$\def\span{\mathrm{Span}_{\mathbb{R}}}$ $\span(e_1, e_2)$, $\span(e_3, e_4)$, $\span(e_1+e_3, e_2+e_4)$ and $\span(e_1+e_4, e_2-e_3)$; the two $W$'s are $\mathrm{Span}_{\mathbb{C}}(e_1 + ie_2, e_3 + i e_4)$ and its complex conjugate. -Let $\omega_i$ be a degenerate skew-symmetric bilinear form of rank $2$, for which $V_i$ is the kernel. For example, if $V_i$ is $\span(e_1, e_2)$, you could take $\omega_i = e_3^* \wedge e_4^*$. Lemmma: $W$ is $\omega_i$-isotropic if and only if $W$ has nontrivial intersection with $V_i$. -So, with the $V_i$ as above, there are two complex isotropic $2$-planes, but no real ones.<|endoftext|> -TITLE: Who streamlined Kontsevich's count of rational curves? -QUESTION [13 upvotes]: Let $N_d$ denote the number of rational curves in $\mathbf P^2$ passing through $3d-1$ points in general position. Maxim Kontsevich discovered a famous recursion for these numbers: -$$ N_d = \sum_{k+l = d} N_k N_l k^2 l \left( l \binom{3d-4}{3k-2} - k \binom{3d-4}{3k-1}\right).$$ -The proof of this recursion goes by interpreting $N_d$ as a Gromov-Witten invariant: one looks at the moduli space $\overline{M}_{0,3d-1}(\mathbf P^2, d)$ and pulls back the class of a point along each evaluation map. Taking the product of all these classes in the Chow ring produces a number. Using that $\mathbf P^2$ is homogeneous one can show that this number actually counts the number of stable maps where the markings are sent to the given points, and it is not hard to see that counting stable maps is the same thing as counting rational curves. Finally, the associativity of the quantum product can be translated to the WDVV differential equations for the Gromov-Witten potential, i.e. the generating function of all Gromov-Witten invariants. These differential equations translate into the above recursion. -At least, this is how the proof is stated in Kontsevich's "Enumeration of rational curves via torus actions" and Konstevich-Manin "Gromov-Witten classes, quantum cohomology and enumerative geometry", which seem to be the earliest published sources. -However, there is also a beautiful streamlined proof which avoids the use of the quantum product. Here one instead works with $\overline M_{0,3d}(\mathbf{P}^2,d)$ (one more marking) and takes the pullback of the classes of two lines in $\mathbf P^2$ along the first two markings and $3d-2$ classes of points for the remaining. The intersection of these are a curve in the moduli space and one then computes the intersection of this curve with two different linearly equivalent boundary divisors. These two intersection numbers can easily be computed by hand, producing the recursion. The proof that these two boundary divisors are linearly equivalent uses the forgetful map to $\overline{M}_{0,4}$ which is also used in the proof of the WDVV equations so in some sense it seems like this proof inlines the particular case of WDVV that is needed in a very clever way. -This latter version of the proof appears for instance in the book of Kock and Vainsencher, in Abramovich's "Lectures on Gromov-Witten invariants on orbifolds", and in the lecture notes I found here. But where is it from originally? All these sources just refer to it as "Kontsevich's proof" without attribution. Did Kontsevich also come up with this streamlined version but did not see it as worth publishing? - -REPLY [11 votes]: In Manin's book "Frobenius manifolds, quantum cohomology, and moduli spaces", section 0.6.2, the argument ("an old trick of enumerative geometry") is attributed to Kontsevich. I would guess that Kontsevich probably came up with the "streamlined proof" but preferred to talk about it within the framework of WDVV because it more readily generalizes this way, and because it leads to nice-sounding statements like "the enumerative formula is equivalent to associativity of quantum multiplication". That's what I imagine at least. Also, I would be inclined to trust Manin's attribution, as he collaborated with Kontsevich on these early works.<|endoftext|> -TITLE: Picard group, Fundamental group, and deformation -QUESTION [8 upvotes]: One of the most elementary theorems about Picard group is probably $\mathrm{Pic} (X \times \mathbb{A}^n) \cong \mathrm{Pic} X$ and $\mathrm{Pic} (X \times \mathbb{P}^n) \cong \mathrm{Pic} X \times \mathbb{Z}$ (we probably need some restriction for $X$ but let's forget about it for now). This looks very similar to the formulas for $\pi_1$ (the fundamental group). So, my question is whether the who has any relationship and whether one can prove those formulas of the Picard groups using some kind of deformation (as in Topology). - -REPLY [3 votes]: First: The identity map on $X$ factors through the inclusion from $X$ to $X\times {\mathbb A}^1$. Therefore $Pic(X)$ is a direct summand of $Pic(X\times{\mathbb A}^1)$. This is the same argument you'd use in topology to show that $\pi_1(X)$ is a direct summand of $\pi_1(X\times I)$ ($I$ being the unit interval). -To show that $Pic(X)$ is equal to all of $Pic(X\times {\mathbb A}^1)$, you can (as you surmise) take a class in $Pic(X\times {\mathbb A}^1)$ and ``deform'' it until it evidently comes from $X$. -Here (roughly) is one way to do this: Represent your class by a divisor $D$. Pull $D$ back to a divisor on $X\times {\mathbb A}^1 \times {\mathbb A}^1$ along the map that sends $(x,s,t)$ to $(x,st)$. Call the pullback $D'$. Then $D'$ restricts to $D$ when $t=1$ and to a divisor that comes from $X$ when $t=0$. Call that latter divisor $D_0$. -You can think of $D'$ as deforming $D$ into $D_0$, which completes the proof. The reason this deformation sets $D$ equal to $D_0$ in the Picard group is that you can think of $D'$ as (roughly) the graph of a function with zeros along $D_0$ and poles along $D$ (after a transformation that takes $1$ to infinity). That's exactly what it takes to make $D-D_0$ trivial in the divisor class group, which, given appropriate assumptions, is the same as the Picard group. So yes, this kind of deformation is just what you need, and just what you've got.<|endoftext|> -TITLE: Complementation of $\omega$-regular languages in reverse mathematics -QUESTION [14 upvotes]: Does anyone know where Büchi's theorem that $\omega$-regular languages are closed under complementation fits into the reverse-mathematics classification scheme? That is, is it equivalent over $\mathrm{RCA}_0$ to one of the usual subsystems of second-order arithmetic? Or, if not known to be equivalent, what is known about where it fits in? -The formulation I have in mind is, for a fixed finite signature $\Sigma$, the statement: for every finite automaton $M$ (over $\Sigma$), there exists a finite automaton $M^c$, such that, for every $\omega$-word $\alpha$ over $\Sigma$, it holds that $\alpha$ is (Büchi-)accepted by $M$ if and only if $\alpha$ is not accepted by $M^c$. - -REPLY [4 votes]: There is a new paper addressing exactly this question: -L. A. Kołodziejczyk, H. Michalewski, P. Pradic, M. Skrzypczak, The logical strength of Büchi's decidability theorem -accepted to CSL 2016. The paper is available at the first author's website: -http://www.mimuw.edu.pl/~lak/buchi_strength.pdf. -The abstract states: - -We prove that the following are equivalent over the weak second-order - arithmetic theory $\text{RCA}_0$: - -Büchi’s complementation theorem for nondeterministic automata on infinite words, -the decidability of the depth-n fragment of the MSO theory of (N, ≤), for each n ≥ 5, -the induction scheme for $\Sigma^0_2$ formulae of arithmetic.<|endoftext|> -TITLE: Are all Shimura Varieties Special Subvarieties of the Siegel modular Variety? -QUESTION [5 upvotes]: Given a Shimura variety $S$, is it possible to imbed $S$ as a special Subvariety -of the Siegel modular variety $A_{g,N}$, for some $g$ and level $N$? I expect that the answer is yes, essentially since every -semisimple group over $\mathbb{Q}$ should imbed into $GL_n$ via its adjoint representation, -and $GL_n$ imbeds into $SP_{2n}$. However, I'm a bit worried about the business regarding weights. -Thank you, -Jacob - -REPLY [4 votes]: The answer is no, in general. The problem is to find an embedding so that the minuscule character corresponding to the Shimura datum for $S$ induces the minuscule character of $GSp_{2n}$ corresponding to a decomposition into Lagrangians. -In the affirmative direction, for most classical, simply connected groups (and only for classical groups, i.e. of types $A$,$B$,$C$ and $D$), the answer is yes; some subtleties crop up for $Spin^*(2n)$ (this is the so called $D^{\mathbb{H}}$ case), for which only the quotient by an order 2 central sub-group admits a symplectic embedding (of Shimura data). -This is all beautifully laid out in Deligne's article 'Varietes de Shimura...' here, following Satake here. See also Proposition 1.21 in Milne's article here<|endoftext|> -TITLE: Characterizing intersection of zero sets of elementary symmetric polynomials on R^n -QUESTION [7 upvotes]: Stated simply, the question is: - -Consider two elementary symmetric polynomials $\sigma_{k}$ and $\sigma_{k+1}$ on $\mathbb{R}^{n}$ with zero sets $U_{k}$ and $U_{k+1}$. Let $V_{i_{1}i_{2}\dotsb i_{j}}$ be the coordinate linear space $\{x\in\mathbb{R}^{n}: x_{i_{1}} = \dotsb = x_{i_{j}} = 0\}$ and $W_{n-k+1} = \cup_{i_{1} < i_{2} < \dotsb < i_{n-k+1}}V_{i_{1}\dotsb i_{n-k+1}}$, the set of all points in $\mathbb{R}^{n}$ with at least $n-k+1$ coordinates equal to zero. Does - \begin{equation} -(\*) \qquad U_{k} \cap U_{k+1} = W_{n-k+1} -\end{equation} - ? Why? - -Note: I am interested in a slightly more specific question. Namely, if $\Gamma_{k}^{+}$ is the component of $\sigma_{k} > 0$ containing the point $(1,1,\dotsb,1)$, then is - -\begin{equation} -(\*\*) \qquad \overline{\Gamma_{k}^{+}} \cap \overline{\Gamma_{k+1}^{+}} \subset W_{n-k+1} -\end{equation} - -? It is well known (see "An Inequality for Hyperbolic Polynomials", Lars Garding) that -\begin{equation} -(\*\*\*)\qquad \Gamma_{k}^{+} \supset \Gamma_{k+1}^{+} -\end{equation} -and that -\begin{equation} -(****)\qquad \{\sigma_{k+1} > 0\} \cap \Gamma_{k}^{+} = \Gamma_{k+1}^{+} -\end{equation} -If $(\*)$ is true, then $(\*\*)$ follows. The equation $(\*)$ is simpler to state and appears to be true. Of course $(\*\*)$ may hold with $(\*)$ failing, but I think this is unlikely. -Background and motivation: -Numerical simulations in mathematica suggest that $(\*)$ holds. -Certainly $W_{n-k+1} \subset U_{k} \cap U_{k+1}$. -I can prove the reverse inclusion in certain cases. For example, denoting $x^{j} = (x_{1}^{j},\dotsb,x_{n}^{j})$ for $x\in\mathbb{R}^{n}$, notice that $W_{n-k+1}$ is exactly the zero set of $\sigma_{k}(x^{2})$. So we have -n arbitrary, k=1: From Newton's identities (see wikipedia), $\sigma_{1}^{2}(x) = \sigma_{1}(x^{2})+2\sigma_{2}(x)$. So if $\sigma_{1}(x) = \sigma_{2}(x) = 0$, then $\sigma_{1}(x^{2}) = 0$ which means $x = (0,\dotsb,0)$ so that $x \in W_{n}$. -n arbitrary, k = 2, $(\*\*)$ only: can be proved similarly to $k=1$ by writing $0 \le \sigma_{2}(x^{2}) = \sigma_{2}^{2}(x) -2\sigma_{1}(x)\sigma_{3}(x)+2\sigma_{4}(x)$. So if $\sigma_{2}(x) = \sigma_{3}(x) = 0$, then $\sigma_{2}(x^{2}) = 2\sigma_{4}(x)$. Now, $\sigma_{4}(x) \le 0$ because of $(\*\*\*)$ and $(\*\*\*\*)$. Hence $\sigma_{2}(x^{2}) = 0$ so that $x \in W_{n-1}$. -n arbitrary, $k=n-2$ and $k =n-1$ can be proved in a similar fashion. However, because of the additional terms in the equation -\begin{equation} -\sigma_{k}(x^{2}) = \sigma_{k}^{2}(x)-2\sigma_{k-1}(x)\sigma_{k+1}(x)+2\sigma_{k-2}(x)\sigma_{k+2}(x)+\dotsb+(-1)^{k}2\sigma_{0}(x)\sigma_{2k}(x) -\end{equation} -for other values of $k$, the above approach fails in general. - -I'm actually trained as a differential geometer, so I may be approaching this problem in the wrong way (perhaps there's a technique in real algebraic geometry?). If you can suggest an alternative approach I would be grateful to hear it. This result seems rather elementary to state so I would be surprised if the result is not known. Thank you. - -REPLY [7 votes]: To establish that $U_k \cap U_{k+1} \subset W_{n-k+1}$: -Consider the polynomial $P(t)=\prod (t-x_i)$. The elementary symmetric polynomials $\sigma_i$ are its coefficients, up to the sign. -Suppose that $\sigma_k=\sigma_{k+1}=0$. This means that $0$ is a multiple root of the derivative of order $(n-k-1)$ of $P$. Now you can conclude that $0$ is a root of $P$ of multiplicity at least $n-k+1$ (that is: at least $n-k+1$ of the numbers $x_i$ are zero) by means of the following facts: - -If a non-constant polynomial $Q$ has - all its roots real, then so does its - derivative $Q'$. Then the roots of - $Q'$ are the multiple roots of $Q$, - plus one root between each pair of - consecutive distinct roots of $Q$, - necessarily simple (otherwise the sum - of the multiplicities of the roots of - $Q'$ would exceed its degree). In - particular, if $x$ is a multiple root - of $Q'$ then it is also necessarily a - root of $Q$.<|endoftext|> -TITLE: Artin Jacobson-semisimple rings are semisimple. Constructively, too? -QUESTION [6 upvotes]: Notation. When I say "ring", I mean "ring with unity" (not necessarily commutative). -Definition. A ring $R$ is said to be left-Artinian if for every sequence $I_0\supseteq I_1\supseteq I_2\supseteq I_3\supseteq ...$ of left ideals of $R$, there exists an $n\in\mathbb N$ such that $I_n=I_{n+1}$. -Definition. A ring $R$ is said to be right-Artinian if for every sequence $I_0\supseteq I_1\supseteq I_2\supseteq I_3\supseteq ...$ of right ideals of $R$, there exists an $n\in\mathbb N$ such that $I_n=I_{n+1}$. -Definition. A ring $R$ is said to be Artinian if it is both left-Artinian and right-Artinian. -Definition. The Jacobson radical $\operatorname{Ra}R$ of a ring $R$ is defined by one of the following equivalent definitions: -$\operatorname{Ra}R = \left\lbrace r\in R\mid \text{for every }s\in R\text{, the element }1-rs\text{ of }R\text{ is invertible}\right\rbrace$; -$\operatorname{Ra}R = \left\lbrace r\in R\mid \text{for every }s\in R\text{, the element }1-sr\text{ of }R\text{ is invertible}\right\rbrace$; -$\operatorname{Ra}R = \left\lbrace r\in R\mid \text{for every }\left(s,t\right)\in R^2\text{, the element }1-srt\text{ of }R\text{ is invertible}\right\rbrace$; -$\operatorname{Ra}R = \left\lbrace r\in R\mid \text{every right ideal }I\text{ of }R\text{ satisfying }I+rR=R\text{ satisfies }I=R\right\rbrace$; -$\operatorname{Ra}R = \left\lbrace r\in R\mid \text{every left ideal }I\text{ of }R\text{ satisfying }I+Rr=R\text{ satisfies }I=R\right\rbrace$; -$\operatorname{Ra}R = \left\lbrace r\in R\mid \text{every f.g. right }R\text{-module }M\text{ satisfying }MrR=M\text{ satisfies }M=0\right\rbrace$; -$\operatorname{Ra}R = \left\lbrace r\in R\mid \text{every f.g. left }R\text{-module }M\text{ satisfying }RrM=M\text{ satisfies }M=0\right\rbrace$ -(where "f.g." means "finitely generated"). (Note that the equivalences are constructive; I have written up the proofs in German a while ago (search for "Jacobson-Radikal") and will translate when I have the time.) -Definition. A ring $R$ is said to be von Neumann regular if for every $r\in R$, there exists some $x\in R$ such that $rxr=r$. -Question: Can we constructively prove that every Artinian ring $R$ satisfying $\operatorname{Ra}R=0$ is von Neumann regular? (This is proven classically using the AC in Lam, "A first course in noncommutative rings", Theorem (4.14) + Corollary (4.24).) -Normally, theorems in algebra can be proved constructively if we know a classical proof. There are methods for this (scindage a la Lombardi; dynamic proofs; Gödel-Gentzen etc.). Unfortunately, whenever chain conditions (such as Artinianity) are involved, these methods break down. The constructive Artinian condition is neither easy to use nor easy to satisfy, so I am not completely sure whether the question is the right one to ask - but I don't know of a better one. -While constructive Artinianity is far less useful than classical Artinianity, it can still be applied to chains of ideals such as $R\supseteq rR\supseteq r^2R\supseteq r^3R\supseteq ...$ to conclude that for every $r\in R$ there exists some $n\in\mathbb N$ and some $y\in R$ such that $r^n=r^{n+1}y$. This can then be juggled with (for example, we can conclude that $r^n=r^ayr^b$ for any two nonnegative integers $a$ and $b$ with $a+b=n+1$; here we use $\operatorname{Ra}R=0$). This is, at the moment, my main reason to believe that the Question above has a positive answer (we mainly have to bring the $n$ down to $1$). But, as I said, I am far from sure about this. -Meta-question: What is the (morally) right constructive analogue of the notions "Artinian" and "Noetherian"? Given the definition of "Artinian" above, I am not sure if $\mathbb F_2$ is Artinian, because I could take a sequence $S_0,S_1,S_2,...$ of independent statements which are independent of each other too (is this possible?) and then let $I_n$ be the ideal containing $1$ if $S_0, S_1, ..., S_{n-1}$ hold. So let me pose a different question, which is actually interesting without relying on constructivity: -Concrete question. Let $R$ be a ring with $\operatorname{Ra} R = 0$. Assume that, for every $r \in R$, there exists an $n \in \mathbb{N}$ and a $y \in R$ such that $r^n = r^{n+1} y$. Also assume that, for every $r \in R$, there exists an $n \in \mathbb{N}$ and a $y \in R$ such that $r^n = y r^{n+1}$. Must $R$ then be von Neumann regular? -Notice that this is NOT a constructive translation of the classical theorem. It is a stronger conjecture which has the advantage of not requiring a constructive translation to begin with. - -REPLY [4 votes]: The question has been changed, so I'm submitting a second answer. -The answer to the new concrete question is no. Your question boils down to whether there exists a semi-primitive strongly $\pi$-regular ring which is not von Neumann regular. The answer is yes. For instance, you can use the subring of $\prod_{\omega}\mathbb{M}_2(\mathbb{F}_2)$ consisting of sequences of matrices which are eventually stable (meaning the matrices stop changing after some point) and eventually upper-triangular. -By the way, suppose we revert back to the original question of whether it is possible to find a constructive proof that left/right artinian and semiprimitivity implies v.N. regularity, and we also assume that semiprimitivity gives us an oracle such that for every element $r\in R$ it gives us $s\in S$ such that $1-rs$ is not invertible. Using a construction similar to this one, I believe I can show that we cannot prove v.N. regularity (essentially because of the example above).<|endoftext|> -TITLE: symplectic 4-manifolds with free circle action -QUESTION [8 upvotes]: Hi. I have a question. -Let $(M,\omega)$ be a closed symplectic 4-manifold equipped with a free circle action which preserves $\omega$ (symplectic circle action). -My question is , is there an example of $M$ which is not homeomorphic to $S^1 \times N$? ($N$ is a closed oriented 3-manifold) -Thank you in advance. - -REPLY [11 votes]: Here's an example, using a construction of Fernandez, Gray and Morgan (1991): -Take a closed surface $S$ with area form $\omega$, let $\phi$ be an area-preserving diffeomorphism, and $p\colon S_\phi \to S^1$ its mapping torus. This carries a closed 2-form $\omega_\phi$ induced by $\omega$, and a closed 1-form $p^\ast dt$. Take a class $e\in H^2(S_\phi;\mathbb{Z})$ which restricts trivially to $H^2(S;\mathbb{Z})$. Then $e$ has a de Rham representative of form $p^\ast dt\wedge a$, where $a$ is a closed 1-form. Take $L\to S_\phi$ be a hermitian line bundle with a connection form $i\eta$ of curvature $(-2\pi i) p^\ast dt\wedge a$, and let $\pi\colon M\to S_\phi$ be the unit circle bundle in $L$. Then the 4-manifold $M$ carries the $S^1$-invariant symplectic form $\Omega:= \pi^* \omega_\phi + \pi^*p^\ast dt\wedge \eta$. -Let's take $S$ to have genus $>1$ and $\phi$ to be a Dehn twist along a non-separating curve. The Wang exact sequence identifies $\ker (H^2(S_\phi)\to H^2(S))$ with the cokernel of $(1-\phi^\ast)$ acting on $H^1(S)$. In this case, the cokernel is $\mathbb{Z}$, and we take $e$ to be the generator. The fibration by $S^1$-orbits is non-trivial, but we need to check that the resulting $M$ is not homeomorphic to $S^1\times N^3$ in some weirder fashion. -Well, $\pi_1(M)$ is a non-trivial central $\mathbb{Z}$-extension of $\pi_1(S_\phi)$, and the latter is a semidirect product of $\pi_1(S)$ with $\mathbb{Z}$, where $1\in \mathbb{Z}$ acts on $\pi_1(S)$ by $\phi^{-1}$. If I'm not mistaken, $\pi_1(S_\phi)$ has trivial centre. Hence the centre of $\pi_1(M)$ is $\mathbb{Z}$, the subgroup generated by the $S^1$-fibre $\gamma$. If $M = S^1\times N$ then $\pi_1(M)$ is a trivial central $\mathbb{Z}$-extension of $\pi_1(N)$. But the central $\mathbb{Z}$-subgroup defined by this product splitting must be generated by a multiple of $\gamma$, and so the extension it defines is not trivial after all.<|endoftext|> -TITLE: Existence of closed manifolds with more than 3 linearly independent complex structures? -QUESTION [7 upvotes]: A Riemannian manifold is hyperkähler, if there are three complex structures $I,J,K$, which are all compatible with the Riemannian metric (i.e., $(v,Iw)$ defines a symplectic form and similarly for $J$ and $K$). Furthermore, we also need the complex structures to satisfy the quaternionic relations $I^2=J^2=K^2=-1$ and $IJ+JI=0=IK+KI=JK+KJ$. -I was able to find only very few examples of closed (compact without boundary) hyperkähler manifolds, basically complex tori and K3-surfaces. -In higher dimensions than complex dimension 2, there are also generalized Kummer varieties of tori. Based on K3-surfaces, there are Hilbert schemes and resolutions of singularities in some moduli spaces. - -I am searching for an example of a slightly more general setting: -Let $X$ be a closed Riemannian manifold and $J_1,..., J_r$ be complex structures, all compatible with the Riemannian metric. Furthermore, assume that the $J_l$ to satisfy the relations $J_l^2=-1$ and $J_lJ_k+J_kJ_l=0$ for $k\neq l$. -If we take $X$ to be a vector space of dimension $2^{4a+b}c$ with $c$ odd, I know the maximal number for $r$ is $8a+2^b-1$, i.e. the maximal number is always odd. This carries over to quotients of vector spaces, i.e. for tori and quotients of tori. These examples are flat. -This yields the question: -Is there a closed, non-flat manifold admitting more than three such complex structures? -If $X$ admits more than three complex structures, it also admits three such structures. Therefore every such manifold has to be a hyperkähler manifold, i.e. one of the above mentioned examples or a product of them. -Are there different examples and does one of them admit more complex structures than the 3 for hyperkähler? - -REPLY [7 votes]: A manifold admitting a triple of complex structures satisfying quaternionic -relations also admits a torsion-free connection (called "Obata connection") preserving the quaternionic structure. Such a connection is unique. For a hyperkaehler manifold, the Obata connection coinsides with the Levi-Civita. Therefore, the manifolds with this structure have Levi-Civita connection which preserves the complex structures $J_i$. From Berger's classification of holonomy it follows that they are products of hyperkaehler manifolds, and -actually flat if $i>3$ (because $J_4$ must exchange the tangent bundles to the factors of the product).<|endoftext|> -TITLE: Idempotent measures on the free binary system? -QUESTION [13 upvotes]: Let $(S,*)$ be the free (non associative) binary system on one generator (so $S$ is just the set of terms in $*$ and $1$). There is an extension of $*$ to the space $P(S)$ of finitely additive probability measures on $S$ defined as follows: -$$\mu * \nu (A) = \int \int \mathbf{1}_{*^{-1}(A)} (x,y)\ d \nu (y)\ d \mu (x)$$ -Here is the question: Is there an idempotent measure in $P(S)$ (i.e. a $\mu$ -such that $\mu * \mu = \mu$)? Has this question been considered in the literature? -Here is the motivation: there is a natural way to identify $(S,*)$ with the positive elements of Thompson's group $F$ (elements of $S$ are "rooted ordered binary trees"). It is not hard to show that an idempotent measure is in fact an invariant measure with respect to the action of $F$. Now, I don't expect someone to produce a positive answer (although I will conjecture the even stronger statement that -every compact convex $C \subseteq P(S)$ which is $*$-closed contains an idempotent). I am mostly asking if someone sees how to refute the existence of such a measure or if this question has appeared in the literature. -Some further observations: the map $(\mu,\nu) \mapsto \mu * \nu$ is NOT continuous -($\mu \mapsto \mu * \nu$ is, for each $\nu$, but this is about the extent of continuity). Thus the map $\mu \mapsto \mu * \mu$ is not continuous (if it were, we could apply a fixed point theorem...). -If one drops the assumption of freeness, then it is possible to find idempotents if any of the following are true: - -$S$ is finite (in this case everything is continuous and so fixed point theorems apply). -$(S,*)$ is associative (i.e. $S$ is a semigroup) (this is "Ellis's Lemma''). -$*$ depends only on one argument (again, fixed point theorems apply). - -An auxiliary question is to characterize when a monogenic binary system $(S,*)$ satisfies -that $P(S)$ contains an idempotent. It should be noted that for associative binary operations, it is possible to find an idempotent ultrafilter (i.e. taking values in $\{0,1\}$) but that this is impossible for the free (non associative) binary system one one generator. - -REPLY [2 votes]: This is not an answer but perhaps it gives an idea where to look for it. I think that existence of idempotent measures on $\mathbb N$ is the key point in Furstenberg's proof of van der Waerden's theorem and its generalizations and that one can view van der Waerden's theorem as a reformulation of existence of idempotent measures (I may be wrong here). An analog of van der Waerden's theorem is true for the free binary system (groupoid) with $n$ generators: for every decomposition of $A$ into $k$ parts there exists a polynomial $p(x)$ (i.e. a term with one variable) such that all elements $p(x_i), i=1,...,n,$ belong to the same partition class (it is not difficult and was proved by Bespamyatnyh and myself in 1982 but could have been proved by somebody else before). I do not know if it implies existence of idempotent measure, but it may be worthwhile looking at it.<|endoftext|> -TITLE: Why not define triangulated categories using a mapping cone functor? -QUESTION [25 upvotes]: Recall that the usual definition of a triangulated category is an additive category equipped with a self equivalence called $[1]$ in which certain diagrams, of the form $X \to Y \to Z \to X[1]$ are called "exact", satisfying certain axioms. Two of these axioms are that -(1) Given $X \to Y$, it can be extended to an exact triangle $X \to Y \to Z \to X[1]$ and -(2) Given a commuting diagram -$$\begin{matrix} X & \to & Y \\ \downarrow & & \downarrow \\ X' & \to & Y' \end{matrix}$$ -and exact triangles $X \to Y \to Z \to X[1]$ and $X' \to Y' \to Z' \to X'[1]$, there is a map $Z \to Z'$ making the obvious diagram commute. -And, as every source on triangulated categories points out, one of the standard problems with the theory is that there is no uniqueness statement in these axioms. So, why not make one? -I envision a definition as follows: For any category $C$, let $Ar(C)$ be the category whose objects are diagrams $X \stackrel{f}{\longrightarrow} Y$ in $C$ and whose morphisms are commuting squares in $C$. Note that there are obvious functors $\mathrm{Source}$ and $\mathrm{Target}$ from $Ar(C) \to C$, and a natural transformation $\mathrm{Source} \to \mathrm{Target}$. Define a conical category to be an additive category $C$ with a self-equivalence $[1]$ and a functor $\mathrm{Cone} : Ar(C) \to C$, equipped with a natural transformations $\mathrm{Target} \to \mathrm{Cone} \to \mathrm{Source}[1]$, obeying certain axioms. -I noticed one place you have to be careful. In a triangulated category, if $X \to Y \to Z \to X[1]$ is exact, then so is $Y \to Z \to X[1] \to Y[1]$ (with a certain sign flip). The most obvious analoguous thing in a conical category would be for $\mathrm{Cone}(Y \to \mathrm{Cone}(X \to Y))$ to equal $X[1]$; the right thing is to ask for a natural isomorphism instead. -But everything else seems work out OK, at least in the case of the homotopy category of chain complexes. And it seems much more natural. What goes wrong if you try this? -I know that this is the sort of subject where people tend to mention $\infty$-categories; please bear in mind that I don't understand those very well. Everything I've said above just used ordinary $1$-categories, as far as I can tell. - -REPLY [15 votes]: Verdier, Astérisque 239, Ch. II, Prop. 1.2.13 (p. 104) says that a triangulated category (with countable coproducts or products) equipped with a cone functor has to split. -Ben Wieland is right - you get a cone functor when working with dérivateurs (or filtered derived categories), but that functor is no longer intrinsic to the base category, and you have to carry diagram categories along.<|endoftext|> -TITLE: Bruhat decomposition for G(R), R local ring or R=Z/p^r -QUESTION [7 upvotes]: Is there an invariant, which encodes the failure of the Bruhat decomposition to hold for a reductive group with coefficients in a local ring like the p-adic integers or the ring $\mathbb{Z}/\mathfrak{p}^r$? -Example $G =GL_2$: Fix a Borel subgroup $B$, e.g. the upper diagonal matrixes. The coset space $B \backslash G$ (or $G/B$) are isomorphic to the projective line. However, the Bruhat decomposition $G = BWB$, where $W$ is the Weyl group, does not hold for the group of $R$ points, where $R$ is not a field. Can we describe $B\backslash G/B$ as a variety over $\mathbb{Z}$ here? - -REPLY [10 votes]: Bruhat decomposition over $\mathbf Z/p^r\mathbf Z$ is precisely the problem we looked at in this paper. We defined several invariants of double cosets, and classified the pairs $(n,k)$ for which, when $G=GL_n(\mathbf Z/p^k\mathbf Z)$, the cardinality of $B\backslash G/B$ does not depend on $p$. Unfortunately, the general question seems to involve wild classification problems.<|endoftext|> -TITLE: Pull-backs which are push-outs -QUESTION [14 upvotes]: Consider a commutative square in a category $\mathcal{C}$ -$$\begin{array}{ccc} -A&\rightarrow&B\\\ -\downarrow&&\downarrow\\\ -C&\rightarrow&D -\end{array}$$ -Suppose $\mathcal{C}$ is abelian. If this square is a pull-back and $B\rightarrow D$ or $C\rightarrow D$ is an epimorphism, then this square is also a push-out square. Dually, if this square is a push-out and $A\rightarrow B$ or $A\rightarrow C$ is a monomorphism, then this square is also a pull-back square. ¿Are there more general kinds of categories were such things happen? - -REPLY [17 votes]: Yes! Pretoposes (and in particular toposes) also have this property. It is a remarkable fact that pretoposes (which you can think of as having the first-order exactness properties of toposes or $Set$-like categories) have "most" of the same exactness properties as abelian categories (see below). -In fact, this is the beginning of a remarkable set of observations due to Peter Freyd, and expounded by him in a discussion at the categories mailing list, which led to a sharp distinction between pretoposes and abelian categories as concentrated particularly in the behavior of the initial object. (In an abelian category, $A \times 0 \cong A$, whereas in a pretopos $A \times 0 \cong 0$. But this is practically the only essential difference.) In fact, Freyd showed that abelian categories and pretoposes are special cases of what he dubbed "AT categories", which contain the core exactness properties which are common to abelian categories and pretoposes. AT categories cut so close to the essence of each of these two special cases that in fact every AT category splits cleanly as a product of an abelian category and a pretopos! -I wrote up my own account of this in the nLab, here.<|endoftext|> -TITLE: how irregular can a $p$-adic Galois representation be? -QUESTION [13 upvotes]: Fix $p>0$ a rational prime, and $K$ an algebraic number field with Galois group $G_K:=Gal(\bar{\mathbb{Q}}/K) $. The Fontaine-Mazur conjecture predicts that if $\rho:G_K\rightarrow GL(V)$ is a finite dimensional $\mathbb{Q}_p$-representation, then it comes from a motive over $K$ (like a subquotient of $H^i_c(X\times_K\bar{\mathbb{Q}}, \mathbb{Q}_p)$) exactly when it is unramified over almost every finite place, and potentially semi-stable over those finite prime dividing $p$. -My question is the contrary: how many examples do we have for $p$-adic Galois representations having infinite images but for which the conditions of Fontaine-Mazur fails? Maybe it should be difficult to construct them when the dimension of the representation is large? Could one get continuous $p$-adic representations that ramifies at infinitely many places? - -REPLY [13 votes]: On the contrary, it is not so hard by deformation theoretic arguments to create $p$-adic Galois representations which are not even Hodge--Tate at $p$ (and hence not de Rham (equivalently, pst) at $p$). So a reasonable intuition is that "most" $p$-adic Galois representations are not de Rham at $p$.<|endoftext|> -TITLE: Selmer of an abelian variety versus that of its dual. -QUESTION [5 upvotes]: What is the precise relationship between the Selmer group of an abelian variety and that of its dual? For instance, does the vanishing of one not imply the same for the other? -To fix ideas, let $A$ be an abelian variety defined over a number field $K$, with $A^t$ -denote the corresponding dual abelian variety. Fix a rational prime $p$. Suppose for instance that we consider the compactified Selmer group \begin{align*} \mathfrak{S}(A/K) &= -\ker \left( H^1(G_S(K), A_{p^n}) \longrightarrow \bigoplus_{v \in S} H^1(K_v, A(\overline{K}_v) )(p) \right).\end{align*} Here, $S$ is any finite set of primes -of $K$ containing the primes above $p$ and the primes where $A$ has bad reduction; - $G_S(K) = \operatorname{Gal}(K^S/K)$, where $K^S$ is the maximal extension of -$K$ unramified outside of $S$ and the archimedean primes of $K$, and -$A_{p^n} = \ker \left([p^n]:A \longrightarrow A\right)$ denotes the $p^n$ torsion of $A$. -Does $\mathfrak{S}(A/K) =0$ if and only if $\mathfrak{S}(A^t/K)=0$? Clearly this will be -the case if $A$ is principally polarized (in which case there is an isomorphism $A \cong A^t$). However, the general case seems tricky to prove by inspection. I am aware that if the $p$-primary part of the Tate-Shafarevich group $\operatorname{Sha}(A/K)$ is finite, then its follows from basic properties of the Cassels-Tate pairing that $\mathfrak{S}(A/K) \cong \mathfrak{S}(A^t/K)$ if and only if $A(K)_{p^{\infty}} \cong A^t(K)_{p^{\infty}}$ (where $A(K)_{p^{\infty}} = \bigcup_{n \geq 0} A(K)_{p^n}$). But, when is it the case that this latter condition is known (not) to be true? Is there not a better, perhaps unconditional, deduction? Surely this ought to be classical ... - -REPLY [5 votes]: Let $\varphi:A\to A^t$ be a polarization. Then $\varphi$ is an isogeny. -In order to study the difference between the Selmer groups of $A$ and of $A^t$ you need to study the torsion subgroups of $A(K)$ and $A^t(K)$, and to study the difference between the Tate-Shafarevich groups of $A$ and $A^t$. The comparison of the torsion subgroups is quite straightforward. -If you believe the Birch and Swinnerton-Dyer conjecture then -$$ \frac{ | Sha(A/K)| }{ | Sha(A^t/K)|} = \frac{\Omega_{A^t}}{\Omega_A}\prod_\ell \frac{c_{A^t,\ell}}{c_{A,\ell}} $$ -holds, where $\Omega_A$ is the real period and the $c_{A,\ell}$ are the Tamagawa numbers. -Actually, this formula is proven under the hypothesis that $Sha(A/K)$ is finite. -Now, the quotient of the product of Tamagawa numbers might be different from one. Actually, I expect that this product can be arbitrary large. (A similar result can be used to construct elliptic curves with large Tate-Shafarevich groups.)<|endoftext|> -TITLE: Higher K theory and algebraic cycles in representation theory? -QUESTION [7 upvotes]: Can anybody talk about how Higher K theory and algebraic cycles play roles in representation theory? I am more interested in how they play roles in Kazhdan-Lusztig conjectures. -Of course K_0 plays important roles in representation theory,but I don't know whether Higher K theory is also useful in representation theory. -Any references will be very welcome! -Thanks in advance - -REPLY [9 votes]: The $K_2$ functor has an important role in representation theory, starting from -works of Bloch in which he identified the Kac-Moody central extension of loop algebras in terms of $K_2$ - a more updated form of this is the IHES paper of Deligne-Brylinski about $K_2$ central extensions of reductive groups. The universal class in $H^4(BG,Z)$ for a simple group $G$ which is responsible for the level in Kac-Moody algebras/Chern-Simons theory or the $q$ in quantum groups etc can be interpreted as representing this $K_2$ central extension. Anyway it's a very beautiful story we haven't really fully seen the impact of yet (but is also behind eg the famous constructions of Fock-Goncharov in Teichmuller theory). -Anyway that's not the kind of role you're talking about, in which we take higher K groups of a category of representations. One way to answer that is to say certainly the full homotopy theory of a category of representations is very important - but there we usually ask for something more subtle, ie identifying the entire category itself in some geometric way rather than just its $K$-theory spectrum or the homotopy groups of the latter. To really see higher K-groups arising the way you're asking I think you'd need to consider problems in which FAMILIES of representations play a central role -- single representations (or contractible families thereof) are measured by $K_0$, but if you are interested in measuring families over a circle (ie automorphisms) or over more complicated bases (eg higher spheres - i.e. $n$-automorphisms), these would be measured by invariants in $K_1$ or higher $K$-groups. -[Edit: When I said "something more subtle" I meant the following: n equivalence of (enhanced) derived categories gives rise to an isomorphism of K-theories -- eg Beilinson-Bernstein localization identifies K-groups of categories of Lie algebra representations with those of flag varieties. This certainly doesn't mean in general we know explicitly the K-groups! eg there are few categories we understand as well as $Z$-modules, but the corresponding K-groups are complicated! However it seems that these subtleties are arithmetic, and I don't see a clear representation theoretic role for them, though would be happy to learn otherwise.] -The two other things that come to mind from your question are - -in Jacob Lurie's survey article on elliptic cohomology (available here) there's a theorem describing the K-theory spectrum of categories of integrable loop group representations in terms -of a DAG version of nonabelian theta functions (in the spirit of results of Kac-Peterson, Looijenga and Ando) -Beilinson's work on epsilon factors explains algebraic K-theory beautifully and applies it in a context related to automorphic forms.<|endoftext|> -TITLE: Reference request for manifold learning -QUESTION [10 upvotes]: I am interested in learning about manifold learning (no pun intended) and would like to know of some references that discuss the subject from a more geometric perspective. By manifold learning I mean the idea of studying high dimensional data using techniques from geometry. -I'm interested in knowing how topics from differential geometry and topology such as Hodge theory and Morse theory can be used to study questions in manifold learning. I thought I would ask if people have any recommendations for papers or books that explain these topics more from a more geometric perspective. -Update: -I expect that there is no mythical survey paper that explains all aspects of manifold learning to someone that knows about geometry and topology. Specifically, I would be interested in knowing of some survey papers which explain how tools from Riemannian geometry would be useful in manifold learning. Perhaps how such tools can be used for nonlinear dimensionality reduction. - -REPLY [5 votes]: This is not exactly what you're looking for, but on the subject of Topology in Computer Science, here are two recommendations I can make: - -Topology and its Applications, William F. Basener, Wiley-Interscience, 2006 -Topology for Computing, Afra Zomorodian, Cambridge University Press, 2006 - -They both give some inkling into the Differential Topology aspects of Machine Learning. Also, not sure if you've already seen this, but here are some lectures from the likes of Smale on Machine Learning.<|endoftext|> -TITLE: What is the largest Laver table which has been computed? -QUESTION [21 upvotes]: Richard Laver proved that there is a unique binary operation $*$ on $\{1,\ldots,2^n\}$ which satisfies $$a*1 \equiv a+1 \mod 2^n$$ -$$a* (b* c) = (a* b) * (a * c).$$ -This is the $n$th Laver table $(A_n,*)$. -There is an algorithm for computing $a * b$ in $A_n$, but in general (and especially for small values of $a$), this requires one to compute much of the rest of $A_n$. -What is the largest value for $n$ for which someone can, in a modest amount of time, compute an arbitrary entry in $A_n$? I am able to compute entries in $A_{27}$. -I should note that the map which sends $a$ to $a\ \mathrm{mod}\ 2^m$ defines a homomorphism from $A_n$ to $A_m$ for $m < n$ and hence the problem becomes strictly harder for larger $n$. -Edit: I have actually been able to compute $A_{28}$, not just $A_{27}$. - -REPLY [16 votes]: On Azimuth, on May 6, 2016, Joseph van Name wrote: - -The largest classical Laver table computed is actually $A_{48}$. The 48th table was computed by Dougherty and the algorithm was originally described in Dougherty's paper here. With today's technology I could imagine that one could compute $A_{96}$ if one has access to a sufficiently powerful computer. -One can compute the classical Laver tables up to the 48th table on your computer here at my website.<|endoftext|> -TITLE: Testing whether an integer is the sum of two squares -QUESTION [46 upvotes]: Is there a fast (probabilistic or deterministic) algorithm for determining whether an integer $n$ is a sum of two squares? -By "fast" here I mean polynomial time (i.e. time $O((\log n)^{O(1)})$). Note that I am interested only in whether the integer can be represented in such a way, not in how it is represented. -Since a fast algorithm is required, it will not do to use factorization. -It would be odd if this turned out to be harder than detecting primality, since prime numbers are rarer. -(This is a question that came up in a talk I just gave.) - -REPLY [12 votes]: There are situations where we know that a number is a sum of two squares although we do not know the factorization, and in fact know very little about the factors. This question, for example, notes that $F_{2k+1}=F_k^2+F_{k+1}^2$ (the $F_k$ are the Fibonacci numbers). On the flip side, though, while Fibonacci numbers are easier to factor than most, they are not trivial (I think Blair Kelly maintains a list of factors). -To be specific, $F_{1049}$ is a sum of two squares, but has not yet been factored.<|endoftext|> -TITLE: Tensor product of simple representations -QUESTION [7 upvotes]: Let $G$ be a linear algebraic group over some field, and let $V$ and $W$ be two simple rational representations of $G.$ Is $V\otimes W$ semi-simple? -I was trying to convince myself that if $G$ has a faithful semi-simple representation, then $G$ is linearly reductive, and was reduced to the question above. The problem I have in mind is over characteristic 0, but answers addressing char. $p$ is equally appreciated too! - -REPLY [8 votes]: If $G$ is a(ny) group, if $k$ is a field of characteristic 0, and if $V$ and $W$ are semisimple finite dimensional $kG$ modules, then $V \otimes_k W$ is indeed semisimple as a $kG$-module. This is due to Chevalley, and (I think I'm not off-base in saying this) inspired the characteristic $p>0$ result of Serre mentioned in other answers/comments. -The argument goes as follows: it is enough to prove the result after replacing $k$ by an algebraic closure. Now replace $G$ by the Zariski closure of its image in $GL(V) \times GL(W)$ -- this Zariski closure leaves invariant the same subspaces of $V \otimes_k W$ as does $G$, so we may suppose $G$ to be a linear algebraic group over $k$. -Since representations of finite groups in char. 0 are semisimple, a $G$-representation -is semisimple just in case that is true upon restriction to the connected component $G^0$. Thus we may and will suppose $G$ to be connected. -Finally, note that $G$ has a faithful semisimple representation, namely $V \oplus W$. Thus -the unipotent radical of $G$ is trivial so that $G$ is a connected and reductive group over $k$. Now the semisimplicity of $V \otimes W$ follows (every finite dimensional rational representation of $G$ is semisimple).<|endoftext|> -TITLE: Polynomial contact structures on $RP^3$ -QUESTION [13 upvotes]: Let us consider polynomial contact structures on $\mathbb RP^3$, i.e. contact structures on $\mathbb R^3$ defined by a form $w=Pdx+Qdy+Rdz,\ P,Q,R\in \mathbb R[x,y,z]\ $ in an affine part and then extended to $\mathbb RP^3$, and $ w \wedge dw \ne 0$ everywhere. -One can find all such forms $w$ that $deg P, deg Q, deg R \leq 1$ by direct calculation: -$w=(qy-rz+a)dx+ (pz-qx+b)dy + (rx-py+c)dz,\ a,b,c,p,q,r\in \mathbb R;$ $ap+br+cq \ne 0$. -But I can't do anything for greater degrees. Do you know any criteria for coefficients of $P,Q,R$? -Does anybody know any contact polynomial form with $deg P, deg Q, deg R \geq 2$? -Added: What is the form (I mean form coefficients in $\mathbb R^3\subset \mathbb RP^3$) defining the polynomial contact structure constructed by -plurisubharmonic function $f=x^4+y^4+z^4+t^4$? -Answer: $f=x^4+y^4+z^4+t^4$ is not strictly plurisubharmonic (see on the plane $x=y=0$ on subspace generated by $dx,dy$). So it does not produce a contact structure. - -REPLY [3 votes]: There is a very good article "Complex contact threefolds and their contact curves" of Yun-Gang Ye where on can find a classification of complex contact structures on threefolds<|endoftext|> -TITLE: non-locally trivial A^n bundles -QUESTION [13 upvotes]: Let $f: X \to Y$ be a morphism of varieties such that its fibres are isomorphic to $\mathbb{A}^n$. Since the definition of a vector bundle stipulates that $f$ be locally the projection $U \times \mathbb{A}^n \to U$, it is likely that there exist morphisms that are not locally of that form, but I can't come up with an example. -So the question is: what is an example of a morphism with fibres $\mathbb{A}^n$ that is not locally trivial? not locally isotrivial? -UPDATE: what if one assumes vector space structure on the fibres? - -REPLY [17 votes]: In Jack's example the fiber is not scheme-theoretically $\mathbb A^1$. You can get a counterexample by taking $Y$ to be a nodal curve, $Y'$ its the normalization, with one of the two points in the inverse image of the node removed, and $X = Y' \times \mathbb A^1$. -If we assume that the map is smooth, this becomes quite subtle. It is false in positive characteristic. Let $k$ be a field of characteristic $p > 0$. Take $Y = \mathbb A^1 = \mathop{\rm Spec}k[t]$, $Y' = \mathop{\rm Spec} k[t,x]/(x^p - t)$. Of course $Y' \simeq \mathbb A^1$, but the natural map $Y' \to Y$ is an inseparable homemorphism. Now embed $Y'$ in $\mathbb P^1 \times Y$ over $Y$, and take $X$ to be the difference $(\mathbb P^1 \times Y) \smallsetminus Y'$. -On the other hand, it is not so hard to show that in characteristic 0 the answer is positive for $n = 1$ (if $Y$ is reduced), and I believe it is known to be true $n = 2$. The general case seems estremely hard. -I am afraid that Sasha'a argument does not work; if the fiber does not have a vector spaces structure, there is not reason that choosing points gives you a trivialization. -[Edit] The question has been updated with "what if one assumes vector space structure on the fibres?"? -Well, $\mathbb A^n$ can always be given a vector space structure. In my first example, the fibers are canonically isomorphic to $\mathbb A^1$, so they have a natural vector space structure. -However, if the map $X \to Y$ is smooth, and the vector space stucture is allowed to "vary algebraically" that is, if the zero section $Y \to X$ is a regular function, the addition gives a regular function $X \times_Y X \to X$, and scalar multiplication gives a regular function $\mathbb A^1 \times X \to X$, then $X$ is in fact a vector bundle. The proof uses some machinery: one uses smoothness to construct bases locally in the étale topology, showing that $X$ is étale locally trivial over $Y$, and descent theory to show that in fact $X$ is Zariski locally trivial. - -REPLY [2 votes]: Does your definition of varieties allow them to be disconnected? If so, let $Y$ be any variety, $P\in Y$ a closed point, $U$ the complement of $P$, $X_1$ a vector bundle over $U$, $X_2$ a vector bundle over $P$, and $X=X_1\cup X_2$.<|endoftext|> -TITLE: Appropiate models of numerical computation -QUESTION [5 upvotes]: Hello, -in contrast to the more discrete part of computational mathematics (cryptography, combinatorial computation), numerical mathematics seems to ignore typical questions of theoretical computer science -- what does 'algorithm' or 'computation' mean, what is the model of computation. -This is far from fallacious. For example, a finite element theorist mostly investigates only approximation schemes and convergence rates, which in principle do not demand any computation at all. Algorithms are typically neat and short, and any exceptions to these are not rarely just combinatorial insertions for mesh management. The other end of the spectrum compromises very technical numerical mathematics. - In either case, the 'deep-down' part is largely abandoned as soon as possible, because it is largely irrelevant. Just like no cryptographer enjoys talking about Turing machines. -Has there been a rigourous treatment of a numerical model, a justification why (or for whom) a certain model might be appropiate? -I am aware of computable analysis and numerical mathematicians who participate in this field. But I am not aware of a numerical model like, say, a numerical random access machine. I even suppose there different models appropiate for researchers in fundamental numerical algoriths or a FEM reasearcher, depending on which level of detail is needed. - -REPLY [4 votes]: There are mainly three approaches to deal with computational complexity of continuous problems. -1. Information Based Complexity (analytical complexity). It is a very general framework that describes the complexity of a problem in terms of the number of 'operations' (specified by the problem itself) needed to solve it on condition that we are given only rough information about the initial problem. This theory is mostly about the 'real' complexity --- which is independent of any particular model of computation --- it gives lower bounds on all possible models. See 'Information-Based Complexity' by Traub, Wasilowski and Woźniakowski for more details. -2. Blum–Shub–Smale machine and that like (algebraic complexity). The idea comes from the good old days when people believed that it was possible to build analog computers that are more powerful than turing machines. I was to say that the idea of stroing infinite information in a single cell, and comparing two such cells in a finite time is a bit crazy from both practical and theoretical point of view, but I guess it is prudent to refrain from making such comments. These models are inconsistent with physical laws, so it should not be strange that they allow some 'dirty hacks' (for example there are uncomputable problems easily solvable on Blum–Shub–Smale machine; under some definitions, one may show that both $P/Poly$ and $NP$ are solvable in polynomial time). -3. Turing machines (discrete complexity). If you really want to solve a problem on a computer, then you really have to transform it into a discrete one (either symbolic or numeric). But then, there is nothing left but the classic complexity :-) -David, I do agree with your reasoning, and (so) with your conclusion. However, notice that it is always easy to falsify statements such as my remarks. Simply, the reality is so complex, that in every such proclamation there must be much more things that we have to ignore, than we are able to take into consideration. The crucial point here is that BSS allows us to perform dirty tricks and reach something paradoxical, which otherwise we would have not accepted; and that the extra power does not give us anything besides these paradoxes. Every algorithm for BSS either: has its counterpart in the standard model, or is ‘unrealizable’ (we may have an infinite precision, and infinite memory, but when we cut these infinities at any stage, we get a Turing machine). Put it differently, according to our current knowledge every ‘sensible’ ‘realization’ of a BSS has to factor through a Turing machine.<|endoftext|> -TITLE: Is it possible for P(N) to be larger than Aleph_omega? -QUESTION [5 upvotes]: I have seen a proof that $|\mathcal{P}(\mathbb{N})| \neq \aleph_\omega$ using the fact that $\aleph_\omega$ is the union of countably many smaller cardinals, while $|\mathcal{P}(\mathbb{N})|$ is not. Is it consistent with ZFC that $|\mathcal{P}(\mathbb{N})| > \aleph_\omega$? - -REPLY [13 votes]: Yes. In fact, the only requirement on $2^{\aleph_0}$ is that $cf(2^{\aleph_0})>\aleph_0$. (Cohen's argument can make the power set any regular cardinal, and I think it requires at most small modifications to handle the general case.)<|endoftext|> -TITLE: Degree of commutativity of finite groups and subgroups -QUESTION [6 upvotes]: Recently I started reading some articles about the -degree of commutativity of finite groups. I have some questions: - -In "Subgroup commutativity degrees of finite groups" Tarnauceanu -proposes the following formula for calculating the degree of -commutativity of subgroups of a finite group G: - -$$ \mathrm{sd}(G) = \frac{\left| \left\{(H,K) \in \mathscr{L}(G)^2:HK=KH\right\}\right|}{\left|\mathscr{L}(G)\right|^2}, $$ -He proves that if $G_1, G_2, \ldots , G_n$ are finite groups of coprime order, then -$$ \mathrm{sd}\left(\times_{i=1}^{n}G_{i}\right)=\prod_{i=1}^{n}\mathrm{sd}(G_i). $$ -My first question is about what happens if we omit the hypothesis that $G_i$ have coprime orders, that is, if there exists some estimate for -$$\mathrm{sd}\left(\times_{i=1}^{n}G_{i}\right)$$ -in terms of $\mathrm{sd}(G_i)$. - -In "Central Extensions and Commutativity Degree" Lescot proposes -the following formula for calculating the degree of commutativity of a -finite group $G$: - -$$\mathrm{d}(G)=\frac{1}{|G|^2} \left|\left\{(x,y) \in G^2:xy=yx\right\}\right|.$$ -My second question is about the order of the group $G$. Is there -any theory for the case when $G$ is infinite? For example, $G$ might be a -group equipped with a Haar measure. I have found no literature about this -case. -Remark: Sorry for the obscurity of my question, the first question is: For $G_1,\ldots,G_n$ arbitrary groups, is there any estimate for the degree of commutativity of subgroups of $G=direct~product~of~the~groups~ G_i$ in terms of the degree of commutativity of the groups $G_i$? My second question was partially answered. A friend showed me the following article: tmu.ac.ir/salg20/talks/Rezaei.pdf. - -REPLY [3 votes]: I am quite late on this one, but here's a partial answer to your first question. -First, a couple of comments: - -I take the view that the term "commutativity" is unfortunate when applied to subgroups; permutability is more appropriate. -There is really no need to talk about the subgroup lattice of the group $G$. Lattices have structure, but here we only need their size, so set of subgroups, -which you may denote by $\mathrm{s}(G)$ for instance, would be just as good. - -The coprime orders case you mention (Proposition 2.2 of the original article) is, of course, valid. If $\gcd(|A|,|B|)=1$, then every subgroup of $A \times B$ is -a subproduct, i.e., a direct product of subgroups of the two factors. This follows directly from Goursat's lemma; you can have a look at - -D.D. Anderson and V. Camillo, Subgroups of direct products of groups, ideals and subrings of direct products of rings, and Goursat’s lemma, Rings, modules and representations. International conference on rings and things in honor of Carl Faith and Barbara Osofsky, Zanesville, OH, USA, June 15–17, 2007, American Mathematical Society, 2009, pp. 1–12. - -for an excellent account of this useful result (and some generalisations). As far as I can tell, Tarnauceanu does not offer any justification for Proposition 2.2 in his paper, that is why I mention this. - -The following properties are all satisfied by $\mathrm{d}$, but not by $\mathrm{sd}$: - -If $G_1$, $G_2$ are groups then $\mathrm{d}(G_1 \times G_2) = \mathrm{d}(G_1) \cdot \mathrm{d}(G_2)$. -If $N$ is a normal subgroup of $G$ then $\mathrm{d}(G) \leq \mathrm{d}(N) \cdot \mathrm{d}(G/N)$. -If $H$ is a subgroup of $G$ then $\mathrm{d}(H) \geq \mathrm{d}(G)$. - - -For instance, the dicyclic group of order 12 with presentation -$$\mathrm{Dic}_3 = \left\langle a, x : a^{6} = 1, x^2 = a^3, x^{-1}ax = a^{-1}\right\rangle,$$ -has $\mathrm{sd}\left( \mathrm{Dic}_3\right) = 29/32$ and a normal subgroup of order 2 with quotient isomorphic to $\Sigma_3$, -but $\mathrm{sd}\left( \Sigma_3\right) = 5/6$, thus 2. is not true. Then the group of order 16 which is a central product of $D_8$ and $C_4$ over a common cyclic central subgroup of order 2 has subgroup permutability degree equal to $505/529$, but $\mathrm{sd}\left( D_8 \right) = 23/25$, thereby disproving 3. -Finally, $\mathrm{sd}\left( C_2 \times \Sigma_3\right) = \mathrm{sd}\left(D_{12} \right) = 101/128$, but $\mathrm{sd}\left( \Sigma_3\right) = 5/6$, -hence 1. does not hold either.<|endoftext|> -TITLE: Groups, quantum groups and (fill in the blank) -QUESTION [39 upvotes]: In the study of special functions there are three levels of objects, classical, basic and elliptic. These correspond to classical hypergeometric functions, basic (q-) hypergeometric functions, and elliptic hypergeometric functions. -In combinatorics these notions are related to enumeration, q-enumeration and "elliptic enumeration" (see this article of Schlosser). -Now, I always related the passing to q-analogs by analogy to the "way" one passes from groups to quantum groups. And indeed, q-analogs and quantum groups are not entirely unrelated concepts. But this makes me ask the question in the title, whether someone has considered quantum groups at the "elliptic level", and if so what are they? - -REPLY [3 votes]: Browsing through the answers to this old question, I am surprised that nobody mentions the Sklyanin algebra, which is surely the first and most fundamental example of an elliptic quantum group. It was introduced by Sklyanin in 1982, before Drinfeld explained quantum groups to the mathematics community. The Sklyanin algebra is constructed from the R-matrix of the eight-vertex model in much the same way as the standard SL(2) quantum group is constructed from the six-vertex model. In Baxter's solution of the eight-vertex model, a key step is that the R-matrix is equivalent (via a vertex-face transformation) to a dynamical R-matrix. Starting with this dynamical R-matrix leads to a Felder-type elliptic quantum group as mentioned in Adrien's answer. -You mention elliptic hypergeometric functions. They are linked to both Sklyanin- and Felder-type elliptic quantum groups in a completely similar way as classical hypergeometric functions are linked to Lie groups and basic hypergeometric functions to standard (non-elliptic) quantum groups.<|endoftext|> -TITLE: Put as many points as possible in an equilateral triangle of side 1 with their minimal distance greater than 1/n -QUESTION [13 upvotes]: It is known by the pigeon-hole principle that: - -If we select $5$ points within an equilateral triangle with side $1$, there must be at least two whose distance apart is less than or equal to $1/2$. -And if we select $10$ points, there must be at least two whose distance apart is $\leq 1/3$. -Generally, if we select $n^2+1$ points, there must be at least two with distance $\leq 1/n$. - -But $n^2+1$ seems not to be a tight bound. My question is: - -To determine a minimum integer $m(n)$ such that if we select $m(n)$ points within an equilateral triangle with side $1$, there must be at least two points having distance $\leq 1/n$. - -Equivalently, - -To determine the maximum integer $m(n)$ satisfying that there exists a configuration of $m(n)-1$ points within an equilateral triangle with side $1$ such that the minimal distance among these points is greater than $1/n$. - -It is clear that $m(1)=2$ and $m(2)=5$, both matching $n^2+1$. But a roughly pencil-and-paper work shows that $m(3)$ is not $10$ anymore. -Note: points can be located on the three sides of the triangle. - -REPLY [5 votes]: This is related to problems of packing circles into an equilateral triangle, and covering an equilateral triangle by circles. Some data on these problems is available at https://erich-friedman.github.io/packing/cirintri/ and https://erich-friedman.github.io/packing/circovtri/ -Since the Heilbronn triangle problem has been mentioned, that has data at https://erich-friedman.github.io/packing/heiltri/<|endoftext|> -TITLE: Explicitly describable maximal unramified extension of a number field -QUESTION [17 upvotes]: This is an idle question. Is there an example of a number field $K$ for which the maximal everywhere-unramified extension $M|K$ is of infinite degree, and yet the group $\mathrm{Gal}(M|K)$ has been completely determined ? - -REPLY [5 votes]: It seems the question is not so idle after all, for the Fontaine-Mazur conjecture has something to say about it, as I discovered today in Neukirch-Schmidt-Wingberg (the NSW of Cam's answer). To paraphrase their (10.8.13), every representation -$$ -\operatorname{Gal}(M|K)\to -\operatorname{GL}_n(\mathbf{Q}_p) -$$ -of the group of automorphisms of the maximal unramified extension $M$ of a number field $K$ has finite image, even if the degree $[M:K]$ is infinite. This is certainly a very strong restriction.<|endoftext|> -TITLE: How can there be topological 4-manifolds with no differentiable structure? -QUESTION [89 upvotes]: This is a very naive question, and I'm hoping that it will be matched by a correspondingly elementary answer. It is well known that not every topological 4-manifold admits a smooth structure. So what's wrong with the following very sketchy proof that, actually, a topological 4-manifold does admit a smooth structure (apart from the sketchiness)? -Step 1: Embed the manifold into $\mathbb{R}^9$, which, as I understand it, can be done. -Step 2: "Iron out the kinks" in the embedded manifold. -Step 3: Once the embedded manifold looks nice enough, give it an obvious smooth structure coming from $\mathbb{R}^9$. -The second step looks the dodgiest to me, because my intuition comes from cases that are presumably much too special, such as a 2-dimensional manifold sitting in $\mathbb{R}^3$. Take, for instance, the surface of a cube. We can easily smooth off the corners and edges and obtain a smooth manifold. There are many smoothing methods around (such as convolving with nice objects). So why can't we find one that works in general? -When I try to think how I would actually go about it, then I do of course run into difficulties. For instance, in the cube case I could take all points outside the cube of some fixed small distance from the cube. That would give me a smoother version. But if I try a trick like that when the codimension is not 1, then I get a set of the wrong dimension. That suggests that I have to make a clever choice of direction, and I don't see an obvious way of doing that. -I have similar questions about other wacky (as they seem to me) facts about manifolds, such as the existence of topological manifolds that cannot be given piecewise linear triangulations. I'm not looking for an insight into why such results are true. All I want to understand is why they are not obviously false. Can anyone say anything that might be helpful? - -REPLY [26 votes]: I am not sure how relevant it is, but here is something that made me better understand the possible problems when one tries to smooth out an embedded topological manifold. -Take your favorite knot $K$ in $S^3$ (or, if you are minimalist, take your second favorite knot so that it is not trivial), view $S^3$ as an equator in $S^4$, and consider the cone over $K$, based on the south pole of $S^4$ (using geodesics of the round metric, say). You get what is called a non-tame embedding of a disc in $S^4$. Now it is intuitive that if you deform continuously the embedding, then you will deform continuously its trace on a small sphere centered at the south pole (called, by Milnor I think, the link of the embedding at the point), which starts from a copy of $K$. So any close embedding must also be non-tame, while if it where differentiable it would have a trivial knot as its link.<|endoftext|> -TITLE: A Query regarding the Halting Problem (Omega): Halting Probability for Given Input Size -QUESTION [6 upvotes]: I was studying the Halting Problem in context of the Probability and had a few doubts regarding it. Hope someone could help me out. -I am aware of the probability of a Random program halting on a Universal Turing Machine (with a randomly chosen input) is given by Chatin's Constant (which is normal and transcendental) number thus cannot be exactly computed. -My Query is the following: -Suppose one is given a Specific Universal Turing Machine Model (T) for doing all the computations. Also the following facts are given: - -The size of the randomly chosen Program (P) is 'at most' (Pk) bits. -The size of the randomly chosen Input (I) for the Program (P) is 'at most' (Ik) bits. - -**The Program and Input are nothing but two randomly chosen strings. -Now given a random instance of a program and the Input for the Turing Machine (T) [with the size limits for Pk and Ik known]: -a. Is it possible for one to calculate the probability of any random Pair halting on the Turing machine (T) ? -b. If not possible to calculate the exact probability (the value being transcendental) then the first few Most Significant Bits, let us say first Min(Pk, Ik) bits of the Probability i.e. just an aproximation ? -c. Is is possible to comment on a rather modified Probability Statement (and find an aproximation) which is as follows: -Probability(P, I, Lk) : Its defined as the Probability that a random Pair will halt after exactly Lk Steps/Operations on the Turing Machine T ? -d. How do we go about finding it as the Chatin's formula is for an infinite Series? -Hope someone could help me out here. - -REPLY [2 votes]: " -Hi Tarandeep, -The halting probability for a random n-bit program certainly won't be transcendental -- in fact it's rational ( # halting progs / 2^n )! On the other hand, that probability clearly isn't computable given n as input, since if it was then you could use it to solve the halting problem. -Hope that helps, -Scott -" -Thanks to Prof. Scott Aaronson.<|endoftext|> -TITLE: Is it always possible to extend a closed (1, 1)-form on a divisor to a closed (1, 1)-form on a tubular neighbourhood? -QUESTION [7 upvotes]: Let $X$ be a compact Kahler manifold, let $D$ be a smooth divisor in $X$, and let $U$ be a tubular neighbourhood of $D$ in $X$. Suppose that $D$ is Fano. Is it possible to extend every closed (1, 1)-form on $D$ to a closed (1, 1)-form on $U$? - -REPLY [5 votes]: Yes. For the proof, see e.g. http://arxiv.org/abs/math/0609617 -Theorem 4.1. Here a stronger result is actually proven: -Theorem: Let $(M, \omega)$ be a compact Kahler manifold, -and $Z\subset M$ a closed complex submanifold. -Denote by $[\omega]\in H^2(M)$ the Kahler class of $M$. -Consider a Kahler form $\omega_0$ on $Z$ such that -its Kahler class coinsides with the restriction -$[\omega]| Z$. Then there exists a -Kahler form $\omega'$ on $M$ in the same Kahler -class as $\omega$, such that $\omega| -Z=\omega_0$. -An additional cohomological assumption is needed, because we build a global extension, and for an extension to a local neighbourhood you don't need it; the positivity (needed in assumption) is achieved by adding a big multiple of a Kaehler form.<|endoftext|> -TITLE: Why are hypergeometric series important and do they have a geometric or heuristic motivation? -QUESTION [42 upvotes]: Apart from telling that the hypergeometric functions (or series) are the solutions to the (essentially unique?) fuchsian equation on the Riemann sphere with 3 "regular singular points", the wikipedia article doesn't illuminate much about why this kind of special functions should form such a natural topic in mathematics (and in fact have been throughout 19th century). -Simply: - -What are hypergeometric series really, and why they should be (or have been in the past centuries) important/interesting? - -REPLY [13 votes]: Hypergeometric functions arise as matrix coefficients of representations of Lie groups. -This is my formal answer, but I will also describe very informally why I believe this is an answer to your question. (By this I roughly mean: why this might have triggered interest of 19th century mathematicians who didn't have the language to be aware that this is what they were looking at. However: I do no not know a thing about the history of the subject so this is just a mathematical remark from a modern perspective.) -Interesting geometric spaces tend to have a rich group of symmetries (either because that is what makes them interesting (e.g. the circle), or because the symmetry is the only thing that gives us any grip on the object (e.g. spacetime)). Also studying a space through the functions on it has proved to be a powerful way to do mathematics. Hence representations of Lie groups on vector spaces of functions arise naturally. -Now it turns out that the matrix coefficient functions (of smaller representations) really give you a grip on these rather huge representations. The canonical example is when the group is compact and the interesting space on which the functions live is the group itself. In this case the Peter-Weyl theorem states that matrix coefficient functions form a basis. (Which at least makes it kind of credible that they are also useful for understanding functions on other compact homogeneous spaces.) -Now not everything is compact. (This is relevant here as Gauss' hypergeometric function appears as a matrix coefficient of an $SL(2, \mathbb{R})$-representation. (Incidentally this was the answer to prof. Koornwinder's question at my own thesis defense, see my comment to the original post above.)) However, also in this case matrix coefficient functions help you understand general representations on function spaces. For instance, in the infinite dimensional case (for non-compact groups irreducible representations need not be finite dimensional) it is not immediately clear (to say the least) that a given Lie algebra representation integrates to group representation. One way to make this work (given some extra conditions) is to first construct the matrix coefficients of the hypothesized group representation and then construct the actual representation from that (see for instance [1]). To me this always felt a bit like proving the existence of the Yeti from its footprint, but the difference is that it actually works. -Final remark: this is just the first part of JME's answer in disguise since in the setting of a Lie group acting on a space of functions on a homogeneous space (for that Lie-group), the differential equations JME is talking about come from the action of the Lie algebra. -[1] v.d. Ban: Induced representations and the Langlands classification [http://www.staff.science.uu.nl/~ban00101/manus/edinb.pdf]<|endoftext|> -TITLE: Ordinal Analysis of Peano Arithmetic with Restricted Induction -QUESTION [8 upvotes]: If we take Peano Arithmetic and restrict induction to formulas over various fragments of the arithmetic hierarchy, say to the $\Sigma^0_n$ formulas for various $n$ or some other interesting fragments, how does the proof theoretic ordinal for the theory vary? - -REPLY [7 votes]: The proof-theoretic ordinal of $I\Sigma^0_n$ (for $n > 0$) is well-known to be $\omega_{n+1}$, where $\omega_1:=\omega$, $\omega_{n+1}:=\omega^{\omega_n}$. See e.g. Avigad & Sommer. - -REPLY [5 votes]: Emil's answer gives you the ordinals you are asking for, but it may be worth to add a complementary remark: With restricted induction, one needs to be slightly careful about how the ordinals are computed. -For example, the provably recursive functions of $I\Sigma^0_1$ are precisely the primitive recursive functions. -However, these are precisely the $\omega^2$-recursive functions, i.e., those that can be proved total using the infinitary proof-system $\vdash^{\omega^2}_0$ of Tait "Normal derivability in classical logic", in "The syntax and semantics of infinitary languages", Lecture Notes in Mathematics 72, Springer, pp. 204-236. -Given $f:{\mathbb N}\to{\mathbb N}$, let $E(f)$ consist of all those functions "explicitly definable" using 0,1,$f$,$+$, restricted subtraction, and bounded sums and products. -There are several fast-growing hierarchies of recursive functions one uses to analyse fragments of arithmetic. The functions $B_\alpha$ are defined inductively: $B_0(n)=n+1$, $B_{\alpha+1}(n)=B_\alpha(B_\alpha(n))$, $B_\lambda(n)=B_{\lambda_n}(n)$ for $\lambda$ limit, where the $\lambda_n$ are the "natural" strictly increasing sequence of ordinals converging to $\lambda$. -The functions $F_\alpha$ are defined similarly, except that $F_{\alpha+1}(n)=F^{n+1}_\alpha(n)$, where the superindex denotes iterated composition ($n+1$ times). This is the sequence of functions most used in this context. -The functions of the Hardy hierarchy are defined by $H_0(n)=n$, $H_{\alpha+1}(n)=H_\alpha(n+1)$, and $H_\lambda(n)=H_{\lambda_n}(n)$. -Then the primitive recursive functions are precisely the functions in $$\bigcup_{\alpha\prec\omega^2}E(B_\alpha)=\bigcup_{\alpha\prec\omega}E(F_\alpha)=\bigcup_{\alpha\prec\omega^\omega}E(H_\alpha);$$ here, $\prec$ is a partial subordering of the ordinals, but at this level we may identify it with the usual $\lt$. -All this is discussed in great detail in the nice paper by Fairtlough and Wainer, "Hierarchies of provably recursive functions", in "Handbook of Proof Theory", Elsevier, pp. 149-207.<|endoftext|> -TITLE: Maximal Ideals in Formal Laurent Series Rings? -QUESTION [27 upvotes]: Setup: Let $k$ be a field, let $n$ be a positive integer, and let $R := k[[x_1,\ldots,x_n]]$ denote the commutative ring of formal power series over $k$ in $x_1,\ldots,x_n$. We know that there is exactly one maximal ideal for $R$, namely $\langle x_1,\ldots,x_n \rangle$. -By localizing at the multiples of the $x_1,\ldots,x_n$, we can construct a multivariable Laurent series ring $L$. In particular, $L$ is equal to the ring of series of the form $$\sum _{m_1,\ldots,m_n \in \mathbb{Z}} \quad \lambda _{(m_1,\ldots,m_n)} \; x^{m_1}\cdots x^{m_n},$$ for $\lambda_{(m_1,\ldots,m_n)} \in k$, but with $\lambda _{(m_1,\ldots,m_n)} = 0$ when the minimum of the $m_1,\ldots,m_n$ is $\ll 0$. -When $n = 1$ it is well known that $L$ is a field. However, for $n > 1$ the situation is more subtle. For example, when $n=2$, the ideals of the form $\langle x_1 - \mu x_2 \rangle$, for nonzero $\mu \in k$, are maximal ideals. -My general question: Does anyone know of an explicit description of the maximal ideals of $L$, for $n > 1$? -A more refined question: Suppose $k$ is algebraically closed. Consider the Laurent polynomial ring $P := k[x_1^{\pm 1},\ldots,x_n^{\pm 1}]$, and let $H$ denote the multiplicative $n$-torus $(k^\times)^n$. -There is a natural action of $H$ on $P$, obtained via $$(\alpha_1,\ldots,\alpha_n)\cdot x_i = \alpha_ix_i,$$ for $(\alpha_1,\ldots,\alpha_n) \in H$. The action of $H$ on $P$ induces an action of $H$ on the maximal spectrum of $P$, and it follows from Hilbert's Nullstellensatz that there is exactly one $H$-orbit of maximal ideals. -So the refined question is: Consider the analogous action of $H$ on $L$. Are there finitely or infinitely many $H$-orbits of maximal ideals of $L$? -My own motivation: These commutative Laurent series rings show up as central subalgebras of $q$-commutative Laurent series rings (i.e., formal Laurent series where $x_i x_j = q_{ij}x_j x_i$ for suitable scalars $q_{ij}$). In joint work with Linhong Wang, I have been studying $q$-commutative power and Laurent series rings. -Properties of the prime and primitive ideals in these noncommutative algebras are strongly influenced by the behavior of these central subalgebras. -Thank you for your time! Any hints or references greatly appreciated. - -REPLY [6 votes]: I think that the question difficult as illustrated by Hailong's answer. I suspect that it will be hard to even find a nice parameterisation of the $H$-orbits of maximal ideals in your refined question. Certianly, as Hailong implies there will be infinitely many such orbits. -You might find in helpful to consider a valuation theoretic approach. Konstantin Ardakov has done some work on related --- although not quite similar --- questions that I believe he has nearly finished writing up. Maybe he will appear and say more. -Edit: For the record Konstantin's work has now appeared here http://arxiv.org/abs/1108.0371<|endoftext|> -TITLE: Computations in group cohomology -QUESTION [7 upvotes]: Hello, -Given a finitely presentable group $G$, I'm interested in the cup-product from $H^1$ to $H^2$ with real coefficients. I want to know if this is explicitly computable (with a computer) with a presentation of the group. -More precisely, I want a program that takes the generators and relations as entries and returns the dimension of the $H^1$ and a finite generating set of linear relations between the cup-products of every couple of elements in a basis of $H^1$. (I am not really interested in all the $H^2$) -Does this seem possible ? -I precise that I am not really familiar with group cohomology and I ask this question because it is certainly known if such a problem cannot be resolved with an efficient algorithm. -The problem comes in the study of Kähler groups where this cup-product plays an important role. -Thank you. - -REPLY [3 votes]: Using Andy's comment and Theorem 6.1 in this paper you can easily work out a computer program to calculate what you wish from a presentation. You will only need to work with groups of nilpotency class $2$.<|endoftext|> -TITLE: minimal resolution of singularities -QUESTION [7 upvotes]: What is the minimal resolution of singularities of the surface -$S^2(X^3+Y^3+Z^3)-3(S^2+T^2)XYZ=0$ which is a subset of $\mathbb{P}^1\times\mathbb{P}^2$ - Please note that in this equation $[S:T]\in{\mathbb{P^1}}$ and $[X:Y:Z]\in{\mathbb {P^2}}$ and by $\mathbb{P^n}$ we mean n-dimensional complex projective space. - -REPLY [7 votes]: A simple computation shows that the equation -$$ -u(x^3+y^3+z^3)-3vxyz=0 -$$ -defines a non-singular surface $F\subset\mathbb P^1\times \mathbb P^2$. The projection to $\mathbb P^1$ gives an elliptic fibration $\sigma:F\to \mathbb P^1$. This has exactly two singular fibers, over $[0:1]$ and $[1:1]$, each consisting of three lines not going through a common point. The local equation for the projection at the singular points of the fibers is $$(\xi,\eta)\mapsto \zeta=\xi\eta.$$ - -Details (to satisfy popular demand): -a) Near the point $[0:1]\times[0:0:1]$, let $\zeta=\dfrac uv$, $\xi=\left(\dfrac{3vz}{x^3+y^3+z^3}\right)\cdot x$, and $\eta=y$. Notice that $\dfrac{3vz}{x^3+y^3+z^3}$ is a unit near that point. Near the other singular points of the fiber over $[0:1]$ permute the variables accordingly. -b) Near the point $[1:1]\times [1:1:1]$, let $\zeta=\dfrac{v-u}u$, $\xi=\left(\dfrac{x+y+z}{3xyz}\right)\cdot(x+\omega y+\omega^2 z)$, and $\eta=x+\omega y+\omega^2 z$ where $\omega\neq 1$ is a $3$rd root of unity. In particular $1+\omega+\omega^2=0$. Notice that $x^3+y^3+z^3-3xyz=(x+y+z)(x+\omega y+\omega^2 z)(x+\omega y+\omega^2 z)$. Permute the three linear factors accordingly for the other two singular points of the fiber. - -Note: Actually one can conclude the stated local condition without doing this explicit calculation. The point is this: we know that the singular fiber is three lines in the plane intersecting in three separate points. Therefore, locally each of the singularities of the fiber is defined by $\zeta=\xi\eta$. Since the nearby fibers are smooth, the family, locally, is a smoothing of a node. The versal deformation space of a node is one dimensional (it's exactly what the displayed equation claims) and hence this smoothing has to be locally isomorphic to that. - - -Now consider a base change of $\sigma$ by taking square roots $\mathbb P^1\to \mathbb P^1$, $[s:t]\mapsto [s^2:s^2+t^2]$. The new surface $G=F\times_{\mathbb P^1}\mathbb P^1$ is the surface in the question. This will acquire singularities over the points where $\sigma$ was not a smooth morphism. We saw above that the local equation of the map at those points is given by -$$(\xi,\eta)\mapsto \zeta=\xi\eta.$$ -The base change replaces $\zeta$ by $\zeta^2$, so the local equation of the surface becomes $$\zeta^2=\xi\eta.$$ - -Details: -a) near $[0:1]$ we had above $\zeta=\dfrac uv$, so the base change makes it $\zeta=\dfrac{s^2}{s^2+t^2}=\left(\dfrac 1{1+\tau^2}\right)\rho^2$. Replace $\zeta$ with $\rho$ and $\xi$ with $\xi\cdot(1+\tau^2)$. -b) near $[1:1]$ we had $\zeta=\dfrac{v-u}u$, so the base change makes it $\zeta=\dfrac{t^2}{s^2}=\rho^2$. Replace $\zeta$ with $\rho$. - -Note: Again, this can be done without the explicit computation. Any two-to-one map $\mathbb P^1\to \mathbb P^1$ is simply taking roots of the local coordinates defining the points where that map is branched. Therefore if $\zeta$ is the local equation of the branch point, then the cover replaces $\zeta$ with $\zeta^2$. - - -In other words, the surface has exactly $6$ singular points, each locally analytically isomorphic to the vertex of a quadratic cone, and hence blowing up these points (once) yields the minimal resolution. - -Edit history: - 1) Thanks to JME for pointing out the typo in the definition of the base change map. -2) Edit 1: added the local calculation for the description of the map near the singular points. -3) Edit 2: added the theoretical argument (which in my mind actually preceded the calculation) that implies the same result as the calculation.<|endoftext|> -TITLE: Kronecker Approximation theorem and Fibonacci numbers -QUESTION [7 upvotes]: There is a famous old theorem by Kronecker that for every positive real $\alpha$ and $\epsilon>0$ there exists a positive integer n such that $\alpha n$ is within $\epsilon$ of an integer. -Recently I found that the same result is true if we replace $\alpha n$ by $\alpha n^2$ or any polinomial p such that $p(0)=0$. -Could this result be generalised to other functions? Particularly I'm curious about sequences $\alpha 2^n$ and $\alpha F_n$ where by $F_n$ I denote n-th Fibonacci number. -Does anyone know anything about it? - -REPLY [4 votes]: You've received two good answers, but I'll elaborate a bit. -Usually equidistribution on the torus (or more general, compact groups) wrt the Haar measure is achieved by computing the Weyl sums and showing that there are some cancellations. -The question you are referring to is studied in the area of so-called "sparse equidistribution" (although it is more of sparse density if you would like). -The problem with the harmonic-analytic approach is by summing (or integrating) over very sparse part of your period. It is usually not stright forward to bound such exponential sums. -For example, Vinogradov proved for example that $ \{p_{n}x\} $ is equidistributed mod 1 for all irrational x, to bound the Weyl sums, he used sieves with what is called now Vinogradov sums, and a result about the odd Goldbach conjecture. -Now if you are interested in a metric result (i.e. a.e. x), then it is a very classical result that for every increasing unbounded sequence \$ {a_{n}}$ and for a.e. x, one have that ${a_{n}x}$ is equi. mod 1, this is done by taking the Weyl sum, computing its L^2 norm, and then sub. limit and integration by the DCT. -Now the question if such a result follows for every x is very subtle, and not always amenable to harmonic-analytic approach, and the current state of the art actually lies in the ergodic approaches. -If you have a sequence which is contained inside a geometric progression, then there exists x's for which $\{a_{n}x\}$ is not equi. more generally, for $\{q^{n}\}$ say, you can find x's whose orbit closure is with any Hausdorff dimension you want (the reason here that as a dynamical system, this is isomorphic to Bernoulli shift on $q$ letters). -More generally, a result due to Boshernitzan says that if you have a lacunary sequence (the limit of the ratios of consecutive elements is larger than 1), then the Hausdorff dimension of the set of exceptional x's (such that $\{a_{n}x\}$ is not dense/equidistributed) is 1. -On the contrary, Boshernitzan shown that if the sequence is non-lacunary (the ratio tends to 1, you should think about it as having sub-exp. growth), then the Hausdorff dimension of the set for which $\{a_{n}x\}$ is equi. is 1. -There were even some old results due to Erdos from the 1950's about it (he talked about convolution of Bernoulli measures, which can be interpreted in this sense as well). -A very peculiar discovery by Furstenberg (67) shown that if you have a non-lacunary semigroup, say $\{2^{n}3^{m}\}$ then for every irrational x you get that $\{2^{n}3^{m}x\}$ is dense mod 1 (certainly not equidistributed). -This result is very interesting, because you have density for every x. Moreover, recently, Bourgain-Lindenstrauss-P. Michel and Venkatesh proved an effective version of that theorem (meaning that you fix some epsilon, you can estimate how far you need to go in-order to find an element which is epsilon-close to an integer). -An even recent work (by myself, still preprint), generalizing the Bourgain-Lindenstrauss paper, and I shown that sets like $\{2^{n}3^{3^{m}}3^{3^{k^2}}x\}$ are dense for every x. -About Fibonacci sequences, it follows from my work (based on other work of D. Meiri whith Yuval Peres and Elon Lindnstrauss), that you can prove density of sequences such as $\{2^{n}3^{3^{m}}F_{k}x\}$ for every irrational x. -For the general Fibonachi sequences, you are basically in the lacunary case, which Boshernitzan already covered in certain sense.<|endoftext|> -TITLE: Slick Proof of Kudo Transgression Theorem -QUESTION [17 upvotes]: The Kudo Trangression Theorem has to do with the transgression in the Leray-Serre spectral sequence for cohomology in $\mathbb{Z}/p$ ($p$ odd). It can be proved by the method of the universal example, once it is shown that in the path-loop fibration sequence $K(\mathbb{Z}/p,2n) \to P(K(\mathbb{Z}/p,2n+1)) \to K(\mathbb{Z}/p,2n+1)$ - -the fundamental class $v$ of the fiber transgresses to $u$, that of the base -this forces a zig-zag of cancellation, up to $v^{p-1}\mapsto u \otimes v^{p-2}$ -also $v^p$ transgresses to $P^n(u)$, and -$u\otimes v^{p-1}$ "transgresses" to $\beta P^n(u)$. - -Parts (1), (2) and (3) are easy, but part (4) seems difficult. There is a proof along these lines in a paper of Browder from the mid 1960's (he attributes the proof to Milgram), but the proof of (4) is actually quite hard and leans heavily on algebraic mucking around in the spectral sequence. -Does anyone know of a clever way to prove (4)? -Edit: Let's say we know by induction that the cohomology of the fiber is what it has to be. Then I think the behavior of the spectral sequence is forced in dimensions below that of $u\otimes v^{p-1}$. Does this show that $u\otimes v^{p-1}$ "transgresses"? Suppose it does; then its image is $Q(u)$, where $Q$ is a cohomology operation that vanishes when looped (since it is not the transgression of a class in the fiber). Perhaps we can argue that $Q$ must be $\beta P^n$, up to sign? - -REPLY [10 votes]: I didn't think this was all that hard when I wrote -``A general algebraic approach to Steenrod operations'' -which appeared in 1970. It is available on my web page. -The result in question is Theorem 3.4 there. As proved -there, the result applies to quite general situations -and not just cohomology. For example, it applies to -homology spectral sequences of iterated loop spaces, -with Dyer-Lashof operations replacing Steenrod operations. -The proof does use some mucking around with chain level -operations, but that is perhaps easier than trying to do -it without such operations.<|endoftext|> -TITLE: Cops and drunken robbers -QUESTION [17 upvotes]: Consider a game of cops and robbers on a finite graph. The robber, for reasons left to the imagination, moves entirely randomly: at each step, he moves to a randomly chosen neighbour of his current vertex. The cop's job is to catch the robber as quickly as possible: - -How do we find a strategy for the cop which minimizes the expected number of steps before she catches the robber? - -If I'm reading this paper correctly, the minimum expected catch time is finite even if the graph is not cop-win. [Edit: as mentioned in the comments below, the cop and robber will never meet if they move at the same time and are in opposite parts of a bipartite graph. Provided this is not the case, I think the paper's argument can be modified to show that the expected catch time is finite. If the cop and robber take turns, there is no need for an additional assumption and the paper can be used directly. I'd be interested in either setup.] -For example, chasing a robber around the cycle $C_n$ gives an expected catch time of $$\sum_{k=0}^\infty\ k\cdot\frac{\binom{k}{d/2}}{2^k},$$ where $d$ is the initial distance between the cop and the robber. (Since the cycle has so much symmetry, it can be shown that this strategy is the best possible.) -I'm curious as to whether, for instance, the usual optimal strategy in a cop-win graph is optimal in this sense. I'm also interested in some generalizations of this problem (by giving weights to various things). But I don't know whether the basic problem is open, trivial, or somewhere in between, so I'll ask the catch-all question: - -What is known about this problem? - -REPLY [5 votes]: Here are some recent papers on the topic: - -Natasha Komarov, Peter Winkler, Capturing the Drunk Robber on a Graph -Athanasios Kehagias, Paweł Prałat: Some remarks on cops and drunk robbers -Athanasios Kehagias, Dieter Mitsche, Paweł Prałat: Cops and invisible robbers: The cost of drunkenness - -There are also some related slides: here and here (by the OP).<|endoftext|> -TITLE: Parallelizability of the Milnor's exotic spheres in dimension 7 -QUESTION [79 upvotes]: Are the Milnor's seven dimensional exotic spheres parallelizable? - -REPLY [40 votes]: Here's another way to answer the original question. There is a theorem of Bredon and Kosinski (Annals, 1966) which says that if a manifold $M^n$ is stably parallelizable, then either $M^n$ is parallelizable or the maximum number of linearly independent vector fields on $M^n$ is the same as on $S^n$. Since $S^7$ is parallelizable, this implies that exotic 7-spheres are parallelizable (since they are stably parallelizable).<|endoftext|> -TITLE: Can a smooth, immersed loop in R^2 become not nullhomotopic by removing a point? -QUESTION [8 upvotes]: ATT -More precisely, let $\gamma :S^1\rightarrow R^2$ be a smooth immersed loop, the question is whether it is true that there is a point $p\in R^2-\gamma(S^1)$ such that $\gamma$ is not homotopic to constant map. -Actually I'm not sure whether I choose the right tag. Tell me if I choose wrongly. -I hope it won't turn out to be trivial. -(Does the tex turn out all right? I don't seem to have the plug-in to display it.) - -REPLY [3 votes]: For a different way of looking at the same issue see http://en.wikipedia.org/wiki/Pochhammer_contour . The contour is set up to have winding number 0 around any point.<|endoftext|> -TITLE: Polynomial group Laws on $\mathbb{R}^2$ -QUESTION [5 upvotes]: When students are first learning about groups, a classic example of a group that is not defined as a set of functions is the group whose underlying set is $\mathbb{R}\setminus-1$, and whose operation is $x*y=x+y+xy$. This naturally leads one to wonder about what other polynomials in two variables give rise to a group law on $\mathbb{R}$. Is there any nice criteria for such polynomials, or, in the case that there is not, are there any nice classes of polynomials that are group laws? - -REPLY [3 votes]: All the above observations and much much more can be found in the classic paper: -Hinrichs, L.; Niven, I.; Eynden, C.L.V., Fields defined by polynomials, Pac. J. Math. 14, 537-545 (1964). ZBL0122.29103. -Here the authors prove that $f(x,y)$ is a polynomially defined group law on an infinite field $k$ if and only if $f(x,y) = x+y+c$ for some constant $c$ in $k$. They go on to characterize polynomially defined field operations on $R\times R$ as well. -Very useful paper from the point of view of this topic especially at a beginning level. -- R. Padmanabhan, University of Manitoba.<|endoftext|> -TITLE: The whole plethora of topology -QUESTION [16 upvotes]: In his answer to a recent MO question, Johannes Ebert sketches the proof of a very nice result (implying that homotopy spheres are parallelizable) which, as he says, involves the whole plethora of topology of the 1950s: Bott periodicity, Hirzebruch signature formula and Adams's results on the Hopf invariant and the J-homomorphism... -What is the equivalent plethora of topology of the end of the XXth century? - -REPLY [17 votes]: Both John and Charles are right, there is a lot more topology now then there was then. New areas have popped up, older areas have expanded and forked even. I don't know about most of the different areas, but Charles and John haven't offered an answer yet, so here we go. Also, everything below is just what I have gleaned from conversations, personal reading and inference. That being said I will not include the "It seems to me..." etc. but you should certainly feel free to include it in your reading. -The material that Johannes referred to is a lot of hardcore algebraic topology from the 1950's. A lot of this stuff was concerned with proving theorems in geometry that could be attacked using homotopy theory, think vector fields on spheres and Hopf invariant one. These proofs were culminations of an immense amount of work. A lot of very hard computational work went into this. Specifically, a deep understanding of the structure of the Steenrod algebra and its cohomology. The Adams spectral sequence was invented around this time, and it organized a lot of the homotopical work into one frame, but not necessarily the geometry implied by the homotopy theory. -Jumping off from the Adams SS we have Novikov "taking Adams suggestion" and working with $MU$ instead of $H\mathbb{F}_2$ as a base for a Adams SS. I am not familiar with Novikov's work, but after Quillen proved his theorem relating $\pi_*MU$ to the Lazard ring a lot of cool things started happening (this is where number theory comes in!). Quillen also proved that $MU_p$ (localized at p) splits as a wedge of suspensions of $BP$. As everyone knows, you have to do homotopy theory one prime at a time now, so we should be working with the Adams-Novikov SS based on $BP$ (from know on some prime $p$ is fixed). Miller, Ravenel, and Wilson did a lot of hard work on this. I am not sure about the timing, but at some point Morava had some preprints floating around that were a huge influence on the work to come. $E_2$ of this ANSS is really hard to compute, so you need to (like May did) have a spectral sequence converging to it. This is called the Chromatic spectral sequence. Anyway, the upshot is that people started finding families of elements in the homotopy groups of spheres, the greek letter elements for example. -This at some point Ravenel saw the work of Bousfield on localization and wrote a paper on "Localization with respect to certain periodic homology theories." This paper ends with the Ravenel Conjectures which shaped a lot of modern homotopy theory. All but one of the conjectures were solved by Devinatz, Hopkins, and Smith (the unsolved conjecture is called the Telescope conjecture, which people believe to be false nowadays). These were solved in the early 90's. I think most people that understand any of this stuff understand a lot more than I do, so I won't say anymore about that. -After the solution to the Ravenel conjectures (at least historically) people started to make computations in the $K(n)$-local setting ($K(n)$ is like a graded field in homotopy theory, $\pi_*(K(n))=\mathbb{F}_p[v_n,v_n^{-1}]$). These theories are called Morava $K$-theories and they have the effect of isolating type $n$ phenomena (type 0 is what rational homotopy can see, type 1 is what $p$-local $K$ theory can see...). There are also Morava $E$-theories (also called Lubin-Tate theories because of their coefficients and denoted $E_n$ for chromatic level n) and Johnson-Wilson $E$-theories being used (denoted $E(n)$ for chromatic level n), they see more of the chromatic picture, in that $E_n$ should see all phenomena of degree $n$ and lower (I am not really clear on the difference between the two, but it seems fashionable to spend more time talking about the Morava $E$-theories these days, for good reasons that I do not know). It is a theorem that the $K(n)$ (or maybe $E_n$) localizations of a finite CW complex tell you what you want to know about the $p$-localization of a finite CW complex. So the program was initiated to understand the $p$-local sphere by first understand the its localizations with respect to $K(n)$ for each $n$. -The construction of these theories, prior to a good smash product on the category of spectra, were all a little ad hoc. If I have a formal group law over a ring $R$ then I get a map from $\phi: R \to pi_*MU$. Now the Landweber exact functor theorem will tell me when $MU_*(X)\otimes_{\phi} R$ is a cohomology theory. Then Brown representability gives me a spectrum. This is not such a good construction, because the spectrum is coming "out of the vacuum". But along with a good construction of the smash product we got a good construction of spectra related to $MU$. (There is also the Baas-Sullivan construction that is related to manifolds with singularity, but I know even less about this). Using this perspective though, about formal group laws, we can say that $\pi_*E_n$ is the ring that classifies the deformations of the universal height $n$ formal group law over $\mathbb{F}_p$ (this ring comes out of work of Lubin and Tate on formal group laws, hence the alternate name). This height $n$ formal group law has automorphisms and so we get an action of that automorphism group, called the Morava stabilizer group $\mathbb{S}_n$ on $\pi_*E(_n$. Let $\mathbb{G}$ be a maximal finite subgroup of $\mathbb{S}_n$, then one can lift that action to get an action the spectrum $E_n$. Hopkins and Miller were able to show that the the spectra involved were $A_{infty}$. Goerss and Hopkins were then able to improve this to get the spectra to be $E_{\infty}$ (I think this is also related to the action of the Morava stabilizer groups acting by ring maps, but I am no sure). If we take the homotopy fixed points of the above action we get what are called higher real $K$ theories (the name comes from the fact that $KO$ can be gotten from $KU$ by looking at the (homotopy) fixed points of $KU$ under the obvious $C_2$ action). -The above construction gives you some very interesting cohomology theories, but it is hard to understand what geometric implication they might have, and any higher order structure they might have. For example, $BP$ has only recently been shown to be $E_4$ by Basterra and Mandel. Also, Niles Johnson and Justin Noel have recently shown that the natural complex orientation of $BP$ can not be an $E_\infty$ map. Anyway, the point is that a more "geometric" construction is needed here, and this is where TMF and TAF enter the picture. And that is a little sliver of what one might call the modern plethora of topology. -Let me briefly clarify what I mean by geometric construction: it is not that we can relate the spectra above to geometry in the sense of manifolds or vector bundles (there is a construction of "$K$"-theory of two vector bundles that is $v_2$ periodic (chromatic level 2) but I do not think that this theory is complex orientable, or it has some other deficiency that keeps it from being an example of an "elliptic" cohomology theory). We do have the construction directly in the sense that it does not come from an abstract existence result. Tyler is certainly right when he says we do not know (we the community, maybe individuals have yet to post preprints) what representatives of elements of, for example, $TMF^*(X)$ look like. -Also, I feel like I should mention that Davis and Mahowald have a way of getting embedding theorems out of these stable homotopy theory computations. By embedding theorems I mean results concerning when you can embed $\mathbb{R}P^n$'s. -Here are some topics I did not mention: LS Category, Goodwillie Calculus or Functor Calculus in general, Rational Homotopy Theory, Equivariant homotopy theory, Embedding theory, Surgery, Models of the category of spectra with good smash products, THH, TC, and Algebraic K-theory. -Some people I did not make mention of that played a large role (I am sure I am forgetting some): Mark Behrens, Tyler Lawson, Mark Mahowald, Peter Landweber, Mike Hill, Jacob Lurie, Paul Goerss, Mark Hovey, and so many more. -I really do apologize if I have made any aweful errors, please let me know so I can fix them.<|endoftext|> -TITLE: Cup products and the transfer map -QUESTION [5 upvotes]: Let $G_1$ be a finite-index subgroup of $G_2$. Let $i : H^{\ast}(G_2) \rightarrow H^{\ast}(G_1)$ be the induced map of rings. There is then a transfer homomorphism $\tau : H^{\ast}(G_1) \rightarrow H^{\ast}(G_2)$ whose key property is that $\tau(i(x)) = [G_2:G_1] \cdot x$ for all $x \in H^{\ast}(G_2)$. I have two questions. - -If $\tau$ a map of rings? In other words, if $x,y \in H^{\ast}(G_1)$, then must we have $\tau(x \cup y) = \tau(x) \cup \tau(y)$? My guess is that the answer is "no". -Assuming that the answer to the first question is "no", does there exist explicit examples of groups $G_1$ and $G_2$ as above and elements $x_1,\ldots,x_k \in H^1(G_1)$ such that $\tau(x_i)=0$ for all $i$ but $\tau(x_1 \cup \cdots \cup x_k) \neq 0$? - -REPLY [8 votes]: What is true is that $\tau$ is a map of modules; that is, -$$\tau(i^*(x)\cup y) = x\cup \tau(y)$$ -for $x\in H^*(G_2)$ and $y\in H^*(G_1)$. -In particular, the kernel of $\tau$ is a sub-$i^*(H^*(G_2))$-module of $H^*(G_1)$. -For an example, consider $G_1=C_p$ (cyclic group) and $G_2=\Sigma_p$ (symmetric group), where $p$ is an odd prime. The generator $x\in H^2(C_p)$ satisfies $\tau(x)=0$ (since $H^2(\Sigma_p)=0$), but $\tau(x^{p-1})\neq 0$. -Added. As Neil points out, I'm using cohomology with mod $p$ coefficients here.<|endoftext|> -TITLE: Automorphism Group of some Classical groups -QUESTION [5 upvotes]: Hi All, -I would like to know the Automorphism group of some simple classical groups, such as PSL(n,q) or some PSU or PSp groups. Could you please give me some recommended books or papers then? I could not find one. -Thanks in advance. - -REPLY [9 votes]: As Tom points out, the Springer survey by Dieudonne (in French) is a standard source, with lots of references to the primary literature since at least the older work of Schreier and -van der Waerden. Later work by O'Meara and others has mainly been concerned with more general settings over fields and commutative rings, going in the direction of algebraic K-theory. By now the specific groups mentioned in the question have also been treated in other sources, such as the 2009 Springer graduate text GTM 251 *The Finite Simple Groups" by Robert A. Wilson. There are also some relevant MO discussions, if you search for "automorphism group". -The algebraic group viewpoint on automorphism groups was developed by Steinberg for linear groups over finite fields, while I showed how to adapt those methods to infinite fields in a 1969 paper freely available here.<|endoftext|> -TITLE: Good chalk in the UK -QUESTION [11 upvotes]: Sometime ago it was asked in Mathoverflow about good chalk in the US Where to buy premium white chalk in the U.S., like they have at RIMS?. I will be grateful for any recommendations on good chalk in the UK and where to get it. - -REPLY [2 votes]: I was recently pointed to this site for buying Japanese chalk (white and color, Rikagaku brand), as well as other Japanese pens and office supplies. -http://www.jetpens.com/ -They will ship internationally, though the shipping looks rather expensive.<|endoftext|> -TITLE: Why do A_\infty functors form an A_\infty category? -QUESTION [12 upvotes]: I am in a reading group studying Seidel's book (Fukaya Categories and Picard-Lefschetz Theory). All of the participants have backgrounds in symplectic topology/pseudoholomorphic curve methods. We are stuck in trying to understand the chapter presenting the algebraic background for Fukaya Categories. -Seidel makes the following claim: Non-unital $A_\infty$ functors $\mathcal F: \mathcal A \rightarrow \mathcal B$ are themselves the objects of a non-unital $A_\infty$ category. The morphisms $\mathrm{hom}( \mathcal F_0, \mathcal F_1)$ are something he calls (following Fukaya) pre-natural transformations. (The morphisms $T$ for which $\mu_1(T) = 0$ are the natural transformations.) Seidel then provides the formulae for the compositions $\mu_d$. -(This is discussed in Section (1d) of the book [page 10].) -In our working group, we tried to check that these formulae for the compositions satisfied the $A_\infty$ associativity equations, but were unable to do so beyond $\mu_1$. -I have two questions (that may be the same question): - -Why do these composition maps satisfy the $A_\infty$ associativity equations? Is there a way of understanding this geometrically? - -REPLY [7 votes]: I can explain the pictures I usually draw to think of $A_\infty$ functors, -but I don't know if they're standard. Anyway, I'll describe what is -just a rubric for ingesting the long formulas, nothing more. -Let's consider first the Yoneda embedding $Y$, which re-thinks an object $L$ -in an $A_\infty$-category $A$ as an $A$-module, or functor from $A^{op}$ to -chain complexes. So $Y_L(M) = hom_A(M,L).$ -I confess that when I confront these formulas/concepts, I always think -in terms of the Fukaya category, which is very amenable to pictures -and for which the $A_\infty$ structures are geometric. -So I draw a curve on a piece of paper and label it $L$. (The curve -is literally a Lagrangian submanifold of my ${\mathbb R}^2$ piece of -paper.) When I want -to think of $L$ in terms of its Yoneda image, I draw the SAME curve, but as a -squiggly line. -So what is the data that the squiggly line gives us? For each object $M$ -(a regular curve on my paper), we have the intersection points, which form a -graded vector space $hom_A^*(M,L).$ This vector space has the -structure of a chain complex (Floer), with differential given by -football-shaped bi-gons with one regular side and one squiggly side. -For a pair of other objects, $M_1, M_2,$ we get a map -$$\mu^2: hom_A(M_2,L)\otimes hom_A(M_1,M_2) \rightarrow hom_A(M_1,L),$$ -and so on for all the structure of a module (section 1j, p. 19). -For the Fukaya category, -the equations 1.19 follow (for non-squiggly lines) from studying degenerations -of 1-parameter families of holomorphic polygons. Now squigglifying those same pictures -gives 1.19 for an arbitrary module, and the equations are similar for -not just modules but arbitrary functor between two $A_\infty$-categories. -What data do we have if we have two squiggly lines $L_1$ and $L_2$? -They should intersect at a morphism between functors (and it should have a degree). -This morphism of functores gives more data, using the Fukaya perspective. -If we added one normal line $M$, we'd have the spaces $Y_{L_1}(M)$ and $Y_{_2}(M)$, -and have a triangle which is a map between them. Higher polygons and the -relations between them (by considering one-parameter families) should -give you all the equations and give you a hint as to verify them. -(But no promises!) -Hope that lengthy and pretty vague description was worth our time. -(Oh, geez, this was a March 11 question? Probably stale by now!)<|endoftext|> -TITLE: Are nontrivial integer solutions known for $x^3+y^3+z^3=3$? -QUESTION [33 upvotes]: The Diophantine equation -$$x^3+y^3+z^3=3$$ -has four easy integer solutions: $(1,1,1)$ and the three permutations of $(4,4,-5)$. Elsenhans and Jahnel wrote in 2007 that these were all the solutions known at that time. - -Are any other solutions known? - -By a conjecture of Tyszka, it would follow that if this equation had finitely many roots, then each component of a solution tuple would be at most $2^{2^{12}/3} \lt 2^{1365.34}$ in absolute value. (To see this, it is enough to express the equation using a Diophantine system in 13 variables in the form considered by Tyszka.) This leaves a large gap, since Elsenhans and Jahnel only considered solutions with components up to $10^{14} \approx 2^{46.5}$ in absolute value. It is also not obvious whether Tyszka's conjecture is true. -OEIS sequence A173515 refers to equations of the form $x^3+y^3=z^3-n$, for $n$ a positive integer, as "Fermat near-misses". Infinite families of solutions are known for $n=\pm 1$, including one constructed by Ramanujan from generating functions (see Rowland's survey). - -Andreas-Stephan Elsenhans and Jörg Jahnel, New sums of three cubes, Math. Comp. 78 (2009), 1227–1230. DOI: 10.1090/S0025-5718-08-02168-6. (preprint) -Apoloniusz Tyszka, A conjecture on integer arithmetic, Newsletter of the European Mathematical Society (75), March 2010, 56–57. (issue) -Eric S. Rowland, Known Families of Integer Solutions of $x^3+y^3+z^3=n$, 2005. (manuscript) - -REPLY [9 votes]: The solution given in the comment of Y. Zhao: -$569936821221962380720^3 + (-569936821113563493509)^3 + (-472715493453327032)^3 = 3$. -Note that each of these numbers is larger than the limit of $10^{14}$ at which Elsenhans and Jahnel stopped their search in 2007. The smallest has 18 digits, while the other two have 21.<|endoftext|> -TITLE: Leibnizian calculus textbook -QUESTION [11 upvotes]: Where can I find a calculus textbook that emphasizes differentials? -Is there such a book that I could realistically require my calculus students to use? -I want a textbook that supports me when I tell my students something like: -$\Delta((x^2+1)^5)\approx5(x^2+1)^4\Delta(x^2+1)\approx5(x^2+1)^4(2x\Delta x)$ -$d((x^2+1)^5)=5(x^2+1)^4d(x^2+1)=5(x^2+1)^4(2x\ dx)$ -Or: -$\Sigma_{k=1}^n 3x_k^2\Delta x_k\approx\Sigma_{k=1}^n\Delta(x_k^3)=x_n^3-x_1^3$ -$\int_{x=0}^{x=4}3x^2\ dx=\int_{x=0}^{x=4}d(x^3)=4^3-0^3=64$ -Perhaps I could write this book someday, but it'd be a lot easier for me -if my students and I could just buy and/or download a book that takes -this approach without neglecting to provide a cornucopia of exercises, examples, and applications similar to what's available in today's most popular calculus textbooks. - -REPLY [4 votes]: I have written a textbook called Intuitive Infinitesimal Calculus, which teaches infinitesimal calculus the classical, informal way, informed by my Ph.D. research on the history of the Leibnizian calculus.<|endoftext|> -TITLE: Why is "P = NP implies EXP has circuit of $2^n/n^" interesting? [Soft, Philosophical] -QUESTION [7 upvotes]: For example, "P=NP implies PH=P" is interesting ... because most of us don't believe PH=P, so it provides strong evidence P != NP. -On other hand, "P=NP implies EXP has circuit of $2^n/n$ size" seems fairly weak for the following reasons: -(1) we know that there are circuits of size $2^n/n$ -(2) we know that most random functions have circuits of size $2^n/n$ -(3) EXPTIME is really really big -so what's the big deal if EXP has a circuit of size $2^n/n$? [ In fact, I would not surprised if EXP has a circuit of size $2^n/n$ even if P != NP ]. -Question: What am I missing here? Am I misunderstanding the size of EXP? Is there evidence to suggest EXP mostly has small circuits? Is there something brilliant in the proof technique? -Thanks! -EDIT: Marked as community wik due to "no right answer." - -REPLY [13 votes]: Everybody believes that EXP contains problems of exponential circuit complexity, but we are very far from proving it, and in fact we know that any proof of such a result cannot be a relativizing argument, cannot be an ''algebraizing'' argument in the sense of Aaronson and Wigderson, and cannot be a ``natural proof'' in the sense of Razborov and Rudich. So it is a very challenging open question, and an important one, because if we are every going to prove that NP contains problems of exponential circuit complexity, we first have to be able to at least prove it for EXP. -The only circuit lower bound technique that can hopefully overcome the relativization, algebraization and natural proof barriers is Ryan Williams idea of using satisfiability algorithms that are faster than brute-force search. The fact that P=NP implies exponential circuit lower bounds for EXP is a toy version of Williams's approach (which requires much weaker conditions than P=NP to work, and in fact it requires conditions that are quite possibly true). -Another way to look at this result is that, if one could prove that $P\neq NP$ implies that EXP has problems of superpolynomial circuit complexity, then we would have, unconditionally, then EXP has problems of superpolynomial circuit complexity. (Such two-parts proof, in which we use different arguments depending on the answer to an open question, exist, for example there is Kannan's proof that $NP^{NP}$ has problems of circuit complexity at least $n^k$ for every k. The proof has two parts depending on whether or not SAT can be solved by circuits of size $O(n^k)$.)<|endoftext|> -TITLE: Erdos distance problem n=12 -QUESTION [15 upvotes]: The recent paper On the Erdos distinct distance problem in the plane -Authors: Larry Guth, Nets Hawk Katz prodded me to get a non-trivial example. Here is what I cannot find: an example of 12 distinct points in the plane with only 5 different distances between points. The regular 12-polygon has 6 different lengths but I cannot do better. http://oeis.org/A186704 implies that there is one> - -REPLY [5 votes]: Thought it might be nice to show the set placed on the hexagonal lattice. - -Here’s a Desmos link if you wish to play around with such point configurations.<|endoftext|> -TITLE: When are complex polynomial maps almost surjective? -QUESTION [13 upvotes]: Consider a complex polynomial map $f: \mathbb{C}^n \rightarrow \mathbb{C}^n$. -For $n = 1$, the fundamental theorem of algebra says that, for any $y \in \mathbb{C}$ there exists $x \in \mathbb{C}$ such that $y = f(x)$. Thus for $n = 1$, $f(\mathbb{C}) = \mathbb{C}$ if and only if $f$ is a non-constant polynomial. Can we generalize this statement to $n > 1$? - -For arbitrary $n$, what is a necessary and sufficient condition to say that closure$\left(f(\mathbb{C}^n)\right)$ $ = \mathbb{C}^n$? - -Suppose the polynomials $f_1,f_2,\cdots,f_n$ are algebraically dependent, i.e., there exists an annihilating polynomial $F$ such that $F(f_1,f_2,\cdots,f_n) = 0$, then the image $f(\mathbb{C}^n)$ is a subset of the affine variety $V(F)$ of dimension $n-1$. Hence a necessary condition is that the polynomials $f_1,f_2,\cdots,f_n$ must be algebraically independent. Can we show that it a sufficient condition as well? - -REPLY [22 votes]: Being algebraically independent is indeed a necessary and sufficient condition for the image of $f$ to be dense. -As $f\colon\mathbb{C}^n\to\mathbb{C}^n$ is regular, its image is constructible and, in particular, contains a non-empty open subset of its closure (under the Zariski topology). See Theorem 10.2 of J.S. Milne's algebraic geometry notes. If the Zariski closure of $f(\mathbb{C}^n)$ is all of $\mathbb{C}^n$, then $f(\mathbb{C}^n)$ contains a Zariski open set and is dense in the standard topology. Otherwise, the Zariski closure of $f(\mathbb{C}^n)$ is the zero set of a nontrivial ideal $I\subset\mathbb{C}[X_1,\ldots,X_n]$ and you can take any $F\in I\setminus\{0\}$ to see that $f_i$ are algebraically dependent. In fact, this argument works for regular maps $f\colon V\to\mathbb{C}^n$ for any variety $V$. There is no need to restrict the domain to be $\mathbb{C}^n$.<|endoftext|> -TITLE: Possibility of an Elementary Differential Geometry Course -QUESTION [47 upvotes]: I have to admit I'm not sure if this is an appropriate question. It's related to research in math education, but not directly to math. -I've found that in talking to professional physicists and engineers, most of them find some use for differential geometry nowadays. One theoretical physicist went as far as to say you could "do nothing serious without it." Yet at most schools (at least the few I've looked at) differential geometry is reserved for graduate students in math and advanced math undergraduates. No schools I looked at had an elementary differential geometry class in, say, a similar style as the calculus sequence. Some of the people I talked to also expressed a lot of difficulty in learning it for the first time on their own. I myself am taking an advanced graduate course in General Relativity, and a good portion of the difficulty of the students is in misuderstanding the fundamental concepts of differential geometry. -To cover differential geometry rigorously, of course one needs quite a bit of advanced mathematics, including topology and analysis. But universities teach elementary calculus classes, most of which are not terribly rigorous, but are sufficient for the purposes of non-mathematicians. Linear algebra, multivariate calculus, and a bit of differential equations would (in my mind) be sufficient to teach a course for engineers. You might argue that one needs to know the theory of manifolds first, but I see this as analagous to studying calculus without really knowing the structure of $\mathbb{R}$. -From my viewpoint, differential geometry is the logical extension of calculus. Based on it's huge (and growing) impact on applied disciplines, It seems logical to have a course in it for engineers and physicists, which I would put immediately after the final semester of calculus (assuming the students have also had linear algebra). -So my question is this: Are there specific instances, either textbooks or courses at a university, of differential geometry classes taught with the intent of being useful for engineers and scientists, which assume only basic calculus knowledge and linear algebra? (Obviously, there are books like "Differential Geometry for Physicists," but I really mean something that would be used by mathematicians teaching such a course). If so, how successful have these courses/books been? If not, or if the attempts have been unsuccessful, is there any particular reason as to why it is not feasable/common? - -REPLY [3 votes]: From my experience (Italian university) the situation regarding physics and engeneering is quite different. -In physics elementary differential geometry is partly taught inside what can be called Calculus II (mainly computations of length of curves and area of surfaces) and partly inside more advanced math courses (typical denomination: Math Methods for physics) which are, however, usually centered towards manifold thoery and the like, no need to explain why. So it is not exceptional to meet someone that knows the definition of Riemannian curvature tensor but does not know what the torsion of a space curve is. Textbooks are usualy those already mentioned in the discussion. -In engeneering there was almost nothing of this kind until few years ago. Recently some course denominated "elementary differential geometry for..." appeared (dots can be: industrial design, mechanical eng. etc.). Non standard textbooks here, however. Gray "Modern Differential geometry of curves and surfaces" was quite considered for a while due to its extensive use of Mathematica, I personally got lot of inspiration from Galliers' "Geometrich methods and applications for computer science and engeneering" Springer, whih contains a full chapter on Elementary differential geometry.<|endoftext|> -TITLE: Cut Locus in a Graph -QUESTION [6 upvotes]: I am wondering if the concept of a cut locus has been defined and explored in discrete graphs, rather than their usual home on manifolds? -The Wikipedia definition (which I believe I (co-?)authored) is: - -The cut locus of $S$ is the closure of the set of all points $p\in X$ that have two or more distinct shortest paths in $X$ from $S$ to $p$. - -For my application, $S$ is a single vertex $x$ of a graph $G$, -and path length is measured by the number of edges in a path. -One possible defintion is: - -The cut locus $C(x)$ of a vertex $x$ in a graph $G$ is (a) the set of - all the vertices $v$ that have two or more distinct paths from $x$, - unioned with (b) all pairs of vertices $(u,v)$—and the edge between - them—such - that $u$ and $v$ have distinct shortest paths from $x$ of the same - length, and $(u,v)$ is an edge of $G$. - -This definition is a bit cumbersome, but I want to capture both -even (a) and odd (b) cycles. -Here is an example, with the even-cycle vertices one color, the odd-cycle edges another: - -I see two possible interpretations of the phrase "distinct shortest paths": - - Two paths are distinct if they are not identical. - Two paths are distinct if they are disjoint, except for the start and end vertices. - -The figure above uses the first definition, whereas the second definition would remove -the two *-ed vertices from the cut locus (because the paths are not identical; rather -they share interior vertices and/or edges and so they are not disjoint). -Again, my main question is: Has this this or similar notions been -studied? I am hoping to find theorems in the literature of the form: - -If $G$ satisfies properties $\{ ... \}$, then $C(x)$ satisfies - properties $\{ ... \}$. - -For example, under what conditions on $G$ is the cut locus a forest -(i.e., devoid of cycles)? -I know this is a fishing expedition, but: Thanks for any pointers or ideas! - -REPLY [2 votes]: Hopefully, I didn't completely misread the definitions. So here goes. -If we use definition 1. of distinct, (b) could be simplified to "all edges between vertices at the same distance from x" (and the vertices these edge induce). Indeed, since the endpoints are different, the paths are distinct in the sense 1. -To determine if $C(x)$ contains cycles, we do not need to look at (a) (it only contains vertices, but no edges). Thus, with definition 1., it boils down to determining if the graph induced by vertices at distance $i$ from $x$ contains a cycle for each $i$. -This seems make the question slightly "less interesting" since we can add a universal vertex $x$ to any graph $G$ (to make $G$ the cut locus of $x$). We could also subdivide the edge incident to $x$ in $G+x$ to make $G$ appear in the cut locus at higher distances. -In fact, for definition 1., any question about the cut locus involving edges would only involve the graph induced by vertices at distance $i$ from $x$ for each value of $i$ (i.e., the $i$th neighbourhood of $x$). -Actually, even for definition 2., we only need to look at (b) for any question involving edges of $C(x)$. -As for definition 2., as Gjergi pointed out, the only vertices that can be in the cut locus are those in the same 2-connected component (block) as $x$. If $x$ is a cut vertex (and thus in multiple blocks), we get a different component of $C(x)$ for each block so we may want to consider them separately anyway. Again, we get different components for vertices at each distance in $C(x)$, so we may as well consider them separately. Let $G_i(x)$ be the graph induced by vertices at distance $i$ from $x$ in $C(x)$. -Then $G_1(x)$ is the graph induced by all vertices in the block of $x$ at distance 1 from $x$. -$G_2(x)$ is the graph induced by all vertices of $G-E(G_1(x))$ in the block of $x$ at distance 2 from $x$. -$G_3(x)$ is the graph induced by all vertices of $G-E(G_1(x))-E(G_2(x))$ in the block of $x$ at distance 3 from $x$. -and so on.<|endoftext|> -TITLE: How to solve the linearized Navier-Stokes equations in L^P? -QUESTION [6 upvotes]: Let $\Omega\subset \mathbb{R}^3$ be an open set with smooth boundary $\partial \Omega$. -Consider the following linearized Navier-Stokes equations in $Q_T=\Omega\times (0,T)$ for an arbitrarily fixed $T\in (0,\infty)$, -$$ -u_t-\Delta u+a(x,t)u+b\cdot \nabla u+\nabla p=f(x,t),\text{div } u=0 -$$ -with the initial and boundary conditions $u(x,0)=0, \left.u(x,t)\right|_{\partial \Omega\times (0,T)}=0$. Here $u(x,t)=(u^1(x,t),u^2(x,t),u^3(x,t))$ and $p(x,t)$ denote the unknown velocity and pressure respectively, $a(x,t)$ and $b(x,t)$ denote the given coefficients. -Question: Suppose that $$a\in L^r(0,T; L^s(\Omega)), b\in L^{r_1}(0,T; L^{s_1}(\Omega)),$$ where $2/r+3/s<2$, $2/r_1+3/s_1<1$, -and $f(x,t)\in C_0^\infty(\Omega\times (0,T))$, can we solve the above equations in arbitrary $L^p$? Can we get the estimates such as - $$\|u_t\|_{L^p(Q_T)}+\|D^2 u\|_{L^p(Q_T)}+\|u\|_{L^p(Q_T)}\leq \|f\|_{L^p(Q_T)}?$$ -Solonnikov dealed with this problem in his paper "Estimates for solution of nonstationary Navier-Stokes equations" (http://www.springerlink.com/index/N8374858XNT22P11.pdf). -However, I can not verify his proof (Page 487 to Page 489). -Who can help me? Any comment will be deeply appreciated. - -REPLY [2 votes]: Y. Giga, H. Sohr, Abstract L^{p} estimates for the Cauchy problem with applications to the Navier-Stokes equations in exterior domains, J. Funct. Anal., 102 (1991), 72-94. -Also: The Navier-Stokes equations. An elementary analytic approach. Hermann Sohr.<|endoftext|> -TITLE: About the Serre-Tate theorem -QUESTION [8 upvotes]: It is somehow a general principle that the (infinitesimal) local behavior of a representable moduli functor $X$ at some point $x$ is closely related to the deformation problem of the structure parameterized by $x$: the deformation functor should be pro-represented by the formal completion of $X$ at $x$ ($X$ representable as assumed). In particular, if $x$ is a smooth geometric point, then the associated deformation functor is pro-represented by a ring formal power series of $d$ variables, $d=\dim_xX$. -What does one find when linking this principle with the Serre-Tate theorem? The theorem affirms that the deformation functor of an ordinary abelian variety of dimension $g$ over an algebraically closed field of char.$p$ is pro-represented by a formal torus of dimension $g^2$. does this functor corresponds to the formal completion of some muduli functor of abelian scheme? One cannot expect the Siegel moduli (with level structures) of genus $g>1$ to work as the latter is of rel.dimension $g(g+1)/2$. Does the dimension jump to $g^2$ because one is looking "purely locally" through the $p$-divisible group forgetting the polarization? -On the other hand, let $\mathcal{M}$ be the moduli functor of principally polarized $g$-dimensional abelian scheme with $\Gamma_0(p)$ level structure, $p$ being a rational prime (say large enough). It admits a coarse moduli scheme $M$, and according to a theorem of Chai and Norman ("Bad reduction of Siegel moduli scheme of genus two ...") the formal completion of $M$ at a geometric point $x$ corresponding to an ordinary abelian variety of char.$p$ is a formal torus of dimension $g(g+1)/2$, which agrees with the principle mentioned in the beginning. - -REPLY [7 votes]: Concerning the first question, the answer is yes. -Given an abelian variety $A$ over $k$ of genus $g$, we have that the first cohomology of its tangent space is of dimension $g^2$. -Moreover, by a theorem of Grothendieck, we know that the deformations of $A$ are unobstructed. This yields that the base ring of the universal deformation is isomorphic to -$k[t_1, \dots, t_{g^2}]$. -Using the theory of p-divisible groups, one can show that in the case where $A$ is ordinary, the formal scheme representing the deformation functor can be equipped with the structure of a formal group scheme. This corresponds to a more canonical choice of parameters compared to the above one. In any case, the dimensions are the same. -In this deformation-theoretic argument, polarizations are not involved. If you add them as additional data to the moduli functor, dimensions can change.<|endoftext|> -TITLE: Group ring and left zero divisor -QUESTION [12 upvotes]: Let $K$ be a finite field and $G$ be a discrete group. - -Is it true that for every $a,b\in K[G]$ the condition $ab=0$ implies $ba=0$? - -It does not seem to be related to zero divisor problem, any ideas if this can be true and for which fields? - -REPLY [2 votes]: If $G$ is torsion-free then the question of reversibility of $K[G]$ (that is, does $ab = 0$ imply $ba = 0$) is in fact equivalent to the zero divisor conjecture, for any field $K$. -Connell showed that such a $K[G]$ is prime, so given non-zero $a, b \in K[G]$ there exists $c \in K[G]$ such that $bca \neq 0$. If it were the case that $ab = 0$ then we have $a (bc) = 0$ but $(bc) a \neq 0$ and $K[G]$ is not reversible. This trick is how you show that the unit conjecture implies the zero divisor conjecture. -(In the other direction, if a ring has no zero divisors then reversibility holds trivially.) -Connell, I. G., On the group ring, Can. J. Math. 15, 650-685 (1963). ZBL0121.03502.<|endoftext|> -TITLE: Historical Articles about zeta functions of curves -QUESTION [11 upvotes]: Are there any historical articles about the origins of zeta functions of curves over global fields (undoubtedly starting with $\mathbb{Q}$)? In particular who (and when did this happen) first have the idea of creating such a thing? Was it by analogy with the zeta functions of number fields? - -REPLY [3 votes]: This does not adress the actual question, but might be useful, if -somebody wants to trace back the literature as aluded to in another answer. -Peter Roquette has writteen a series of articles entitled -The Riemann hypothesis in characteristic p, its origin and development -several parts, total length close to 200 pages. -Bibliographic details and pdf are avalaible here: -http://www.rzuser.uni-heidelberg.de/~ci3/manu.html -(see papers 26,27,36). -In particular, in later parts various of Hasse's paper related to this are discussed.<|endoftext|> -TITLE: invariant polynomials on 3 by 3 matrices -QUESTION [12 upvotes]: Hi there: -A freshman level question here. A polynomial p on the entries of n by n matrices is said to be invariant if p(A)=p(sAs^{-1}) for every invertible matrix s. For example, for 3 by 3 matrices, tr(A^2) and (trA)^2 are two linearly independent invariant polynomials of degree 2. But are there other degree 2 invariant polynomials on 3 by 3 matrices? Any reference is very appreciated. -Ron - -REPLY [13 votes]: I guess my comment is worth expanding into an answer. Over an arbitrary field $k$, an invariant polynomial on $\mathcal{M}_n(k)$ extends to an invariant polynomial on $\mathcal{M}_n(\bar{k})$. Since the diagonalizable matrices are Zariski dense in $\mathcal{M}_n(\bar{k})$, such a polynomial is determined by what it does to diagonal matrices, and it must be a symmetric polynomial of the entries of any diagonal matrix. Conversely, over an arbitrary field (in fact over an arbitrary commutative ring $R$) the elementary symmetric polynomials (all of which are coefficients of the characteristic polynomial, hence all of which really do come from invariant polynomials) generate the ring of symmetric polynomials in $n$ variables. -Hence the invariants of degree $d$ are precisely the symmetric polynomials of degree $d$ over $k$. A basis of the symmetric polynomials of degree $2$ is always given by $\{ e_1^2, e_2 \}$ where $e_i$ is the $i^{th}$ elementary symmetric polynomial. When $\text{char}(k) \neq 2$ we can instead use $\{ p_1^2, p_2 \}$ where $p_i$ is the $i^{th}$ power sum, since $p_1 = e_1$ and $p_2 = \frac{e_1^2 - e_2}{2}$, but if $\text{char}(k) = 2$ this change of coordinates is not well-defined. -Wikipedia should have proofs of the statements I made above about symmetric polynomials; alternately, see for example Chapter 7 of Stanley's Enumerative Combinatorics Vol. II. Note that on the one hand this result constrains what the possible characters of a "nice" representation of $\text{GL}_n$ can look like, and on the other hand suggests that the characters of the "nice" irreducible representations of $\text{GL}_n$ form a distinguished basis for the symmetric polynomials. These are precisely the Schur polynomials.<|endoftext|> -TITLE: Parts of Set Theory immune to independence -QUESTION [14 upvotes]: The motivation for asking this question is a passage (3.2) in an article by Greg Hjorth where he said that "...it is also an attractive feature of the theory of Borel cardinalities and of the theory of $L(\mathbb{R})$ cardinality that these are largely immunized against independence." -My question is: What other parts of Set Theory are so immunised? Assuming something extra is okay(say Determinacy), but whatever it is, should settle 'most' of the questions. -The wording above is not too clear, but I'm not sure how I can make a stronger statement(suggestions about this would be great). I guess that if you ask high-level enough questions(definability hierarchy wise), independence will come in sooner or later. However, I'm not really asking this from the point of view of absoluteness, but from the point of view of what large class of questions can be settled by what small set of tools. -A bonus question is: Why is the part about Borel cardinalities true? I guess this might have some simple absoluteness explanation, but the 'largely' tells me there is more to it than I think. -Thanks in advance. -P.S. I'm not sure if I've tagged this appropriately, perhaps a 'soft-question' or 'big-list' one would be okay (although I wonder how big the list would be!). I hope people will retag it if they find it suitable. - -REPLY [3 votes]: Some of Shelah's work has shown that parts of set theory that seemed to be dominated by independence results (for example, cardinal arithmetic) actually have a lot of nontrivial structure that is provable in ZFC. I don't know enough about the area to say any more, but I can point to Shelah's survey paper "Cardinal arithmetic for skeptics."<|endoftext|> -TITLE: Is it possible to improve on Siegel's theorem for exceptional zeroes? -QUESTION [7 upvotes]: Let $\chi$ be a real nonprincipal character modulo $q$. Siegel's theorem on exceptional zeroes states that for any $\epsilon >0$ there exists a positive number $C(\epsilon)$ such that, for any real zero $\beta$ of $L(s,\chi)$, -$\displaystyle\beta \leqslant 1-\frac{C(\epsilon)}{q^\epsilon}.$ -This is a superior estimate to anything that can currently be said about exceptional zeroes (though it is ineffective, in the sense that the method of proof makes it impossible to assign a numerical value to $C(\epsilon)$ for given $\epsilon$). Regardless, it is still very useful in applications; for example to put a bound on $\psi(x,\chi)$ for primitive $\chi$, which one needs to prove the Bombieri-Vinogradov theorem. Improving Siegel's estimate may even pave a way to a proof of the (weak) Elliott-Halberstam conjecture, if a few other theoretical leaps are made. -Question: What is the next logical step (if any) after this theorem -- is it true that the only way we can move forward is to disprove the existence of Siegel zeroes? -Also, what is a 'reasonable' conjecture to make? Is there any way to improve on Siegel's estimate? - -REPLY [4 votes]: As Matt's comment indicates, this is a very broad question. One possible 'way to move forward' is considered in the paper of Sarnak and Zaharescu titled Some remarks on Landau-Siegel zeros, Duke Math. J. v. 111 (2002), pp. 495–507. They show the surprising (to me) result that if the only zeros off the critical line are real (ie allowing Landau-Siegel zeros) one can still get good lower bounds on the class number. -A general survey of the problem of the Siegel zero, discussing the Sarnak-Zarahescu result above and many other interesting developments, can be found in Iwaniec's Conversations on the Exceptional Character -http://www.springerlink.com/content/j662217v66667l10/<|endoftext|> -TITLE: A question about an application of Molien's formula to find the generators and relations of an invariant ring -QUESTION [9 upvotes]: In the very beginning of the book "Introduction to Invariants and Moduli" Shigeru Mukai -proves Molien's formula for the Hilbert series of the invariant ring of a finite group action on $\mathbb C^n$. For example, in the case of the standard action of Quaternions on $\mathbb C^2$ the Hilbert series is $\frac{1-t^{12}}{(1-t^4)^2(1-t^6)}$. -After this Mukai explains that this formula hints us that the ring of invariants can be generated by two elements of order $4$ and one element of order $6$, and indeed such elements can be found: -$A=x^4+y^4$, $B=x^2y^2$, and $C=xy(x^4-y^4)$. Then one finds a relation -$C^2=A^2B -4B^3$ and this gives a complete description of the ring of invariants. -Moreover the same procedure is shown to work in several other cases (e.g. binary icosahedral group). -My question is a follows: Is there some theorem that say that this heuristics works often? Namely, if we have an action of a finite group $G$ on $\mathbb C^n$, in order to describe the ring of invariants, we first look on the denominator of the Hilbert series (given by Molien's formula) and try to associate an invariant polynomial of degree $n$ to each factor $(1-t^n)$ (so that this gives us a full set of generators). Or at least, in practice, is this the first thing that one tries to do? - -REPLY [9 votes]: As pointed out by Richard Stanley, there are some subtleties in finding the invariant generators. -The Molien series only tells you the Hilbert series of the ring of invariants up to reduced rational forms. So some information is lost, as illustrated in Richard's answer. -One can gain some information by looking at the actual series and a bit of work. For example, if your Molien series looks like: -$$1 + 2t^2 + 4t^4 +...$$ -Then you know there are two invariant forms of degrees $2$. They generate three elements in degree $4$ ($f^2,fg,g^2$) so there must be another algebraically independent form in degree $4$, etc... -Once you know the degree of the generators, a decent way to guess them is to use the Reynolds operator (I assume the group is finite of order prime to characteristic of the field): -$$f \mapsto 1/|G|\sum_{g\in G} gf$$ -Starting with any polynomial $f$, this always lands in the invariant subring, as you can easily check. -Once you know the first few generators, try to match the Hilbert series of the subalgebra generated by those with your Hilbert series. If they don't match, there must be another generator of degree equals to the first point the series mismatched, and on you go... -ADDED: I originally tried not to mention the word "Cohen-Macaulay", but it is particularly relevant here, so perhaps I should. -The Reynolds operator mentioned above give you a splitting $R^G \to R= \mathbb C[x_1,\cdots, x_n]$. Such splitting implies that $R^G$ is Cohen-Macaulay. -Why is that relevant? The point is that once you come up with $n$ invariant forms, say $f_1,\cdots f_n$ of degrees $d_1,\cdots d_n$, which you happen to know is a regular sequence on $R$ (hence also regular on $R^G$). Then as $R^G$ is Cohen-Macaulay, it is free over $R' = \mathbb C[f_1, \cdots, f_n]$. Assuming you can write -$$R^G = \oplus_1^m h_jR'$$ -with each $h_j$ of degree $e_j$, then the Hilbert series of $R^G$ must be -$$\frac{\sum t^{e_j}}{\prod(1-t^{d_i})} $$ -So by comparing with the Molien series you know the degree of each $h_j$. (See this survey by Richard Stanley!)<|endoftext|> -TITLE: Does a degeneration always have a larger-dimensional automorphism group? -QUESTION [14 upvotes]: Suppose $\newcommand{\X}{\mathcal{X}}\X$ is an algebraic stack over a field $k$, $\xi$ is a $k$-point which has another $k$-point $x$ in its closure ($x$ is an isotrivial degeneration of $\xi$). Must the dimension of $Aut(x)$ be larger than the dimension of $Aut(\xi)$? - -If $\X=[X/G]$, where $G$ is an algebraic group over $k$, then the answer is yes. An isotrivial degeneration corresponds to a $G$-orbit with another $G$-orbit in its closure. The closure of a $G$-orbit can only contain smaller dimensional orbits, and the dimension of the orbit is complementary to the dimension of the stabilizer of a point in that orbit, so when one orbit degenerates to another, there is always a jump in the dimension of the stabilizer. -Almost every algebraic stack I can think of is étale-locally of the form $[X/G]$, and étale maps preserve the dimensions of automorphism groups. This suggests that the answer is probably "yes." - -Example -Consider the 1-parameter family of $2\times 2$ matrices $\begin{pmatrix}1&t\\ 0&1\end{pmatrix}$. If we are studying matrices up to conjugation, then away from $t=0$, the family is (isomorphic to) the constant family $\begin{pmatrix}1&1\\ 0&1\end{pmatrix}$. However, at $t=0$ you get a different Jordan type. So we say that this family is an isotrivial degeneration of Jordan types $\begin{pmatrix}1&1\\ 0&1\end{pmatrix}\rightsquigarrow \begin{pmatrix}1&0\\ 0&1\end{pmatrix}$. When you pass to the more degenerate Jordan type, the automorphism group (i.e. the group of ways in which the matrix is self-conjugate) jumps from something 2-dimensional to something 4-dimensional. - -Motivation -It sometimes happens that you want to determine all degeneration relations among a collection of points in an algebraic stack. For example, you may be trying to determine if some map is weakly proper.† If the answer to this question is "yes," then you can rule out certain degenerations by looking at dimensions of automorphism groups. -†See Weakly proper moduli stacks of curves by Alper, Smyth, and van der Wyck. - -REPLY [5 votes]: This is not a complete answer but a suggestion towards a "yes" answer, too long for a comment. -You can represent $\xi$ and $x$ by a morphism $\varphi: S\to\mathcal{X}$, where $S$ is a normal affine $k$-scheme of finite type with a point $s\in S(k)$, such that $\varphi(s)=x$ and $\varphi_{\mid U}$ factors through $\xi$, where $U:=S\setminus\{s\}$. On $S\times S$, consider the two objects $\varphi_1$ and $\varphi_2$ of $\mathcal{X}$ obtained by pullback via the projections. Put $Y:=\underline{\mathrm{Isom}}_{\mathcal{X}}(\varphi_1,\varphi_2)$. This is an algebraic space of finite type over $S\times S$, and a pseudo-torsor under the group we are interested in. The assumptions on $x$ and $\xi$ imply that the image of $f:Y\to S\times S$ is $(U\times U)\cup\{(s,s)\}$. Hence $f$ is not open, and therefore not equidimensional by EGA IV (14.4.4) since $S\times S$ is geometrically unibranch. Hopefully, this implies the result on the group but we have to be careful; for instance our group may have constant dimension but extra components at $x$, which would account for the lack of equidimensionality at some points of the fibre at $(s,s)$ (but not all?).<|endoftext|> -TITLE: A-infinity structure of E-infinity algebras -QUESTION [5 upvotes]: This is perhaps somewhat related to this question. Fix a field $k$ of characteristic $p>0$. Suppose that $A$ is an $E_\infty$-algebra over $k$. Then $A$ also has an $A_\infty$-algebra structure, and therefore so does its homology $HA$. Its homology is also graded commutative. -I'm looking for an extended version of graded commutativity, a result like this: if $m_n$ is any of the higher multiplications in the $A_\infty$ structure on $HA$, then for any $1 \leq j \leq n$, -$$ -x m_n(a_1 \otimes \cdots \otimes a_n) = \pm m_n(a_1 \otimes \cdots \otimes x a_j \otimes \cdots \otimes a_n). -$$ -This is certainly true if $n=2$ by graded commutativity. What about for larger values? -(I'm tempted to tag any question about $E_\infty$-algebras as "commutative algebra", but I suppose that would be misleading...) -Edit: as Fernando points out in his comment, this is too much to expect in general. The $A_\infty$-algebra structure on $HA$ is not unique, so perhaps the right question is, are there conditions on $x$ and the $a_i$ so that, for some choice of $m_n$, $x m_n(\dots) = \dots$? -Along with Fernando's example, another one to consider is the mod $p$ cohomology of a cyclic group of order $p$, with $p$ odd. If $x$ is the generator of $H^1$ and $y$ is the generator of $H^2$, then $m_p(x^{\otimes p}) = \pm y$, so $x m_p(x^{\otimes p}) = \pm xy \neq 0$ while I think $m_p(x^2 \otimes x^{\otimes p-1}) = 0$, since $x^2=0$. - -REPLY [2 votes]: as Fernando points out in his comment, - this is too much to expect in general. - The A∞-algebra structure on HA is not - unique - -There is a unique $A_\infty$-algebra structure on $H(A)$ such that -$H(A)$ and $A$ are weakly equivalent (i.e. $A_\infty$-quasi-isomorphic). -About your question, in characteristic 0 you could replace your $E_\infty$-algebra by a weakly equivalent DG commutative algebra... then on its cohomology you would get a $C_\infty$-structure (which is a strictly commutative $A_\infty$-structure). -In positive characteristic this is more difficult. But there is still the possibility to deal with divided power algebras (see e.g. http://math.univ-lille1.fr/~fresse/PartitionHomology.pdf on page 18 for a hint). -I know this is not really an answer to your question. But I hope to can help.<|endoftext|> -TITLE: Is there an interpretation of the "anticommutative" symmetric monoidal structure on $\mathbb{Z}$-graded abelian groups in terms of $\mathbb{G}_m$ actions? -QUESTION [6 upvotes]: The category of $\mathbb{Z}$-graded abelian groups is equivalent to the category of comodules over the commutative Hopf algebra (over $\mathbb{Z}$) $A=\mathbb{Z}[t,t^{-1}]$, with comultiplication $t\mapsto t\otimes t$ and counit $t\mapsto 1$. Explicitly, the correspondence is given as follows: given an $A$-comodule $M$ with structure map $\psi:M\to A\otimes M$, let $M_n=\{x\in M:\psi(x)=t^n\otimes x\}$. Then the coassociativity and counitality of $\psi$ can be easily checked to imply that $M$ is actually the direct sum of these $M_n$, and it can similarly be checked that a map $M\to M'$ of two such comodules commutes with the coaction of $A$ iff it maps $M_n$ to $M_n'$ for all $n$. The Hopf algebra $A$ is the ring of functions on the algebraic group $\mathbb{G}_m$, and this equivalence of categories is closely related to the correspondence between graded algebras and projective varieties (through the fact that projective varieties are quotients of certain quasiaffine varieties under an action of $\mathbb{G}_m$). -The commutativity of $A$ gives a natural symmetric monoidal structure on the category of $A$-comodules, given by $M\otimes_{\mathbb{Z}}M'$ with coaction $M\otimes M'\to(A\otimes M)\otimes(A\otimes M')\cong(A\otimes A)\otimes(M\otimes M')\to A\otimes M\otimes M'$, where the first map is given by the coactions on $M$ and $M'$ and the last map is given by the multiplication on $A$. This tensor product corresponds to the standard tensor product of graded modules, in which $(M\otimes M')_n=\bigoplus_{i+j=n}M_i\otimes M_j'$. -In this symmetric monoidal structure (whose existence only depends on the fact that $A$ is a commutative Hopf algebra), the swap isomorphism $M\otimes M'\to M'\otimes M$ is the standard swap map coming from the underlying symmetric monoidal tensor product of $\mathbb{Z}$-modules. This corresponds to the "even" symmetric monoidal structure on graded objects, in which the swap map comes from the standard swap map $M_i\otimes M_j'\to M_j'\otimes M_i$. Commutative monoid objects under this structure are what an algebraic geometer might call graded commutative rings. However, there is a second symmetric monoidal structure on graded abelian groups with the same tensor product but for which the swap map $M_i\otimes M_j'\to M_j'\otimes M_i$ is multiplied by a sign $(-1)^{ij}$. In this symmetric monoidal structure, a commutative monoid object is a "skew-commutative graded ring", or what a topologist would just call a graded commutative ring: even degree elements commute with everything, and odd degree elements anticommute with each other. -My question is: is there a natural interpretation of this second symmetric monoidal structure when we view graded abelian groups as $A$-comodules? Morally, I would say it can't come naturally just from the Hopf algebra structure of $A$, because we could reinterpret $A$ as the ring $\mathbb{Z}[s^2,s^{-2}]$ so that everything ought to be treated as having even grading, and then we get the original symmetric monoidal structure (this is like reinterpreting a $\mathbb{Z}$-graded ojbect as actually being $2\mathbb{Z}$-graded). Thus, I suppose I'm really asking: is there some additional structure you can put on $A$ that makes the "anticommutative" symmetric monoidal structure pop out naturally (and have a purely categorical description)? - -REPLY [5 votes]: As Qiaochu points out, you can include a little extra data in your Hopf algebra $\mathbb G_m$, and use it to realize the braiding. Here's the best way to say this, since you've decided to work with comodules over $\mathbb Z[t,t^{-1}]$. The point is that the comultiplication on $\mathbb Z[t,t^{-1}]$ makes the space of linear maps $\hom(\mathbb Z[t,t^{-1}],\mathbb Z)$ into an algebra under "convolution", and also the space of linear maps $\hom(\mathbb Z[t,t^{-1}]^{\otimes 2},\mathbb Z)$; also, $\hom(\mathbb Z[t,t^{-1}]^{\otimes 2},A)$ is a bimodule for this algebra for any abelian group $A$. A triangular structure is an element $\beta \in \hom(\mathbb Z[t,t^{-1}]^{\otimes 2},\mathbb Z)$ which squares to identity under convolution, and conjugates the multiplication to its opposite (which is, in this case, the same coproduct), and also should satisfy some other conditions. In your case, the map $\beta$ on a basis is $\beta(t^n,t^m) = (-1)^{nm}$, which should look familiar. (A quasitriangular structure is the same except that $\beta$ is required to be invertible but not square-$1$.) -The funny thing, of course, is why you would have to put this in by hand. Since $\mathbb Z[t,t^{-1}]$ is commutative, there is a canonical choice of $\beta$, the identity element in $\hom(\mathbb Z[t,t^{-1}]^{\otimes 2},\mathbb Z)$ (the tensor square of the counit). But you could have picked some other category (say, super-modules), and then built a commutative Hopf algebra where the canonical triangular structure was this one. -Note that $\beta$ is not an algebra homomorphism, so it's hard to see it from the geometric side. This is not surprising. The corresponding notion for groups and sets would be: let $G$ be a group; then a triangular structure for $G$ is a map of sets $1 \to G^{\times 2}$ that commutes with the (unique!) comultiplication $G \to G^{\times 2}$; such a map is just an ordered pair $(a,b)$ of central elements of $G$; but then the other conditions come in, which say that $(a,b,b) = (a^2,b,b)$ and something similar on the other side, and also that $a,b$ are invertible (self-inverse if you want triangular rather than just _quasitriangular), and clearly this means that $(a,b) = (1,1)$. So a group admits only one quasitriangular structure --- its canonical one. From the category-theory side, all I said was that the cartesian product (in $G$-sets), thought of as a monoidal structure, is braided in a unique way. -In the physics world, one does imagine that $\mathbb G_m$ acts on one's graded vector spaces; for a physicists, an action is the same as a conserved quantitiy, and there are many formulas that include explicit mention of this one, called "ghost number" in most situations. The action is a funny one — physicists work over $\mathbb C$ with its unitary structure, and this action is antiunitary rather than unitary. I don't know of a similar discussion in TQFT (of course, supermanifolds are very geometric).<|endoftext|> -TITLE: A partial converse to Bertrand's Postulate -QUESTION [8 upvotes]: Sloane's A077463 obviously suggests that for any positive integer $n$ there exist $n$ consecutive primes and only them in between $m$ and $2m$ for some natural number $m$. -For instance, for -$n=1$, take $m=2$; $\hspace{.2in}$$2<3<4$; -$n=2$, take $m=7$; $\hspace{.2in}$$7<11,13<14$; -$n=3$, take $m=9$; $\hspace{.2in}$$9<11,13,17<18$; -$n=4$, take $m=15$; $\hspace{.1in}$$15<17,19,23,29<30$,$\hspace{.1in}$ etc. -This problem offers a partial converse(given the number of primes, one seeks an exact interval $[m,2m]$) to Bertrand's Postulate(given an interval, one seeks at least one prime in it). -I would like to know - -whether this problem is solved, or -whether there are stronger known conjectures to which it is a consequence. - -Thanks, as always - -REPLY [18 votes]: Consider the function $f(m):=\pi(2m)-\pi(m)$ which counts the number of primes $m< p \leq 2m$. It is easy to see that $f(m+1)-f(m)$ equals $-1$ or $0$ or $1$ depending whether $m+1$ and $2m+1$ are primes. On the other hand, by simple estimates for prime numbers, it can be seen that $f(m)$ tends to infinity. Therefore $f(m)$ takes all positive integer values. -EDIT: Of course the same holds for $\pi(2m)-\pi(m-1)$ which counts the primes in $[m,2m]$.<|endoftext|> -TITLE: complex fourier coefficients, introduced by? -QUESTION [16 upvotes]: I remember reading somewhere that the complex Fourier coefficients were introduced by an engineer sometime around 1900, but I can't find anymore this information. -Does anyone know the name of this person and where I can find a reference to it? -EDIT: I state the question more clearly: "Who was it that first wrote a Fourier series not as a sum of sines and cosines but as a sums of complex exponentials, with the relative formula for the coefficients?". I may be totally wrong about this all, since I don't remember well and that's why I'm asking. Also don't take the 1900 thing seriously, I may be off by 50+ years. - -REPLY [3 votes]: As another data point, my copy of Whitaker and Watson A course of Modern Analysis, 4th edition from 1927 uses the trig versions for basically everything, but Example 1, $$f(z) = \sin(z) - 1/2 \sin 2z + 1/3 \sin(3x) -\ldots$$ is immediately converted to $$f(z) = \frac{1}{2}i(e^{iz}-1/2e^{2iz}+\ldots)+\frac{1}{2}i(e^{-iz}-1/2e^{2iz}+\ldots)$$ The references given "for a fuller account of investigations subsequent to Riemann" are Hobson's Functions of a Real Variable and de la Vallée Poussin's Cours d'Analyse Infinitésimale.<|endoftext|> -TITLE: Partitions to different parts not exceeding $n$ -QUESTION [25 upvotes]: Consider the polynomial $(1+x)(1+x^2)\dots (1+x^n)=1+x+\dots+x^{n(n+1)/2}$, which enumerates subj. How to prove that it's coefficients increase up to $x^{n(n+1)/4}$ (and hence decrease after this)? Or maybe this is false? -This problem was proposed long ago on some Russian high school competition, but nobody managed to solve, including jury. - -REPLY [34 votes]: In fact, you cannot prove unimodality of the coefficients of $P_n(x)=(1+x)\cdots (1+x^n)$ using the result about $B_m/G$ mentioned by Qiaochu. The $P_n(x)$ result is implicit in work of Dynkin on the principal sl(2) subalgebra of a complex semisimple Lie algebra. Hughes was the first to realize that a special case of Dynkin's result implied the unimodality of $P_n(x)$. Proctor removed all the Lie algebra theory from the proof, yielding a proof involving only elementary linear algebra. See http://math.mit.edu/~rstan/pubs/pubfiles/72.pdf, esp. equation (23). See also http://math.mit.edu/~rstan/algcomb/algcomb.pdf, beginning on page 90. It is a long-standing open problem to find a combinatorial proof of this unimodality result. I would be extremely impressed if a high school student could prove unimodality by any method.<|endoftext|> -TITLE: Hodge theory on complex spaces -QUESTION [9 upvotes]: If $X$ is a compact Kahler manifold, then Hodge theory says that its cohomology decomposes as a direct sum -$$ H^{p+q}(X,\mathbb C) = \bigoplus_{p,q} H^{p,q}(X,\mathbb C) $$ -where $H^{p,q}(X,\mathbb C) = H^q(X,\Omega_X^p)$ are the Dolbeault cohomology groups and $H^{p,q}$ and $H^{q,p}$ are conjugate isomorphic. One can prove that the same holds for a Fujiki manifold, i.e. a complex manifold bimeromorphic to a Kahler manifold. -Q: What happens for compact complex spaces? -Here "complex space" should mean singular and non-reduced. Is there a notion of a "Kahler" complex space, on which some predictions of Hodge theory hold? The question should make sense, as we can define $H^{p+q}(X,\mathbb C)$ topologically and the $H^q(X,\Omega_X^p)$ exist for a complex space $X$, but do we even know that the latter are subgroups of the former for some special spaces $X$? -One can imagine naively defining such spaces as those which admit a Fujiki desingularisation (Kahler to begin with, but desingularisations are not unique so we need to compare them somehow), and then hoping that the "upstairs" decomposition induces one "downstairs". Of course, if it were that simple someone would have done it ages ago. - -REPLY [14 votes]: Let me add to Donu's mentioning Du Bois's Hodge decomposition. First of all, many feel that part of the credit is due to Deligne as Du Bois built heavily on his ideas. Then again that is probably true for many things in Hodge theory. -Anyway, Du Bois's main idea was that one can do Deligne's construction "one step earlier" in the sense that Deligne used simplicial resolutions to build his Hodge structure on the cohomology groups, and Du Bois does this in the derived category of coherent sheaves (with some conditions...) so he obtains a filtered complex (now mostly called the Deligne-Du Bois complex) which is quasi-isomorphic to the constant sheaf $\mathbb C$ and whose associated graded quotients (these are actual quotients in the category of complexes, before one passes to the derived category) are the objects $\underline{\Omega}_X^{p}$ Donu mentions. -It turns out that it follows directly from the construction that there is a Hodge-to-de-Rham (a.k.a. Frölicher) spectral sequence and from Deligne's work that it degenerates at the $E_1$ level. So, a lot of things actually work out the same way as in the smooth case if one uses $\underline{\Omega}_X^{p}$ in place of ${\Omega}_X^{p}$. For instance, the Kodaira-Akizuki-Nakano vanishing theorem also holds (as well as the existence of the Gysin map, etc) that can be used to prove a singular version of the Lefschetz hyperplane theorem cf. here. -As Donu mentioned, it is not known in general when the natural map ${\Omega}_X^{p}\to\underline{\Omega}_X^{p}$ is an isomorphism, except for $p=0$ which is the definition of Du Bois singularities. I think in general (that is, for arbitrary $p$) it is true that for toroidal singularities this is an isomorphism, or at least there is a good description of each object and the map between them. - We have a pretty good understanding of Du Bois singularities, although not complete and there are still many interesting open questions. Rational singularities are Du Bois by this, log canonical singularities are Du Bois by this. For an intro to Du Bois singularities and related stuff you can try this. -There is also an intriguing connection to singularities defined via the action of Frobenius in positive characteristic. For more on this see this and other works of Karl Schwede. -Regarding the question on having something that reflects the non-reduced structure, I -am afraid that the topological $H^i$ do not see the non-reduced structure, so I don't think you can expect any reasonable Hodge theory that remembers that.<|endoftext|> -TITLE: Is a space with no covering spaces simply connected? -QUESTION [10 upvotes]: Suppose $X$ is a path connected space such that every connected covering space of $X$ is trivial (1-fold.) Must $X$ be simply connected? -Intuitively, the answer seems to be no (imagine taking a disk, cutting out a square, and gluing in $T\times T$ where $T$ is the topologist's sine curve.) But this is a rather weak intuition. - -REPLY [14 votes]: No, the harmonic archipelago (An illustration is on pg 7 of W. A. Bogley and A. J. Sieradski, Universal path spaces, preprint) is a locally path connected subspace of $\mathbb{R}^{3}$ and has uncountable fundamental group but every connected cover is trivial. -This is part of a more general phenomenon. Let $\pi_{1}^{top}(X)$ be the fundamental group of a space $X$ with the quotient topology of the loop space $\Omega(X)$ with the compact-open topology (sometimes called the "topological fundamental group"). This is a quasitopological group (in that inversion is continuous and multiplication is continuous in each variable) but is not always a topological group. If $p:X\rightarrow Y$ is a covering map the induced homomorphism $p_{\ast}:\pi_{1}^{top}(X)\rightarrow \pi_{1}^{top}(Y)$ is an open embedding of quasitopological groups (i.e. $\pi_{1}^{top}(X)$ embeds as an open subgroup). One consequence of this is that if $\pi_{1}^{top}(X)$ has the indiscrete topology, then either $\pi_{1}(X)=1$ or every connected covering of $X$ is trivial. There are lots of examples of spaces where this occurs other than the harmonic archipelago.<|endoftext|> -TITLE: Number of n-th roots of unity over finite fields -QUESTION [5 upvotes]: How many $n$-th roots of unity does a finite field $\mathbb{GF}\left(p^k\right)$ have? ($p$ is prime). And then kind of related to that, is there a finite field with exactly $n_1,\ldots,n_N$ $n_1$-th,...$n_N$-th roots of unity? - -REPLY [9 votes]: I can quickly answer your first question. The multiplicative group of $\mathbb{GF}(p^k)$ is cyclic, let $g$ be a generator. For an element $x$ of the group $x^n=1$ holds iff $x=g^m$ with $nm$ divisible by $p^k-1$. The latter is equivalent to $m$ divisible by $(p^k-1)/d$, where $d:=\gcd(n,p^k-1)$, hence the $n$-th roots of unity form the subgroup generated by $g^{(p^k-1)/d}$. This subgroup clearly has $d$ elements, so the number of $n$-th roots equals $\gcd(n,p^k-1)$. -EDIT: I did not see KConrad's comment (I was typing slowly). -EDIT: The second question is also easy. It is clearly necessary that the $n_i$'s be coprime with $p$. If this condition holds, then there is a $k>0$ such that $p^k-1$ is divisible by each of the $n_i$'s, and such a $k$ will do by the above (while no other $k$ will do).<|endoftext|> -TITLE: Tamagawa numbers of abelian varieties and torsion. -QUESTION [5 upvotes]: Let $A$ be an abelian variety defined over a number field $K$. Fix a prime $v \subset \mathcal{O}_K$, with underlying rational prime $p$. What relationship, known or conjectural (if any), should there be between the local Tamagawa number $c_v(A) = [A(K_v): A_0(K_v)]$ and the cardinality of the $p$-primary subgroup $A(K_v)(p)$ of $A(K_v)$? -I am aware of the recent work of Lorenzini, which -considers possible cancellations of the ratio \begin{align*} -\frac{\prod_{v \subset \mathcal{O}_K} c_v(A)}{\vert A(K)_{\operatorname{tors}}\vert}, -\end{align*} and shows for instance this ratio for $A$ an elliptic curve defined -over $K ={\bf{Q}}$ is always greater than or equal to $\frac{1}{5}$. -The question comes up naturally in a certain Euler characteristic computations (e.g. for the -$p^{\infty}$-Selmer group of $A$ defined over some profinite Galois extension $K_{\infty}$ of $K$, where certain primes of $K$ are known to split completely, and hence where the torsion subgroup $A(K_{\infty})_{\operatorname{tors}}$ is known to be finite). In particular, granted the refined conjecture of Birch and Swinnerton-Dyer, it is apparent from these computations that $c_v(A)$ for a prime of bad (multiplicative) reduction $v$ is given essentially by the quotient \begin{align*}\frac{\vert H^1(G_w, A(K_{\infty, w}))(p)\vert }{\vert H^2(G_w, A(K_{\infty, w}))(p)\vert }. \end{align*} Here, $K_{\infty, w}$ is the union of all completions at primes above $v$ in $K_{\infty}$, and $G_w$ denotes the Galois group $\operatorname{Gal}(K_{\infty, w}/K_v)$. -If $A$ has good ordinary reduction at all primes above $p$ in $K$, then it is possible to use the Coates-Greenberg theory of deeply ramified extensions to characterize the local factor at $v$ in the Euler characteristic formula coming from this quotient as $\vert \widetilde{A}(\kappa_v)(p)\vert^2$, where $\widetilde{A}$ is the reduction of $A$ mod $v$, and $\kappa_v$ the residue field at $v$ (which is consistent with B+S-D). But, this characterization seems to break down when $A$ does not have good ordinary reduction at $v$, and I have not found any calculations in the literature for this case of bad reduction. Hence why I ask about any possible known relation to torsion. Sorry if the question is silly! - -REPLY [4 votes]: There is no general relation between the local $p$-primary torsion and the Tamagawa numbers. I believe one can have $p$-torsion points that map to non-trivial or to the trivial element in the group of components of the Neron model. -This should indicate you that, in your Euler characteristic formula, you can not hope to replace the square of the local $p$-primary part with just the Tamagawa numbers when the reduction is not good ordinary. -If I understand correctly what you want to do is to compute the cokernel of the universal norms on the local points at a place above $p$. I would start by looking at papers about the case of a $\mathbb{Z}_p$-extension. If the reduction is multiplicative, then the computations were carried out by John W. Jones in Compositio 73 (1990). He has also a paper about the additive case. If your reduction is good but not ordinary, then it is a completely different story since the cokernel is not finite. I am sure there are many other references<|endoftext|> -TITLE: Questions from Chern-Simons theory -QUESTION [13 upvotes]: I am currently reading an article about TFTs (DW - Group Cohomology and TFTs), and I have -a few questions: -(1) Let $M$ be a 3-man., then we know there exists some 4-man. $B$ such that $\partial B = M$. Now, let $E$ be a $G$-bundle over $M$, when can we extend $E$ to a $G$-bundle over $B$ and a connection $A$ over $E$ to a connection $A'$ over $E'$? -(2) If we can construct such an $E'$ over $B$ then we can re-define the CS action as -$CS(A') = \frac{k}{8\pi^2}\int_B Tr(F'\wedge F')$, where $F'$ is the curvature of $A'$. -According to the paper, if $k$ is an integer then $CS(A')$ is independent, mod 1, of the -choice of $B$ and of the extension $E$ and $A$. How do you see this? -(3) Later on they give a better def of the action, $CS = \frac{1}{n}\left(\int_B \frac{k}{8\pi^2}Tr(F'\wedge F') - \langle \gamma^*\omega,B\rangle\right)$, where $\gamma : B -\rightarrow BG$ and $\omega$ is an integer-valued cocycle. They then go on to say that -if $B$ is closed then $\int_B \frac{k}{8\pi^2}Tr(F'\wedge F') = \langle \gamma^*\omega,B\rangle$ and so $CS$ is independent of $B$ and the way we have continued the bundle and the connection. Why is this so? - -REPLY [8 votes]: This is not an answer to your questions, but rather a way to avoid them. Additionally, the assumption that the group $G$ is connected or simply-connected becomes unnecessary. -For a general (compact) Lie group $G$, the level $k$ is not an integer but a class $k \in H^4(BG,\mathbb{Z})$. (If the group is simple, connected and simply-connected, then $H^4(BG,\mathbb{Z}) = \mathbb{Z}$; this reproduces the integer.) -Associated to the class $k$ and the connection $A$ on the bundle $E$ is a bundle 2-gerbe with connection over $M$, the Chern-Simons 2-gerbe $\mathbb{CS}(k,A)$. Connections on bundle 2-gerbes have $U(1)$-valued holonomies around closed oriented 3-manifolds, and so one defines the (exponentiated) Chern-Simons action -$$ -S_{CS}(k,A) := Hol_{\mathbb{CS}(k,A)}(M) \in U(1). -$$ -With this definition, it is not necessary to extend the manifold $M$ or the bundle $E$. -However, if extensions of $M$, $E$ and $A$ exist one has the following: there is a Stokes' Theorem for 2-gerbe holonomy, namely -$$ -Hol_{\mathbb{G}}(M) = \exp \left ( \int_B \mathrm{curv}(\mathbb{G}) \right ). -$$ -Now, the curvature of the Chern-Simons 2-gerbe $\mathbb{CS}(k,A)$ is $\frac{1}{8\pi^2} b_{k} ( F \wedge F )$, where the bilinear form $b_{k}: \mathfrak{g} \times \mathfrak{g} \to \mathbb{R}$ corresponds to $k$ under the Chern-Weil homomorphism. So we have -$$ -S_{CS}(k,A) = \exp \left ( \frac{1}{8\pi^2} \int_B b_{k}(F \wedge F) \right ) -$$ -which is the formula you have asked for in (2).<|endoftext|> -TITLE: Does there exist a contractible fiber bundle with fiber $G(\infty)$ and base $SU(\infty)$? -QUESTION [9 upvotes]: Bott periodicity implies that $\Omega(SU)\simeq G(\infty)$. Here, by $G(\infty)$, I mean the direct limit $\underset{m\to \infty}{\lim} G_m(\mathbb{C}^{2m})$ where $G_m(\mathbb{C}^{2m})\subset G_{m+1}(\mathbb{C}^{2m+2})$ by stabilization (or some similar nice model for $BU$). This is the classifying space for $U = \underset{m\to\infty}{\lim} U(m)$, and may be identified with $U/(U\times U)$, where the coordinates of each factor of the product subgroup alternate. From the path space construction, one knows that there is a contractible Serre fibration $\Omega SU \to E \to SU$, where $E$ is the space of paths in $SU$ from the identity. The fibers are each homotopic to $G(\infty)$. -The question I have is whether this may be realized by a contractible fiber bundle $G(\infty)\to E \to SU$ ? Other quasi-fibrations were given by Aguilar and Prieto, and by Behrens. There are also fibrations of this sort constructed using symplectic reduction by Latour and by Giroux, where the fibers are homotopic to $G(\infty)$. What I'm asking for is whether there exists a contractible fiber bundle rather than just a fibration or quasi-fibration? -Presumably such a bundle would arise from a map $SU \to BDiff(G(\infty))$. The isometry group of $G(\infty)$ contains $SU$, so one could ask a fortiori whether there is a map $SU \to BSU$, that is a map $f: SU \to G(\infty)$, such that the induced fiber bundle is contractible? Added clarification: The induced bundle would come from the fiber product of the pull-back of the $U$ bundle $U\to E \to G(\infty)$ with the action of $U$ on $G(\infty)$: $f^{*}(E) \times_{U} G(\infty)$. - -REPLY [7 votes]: I'm not completely sure that I understand your notation, so this may be not what you want, but in case it is close enough, I'll have a go. The bit I'm assuming is that $G_m$ is the Grassmannian of $m$-places in $\mathbb{C}^{2m}$. That seems fairly safe, but my brain is refusing to check the homotopy types of everything involved at this hour. -I think that you can do this if you combine a construction of mine from The Co-Riemannian Structure of Smooth Loop Spaces with some bits from Loop Groups by Pressley and Segal, and then check a few details about how stuff holds together in the limit. -The bit that you need from Loop Groups is that the polynomial loop group, $\Omega_{\text{pol}} S U(n)$ acts on the finite restricted Grassmannian $Gr_0(H)$. Let me remind you what that space is: we start with the polarised Hilbert space, $H = L^2(S^1,\mathbb{C}^n)$, polarised as $H = H_+ \oplus H_-$ where $H_+$ are those functions with only positive Fourier coefficients and $H_-$ with strictly negative Fourier coefficients. From this, we define - -$$ -Gr_0(H) = \{W \subseteq H : \exists k : z^k H_+ \subseteq W \subseteq z^{-k} H_+\} -$$ - -This is the union of $Gr(H_{-k,k})$ where $H_{-k,k} = z^{-k}H_+/z^kH_+$ so this (if I'm reading things aright) is the $G(\infty)$ of your question (Loop Groups, section 7.2). -The next bit that we need is that $\Omega_{\text{pol}} SU(n)$ acts on this space. This is from Theorem 8.3.2 and Proposition 8.3.3 in Loop Groups. -Putting these two together, if we have a principal $\Omega_{\text{pol}} SU(n)$-bundle over a space then we get a $G(\infty)$-fibre bundle over said space. -So now comes the bit from my work. In Section 3.2.3 of The co-Riemannian Structure of Smooth Loop Spaces, I construct a principal $\Omega_{\text{pol}} SU(n)$-bundle over $SU(n)$, with the property that under the obvious inclusion $\Omega_{\text{pol}} SU(n) \to \Omega SU(n)$ then this becomes the bundle coming from the usual path construction (so, although I don't need this in that paper, Bott periodicity implies that it is contractible). -So now we have a fibre bundle with fibre $G(\infty)$ over $SU(n)$. The last bit that you need is to show that under the inclusion $SU(n) \to SU(n+1)$ then these fibre bundles are compatible. The bit where this needs care is in the action of $\Omega_{\text{pol}} SU(n)$ on $Gr_0(H)$. But I think that if you include $L^2(S^1, \mathbb{C}^n)$ in to $L^2(S^1, \mathbb{C}^{n+1})$ at the same time, then you should get a diagram that works. The resulting Grassmannian will be $\bigcup Gr_0(L^2(S^1,\mathbb{C}^n))$ but that will still be $G(\infty)$ (assuming that I've understood the question correctly). -As I said, there's a few ifs and buts here: if I understood the question correctly, and if everything holds together in the limit (but I'm pretty sure that the second "if" is fine), but obviously as I'm less sure about the first if I haven't checked all the details.<|endoftext|> -TITLE: Homology of homotopy fixed point spaces -QUESTION [7 upvotes]: This is a general question for the homotopy theory crowd: How does one go about computing the homology and homotopy groups of homotopy fixed point spaces $X^{hG}:= Map^G(EG, X)$ for the action of a group $G$ on a space $X$? There seem to be some tools: - -Lannes' theory: which allows you to compute (or at least say something about) $H_*(X^{hG}, \mathbb{F}_p)$ when $G$ is a $p$-group. -Homotopy fixed point spectral sequences, which allow you to compute the stable homotopy groups of homotopy fixed point spectra. - -Are there other tools out there? I feel like (1.) should be the harder version of a fact that I'm missing about computing $H_*(X^{hG}, \mathbb{F}_p)$ when $|G|$ is coprime to $p$. Regarding (2.), is there any hope of an unstable homotopy fixed point spectral sequence? - -REPLY [4 votes]: Hej Craig, -Re (2) as Tilman says in his comment, there is an unstable homotopy fixed point spectral sequence, a special case of the spectral sequence of a homotopy limit as described by Bousfield and others. -Re (1) when X is finite (and more generally), Lannes theory should be seen as generalization of ordinary Smith theory. Smith theory only works for p-groups, so I don't think it is a harder version of a prime-to-p statement.<|endoftext|> -TITLE: On MacMahon's conjecture and a Schur function identity -QUESTION [5 upvotes]: Recently I am reading Professor Bressoud's book "Proofs and confirmations"。And chapter 4 of his book is about using Schur functions to prove Macmahon's conjecture on symmetric plane partitions: compute the genetaring function for plane partitions in box $B(r,r,t)$ that are symmetric about their first two coorbinates. His method is Macdonald's, which employs Weyl's determinant on root systems. The crucial step is to prove the following equation and put the right end into a product of the $x_i$'s. -$$\sum_{\lambda\subseteq \{t^r\}} s_{\lambda}(x_1,\cdots,x_r) = \frac{\det(x_i^{j-1}-x_i^{t+2r-j})}{\det(x_i^{j-1}-x_i^{2r-j})}.$$ -where $\lambda$ runs over all partions contained in a $r\times t$ rectangle. -The proof in quite "tedious",because it involves a determinant expansion and their are a bunch of transformations of the summations. But with the RSK correspondence, its not hard to prove this identity: -$$\sum_{\lambda} s_{\lambda}(x_1,\cdots,x_r)=\prod_{i=1}^r\frac{1}{1-x_i}\frac{1}{1-x_ix_j}$$ -here $\lambda$ runs over all partitions whose length is $\leq r$. -(I learned this in Stanley's book ) -and what we really want is the following -$$\sum_{\lambda\subseteq \{t^r\}} s_{\lambda}(x_1,\cdots,x_r) =?$$ -(it's good to have a product of the $x_i$'s at the right hand.) -So my questions is, can we find a more direct way to proof Macmahon's conjecture with the help of RSK correspondence, especially find a nicer way to express the -$$\sum_{\lambda} s_{\lambda}(x_1,\cdots,x_r)?$$ - -REPLY [3 votes]: I doubt it... as far as I know, the RSK correspondence isn't very well-behaved on the set of Young tableaux that you need. -This doesn't entirely rule out the possibility of other "nice" proofs, though with current technology, I think you'll need to do a determinant evaluation at some point for this particular problem. That's not necessarily a bad thing, as there are some rather astonishing modern ways of evaluating determinants, especially those coming from tiling problems, plane partitions and the like. -I'd suggest you take a look at C. Krattenthaler's inspiring papers "Advanced Determinant Calculus", http://arxiv.org/pdf/math.CO/9902004, and "Advanced Determinant Calculus: A Complement", http://arxiv.org/pdf/math/0503507.<|endoftext|> -TITLE: The Galois group of a random polynomial -QUESTION [26 upvotes]: Intuitively, the Galois group (of a splitting field over $\mathbb Q$ of) a polynomial $f\in\mathbb Q[X]$ taken at random is most probably the full permutation group on the roots of $f$. This intuition can be made precise as follows. -Elementary: For inegers $d \geq 1$ and $N \geq 1$, consider the set $P_{d,N}$ of polynomials in one variable of degree $\leq d$ whose coefficients have absolute value $\leq N$. The set $P_{d,N}$ is finite, and we can consider the subset $Q_{d,N}$ of those $f\in P_{d,N}$ whose Galois group is the full symmetric group $S_d$. Then, is it true that the ratio $|Q_{d,N}|/|P_{d,N}|$ goes to $1$ as $N$ goes to infinity? If so, are there good estimates for the speed of convergence? -Less elementary, but still the same: Let $k$ be a number field. For an integer $d\geq 1$, indentify $k$--points of the $d$--dimensional projective space $\mathbb P^d$ with nonzero polynomials of degree $\leq d$ over $k$ up to scalars. This way, we can speak about the Galois group of a point $f\in \mathbb P^d(k)$. For an integer $N$, let $p_{d,N}$ be the number of points of $P^d(k)$ of height $\leq N$, and let $q_{d,N}$ be the number of points of height $\leq N$ and with Galois group $S_d$. Then, is it true that -$$\frac{q_{d,N}}{p_{d,N}}\to 1$$ -as $N$ goes to infinity? And again, are there good estimates for the speed of convergence? -Remark: The subset of $\mathbb Q^{d+1} = \mathbb A^{d+1}(\mathbb Q)$ of those elements $(a_0,\ldots, a_d)$ such that the Galois group of the polynomial $f = a_0+a_1X+\cdots+a_dX^d$ is $S_d$ is Zariski dense in $\mathbb A^{d+1}$. This is so because the Galois group of the generic polynomial $f = t_0+t_1X+\cdots+t_dX^d$ with indeterminates $t_0,\ldots, t_d$ over $\mathbb Q(t_0,\ldots, t_d)$ is $S_d$ and the fact that $\mathbb Q$ (or any number field) is a Hilbertian field. - -REPLY [15 votes]: As Torsten remarks, the answer is yes and the result is well-known. P. X. Gallagher proved the following: If $E_d(N)$ denotes the set of degree $d$ polynomials with coefficients $|a_i|\le N$ with Galois group not equal to $S_d$, then -$$ -|E_d(N)|\ll N^{d-1/2}\log N -$$This bound gives you what you want. There is a discussion and a list of references in this sci.math post.<|endoftext|> -TITLE: Finitely generated subgroups of a product of free groups -QUESTION [9 upvotes]: Is it true that a finitely generated subgroup of a cartesian product of free groups has a finite cohomological dimension? -The same question about pro-$p$ groups: -Is it true that a finitely generated closed subgroup of a cartesian product of free pro $p$-groups has a finite cohomological dimension? - -REPLY [10 votes]: Regarding your first question, the answer is 'yes'. Consider an arbitrary direct product of free groups $\prod_\alpha F_\alpha$ and $H$ a finitely generated subgroup. Then $H$ is residually free. It follows from work of Baumslag--Remeslennikov--Miasnikov (I think, originally - there are now many proofs of this fact) that $H$ is a subgroup of a finite direct product of limit groups. Sela and Kharlampovich--Miasnikov proved that limit groups have finite $K(G,1)$'s, so the same is true of a finite direct product of limit groups. The covering space corresponding to $H$ is then a $K(H,1)$ with cells in only finitely many dimensions. -Let me add that it's not obvious how to prove the analogue of this in the pro-p world, as the pro-p analogues of limit groups are not well understood (or possibly even defined).<|endoftext|> -TITLE: Mono- and epi-morphisms for C*-algebras -QUESTION [10 upvotes]: This question is motivated by Yemon Choi's answer here: Epimorphisms have dense range in TopHausGrp? -It's well-known that the category of unital commutative C*-algebras and $*$-homomorphisms is dual to the category of compact Hausdorff spaces and continuous maps. One finds that for all C*-algebras (with $*$-homomorphisms as morphisms) that monomorphisms are just injective maps, and epimorphisms are surjections (the latter point is non-trivial-- see the final paper which Yemon suggests in the link above). -Suppose instead we look at locally compact Hausdorff spaces, with continuous maps as morphisms. Then dually, we get all commutative C*-algebras, but now the notion of a $*$-homomorphism is too restrictive (it corresponds to proper continuous maps). Instead we say that a morphism between C*-algebras $A$ and $B$ is a non-degenerate $*$-homomorphism $A\rightarrow M(B)$ form $A$ to the multiplier algebra of $B$. Such a map extends uniquely to a strictly continuous $*$-homomoprhisms $M(A)\rightarrow M(B)$, and so we can compose such maps. Hence we get a category. A little checking shows that the full subcategory of commutative C*-algebras, with these morphisms, is now dual to the category of locally compact Hausdorff spaces with continuous maps. (I think Woronowicz was the first person to articulate this view). - -For C*-algebras, with morphisms as arrows, what are epimorphisms and monomorphisms? - -Restricting to the commutative case, we can instead look at locally compact Hausdorff spaces, and reverse the arrows. So working through, a monomorphism remains just an injective map; but I see no simple description of epimorphisms (at the level of algebras-- for spaces, it's just injective continuous maps). -Edit: Maybe this notion of "non-degenerate" is confusing. If $f:X\rightarrow Y$ is a continuous map between locally compact Hausdorff spaces, then we define $f_*:C_0(Y)\rightarrow C^b(X); a \mapsto a\circ f$. Notice that we really do need the codomain to be all bounded continuous functions-- but that's okay, as $C^b(X)$ is just the multiplier algebra of $C_0(X)$, and $f_*$ turns out to be non-degenerate. Conversely, every non-degenerate $*$-homomorphism $C_0(Y)\rightarrow C^b(X)$ arises in this way (but a general $*$-homomorphism $C_0(Y)\rightarrow C^b(X)$ can be much more complicated). - -REPLY [4 votes]: I'm not sure why $*$-homomorphisms would be too restrictive a choice, but taking those as morphisms between C*-algebras as objects, the epimorphisms are precisely the surjective $*$-homomorphisms. This is proposition 2 in G. A. Reid's "Epimorphisms and surjectivity", Inventiones Mathematicae 9:295-307, 1970. (EuDML)<|endoftext|> -TITLE: Groups whose normal subgroups form a chain with respect to inclusion -QUESTION [7 upvotes]: Let G be a finite group. In general, given two normal subgroups N and K of G, we need not have N < K or K < N. The easiest example is the Klein 4-group V4 and its subgroups of order 2. So assume that G has such a property, that is, the normal subgroups of G constitute a chain with respect to inclusion. For example, simple groups, cyclic groups and symmetric groups satisfy this property. Certainly, if G has that property, then the normal subgroups are necessarily characteristic. Furthermore, the center of G must be cyclic. Indeed, every abelian group with this property must be a cyclic p-group (and vice-versa). This also shows that G/G' is cyclic, for the property is hereditary under quotients. -I would like to know if these groups have been studied before. If so, can you please provide some references? - -REPLY [6 votes]: For solvable groups without Frattini chief factors, this is equivalent to each of the following (individually): - -having a unique chief series, -every quotient group having a faithful primitive permutation action, -the upper Fitting series being a chief series -the lower Fitting series being a chief series - -This is shown in: - -Hawkes, Trevor O. - "Two applications of twisted wreath products to finite soluble groups." - Trans. Amer. Math. Soc. 214 (1975), 325–335. - MR379657 - DOI:10.2307/1997110 - -You might also be interested in the safari for zebra groups. -However, there are solvable groups with Frattini factors whose normal subgroups form a chain: SL(2,3) for example.<|endoftext|> -TITLE: Basic questions about stacks -QUESTION [10 upvotes]: I'm trying to understand some basics of stacks in algebraic geometry and have three questions: -1) As far as I understand, the moduli stack of vector bundles over a scheme $X$ is a replacement for the non-existent moduli space of vector bundles over $X$. Is this the only reason for the study of this stack or isn't it actually important to remember the isomorphisms between vector bundles? -2) What about replacing schemes by manifolds and define moduli stacks in a similar way? Does it make sense to talk about about stacks in the "euclidean" topology on the site of open subsets of euclidean spaces? Of course it makes sense, but I'm wondering if there is any literature about it. For example, is there a "generalized manifold" $BGL_n$ such that for every manifold $X$ we have an equivalence of categories between $Hom(X,BGL_n)$ and the category of vector bundles on $X$? -3) The objects of algebraic geometry have made a great evolution in the 20th century. Projective varieties, schemes, algebraic spaces, stacks. Is there an "upper bound" of this process of abstraction? I think that each abstraction was motivated by concrete geometric problems, but it might be argued if we actually solve these problems just by enlarging the category of geometric objects in consideration. This leads to the vague question: What are the fundamental ideas of algebraic geometry which will hopefully survive the next abstraction? - -REPLY [8 votes]: The notion of fine moduli space requires the existence of a universal family. In this case, you want a scheme $M$ equipped with a rank $n$ vector bundle $V$ on $M \times X$, such that pullback induces a natural bijection between the set of maps from any other scheme $Y$ to $M$ and the set of rank $n$ vector bundles on $X \times Y$. You can view $V$ as a family of vector bundles on $X$, parametrized by $M$. Vector bundles of positive rank do not admit universal families, in part due to the existence of automorphisms (and the existence of schemes with nontrivial fundamental group that can act as bases of nontrivial families). I don't have a precise grasp on what your question is asking, but depending on the application of choice, one can sometimes work with a coarse moduli space (which is roughly a way to ignore automorphisms), and one can sometimes rigidify the moduli problem to get a natural cover of the stack by a scheme. If $X$ is a general scheme (instead of, e.g., a point or a projective curve) the stack of vector bundles is unlikely to be algebraic, and nether simplifying option looks promising. -My wild guess is that you intend the Euclidean site to be an analogue of the category of affine schemes of finite type. You can form a notion of stack in topological spaces, by following the usual fibered category route, and you can certainly restrict to the subcategory of open subsets of Euclidean space. As Johannes Ebert mentioned in the comments, Noohi has some papers online that describe topological stacks. Some names that show up in the smooth setting include Alan Weinstein, Cristian Blohmann, and Chenchang Zhu (but I am relatively unfamiliar with this area). -Right now, there is an upper bound on the information content in mathematical abstraction given by the finite size of the human brain. Even if the robots take over, there is the finite size of the observable universe. More to the point at hand, objects more abstract than stacks were already considered in algebraic geometry during the 20th century. For example Grothendieck's Pursuing Stacks is one of the early attempts to apply homotopy theory techniques to work with more abstract objects like $n$-categories and $n$-stacks. I am not qualified to answer your revised question about fundamental ideas.<|endoftext|> -TITLE: Degree zero zero-cycles on the square of a curve -QUESTION [7 upvotes]: A well-known mathematician once explained the following conjecture to me, as "an example of how little we know about cycles of codimension $\geq 2$." Let $C$ be a curve defined over a number field $k$, and let $x_1,x_2 \in C(\overline{k})$ be distinct points. Then the zero-cycle $\gamma=(x_1,x_1)+(x_2,x_2)-(x_1,x_2)-(x_2,x_1)\subset C\times C$ has degree zero, and the conjecture is that this cycle is torsion in the Chow group $Ch^2(C\times C)_0$ of degree-zero zero-cycles modulo rational equivalence. In other words, there should be an integer $n\geq 1$, curves $C_i \subset C\times C$ and rational functions $f_i$ on $C_i$, such that -$n\gamma=\sum_{C_i}\mathrm{div}(f_i)$. -The significance of $k$ being a number field is that $\gamma$ maps to zero in the relevant intermediate Jacobian, so apparently some conjectures of Bloch on height pairings suggest that it is torsion. (My apologies for the vagueness here; I heard this years ago!) When $C$ has genus one, the conjecture is easy to prove. My questions: - -Is anything known about this conjecture in higher genus? -Is this conjecture related to, or implied by, any "mainstream" conjectures in arithmetic geometry? -Is there an in-print-somewhere reference for this conjecture? Should it be attributed to Bloch, or to somebody else? - -REPLY [9 votes]: I think this conjecture is normally attributed to Bloch and Beilinson, and it is a special case of their general conjecture that Albanese equivalence coincides with rational equivalence (up to torsion) on smooth projective varieties over number fields. -(For varieties over any field --- of char. zero say --- the Chow groups are expected to -have a filtration, whose first steps are homological equivalence, then albanese equivalence. -The higher steps in the filtration are conjecturally related to how the two basic conditions --- homological or Albanese triviality --- interact with specialization of the variety. Since a variety over a number field can't be specialized, the filtration should stop after -Albanese equivalence, hence the conjecture.) -As far as I know there is no literature on this question to speak of, and it is wide open. -(I would love to hear something to the contrary!) There is literature on the conjectural filtration on Chow groups --- look in the motives volumes, at some of the papers of Green and Girffiths on arithmetic Hodge structures, and at some of Morihiko Saito's papers. (In both cases, the papers --- at least the ones I remember --- are from around ten years ago. I think there might be '98 ICM talk, for example.)<|endoftext|> -TITLE: Integrality of the canonical trace and topology -QUESTION [7 upvotes]: Let $G$ be a discrete group and consider the reduced group C* algebra $C_r^\ast(G)$, viewed as an algebra of bounded operators on $\ell^2(G)$ by the regular representation. The canonical trace on $C_r^\ast(G)$ is simply defined to be: $\tau(T) = (T \delta_e, \delta_e)$ where $\delta_e$ is the basis vector of $\ell^2(G)$ corresponding to the identity. This trace gives rise to a homomorphism $\tau_\ast: K_0(C_r^\ast(G)) \to \mathbb{R}$ on K-theory. -It is a very deep topological fact that the image of $\tau_\ast$ consists of integers for a huge class of groups, specifically those groups for which the Baum-Connes conjecture is true. To see this, let $M$ be a closed manifold with fundamental group $G$, let $\mu: K_0(M) \to K_0(C_r^\ast(G))$ be the assembly map on K-homology, and note that $\tau_\ast \circ \mu$ agrees with the (integer valued) index map $K_0(M) \to \mathbb{R}$. Thus $\tau_\ast$ is integer-valued so long as the assembly map is surjective, and the Baum-Connes conjecture asserts that it is an isomorphism. -Here are my questions: - -Are there easy proofs of the integrality of $\tau_\ast$ in (nontrivial) special cases, or at least proofs that don't resort to heavy-duty topological machinery? - -Even better, - -Are there groups for which the Baum-Connes conjecture is not known but $\tau_\ast$ is known to be integer-valued? - -These two questions stem from my intuition that this should be a problem in analysis or possibly representation theory rather than topology. But in the likely event that this intuition is wrong... - -Is the integrality of $\tau_\ast$ related (or equivalent) to any more transparently topological phenomena? - -REPLY [4 votes]: For abelian groups $G$ (maybe a trivial case?), the canonical trace is given by integration over the Pontryagin dual $\hat{G}$ (with respect to normalized Haar measure), so we have equivalences: -(1) $\tau_*$ is integer-valued on $K_0(C(\hat{G}))$; -(2) the Kaplansky-Kadison conjecture holds for $C^*_r(G)$; -(3) $\hat{G}$ is connected; -(4) $G$ is torsion-free. -In general, for non-abelian $G$, proving the integrality of the trace is done by means of an index theorem, so that it is deeply related to surjectivity of the Baum-Connes map. This is to say that, to my knowledge, there are no examples of groups where integrality of the trace is known, but Baum-Connes is not. There are a few cases however, where the Kaplansky-Kadison conjecture has been established independently of the Baum-Connes conjecture. This is the case for torsion-free hyperbolic groups, see Michael Puschnigg, The Kadison-Kaplansky conjecture for word-hyperbolic groups. -Invent. Math. 149, no. 1, 153--194, 2002.<|endoftext|> -TITLE: Analytic Torsion in the Derived Category -QUESTION [12 upvotes]: I recently learned about analytic torsion and about the amazing Cheeger-Muller theorem identifying analytic and Reidemeister torsion for compact Riemannian manifolds. -Now analytic torsion is defined in terms of the eigenvalues of the Laplacian, acting on the de Rham complex; essentially, one takes the alternating product of the zeta-regularized determinants of the Laplacian, suitably normalized. This is all reasonably natural. -What bothers me about this definition is that it depends so strongly on properties of the de Rham complex, which I see as just a (particularly nice) choice of acyclic resolution for the constant sheaves $\underline{\mathbb{R}}$ or $\underline{\mathbb{C}}$. So my question is: - -Is it possible to recover the analytic torsion from the derived category of sheaves over (a compact Riemannian manifold) $M$? - -I would also be happy with any other "intrinsic" characterization of analytic torsion. - -EDIT (3/15/2011): This is profilesdroxford54's suggested recipe as I understand it: (1) Take the determinant of any perfect complex resolving the constant sheaf, with its natural integral structure coming from the inclusion of the constant sheaf $\underline{\mathbb{Z}}$ in $\underline{\mathbb{R}}$; (2) Take the determinant of the de Rham complex with the integral structure given by zeta regularization; (3) The difference in the two integral structures is the analytic torsion (though I do not immediately see why there is a natural map between the two determinant lines, which is essential). Of course this recipe does not free us from the de Rham complex, but it suggests a way to do so, which I'll flesh out in three more focused questions. - -(1) Is there a good theory of the derived category of sheaves on a compact Riemannian manifold, with some extra structure (i.e. self-dual, as in the case of the de Rham complex over compact Riemannian manifolds, or equivalently, with an inner product)? This seems to me like a very natural thing to study, so I am not as pessimistic as profilesdroxford54. -(2) In the framework of such a derived category (which must contain lots of infinite-dimensional objects, e.g. the de Rham complex) is there a good theory of the determinant, and does it agree with the zeta-regularized determinant? Someone recently mentioned some determinantal theory in the infinite-dimensional derived setting to me, but I know next to nothing about this sort of thing; references are welcome. -(3) Why is there a canonical map (up to sign?) between this new determinant line and the usual one (which is necessary to compare integral structure)? - -REPLY [4 votes]: What does depend on the "derived category" is the notion of determinant line. Thus take any local system $\mathcal E$ of $R$-modules on $M$ (though one should probably work with constructible sheaves of some kind), then compute cohomology any way you want -- de Rham, etc. So long as the the cohomology is a perfect complex of $R$ modules, it has a determinant -- a free rank one projective $R$-module. -So far this has nothing to do with the Riemannanian structure. Now suppose there is a second structure -- say $M$ is Riemannian and $\mathcal E$ is a local system of inner product spaces, where $R$ is a subring of the reals. Then you can view the DeRham cx of $\mathcal E$ as a perfect complex of (infinite dimensional) inner-product spaces, and the zeta normalization gives a way of defining the determinant of this as a one dimensional inner product space. -The torsion is the relationship between these two structures. EG if $R={\mathbf Z}$ then it is the real number given by the length of a generator of the line. -To give this part of the construction a more "derived category" feel, one would need a notion of quasi-isomorphism "preserving inner products". I am not sure anyone has really thought hard about this.<|endoftext|> -TITLE: Does Ricci flow depend continuously on the initial metric? -QUESTION [25 upvotes]: Consider a version of Ricci flow for which short time existence and uniqueness are known, -e.g. the Ricci flow on a closed manifold. Does the solution $g_t$ for small $t$ depend continuously on the initial metric? -I thought the answer is "yes" for Ricci flow on a closed manifold but I cannot see why. -My immediate interest is the same question for the instantaneously complete Ricci flow on $\mathbb R^2$ studied by Giesen and Topping. - -REPLY [5 votes]: A complete proof of continuous dependence of Ricci flow was published recently by Eric Bahuaud, Christine Guenther & James Isenberg. Let me record their main theorem: - -Theorem A (Continuous Dependence of the Ricci Flow) Let $(M, g_0)$ be a smooth -compact Riemannian manifold. Let $g_0(t)$ be the maximal solution of the Ricci flow -(1) for $t\in [0,\tau (g_0))$, $\tau (g_0) \le \infty$. Choose $0 < \tau < \tau(g_0)$. Let $k \ge 2$. There exist positive constants $r$ and $C$ depending only on $g_0$ and $\tau$ such that if -$$\|g_1 − g_0\|_{h^{k+2,\alpha}} \le r,$$ -then for the unique solution $g_1(t)$ of the Ricci flow starting at $g_1$, the maximal existence time satisfies $\tau (g_1)\ge \tau$, and -$$\|g_1(t) − g_0(t)\|_{h^{k,\alpha}} \le C\|g_1 − g_0\|_{h^{k+2,\alpha}}$$ -for all $t \in [0,\tau ]$. - -Their proof involves the De-Turck trick and standard semi-group theory in parabolic PDE. They also cite this MO post in their paper.<|endoftext|> -TITLE: Looking for a simple proof that R^2 has only one smooth structure -QUESTION [16 upvotes]: So not so long ago, I asked for a simple proof that $\mathbf{R}$ has only one smooth structure. A proof that was communicated to me by Ryan Budney (link text) was the following: -So let me recall his argument: So let $X$ be a real line endowed with a "potentially" -exotic smooth structure. We know that $X$ is Hausdorff and paracompact so -for every open covering $\mathcal{U}$ of $X$ we have a partition of unity dominated by -$\mathcal{U}$. Using this we can endow $\mathbf{R}$ with a Riemanian metric $ds^2$ (choose your favorite open covering which is locally finite!). -Let $x_0$ be a point of $X$ so that $X-x_0=X^+\bigcup X^{-}$ is the disjoint union -of the two components. Finally, -note that one may integrate this metric against the Haar measure of $X$ using the velocity vectors $1$ and $-1$ in the fiber above $x_0$ to get two bijections -$f^+:X^+\rightarrow\mathbf{R}_{>0}$ -and -$f^-:X^-\rightarrow\mathbf{R}_{<0}$. -Since the metric $ds^2$ is smooth we see that -$f^+$ and $f^-$ are smooth and that they glue in a smooth way. So basically, the proof -works because we can think of $\mathbf{R}$ as the union of two geodesics. -Q: Is there somekind of similar argument for $\mathbf{R}^2$ and $\mathbf{R}^3$ ? -Any simple proof along different lines is welcome! - -REPLY [14 votes]: Some comments alluded to the possibility to show this using the Riemann uniformization theorem (by paracompactness, any oriented $2$-manifold has an almost complex structure, which is integrable by Newlander-Nirenberg and by the uniformization theorem, it will be biholomorphically equivalent to the plane or the unit disc, hence diffeomorphic to $R^2$). This is not circular, but to claim that this is "simple" would be utterly absurd. The complete proof of the uniformization theorem is one of the hardest mathematical achievements of the early 20th century; the proof uses a lot of analysis and also a bit of algebraic topology. -Using Morse theory, you can argue as follows: let $U$ be a connected noncompact surface, pick a Morse function $f: U \to \mathbb{R}$. One can modify $f$ so that it has no critical points of index $2$ and precisely one critical point of index $0$, so let us assume that $f$ has this property. This is the most basic case of the handle-cancellation technique. -Now let $C_{\ast}(f)$ be the chain complex of the Morse function. $C_k (f)$ has the critical points of index $k$ as a basis. If $f$ is above, it follows that $C_0 (f)=Z$, $C_k (f)=0$ if $k \geq 2$. The differential $C_1 \to C_0$ will be zero and so $H_1 (U)= C_1 (f)$. -If $H_1 (U)=0$, we see that there is a Morse function $f:U \to \mathbb{R}$ with precisely one minimum. Use the flow lines of $f$ to cook up a diffeomorphism $f: U \to \mathbb{R}^2$. -I don't think you get this result much cheaper.<|endoftext|> -TITLE: Is there a high-concept explanation for why "simplicial" leads to "homotopy-theoretic"? -QUESTION [90 upvotes]: My (limited) understanding is that simplicial methods tend to be used whenever you want some kind of nontrivial homotopy theory -- for instance, to get a nice model structure, you use simplicial sets and not just plain sets; to make $\mathbb{A}^1$-homotopy work, you work with simplicial (pre?)sheaves and not just plain sheaves or schemes; to construct the cotangent complex (which if I understand correctly is a homotopical construction, hopefully a Quillen derived functor on the category of simplicial algebras), you use simplicial commutative rings. -But why does "simplicial" make everything work so well? For instance, a simplicial set is a contravariant functor $\Delta \to \mathbf{Sets}$ for $\Delta$ the simplex category: what is so wonderful about $\Delta$ that allows a model structure (and one, moreover, Quillen equivalent to topological spaces) appear? - -REPLY [4 votes]: There are many ways in which the categories of simplicial sets and of simplicial objects work very well, as mentioned above. More recently there has been a revival of the use of cubical sets, but with the additional structure of connections, derived from the monoid structures $\max,\min: I^2 \to I$. Dan Kan's first paper was cubical, but it was then realised that cubical groups were not Kan complexes, and there was a serious problem with realisation of cartesian products. In 1996 A. Tonks proved cubical groups with connections are Kan and G. Maltsiniotis has proved that cubical sets with connections form a test category in the sense of Grothendieck. -Thee is more on cubical sets in my answer to this mathoverflow question, in the book Nonabelian Algebraic Topology, and in this recent article. -There are lots of areas which have been well worked over in the simplicial context but not in the cubical (with connection) context, e.g. cubical groups. So there is lots more evaluation work to be done.<|endoftext|> -TITLE: How was the importance of the zeta function discovered? -QUESTION [21 upvotes]: This question is similar to Why do zeta functions contain so much information? , but is distinct. If the answers to that question answer this one also, I don't understand why. -The question is this: with the benefit of hindsight, the zeta function had become the basis of a great body of theory, leading to generalizations of CFT, and the powerful Langlands conjectures. But what made the 19th century mathematicians stumble on something so big? After all $\sum \frac{1}{n^s}$ is just one of many possible functions one can define that have to do with prime numbers. How and why did was the a priori fancifully defined function recognized as being of fundamental importance? - -REPLY [14 votes]: Andre Weil has an article called "Prehistory of the zeta function" (reviewed by Jutila on mathscinet). I read this article many years ago, but this is basically what I remember of its content. Apparently the divergence of the harmonic series was known in 1650. Euler computed the special values at even integers and derived some kind of a functional equation. He also proved the Euler product formula and gave a proof of the infinitude of prime numbers using the Euler product. Dirichlet defined general L functions that now bear his name but only for real s>1. Riemann extended the definition of the zeta function to all complex values and proved the functional equation. According to Weil there were other people who had proved functional equations for functions that were closely related to the zeta function (namely, Malmstén, Schlömilch and Clausen from the review), but perhaps Riemann's contribution is the singular paper that established the importance of the zeta function as an important object to study. Weil believes that Riemann was influenced by his discussion with Eisenstein.<|endoftext|> -TITLE: Structure of cuspidal Bernstein components---do non-commutative endomorphism rings ever really show up? -QUESTION [9 upvotes]: Let $F$ be a finite extension of $\mathbf{Q}_p$ with integers $\mathscr{O}$, let $\mathbb{G}$ be a connected reductive group over $F$ and let $G=\mathbb{G}(F)$ be its $F$-points. Let $X(G)=Hom(G,GL(1))$ denote the group of algebraic characters of $G$ and let $G^0$ denote -the intersection of $\chi^{-1}(\mathscr{O}^\times)$ as $\chi$ runs over $X(G)$. For example, if $G=GL(n,F)$ then $G^0$ is the elements of $G$ with determinant equal to a unit. Informally, $G^0$ is generated by the derived subgroup of $G$ and the maximal compact subgroup of the centre of $G$. -Let $\pi$ be an irreducible supercuspidal representation of $G$ (over the complexes, as usual) and consider the restriction of $\pi$ to $G^0$. If $Z$ is the centre of $G$ then $G^0Z$ is a normal subgroup of finite index in $G$, and $G/G^0Z$ is abelian. Hence the restriction of $\pi$ to $G^0$ can be written as a finite direct sum of irreduibles $\pi_i$. -My question is: is there an example known where the isomorphism class of some $\pi_i$ occurs with multiplicity greater than one in $\pi$? That is---is any supercuspidal $\pi$ always "multiplicity-free" as a representation of $G^0$? - -Why do I want to know this? -Those who know about Bernstein components will know the motivation behind this question. -When analysing the cuspidal Bernstein component of the category of smooth representations of $G$ associated to $\pi$, life can be very easy (Schur orthogonality relations etc) if all matrix coefficients of $\pi$ are actually compactly-supported. But a general connected reductive group may have a non-trivial centre, whose $F$-points are typically non-compact, which means compactly-supported matrix coefficients are very rare. The point about $G^0$ is that it has compact centre so these problems go away, but the representation theory of $G^0$ is very close to that of $G$. However, when moving from the study of the representation theory of $G^0$ to $G$ one has to induce back up; a typical ring that shows up in this procedure is $R:=End_G(Ind_{G^0}^G(\pi_i))$ for a $\pi_i$ as in the question. If $\pi_i$ shows up (up to isomorphism) with multiplicity greater than one then this ring will be non-commutative (in fact this is an iff). The Bernstein component corresponding to $\pi$ is isomorphic to the category of right $R$-modules, and if $R$ is commutative then it seems to me that it's always isomorphic to the algebraic functions on a product of $GL(1,\mathbf{C})$'s so it's a really "easy" ring, but the literature seems to leave open the possibility that non-commutative $R$s can occur. - -What do I know? -If $G/G^0$ is isomorphic to the integers then multiplicities greater than one cannot occur. In particular if $G=GL_n$ then multiplicities greater than one cannot occur. If $G$ is a product of $GL_n$'s then the same arguments apply and multiplicities greater than one cannot occur. If $\pi$ admits a Whittaker model then $\pi$ is multiplicity-free as a representation of $G_{der}$ and hence as a representation of $G^0$, and so again multiplicites cannot occur. I learnt this from Remark 1.6.1.3 of "The Bernstein decomposition and the Bernstein centre" by Alan Roche. Beyond this I know nothing. - -REPLY [5 votes]: For classical groups, V. Heiermann has proved that $Res^G_{G^0}(\pi)$ has no -multiplicity, see : - -Opérateurs d'entrelacement et algèbres de Hecke avec paramètres d'un groupe réductif p-adique - le cas des groupes classiques - Selecta Mathematica. - -In general, the endomorphism algebra $R$ is a free module of rank $m^2$ over its center Z, where $m$ is the multiplicity of an irreducible in $Res^G_{G^0}(\pi)$. See -p. 181 in my book "Représentations des groupes réductifs $p$-adiques", SMF, cours spécialisés 17. I suspect that $R$ and $Z$ could be Morita equivalent. -$Z$ is always indeed isomorphic to the ring of algebraic functions on a product of GL(1,C)'s.<|endoftext|> -TITLE: Subgroup structure of $\mathrm{SO}(1,n)_0$ -QUESTION [5 upvotes]: A classification (up to conjugacy) of all closed subgroups of the (identity component of the) Lorentz group $\mathrm{SO}(1,3)_0$ in terms of the subalgebras of its Lie algebra was given in - -R. Shaw. The subgroup structure of the homogeneous Lorentz group. The Quaterly Journal of Mathematics, Oxford 21 (1970) 101-124 - -(see also the book of Hall: Symmetry and Curvature Structure in General Relativity). -My question is whether there exists a similar classification for $\mathrm{SO}(1,n)_0$, $n\ge 4$. - -REPLY [2 votes]: I'm not certain of the complete answer, but I think there's a rough classification as follows. If $H < SO_0(n,1)$, then either - -$H$ is compact and is conjugate into $O(n)$ -$H$ is solvable and preserves a 1-dimensional subspace on the light cone (i.e. an isotropic subspace). In this case, $H$ acts by conformal affine transformations of $\mathbb{R}^n$. -$H$ preserves a 2-dimensional subspace of signature $(1,1)$ (this could actually be subsumed in the previous case). Here, the action restricted to this subspace is $SO_0(1,1)\cong \mathbb{R}$. Then $H$ acts on the orthogonal space by a subgroup of $O(n-1)$. The action is a direct product of a closed subgroup of $O(n-1)$ and $\mathbb{R}$. However, this product is not canonical, since any generator of the $\mathbb{R}$ factor may be modified by a 1-parameter subgroup of the $O(n-1)$ factor. -$H$ fixes a higher dimensional subspace of signature $(k,1)$, $k>1$. Then $H$ splits as a product of $SO_0(k,1)$ and a compact subgroup of $O(n-k)$. - -Classifying the compact subgroups of $O(n)$ is complicated, so I think this might the best one can say in general. I don't know a reference, and I hope I haven't overlooked any possibilities. The way I think about this is the action on hyperbolic $n$-space and its compactification by the sphere at infinity. Either the action fixes a point in $\mathbb{H}^n$, or it fixes a point at infinity, or it preserves a totally geodesic subspace of some dimension.<|endoftext|> -TITLE: Classifying spaces of E_1 - spaces -QUESTION [6 upvotes]: Hello, -I try to understand aspects of homotopy coherence, in particular "recognition principle" of May. -About the following I did not think a lot, but I decided to ask here anyway, so to save reinvention of the wheel and get clarifying comments. -Consider $E_1$ - the topological operad of small $1$-cubes. An $E_1$-space for me is a space together with an action of $E_1$. I think of an $E_1$-space as of a space together with a multiplication, associative up to "coherent homotopies". -Questions: -1) What will be the definition of a torsor for an $E_1$-space over some base, i.e. the analog of a principal homogeneous space for a topological group. -2) What will be the definition of a classifying space of a particular $E_1$-space. -3) If our $E_1$-space is the loop space $\Omega X$ of some space $X$ (with the satndard $E_1$-action), is true then that $X$ is the classifying space of $\Omega X$. -Probably in the above I did not insert some technical issues involving perhaps words like "group-like" or "fibrant", which I will be happy to hear about. -Thank you, -Sasha - -REPLY [14 votes]: From the horse's mouth. - -I would think a good theory of parametrized $E_1$-spaces should not be too -hard to develop, along the general lines of parametrized spaces (and spectra) -as developed ad nauseum in - -J.P. May and J. Sigurdsson. Parametrized homotopy theory. -Presumably the fibers should be grouplike. A current student, John Lind, could -answer better. He is working on classification theorems in a more sophisticated -context of parametrized spectra. - -There are several constructions. My original machine in Geo (The Geometry of -iterated loop spaces), Thm 13.1, gave $B(\Sigma,E_1,X)$ as a delooping of an -$E_1$-space $X$, using your notation. (The cited result works for $E_n$-spaces -for all $n$. One can also convert $X$ to an equivalent topological monoid $B(M,E_1,X)$, -by Thm 13.4 of Geo, and take the ordinary classifying space of that. These two constructions -are compared in papers by Thomason and Fiedorowicz, circa 1980, or maybe earlier. -This is answered affirmatively for all $n$ in my original work, in part (vi) of -Thm 13.1: $B(\Sigma^n,E_n,\Omega^nY)$ is weakly equivalent to $Y$ if $Y$ is $n$-connected. -The proviso can be improved to $n-1$-connected. It is then obviously necessary, since applying $\Omega^n$ loses any information about $\pi_0$ through $\pi_{n-1}$.<|endoftext|> -TITLE: Understanding "infinite" relations in groups -QUESTION [27 upvotes]: Consider the matrices $A = \frac{1}{5}\begin{pmatrix}5&0&0\\\ 2&2&1\\\ 2&1&2\end{pmatrix}$, $B = \frac{1}{5}\begin{pmatrix}2&2&1\\\ 0&5&0\\\ 1&2&2\end{pmatrix}$, and $C = \frac{1}{5}\begin{pmatrix}2&1&2\\\ 1&2&2\\\ 0&0&5\end{pmatrix}$. -The group $G\subset GL_3(\mathbb{C})$ they generate is free of rank 3. However, one also has: -$A^\infty := \lim_{k\rightarrow\infty}A^k = \begin{pmatrix}1&0&0\\\ 1&0&0\\\ 1&0&0\\\ \end{pmatrix}$, and similarly $B^\infty = \begin{pmatrix}0&1&0\\\ 0&1&0\\\ 0&1&0\\\ \end{pmatrix}$ and $C^\infty = \begin{pmatrix}0&0&1\\\ 0&0&1\\\ 0&0&1\\\ \end{pmatrix}$. -This leads to three "infinite" relations among $A$,$B$, and $C:$ -$A^\infty B = B^\infty A$ -$B^\infty C = C^\infty B$ -$C^\infty A = A^\infty C$ -Of course these are not true group relations since they involve infinite products and furthermore these infinite products are singular, hence do not actually lie in the group generated by $A$, $B$, and $C$, but rather are relations among elements of the "boundary" of this group (in an appropriate sense of the word boundary). - -Question 1: I am wondering if anyone has studied such infinite/boundary relations on groups, and in particular if they have any usefulness in understanding any properties of the original group. -Question 2: Suppose $H$ is another free group of rank 3 in $GL_3(\mathbb{C})$. $H$ is of course isomorphic to $G$, but may not have any boundary relations. Is there a way to use the infinite relations to define a "stronger" notion of isomorphism which says that $G$ and $H$ are not isomorphic since they do not have the same infinite relations? Edit: Upon further thought, this latter question is more of a question about the particular representation of $G$ chosen, so perhaps it is not so interesting? Or maybe something aside from the usual stuff about isomorphism of representations might be relevant? - -Motivation: Awhile back I was studying harmonic functions on the Sierpinski Gasket; these three matrices arise naturally in the evaluation of such functions. In this context the infinite relations above are equivalent to the statement that such functions are well-defined at each point of the Gasket. The relations above were occasionally useful for proving various facts, but at the time I never bothered to consider them in the context of group relations. -More generally they can arise in studying subgroups of $GL_n(R)$ for $R$ a commutative unital ring. In this case the "boundary" of the subgroup may be a subset of the singular elements of $Mat_n(R)$ and one can ask questions about relations between elements of this "boundary". - -REPLY [17 votes]: This answer is partly an expansion of the comments so far. What you are looking for is a type of decoration on groups that (1) lets you evaluate infinite products, and (2) also creates extra points that are the value of some or all of those products. There is one fairly inevitable answer to (1): You should consider topological groups or at least topological spaces, since taking limits in a topology is the main way to extend finite products to infinite products. But for (2), a topology isn't good enough. You get more information from a metric than from a topology, because a metric space has a completion. On the other hand, a metric isn't entirely satisfactory either, among other reasons because it has extra, non-canonical data. A good alternative to a metric is a uniformity. A uniform space also has a completion (in which all Cauchy nets converge) and every metric space is canonically a uniform space. -In fact, every well-behaved topological group has a canonical associated uniformity, because you can use the neighborhoods of the identity to make uniform neighborhoods. (I am a little hazy on how exactly the topological group should be well-behaved. It should be sufficient for it to be Hausdorff and for the left and right uniformities to agree.) Thus every such topological group has a completion which is also a topological group, and which then has infinite products which are group elements. As Mark Sapir suggests, one such uniformity is the profinite uniformity on a residually finite group. -At the other end, I think that a word-hyperbolic group $G$ has a uniformity defined by Gromov whose completion points are called the Gromov boundary. However, multiplication is not uniformly continuous in this construction; rather you only have that the uniformity is left-invariant. Thus, the completion is not a topological group, but it does have a left action of $G$. You can only define right-infinite products, and you can only multiply them by group elements on the left. Still, you can do that much, which still lets you consider many interesting infinite relations. -Your example of groups of matrices is modeled by a uniformity of intermediate quality. The uniformity of the additive Lie group structure on $M_3(\mathbb{C})$ is not the same as the uniformity of the multiplicative Lie group structure on $GL_3(\mathbb{C})$, even though the topologies are the same. Multiplication extends to the completion, but inversion does not. (Toy model: The function $1/x$ on $\mathbb{R}^\times$ is not uniformly continuous in the standard metric on $\mathbb{R}$ and obviously does not extend to $0$.) The completion of $GL_3(\mathbb{C})$ is obviously $M_3(\mathbb{C})$, and any subgroup has an inherited uniformity and a completion which is then a semigroup. So an answer to your second question is that a discrete group can have many different semigroup-completed uniformities, some of which come from matrix representations.<|endoftext|> -TITLE: dualizing sheaf of a nodal curve -QUESTION [19 upvotes]: I'm trying to understand the dualizing sheaf $\omega_C$ on a nodal curve $C$, in particular why is $H^1(C,\omega_C)=k$, where $k$ is the algebraically closed ground field. I know this sheaf is defined as the push-forward of the sheaf of rational differentials on the normalization $\tilde{C}$ of $C$ with at most simple poles at the points lying over the nodal points of $C$ and such that the sum of residues at the two points lying over the node will be zero. I can show that this is indeed an invertible sheaf on $C$, but I have no clue, despite my many attempts, how to show that $H^1(C,\omega_C)=k$. I've been able to show it in some very simple cases using Cech cohomology, but can someone explain to me how to do it in general? - -REPLY [27 votes]: If $\tilde{C}$ is the normalization, with two points $x$ and $y$ being identified under the map $\pi: \tilde{C} \to C$ to the node $z$ of $C$, then we have an exact sequence -$$0 \to \Omega^1_{\tilde C} \to \Omega^1_{\tilde C}(x + y) \to k_x \oplus k_y \to 0,$$ -where $k_x$ and $k_y$ are the skyscraper sheaves at the points $x$ and $y$. -Pushing forward (which is exact because the map $\pi$ is finite, and so in particular affine) -we get an exact sequence -$$0 \to \pi_* \Omega^1_{\tilde C} \to \pi_*\Omega^1_{\tilde C}(x+y) \to k_z^{\oplus 2} \to 0.$$ -Now there is a short exact sequence $0 \to k_z \to k_z^{\oplus 2} \to k_z \to 0$, -where the third arrow is just given by adding the two components, and -$\omega_C$ is the preimage of (the first copy of) $k_z$ under the surjection -$\pi_* \Omega^1_{\tilde C}(x+y) \to k_z^{\oplus 2}$. -In conclusion, we have an exact sequence -$$0 \to \pi_* \Omega^1_{\tilde C} \to \omega_{C} \to k_z \to 0.$$ -Now taking cohomology (and recalling that $H^i(C,\pi_*\mathcal F) = H^i(\tilde{C},\mathcal F)$ for -a coherent sheaf on $\tilde{C}$), we obtain -$$0 \to H^0(\tilde{C},\Omega^1_{\tilde C}) \to H^0(C,\omega_C) \to -H^0(C,k_z) \to H^1(\tilde{C},\Omega^1_{\tilde C}) \to H^1(C,\omega_C) \to 0.$$ -(The point here being that $H^1$ of a skyscraper sheaf such as $k_z$ vanishes.) -I claim that in this exact sequence the map $H^1(\tilde{C},\Omega^1_{\tilde C}) -\to H^1(C,\omega_C)$ is an isomorphism, and hence that the latter is one-dimensional, since the former is. -For this, it is equivalent to show that the map -$H^0(C,\omega_C) \to H^0(C,k_Z) = k$ is surjective. -Now $H^0(C,\omega_C) \subset H^0(C,\pi_*\Omega^1_C(x+y)) = H^0(\tilde{C},\Omega^1(x+y)).$ -The residue theorem shows that we may find a differential $\omega \in -H^0(\tilde{C},\Omega^1(x+y))$ whose residues at $x$ and $y$ are non-zero. (These residues -are then negative to one another.) Thought of as a section of -$H^0(C,\pi_*\Omega^1_C(x+y))$, this differential $\omega$ clearly lies in -$H^0(C,\omega_C)$. Its image under the map $H^0(C,\omega_C)$ is non-zero (equal to -the residue at either $x$ or at $y$, depending on a choice that was implicitly made above), -and so indeed $H^0(C,\omega_C) \to k$ is surjective. -Summary: The residue theorem guarantees the existence of sections of $H^0(C,\omega_C)$ -which have non-zero residues at $x$ and $y$ when pulled back to $\tilde{C}$, and -this in turn shows that $H^1(C,\omega_C)$ is isomorphic to $H^1(\tilde{C},\Omega^1_C)$, -and hence is one-dimensional.<|endoftext|> -TITLE: Question related to Diophantine approximations and Roth's theorem -QUESTION [6 upvotes]: The following question came up in my arithmetic geometry course yesterday. Suppose $\alpha$ is an irrational real algebraic integer, and suppose $\epsilon >0$ is given. Then by Roth's theorem there are at most finitely many rational numbers $\frac{h}{q}$ with $\gcd(h,q)=1$, $q>1$, such that -$$ -\left| \alpha - \frac{h}{q}\right| < \frac{1}{q^{2+\epsilon}}. -$$ -Are there any results on how large such $q$ can be? Thanks. - -REPLY [3 votes]: My other answer was for the first version of this question. The question has now been changed completely. As Antoine mentioned in his comment, an effective Roth's theorem is not known in general. Finding such a result is the probably the main open problem in Diophantine approximation. There some instances in which a non-trivial effective result can be proved. Most notable is the Baker-Feldman theorem which provides such a result with an exponent of the form $\deg \alpha - \epsilon$ (instead of $2+\epsilon$) for a suitable small positive $\epsilon$.<|endoftext|> -TITLE: About a Delzant polytope. (In particular dodecahedron) -QUESTION [19 upvotes]: Hi. I have a question. -Definition. Delzant polytope $P$ is a rational convex simple polytope with the smooth condition. Here, "smooth" means that for each vertex $v$, the $n$ edges containing $v$ form an element of $SL(n,\mathbb{Z})$, where $n$ is a dimension of $P$. -(If you wonder why this condition is called smooth, See Fulton. Introduction to toric variety chap I) -My question is as follow. -Can dodecahedron be the Delzant polytope? -I mean, is there a symplectic toric manifold whose moment map image is combinatorially equivalent to a dodecahedron? -Delzant's classfication theorem of compact symplectic toric manifold is surely very strong. But I think it is very hard to check whether the given polytope (having many faces) -is of Delzant type or not. If you know any reference of give me any comment, I really appriciate for your help. -Thank you. - -REPLY [7 votes]: Here is the answer to your question: -Elisa Prato, Symplectic Toric Geometry and the Regular Dodecahedron (2015) -Sorry it took so long. Hope you find this helpful. Another article covering all the platonic solids, rational or not, is on its way.<|endoftext|> -TITLE: Why is the zeta function of a variety over a finite field not a polynomial? (question about motives) -QUESTION [8 upvotes]: I've been doing some light(?) reading on motives and the standard conjectures in an attempt to put various things that I tangentially know in perspective. -The question is this: the Weil conjectures assert that $Z=\frac{P_1(t)...P_{2r-1}(t)}{P_0(t)...P_{2r}(t)}$ where the $P_i$'s are certain polynomials. (the assertion is of course stronger, but the rest of it is besides the point) What is the deep reason that these $P_i$'s alternate between numerator and denominator? -In Milne's notes about motives (http://www.jmilne.org/math/xnotes/MOT.pdf), he asserts the following: let $hX$ be the motive corresponding to $X$, and let $h^0X$,...,$h^{2r}X$ be the conjectured decomposition into pure motives (conjecture C in Milne). He then says: define for a pure motive of weight $k$ the zeta function as the characteristic polynomial of the Frobenius if $k$ is odd, and its inverse if $k$ is even. Then extend the definition to motives by: the zeta function of a direct product of motives goes to the product of the zeta functions of the individual motives. Then indeed: -$Z(X,s)=Z(hX,s)=Z(h^0X,s)...Z(h^{2r}X,s)$, where $Z(h^kX,s)=P_k(t)$ for $k$ odd and $\frac{1}{P_k(t)}$ for $k$ even. -So one can reduce this question to: why are we defining the zeta function of a motive of weight $k$ to be the characteristic polynomial of the Frobenius or its inverse depending on the parity? - -REPLY [23 votes]: The zeta function of a variety $X$ over a finite field is a priori defined to be a point counting function, i.e. it is the following product over the closed points of $X$ (thought of as a scheme): -$$\zeta_X(s) = \prod_{x}(1 - | \kappa(x)|^{-s})^{-1},$$ -where $\kappa(x)$ is the residue field of $x$ and $|\kappa(x)|$ denotes its order. (This is motivated by analogy with the Riemann zeta function, which is what we get if we apply the same definition with $X$ replaced by Spec $\mathbb Z$.) -Now this will be a Dirichlet series involving only powers of $p^{-s}$ (if $p$ is the char. of the finite field), and so replacing $p^{-s}$ by $T$, we obtain a power series in $T$, whose -log can be reinterpreted in the usual way as a generating function counting the number of points of $X$ with values in the various extensions of $\mathbb F_p$. -Now one can count these points by the Lefschetz fixed point formula (applied to the $\ell$-adic cohomology), and this gives the alternating product of char. polys. of Frobenius that you write down in your question. -Of course, one could write down their product, rather than their alternating product, but the resulting power series would not have any particular interpretation; in particular, it wouldn't be related to counting points of $X$ in the same way that the zeta function is. -Milne's definition of the $\zeta$-function directly in terms of $\ell$-adic cohomology is to some extent putting the cart before the horse; as Stopple notes, it is a reasonable definition only because of the back story about counting points and so on. -Nevertheless, if you want to take the definition in terms of cohomology as the basic one, then you can ask yourself: how should you define such a quantity if you want it to behave well under chopping up varieties (which is what motives essentially are --- pieces of varieties cut out by correspondences). -The basic quantity that is defined in terms of cohomology and which is additive with respect to cutting up spaces is the Euler characteristic. And for this additivity to hold, it is crucial that involve an alternating sum, with the sign being dictated by the cohomogical degree. The reason is that the behaviour of cohomology under chopping up and/or gluing is given by the excision and Mayer--Vietoris long exact sequences, and it is the alternating sum of the dimensions which is additive in exact sequences. -Viewed cohomologically, the zeta function is like an enhanced, multiplicative version of the Euler characteristic, and like the Euler characteristic, for it to be multiplicative with respect to cutting up varieties, we must form it via an alternating product. -In conclusion: I think that the "deep reason" that you are looking for is the yoga of Euler characteristics.<|endoftext|> -TITLE: Existence of connections on principal bundles -QUESTION [9 upvotes]: Is it always true that a principal $G$-bundle $E$ admits a connection (on the total space, not a local connection on the base manifold $M$)? I know that it must be true, since almost every construction starts off with ...fix a connection on $E$..., I just don't know how to show this rigorously. The only proof I can find is: -Let $U_i\subset M$ be an open subset of $M$. Then $E$ restricted to $U_i$ is trival and we can -construct a connection, denoted $\omega_i$, in this case. Now, let $\big( U_\alpha -\big)$ be an open covering of $M$, and let $\big(f_\alpha\big)$ be a partition of unity subordinate to the cover. Then we can define a connection $\omega = \sum_{\alpha} (f_\alpha \circ \pi) \omega_\alpha$, where $\pi: E\rightarrow M$ is the projection. -However, doesn't the right-hand side of this expression live on $M$? Does this give a connection on $E$? - -REPLY [7 votes]: Another point of view can be found in Atiyah's "Complex analytic connections in fibre bundles". -If $\pi: P \to X$ is principal bundle with fibre a complex (or real) Lie group $G$ on a complex (or differential) manifold $X$, a connection is a $G$-invariant splitting of the following short exact sequence of vector bundles over $P$: -$0 \to T_F P \to TP \to \pi^{-1}TX \to 0$ -Here $T_F P$ denotes the bundle of tangent vectors tangent to the fibre. $G$ acts on all these bundles. One can construct an associated sequence of $G$-invariant sections of these bundles to get a sequence of vector bundles on $X$: -$0 \to (T_F P\)^G \to TP^G \to TX \to 0$ -This is an extension of the vector bundle $TX$ by the vector bundle $T_F P^G$. A connection is now just a splitting of this sequence. By a general result of homological algebra, extensions are classified by -$H^1(X, Hom(TX, T_F P^G))$ -In the differentiable case, $Hom(TX, T_F P^G)$ is a fine sheaf and the cohomology vanishes. So the sequence above is split and we have connections.<|endoftext|> -TITLE: Uniqueness of weak solution L[u]=0 -QUESTION [7 upvotes]: Suppose L is a partial differential operator of arbitrary order with constant coefficients. -If u is in $L^p(\mathbb{R}^n)$ and Lu=0 in distributions, is it necessarily the case that u=0? Does the answer depend on p? -Also, if u is a compactly supported distribution in $\mathbb{R}^n$ with Lu=0 (in the usual sense, i.e. strongly), is it necessarily the case that u=0? -(Suggested reference material appreciated) - -REPLY [3 votes]: Strichartz estimates show that the solution $u$ of the Cauchy problem for several equations of physical interest do belong to an $L^p_t(L^q_x)$ if the initial data is appropriate. When $p=q$, this just means that $u\in L^p$. -For instance, consider the wave equation -$$\partial_t^2u-\Delta_xu=0,\qquad t\in\mathbb R,x\in\mathbb R^d,$$ -in which $n=d+1$. Say that $d\ge3$. Let the initial data be -$$u(0,x)=a(x),\qquad \partial_tu(0,x)=b(0,x),$$ -where $a\in H^1(\mathbb R^d)$ and $b\in L^2(\mathbb R^d)$. Then $u\in L^p(\mathbb R^{1+d})$ with -$$p=2\frac{d+1}{d-2}.$$ -There are variants of this result, but this is too a rich topic to be developped here. -Edit. This phenomenon is called a dispersion effect. It is related to the fact that the curvature of the characteristic cone of $L$ (here $\xi_0^2=\xi_1^2+\cdots+\xi_d^2$) is non-zero.<|endoftext|> -TITLE: Calabi-Yau manifolds and polygonal linkage configuration spaces: related? -QUESTION [9 upvotes]: I was reading about Calabi-Yau manifolds, about which I know little, and was wondering -if these (or related complex manifolds, perhaps K3 surfaces) can be viewed as configuration -spaces (or moduli spaces) of articulated polygonal linkages, say with fixed edge lengths and -vertex joints? This question is a shot in the dark; apologies for its naiveté. But it might help -me understand complex manifolds if I can view them as polygon configuration spaces. -Any references in this general direction would be appreciated. Thanks! - -REPLY [6 votes]: Tim's example has been generalised somewhat by Kapovich and Millson. The space $\mathcal{M}(\mathbf{r})$ of $n$-gons in $\mathbb{E}^3$ with a fixed vector $\mathbf{r}\in\mathbb{R}_+^n$ of side lengths carries the structure of a complex-analytic space with at worst quadratic singularities. This is because $\mathcal{M}(\mathbf{r})$ is a weighted symplectic quotient of the Kähler manifold $(S^2)^n$ by $SO(3)$. -Reference: Michael Kapovich and John J. Millson, The symplectic geometry of polygons in Euclidean space. J. Differential Geom. Volume 44, Number 3 (1996), 479-513. \ No newline at end of file